Signals and Systems Analysis Using Transform Methods and MATLAB

M J ROBERTS Signals and Systems Analysis Using Transform Methods and MATLAB Michael J Roberts Professor Emeritus Department of Electrical and Computer Engineering University of Tennessee Third Edition rob28124fmixxindd 1 061216 208 pm This page intentionally left blank SIGNALS AND SYSTEMS ANALYSIS USING TRANSFORM METHODS AND MATLAB THIRD EDITION Published by McGrawHill Education 2 Penn Plaza New York NY 10121 Copyright 2018 by McGrawHill Education All rights reserved Printed in the United States of America Previous editions 2012 and 2004 No part of this publication may be reproduced or distributed in any form or by any means or stored in a database or retrieval system without the prior written consent of McGrawHill Education including but not limited to in any network or other electronic storage or transmission or broadcast for distance learning Some ancillaries including electronic and print components may not be available to customers outside the United States This book is printed on acidfree paper 1 2 3 4 5 6 7 8 9 QVS 22 21 20 19 18 17 ISBN 9780078028120 MHID 0078028124 Chief Product Officer SVP Products Markets G Scott Virkler Vice President General Manager Products Markets Marty Lange Vice President Content Design Delivery Betsy Whalen Managing Director Thomas Timp Brand Manager Raghothaman Srinivasan Thomas Scaife PhD Director Product Development Rose Koos Product Developer Christine Bower Marketing Manager Shannon ODonnell Director of Digital Content Chelsea Haupt PhD Director Content Design Delivery Linda Avenarius Program Manager Lora Neyens Content Project Managers Jeni McAtee Emily Windelborn Sandy Schnee Buyer Jennifer Pickel Content Licensing Specialists Carrie Burger photo Lorraine Buczek text Cover Image Lauree FeldmanGetty Images Compositor MPS Limited Printer Quad Versailles All credits appearing on page or at the end of the book are considered to be an extension of the copyright page Library of Congress CataloginginPublication Data Roberts Michael J Dr Signals and systems analysis using transform methods and MATLAB Michael J Roberts professor Department of Electrical and Computer Engineering University of Tennessee Third edition New York NY McGrawHill Education 2018 Includes bibliographical references p 786787 and index LCCN 2016043890 ISBN 9780078028120 alk paper LCSH Signal processing System analysis MATLAB LCC TK51029 R63 2018 DDC 6213822dc23 LC record available at httpslccnlocgov2016043890 The Internet addresses listed in the text were accurate at the time of publication The inclusion of a website does not indicate an endorsement by the authors or McGrawHill Education and McGrawHill Education does not guarantee the accuracy of the information presented at these sites mheducationcomhighered rob28124fmixxindd 2 061216 208 pm To my wife Barbara for giving me the time and space to complete this effort and to the memory of my parents Bertie Ellen Pinkerton and Jesse Watts Roberts for their early emphasis on the importance of education rob28124fmixxindd 3 061216 208 pm Preface xii Chapter 1 Introduction 1 11 Signals and Systems Defined 1 12 Types of Signals 3 13 Examples of Systems 8 A Mechanical System 9 A Fluid System 9 A DiscreteTime System 11 Feedback Systems 12 14 A Familiar Signal and System Example 14 15 Use of MATLAB 18 Chapter 2 Mathematical Description of ContinuousTime Signals 19 21 Introduction and Goals 19 22 Functional Notation 20 23 ContinuousTime Signal Functions 20 Complex Exponentials and Sinusoids 21 Functions with Discontinuities 23 The Signum Function 24 The UnitStep Function 24 The UnitRamp Function 26 The Unit Impulse 27 The Impulse the Unit Step and Generalized Derivatives 29 The Equivalence Property of the Impulse 30 The Sampling Property of the Impulse 31 The Scaling Property of the Impulse 31 The Unit Periodic Impulse or Impulse Train 32 A Coordinated Notation for Singularity Functions 33 The UnitRectangle Function 33 24 Combinations of Functions 34 25 Shifting and Scaling 36 Amplitude Scaling 36 Time Shifting 37 Time Scaling 39 Simultaneous Shifting and Scaling 43 26 Differentiation and Integration 47 27 Even and Odd Signals 49 Combinations of Even and Odd Signals 51 Derivatives and Integrals of Even and Odd Signals 53 28 Periodic Signals 53 29 Signal Energy and Power 56 Signal Energy 56 Signal Power 58 210 Summary of Important Points 60 Exercises 61 Exercises with Answers 61 Signal Functions 61 Shifting and Scaling 62 Derivatives and Integrals of Functions 66 Generalized Derivative 67 Even and Odd Functions 67 Periodic Signals 69 Signal Energy and Power of Signals 70 Exercises without Answers 71 Signal Functions 71 Scaling and Shifting 71 Generalized Derivative 76 Derivatives and Integrals of Functions 76 Even and Odd Functions 76 Periodic Functions 77 Signal Energy and Power of Signals 77 Chapter 3 DiscreteTime Signal Description 79 31 Introduction and Goals 79 32 Sampling and Discrete Time 80 33 Sinusoids and Exponentials 82 Sinusoids 82 Exponentials 85 34 Singularity Functions 86 The UnitImpulse Function 86 The UnitSequence Function 87 CONTENTS iv rob28124fmixxindd 4 061216 208 pm Contents v The Signum Function 87 The UnitRamp Function 88 The Unit Periodic Impulse Function or Impulse Train 88 35 Shifting and Scaling 89 Amplitude Scaling 89 Time Shifting 89 Time Scaling 89 Time Compression 90 Time Expansion 90 36 Differencing and Accumulation 94 37 Even and Odd Signals 98 Combinations of Even and Odd Signals 100 Symmetrical Finite Summation of Even and Odd Signals 100 38 Periodic Signals 101 39 Signal Energy and Power 102 Signal Energy 102 Signal Power 103 310 Summary of Important Points 105 Exercises 105 Exercises with Answers 105 Functions 105 Scaling and Shifting Functions 107 Differencing and Accumulation 109 Even and Odd Functions 110 Periodic Functions 111 Signal Energy and Power 112 Exercises without Answers 113 Signal Functions 113 Shifting and Scaling Functions 113 Differencing and Accumulation 114 Even and Odd Functions 114 Periodic Signals 115 Signal Energy and Power 116 Chapter 4 Description of Systems 118 41 Introduction and Goals 118 42 ContinuousTime Systems 119 System Modeling 119 Differential Equations 120 Block Diagrams 124 System Properties 127 Introductory Example 127 Homogeneity 131 Time Invariance 132 Additivity 133 Linearity and Superposition 134 LTI Systems 134 Stability 138 Causality 139 Memory 139 Static Nonlinearity 140 Invertibility 142 Dynamics of SecondOrder Systems 143 Complex Sinusoid Excitation 145 43 DiscreteTime Systems 145 System Modeling 145 Block Diagrams 145 Difference Equations 146 System Properties 152 44 Summary of Important Points 155 Exercises 156 Exercises with Answers 156 System Models 156 Block Diagrams 157 System Properties 158 Exercises without Answers 160 System Models 160 System Properties 162 Chapter 5 TimeDomain System Analysis 164 51 Introduction and Goals 164 52 Continuous Time 164 Impulse Response 164 ContinuousTime Convolution 169 Derivation 169 Graphical and Analytical Examples of Convolution 173 Convolution Properties 178 System Connections 181 Step Response and Impulse Response 181 Stability and Impulse Response 181 Complex Exponential Excitation and the Transfer Function 182 Frequency Response 184 53 Discrete Time 186 Impulse Response 186 DiscreteTime Convolution 189 rob28124fmixxindd 5 061216 208 pm Contents vi Derivation 189 Graphical and Analytical Examples of Convolution 192 Convolution Properties 196 Numerical Convolution 196 DiscreteTime Numerical Convolution 196 ContinuousTime Numerical Convolution 198 Stability and Impulse Response 200 System Connections 200 UnitSequence Response and Impulse Response 201 Complex Exponential Excitation and the Transfer Function 203 Frequency Response 204 54 Summary of Important Points 207 Exercises 207 Exercises with Answers 207 Continuous Time 207 Impulse Response 207 Convolution 209 Stability 213 Frequency Response 214 Discrete Time 214 Impulse Response 214 Convolution 215 Stability 219 Exercises without Answers 221 Continuous Time 221 Impulse Response 221 Convolution 222 Stability 224 Discrete Time 225 Impulse Response 225 Convolution 225 Stability 228 Chapter 6 ContinuousTime Fourier Methods 229 61 Introduction and Goals 229 62 The ContinuousTime Fourier Series 230 Conceptual Basis 230 Orthogonality and the Harmonic Function 234 The Compact Trigonometric Fourier Series 237 Convergence 239 Continuous Signals 239 Discontinuous Signals 240 Minimum Error of FourierSeries Partial Sums 242 The Fourier Series of Even and Odd Periodic Functions 243 FourierSeries Tables and Properties 244 Numerical Computation of the Fourier Series 248 63 The ContinuousTime Fourier Transform 255 Extending the Fourier Series to Aperiodic Signals 255 The Generalized Fourier Transform 260 Fourier Transform Properties 265 Numerical Computation of the Fourier Transform 273 64 Summary of Important Points 281 Exercises 281 Exercises with Answers 281 Fourier Series 281 Orthogonality 282 Forward and Inverse Fourier Transforms 286 Relation of CTFS to CTFT 293 Numerical CTFT 294 System Response 294 Exercises without Answers 294 Fourier Series 294 Forward and Inverse Fourier Transforms 300 System Response 305 Relation of CTFS to CTFT 306 Chapter 7 DiscreteTime Fourier Methods 307 71 Introduction and Goals 307 72 The DiscreteTime Fourier Series and the Discrete Fourier Transform 307 Linearity and ComplexExponential Excitation 307 Orthogonality and the Harmonic Function 311 Discrete Fourier Transform Properties 315 The Fast Fourier Transform 321 73 The DiscreteTime Fourier Transform 323 Extending the Discrete Fourier Transform to Aperiodic Signals 323 Derivation and Definition 324 The Generalized DTFT 326 Convergence of the DiscreteTime Fourier Transform 327 DTFT Properties 327 rob28124fmixxindd 6 061216 208 pm Contents vii Numerical Computation of the DiscreteTime Fourier Transform 334 74 Fourier Method Comparisons 340 75 Summary of Important Points 341 Exercises 342 Exercises with Answers 342 Orthogonality 342 Discrete Fourier Transform 342 DiscreteTime Fourier Transform Definition 344 Forward and Inverse DiscreteTime Fourier Transforms 345 Exercises without Answers 348 Discrete Fourier Transform 348 Forward and Inverse DiscreteTime Fourier Transforms 352 Chapter 8 The Laplace Transform 354 81 Introduction and Goals 354 82 Development of the Laplace Transform 355 Generalizing the Fourier Transform 355 Complex Exponential Excitation and Response 357 83 The Transfer Function 358 84 CascadeConnected Systems 358 85 Direct Form II Realization 359 86 The Inverse Laplace Transform 360 87 Existence of the Laplace Transform 360 TimeLimited Signals 361 Right and LeftSided Signals 361 88 LaplaceTransform Pairs 362 89 PartialFraction Expansion 367 810 LaplaceTransform Properties 377 811 The Unilateral Laplace Transform 379 Definition 379 Properties Unique to the Unilateral Laplace Transform 381 Solution of Differential Equations with Initial Conditions 383 812 PoleZero Diagrams and Frequency Response 385 813 MATLAB System Objects 393 814 Summary of Important Points 395 Exercises 395 Exercises with Answers 395 LaplaceTransform Definition 395 Direct Form II System Realization 396 Forward and Inverse Laplace Transforms 396 Unilateral LaplaceTransform Integral 399 Solving Differential Equations 399 Exercises without Answers 400 Region of Convergence 400 Existence of the Laplace Transform 400 Direct Form II System Realization 400 Forward and Inverse Laplace Transforms 401 Solution of Differential Equations 403 PoleZero Diagrams and Frequency Response 403 Chapter 9 The z Transform 406 91 Introduction and Goals 406 92 Generalizing the DiscreteTime Fourier Transform 407 93 Complex Exponential Excitation and Response 408 94 The Transfer Function 408 95 CascadeConnected Systems 408 96 Direct Form II System Realization 409 97 The Inverse z Transform 410 98 Existence of the z Transform 410 TimeLimited Signals 410 Right and LeftSided Signals 411 99 zTransform Pairs 413 910 zTransform Properties 416 911 Inverse zTransform Methods 417 Synthetic Division 417 PartialFraction Expansion 418 Examples of Forward and Inverse z Transforms 418 912 The Unilateral z Transform 423 Properties Unique to the Unilateral z Transform 423 Solution of Difference Equations 424 913 PoleZero Diagrams and Frequency Response 425 914 MATLAB System Objects 428 In MATLAB 429 915 Transform Method Comparisons 430 916 Summary of Important Points 434 rob28124fmixxindd 7 061216 208 pm Contents viii Exercises 435 Exercises with Answers 435 DirectForm II System Realization 435 Existence of the z Transform 435 Forward and Inverse z Transforms 435 Unilateral zTransform Properties 438 Solution of Difference Equations 438 PoleZero Diagrams and Frequency Response 439 Exercises without Answers 441 Direct Form II System Realization 441 Existence of the z Transform 441 Forward and Inverse zTransforms 441 PoleZero Diagrams and Frequency Response 443 Chapter 10 Sampling and Signal Processing 446 101 Introduction and Goals 446 102 ContinuousTime Sampling 447 Sampling Methods 447 The Sampling Theorem 449 Qualitative Concepts 449 Sampling Theorem Derivation 451 Aliasing 454 Timelimited and Bandlimited Signals 457 Interpolation 458 Ideal Interpolation 458 Practical Interpolation 459 ZeroOrder Hold 460 FirstOrder Hold 460 Sampling Bandpass Signals 461 Sampling a Sinusoid 464 Bandlimited Periodic Signals 467 Signal Processing Using the DFT 470 CTFTDFT Relationship 470 CTFTDTFT Relationship 471 Sampling and PeriodicRepetition Relationship 474 Computing the CTFS Harmonic Function with the DFT 478 Approximating the CTFT with the DFT 478 Forward CTFT 478 Inverse CTFT 479 Approximating the DTFT with the DFT 479 Approximating ContinuousTime Convolution with the DFT 479 Aperiodic Convolution 479 Periodic Convolution 479 DiscreteTime Convolution with the DFT 479 Aperiodic Convolution 479 Periodic Convolution 479 Summary of Signal Processing Using the DFT 480 103 DiscreteTime Sampling 481 PeriodicImpulse Sampling 481 Interpolation 483 104 Summary of Important Points 486 Exercises 487 Exercises with Answers 487 Pulse Amplitude Modulation 487 Sampling 487 Impulse Sampling 489 Nyquist Rates 491 TimeLimited and Bandlimited Signals 492 Interpolation 493 Aliasing 495 Bandlimited Periodic Signals 495 CTFTCTFSDFT Relationships 495 Windows 497 DFT 497 Exercises without Answers 500 Sampling 500 Impulse Sampling 502 Nyquist Rates 504 Aliasing 505 Practical Sampling 505 Bandlimited Periodic Signals 505 DFT 506 DiscreteTime Sampling 508 Chapter 11 Frequency Response Analysis 509 111 Introduction and Goals 509 112 Frequency Response 509 113 ContinuousTime Filters 510 Examples of Filters 510 Ideal Filters 515 Distortion 515 Filter Classifications 516 Ideal Filter Frequency Responses 516 Impulse Responses and Causality 517 The Power Spectrum 520 Noise Removal 520 Bode Diagrams 521 rob28124fmixxindd 8 061216 208 pm Contents ix The Decibel 521 The OneRealPole System 525 The OneRealZero System 526 Integrators and Differentiators 527 FrequencyIndependent Gain 527 Complex Pole and Zero Pairs 530 Practical Filters 532 Passive Filters 532 The Lowpass Filter 532 The Bandpass Filter 535 Active Filters 536 Operational Amplifiers 537 The Integrator 538 The Lowpass Filter 538 114 DiscreteTime Filters 546 Notation 546 Ideal Filters 547 Distortion 547 Filter Classifications 548 Frequency Responses 548 Impulse Responses and Causality 548 Filtering Images 549 Practical Filters 554 Comparison with ContinuousTime Filters 554 Highpass Bandpass and Bandstop Filters 556 The Moving Average Filter 560 The Almost Ideal Lowpass Filter 564 Advantages Compared to ContinuousTime Filters 566 115 Summary of Important Points 566 Exercises 567 Exercises with Answers 567 ContinuousTime Frequency Response 567 ContinuousTime Ideal Filters 567 ContinuousTime Causality 567 Logarithmic Graphs Bode Diagrams and Decibels 568 ContinuousTime Practical Passive Filters 570 ContinuousTime Practical Active Filters 574 DiscreteTime Frequency Response 575 DiscreteTime Ideal Filters 576 DiscreteTime Causality 576 DiscreteTime Practical Filters 577 Exercises without Answers 579 ContinuousTime Frequency Response 579 ContinuousTime Ideal Filters 579 ContinuousTime Causality 579 Bode Diagrams 580 ContinuousTime Practical Passive Filters 580 ContinuousTime Filters 582 ContinuousTime Practical Active Filters 582 DiscreteTime Causality 586 DiscreteTime Filters 587 Chapter 12 Laplace System Analysis 592 121 Introduction and Goals 592 122 System Representations 592 123 System Stability 596 124 System Connections 599 Cascade and Parallel Connections 599 The Feedback Connection 599 Terminology and Basic Relationships 599 Feedback Effects on Stability 600 Beneficial Effects of Feedback 601 Instability Caused by Feedback 604 Stable Oscillation Using Feedback 608 The RootLocus Method 612 Tracking Errors in UnityGain Feedback Systems 618 125 System Analysis Using MATLAB 621 126 System Responses to Standard Signals 623 UnitStep Response 624 Sinusoid Response 627 127 Standard Realizations of Systems 630 Cascade Realization 630 Parallel Realization 632 128 Summary of Important Points 632 Exercises 633 Exercises with Answers 633 Transfer Functions 633 Stability 634 Parallel Cascade and Feedback Connections 635 Root Locus 637 Tracking Errors in UnityGain Feedback Systems 639 System Responses to Standard Signals 640 System Realization 641 Exercises without Answers 642 Stability 642 Transfer Functions 642 Stability 643 rob28124fmixxindd 9 061216 208 pm Contents x Parallel Cascade and Feedback Connections 643 Root Locus 646 Tracking Errors in UnityGain Feedback Systems 647 Response to Standard Signals 647 System Realization 649 Chapter 13 zTransform System Analysis 650 131 Introduction and Goals 650 132 System Models 650 Difference Equations 650 Block Diagrams 651 133 System Stability 651 134 System Connections 652 135 System Responses to Standard Signals 654 UnitSequence Response 654 Response to a Causal Sinusoid 657 136 Simulating ContinuousTime Systems with DiscreteTime Systems 660 zTransformLaplaceTransform Relationships 660 Impulse Invariance 662 SampledData Systems 664 137 Standard Realizations of Systems 670 Cascade Realization 670 Parallel Realization 670 138 Summary of Important Points 671 Exercises 672 Exercises with Answers 672 Stability 672 Parallel Cascade and Feedback Connections 672 Response to Standard Signals 673 Root Locus 674 LaplaceTransformzTransform Relationship 675 SampledData Systems 675 System Realization 676 Exercises without Answers 677 Stability 677 Root Locus 677 Parallel Cascade and Feedback Connections 677 Response to Standard Signals 677 LaplaceTransformzTransform Relationship 679 SampledData Systems 679 System Realization 679 General 679 Chapter 14 Filter Analysis and Design 680 141 Introduction and Goals 680 142 Analog Filters 680 Butterworth Filters 681 Normalized Butterworth Filters 681 Filter Transformations 682 MATLAB Design Tools 684 Chebyshev Elliptic and Bessel Filters 686 143 Digital Filters 689 Simulation of Analog Filters 689 Filter Design Techniques 689 IIR Filter Design 689 TimeDomain Methods 689 ImpulseInvariant Design 689 StepInvariant Design 696 FiniteDifference Design 698 FrequencyDomain Methods 704 The Bilinear Method 706 FIR Filter Design 713 Truncated Ideal Impulse Response 713 Optimal FIR Filter Design 723 MATLAB Design Tools 725 144 Summary of Important Points 727 Exercises 727 Exercises with Answers 727 ContinuousTime Filters 727 FiniteDifference Filter Design 728 Matchedz Transform and Direct Substitution Filter Design 729 Bilinear zTransform Filter Design 730 FIR Filter Design 730 Digital Filter Design Method Comparison 731 Exercises without Answers 731 Analog Filter Design 731 ImpulseInvariant and StepInvariant Filter Design 732 FiniteDifference Filter Design 733 Matched zTransform and Direct Substitution Filter Design 733 Bilinear zTransform Filter Design 733 FIR Filter Design 733 Digital Filter Design Method Comparison 734 rob28124fmixxindd 10 061216 208 pm Contents xi Appendix I Useful Mathematical Relations A1 II ContinuousTime Fourier Series Pairs A4 III Discrete Fourier Transform Pairs A7 IV ContinuousTime Fourier Transform Pairs A10 V DiscreteTime Fourier Transform Pairs A17 VI Tables of Laplace Transform Pairs A22 VII zTransform Pairs A24 Bibliography B1 Index I1 rob28124fmixxindd 11 061216 208 pm PREFACE MOTIVATION I wrote the first and second editions because I love the mathematical beauty of signal and system analysis That has not changed The motivation for the third edi tion is to further refine the book structure in light of reviewers comments correct a few errors from the second edition and significantly rework the exercises AUDIENCE This book is intended to cover a twosemester course sequence in the basics of signal and system analysis during the junior or senior year It can also be used as I have used it as a book for a quick onesemester Masterslevel review of trans form methods as applied to linear systems CHANGES FROM THE SECOND EDITION 1 In response to reviewers comments two chapters from the second edition have been omitted Communication Systems and StateSpace Analysis There seemed to be very little if any coverage of these topics in actual classes 2 The second edition had 550 endofchapter exercises in 16 chapters The third edition has 710 endofchapter exercises in 14 chapters OVERVIEW Except for the omission of two chapters the third edition structure is very similar to the second edition The book begins with mathematical methods for describing signals and systems in both continuous and discrete time I introduce the idea of a transform with the continuoustime Fourier series and from that base move to the Fourier trans form as an extension of the Fourier series to aperiodic signals Then I do the same for discretetime signals I introduce the Laplace transform both as a generalization of the continuoustime Fourier transform for unbounded signals and unstable systems and as a powerful tool in system analysis because of its very close association with the ei genvalues and eigenfunctions of continuoustime linear systems I take a similar path for discretetime systems using the z transform Then I address sampling the relation between continuous and discrete time The rest of the book is devoted to applications in frequencyresponse analysis feedback systems analog and digital filters Through out the book I present examples and introduce MATLAB functions and operations to implement the methods presented A chapterbychapter summary follows CHAPTER SUMMARIES CHAPTER 1 Chapter 1 is an introduction to the general concepts involved in signal and system analysis without any mathematical rigor It is intended to motivate the student by xii rob28124fmixxindd 12 061216 208 pm xiii Preface demonstrating the ubiquity of signals and systems in everyday life and the impor tance of understanding them CHAPTER 2 Chapter 2 is an exploration of methods of mathematically describing continuous time signals of various kinds It begins with familiar functions sinusoids and exponentials and then extends the range of signaldescribing functions to include continuoustime singularity functions switching functions Like most if not all signals and systems textbooks I define the unitstep the signum the unitimpulse and the unitramp functions In addition to these I define a unit rectangle and a unit periodic impulse function The unit periodic impulse function along with convolution provides an especially compact way of mathematically describing arbitrary periodic signals After introducing the new continuoustime signal functions I cover the common types of signal transformations amplitude scaling time shifting time scaling differentiation and integration and apply them to the signal functions Then I cover some characteristics of signals that make them invariant to certain transformations evenness oddness and periodicity and some of the implications of these signal characteristics in signal analysis The last section is on signal energy and power CHAPTER 3 Chapter 3 follows a path similar to Chapter 2 except applied to discretetime signals instead of continuoustime signals I introduce the discretetime sinu soid and exponential and comment on the problems of determining period of a discretetime sinusoid This is the first exposure of the student to some of the implications of sampling I define some discretetime signal functions analo gous to continuoustime singularity functions Then I explore amplitude scaling time shifting time scaling differencing and accumulation for discretetime signal functions pointing out the unique implications and problems that occur especially when time scaling discretetime functions The chapter ends with definitions and discussion of signal energy and power for discretetime signals CHAPTER 4 This chapter addresses the mathematical description of systems First I cover the most common forms of classification of systems homogeneity additivity linearity time invariance causality memory static nonlinearity and invertibility By example I present various types of systems that have or do not have these properties and how to prove various properties from the mathematical description of the system CHAPTER 5 This chapter introduces the concepts of impulse response and convolution as components in the systematic analysis of the response of linear timeinvariant systems I present the mathematical properties of continuoustime convolution and a graphical method of understanding what the convolution integral says I also show how the properties of convolution can be used to combine subsystems that are connected in cascade or parallel into one system and what the impulse response of the overall system must be Then I introduce the idea of a transfer rob28124fmixxindd 13 061216 208 pm xiv Preface function by finding the response of an LTI system to complex sinusoidal exci tation This section is followed by an analogous coverage of discretetime impulse response and convolution CHAPTER 6 This is the beginning of the students exposure to transform methods I begin by graphically introducing the concept that any continuoustime periodic signal with engineering usefulness can be expressed by a linear combination of continuoustime sinusoids real or complex Then I formally derive the Fourier series using the concept of orthogonality to show where the signal description as a function of discrete harmonic number the harmonic function comes from I mention the Dirichlet conditions to let the student know that the continuoustime Fourier series applies to all practical continuoustime signals but not to all imaginable continuoustime signals Then I explore the properties of the Fourier series I have tried to make the Fourier series notation and properties as similar as possible and analogous to the Fourier transform which comes later The harmonic function forms a Fourier series pair with the time function In the first edition I used a notation for har monic function in which lowercase letters were used for timedomain quantities and uppercase letters for their harmonic functions This unfortunately caused some confusion because continuous and discretetime harmonic functions looked the same In this edition I have changed the harmonic function notation for continuoustime signals to make it easily distinguishable I also have a section on the convergence of the Fourier series illustrating the Gibbs phenomenon at function discontinuities I encourage students to use tables and properties to find harmonic functions and this practice prepares them for a similar process in find ing Fourier transforms and later Laplace and z transforms The next major section of Chapter 6 extends the Fourier series to the Fourier transform I introduce the concept by examining what happens to a continuoustime Fourier series as the period of the signal approaches infinity and then define and derive the continuoustime Fourier transform as a gener alization of the continuoustime Fourier series Following that I cover all the important properties of the continuoustime Fourier transform I have taken an ecumenical approach to two different notational conventions that are commonly seen in books on signals and systems control systems digital signal processing communication systems and other applications of Fourier methods such as image processing and Fourier optics the use of either cyclic frequency f or radian fre quency ω I use both and emphasize that the two are simply related through a change of variable I think this better prepares students for seeing both forms in other books in their college and professional careers CHAPTER 7 This chapter introduces the discretetime Fourier series DTFS the discrete Fou rier transform DFT and the discretetime Fourier transform DTFT deriving and defining them in a manner analogous to Chapter 6 The DTFS and the DFT are almost identical I concentrate on the DFT because of its very wide use in digital signal processing I emphasize the important differences caused by the differences between continuous and discretetime signals especially the finite summation range of the DFT as opposed to the generally infinite summation range in the CTFS I also point out the importance of the fact that the DFT relates rob28124fmixxindd 14 061216 208 pm xv Preface a finite set of numbers to another finite set of numbers making it amenable to direct numerical machine computation I discuss the fast Fourier transform as a very efficient algorithm for computing the DFT As in Chapter 6 I use both cyclic and radian frequency forms emphasizing the relationships between them I use F and Ω for discretetime frequencies to distinguish them from f and ω which were used in continuous time Unfortunately some authors reverse these symbols My usage is more consistent with the majority of signals and systems texts This is another example of the lack of standardization of notation in this area The last major section is a comparison of the four Fourier methods I emphasize particu larly the duality between sampling in one domain and periodic repetition in the other domain CHAPTER 8 This chapter introduces the Laplace transform I approach the Laplace trans form from two points of view as a generalization of the Fourier transform to a larger class of signals and as result which naturally follows from the excitation of a linear timeinvariant system by a complex exponential signal I begin by defining the bilateral Laplace transform and discussing significance of the re gion of convergence Then I define the unilateral Laplace transform I derive all the important properties of the Laplace transform I fully explore the method of partialfraction expansion for finding inverse transforms and then show examples of solving differential equations with initial conditions using the uni lateral form CHAPTER 9 This chapter introduces the z transform The development parallels the devel opment of the Laplace transform except applied to discretetime signals and systems I initially define a bilateral transform and discuss the region of con vergence Then I define a unilateral transform I derive all the important prop erties and demonstrate the inverse transform using partialfraction expansion and the solution of difference equations with initial conditions I also show the relationship between the Laplace and z transforms an important idea in the approximation of continuoustime systems by discretetime systems in Chapter 14 CHAPTER 10 This is the first exploration of the correspondence between a continuoustime signal and a discretetime signal formed by sampling it The first section covers how sampling is usually done in real systems using a sampleandhold and an AD converter The second section starts by asking the question of how many samples are enough to describe a continuoustime signal Then the question is answered by deriving the sampling theorem Then I discuss interpolation methods theoret ical and practical the special properties of bandlimited periodic signals I do a complete development of the relationship between the CTFT of a continuoustime signal and DFT of a finitelength set of samples taken from it Then I show how the DFT can be used to approximate the CTFT of an energy signal or a periodic signal The next major section explores the use of the DFT in numerically approx imating various common signalprocessing operations rob28124fmixxindd 15 061216 208 pm xvi CHAPTER 11 This chapter covers various aspects of the use of the CTFT and DTFT in fre quency response analysis The major topics are ideal filters Bode diagrams prac tical passive and active continuoustime filters and basic discretetime filters CHAPTER 12 This chapter is on the application of the Laplace transform including block dia gram representation of systems in the complex frequency domain system stability system interconnections feedback systems including root locus system responses to standard signals and lastly standard realizations of continuoustime systems CHAPTER 13 This chapter is on the application of the z transform including block diagram representation of systems in the complex frequency domain system stability sys tem interconnections feedback systems including rootlocus system responses to standard signals sampleddata systems and standard realizations of discretetime systems CHAPTER 14 This chapter covers the analysis and design of some of the most common types of practical analog and digital filters The analog filter types are Butterworth Chebyshev Types 1 and 2 and Elliptic Cauer filters The section on digital filters covers the most common types of techniques for simulation of analog filters includ ing impulse and stepinvariant finite difference matched z transform direct sub stitution bilinear z transform truncated impulse response and ParksMcClellan numerical design APPENDICES There are seven appendices on useful mathematical formulae tables of the four Fourier transforms Laplace transform tables and z transform tables CONTINUITY The book is structured so as to facilitate skipping some topics without loss of continuity Continuoustime and discretetime topics are covered alternately and continuoustime analysis could be covered without reference to discrete time Also any or all of the last six chapters could be omitted in a shorter course REVIEWS AND EDITING This book owes a lot to the reviewers especially those who really took time and criticized and suggested improvements I am indebted to them I am also indebted to the many students who have endured my classes over the years I believe that our relationship is more symbiotic than they realize That is they learn signal and system analysis from me and I learn how to teach signal and system analysis from them I cannot count the number of times I have been asked a very perceptive question by a student that revealed not only that the students were not understand ing a concept but that I did not understand it as well as I had previously thought Preface rob28124fmixxindd 16 061216 208 pm xvii WRITING STYLE Every author thinks he has found a better way to present material so that students can grasp it and I am no different I have taught this material for many years and through the experience of grading tests have found what students generally do and do not grasp I have spent countless hours in my office oneonone with students explaining these concepts to them and through that experience I have found out what needs to be said In my writing I have tried to simply speak directly to the reader in a straightforward conversational way trying to avoid offputting formality and to the extent possible anticipating the usual misconceptions and revealing the fallacies in them Transform methods are not an obvious idea and at first exposure students can easily get bogged down in a bewildering morass of abstractions and lose sight of the goal which is to analyze a systems response to signals I have tried as every author does to find the magic combination of ac cessibility and mathematical rigor because both are important I think my writing is clear and direct but you the reader will be the final judge of whether or not that is true EXERCISES Each chapter has a group of exercises along with answers and a second group of exercises without answers The first group is intended more or less as a set of drill exercises and the second group as a set of more challenging exercises CONCLUDING REMARKS As I indicated in the preface to first and second editions I welcome any and all criticism corrections and suggestions All comments including ones I disagree with and ones which disagree with others will have a constructive impact on the next edition because they point out a problem If something does not seem right to you it probably will bother others also and it is my task as an author to find a way to solve that problem So I encourage you to be direct and clear in any re marks about what you believe should be changed and not to hesitate to mention any errors you may find from the most trivial to the most significant Michael J Roberts Professor Emeritus Electrical and Computer Engineering University of Tennessee at Knoxville mjrutkedu Preface rob28124fmixxindd 17 061216 208 pm RequiredResults McGrawHill Connect Learn Without Limits Connect is a teaching and learning platform that is proven to deliver better results for students and instructors Connect empowers students by continually adapting to deliver precisely what they need when they need it and how they need it so your class time is more engaging and effective Connect Insight Connect Insight is Connects new one ofakind visual analytics dashboard that provides ataglance information regarding student performance which is immediately actionable By presenting assignment assessment and topical performance results together with 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driving the student toward comprehension and retention of the subject matter Available on smartphones and tablets SmartBook puts learning at the students fingertipsanywhere anytime Adaptive Over 57 billion questions have been answered making McGrawHill Education products more intelligent reliable and precise THE ADAPTIVE READING EXPERIENCE DESIGNED TO TRANSFORM THE WAY STUDENTS READ More students earn As and Bs when they use McGrawHill Education Adaptive products wwwmheducationcom Getty ImagesiStockphoto rob28124fmixxindd 19 061216 209 pm 1 C H A P T E R 1 Introduction 11 SIGNALS AND SYSTEMS DEFINED Any timevarying physical phenomenon that is intended to convey information is a signal Examples of signals are the human voice sign language Morse code traffic signals voltages on telephone wires electric fields emanating from radio or television transmitters and variations of light intensity in an optical fiber on a telephone or com puter network Noise is like a signal in that it is a timevarying physical phenomenon but usually it does not carry useful information and is considered undesirable Signals are operated on by systems When one or more excitations or input signals are applied at one or more system inputs the system produces one or more responses or output signals at its outputs Figure 11 is a block diagram of a singleinput singleoutput system Input System Output Excitation or Input Signal Response or Output Signal Figure 11 Block diagram of a singleinput singleoutput system Transmitter Channel Receiver Information Signal Noisy Information Signal Noise Noise Noise Figure 12 A communication system In a communication system a transmitter produces a signal and a receiver acquires it A channel is the path a signal takes from a transmitter to a receiver Noise is inevitably introduced into the transmitter channel and receiver often at multiple points Figure 12 The transmitter channel and receiver are all components or subsystems of the overall system Scientific instruments are systems that measure a physical phenom enon temperature pressure speed etc and convert it to a voltage or current a sig nal Commercial building control systems Figure 13 industrial plant control systems Figure 14 modern farm machinery Figure 15 avionics in airplanes ignition and fuel pumping controls in automobiles and so on are all systems that operate on signals rob28124ch01001018indd 1 051216 221 pm C h a p t e r 1 Introduction 2 Figure 13 Modern office buildings Vol 43 PhotoDiscGetty Figure 14 Typical industrial plant control room RoyaltyFreePunchstock rob28124ch01001018indd 2 051216 221 pm 12 Types of Signals 3 The term system even encompasses things such as the stock market government weather the human body and the like They all respond when excited Some systems are readily analyzed in detail some can be analyzed approximately but some are so complicated or difficult to measure that we hardly know enough to understand them 12 TYPES OF SIGNALS There are several broad classifications of signals continuoustime discretetime continuousvalue discretevalue random and nonrandom A continuoustime sig nal is defined at every instant of time over some time interval Another common name for some continuoustime signals is analog signal in which the variation of the signal with time is analogous proportional to some physical phenomenon All analog sig nals are continuoustime signals but not all continuoustime signals are analog signals Figure 16 through Figure 18 Sampling a signal is acquiring values from a continuoustime signal at discrete points in time The set of samples forms a discretetime signal A discretetime signal Figure 15 Modern farm tractor with enclosed cab RoyaltyFreeCorbis Figure 16 Examples of continuoustime and discretetime signals n xn DiscreteTime ContinuousValue Signal t xt ContinuousTime ContinuousValue Signal rob28124ch01001018indd 3 051216 221 pm C h a p t e r 1 Introduction 4 can also be created by an inherently discretetime system that produces signal values only at discrete times Figure 16 A continuousvalue signal is one that may have any value within a continuum of allowed values In a continuum any two values can be arbitrarily close together The real numbers form a continuum with infinite extent The real numbers between zero and one form a continuum with finite extent Each is a set with infinitely many mem bers Figure 16 through Figure 18 A discretevalue signal can only have values taken from a discrete set In a discrete set of values the magnitude of the difference between any two values is greater than some positive number The set of integers is an example Discretetime signals are usually transmitted as digital signals a sequence of values of a discretetime signal in the form of digits in some encoded form The term digital is also sometimes used loosely to refer to a discretevalue signal that has only two possible values The digits in this type of digital signal are transmitted by signals that are continuoustime In this case the terms continuoustime and analog are not synonymous A digital signal of this type is a continuoustime signal but not an analog signal because its variation of value with time is not directly analogous to a physical phenomenon Figure 16 through Figure 18 A random signal cannot be predicted exactly and cannot be described by any math ematical function A deterministic signal can be mathematically described A com mon name for a random signal is noise Figure 16 through Figure 18 In practical signal processing it is very common to acquire a signal for processing by a computer by sampling quantizing and encoding it Figure 19 The original signal is a continuousvalue continuoustime signal Sampling acquires its values at discrete times and those values constitute a continuousvalue discretetime signal Quantization approximates each sample as the nearest member of a finite set of dis crete values producing a discretevalue discretetime signal Each signal value in the set of discrete values at discrete times is converted to a sequence of rectangular pulses that encode it into a binary number creating a discretevalue continuoustime signal commonly called a digital signal The steps illustrated in Figure 19 are usually carried out by a single device called an analogtodigital converter ADC Figure 18 Examples of noise and a noisy digital signal Noisy Digital Signal ContinuousTime ContinuousValue Random Signal t xt xt Noise t Figure 17 Examples of continuoustime discretevalue signals ContinuousTime DiscreteValue Signal ContinuousTime DiscreteValue Signal t xt xt t Digital Signal rob28124ch01001018indd 4 051216 221 pm 12 Types of Signals 5 Figure 19 Sampling quantization and encoding of a signal to illustrate various signal types t kΔt k1Δt k2Δt k1Δt kΔt k1Δt k2Δt k1Δt xsn n k k1 k2 k1 xsqn n k k1 k2 k1 xsqet t 111 001 111 011 ContinuousValue ContinuousTime Signal ContinuousValue DiscreteTime Signal DiscreteValue DiscreteTime Signal DiscreteValue ContinuousTime Signal Sampling Quantization Encoding xt Figure 110 Asynchronous serial binary ASCIIencoded voltage signal for the word SIGNAL 0 1 2 3 4 5 6 7 1 0 1 2 3 4 5 6 Time t ms Voltage vt V Serial Binary Voltage Signal for the ASCII Message SIGNAL S I G N A L One common use of binary digital signals is to send text messages using the American Standard Code for Information Interchange ASCII The letters of the al phabet the digits 09 some punctuation characters and several nonprinting control characters for a total of 128 characters are all encoded into a sequence of 7 binary bits The 7 bits are sent sequentially preceded by a start bit and followed by 1 or 2 stop bits for synchronization purposes Typically in directwired connections between digital equipment the bits are represented by a higher voltage 2 to 5 V for a 1 and a lower voltage level around 0 V for a 0 In an asynchronous transmission using one start and one stop bit sending the message SIGNAL the voltage versus time would look as illustrated in Figure 110 rob28124ch01001018indd 5 051216 221 pm C h a p t e r 1 Introduction 6 In 1987 ASCII was extended to Unicode In Unicode the number of bits used to represent a character can be 8 16 24 or 32 and more than 120000 characters are cur rently encoded in modern and historic language characters and multiple symbol sets Digital signals are important in signal analysis because of the spread of digital systems Digital signals often have better immunity to noise than analog signals In binary signal communication the bits can be detected very cleanly until the noise gets very large The detection of bit values in a stream of bits is usually done by comparing the signal value at a predetermined bit time with a threshold If it is above the thresh old it is declared a 1 and if it is below the threshold it is declared a 0 In Figure 111 the xs mark the signal value at the detection time and when this technique is applied to the noisy digital signal one of the bits is incorrectly detected But when the signal is processed by a filter all the bits are correctly detected The filtered digital signal does not look very clean in comparison with the noiseless digital signal but the bits can still be detected with a very low probability of error This is the basic reason that digital signals can have better noise immunity than analog signals An introduction to the analysis and design of filters is presented in Chapters 11 and 15 In this text we will consider both continuoustime and discretetime signals but we will mostly ignore the effects of signal quantization and consider all signals to be continuousvalue Also we will not directly consider the analysis of random signals although random signals will sometimes be used in illustrations The first signals we will study are continuoustime signals Some continuoustime signals can be described by continuous functions of time A signal xt might be described by a function xt 50 sin 200 πt of continuous time t This is an exact description of the signal at every instant of time The signal can also be described graphically Figure 112 Many continuoustime signals are not as easy to describe mathematically Consider the signal in Figure 113 Waveforms like the one in Figure 113 occur in various types of instrumentation and communication systems With the definition of some signal functions and an operation called convolution this signal can be compactly described analyzed and manipulated mathematically Continuoustime signals that can be described by math ematical functions can be transformed into another domain called the frequency domain through the continuoustime Fourier transform In this context transformation means transformation of a signal to the frequency domain This is an important tool in signal analysis which allows certain characteristics of the signal to be more clearly observed Figure 111 Use of a filter to reduce bit error rate in a digital signal xt 1 2 Noiseless Digital Signal t 26 1 2 t 26 1 2 t 26 1 1 0 1 0 0 0 1 0 0 1 0 0 0 1 1 1 1 0 1 0 0 0 1 0 1 1 0 0 0 1 1 1 1 0 1 0 0 0 1 0 0 1 0 0 0 1 1 Bit Error Bit Detection Threshold xft xt nt Noisy Digital Signal Filtered Digital Signal rob28124ch01001018indd 6 051216 221 pm 12 Types of Signals 7 and more easily manipulated than in the time domain In the frequency domain signals are described in terms of the frequencies they contain Without frequencydomain analy sis design and analysis of many systems would be considerably more difficult Discretetime signals are only defined at discrete points in time Figure 114 illustrates some discretetime signals Figure 112 A continuoustime signal described by a mathematical function 50 50 t 10 ms t xt Figure 113 A second continuoustime signal 5 t 20 μs t xt Figure 114 Some discretetime signals n xn n xn n xn n xn So far all the signals we have considered have been described by functions of time An important class of signals is functions of space instead of time images Most of the theories of signals the information they convey and how they are processed by systems in this text will be based on signals that are a variation of a physical phenome non with time But the theories and methods so developed also apply with only minor modifications to the processing of images Time signals are described by the variation of a physical phenomenon as a function of a single independent variable time Spa tial signals or images are described by the variation of a physical phenomenon as a rob28124ch01001018indd 7 051216 221 pm C h a p t e r 1 Introduction 8 function of two orthogonal independent spatial variables conventionally referred to as x and y The physical phenomenon is most commonly light or something that affects the transmission or reflection of light but the techniques of image processing are also applicable to anything that can be mathematically described by a function of two independent variables Historically the practical application of imageprocessing techniques has lagged behind the application of signalprocessing techniques because the amount of infor mation that has to be processed to gather the information from an image is typically much larger than the amount of information required to get the information from a time signal But now image processing is increasingly a practical technique in many situ ations Most image processing is done by computers Some simple imageprocessing operations can be done directly with optics and those can of course be done at very high speeds at the speed of light But direct optical image processing is very limited in its flexibility compared with digital image processing on computers Figure 115 shows two images On the left is an unprocessed Xray image of a carryon bag at an airport checkpoint On the right is the same image after being pro cessed by some imagefiltering operations to reveal the presence of a weapon This text will not go into image processing in any depth but will use some examples of image processing to illustrate concepts in signal processing An understanding of how signals carry information and how systems process sig nals is fundamental to multiple areas of engineering Techniques for the analysis of sig nals processed by systems are the subject of this text This material can be considered as an applied mathematics text more than a text covering the building of useful devices but an understanding of this material is very important for the successful design of useful devices The material that follows builds from some fundamental definitions and concepts to a full range of analysis techniques for continuoustime and discretetime signals in systems 13 EXAMPLES OF SYSTEMS There are many different types of signals and systems A few examples of systems are discussed next The discussion is limited to the qualitative aspects of the system with some illustrations of the behavior of the system under certain conditions These systems will be revisited in Chapter 4 and discussed in a more detailed and quantitative way in the material on system modeling Figure 115 An example of image processing to reveal information Original Xray image and processed version provided by the Imaging Robotics and Intelligent Systems IRIS Laboratory of the Department of Electrical and Computer Engineering at the University of Tennessee Knoxville rob28124ch01001018indd 8 051216 221 pm 13 Examples of Systems 9 A MECHANICAL SYSTEM A man bungee jumps off a bridge over a river Will he get wet The answer depends on several factors 1 The mans height and weight 2 The height of the bridge above the water 3 The length and springiness of the bungee cord When the man jumps off the bridge he goes into free fall caused by the force due to gravitational attraction until the bungee cord extends to its full unstretched length Then the system dynamics change because there is now another force on the man the bungee cords resistance to stretching and he is no longer in free fall We can write and solve a differential equation of motion and determine how far down the man falls before the bungee cord pulls him back up The differential equation of motion is a mathematical model of this mechanical system If the man weighs 80 kg and is 18 m tall and if the bridge is 200 m above the water level and the bungee cord is 30 m long unstretched with a spring constant of 11 Nm the bungee cord is fully extended be fore stretching at t 247 s The equation of motion after the cord starts stretching is xt 1685 sin 03708t 9525 cos 03708t 1013 t 247 11 Figure 116 shows his position versus time for the first 15 seconds From the graph it seems that the man just missed getting wet Figure 116 Mans vertical position versus time bridge level is zero 0 5 10 15 200 180 160 140 120 100 80 60 40 20 0 Time t s Elevation m Bridge Level Water Level Free Fall Bungee Stretched A FLUID SYSTEM A fluid system can also be modeled by a differential equation Consider a cylindrical water tank being fed by an input flow of water with an orifice at the bottom through which flows the output Figure 117 The flow out of the orifice depends on the height of the water in the tank The vari ation of the height of the water depends on the input flow and the output flow The rate rob28124ch01001018indd 9 051216 221 pm C h a p t e r 1 Introduction 10 of change of water volume in the tank is the difference between the input volumetric flow and the output volumetric flow and the volume of water is the crosssectional area of the tank times the height of the water All these factors can be combined into one differential equation for the water level h1 t A 1 d dt h 1 t A 2 2g h 1 t h 2 f 1 t 12 The water level in the tank is graphed in Figure 118 versus time for four volumetric inflows under the assumption that the tank is initially empty Figure 117 Tank with orifice being filled from above h t 1 h 2 v t 2 f t 2 f t 1 A1 A2 Figure 118 Water level versus time for four different volumetric inflows with the tank initially empty 0 1000 2000 3000 4000 5000 6000 7000 8000 0 05 1 15 2 25 3 35 Volumetric Inflow 0001 m3s Volumetric Inflow 0002 m3s Volumetric Inflow 0003 m3s Volumetric Inflow 0004 m3s Tank CrossSectional Area 1 m2 Orifice Area 00005 m2 Time t s Water Level h1t m As the water flows in the water level increases and that increases the water out flow The water level rises until the outflow equals the inflow After that time the water level stays constant Notice that when the inflow is increased by a factor of two the final water level is increased by a factor of four The final water level is proportional to the square of the volumetric inflow That fact makes the differential equation that models the system nonlinear rob28124ch01001018indd 10 051216 221 pm 13 Examples of Systems 11 A DISCRETETIME SYSTEM Discretetime systems can be designed in multiple ways The most common practical example of a discretetime system is a computer A computer is controlled by a clock that determines the timing of all operations Many things happen in a computer at the integrated circuit level between clock pulses but a computer user is only interested in what happens at the times of occurrence of clock pulses From the users point of view the computer is a discretetime system We can simulate the action of a discretetime system with a computer program For example yn 1 yn1 0 while 1 yn2 yn1 yn1 yn yn 197yn1 yn2 end This computer program written in MATLAB simulates a discretetime system with an output signal y that is described by the difference equation yn 197 yn 1 yn 2 13 along with initial conditions y0 1 and y1 0 The value of y at any time index n is the sum of the previous value of y at discrete time n 1 multiplied by 197 minus the value of y previous to that at discrete time n 2 The operation of this system can be diagrammed as in Figure 119 In Figure 119 the two squares containing the letter D are delays of one in discrete time and the arrowhead next to the number 197 represents an amplifier that multiplies the signal entering it by 197 to produce the signal leaving it The circle with the plus sign in it is a summing junction It adds the two signals entering it one of which is negated first to produce the signal leaving it The first 50 values of the signal produced by this system are illustrated in Figure 120 The system in Figure 119 could be built with dedicated hardware Discretetime delay can be implemented with a shift register Multiplication by a constant can be done with an amplifier or with a digital hardware multiplier Summation can also be done with an operational amplifier or with a digital hardware adder Figure 120 Signal produced by the discretetime system in Figure 119 n 50 yn 6 6 Figure 119 Discretetime system example yn yn2 yn1 197 D D rob28124ch01001018indd 11 051216 221 pm C h a p t e r 1 Introduction 12 FEEDBACK SYSTEMS Another important aspect of systems is the use of feedback to improve system perfor mance In a feedback system something in the system observes its response and may modify the input signal to the system to improve the response A familiar example is a thermostat in a house that controls when the air conditioner turns on and off The thermostat has a temperature sensor When the temperature inside the thermostat ex ceeds the level set by the homeowner a switch inside the thermostat closes and turns on the home air conditioner When the temperature inside the thermostat drops a small amount below the level set by the homeowner the switch opens turning off the air conditioner Part of the system a temperature sensor is sensing the thing the system is trying to control the air temperature and feeds back a signal to the device that actually does the controlling the air conditioner In this example the feedback signal is simply the closing or opening of a switch Feedback is a very useful and important concept and feedback systems are every where Take something everyone is familiar with the float valve in an ordinary flush toilet It senses the water level in the tank and when the desired water level is reached it stops the flow of water into the tank The floating ball is the sensor and the valve to which it is connected is the feedback mechanism that controls the water level If all the water valves in all flush toilets were exactly the same and did not change with time and if the water pressure upstream of the valve were known and constant and if the valve were always used in exactly the same kind of water tank it should be possible to replace the float valve with a timer that shuts off the water flow when the water reaches the desired level because the water would always reach the desired level at exactly the same elapsed time But water valves do change with time and water pres sure does fluctuate and different toilets have different tank sizes and shapes Therefore to operate properly under these varying conditions the tankfilling system must adapt by sensing the water level and shutting off the valve when the water reaches the desired level The ability to adapt to changing conditions is the great advantage of feedback methods There are countless examples of the use of feedback 1 Pouring a glass of lemonade involves feedback The person pouring watches the lemonade level in the glass and stops pouring when the desired level is reached 2 Professors give tests to students to report to the students their performance levels This is feedback to let the student know how well she is doing in the class so she can adjust her study habits to achieve her desired grade It is also feedback to the professor to let him know how well his students are learning 3 Driving a car involves feedback The driver senses the speed and direction of the car the proximity of other cars and the lane markings on the road and constantly applies corrective actions with the accelerator brake and steering wheel to maintain a safe speed and position 4 Without feedback the F117 stealth fighter would crash because it is aerodynamically unstable Redundant computers sense the velocity altitude roll pitch and yaw of the aircraft and constantly adjust the control surfaces to maintain the desired flight path Figure 121 Feedback is used in both continuoustime systems and discretetime systems The system in Figure 122 is a discretetime feedback system The response of the system yn is fed back to the upper summing junction after being delayed twice and multi plied by some constants rob28124ch01001018indd 12 051216 221 pm 13 Examples of Systems 13 Let this system be initially at rest meaning that all signals throughout the system are zero before time index n 0 To illustrate the effects of feedback let a 1 let b 15 let c 08 and let the input signal xn change from 0 to 1 at n 0 and stay at 1 for all time n 0 We can see the response y n in Figure 123 Now let c 06 and leave a and b the same Then we get the response in Figure 124 Now let c 05 and leave a and b the same Then we get the response in Figure 125 The response in Figure 125 increases forever This last system is unstable because a bounded input signal produces an unbounded response So feedback can make a system unstable Figure 121 The F117A Nighthawk stealth fighter Vol 87Corbis Figure 122 A discretetime feedback system xn yn D D b a c Figure 123 Discretetime system response with b 15 and c 08 n 60 yn 6 a 1 b 15 c 08 Figure 125 Discretetime system response with b 15 and c 05 n 60 yn 140 a 1 b 15 c 05 Figure 124 Discretetime system response with b 15 and c 06 n 60 yn 12 a 1 b 15 c 06 The system illustrated in Figure 126 is an example of a continuoustime feedback system It is described by the differential equation yʺt ayt xt The homoge neous solution can be written in the form y h t K h1 sin a t K h2 cos a t 14 If the excitation xt is zero and the initial value y t 0 is nonzero or the initial deriva tive of yt is nonzero and the system is allowed to operate in this form after t t 0 yt Figure 126 Continuoustime feedback system xt yt a rob28124ch01001018indd 13 051216 222 pm C h a p t e r 1 Introduction 14 will oscillate sinusoidally forever This system is an oscillator with a stable amplitude So feedback can cause a system to oscillate 14 A FAMILIAR SIGNAL AND SYSTEM EXAMPLE As an example of signals and systems lets look at a signal and system that everyone is familiar with sound and a system that produces andor measures sound Sound is what the ear senses The human ear is sensitive to acoustic pressure waves typically between about 15 Hz and about 20 kHz with some sensitivity variation in that range Below are some graphs of airpressure variations that produce some common sounds These sounds were recorded by a system consisting of a microphone that converts airpressure variation into a continuoustime voltage signal electronic circuitry that processes the continuoustime voltage signal and an ADC that changes the continuoustime voltage signal to a digital signal in the form of a sequence of binary numbers that are then stored in computer memory Figure 127 Figure 127 A sound recording system Microphone ADC Electronics Acoustic Pressure Variation Voltage Processed Voltage Binary Numbers Computer Memory Figure 128 The word signal spoken by an adult male voice 005 01 015 02 025 03 035 04 045 1 05 05 1 Time t s Delta pt Arbitrary Units Adult Male Voice Saying the Word Signal 007 0074 0078 02 01 0 01 02 Time t s Delta pt 015 0155 016 08 06 04 02 0 02 04 06 Time t s Delta pt 03 0305 031 01 005 0 005 Time t s Delta pt Consider the pressure variation graphed in Figure 128 It is the continuoustime pressure signal that produces the sound of the word signal spoken by an adult male the author rob28124ch01001018indd 14 051216 222 pm 14 A Familiar Signal and System Example 15 Analysis of sounds is a large subject but some things about the relationship between this graph of airpressure variation and what a human hears as the word signal can be seen by looking at the graph There are three identifiable bursts of signal 1 from 0 to about 012 seconds 2 from about 012 to about 019 seconds and 3 from about 022 to about 04 seconds Burst 1 is the s in the word signal Burst 2 is the i sound The region between bursts 2 and 3 is the double consonant gn of the word signal Burst 3 is the a sound terminated by the l consonant stop An l is not quite as abrupt a stop as some other consonants so the sound tends to trail off rather than stopping quickly The variation of air pressure is generally faster for the s than for the i or the a In signal analysis we would say that it has more highfrequency content In the blowup of the s sound the airpressure variation looks almost random The i and a sounds are different in that they vary more slowly and are more regular or predictable although not exactly predictable The i and a are formed by vibrations of the vocal cords and therefore exhibit an approximately oscillatory behavior This is described by saying that the i and a are tonal or voiced and the s is not Tonal means having the basic quality of a single tone or pitch or frequency This description is not mathematically precise but is useful qualitatively Another way of looking at a signal is in the frequency domain mentioned above by examining the frequencies or pitches that are present in the signal A common way of illustrating the variation of signal power with frequency is its power spectral density a graph of the power density in the signal versus frequency Figure 129 shows the three bursts s i and a from the word signal and their associated power spectral densities the G f functions Figure 129 Three sounds in the word signal and their associated power spectral densities t Delta pt f 22000 22000 t Delta pt f 22000 22000 t Delta pt f 22000 22000 G f G f G f 016 s 01 s 012 s s Sound i Sound a Sound Power spectral density is just another mathematical tool for analyzing a signal It does not contain any new information but sometimes it can reveal things that are dif ficult to see otherwise In this case the power spectral density of the s sound is widely distributed in frequency whereas the power spectral densities of the i and a sounds are narrowly distributed in the lowest frequencies There is more power in the s sound at rob28124ch01001018indd 15 051216 222 pm C h a p t e r 1 Introduction 16 higher frequencies than in the i and a sounds The s sound has an edge or hissing quality caused by the high frequencies in the s sound The signal in Figure 128 carries information Consider what happens in conver sation when one person says the word signal and another hears it Figure 130 The speaker thinks first of the concept of a signal His brain quickly converts the concept to the word signal Then his brain sends nerve impulses to his vocal cords and di aphragm to create the air movement and vibration and tongue and lip movements to produce the sound of the word signal This sound then propagates through the air between the speaker and the listener The sound strikes the listeners eardrum and the vibrations are converted to nerve impulses which the listeners brain converts first to the sound then the word then the concept signal Conversation is accomplished by a system of considerable sophistication How does the listeners brain know that the complicated pattern in Figure 128 is the word signal The listener is not aware of the detailed airpressure varia tions but instead hears sounds that are caused by the airpressure variation The eardrum and brain convert the complicated airpressure pattern into a few simple features That conversion is similar to what we will do when we convert signals into the frequency domain The process of recognizing a sound by reducing it to a small set of features reduces the amount of information the brain has to process Signal processing and analysis in the technical sense do the same thing but in a more math ematically precise way Two very common problems in signal and system analysis are noise and interfer ence Noise is an undesirable random signal Interference is an undesirable nonran dom signal Noise and interference both tend to obscure the information in a signal Figure 131 shows examples of the signal from Figure 128 with different levels of noise added As the noise power increases there is a gradual degradation in the intelligibility of the signal and at some level of noise the signal becomes unintelligible A measure of the quality of a received signal corrupted by noise is the ratio of the signal power to the noise power commonly called signaltonoise ratio and often abbreviated SNR In each of the examples of Figure 131 the SNR is specified Sounds are not the only signals of course Any physical phenomenon that is mea sured or observed is a signal Also although the majority of signals we will consider in this text will be functions of time a signal can be a function of some other independent Figure 130 Communication between two people involving signals and signal processing by systems Signal Signal rob28124ch01001018indd 16 051216 222 pm 14 A Familiar Signal and System Example 17 SignaltoNoise Ratio 237082 Original Signal Without Noise SignaltoNoise Ratio 37512 SignaltoNoise Ratio 095621 Figure 131 Sound of the word signal with different levels of noise added variable like frequency wavelength distance and so on Figure 132 and Figure 133 illustrate some other kinds of signals Just as sounds are not the only signals conversation between two people is not the only system Examples of other systems include the following 1 An automobile suspension for which the road surface excites the automobile and the position of the chassis relative to the road is the response 2 A chemical mixing vat for which streams of chemicals are the input signals and the mixture of chemicals is the output signal 3 A building environmental control system for which the exterior temperature is the input signal and the interior temperature is the response FarField Intensity of Light Diffracted Through a Slit Tt Ft t Outside Air Temperature 24 hours t Neutron Flux in a Nuclear Reactor Core 1 ms Sλ λ Optical Absorption Spectrum of a Chemical Mixture 400 nm 700 nm Iθ θ 30 TwoDimensional Image Correlation x y Cxy Figure 132 Examples of signals that are functions of one or more continuous independent variables rob28124ch01001018indd 17 051216 222 pm C h a p t e r 1 Introduction 18 4 A chemical spectroscopy system in which white light excites the specimen and the spectrum of transmitted light is the response 5 A telephone network for which voices and data are the input signals and reproductions of those voices and data at a distant location are the output signals 6 Earths atmosphere which is excited by energy from the sun and for which the responses are ocean temperature wind clouds humidity and so on In other words the weather is the response 7 A thermocouple excited by the temperature gradient along its length for which the voltage developed between its ends is the response 8 A trumpet excited by the vibration of the players lips and the positions of the valves for which the response is the tone emanating from the bell The list is endless Any physical entity can be thought of as a system because if we excite it with physical energy it has a physical response 15 USE OF MATLAB Throughout the text examples will be presented showing how signal and system analy sis can be done using MATLAB MATLAB is a highlevel mathematical tool available on many types of computers It is very useful for signal processing and system analysis There is an introduction to MATLAB in Web Appendix A n Nn Dn Nn Pn Number of Cars Crossing an Intersection Between Red Lights United States Population 2 4 6 8 n BallBearing Manufacturers Quality Control Chart for Diameter 1 cm 101 cm 099 cm n 1800 1900 2000 300 Million World War II Great Depression World War I US Civil War n 1950 2000 2500 Number of Annual Sunspots Figure 133 Examples of signals that are functions of a discrete independent variable rob28124ch01001018indd 18 051216 222 pm 19 21 INTRODUCTION AND GOALS Over the years signal and system analysts have observed many signals and have real ized that signals can be classified into groups with similar behavior Figure 21 shows some examples of signals t xt AmplitudeModulated Carrier in a Communication System t xt Car Bumper Height after Car Strikes a Speed Bump t xt Light Intensity from a QSwitched Laser t xt Step Response of an RC Lowpass Filter t xt FrequencyShiftKeyed Binary Bit Stream t xt Manchester Encoded Baseband Binary Bit Stream Figure 21 Examples of signals In signal and system analysis signals are described by mathematical functions Some of the functions that describe real signals should already be familiar exponentials and sinusoids These occur frequently in signal and system analysis One set of functions has been defined to describe the effects on signals of switching operations that often occur in systems Some other functions arise in the development of certain system analysis tech niques which will be introduced in later chapters These functions are all carefully chosen to be simply related to each other and to be easily changed by a wellchosen set of shifting andor scaling operations They are prototype functions which have simple definitions and are easily remembered The types of symmetries and patterns that most frequently occur in real signals will be defined and their effects on signal analysis explored C H A P T E R 2 Mathematical Description of ContinuousTime Signals rob28124ch02019078indd 19 041216 115 pm C h a p t e r 2 Mathematical Description of ContinuousTime Signals 20 CHAPTER GOA L S 1 To define some mathematical functions that can be used to describe signals 2 To develop methods of shifting scaling and combining those functions to represent real signals 3 To recognize certain symmetries and patterns to simplify signal and system analysis 22 FUNCTIONAL NOTATION A function is a correspondence between the argument of the function which lies in its domain and the value returned by the function which lies in its range The most familiar functions are of the form gx where the argument x is a real number and the value returned g is also a real number But the domain andor range of a function can be complex numbers or integers or a variety of other choices of allowed values In this text five types of functions will appear 1 DomainReal numbers RangeReal numbers 2 DomainIntegers RangeReal numbers 3 DomainIntegers RangeComplex numbers 4 DomainReal numbers RangeComplex numbers 5 DomainComplex numbers RangeComplex numbers For functions whose domain is either real numbers or complex numbers the argument will be enclosed in parentheses For functions whose domain is integers the argu ment will be enclosed in brackets These types of functions will be discussed in more detail as they are introduced 23 CONTINUOUSTIME SIGNAL FUNCTIONS If the independent variable of a function is time t and the domain of the function is the real numbers and if the function gt has a defined value at every value of t the function is called a continuoustime function Figure 22 illustrates some continuoustime functions Figure 22 Examples of continuoustime functions t gt t gt t gt t gt a b c d Points of Discontinuity of gt rob28124ch02019078indd 20 041216 115 pm 23 ContinuousTime Signal Functions 21 Figure 22d illustrates a discontinuous function for which the limit of the func tion value as we approach the discontinuity from above is not the same as when we approach it from below If t t 0 is a point of discontinuity of a function gt then lim ε0 g t 0 ε lim ε0 g t 0 ε All four functions ad are continuoustime functions because their values are de fined for all real values of t Therefore the terms continuous and continuoustime mean slightly different things All continuous functions of time are continuoustime func tions but not all continuoustime functions are continuous functions of time COMPLEX EXPONENTIALS AND SINUSOIDS Realvalued sinusoids and exponential functions should already be familiar In gt A cos2πt T 0 θ A cos2π f 0 t θ A cos ω 0 t θ and gt A e σ 0 j ω 0 t A e σ 0 t cos ω 0 t j sin ω 0 t A is the amplitude T 0 is the fundamental period f 0 is the fundamental cyclic frequency and ω 0 is the fundamental radian frequency of the sinusoid t is time and σ 0 is the decay rate of the exponential which is the reciprocal of its time constant τ Figure 23 and Figure 24 All these parameters can be any real number Figure 23 A real sinusoid and a real exponential with parameters indicated graphically t A t A τ gt A cos2πf0tθ gt Aetτ θ2πf0 T0 4 4 t 10 ms t 4sin200πt μA 10 10 t 2 μs t 10cos106πt nC 2 t 01 s t 2e10tm 5 5 t 1 s t 5etsin2πt m s2 Figure 24 Examples of signals described by real sines cosines and exponentials In Figure 24 the units indicate what kind of physical signal is being described Very often in system analysis when only one kind of signal is being followed through a system the units are omitted for the sake of brevity Exponentials exp and sinusoids sin and cos are intrinsic functions in MATLAB The arguments of the sin and cos functions are interpreted by MATLAB as radians not degrees exp1sinpi2cospi ans 27183 10000 10000 pi is the MATLAB symbol for π rob28124ch02019078indd 21 041216 115 pm C h a p t e r 2 Mathematical Description of ContinuousTime Signals 22 Sinusoids and exponentials are very common in signal and system analysis because most continuoustime systems can be described at least approximately by linear constantcoefficient ordinary differential equations whose eigenfunctions are complex exponentials complex powers of e the base of the natural logarithms Eigenfunction means characteristic function and the eigenfunctions have a particularly important relation to the differential equation If the exponent of e is real complex exponentials are the same as real exponentials Through Eulers identity e jx cosx j sinx and the relations cosx 12 e jx e jx and sinx 1j2 e jx e jx complex expo nentials and realvalued sinusoids are closely related If in a function of the form e jx x is a realvalued independent variable this special form of the complex exponential is called a complex sinusoid Figure 25 t 2 Ree j2πt 1 Ime j2πt 1 1 1 t 2 Reej2πt 1 Imej2πt 1 1 1 t 2 Re 2 Im 1 1 2 t 2 Re 1 Im 2 2 1 Figure 25 The relation between real and complex sinusoids In signal and system analysis sinusoids are expressed in either the cyclic fre quency f form A cos2π f 0 t θ or the radian frequency ω form A cos ω 0 t θ The advantages of the f form are the following 1 The fundamental period T 0 and the fundamental cyclic frequency f 0 are simply reciprocals of each other 2 In communication system analysis a spectrum analyzer is often used and its display scale is usually calibrated in Hz Therefore f is the directly observed variable 3 The definition of the Fourier transform Chapter 6 and some transforms and transform relationships are simpler in the f form than in the ω form The advantages of the ω form are the following 1 Resonant frequencies of real systems expressed directly in terms of physical parameters are more simply expressed in the ω form than in the f form The resonant frequency of an LC oscillator is ω 0 2 1LC 2π f 0 2 and the halfpower corner frequency of an RC lowpass filter is ω c 1RC 2π f c 2 The Laplace transform Chapter 8 is defined in a form that is more simply related to the ω form than to the f form rob28124ch02019078indd 22 041216 115 pm 23 ContinuousTime Signal Functions 23 3 Some Fourier transforms are simpler in the ω form 4 Use of ω in some expressions makes them more compact For example A cos ω 0 t θ is a little more compact than A cos2π f 0 t θ Sinusoids and exponentials are important in signal and systems analysis because they arise naturally in the solutions of the differential equations that often describe sys tem dynamics As we will see in the study of the Fourier series and Fourier transform even if signals are not sinusoids most of them can be expressed as linear combinations of sinusoids FUNCTIONS WITH DISCONTINUITIES Continuoustime sines cosines and exponentials are all continuous and differentiable at every point in time But many other types of important signals that occur in practical systems are not continuous or differentiable everywhere A common operation in sys tems is to switch a signal on or off at some time Figure 26 Figure 26 Examples of signals that are switched on or off at some time 20 V 20 V t 50 ns t 3 W t t 10 s t 2 ms t xt xt xt 7 Pa t xt 4 C 0 t 0 3W t 0 0 t 10 s 4e01tC t 10 s 7 Pa t 2 ms 0 t 2 ms 0 t 0 20sin4π107t V t 0 The functional descriptions of the signals in Figure 26 are complete and accurate but are in a cumbersome form Signals of this type can be better described mathemat ically by multiplying a function that is continuous and differentiable for all time by another function that switches from zero to one or one to zero at some finite time In signal and system analysis singularity functions which are related to each other through integrals and derivatives can be used to mathematically describe signals that have discontinuities or discontinuous derivatives These functions and functions that are closely related to them through some common system operations are the subject of this section In the consideration of singularity functions we will extend modify andor generalize some basic mathematical concepts and operations to allow us to efficiently analyze real signals and systems We will extend the con cept of what a derivative is and we will also learn how to use an important mathe matical entity the impulse which is a lot like a function but is not a function in the usual sense rob28124ch02019078indd 23 041216 115 pm C h a p t e r 2 Mathematical Description of ContinuousTime Signals 24 t sgnt 1 1 t sgnt 1 1 Figure 27 The signum function The Signum Function For nonzero arguments the value of the signum function has a magnitude of one and a sign that is the same as the sign of its argument sgn t 1 t 0 0 t 0 1 t 0 21 See Figure 27 The graph on the left in Figure 27 is of the exact mathematical definition The graph on the right is a more common way of representing the function for engineering purposes No practical signal can change discontinuously so if an approximation of the signum function were generated by a signal generator and viewed on an oscilloscope it would look like the graph on the right The signum function is intrinsic in MATLAB and called the sign function The UnitStep Function The unitstep function is defined by ut 1 t 0 12 t 0 0 t 0 22 See Figure 28 It is called the unit step because the step is one unit high in the system of units used to describe the signal1 1 Some authors define the unit step by ut 1 t 0 0 t 0 or ut 1 t 0 0 t 0 or ut 1 t 0 0 t 0 In the middle definition the value at t 0 is undefined but finite The unit steps defined by these definitions all have an identical effect on any real physical system t ut 1 t ut 1 1 2 Figure 28 The unitstep function rob28124ch02019078indd 24 041216 115 pm 23 ContinuousTime Signal Functions 25 The unit step can mathematically represent a common action in real physical sys tems fast switching from one state to another In the circuit of Figure 29 the switch moves from one position to the other at time t 0 The voltage applied to the RC network is v RC t V b ut The current flowing clockwise through the resistor and capacitor is it V b R e tRC ut and the voltage across the capacitor is vt V b 1 e tRC ut There is an intrinsic function in MATLAB called heaviside2 which returns a one for positive arguments a zero for negative arguments and an NaN for zero arguments The MATLAB constant NaN is not a number and indicates an undefined value There are practical problems using this function in numerical computations because the re turn of an undefined value can cause some programs to prematurely terminate or return useless results We can create our own functions in MATLAB which become functions we can call upon just like the intrinsic functions cos sin exp etc MATLAB functions are defined by creating an m file a file whose name has the extension m We could create a function that finds the length of the hypotenuse of a right triangle given the lengths of the other two sides Function to compute the length of the hypotenuse of a right triangle given the lengths of the other two sides a The length of one side b The length of the other side c The length of the hypotenuse function c hypab function c hypab c sqrta2 b2 The first nine lines in this example which are preceded by are comment lines that are not executed but serve to document how the function is used The first execut able line must begin with the keyword function The rest of the first line is in the form result namearg1 arg2 2 Oliver Heaviside was a selftaught English electrical engineer who adapted complex numbers to the study of electrical circuits invented mathematical techniques for the solution of differential equations and reformulated and simplified Maxwells field equations Although at odds with the scientific establishment for most of his life Heaviside changed the face of mathematics and science for years to come It has been reported that a man once complained to Heaviside that his writings were very difficult to read Heavisides response was that they were even more difficult to write Vb R C t 0 vRCt Figure 29 Circuit with a switch whose effect can be represented by a unit step rob28124ch02019078indd 25 041216 115 pm C h a p t e r 2 Mathematical Description of ContinuousTime Signals 26 where result will contain the returned value which can be a scalar a vector or a ma trix or even a cell array or a structure which are beyond the scope of this text name is the function name and arg1 arg2 are the parameters or arguments passed to the function The arguments can also be scalars vectors or matrices or cell arrays or structures The name of the file containing the function definition must be namem Below is a listing of a MATLAB function to implement the unitstep function in numerical computations Unitstep function defined as 0 for input argument values less than zero 12 for input argument values equal to zero and 1 for input argument values greater than zero This function uses the sign function to implement the unitstep function Therefore value at t 0 is defined This avoids having undefined values during the execution of a program that uses it function y usx function y usx y signx 12 This function should be saved in a file named usm The UnitRamp Function Another type of signal that occurs in systems is one that is switched on at some time and changes linearly after that time or changes linearly before some time and is switched off at that time Figure 210 Signals of this kind can be described with the use of the ramp function The unitramp function Figure 211 is the integral of the unitstep function It is called the unitramp function because for positive t its slope is one amplitude unit per time unit rampt t t 0 0 t 0 t uλdλ tu t 23 20 t 100 ms t 1V t 6 s t t 10 s t 20 μs t xt xt xt t xt 4 mA 12 N cm s Figure 210 Functions that change linearly before or after some time or are multiplied by functions that change linearly before or after some time t rampt 1 1 Figure 211 The unitramp function rob28124ch02019078indd 26 041216 115 pm 23 ContinuousTime Signal Functions 27 The ramp is defined by rampt t uτdτ In this equation the symbol τ is the independent variable of the unitstep function and the variable of integration But t is the independent variable of the ramp function The equation says to find the value of the ramp function at any value of t start with τ at negative infinity and move in τ up to τ t while accumulating the area under the unitstep function The total area accumulated from τ to τ t is the value of the ramp function at time t Figure 212 For t less than zero no area is accumulated For t greater than zero the area accumulated equals t because it is the area of a rectangle with width t and height one τ uτ uτ uτ uτ 1 t Rampt 1 2 3 4 5 5 4 3 2 1 1 2 3 4 5 54321 τ 1 1 2 3 4 5 54321 τ 1 1 2 3 4 5 54321 τ 1 1 2 3 4 5 54321 12345 t 1 t 1 t 3 t 5 Figure 212 Integral relationship between the unit step and the unit ramp Some authors prefer to use the expression tut instead of rampt Since they are equal the use of either one is correct and just as legitimate as the other one Below is a MATLAB m file for the ramp function Function to compute the ramp function defined as 0 for values of the argument less than or equal to zero and the value of the argument for arguments greater than zero Uses the unitstep function usx function y rampx function y rampx y xusx The Unit Impulse Before we define the unit impulse we will first explore an important idea Consider a unitarea rectangular pulse defined by Δt 1a t a2 0 t a2 See Figure 213 Let this function multiply a function gt that is finite and continuous at t 0 and find the area A under the product of the two functions A Δt gtdt Figure 214 rob28124ch02019078indd 27 041216 115 pm C h a p t e r 2 Mathematical Description of ContinuousTime Signals 28 Using the definition of Δt we can rewrite the integral as A 1 a a2 a2 gtdt The function gt is continuous at t 0 Therefore it can be expressed as a McLaurin series of the form gt m0 g m 0 m t m g0 g 0t g 0 2 t 2 g m 0 m t m Then the integral becomes A 1 a a2 a2 g0 g 0t g 0 2 t 2 g m 0 m t m dt All the odd powers of t contribute nothing to the integral because it is taken over sym metrical limits about t 0 Carrying out the integral A 1 a ag0 a 3 12 g 0 2 a 5 80 g 4 0 4 Take the limit of this integral as a approaches zero lim a0 A g0 In the limit as a approaches zero the function Δt extracts the value of any continuous finite function gt at time t 0 when the product of Δt and gt is integrated over any range of time that includes time t 0 Now try a different definition of the function Δt Define it now as Δt 1a 1 t a t a 0 t a See Figure 215 If we make the same argument as before we get the area A Δtgtdt 1 a a a 1 t a gtdt Δt t 1 a 2 a 2 a Figure 213 A unitarea rectangular pulse of width a t Δt Δtgt 1 a 2 a 2 a gt Figure 214 Product of a unitarea rectangular pulse centered at t 0 and a func tion gt that is continuous and finite at t 0 rob28124ch02019078indd 28 041216 115 pm 23 ContinuousTime Signal Functions 29 Taking the limit as a approaches zero we again get g0 exactly the same result we got with the previous definition of Δt The two definitions of Δt have the same effect in the limit as a approaches zero but not before The shape of the function is not what is important in the limit but its area is important In either case Δt is a function with an area of one independent of the value of a As a approaches zero these functions do not have a shape in the ordinary sense because there is no time in which to develop one There are many other definitions of Δt that could be used with exactly the same effect in the limit The unit impulse δt can now be implicitly defined by the property that when it is multiplied by any function gt that is finite and continuous at t 0 and the product is integrated over a time range that includes t 0 the result is g0 g0 α β δtgtdt α 0 β In other words δtgtdt lim a0 Δtgtdt 24 where Δt is any of many functions that have the characteristics described above The notation δt is a convenient shorthand notation that avoids having to constantly take a limit when using impulses The Impulse the Unit Step and Generalized Derivatives One way of introducing the unit impulse is to define it as the derivative of the unitstep function Strictly speak ing the derivative of the unit step ut is undefined at t 0 But consider a function gt of time and its time derivative g t in Figure 216 t gt 1 a 2 a 2 1 2 t gʹt a 2 1a a 2 Figure 216 Functions that approach the unit step and unit impulse Δt t 1 a a a Figure 215 A unitarea triangular pulse of base halfwidth a rob28124ch02019078indd 29 041216 115 pm C h a p t e r 2 Mathematical Description of ContinuousTime Signals 30 The derivative of gt exists for all t except at t a2 and at t a2 As a ap proaches zero the function gt approaches the unit step In that same limit the nonzero part of the function g t approaches zero while its area remains the same one So g t is a shortduration pulse whose area is always one the same as the initial definition of Δt above with the same implications The limit as a approaches zero of g t is called the generalized derivative of ut Therefore the unit impulse is the generalized derivative of the unit step The generalized derivative of any function gt with a discontinuity at t t 0 is d dt gt d dt gt t t 0 lim ε0 gt ε gt ε Size of the discontinuity δt t 0 ε 0 The unit step is the integral of the unit impulse ut t δλdλ The derivative of the unit step ut is zero everywhere except at t 0 so the unit impulse is zero everywhere except at t 0 Since the unit step is the integral of the unit impulse a definite integral of the unit impulse whose integration range includes t 0 must have the value one These two facts are often used to define the unit impulse δt 0 t 0 and t 1 t 2 δt dt 1 t 1 0 t 2 0 otherwise 25 The area under an impulse is called its strength or sometimes its weight An impulse with a strength of one is called a unit impulse The exact definition and characteristics of the impulse require a plunge into generalized function theory It will suffice here to consider a unit impulse simply to be a pulse of unit area whose duration is so small that making it any smaller would not significantly change any signals in the system to which it is applied The impulse cannot be graphed in the same way as other functions because its value is undefined when its argument is zero The usual convention for graphing an impulse is to use a vertical arrow Sometimes the strength of the impulse is written beside it in parentheses and sometimes the height of the arrow indicates the strength of the impulse Figure 217 illustrates some ways of representing impulses graphically t δt 1 t δt 1 t 9δt1 9 t 3δt2 3 1 2 Figure 217 Graphical representations of impulses The Equivalence Property of the Impulse A common mathematical operation in signal and system analysis is the product of an impulse with another function gt Aδt t 0 Consider that the impulse Aδt t 0 is the limit of a pulse with area A centered at t t 0 with width a as a approaches zero Figure 218 The product is a pulse whose height rob28124ch02019078indd 30 041216 115 pm 23 ContinuousTime Signal Functions 31 at the midpoint is Ag t 0 a and whose width is a As a approaches zero the pulse be comes an impulse and the strength of that impulse is Ag t 0 Therefore gtAδt t 0 g t 0 Aδt t 0 26 This is sometimes called the equivalence property of the impulse The Sampling Property of the Impulse Another important property of the unit im pulse that follows from the equivalence property is its sampling property gtδt t 0 dt g t 0 27 According to the equivalence property the product gt δt t 0 is equal to g t 0 δt t 0 Since t 0 is one particular value of t it is a constant and g t 0 is also a constant and gtδt t 0 dt g t 0 δt t 0 dt 1 g t 0 Equation 27 is called the sampling property of the impulse because in an integral of this type it samples the value of the function gt at time t t 0 An older name is sifting property The impulse sifts out the value of gt at time t t 0 The Scaling Property of the Impulse Another important property of the impulse is its scaling property δat t 0 1 a δt t 0 28 which can be proven through a change of variable in the integral definition and sepa rate consideration of positive and negative values for a see Exercise 35 Figure 219 illustrates some effects of the scaling property of the impulse There is a function in MATLAB called dirac that implements the unit impulse in a limited sense It returns zero for nonzero arguments and it returns inf for zero argu ments This is not often useful for numerical computations but it is useful for symbolic analysis The continuoustime impulse is not an ordinary function It is sometimes t Aa t0 t0 Aδtt0 t a 2 t0 a 2 t0 a 2 t0 a 2 t0 gtAδtt0 Agt0 a gt0 gt Figure 218 Product of a function gt and a rectangular function that becomes an impulse as its width approaches zero rob28124ch02019078indd 31 041216 115 pm C h a p t e r 2 Mathematical Description of ContinuousTime Signals 32 possible to write a MATLAB function that can in certain types of computations be used to simulate the impulse and obtain useful numerical results But this must be done with great care based on a complete understanding of impulse properties No MATLAB function will be presented here for the continuoustime impulse because of these complications The Unit Periodic Impulse or Impulse Train Another useful generalized function is the periodic impulse or impulse train Figure 220 a uniformly spaced infinite sequence of unit impulses δ T t n δt nT 29 t δ3t t 2 1 1 3 δ t1 2 t 2 2 δ t 1 2 Figure 219 Examples of the effect of the scaling property of impulses t δTt δTt 1 1 1 1 1 T 2T T 2T t 1 T 2T T 2T Figure 220 The periodic impulse We can derive a scaling property for the periodic impulse From the definition δ T at t 0 k δat t 0 kT Using the scaling property of the impulse δ T at t 0 1 a k δt t 0 kT a and the summation can be recognized as a periodic impulse of period T a δ T at t 0 1a δ Ta t t 0 The impulse and periodic impulse may seem very abstract and unrealistic The impulse will appear later in a fundamental operation of linear system analysis the convolution integral Although as a practical matter a true impulse is impossible to generate the mathematical impulse and the periodic impulse are very useful in signal and system analysis Using them rob28124ch02019078indd 32 041216 115 pm 23 ContinuousTime Signal Functions 33 and the convolution operation we can mathematically represent in a compact notation many useful signals that would be more cumbersome to represent in another way3 A Coordinated Notation for Singularity Functions The unit step unit impulse and unit ramp are the most important members of the sin gularity functions In some signal and system literature these functions are indicated by the coordinated notation u k t in which the value of k determines the function For example u 0 t δt u 1 t ut and u 2 t rampt In this notation the subscript indicates how many times an impulse is differentiated to obtain the function in question and a negative subscript indicates that integration is done instead of differentiation The unit doublet u 1 t is defined as the generalized derivative of the unit impulse the unit triplet u 2 t is defined as the generalized derivative of the unit doublet and so on Even though the unit doublet and triplet and higher generalized derivatives are even less practical than the unit impulse they are sometimes useful in signal and system theory The UnitRectangle Function A very common type of signal occurring in systems is one that is switched on at some time and then off at a later time It is convenient to define the unit rectangle function Figure 221 for use in describing this type of signal rectt 1 t 12 12 t 12 0 t 12 ut 12 ut 12 210 It is a unitrectangle function because its width height and area are all one Use of the rectangle function shortens the notation when describing some signals The unitrectangle function can be thought of as a gate function When it multiplies an other function the product is zero outside its nonzero range and is equal to the other function inside its nonzero range The rectangle opens a gate allowing the other function through and then closes the gate again Table 21 summarizes the functions and the impulse and periodic impulse described above Unit rectangle function Uses the unitstep function usx function y rectx function y rectx y usx05 usx 05 3 Some authors prefer to always refer to the periodic impulse as a summation of impulses n δt nT This notation is less compact than δ T t but may be considered easier than remembering how to use the new function name Other authors may use different names t rectt 1 t rectt 1 1 2 1 2 1 2 1 2 1 2 Figure 221 The unitrectangle function rob28124ch02019078indd 33 041216 115 pm C h a p t e r 2 Mathematical Description of ContinuousTime Signals 34 Table 21 Summary of continuoustime signal functions the impulse and the periodic impulse Sine sin2π f 0 t or sin ω 0 t Cosine cos2π f 0 t or cos ω 0 t Exponential e st Unit Step ut Signum sgn t Unit Ramp rampt t ut Unit Impulse δt Periodic Impulse δ T t n δt nT Unit Rectangle rectt ut 12 ut 12 24 COMBINATIONS OF FUNCTIONS Standard functional notation for a continuoustime function is gt in which g is the function name and everything inside the parentheses is called the argument of the function The argument is written in terms of the independent variable In the case of gt t is the independent variable and the expression is the simplest possible expression in terms of t t itself A function gt returns a value g for every value of t it accepts In the function gt 2 4t2 for any value of t there is a corresponding value of g If t is 1 then g is 6 and that is indicated by the notation g1 6 The argument of a function need not be simply the independent variable If gt 5e2t what is gt 3 We replace t by t 3 everywhere on both sides of gt 5e2t get gt 3 5e2t3 Observe that we do not get 5e2t3 Since t was multiplied by minus two in the exponent of e the entire expression t 3 must also be multiplied by minus two in the new exponent of e Whatever was done with t in the function gt must be done with the entire expression involving t in any other function gex pression If gt 3 t2 2t3 then g2t 3 2t2 22t3 3 4t2 16t3 and g1 t 3 1 t2 21 t3 2 4t 5t2 2t 3 If gt 10 cos20πt then gt4 10 cos20πt4 10 cos5πt and get 10 cos20πet If gt 5e10t then g2x 5e20x and gz 1 5e10e10z In MATLAB when a function is invoked by passing an argument to it MATLAB evaluates the argument then computes the function value For most functions if the argument is a vector or matrix a value is returned for each element of the vector or matrix Therefore MATLAB functions do exactly what is described here for arguments that are functions of the independent variable They accept numbers and return other numbers exp15 ans 27183 73891 200855 545982 1484132 us1051 ans 0 0 05000 10000 10000 rect 080408 ans 0 1 1 1 0 rob28124ch02019078indd 34 041216 115 pm 24 Combinations of Functions 35 In some cases a single mathematical function may completely describe a signal But often one function is not enough for an accurate description An operation that allows versatility in the mathematical representation of arbitrary signals is combining two or more functions The combinations can be sums differences products andor quotients of functions Figure 222 shows some examples of sums products and quo tients of functions The sinc function will be defined in Chapter 6 1 1 3 3 1 1 10 10 1 1 1 1 1 2 2 t t t t sin4πt2 cos40πt e2tcos10πt cos20πt cos22πt sin4πt 4πt sinc4t Figure 222 Examples of sums products and quotients of functions ExamplE 21 Graphing function combinations with MATLAB Using MATLAB graph the function combinations x 1 t e t sin20πt e t2 sin19πt x 2 t rect t cos20πt Program to graph some demonstrations of continuoustime function combinations t 012406 Vector of time points for graphing x1 Generate values of x1 for graphing x1 exptsin20pit expt2sin19pit subplot211 Graph in the top half of the figure window p plottx1k Display the graph with black lines setpLineWidth2 Set the line width to 2 Label the abscissa and ordinate xlabelittFontNameTimesFontSize24 rob28124ch02019078indd 35 041216 115 pm C h a p t e r 2 Mathematical Description of ContinuousTime Signals 36 ylabelx1ittFontNameTimesFontSize24 setgcaFontNameTimesFontSize18 grid on t 212402 Vector of time points for graphing x2 Generate values of x2 for graphing x2 recttcos20pit subplot212 Graph in the bottom half of the figure window p plottx2k Display the graph with black lines setpLineWidth2 Set the line width to 2 Label the abscissa and ordinate xlabelittFontNameTimesFontSize24 ylabelx2ittFontNameTimesFontSize24 setgcaFontNameTimesFontSize18 grid on The graphs that result are shown in Figure 223 0 1 2 3 4 5 6 2 0 2 t 2 15 1 05 0 05 1 15 2 1 0 1 t x1t x2t Figure 223 MATLAB graphical result 25 SHIFTING AND SCALING It is important to be able to describe signals both analytically and graphically and to be able to relate the two different kinds of descriptions to each other Let gt be defined by Figure 224 with some selected values in the table to the right of the figure To com plete the function description let gt 0 t 5 AMPLITUDE SCALING Consider multiplying a function by a constant This can be indicated by the notation gt Agt Thus gt Agt multiplies gt at every value of t by A This is called amplitude scaling Figure 225 shows two examples of amplitude scaling the function gt defined in Figure 224 rob28124ch02019078indd 36 041216 115 pm 25 Shifting and Scaling 37 A negative amplitudescaling factor flips the function vertically If the scaling fac tor is 1 as in this example flipping is the only action If the scaling factor is some other factor A and A is negative amplitude scaling can be thought of as two successive operations gt gt A gt a flip followed by a positive amplitude scaling Amplitude scaling directly affects the dependent variable g The following two sec tions introduce the effects of changing the independent variable t TIME SHIFTING If the graph in Figure 224 defines gt what does gt 1 look like We can under stand the effect by graphing the value of gt 1 at multiple points as in Figure 226 It should be apparent after examining the graphs and tables that replacing t by t 1 shifts the function one unit to the right Figure 226 The change t t 1 can be described by saying for every value of t look back one unit in time get the value of g at that time and use it as the value for gt 1 at time t This is called time shifting or time translation We can summarize time shifting by saying that the change of independent variable t t t 0 where t 0 is any constant has the effect of shifting gt to the right by t 0 units Consistent with the accepted interpretation of negative numbers if t 0 is negative the shift is to the left by t 0 units Figure 227 shows some timeshifted and amplitudescaled unitstep functions The rectangle function is the difference between two unitstep functions timeshifted in opposite directions rectt ut 12 ut 12 1 1 1 2 3 4 5 2 3 4 5 2 3 4 5 54321 t gt t gt 5 0 4 0 3 0 2 0 1 0 0 1 1 2 2 3 3 4 4 5 5 0 Figure 224 Graphical definition of a function gt 1 1 1 2 3 4 5 2 3 4 5 2 3 4 5 54321 t 12gt 1 1 1 2 3 4 5 2 3 4 5 2 3 4 5 54321 t gt t 5 4 3 2 1 0 1 2 3 4 5 gt 0 0 0 0 0 1 2 3 4 5 0 t 5 4 3 2 1 0 1 2 3 4 5 12gt 0 0 0 0 0 12 1 32 2 52 0 a 1 1 1 2 3 4 5 2 3 4 5 2 3 4 5 54321 t gt 1 1 1 2 3 4 5 2 3 4 5 2 3 4 5 54321 t gt t 5 4 3 2 1 0 1 2 3 4 5 gt 0 0 0 0 0 1 2 3 4 5 0 t 5 4 3 2 1 0 1 2 3 4 5 gt 0 0 0 0 0 1 2 3 4 5 0 b Figure 225 Two examples of amplitude scaling rob28124ch02019078indd 37 041216 115 pm C h a p t e r 2 Mathematical Description of ContinuousTime Signals 38 Time shifting is accomplished by a change of the independent variable This type of change can be done on any independent variable it need not be time Our examples here are using time but the independent variable could be a spatial dimension In that case we could call it space shifting Later in the chapters on transforms we will have functions of an independent variable frequency and this change will be called frequency shifting The mathematical significance is the same regardless of the name used for the independent variable Amplitude scaling and time shifting occur in many practical physical systems In ordinary conversation there is a propagation delay the time required for a sound wave to propagate from one persons mouth to the other persons ear If that distance is 2 m and sound travels at about 330 meters per second the propagation delay is about 6 ms a delay that is not noticeable But consider an observer watching a pile driver drive a pile from 100 m away First the observer senses the image of the driver striking the pile There is a slight delay due to the speed of light from the pile driver to the eye but it is less than a microsecond The sound of the driver striking the pile arrives about 03 seconds later a noticeable delay This is an example of a time shift The sound of the driver striking the pile is much louder near the driver than at a distance of 100 m an example of amplitude scaling Another familiar example is the delay between seeing a lightning strike and hearing the thunder it produces As a more technological example consider a satellite communication system Figure 228 A ground station sends a strong electromagnetic signal to a satellite When the signal reaches the satellite the electromagnetic field is much weaker than when it left the ground station and it arrives later because of the propagation delay If the satellite is geosynchronous it is about 36000 km above the earth so if the ground station is di rectly below the satellite the propagation delay on the uplink is about 120 ms For ground 1 1 1 2 3 4 5 2 3 4 5 2 3 4 5 54321 t gt1 1 1 1 2 3 4 5 2 3 4 5 2 3 4 5 54321 t gt t 5 4 3 2 1 0 1 2 3 4 5 gt 0 0 0 0 0 1 2 3 4 5 0 t gt1 5 0 4 0 3 0 2 0 1 0 0 0 1 1 2 2 3 3 4 4 5 t1 6 5 4 3 2 1 0 1 2 3 4 5 Figure 226 Graph of gt 1 in relation to gt illustrating time shifting Figure 227 Amplitudescaled and timeshifted unitstep functions t 2ut4 2 4 t 4ut 4 t 10ut 10 7u6t t 7 6 rob28124ch02019078indd 38 041216 115 pm 25 Shifting and Scaling 39 stations not directly below the satellite the delay is a little more If the transmitted signal is A xt the received signal is B xt t p where B is typically much smaller than A and t p is the propagation time In communication links between locations on earth that are very far apart more than one up and down link may be required to communicate If that communication is voice communication between a television anchor in New York and a reporter in Calcutta the total delay can easily be one second a noticeable delay that can cause significant awkwardness in conversation Imagine the problem of communicating with the first astronauts on Mars The minimum oneway delay when Earth and Mars are in their closest proximity is more than 4 minutes In the case of longrange twoway communication time delay is a problem In other situations it can be quite useful as in radar and sonar In this case the time delay between when a pulse is sent out and when a reflection returns indicates the distance to the object from which the pulse reflected for example an airplane or a submarine TIME SCALING Consider next the change of independent variable indicated by t ta This expands the function gt horizontally by the factor a in gta This is called time scaling As an example lets compute and graph selected values of gt2 Figure 229 Consider next the change t t2 This is identical to the last example except the scaling factor is now 2 instead of 2 Figure 230 Time scaling t ta expands the function horizontally by a factor of a and if a 0 the function is also time reversed Time reversal means flipping the curve horizontally The case of a negative a can be conceived as t t followed by t ta The first step t t timereverses the func tion without changing its horizontal scale The second step t ta timescales the alreadytimereversed function by the scaling factor a Time scaling can also be indicated by t bt This is not really new because it is the same as t ta with b 1a So all the rules for time scaling still apply with that relation between the two scaling constants a and b Figure 228 Communication satellite in orbit Vol 4 PhotoDiscGetty rob28124ch02019078indd 39 041216 115 pm C h a p t e r 2 Mathematical Description of ContinuousTime Signals 40 1 1 1 2 3 4 5 2 3 4 5 2 3 4 5 6 7 8 9 10 10987654321 t gt2 1 1 1 2 3 4 5 2 3 4 5 2 3 4 5 54321 t gt t gt 5 0 4 0 3 0 2 0 1 0 0 1 1 2 2 3 3 4 4 5 5 0 t t2 gt2 10 5 0 8 4 0 6 3 0 4 2 0 2 1 0 0 0 1 2 1 2 4 2 3 6 3 4 8 4 5 10 5 0 Figure 229 Graph of gt2 in relation to gt illustrating time scaling 1 1 1 2 3 4 5 2 3 4 5 2 3 4 5 6 7 8 9 10 10987654321 t gt2 1 1 1 2 3 4 5 2 3 4 5 2 3 4 5 54321 t gt t gt 5 0 4 0 3 0 2 0 1 0 0 1 1 2 2 3 3 4 4 5 5 0 t t2 gt2 10 5 0 8 4 5 6 3 4 4 2 3 2 1 2 0 0 1 2 1 0 4 2 0 6 3 0 8 4 0 10 5 0 Figure 230 Graph of gt2 in relation to gt illustrating time scaling for a negative scaling factor rob28124ch02019078indd 40 041216 115 pm 25 Shifting and Scaling 41 A common experience that illustrates time scaling is the Doppler effect If we stand by the side of a road and a fire truck approaches while sounding its horn as the fire truck passes both the volume and the pitch of the horn seem to change Figure 231 The volume changes because of the proximity of the horn the closer it is to us the louder it is But why does the pitch change The horn is doing exactly the same thing all the time so it is not the pitch of the sound produced by the horn that changes but rather the pitch of the sound that arrives at our ears As the fire truck approaches each successive compression of air caused by the horn occurs a little closer to us than the last one so it arrives at our ears in a shorter time than the previous compression and that makes the frequency of the sound wave at our ear higher than the frequency emitted by the horn As the fire truck passes the opposite effect occurs and the sound of the horn arriving at our ears shifts to a lower frequency While we are hearing a pitch change the firefighters on the truck hear a constant horn pitch Let the sound heard by the firefighters be described by gt As the fire truck ap proaches the sound we hear is Atgat where At is an increasing function of time which accounts for the volume change and a is a number slightly greater than one The change in amplitude as a function of time is called amplitude modulation in com munication systems After the fire truck passes the sound we hear shifts to Btgbt where Bt is a decreasing function of time and b is slightly less than one Figure 232 In Figure 232 modulated sinusoids are used to represent the horn sound This is not precise but it serves to illustrate the important points The Doppler shift also occurs with light waves The red shift of optical spectra from distant stars is what first indicated that the universe was expanding When a star is receding from the earth the light we receive on earth experiences a Doppler shift that reduces the frequency of all the light waves emitted by the star Figure 233 Since the color red has the lowest frequency detectable by the human eye a reduction in frequency is called a red shift because the visible spectral characteristics all seem to Figure 231 Firefighters on a fire truck Vol 94 Corbis rob28124ch02019078indd 41 041216 115 pm C h a p t e r 2 Mathematical Description of ContinuousTime Signals 42 move toward the red end of the spectrum The light from a star has many characteristic variations with frequency because of the composition of the star and the path from the star to the observer The amount of shift can be determined by comparing the spectral patterns of the light from the star with known spectral patterns measured on Earth in a laboratory Time scaling is a change of the independent variable As was true of time shifting this type of change can be done on any independent variable it need not be time In later chapters we will do some frequency scaling Figure 233 The Lagoon nebula Vol 34 PhotoDiscGetty Figure 232 Illustration of the Doppler effect t gt Sound Heard by Firefighters t Atgat Sound As Truck Approaches t Btgbt Sound After Truck Passes rob28124ch02019078indd 42 041216 115 pm 25 Shifting and Scaling 43 SIMULTANEOUS SHIFTING AND SCALING All three function changes amplitude scaling time scaling and time shifting can be applied simultaneously gt Ag t t 0 a 211 To understand the overall effect it is usually best to break down a multiple change like 211 into successive simple changes gt amplitude scaling A Agt tta Agta tt t 0 Ag t t 0 a 212 Observe here that the order of the changes is important If we exchange the order of the timescaling and timeshifting operations in 212 we get gt amplitude scaling A Agt tt t 0 Agt t 0 tta Agta t 0 Ag t t 0 a This result is different from the preceding result unless a 1 or t 0 0 For a different kind of multiple change a different sequence may be better for example Agbt t 0 In this case the sequence of amplitude scaling time shifting and then time scaling is the simplest path to a correct result gt amplitude scaling A Agt tt t 0 Agt t 0 tbt Agbt t 0 Figure 234 and Figure 235 illustrate some steps graphically for two functions In these figures certain points are labeled with letters beginning with a and proceeding t gt 1 a b c d e f g 1 2 1 2 t 2gt 2 a b c d e f g 1 2 1 2 2g Amplitude Scaling Time Scaling Time Shifting t 2 a b c d e f g t 4 2 2 t 2g 2 a b c d e f g t2 4 4 Figure 234 A sequence of amplitude scaling time scaling and time shifting a function rob28124ch02019078indd 43 041216 115 pm C h a p t e r 2 Mathematical Description of ContinuousTime Signals 44 alphabetically As each functional change is made corresponding points have the same letter designation The functions previously introduced along with function scaling and shifting allow us to describe a wide variety of signals A signal that has a decaying exponential shape after some time t t 0 and is zero before that can be represented in the compact mathematical form xt A e tτ ut t 0 Figure 236 A signal that has the shape of a negative sine function before time t 0 and a positive sine function after time t 0 can be represented by xt A sin2π f 0 t sgnt Figure 237 Figure 235 A sequence of amplitude scaling time shifting and time scaling a function Amplitude Scaling Time Scaling Time Shifting t 3gt 3 1 a b c d e t gt 1 1 1 a b c d e 3g2t1 1 t 3 a b c e d 2 t 3gt1 3 a b c d e t t0 A τ Figure 236 A decaying exponential switched on at time t t 0 t xt A Figure 237 Product of a sine and a signum function rob28124ch02019078indd 44 041216 115 pm 25 Shifting and Scaling 45 A signal that is a burst of a sinusoid between times t 1 and t 5 and zero else where can be represented by xt A cos2π f 0 t θ rectt 34 Figure 238 ExamplE 22 Graphing function scaling and shifting with MATLAB Using MATLAB graph the function defined by gt 0 t 2 4 2t 2 t 0 4 3t 0 t 4 16 2t 4 t 8 0 t 8 Then graph the functions 3gt 1 12g3t 2gt 12 We must first choose a range of t over which to graph the function and the space be tween points in t to yield a curve that closely approximates the actual function Lets choose a range of 5 t 20 and a space between points of 01 Also lets use the function feature of MATLAB that allows us to define the function gt as a separate MATLAB program an m file Then we can refer to it when graphing the transformed functions and not have to retype the function description The gm file contains the following code function y gt Calculate the functional variation for each range of time t y1 4 2t y2 4 3t y3 16 2t Splice together the different functional variations in their respective ranges of validity y y12t t 0 y20t t 4 y34t t 8 The MATLAB program contains the following code Program to graph the function gt t2 2t 1 and then to graph 3gt1 g3t2 and 2gt12 tmin 4 tmax 20 Set the time range for the graph dt 01 Set the time between points t tmindttmax Set the vector of times for the graph g0 gt Compute the original gt Figure 238 A sinusoidal burst t xt A 1 5 rob28124ch02019078indd 45 041216 115 pm C h a p t e r 2 Mathematical Description of ContinuousTime Signals 46 g1 3gt1 Compute the first change g2 g3t2 Compute the second change g3 2gt12 Compute the third change Find the maximum and minimum g values in all the scaled or shifted functions and use them to scale all graphs the same gmax maxmaxg0 maxg1 maxg2 maxg3 gmin minming0 ming1 ming2 ming3 Graph all four functions in a 2 by 2 arrangement Graph them all on equal scales using the axis command Draw grid lines using the grid command to aid in reading values subplot221 p plottg0k setpLineWidth2 xlabelt ylabelgt titleOriginal Function gt axistmintmaxgmingmax grid subplot222 p plottg1k setpLineWidth2 xlabelt ylabel3gt1 titleFirst Change axistmintmaxgmingmax grid subplot223 p plottg2k setpLineWidth2 xlabelt ylabelg3t2 titleSecond Change axistmintmaxgmingmax grid subplot224 p plottg3k setpLineWidth2 xlabelt ylabel2gt12 titleThird Change axistmintmaxgmingmax grid The graphical results are displayed in Figure 239 Figure 239 MATLAB graphs of scaled andor shifted functions 0 5 10 15 20 15 10 5 0 5 10 15 20 15 10 5 0 5 10 15 20 t gt Original Function gt 0 5 10 15 20 t 3gt1 First Change 0 5 10 15 20 15 10 5 0 5 10 15 20 15 10 5 0 5 10 15 20 t g3t2 Second Change 0 5 10 15 20 t 2gt12 Third Change rob28124ch02019078indd 46 041216 115 pm 26 Differentiation and Integration 47 Figure 240 shows more examples of amplitudescaled timeshifted and timescaled versions of the functions just introduced 26 DIFFERENTIATION AND INTEGRATION Integration and differentiation are common signal processing operations in practical systems The derivative of a function at any time t is its slope at that time and the in tegral of a function at any time t is the accumulated area under the function up to that time Figure 241 illustrates some functions and their derivatives The zero crossings of all the derivatives have been indicated by light vertical lines that lead to the maxima and minima of the corresponding function There is a function diff in MATLAB that does symbolic differentiation x symx diffsinx2 ans 2cosx2x This function can also be used numerically to find the differences between adjacent values in a vector These finite differences can then be divided by the increment of the independent variable to approximate some derivatives of the function that produced the vector t 3rect t1 4 1 3 3 t 4u3t 4 3 t 5ramp01t 10 5 t 3sgn2t 3 3 Figure 240 More examples of amplitudescaled timeshifted and timescaled functions t 4 xt 4 4 t 4 dxdt 4 4 t 4 4 xt 1 1 t 4 4 dxdt 1 1 t 5 5 xt 1 1 t 5 5 dxdt 1 1 t 4 xt 1 1 t 4 dxdt 1 1 Figure 241 Some functions and their derivatives rob28124ch02019078indd 47 041216 115 pm C h a p t e r 2 Mathematical Description of ContinuousTime Signals 48 dx 01 x 03dx08 expx ans 13499 14918 16487 18221 20138 22255 diffexpxdx ans 14197 15690 17340 19163 21179 Integration is a little more complicated than differentiation Given a function its derivative is unambiguously determinable if it exists However its integral is not unambiguously determinable without some more information This is inherent in one of the first principles learned in integral calculus If a function gx has a derivative g x then the function gx K K a constant has the same derivative g x regardless of the value of the constant K Since integration is the opposite of differentiation what is the integral of g x It could be gx but it could also be gx K The term integral has different meanings in different contexts Generally speaking integration and differentiation are inverse operations An antiderivative of a function of time gt is any function of time that when differentiated with respect to time yields gt An antiderivative is indicated by an integral sign without limits For example sin2πt 2π cos2πtdt In words sin2πt2π is an antiderivative of cos2πt An indefinite integral is an an tiderivative plus a constant For example ht gtdt C A definite integral is an integral taken between two limits For example A α β gtdt If α and β are constants then A is also a constant the area under gt between α and β In signal and system anal ysis a particular form of definite integral ht t gτdτ is often used The variable of integration is τ so during the integration process the upper integration limit t is treated like a constant But after the integration is finished t is the independent variable in ht This type of integral is sometimes called a running integral or a cumulative integral It is the accumulated area under a function for all time before t and that depends on what t is Often in practice we know that a function of time is zero before t t 0 Then we know that t 0 gt dt is zero Then the integral of that function from any time t 1 t 0 to any time t t 0 is unambiguous It can only be the area under the function from time t t 0 to time t t 1 t gτdτ t 1 t 0 gτdτ 0 t 0 t gτdτ t 0 t gτdτ Figure 242 illustrates some functions and their integrals In Figure 242 the two functions on the right are zero before time t 0 and the in tegrals illustrated assume a lower limit on the integral less than zero thereby producing a single unambiguous result The two on the left are illustrated with multiple possible integrals differing from each other only by constants They all have the same derivative and are all equally valid candidates for the integral in the absence of extra information There is a function int in MATLAB that can do symbolic integration symx int11x2 ans atanx rob28124ch02019078indd 48 041216 115 pm 27 Even and Odd Signals 49 This function cannot be used to do numerical integration Another function cumsum can be used to do numerical integration cumsum15 ans 1 3 6 10 15 dx pi16 x 0dxpi4 y sinx y 0 01951 03827 05556 07071 cumsumydx ans 0 00383 01134 02225 03614 There are also other more sophisticated numerical integration functions in MATLAB for example trapz which uses a trapezoidal approximation and quad which uses adaptive Simpson quadrature 27 EVEN AND ODD SIGNALS Some functions have the property that when they undergo certain types of shifting and or scaling the function values do not change They are invariant under that shifting andor scaling An even function of t is invariant under time reversal t t and an odd function of t is invariant under the amplitude scaling and time reversal gt gt An even function gt is one for which gt gt and an odd function is one for which gt gt A simple way of visualizing even and odd functions is to imagine that the ordinate axis the gt axis is a mirror For even functions the part of gt for t 0 and the part of gt for t 0 are mirror images of each other For an odd function the same two parts of the function are negative mirror images of each other Figure 243 and Figure 244 t 1 4 xt 1 1 t 1 4 1 1 t 4 xt 1 1 t 4 1 t 4 xt 4 t 4 4 1 4 xt 1 1 t t 1 4 1 1 xτdτ t xτdτ t xτdτ t xτdτ t Figure 242 Some functions and their integrals rob28124ch02019078indd 49 041216 115 pm C h a p t e r 2 Mathematical Description of ContinuousTime Signals 50 Some functions are even some are odd and some are neither even nor odd But any function gt is the sum of its even and odd parts gt g e t g o t The even and odd parts of a function gt are g e t gt gt 2 g o t gt gt 2 213 If the odd part of a function is zero the function is even and if the even part of a func tion is zero the function is odd ExamplE 23 Even and odd parts of a function What are the even and odd parts of the function gt t t 2 3 They are g e t gt gt 2 t t 2 3 t t 2 3 2 0 g o t t t 2 3 t t 2 3 2 t t 2 3 Therefore gt is an odd function Program to graph the even and odd parts of a function function GraphEvenAndOdd t 5015 Set up a time vector for the graph ge gt gt2 Compute the evenpart values go gt gt2 Compute the oddpart values Graph the even and odd parts subplot211 ptr plottgek setptrLineWidth2 grid on xlabelittFontNameTimesFontSize24 ylabelgeittFontNameTimesFontSize24 subplot212 ptr plottgok setptrLineWidth2 grid on xlabelittFontNameTimesFontSize24 ylabelgoittFontNameTimesFontSize24 function y gx Function definition for gx y xx23 Figure 245 illustrates the graphical output of the MATLAB program t t gt Even Function gt Odd Function Figure 244 Two very common and useful functions one even and one odd t gt Even Function t gt Odd Function Figure 243 Examples of even and odd functions rob28124ch02019078indd 50 041216 115 pm 27 Even and Odd Signals 51 This MATLAB code example begins with the keyword function A MATLAB program file that does not begin with function is called a script file One that does begin with function de fines a function This code example contains two function definitions The second function is called a subfunction It is used only by the main function in this case GraphEvenAndOdd and is not accessible by any functions or scripts exterior to this function definition A function may have any number of subfunctions A script file cannot use subfunctions COMBINATIONS OF EVEN AND ODD SIGNALS Let g 1 t and g 2 t both be even functions Then g 1 t g 1 t and g 2 t g 2 t Let gt g 1 t g 2 t Then gt g 1 t g 2 t and using the evenness of g 1 t and g 2 t gt g 1 t g 2 t gt proving that the sum of two even functions is also even Now let gt g 1 t g 2 t Then gt g 1 tg 2 t g 1 tg 2 t gt proving that the product of two even functions is also even Now let g 1 t and g 2 t both be odd Then gt g 1 t g 2 t g 1 t g 2 t gt proving that the sum of two odd functions is odd Then let gt g 1 tg 2 t g 1 t g 2 t g 1 t g 2 t gt proving that the product of two odd functions is even By similar reasoning we can show that if two functions are even their sum dif ference product and quotient are even too If two functions are odd their sum and difference are odd but their product and quotient are even If one function is even and the other is odd their product and quotient are odd Figure 246 Figure 245 Graphical output of the MATLAB program 5 4 3 2 1 0 1 2 3 4 5 0 t 5 4 3 2 1 0 1 2 3 4 5 150 100 50 0 50 100 150 t got 1 05 1 get 05 Function Types Both Even Both Odd One Even One Odd Sum Even Odd Neither Difference Even Odd Neither Product Even Even Odd Quotient Even Even Odd Figure 246 Combinations of even and odd functions rob28124ch02019078indd 51 041216 115 pm C h a p t e r 2 Mathematical Description of ContinuousTime Signals 52 Figure 247 Product of even and odd functions t g2t g1t g1tg2t t t t t t g2t g1t g1tg2t Figure 248 Product of two even functions The most important even and odd functions in signal and system analysis are co sines and sines Cosines are even and sines are odd Figure 247 through Figure 249 give some examples of products of even and odd functions g2t g1t g2tg1t t t t Figure 249 Product of two odd functions rob28124ch02019078indd 52 041216 115 pm 28 Periodic Signals 53 DERIVATIVES AND INTEGRALS OF EVEN AND ODD SIGNALS The definite integrals of even and odd functions can be simplified in certain common cases If gt is an even function and a is a real constant a a gtdt a 0 gtdt 0 a gtdt 0 a gtdt 0 a gtdt Making the change of variable τ t in the first integral on the right and using gτ gτ a a gtdt 2 0 a gtdt which should be geometrically obvious by looking at the graph of the function Figure 251a By similar reasoning if gt is an odd function then a a gtdt 0 which should also be geometrically obvious Figure 251b Function Type Even Odd Derivative Odd Even Integral OddConstant Even Figure 250 Function types and the types of their derivatives and integrals Let gt be an even function Then gt gt Using the chain rule of differen tiation the derivative of gt is g t g t an odd function So the derivative of any even function is an odd function Similarly the derivative of any odd function is an even function We can turn the arguments around to say that the integral of any even function is an odd function plus a constant of integration and the integral of any odd function is an even function plus a constant of integration and therefore still even because a constant is an even function Figure 250 a a a a t t gt gt Area 1 Area 1 Area 2 Area 2 Area 1 Area 1 Area 2 Area 2 Even Function Odd Function Figure 251 Integrals of a even functions and b odd functions over symmetrical limits 28 PERIODIC SIGNALS A periodic signal is one that has been repeating a pattern for a semiinfinite time and will continue to repeat that pattern for a semiinfinite time A periodic function gt is one for which gt gt nT for any integer value of n where T is a period of the function Another way of saying that a function of t is periodic is to say that it is invariant under the time shift t t nT The function repeats every T seconds Of course it also repeats every 2T 3T or nT seconds n is an integer Therefore 2T or 3T or nT are all periods of the function The minimum positive interval over which a function repeats is called its fundamental period T 0 The fundamental cyclic frequency f 0 is the re ciprocal of the fundamental period f 0 1 T 0 and the fundamental radian frequency is ω 0 2π f 0 2π T 0 rob28124ch02019078indd 53 041216 115 pm C h a p t e r 2 Mathematical Description of ContinuousTime Signals 54 Some common examples of periodic functions are real or complex sinusoids and com binations of real andor complex sinusoids We will see later that other more complicated types of periodic functions with different periodically repeating shapes can be generated and mathematically described Figure 252 gives some examples of periodic functions A function that is not periodic is called an aperiodic function Because of the similarity of the phrase aperiodic function and the phrase a periodic function it is probably better when speaking to use the term nonperiodic or not periodic to avoid confusion t xt T0 t xt T0 t xt T0 Figure 252 Examples of periodic functions with fundamental period T 0 In practical systems a signal is never actually periodic because it did not exist until it was created at some finite time in the past and it will stop at some finite time in the future However often a signal has been repeating for a very long time before the time we want to analyze the signal and will repeat for a very long time after that In many cases approximating the signal by a periodic function introduces negligible error Ex amples of signals that would be properly approximated by periodic functions would be rectified sinusoids in an AC to DC converter horizontal sync signals in a television the angular shaft position of a generator in a power plant the firing pattern of spark plugs in an automobile traveling at constant speed the vibration of a quartz crystal in a wristwatch the angular position of a pendulum on a grandfather clock and so on Many natural phenomena are for all practical purposes periodic most planet satellite and comet orbital positions the phases of the moon the electric field emitted by a Cesium atom at resonance the migration patterns of birds the caribou mating season and so forth Periodic phenomena play a large part both in the natural world and in the realm of artificial systems A common situation in signal and system analysis is to have a signal that is the sum of two periodic signals Let x 1 t be a periodic signal with fundamental period T 01 and let x 2 t be a periodic signal with fundamental period T 02 and let xt x 1 t x 2 t Whether or not xt is periodic depends on the relationship between the two pe riods T 01 and T 02 If a time T can be found that is an integer multiple of T 01 and also an integer multiple of T 02 then T is a period of both x 1 t and x 2 t and x 1 t x 1 t T and x 2 t x 2 t T 214 Time shifting xt x 1 t x 2 t with t t T xt T x 1 t T x 2 t T 215 Then combining 215 with 214 xt T x 1 t x 2 t xt proving that xt is periodic with period T The smallest positive value of T that is an integer multiple of both T 01 and T 02 is the fundamental period T 0 of xt This smallest value of T is called the least common multiple LCM of T 01 and T 02 If T 01 T 02 is a rob28124ch02019078indd 54 041216 115 pm 28 Periodic Signals 55 rational number a ratio of integers the LCM is finite and xt is periodic If T 01 T 02 is an irrational number xt is aperiodic Sometimes an alternate method for finding the period of the sum of two periodic func tions is easier than finding the LCM of the two periods If the fundamental period of the sum is the LCM of the two fundamental periods of the two functions then the fundamen tal frequency of the sum is the greatest common divisor GCD of the two fundamental frequencies and is therefore the reciprocal of the LCM of the two fundamental periods ExamplE 24 Fundamental period of a signal Which of these functions are periodic and if one is what is its fundamental period a gt 7 sin400πt The sine function repeats when its total argument is increased or decreased by any integer multiple of 2π radians Therefore sin400πt 2nπ sin400πt n T 0 Setting the arguments equal 400πt 2nπ 400πt n T 0 or 2nπ 400πn T 0 or T 0 1200 An alternate way of finding the fundamental period is to realize that 7 sin400πt is in the form A sin2π f 0 t or A sin ω 0 t where f 0 is the fundamental cyclic frequency and ω 0 is the fundamental radian frequency In this case f 0 200 and ω 0 400π Since the fundamental period is the reciprocal of the fundamental cyclic frequency T 0 1200 b gt 3 t 2 This is a seconddegree polynomial As t increases or decreases from zero the function value increases monotonically always in the same direction No function that increases monotonically can be periodic because if a fixed amount is added to the argument t the function must be larger or smaller than for the current t This function is not periodic c gt e j60πt This is a complex sinusoid That is easily seen by expressing it as the sum of a cosine and a sine through Eulers identity gt cos60πt j sin60πt The function gt is a linear combination of two periodic signals that have the same fundamental cyclic frequency 60π2π 30 Therefore the fundamental frequency of gt is 30 Hz and the fundamental period is 130 s d gt 10 sin12πt 4 cos18πt This is the sum of two functions that are both periodic Their fundamental periods are 16 second and 19 second The LCM is 13 second See Web Appendix B for a systematic method for finding least common multiples There are two fundamental periods of the first function and three fundamental periods of the second function in that rob28124ch02019078indd 55 041216 115 pm C h a p t e r 2 Mathematical Description of ContinuousTime Signals 56 time Therefore the fundamental period of the overall function is 13 second Figure 253 The two fundamental frequencies are 6 Hz and 9 Hz Their GCD is 3 Hz which is the reciprocal of 13 second the LCM of the two fundamental periods Figure 253 Signals with frequencies of 6 Hz and 9 Hz and their sum x1t 1 1 First Sinusoid t x2t 1 1 Second Sinusoid t x1tx2t 2 2 Two Periods of the Sum t e gt 10 sin12πt 4 cos18t This function is exactly like the function in d except that a π is missing in the second argument The fundamental periods are now 16 second and π9 seconds and the ratio of the two fundamental periods is either 2π3 or 32π both of which are irrational Therefore gt is aperiodic This function although made up of the sum of two periodic functions is not periodic because it does not repeat exactly in a finite time It is sometimes referred to as almost periodic because in looking at a graph of the function it seems to repeat in a finite time But strictly speaking it is aperiodic There is a function lcm in MATLAB for finding least common multiples It is somewhat limited because it accepts only two arguments which can be scalar integers or arrays of integers There is also a function gcd which finds the greatest common divisor of two integers or two arrays of integers lcm3247 ans 1504 gcd93771522 ans 3 11 29 SIGNAL ENERGY AND POWER SIGNAL ENERGY All physical activity is mediated by a transfer of energy Real physical systems respond to the energy of an excitation It is important at this point to establish some terminol ogy describing the energy and power of signals In the study of signals in systems the rob28124ch02019078indd 56 041216 115 pm 29 Signal Energy and Power 57 signals are often treated as mathematical abstractions Often the physical significance of the signal is ignored for the sake of simplicity of analysis Typical signals in electri cal systems are voltages or currents but they could be charge or electric field or some other physical quantity In other types of systems a signal could be a force a tempera ture a chemical concentration a neutron flux and so on Because of the many different kinds of physical signals that can be operated on by systems the term signal energy has been defined Signal energy as opposed to just energy of a signal is defined as the area under the square of the magnitude of the signal The signal energy of a signal xt is E x xt 2 dt 216 The units of signal energy depend on the units of the signal If the signal unit is the volt V the signal energy is expressed in V 2 s Signal energy is proportional to the actual physical energy delivered by a signal but not necessarily equal to that physical energy In the case of a current signal it through a resistor R the actual energy delivered to the resistor would be Energy it 2 R dt R it 2 dt R E i Signal energy is proportional to actual energy and the proportionality constant in this case is R For a different kind of signal the proportionality constant would be differ ent In many kinds of system analysis the use of signal energy is more convenient than the use of actual physical energy ExamplE 25 Signal energy of a signal Find the signal energy of xt 31 t4 t 4 0 otherwise From the definition of signal energy E x xt 2 dt 4 4 31 t4 2 dt Taking advantage of the fact that xt is an even function E x 2 3 2 0 4 1 t4 2 dt 18 0 4 1 t 2 t 2 16 dt 18 t t 2 4 t 3 48 0 4 24 dt 001 Time increment t 8dt8 Time vector for computing samples of xt Compute samples of xt x 31abst4abst4 Compute energy of xt using trapezoidal rule approximation to the integral Ex trapzx2dt dispSignal Energy num2strEx Signal Energy 240001 rob28124ch02019078indd 57 041216 115 pm C h a p t e r 2 Mathematical Description of ContinuousTime Signals 58 SIGNAL POWER For many signals the integral E x xt 2 dt does not converge because the signal energy is infinite This usually occurs because the signal is not time limited The term time limited means that the signal is nonzero over only a finite time An example of a signal with infinite energy is the sinusoidal signal xt A cos2π f 0 t A 0 Over an infinite time interval the area under the square of this signal is infinite For signals of this type it is more convenient to deal with the average signal power instead of the signal energy Average signal power of a signal xt is defined by P x lim T 1 T T2 T2 xt 2 dt 217 The integral is the signal energy of the signal over a time T and it is then divided by T yielding the average signal power over time T Then as T approaches infinity this av erage signal power becomes the average signal power over all time For periodic signals the average signal power calculation may be simpler The average value of any periodic function is the average over any period Therefore since the square of a periodic function is also periodic for periodic signals P x 1 T t 0 t 0 T xt 2 dt 1 T T xt 2 dt where the notation T means the same thing as t 0 t 0 T for any arbitrary choice of t 0 where T can be any period of xt 2 ExamplE 26 Signal power of a sinusoidal signal Find the average signal power of xt A cos2π f 0 t θ From the definition of average signal power for a periodic signal P x 1 T T A cos2π f 0 t θ 2 dt A 2 T 0 T 0 2 T 0 2 cos 2 2πt T 0 θdt Using the trigonometric identity cosx cosy 12cosx y cosx y we get P x A 2 2 T 0 T 0 2 T 0 2 1 cos4πt T 0 2θ dt A 2 2 T 0 T 0 2 T 0 2 dt A 2 2 T 0 T 0 2 T 0 2 cos4πt T 0 2θdt 0 A 2 2 The second integral on the right is zero because it is the integral of a sinusoid over two funda mental periods The signal power is P x A 2 2 This result is independent of the phase θ and the frequency f 0 It depends only on the amplitude A A 1 Amplitude of xt th 0 Phase shift of xt f0 1 Fundamental frequency T0 1f0 Fundamental period dt T0100 Time increment for sampling xt rob28124ch02019078indd 58 041216 115 pm 29 Signal Energy and Power 59 t 0dtT0 Time vector for computing samples of xt Compute samples of xt over one fundamental period x Acos2pif0t th Compute signal power using trapezoidal approximation to integral Px trapzx2dtT0 dispSignal Power num2strPx Signal Power 05 Signals that have finite signal energy are referred to as energy signals and signals that have infinite signal energy but finite average signal power are referred to as power signals No real physical signal can actually have infinite energy or infinite average power because there is not enough energy or power in the universe available But we often analyze signals that according to their strict mathematical definition have in finite energy a sinusoid for example How relevant can an analysis be if it is done with signals that cannot physically exist Very relevant The reason mathematical sinusoids have infinite signal energy is that they have always existed and will always exist Of course practical signals never have that quality They all had to begin at some finite time and they will all end at some later finite time They are actually time limited and have finite signal energy But in much system analysis the analysis is steadystate anal ysis of a system in which all signals are treated as periodic The analysis is still relevant and useful because it is a good approximation to reality it is often much simpler than an exact analysis and it yields useful results All periodic signals are power signals except for the trivial signal xt 0 because they all endure for an infinite time ExamplE 27 Finding signal energy and power of signals using MATLAB Using MATLAB find the signal energy or power of the signals a xt 4 e t10 rect t 4 3 b A periodic signal of fundamental period 10 described over one period by xt 3t 5 t 5 Then compare the results with analytical calculations Program to compute the signal energy or power of some example signals a dt 01 t 7dt13 Set up a vector of times at which to compute the function Time interval is 01 Compute the function values and their squares x 4expt10rectt43 xsq x2 Ex trapztxsq Use trapezoidalrule numerical integration to find the area under rob28124ch02019078indd 59 041216 115 pm C h a p t e r 2 Mathematical Description of ContinuousTime Signals 60 the function squared and display the result dispa Ex num2strEx b T0 10 The fundamental period is 10 dt 01 t 5dt5 Set up a vector of times at which to compute the function Time interval is 01 x 3t xsq x2 Compute the function values and their squares over one fundamental period Px trapztxsqT0 Use trapezoidalrule numerical integration to find the area under the function squared divide the period and display the result dispb Px num2strPx The output of this program is a Ex 215177 b Px 75015 Analytical computations a E x xt 2 dt 25 55 4 e t10 2 dt 16 25 55 e t5 dτ 5 16 e t5 25 55 21888 The small difference in results is probably due to the error inherent in trapezoidalrule integration It could be reduced by using time points spaced more closely together b P x 1 10 5 5 3t 2 dt 1 5 0 5 9 t 2 dt 1 5 3 t 3 0 5 375 5 75 Check 210 SUMMARY OF IMPORTANT POINTS 1 The term continuous and the term continuoustime mean different things 2 A continuoustime impulse although very useful in signal and system analysis is not a function in the ordinary sense 3 Many practical signals can be described by combinations of shifted andor scaled standard functions and the order in which scaling and shifting are done is significant 4 Signal energy is in general not the same thing as the actual physical energy delivered by a signal 5 A signal with finite signal energy is called an energy signal and a signal with infinite signal energy and finite average power is called a power signal rob28124ch02019078indd 60 041216 115 pm 61 Exercises with Answers EXERCISES WITH ANSWERS Answers to each exercise are in random order Signal Functions 1 If g t 7 e 2t3 write out and simplify a g 3 b g 2 t c g t 10 4 d g jt e g jt g jt 2 f g jt 3 2 g jt 3 2 2 Answers 03485 cos 2t 7 cos t 86387 10 4 7 e j2t3 7 e t511 7 e 72t 2 If g x x 2 4x 4 write out and simplify a g z b g u v c g e jt d g g t e g 2 Answers u 2 v 2 2uv 4u 4v 4 t 4 8 t 3 20 t 2 16t 4 0 e jt 2 2 z 2 4z 4 3 Find the magnitudes and phases of these complex quantities a e 3j23 b e 2j6 c 100 8 j13 Answers 10191 radians 65512 00498 23 radians 73891 6 radians 4 Let G f j4f 2 j7f 11 a What value does the magnitude of this function approach as f approaches positive infinity b What value in radians does the phase of this function approach as f approaches zero from the positive side Answer π 2 radians 6285 5 Let X f jf jf 10 a Find the magnitude X 4 and the angle X 4 in radians b What value in radians does X f approach as f approaches zero from the positive side Answers π 2 radians 119 radians 03714 rob28124ch02019078indd 61 041216 115 pm C h a p t e r 2 Mathematical Description of ContinuousTime Signals 62 Shifting and Scaling 6 For each function g t graph g t g t g t 1 and g 2t a b t gt 2 4 t gt 1 1 3 3 Answers t gt 1 1 3 3 t g2t 1 4 t gt 2 4 t gt1 3 1 4 t gt 1 1 3 3 t g2t 3 3 1 2 1 2 t gt 2 4 t gt1 1 2 3 3 7 Find the values of the following signals at the indicated times a x t 2 rect t 4 x 1 b x t 5 rect t 2 sgn 2t x 05 c x t 9 rect t 10 sgn 3 t 2 x 1 Answers 2 9 5 8 For each pair of functions in Figure E8 provide the values of the constants A t 0 and w in the functional transformation g2t Ag1t t0 w 4 2 0 2 4 2 1 0 1 2 t g1t g1t g1t g2t g2t g2t a 4 2 0 2 4 2 1 0 1 2 t a 4 2 0 2 4 2 1 0 1 2 t b 4 2 0 2 4 2 1 0 1 2 t b 4 2 0 2 4 2 1 0 1 2 t c 4 2 0 2 4 2 1 0 1 2 t c rob28124ch02019078indd 62 041216 115 pm 63 Exercises with Answers 4 2 0 2 4 2 1 0 1 2 t g1t g1t g1t g2t g2t g2t a 4 2 0 2 4 2 1 0 1 2 t a 4 2 0 2 4 2 1 0 1 2 t b 4 2 0 2 4 2 1 0 1 2 t b 4 2 0 2 4 2 1 0 1 2 t c 4 2 0 2 4 2 1 0 1 2 t c 4 2 0 2 4 2 1 0 1 2 t g1t g1t g1t g2t g2t g2t a 4 2 0 2 4 2 1 0 1 2 t a 4 2 0 2 4 2 1 0 1 2 t b 4 2 0 2 4 2 1 0 1 2 t b 4 2 0 2 4 2 1 0 1 2 t c 4 2 0 2 4 2 1 0 1 2 t c Figure E8 Answers A 1 2 t 0 1 w 2 A 2 t 0 0 w 1 2 A 2 t 0 1 w 1 9 For each pair of functions in Figure E9 provide the values of the constants A t 0 and a in the functional transformation g 2 t Ag 1 wt t 0 a 10 5 0 5 10 8 4 0 4 8 t g1t g2t 10 5 0 5 10 8 4 0 4 8 t b g1t g2t 10 5 0 5 10 8 4 0 4 8 t 10 5 0 5 10 8 4 0 4 8 t c g1t g2t 10 5 0 5 10 8 4 0 4 8 t 10 5 0 5 10 8 4 0 4 8 t rob28124ch02019078indd 63 041216 115 pm C h a p t e r 2 Mathematical Description of ContinuousTime Signals 64 d g1t g2t 10 5 0 5 10 8 4 0 4 8 t 10 5 0 5 10 8 4 0 4 8 t e g1t g2t 10 5 0 5 10 8 4 0 4 8 t 10 5 0 5 10 8 4 0 4 8 t Figure E9 Answers A 3 w 2 t 0 2 A 2 w 1 3 t 0 2 A 2 w 2 t 0 2 A 3 w 1 2 t 0 2 A 3 w 1 3 or 1 3 t 0 6 or 3 10 In Figure E10 is plotted a function g 1 t that is zero for all time outside the range plotted Let some other functions be defined by g 2 t 3 g 1 2 t g 3 t 2 g 1 t 4 g 4 t g 1 t 3 2 Find these values a g 2 1 b g 3 1 c g 4 t g 3 t t2 d 3 1 g 4 t dt t g1t 3 2 1 4 1 2 3 4 4 3 2 1 1 4 2 3 Figure E10 Answers 3 2 32 35 11 A function G f is defined by G f ej2π f rect f 2 Graph the magnitude and phase of G f 10 G f 10 over the range 20 f 20 rob28124ch02019078indd 64 041216 115 pm 65 Exercises with Answers Answer Gf10G f10 f 20 20 20 20 1 f π π G f10 G f10 12 Let x 1 t 3 rect t 1 6 and x 2 t ramp t u t u t 4 a Graph them in the time range 10 t 10 b Graph x t x 1 2t x 2 t 2 in the time range 10 t 10 Answers t x1t x2t 6 6 6 6 10 10 10 10 10 10 x12tx2t2 8 8 t t 13 Write an expression consisting of a summation of unitstep functions to represent a signal that consists of rectangular pulses of width 6 ms and height 3 that occur at a uniform rate of 100 pulses per second with the leading edge of the first pulse occurring at time t 0 Answer x t 3 n0 u t 001n u t 001n 0006 14 Find the strengths of the following impulses a 3δ 4t b 5δ 3 t 1 Answers 34 53 15 Find the strengths and spacing between the impulses in the periodic impulse 9 δ 11 5t Answers 95 115 rob28124ch02019078indd 65 041216 115 pm C h a p t e r 2 Mathematical Description of ContinuousTime Signals 66 Derivatives and Integrals of Functions 16 Graph the derivative of x t 1 e t u t Answer t 1 4 xt 1 1 t 1 4 dxdt 1 1 17 Find the numerical value of each integral a 2 11 u 4 t dt b 1 8 δ t 3 2δ 4t dt c 12 52 δ 2 3t dt d δ t 4 ramp 2t dt e 3 10 ramp 2t 4 dt f 11 82 3 sin 200t δ t 7 dt g 5 5 sin πt 20 dt h 2 10 39 t 2 δ 4 t 1 dt i e 18t u t δ 10t 2 dt j 2 9 9δ t 4 5 dt k 6 3 5δ 3 t 4 dt l ramp 3t δ t 4 dt m 1 17 δ 3 t cos 2πt 3 dt Answers 45 0 8 0002732 4173 12 6 0 5 0 1 12 64 18 Graph the integral from negative infinity to time t of the functions in Figure E18 which are zero for all time t 0 gt t 1 1 2 3 1 2 gt t 1 1 2 3 Figure E18 Answers gt dt t 1 1 2 3 1 2 gt dt t 1 1 2 3 19 If 4u t 5 d dt x t what is the function x t Answer x t 4 ramp t 5 rob28124ch02019078indd 66 041216 116 pm 67 Exercises with Answers Generalized Derivative 20 The generalized derivative of 18 rect t 2 3 consists of two impulses Find their numerical locations and strengths Answers 35 and 18 05 and 18 Even and Odd Functions 21 Classify the following functions as even odd or neither a cos 2πt tri t 1 b sin 2πt rect t 5 Answers Odd Neither 22 An even function g t is described over the time range 0 t 10 by g t 2t 0 t 3 15 3t 3 t 7 2 7 t 10 a What is the value of g t at time t 5 b What is the value of the first derivative of gt at time t 6 Answers 0 3 23 Find the even and odd parts of these functions a g t 2 t 2 3t 6 b g t 20 cos 40πt π 4 c g t 2 t 2 3t 6 1 t d g t t2 t21 4t2 e g t t 2 t 1 4t f g t 20 4 t 2 7t 1 t Answers 20 4 t 2 1 t and 7t 1 t 20 2 cos 40πt and 20 2 sin 40πt 7 t 2 and t 2 4 t 2 2 t 2 6 and 3t 0 and t 2 t 2 1 4 t 2 6 5 t 2 1 t 2 and t 2 t 2 9 1 t 2 24 Graph the even and odd parts of the functions in Figure E24 a b t gt 1 1 t gt 2 1 1 1 Figure E24 rob28124ch02019078indd 67 041216 116 pm C h a p t e r 2 Mathematical Description of ContinuousTime Signals 68 Answers t get 2 1 1 1 t got 2 1 1 1 t get 1 1 t got 1 1 25 Graph the indicated product or quotient g t of the functions in Figure E25 t 1 1 1 1 t 1 1 1 gt Multiplication 1 1 1 1 1 1 1 1 t t gt Multiplication 1 1 1 1 t t gt Multiplication 1 1 1 1 t t gt Multiplication 1 1 1 1 1 1 1 t t gt Multiplication 1 1 1 1 1 1 t t gt Multiplication 1 1 1 1 t t gt Division 1 1 1 1 1 π t t gt Division Figure E25 rob28124ch02019078indd 68 041216 116 pm 69 Exercises with Answers 26 Use the properties of integrals of even and odd functions to evaluate these integrals in the quickest way a 1 1 2 t dt b 120 120 4 cos 10πt 8 sin 5πt dt c 120 120 4t cos 10πt dt d 110 110 t sin 10πt dt e 1 1 e t dt f 1 1 t e t dt Answers 0 1 50π 4 8 10π 0 1264 Periodic Signals 27 Find the fundamental period and fundamental frequency of each of these functions a g t 10 cos 50πt b g t 10 cos 50πt π 4 c g t cos 50πt sin 15πt d g t cos 2πt sin 3πt cos 5πt 3π 4 e g t 3 sin 20t 8 cos 4t f g t 10 sin 20t 7 cos 10πt g g t 3 cos 2000πt 8 sin 2500πt h g t g1t g2t g 1 t is periodic with fundamental period T 01 15μs g 2 t is periodic with fundamental period T 02 40μs Answers 120 μs and 8333 1 3 Hz 125 s and 25 Hz π2 s and 2π Hz 2 s and 12 Hz 125 s and 25 Hz Not Periodic 04 s and 25 Hz 4 ms and 250 Hz 28 Find a function of continuous time t for which the two successive transformations t t and t t 1 leave the function unchanged Answer Any even periodic function with a period of one Answers gt t 1 1 1 gt t 1 gt t 1 1 1 1 gt t 1 1 1 gt 1 1 1 t gt t 1 1 1 1 gt t 1 1 1 gt t 1 1 1 1 rob28124ch02019078indd 69 041216 116 pm C h a p t e r 2 Mathematical Description of ContinuousTime Signals 70 29 One period of a periodic signal x t with period T 0 is graphed in Figure E29 Assuming x t has a period T 0 what is the value of x t at time t 220 ms 5 ms 10 ms 15 ms t xt 4 3 2 1 1 2 3 4 T0 20 ms Figure E29 Answer 2 Signal Energy and Power of Signals 30 Find the signal energy of each of these signals a x t 2 rect t b x t A u t u t 10 c x t u t u 10 t d x t rect t cos 2πt e x t rect t cos 4πt f x t rect t sin 2πt g x t t 1 t 1 0 otherwise Answers 12 12 12 10 A 2 4 23 31 A signal is described by x t A rect t B rect t 05 What is its signal energy Answer A 2 B 2 AB 32 Find the average signal power of the periodic signal x t in Figure E32 1 4 3 2 1 2 3 4 1 1 2 3 2 3 t xt Figure E32 Answer 89 33 Find the average signal power of each of these signals a x t A b x t u t c x t A cos2πf0t θ rob28124ch02019078indd 70 041216 116 pm 71 Exercises without Answers d x t is periodic with fundamental period four and one fundamental period is described by x t t 1 t 1 t 5 e x t is periodic with fundamental period six This signal is described over one fundamental period by x t rect t 2 3 4 rect t 4 2 0 t 6 Answers A 2 5167 A 2 2 885333 12 EXERCISES WITHOUT ANSWERS Signal Functions 34 Let the unit impulse function be represented by the limit δ x lim a0 1a rect xa a 0 The function 1a rect xa has an area of one regardless of the value of a a What is the area of the function δ 4x lim a0 1a rect 4xa b What is the area of the function δ 6x lim a0 1a rect 6xa c What is the area of the function δ bx lim a0 1a rect bxa for b positive and for b negative 35 Using a change of variable and the definition of the unit impulse prove that δ a t t 0 1 a δ t t 0 36 Using the results of Exercise 35 a Show that δ 1 ax 1 a n δx n a b Show that the average value of δ 1 ax is one independent of the value of a c Show that even though δ at 1 a δ t δ 1 ax 1 a δ 1 x Scaling and Shifting 37 A signal is zero for all time before t 2 rises linearly from 0 to 3 between t 2 and t 4 and is zero for all time after that This signal can be expressed in the form x t A rect t t 01 w 1 tri t t 02 w 2 Find the numerical values of the constants 38 Let x t 3 e t4 u t 1 and let y t 4x 5t a What is the smallest value of t for which y t is not zero b What is the maximum value of y t over all time c What is the minimum value of y t over all time d What is the value of y 1 e What is the largest value of t for which y t 2 39 Graph these singularity and related functions a g t 2u 4 t b g t u 2t c g t 5 sgn t 4 d g t 1 sgn 4 t e g t 5 ramp t 1 f g t 3 ramp 2t rob28124ch02019078indd 71 041216 116 pm C h a p t e r 2 Mathematical Description of ContinuousTime Signals 72 g g t 2δ t 3 h g t 6δ 3t 9 i g t 4δ 2 t 1 j g t 2 δ 1 t 1 2 k g t 8 δ 1 4t l g t 6 δ 2 t 1 m g t 2 rect t 3 n g t 4 rect t 1 2 o g t 3 rect t 2 p g t 01 rect t 3 4 40 Graph these functions a g t u t u t 1 b g t rect t 1 2 c g t 4 ramp t u t 2 d g t sgn t sin 2πt e g t 5 e t4 u t f g t rect t cos 2πt g g t 6 rect t cos 3πt h g t u t 1 2 ramp 1 2 t i g t rect t 1 2 rect t 1 2 j g t t δ λ 1 2δ λ δ λ 1 dλ k g t 2 ramp t rect t 1 2 l g t 3 rect t 4 6 rect t 2 41 A continuoustime signal x t is defined by the graph below Let y t 4x t 3 and let z t 8x t 4 Find the numerical values a y 3 b z 4 c d dt z t t10 t 10 10 10 10 xt 42 Find the numerical values of a ramp 3 2 rect 2 10 b 3δ t 4 cos πt10 dt c d dt 2 sgn t5 ramp t 8 t13 43 Let a function be defined by g t tri t Below are four other functions based on this function All of them are zero for large negative values of t g 1 t 5g 2 t 6 g 2 t 7g 3t 4g t 4 g 3 t g t 2 4g t 4 3 g 4 t 5g t g t 1 2 a Which of these transformed functions is the first to become nonzero becomes nonzero at the earliest time b Which of these transformed functions is the last to go back to zero and stay there c Which of these transformed functions has a maximum value that is greater than all the other maximum values of all the other transformed functions rob28124ch02019078indd 72 041216 116 pm 73 Exercises without Answers Greater than in the strict mathematical sense of more positive than For example 2 5 d Which of these transformed functions has a minimum value that is less than all the other minimum values of all the other transformed functions 44 a Write a functional description of the timedomain energy signal in Figure E44 as the product of two functions of t 1 05 0 05 1 1 05 0 05 1 t xt Energy Signal a f X f b f Phase of X f Figure E44 b Write a functional description of the frequencydomain signal as the sum of two functions of f c Find the numerical values of a and b 45 A function g t has the following description It is zero for t 5 It has a slope of 2 in the range 5 t 2 It has the shape of a sine wave of unit amplitude and with a frequency of 1 4 Hz plus a constant in the range 2 t 2 For t 2 it decays exponentially toward zero with a time constant of 2 seconds It is continuous everywhere a Write an exact mathematical description of this function b Graph g t in the range 10 t 10 c Graph g 2t in the range 10 t 10 d Graph 2g 3 t in the range 10 t 10 e Graph 2g t 1 2 in the range 10 t 10 46 A signal occurring in a television set is illustrated in Figure E46 Write a mathematical description of it Signal in Television t μs 10 60 xt 10 5 Figure E46 Signal occurring in a television set rob28124ch02019078indd 73 041216 116 pm C h a p t e r 2 Mathematical Description of ContinuousTime Signals 74 47 The signal illustrated in Figure E47 is part of a binaryphaseshiftkeyed BPSK binary data transmission Write a mathematical description of it t ms 4 xt 1 1 BPSK Signal Figure E47 BPSK signal 48 The signal illustrated in Figure E48 is the response of an RC lowpass filter to a sudden change in excitation Write a mathematical description of it t ns 20 xt 6 4 RC Filter Signal 13333 4 Figure E48 Transient response of an RC filter 49 Describe the signal in Figure E49 as a ramp function minus a summation of step functions 4 15 xt t Figure E49 50 Mathematically describe the signal in Figure E50 9 9 xt t Semicircle Figure E50 51 Let two signals be defined by x 1 t 1 cos 2πt 0 0 cos 2πt 0 and x 2 t sin 2πt 10 rob28124ch02019078indd 74 041216 116 pm 75 Exercises without Answers Graph these products over the time range 5 t 5 a x 1 2t x 2 t b x 1 t 5 x 2 20t c x 1 t 5 x 2 20 t 1 d x 1 t 2 5 x 2 20t 52 Given the graphical definitions of functions in Figure E52 graph the indicated transformations a 2 2 3 4 5 6 1 1 2 2 t gt g t 0 t 6 or t 2 b 1 1 4 3 2 2 3 4 1 1 2 3 2 3 t xt t t 4 g t 2g t 1 2 g t is periodic with fundamental period 4 Figure E52 53 For each pair of functions graphed in Figure E53 determine what transformation has been done and write a correct functional expression for the transformed function a 2 1 2 3 4 5 6 2 1 t gt 1 1 2 3 4 32 4 2 1 t b 2 1 2 3 4 5 6 2 t gt 2 1 2 3 4 5 6 1 t Figure E53 g t g 2t g t 3g t rob28124ch02019078indd 75 041216 116 pm C h a p t e r 2 Mathematical Description of ContinuousTime Signals 76 54 Graph the magnitude and phase of each function versus f a G f jf 1 jf 10 b G f rect f 1000 100 rect f 1000 100 e jπf500 c G f 1 250 f 2 j3f Generalized Derivative 55 Graph the generalized derivative of g t 3 sin πt 2 rect t 56 Find the generalized derivative of the function described by x t 4 t 3 7t t 3 Derivatives and Integrals of Functions 57 What is the numerical value of each of the following integrals a δ t cos 48πt dt b δ t 5 cos πt dt c 0 20 δ t 8 rect t 16 dt d 8 22 8 e 4t δ t 2 dt e 11 82 3 sin 200t δ t 7 dt f 2 10 39 t 2 δ 4 t 1 dt 58 Graph the time derivatives of these functions a g t sin 2πt sgn t b g t cos 2πt Even and Odd Functions 59 Find the even and odd parts of each of these functions a g t 10 sin 20πt b g t 20 t 3 c x t 8 7 t 2 d x t 1 t e x t 6t f g t 4t cos 10πt g g t cos πt πt h g t 12 sin 4πt 4πt i g t 8 7t cos 32πt j g t 8 7 t 2 sin 32πt 60 Is there a function that is both even and odd simultaneously Discuss rob28124ch02019078indd 76 041216 116 pm 77 Exercises without Answers 61 Find and graph the even and odd parts of the function x t in Figure E61 t xt 1 11 1 2 4 3 2 5 2 3 4 5 Figure E61 Periodic Functions 62 For each of the following signals decide whether it is periodic and if it is find the fundamental period a g t 28 sin 400πt b g t 14 40 cos 60πt c g t 5t 2 cos 5000πt d g t 28 sin 400πt 12 cos 500πt e g t 10 sin 5t 4 cos 7t f g t 4 sin 3t 3 sin 3 t 63 Is a constant a periodic signal Explain why it is or is not periodic and if it is periodic what is its fundamental period Signal Energy and Power of Signals 64 Find the signal energy of each of these signals a 2 rect t b rect 8t c 3 rect t 4 d 2 sin 200πt e δ t Hint First find the signal energy of a signal which approaches an impulse some limit then take the limit f x t d dt rect t g x t t rect λ dλ h x t e 1j8π t u t i x t 2 rect t 4 3 rect t 1 4 j x t 3 rect t 1 6 tri t 4 6 k x t 5 e 4t u t l A signal x t has the following description 1 It is zero for all time t 4 2 It is a straight line from the point t 4 x 0 to the point t 4 x 4 3 It is a straight line from the point t 4 x 4 to the point t 3 x 0 4 It is zero for all time t 3 65 An even continuoustime energy signal x t is described in positive time by x t 3u t 5u t 4 11u t 7 0 t 10 0 t 10 Another continuoustime energy signal y t is described by y t 3x 2t 2 a Find the signal energy E x of x t b Find the signal energy E y of y t rob28124ch02019078indd 77 041216 116 pm C h a p t e r 2 Mathematical Description of ContinuousTime Signals 78 66 Find the average signal power of each of these signals a x t 2 sin 200πt b x t δ 1 t c x t e j100πt d A periodic continuoustime signal with fundamental period 12 described over one fundamental period by x t 3 rect t 3 4 4rect t 2 1 6 t 6 e x t 3 sgn 2 t 4 67 A signal x is periodic with fundamental period T0 6 This signal is described over the time period 0 t 6 by rect t 2 3 4 rect t 4 2 What is the signal power of this signal rob28124ch02019078indd 78 041216 116 pm 79 Figure 31 Examples of discretetime signals n xn Daily Closing NASDAQ Composite Index Weekly Average Temperature n xn Samples from an Exponentially Damped Sinusoid n xn Most of the functions and methods developed for describing continuoustime signals have very similar counterparts in the description of discretetime signals But some operations on discretetime signals are fundamentally different causing phenomena that do not occur in continuoustime signal analysis The fundamental difference between continuoustime and discretetime signals is that the signal values occurring as time passes in a discretetime signal are countable and in a continuoustime signal they are uncountable CH APTER GOAL S 1 To define mathematical functions that can be used to describe discretetime signals 2 To develop methods of shifting scaling and combining those functions to represent practical signals and to appreciate why these operations are different in discretetime than in continuoustime C H A P T E R 3 31 INTRODUCTION AND GOALS In the 20th century digital computing machinery developed from its infancy to its position today as a ubiquitous and indispensable part of our society and economy The effect of digital computation on signals and systems is equally broad Every day operations that were once done by continuoustime systems are being replaced by discretetime systems There are systems that are inherently discrete time but most of the application of discretetime signal processing is on signals that are created by sampling continu oustime signals Figure 31 shows some examples of discretetime signals DiscreteTime Signal Description rob28124ch03079117indd 79 041216 119 pm C h a p t e r 3 DiscreteTime Signal Description 80 3 To recognize certain symmetries and patterns to simplify analysis of discretetime signals 32 SAMPLING AND DISCRETE TIME Of increasing importance in signal and system analysis are discretetime functions that describe discretetime signals The most common examples of discretetime signals are those obtained by sampling continuoustime signals Sampling means acquiring the values of a signal at discrete points in time One way to visualize sampling is through the example of a voltage signal and a switch used as an ideal sampler Figure 32a The switch closes for an infinitesimal time at discrete points in time Only the values of the continuoustime signal xt at those discrete times are assigned to the discretetime signal xn If there is a fixed time Ts between samples the sampling is called uniform sampling in which the sampling times are integer multiples of a sampling period or sampling interval Ts The time of a sample n Ts can be replaced by the integer n which indexes the sample Figure 33 This type of operation can be envisioned by imagining that the switch simply rotates at a constant cyclic velocity f s cycles per second as in Figure 32b with the time between samples being T s 1 f s 2π ω s We will use a simplified notation for a Figure 33 Creating a discretetime signal by sampling a continuoustime signal gt t gn n Ts Figure 32 a An ideal sampler b an ideal sampler sampling uniformly xt xn xt xn ωs fs b a rob28124ch03079117indd 80 041216 119 pm 32 Sampling and Discrete Time 81 discretetime function gn formed by sampling which at every point of continuity of gt is the same as gnTs and in which n is an integer The square brackets enclosing the argument indicate a discretetime function as contrasted with the parentheses that indicate a continuoustime function The independent variable n is called discrete time because it indexes discrete points in time even though it is dimensionless not having units of seconds as t and Ts do Since discretetime functions are only defined for integer values of n the values of expressions like g27 or g34 are undefined Functions that are defined for continuous arguments can also be given discrete time as an argument for example sin2π f 0 nTs We can form a discretetime function from a continuoustime function by sampling for example gn sin2π f 0 nTs Then although the sine is defined for any real argument value the function gn is only defined for integer values of n That is g78 is undefined even though sin2π f 0 78Ts is defined1 In engineering practice the most important examples of discretetime systems are sequentialstate machines the most common example being a computer Computers are driven by a clock a fixedfrequency oscillator The clock generates pulses at regular intervals in time and at the end of each clock cycle the computer has executed an instruction and changed from one logical state to the next The computer has become a very important tool in all phases of the modern economy so understanding how discretetime signals are processed by sequentialstate machines is very important especially to engineers Figure 34 illustrates some discretetime functions that could describe discretetime signals The type of graph used in Figure 34 is called a stem plot in which a dot indicates the functional value and the stems always connect the dot to the discrete time n axis This is a widely used method of graphing discretetime functions MATLAB has a command stem that generates stem plots The use of MATLAB to draw graphs is an example of sampling MATLAB can only deal with finitelength vectors so to draw a graph of a continuoustime function we must decide how many points to put in the time vector so that when MATLAB draws straight lines between the function values at those times the graph looks like a continuoustime function Sampling will be considered in much more depth in Chapter 10 1 If we were to define a function as gn 5sin2πf0nTs the parentheses in gn would indicate that any real value of n would be acceptable integer or otherwise Although this is mathematically legal it is not a good idea because we are using the symbol t for continuous time and the symbol n for discrete time and the notation gn although mathematically defined would be confusing Figure 34 Examples of discretetime functions n gn n gn n gn n gn rob28124ch03079117indd 81 041216 119 pm C h a p t e r 3 DiscreteTime Signal Description 82 33 SINUSOIDS AND EXPONENTIALS Exponentials and sinusoids are as important in discretetime signal and system analysis as in continuoustime signal and system analysis Most discretetime systems can be described at least approximately by difference equations The eigenfunction of linear constantcoefficient ordinary difference equations is the complex exponential and the real exponential is a special case of the complex exponential Any sinusoid is a linear combination of complex exponentials The solution of difference equations will be considered in more detail in Chapter 4 Discretetime exponentials and sinusoids can be defined in a manner analogous to their continuoustime counterparts as gn A e βn or gn A z n where z e β and gn A cos2π F 0 n θ or gn A cos Ω 0 n θ where z and β are complex constants A is a real constant θ is a real phase shift in radians F 0 is a real number Ω 0 2π F 0 and n is discrete time SINUSOIDS There are some important differences between continuoustime and discretetime sinusoids One difference is that if we create a discretetime sinusoid by sampling a continuoustime sinusoid the period of the discretetime sinusoid may not be readily apparent and in fact the discretetime sinusoid may not even be periodic Let a discrete time sinusoid gn A cos2π F 0 n θ be related to a continuoustime sinusoid gt A cos2π f 0 t θ through gn gnTs Then F 0 f 0 Ts f 0 f s where f s 1Ts is the sampling rate The requirement on a discretetime sinusoid that it be periodic is that for some discrete time n and some integer m 2π F 0 n 2πm Solving for F 0 F 0 mn indicating that F 0 must be a rational number a ratio of integers Since sampling forces the relationship F 0 f 0 f s this also means that for a discretetime sinusoid to be periodic the ratio of the fundamental frequency of the continuoustime sinusoid to the sampling rate must be rational What is the fundamental period of the sinusoid gn 4 cos 72πn 19 4 cos2π3619n F 0 is 3619 and the smallest positive discrete time n that solves F 0 n m m an integer is n 19 So the fundamental period is 19 If F 0 is a rational number and is expressed as a ratio of integers F 0 q N 0 and if the fraction has been reduced to its simplest form by canceling common factors in the numerator and denominator then the funda mental period of the sinusoid is N 0 not 1 F 0 N 0 q unless q 1 Compare this result with the fundamental period of the continuoustime sinusoid gt 4 cos72πt19 whose fun damental period T 0 is 1936 not 19 Figure 35 illustrates some discretetime sinusoids When F 0 is not the reciprocal of an integer a discretetime sinusoid may not be immediately recognizable from its graph as a sinusoid This is the case for Figure 35c and d The sinusoid in Figure 35d is aperiodic A source of confusion for students when first encountering a discretetime sinusoid of the form A cos 2π F 0 n or A cos Ω 0 n is the question What are F 0 and Ω 0 In the continuoustime sinusoids A cos2π f 0 t and A cos ω 0 t f 0 is the cyclic frequency in Hz or cyclessecond and ω 0 is the radian frequency in radianssecond The argument of the cosine must be dimensionless and the products 2π f 0 t and ω 0 t are dimensionless because the cycle and radian are ratios of lengths and the second in t and the second 1 rob28124ch03079117indd 82 041216 119 pm 33 Sinusoids and Exponentials 83 in f 0 or ω 0 cancel Likewise the arguments 2π F 0 n and Ω 0 n must also be dimensionless Remember n does not have units of seconds Even though we call it discrete time it is really a time index not time itself If we think of n as indexing the samples then for example n 3 indicates the third sample taken after the initial discrete time n 0 So we can think of n as having units of samples Therefore F 0 should have units of cycles sample to make 2π F 0 n dimensionless and Ω 0 should have units of radianssample to make Ω 0 n dimensionless If we sample a continuoustime sinusoid A cos 2π f 0 t with fundamental frequency f 0 cyclessecond at a rate of f s samplessecond we form the discretetime sinusoid A cos2π f 0 n T s A cos2πn f 0 f s A cos2π F 0 n F 0 f 0 f s and the units are consistent F 0 in cyclessample f 0 in cyclessecond f s in samplessecond So F 0 is a cyclic frequency normalized to the sampling rate Similarly Ω 0 ω 0 f s is a normalized radian frequency in radianssample Ω 0 in radianssample ω 0 in radianssecond f s in samplessecond One other aspect of discretetime sinusoids that will be very important in Chapter 10 in the consideration of sampling is that two discretetime sinusoids g 1 n A cos2π F 1 n θ and g 2 n A cos2π F 2 n θ can be identical even if F 1 and F 2 are different For example Figure 35 Four discretetime sinusoids n xn xn xn xn 1 1 sin2πF0n F0 116 Fundamental Period is 16 sin2πF0n F0 1116 Fundamental Period is 16 sin2πF0n F0 π16 Aperiodic sin2πF0n F0 216 Fundamental Period is 8 n 1 1 n 1 1 n 1 1 a b c d rob28124ch03079117indd 83 041216 119 pm C h a p t e r 3 DiscreteTime Signal Description 84 the two sinusoids g 1 n cos2πn5 and g 2 n cos12πn5 are described by differ entlooking expressions but when we graph them versus discrete time n we see that they are identical Figure 36 The dashed lines in Figure 36 are the continuoustime functions g 1 t cos2πt5 and g 2 t cos12πt5 where n and t are related by t n T s The continuoustime functions are obviously different but the discretetime functions are not The reason the two discretetime functions are identical can be seen by rewriting g 2 n in the form g 2 n cos 2π 5 n 10π 5 n cos 2π 5 n 2πn Then using the principle that if any integer multiple of 2π is added to the angle of a sinusoid the value is not changed g 2 n cos 2π 5 n 2πn cos 2π 5 n g 1 n because discretetime n is an integer Since the two discretetime cyclic frequencies in this example are F 1 15 and F 2 65 that must mean that they are equivalent as frequencies in a discretetime sinusoid That can be seen by realizing that at a frequency of 15 cyclessample the angular change per sample is 2π5 and at a frequency of 65 cyclessample the angular change per sample is 12π5 As shown above those two angles yield exactly the same values as arguments of a sinusoid So in a discrete time sinusoid of the form cos2π F 0 n θ if we change F 0 by adding any integer the sinusoid is unchanged Similarly in a discretetime sinusoid of the form cos Ω 0 n θ if we change Ω 0 by adding any integer multiple of 2π the sinusoid is unchanged One can then imagine an experiment in which we generate a sinusoid sin2πFn and let F be a variable As F changes in steps of 025 from 0 to 175 we get a sequence of Figure 36 Two cosines with different Fs but the same functional behavior 1 1 1 2 3 4 5 6 7 8 9 10 g1n cos 2πn 5 n 1 1 1 2 3 4 5 6 7 8 9 10 12πn 5 g2n cos n rob28124ch03079117indd 84 041216 119 pm 33 Sinusoids and Exponentials 85 Figure 37 Illustration that a discretetime sinusoid with frequency F repeats every time F changes by one nt 4 xn 1 1 F 0 nt 4 xn 1 1 F 025 nt 4 xn 1 1 F 05 nt 4 xn 1 1 F 075 nt 4 xn 1 1 F 1 nt 4 xn 1 1 F 125 nt 4 xn 1 1 F 15 nt 4 xn 1 1 F 175 xn cos2πFn Dashed line is xt cos2πFt Figure 38 Behavior of gn A z n for different real zs 0 z 1 n z 1 n 1 z 0 n z 1 n Figure 39 Behavior of gn A z n for some complex zs n n z 1 z 1 n Regn n Imgn Regn Imgn discretetime sinusoids Figure 37 Any two discretetime sinusoids whose F values differ by an integer are identical EXPONENTIALS The most common way of writing a discretetime exponential is in the form gn A z n This does not look like a continuoustime exponential which has the form gt A e βt because there is no e but it is still an exponential because gn A z n could have been written as gn A e βn where z e β The form A z n is a little simpler and is generally preferred Discretetime exponentials can have a variety of functional behaviors depending on the value of z in gn A z n Figure 38 and Figure 39 summarize the functional form of an exponential for different values of z rob28124ch03079117indd 85 041216 119 pm C h a p t e r 3 DiscreteTime Signal Description 86 34 SINGULARITY FUNCTIONS There is a set of discretetime functions that are analogous to continuoustime singu larity functions and have similar uses THE UNITIMPULSE FUNCTION The unitimpulse function sometimes called the unitsample function Figure 310 is defined by δn 1 n 0 0 n 0 31 n δn 1 Figure 310 The unitimpulse function The discretetime unitimpulse function suffers from none of the mathematical peculiarities that the continuoustime unit impulse has The discretetime unit impulse does not have a property corresponding to the scaling property of the continuoustime unit impulse Therefore δn δan for any nonzero finite integer value of a But the discretetime impulse does have a sampling property It is n Aδn n 0 xn Ax n 0 32 Since the impulse is only nonzero where its argument is zero the summation over all n is a summation of terms that are all zero except at n n 0 When n n 0 xn x n 0 and that result is simply multiplied by the scale factor A We did not have a MATLAB function for continuoustime impulses but we can make one for discretetime impulses Function to generate the discretetime impulse function defined as one for input integer arguments equal to zero and zero otherwise Returns NaN for noninteger arguments function y impDn function y impDn y doublen 0 Impulse is one where argument is zero and zero otherwise I findroundn n Index noninteger values of n yI NaN Set those return values to NaN This MATLAB function implements the functional behavior of δn including returning undefined values NaN for arguments that are not integers The D at the rob28124ch03079117indd 86 041216 119 pm 34 Singularity Functions 87 end of the function name indicates that it is a discretetime function We cannot use the convention of square brackets enclosing the argument in MATLAB to indicate a discretetime function Square brackets in MATLAB have a different meaning THE UNITSEQUENCE FUNCTION The discretetime function that corresponds to the continuoustime unit step is the unitsequence function Figure 311 un 1 n 0 0 n 0 33 n un 1 Figure 311 The unitsequence function For this function there is no disagreement or ambiguity about its value at n 0 It is one and every author agrees Unit sequence function defined as 0 for input integer argument values less than zero and 1 for input integer argument values equal to or greater than zero Returns NaN for noninteger arguments function y usDn function y usDn y doublen 0 Set output to one for non negative arguments I findroundn n Index noninteger values of n yI NaN Set those return values to NaN THE SIGNUM FUNCTION The discretetime function corresponding to the continuoustime signum function is defined in Figure 312 sgnn 1 n 0 0 n 0 1 n 0 34 Signum function defined as 1 for input integer argument values less than zero 1 for input integer argument values greater than zero and zero for input argument values equal to zero Returns NaN for noninteger arguments function y signDn rob28124ch03079117indd 87 041216 119 pm C h a p t e r 3 DiscreteTime Signal Description 88 function y signDn y signn Use the MATLAB sign function I findroundn n Index noninteger values of n yI NaN Set those return values to NaN THE UNITRAMP FUNCTION The discretetime function corresponding to the continuoustime unit ramp is defined in Figure 313 rampn n n 0 0 n 0 nun 35 Unit discretetime ramp function defined as 0 for input integer argument values equal to or less than zero and n for input integer argument values greater than zero Returns NaN for noninteger arguments function y rampDn function y rampDn y rampn Use the continuoustime ramp I findroundn n Index noninteger values of n yI NaN Set those return values to NaN THE UNIT PERIODIC IMPULSE FUNCTION OR IMPULSE TRAIN The unit discretetime periodic impulse or impulse train Figure 314 is defined by δ N n m δn mN 36 n δNn 1 N N 2N Figure 314 The unit periodic impulse function Figure 312 The signum function n sgnn 1 1 n rampn 4 4 8 8 Figure 313 The unitramp function rob28124ch03079117indd 88 041216 119 pm 35 Shifting and Scaling 89 Discretetime periodic impulse function defined as 1 for input integer argument values equal to integer multiples of N and 0 otherwise N must be a positive integer Returns NaN for nonpositive integer values function y impNDNn function y impNDNn if N roundN N 0 y doublenN roundnN Set return values to one at all values of n that are integer multiples of N I findroundn n Index noninteger values of n yI NaN Set those return values to NaN else y NaNn Return a vector of NaNs dispIn impND the period parameter N is not a positive integer end The new discretetime signal functions are summarized in Table 31 Table 31 Summary of discretetime signal functions Sine sin2πF0n Sampled ContinuousTime Cosine cos2πF0n Sampled ContinuousTime Exponential zn Sampled ContinuousTime Unit Sequence un Inherently DiscreteTime Signum sgnn Inherently DiscreteTime Ramp rampn Inherently DiscreteTime Impulse δn Inherently DiscreteTime Periodic Impulse δNn Inherently DiscreteTime 35 SHIFTING AND SCALING The general principles that govern scaling and shifting of continuoustime functions also apply to discretetime functions but with some interesting differences caused by the fundamental differences between continuous time and discrete time Just as a continuoustime function does a discretetime function accepts a number and returns another number The general principle that the expression in gexpression is treated in exactly the same way that n is treated in the definition gn still holds AMPLITUDE SCALING Amplitude scaling for discretetime functions is exactly the same as it is for continuoustime functions TIME SHIFTING Let a function gn be defined by the graph and table in Figure 315 Now let n n 3 Time shifting is essentially the same for discretetime and for continuoustime functions except that the shift must be an integer otherwise the shifted function would have unde fined values Figure 316 TIME SCALING Amplitude scaling and time shifting for discretetime and continuoustime functions are very similar That is not so true when we examine time scaling for discretetime functions There are two cases to examine time compression and time expansion rob28124ch03079117indd 89 041216 119 pm C h a p t e r 3 DiscreteTime Signal Description 90 Time Compression Time compression is accomplished by a scaling of the form n Kn where K 1 and K is an integer Time compression for discretetime functions is similar to time compression for continuoustime functions in that the function seems to occur faster in time But in the case of discretetime functions there is another effect called decimation Consider the time scaling n 2n illustrated in Figure 317 For each integer n in g2n the value 2n must be an even integer Therefore for this scaling by a factor of two the oddintegerindexed values of gn are never needed to find values for g2n The function has been decimated by a factor of two because the graph of g2n only uses every other value of gn For larger scaling constants the decimation factor is obviously higher Decimation does not happen in scaling continuoustime functions because in using a scaling t Kt all real t values map into real Kt values without any missing values The fundamental difference between continuoustime and discretetime functions is that the domain of a continuoustime function is all real numbers an uncountable infinity of times and the domain of discretetime functions is all integers a countable infinity of discrete times Time Expansion The other timescaling case time expansion is even stranger than time compression If we want to graph for example gn2 for each integer value of n we must assign a value to gn2 by finding the corresponding value in the original function definition But when n is one n2 is onehalf and g12 is not defined The value of the timescaled function gnK is undefined unless nK is an integer We could simply leave those values undefined or we could interpolate between them using the values of gnK at the next higher and next lower values of n at which nK is an integer Interpolation is a process of computing functional values between two known values according to some formula Since interpolation begs the question of what interpolation formula to use we will simply leave gnK undefined if nK is not an integer Even though time expansion as described above seems to be totally useless there is a type of time expansion that is actually often useful Suppose we have an original function xn and we form a new function yn xnK nK an integer 0 otherwise as in Figure 318 where K 2 Figure 315 Graphical definition of a function gn gn 0 n 15 n gn 10 n gn 1 1 0 2 1 3 2 4 3 5 4 6 5 7 6 8 7 9 8 10 9 5 10 0 Figure 316 Graph of gn 3 illustrating time shifting n gn3 10 n n3 gn3 4 1 1 3 0 2 2 1 3 1 2 4 0 3 5 1 4 6 2 5 7 3 6 8 4 7 9 5 8 10 6 9 5 7 10 0 rob28124ch03079117indd 90 041216 119 pm 35 Shifting and Scaling 91 Figure 317 Time compression for a discretetime function n 15 15 g2n 10 n 15 15 gn 10 n 1 1 gn 0 2 1 3 2 4 3 5 4 6 5 7 6 8 7 9 8 10 9 5 10 0 n 2n g2n 0 0 2 1 2 4 2 4 6 3 6 8 4 8 10 Figure 318 Alternate form of time expansion n xn 10 n yn 10 a e b cd a e b cd rob28124ch03079117indd 91 041216 119 pm C h a p t e r 3 DiscreteTime Signal Description 92 All the values of the expanded function are defined and all the values of x that occur at discrete time n occur in y at discrete time Kn All that has really been done is to replace all the undefined values from the former time expansion with zeros If we were to timecompress y by a factor K we would get all the x values back in their original positions and all the values that would be removed by decimating y would be zeros ExamplE 31 Graphing shifting and scaling of discretetime functions Using MATLAB graph the function gn 10 08 n sin3πn16un Then graph the functions g2n and gn3 Discretetime functions are easier to program in MATLAB than continuoustime functions because MATLAB is inherently oriented toward calculation of functional values at discrete values of the independent variable For discretetime functions there is no need to decide how close together to make the time points to make the graph look continuous because the function is not continuous A good way to handle graphing the function and the timescaled functions is to define the original function as an m file But we need to ensure that the function definition includes its discretetime behavior for noninteger values of discrete time the function is undefined MATLAB handles undefined results by assigning to them the special value NaN The only other programming problem is how to handle the two different functional descriptions in the two different ranges of n We can do that with logical and relational operators as demonstrated below in gm function y gn Compute the function y 1008nsin3pin16usDn I findroundn n Find all noninteger ns yI NaN Set those return values to NaN We still must decide over what range of discrete times to graph the function Since it is zero for negative times we should represent that time range with at least a few points to show that it suddenly turns on at time zero Then for positive times it has the shape of an exponentially decaying sinusoid If we graph a few time constants of the exponential decay the function will be practically zero after that time So the time range should be something like 5 n 16 to draw a reasonable representation of the original function But the timeexpanded function gn3 will be wider in discrete time and require more discrete time to see the functional behavior Therefore to really see all of the functions on the same scale for comparison lets set the range of discrete times to 5 n 48 Graphing a discretetime function and compressed and expanded transformations of it Compute values of the original function and the timescaled versions in this section rob28124ch03079117indd 92 041216 119 pm 35 Shifting and Scaling 93 n 548 Set the discrete times for function computation g0 gn Compute the original function values g1 g2n Compute the compressed function values g2 gn3 Compute the expanded function values Display the original and timescaled functions graphically in this section Graph the original function subplot311 Graph first of three graphs stacked vertically p stemng0kfilled Stem plot the original function setpLineWidth2MarkerSize4 Set the line weight and dot size ylabelgn Label the original function axis Graph the timecompressed function subplot312 Graph second of three plots stacked vertically p stemng1kfilled Stem plot the compressed function setpLineWidth2MarkerSize4 Set the line weight and dot size ylabelg2n Label the compressed function axis Graph the timeexpanded function subplot313 Graph third of three graphs stacked vertically p stemng2kfilled Stem plot the expanded function setpLineWidth2MarkerSize4 Set the line weight and dot size xlabelDiscrete time n Label the expanded function axis ylabelgn3 Label the discretetime axis rob28124ch03079117indd 93 041216 119 pm C h a p t e r 3 DiscreteTime Signal Description 94 Figure 319 illustrates the output of the MATLAB program 36 DIFFERENCING AND ACCUMULATION Just as differentiation and integration are important for continuoustime functions the analogous operations differencing and accumulation are important for discretetime functions The first derivative of a continuoustime function gt is usually defined by d dt gt lim Δt0 gt Δt gt Δt But it can also be defined by d dt gt lim Δt0 gt gt Δt Δt or d dt gt lim Δt0 gt Δt gt Δt 2Δt In the limit all these definitions yield the same derivative if it exists But if Δt remains finite these expressions are not identical The operation on a discretetime signal that is analogous to the derivative is the difference The first forward difference of a discretetime function gn is gn 1 gn See Web Appendix D for more on differencing and difference equations The first backward difference of a discretetime function is gn gn 1 which is the first forward difference of gn 1 Figure 320 illustrates some discretetime functions and their first forward or backward differences The differencing operation applied to samples from a continuoustime function yields a result that looks a lot like but not exactly like samples of the derivative of that continuoustime function to within a scale factor The discretetime counterpart of integration is accumulation or summation The accumulation of gn is defined by m n gm The ambiguity problem that Figure 319 Graphs of gn g2n and gn3 10 0 10 20 30 40 50 0 2 4 gn 10 0 10 20 30 40 50 2 0 2 4 6 10 0 10 20 30 40 50 2 0 2 4 6 2 6 g2n gn3 Discrete Time n rob28124ch03079117indd 94 041216 119 pm 36 Differencing and Accumulation 95 n 20 xn 1 1 n 20 xn1 xn 1 1 n 5 20 xn 1 n 5 20 1 n 5 20 xn 1 n 5 20 1 n 10 10 xn 1 1 n 10 10 1 1 xn xn1 xn xn1 xn1 xn Backward Differences Forward Differences Figure 320 Some functions and their backward or forward differences occurs in the integration of a continuoustime function exists for accumula tion of discretetime functions The accumulation of a function is not unique Multiple functions can have the same first forward or backward difference but just as in integration these functions only differ from each other by an additive constant Let hn gn gn 1 the first backward difference of gn Then accumulate both sides m n hm m n gm gm 1 or m n h m g1 g2 g0 g1 gn gn 1 Gathering values of gn occurring at the same time m n hm g1 g1 0 g0 g0 0 gn 1 gn 1 0 gn rob28124ch03079117indd 95 041216 119 pm C h a p t e r 3 DiscreteTime Signal Description 96 and m n hm gn This result proves that accumulation and firstbackwarddifference are inverse operations The first backward difference of the accumulation of any function gn is gn Figure 321 illustrates two functions hn and their accumulations gn In each of the graphs of Figure 321 the accumulation was done based on the assumption that all function values of hn before the time range graphed are zero In a manner analogous to the integralderivative relationship between the continuoustime unit step and the continuoustime unit impulse the unit sequence is the accumulation of the unit impulse un m n δm and the unit impulse is the first backward difference of the unit sequence δn un un 1 Also the discretetime unit ramp is defined as the accumulation of a unit sequence function delayed by one in discrete time rampn m n um 1 and the unit sequence is the first forward difference of the unit ramp un rampn 1 rampn and the first backward difference of rampn 1 MATLAB can compute differences of discretetime functions using the diff function The diff function accepts a vector of length N as its argument and returns a vector of forward differences of length N 1 MATLAB can also compute the accumulation of a function using the cumsum cumulative summation function The cumsum function accepts a vector as its argument and returns a vector of equal length that is the accumulation of the elements in the argument vector For example a 110 a 1 2 3 4 5 6 7 8 9 10 diffa ans 1 1 1 1 1 1 1 1 1 cumsuma ans 1 3 6 10 15 21 28 36 45 55 b randn15 n 5 20 hn 2 2 n 10 10 hn 2 n 5 20 gn 2 2 n 10 10 gn 8 Figure 321 Two functions hn and their accumulations gn rob28124ch03079117indd 96 041216 119 pm 36 Differencing and Accumulation 97 b 11909 11892 00376 03273 01746 diffb ans 00018 12268 03649 01527 cumsumb ans 11909 23801 23424 26697 28444 It is apparent from these examples that cumsum assumes that the value of the accumu lation is zero before the first element in the vector ExamplE 32 Graphing the accumulation of a function using MATLAB Using MATLAB graph the accumulation of the function xn cos2πn36 from n 0 to n 36 under the assumption that the accumulation before time n 0 is zero Program to demonstrate accumulation of a function over a finite time using the cumsum function n 036 Discretetime vector x cos2pin36 Values of xn Graph the accumulation of the function xn p stemncumsumxkfilled setpLineWidth2MarkerSize4 xlabelitnFontNameTimesFontSize24 ylabelxitnFontNameTimesFontSize24 Figure 322 illustrates the output of the MATLAB program Figure 322 Accumulation of a cosine 6 4 2 0 2 4 6 8 xn 0 5 10 15 20 25 30 35 40 n Accumulation of a Cosine rob28124ch03079117indd 97 041216 119 pm C h a p t e r 3 DiscreteTime Signal Description 98 Notice that this cosine accumulation looks a lot like but not exactly like a sine function That occurs because the accumulation process is analogous to the integration process for continuoustime functions and the integral of a cosine is a sine 37 EVEN AND ODD SIGNALS Like continuoustime functions discretetime functions can also be classified by the properties of evenness and oddness The defining relationships are completely analogous to those for continuoustime functions If gn gn then gn is even and if gn gn gn is odd Figure 323 shows some examples of even and odd functions Figure 323 Examples of even and odd functions n n gn Even Function gn Odd Function The even and odd parts of a function gn are found exactly the same way as for continuoustime functions g e n gn gn 2 and g o n gn gn 2 37 An even function has an odd part that is zero and an odd function has an even part that is zero ExamplE 33 Even and odd parts of a function Find the even and odd parts of the function gn sin2πn71 n 2 g e n sin2πn71 n2 sin2πn71 n2 2 g e n sin2πn71 n2 sin2πn71 n2 2 0 g o n sin2πn71 n2 sin2πn71 n2 2 sin 2πn 7 1 n 2 The function gn is odd function EvenOdd n 1414 Discretetime vector for graphing gen and gon Compute the even part of gn rob28124ch03079117indd 98 041216 119 pm 37 Even and Odd Signals 99 ge gngn2 Compute the odd part of gn go gngn2 close all figurePosition20201200800 subplot211 ptr stemngekfilled grid on setptrLineWidth2MarkerSize4 xlabelitnFontNameTimesFontSize24 ylabelgiteitnFontNameTimesFontSize24 titleEven Part of gnFontNameTimesFontSize24 setgcaFontNameTimesFontSize18 subplot212 ptr stemngokfilled grid on setptrLineWidth2MarkerSize4 xlabelitnFontNameTimesFontSize24 ylabelgitoitnFontNameTimesFontSize24 titleOdd Part of gnFontNameTimesFontSize24 setgcaFontNameTimesFontSize18 function y gn y sin2pin71n2 Figure 324 n 1 05 0 05 1 gen Even Part of gn 15 10 5 0 5 10 15 15 10 5 0 5 10 15 n 04 02 0 02 04 gon Odd Part of gn rob28124ch03079117indd 99 041216 119 pm C h a p t e r 3 DiscreteTime Signal Description 100 Figure 326 Product of two odd functions n g1n n g2n n g1ng2n Figure 325 Product of two even functions n g1n n g2n n g1ng2n Figure 327 Product of an even and an odd function n g1n n g2n n g1ng2n COMBINATIONS OF EVEN AND ODD SIGNALS All the properties of combinations of functions that apply to continuoustime functions also apply to discretetime functions If two functions are even their sum difference product and quotient are even too If two functions are odd their sum and difference are odd but their product and quotient are even If one function is even and the other is odd their product and quotient are odd In Figure 325 through Figure 327 are some examples of products of even and odd functions SYMMETRICAL FINITE SUMMATION OF EVEN AND ODD SIGNALS The definite integral of continuoustime functions over symmetrical limits is analogous to summation of discretetime functions over symmetrical limits Properties hold for rob28124ch03079117indd 100 041216 119 pm 38 Periodic Signals 101 summations of discretetime functions that are similar to but not identical to those for in tegrals of continuoustime functions If gn is an even function and N is a positive integer nN N gn g0 2 n1 N gn and if gn is an odd function nN N gn 0 See Figure 328 Figure 328 Summations of even and odd discretetime functions n gn n gn Even Function Odd Function N N N N Sum 1 Sum 1 Sum 2 Sum 2 Sum 1 Sum 1 Sum 2 Sum 2 Figure 329 Examples of periodic functions with fundamental period N 0 n gn n gn n gn N0 N0 N0 38 PERIODIC SIGNALS A periodic function is one that is invariant under the time shift n n mN where N is a period of the function and m is any integer The fundamental period N 0 is the minimum positive discrete time in which the function repeats Figure 329 shows some examples of periodic functions The fundamental frequency is F 0 1 N 0 in cyclessample or Ω 0 2π N 0 in radians sample Remember that the units of discretetime frequency are not Hz or radians second because the unit of discretetime is not the second ExamplE 34 Fundamental period of a function Graph the function gn 2cos9πn4 3sin6πn5 over the range 50 n 50 From the graph determine the fundamental period rob28124ch03079117indd 101 041216 119 pm C h a p t e r 3 DiscreteTime Signal Description 102 Figure 330 shows the function gn n 50 50 gn 5 5 N0 40 Figure 330 The function gn 2 cos9πn4 3 sin6πn5 As a check on this graphically determined answer the function can also be written in the form gn 2 cos2π98n 3 sin2π35n The two fundamental periods of the two individual sinusoids are then 8 and 5 and their LCM is 40 which is the fundamental period of gn 39 SIGNAL ENERGY AND POWER SIGNAL ENERGY Signal energy is defined by E x n xn 2 38 and its units are simply the square of the units of the signal itself ExamplE 35 Signal energy of a signal Find the signal energy of xn 12 n un From the definition of signal energy E x n xn 2 n 1 2 n un 2 n0 1 2 n 2 n0 1 2 2n 1 1 22 1 24 This infinite series can be rewritten as E x 1 1 4 1 42 We can use the formula for the summation of an infinite geometric series n0 r n 1 1 r r 1 to get E x 1 1 14 4 3 rob28124ch03079117indd 102 041216 119 pm 39 Signal Energy and Power 103 SIGNAL POWER For many signals encountered in signal and system analysis the summation E x n xn 2 does not converge because the signal energy is infinite and this usually occurs because the signal is not time limited The unit sequence is an example of a signal with infinite energy For signals of this type it is more convenient to deal with the average signal power of the signal instead of the signal energy The definition of average signal power is P x lim N 1 2N nN N1 xn 2 39 and this is the average signal power over all time Why is the upper summation limit N 1 instead of N For periodic signals the average signal power calculation may be simpler The average value of any periodic function is the average over any period and P x 1 N n n 0 n 0 N1 xn 2 1 N nN xn 2 n 0 any integer 310 where the notation nN means the summation over any range of consecutive ns that is N in length where N can be any period of xn 2 ExamplE 36 Finding signal energy and power of signals using MATLAB Using MATLAB find the signal energy or power of the signals a xn 09 n sin2πn4 and b xn 4 δ 5 n 7 δ 7 n Then compare the results with analytical calculations Program to compute the signal energy or power of some example signals a n 100100 Set up a vector of discrete times at which to compute the value of the function Compute the value of the function and its square x 09absnsin2pin4 xsq x2 Ex sumxsq Use the sum function in MATLAB to find the total energy and display the result dispb Ex num2strEx b rob28124ch03079117indd 103 041216 119 pm C h a p t e r 3 DiscreteTime Signal Description 104 N0 35 The fundamental period is 35 n 0N01 Set up a vector of discrete times over one period at which to compute the value of the function Compute the value of the function and its square x 4impND5n 7impND7n xsq x2 Px sumxsqN0 Use the sum function in MATLAB to find the average power and display the result dispd Px num2strPx The output of this program is a Ex 47107 b Px 86 Analytical computations a E x n xn 2 n 09 n sin2πn4 2 E x n0 09 n sin2πn4 2 n 0 09 n sin2πn4 2 x0 2 0 E x n0 09 2n sin 2 2πn4 n 0 09 2n sin 2 2πn4 E x 1 2 n0 09 2n 1 cosπn 1 2 n 0 09 2n 1 cosπn Using the even symmetry of the cosine function and letting n n in the second summation E x n0 09 2n 1 cosπn E x n0 09 2n 09 2n e jπn e jπn 2 n0 081 n 1 2 n0 081 e jπ n n0 081 e jπ n Using the formula for the sum of an infinite geometric series n0 r n 1 1 r r 1 E x 1 1 081 1 2 1 1 081 e jπ 1 1 081 e jπ E x 1 1 081 1 2 1 1 081 1 1 081 1 1 081 1 1 081 47107 Check b P x 1 N 0 nN0 xn 2 1 N 0 n0 N 0 1 xn 2 1 35 n0 34 4 δ 5 n 7 δ 7 n 2 The impulses in the two impulse train functions only coincide at integer multiples of 35 Therefore in this summation range they coincide only at n 0 The net impulse strength at rob28124ch03079117indd 104 041216 119 pm Exercises with Answers 105 n 0 is therefore 3 All the other impulses occur alone and the sum of the squares is the same as the square of the sum Therefore P x 1 35 3 2 n0 4 2 n5 7 2 n7 4 2 n10 7 2 n14 4 2 n15 4 2 n20 7 2 n21 4 2 n25 7 2 n28 4 2 n30 P x 9 6 4 2 4 7 2 35 9 96 196 35 86 Check 310 SUMMARY OF IMPORTANT POINTS 1 A discretetime signal can be formed from a continuoustime signal by sampling 2 A discretetime signal is not defined at noninteger values of discrete time 3 Discretetime signals formed by sampling periodic continuoustime signals may be aperiodic 4 Two differentlooking analytical descriptions can produce identical discretetime functions 5 A timeshifted version of a discretetime function is only defined for integer shifts in discrete time 6 Time scaling a discretetime function can produce decimation or undefined values phenomena that do not occur when time scaling continuoustime functions EXERCISES WITH ANSWERS Answers to each exercise are in random order Functions 1 In Figure E1 is a circuit in which a voltage x t A sin 2π f 0 t is connected periodically to a resistor by a switch The switch rotates at a frequency f s of 500 rpm The switch is closed at time t 0 and each time the switch closes it stays closed for 10 ms xit fs xot Figure E1 a If A 5 and f 0 1 graph the excitation voltage xit and the response voltage xot for 0 t 2 b If A 5 and f 0 10 graph the excitation voltage xit and the response voltage xot for 0 t 1 c This is an approximation of an ideal sampler If the sampling process were ideal what discretetime signal x n would be produced in parts a and b Graph them versus discrete time n rob28124ch03079117indd 105 041216 119 pm C h a p t e r 3 DiscreteTime Signal Description 106 Answers n 20 xn 6 6 t 1 xit 6 6 t 1 xot 6 6 n 8 xn 6 6 t 2 xit 6 6 t 2 xot 6 6 2 Let x 1 n 5 cos 2πn8 and x 2 n 8 e n6 2 Graph the following combinations of those two signals over the discretetime range 20 n 20 If a signal has some defined and some undefined values just plot the defined values a x n x 1 n x 2 n b x n 4 x 1 n 2 x 2 n c x n x 1 2n x 2 3n d x n x 1 2n x 2 n e x n 2 x 1 n2 4 x 2 n3 Answers n 20 20 xn 40 20 n 20 20 xn 40 5 n 20 20 xn 40 40 n 20 20 xn 40 20 n 20 20 xn 50000 10000 rob28124ch03079117indd 106 041216 119 pm Exercises with Answers 107 3 Find the numerical values of a n18 33 38 n 2 δ n 6 b n4 7 12 04 n u n δ 3 n Answers 128172 1368 Scaling and Shifting Functions 4 A discretetime function is defined by g n 3δ n 4 ramp n 1 What is its only nonzero value Answer 15 5 A discretetime signal has the following values n 5 4 3 2 1 0 1 2 3 4 5 6 7 8 9 10 11 12 13 x n 4 2 5 4 10 6 9 9 1 9 6 2 2 2 0 2 9 5 3 For all other n x n is zero Let y n x 2n 1 Find the numerical values y 2 y 1 y 4 y 7 y 12 Answers 3 0 2 4 5 6 Find the numerical values of these functions a ramp 6 u 2 b n 7 u n 9 u n 10 c g 4 where g n sin 2π n 3 8 δ n 3 Answers 6 0707 17 7 For each pair of functions in Figure E7 provide the values of the constants in g 2 n A g 1 a n n 0 10 5 0 5 10 2 1 0 1 2 n a a 10 5 0 5 10 2 1 0 1 2 n 10 5 0 5 10 2 1 0 1 2 n 10 5 0 5 10 2 1 0 1 2 n b b g1n g1n g2n g2n Figure E7 Answers A 12 n 0 0 a 2 or 2 A 1 n 0 1 a 2 or 2 rob28124ch03079117indd 107 041216 119 pm C h a p t e r 3 DiscreteTime Signal Description 108 8 A function g n is defined by g n 2 n 4 n 4 n 1 4 n 1 n Graph g n g 2 n g 2n and g n2 Answers n 10 10 gn 4 4 n 10 10 gn2 4 4 n 10 10 g2n 4 4 n 10 10 g2n 4 4 9 Find the maximum value over all discrete time of x n ramp n 4 ramp 5 n Answer 20 10 A discretetime signal has the following values for times n 8 to n 8 and is zero for all other times n 8 7 6 5 4 3 2 1 0 1 2 3 4 5 6 7 8 x n 9 4 9 9 4 9 9 4 9 9 4 9 9 4 9 9 4 This signal can be expressed in the form x n A B δ N 1 n n 0 u n N 2 u n N 2 1 Find the numerical values of the constants x n 9 5 δ 3 n 1 u n 8 u n 9 or x n 9 5 δ 3 n 2 u n 8 u n 9 Answers 3 5 8 9 1 or 3 2 8 5 9 11 A signal x n is nonzero only in the range 1 n 14 If y n x 3n If y n x 3n how many of the nonzero values of x appear in y Answer 4 12 A discretetime function g 1 n is illustrated in Figure E12 It is zero for all time outside the range graphed below Let some other functions be defined by g 2 n g 1 2n g 3 n 2 g 1 n 2 g 4 n 3 g 1 n 3 Find the following numerical values a g 4 2 b g 4 t g 3 t t2 c g 2 n g 3 n n1 d n1 1 g 2 n rob28124ch03079117indd 108 041216 119 pm Exercises with Answers 109 Figure E12 n g1 n 1 1 3 2 4 2 3 4 1 2 3 4 4 3 2 1 Answers Undefined 32 0 32 Differencing and Accumulation 13 Graph the backward differences of the discretetime functions in Figure E13 a b c n 4 20 gn 1 n 4 20 gn 1 n 4 20 gn n102 4 Figure E13 Answers n 4 20 025 05 gn gn1 n 4 20 1 1 gn gn1 n 4 20 1 1 gn gn1 14 The signal x n is defined in Figure E14 Let y n be the first backward difference of x n and let z n be the accumulation of x n Assume that x n is zero for all n 0 n 20 xn 6 4 Figure E14 a What is the value of y 4 b What is the value of z 6 Answers 8 3 rob28124ch03079117indd 109 041216 119 pm C h a p t e r 3 DiscreteTime Signal Description 110 15 Graph the accumulation g n of each of these discretetime functions h n which are zero for all times n 16 a h n δ n b h n u n c h n cos 2πn16 u n d h n cos 2πn8 u n e h n cos 2πn16 u n 8 Answers n 16 16 gn 3 3 n 16 16 gn 3 3 n 16 16 gn 3 3 n 16 16 gn 16 n 16 16 gn 1 Even and Odd Functions 16 Find and graph the even and odd parts of these functions a g n u n u n 4 b g n e n4 u n c g n cos 2πn4 d g n sin 2πn4 u n Answers n 10 10 gen 1 1 n 10 10 gon 1 1 n 10 10 gen 1 1 n 10 10 gon 1 1 n 10 10 gen 1 1 n 10 10 gon 1 1 n 10 10 gen 1 1 n 10 10 gon 1 1 rob28124ch03079117indd 110 041216 119 pm Exercises with Answers 111 17 Graph g n for the signals in Figure E17 a g1n g2n 10 10 1 1 10 10 1 1 n n gn Multiplication b g1n g2n 4 20 1 1 4 20 1 1 n n gn Multiplication c g1n g2n 4 20 1 1 4 20 1 1 n n gn Multiplication d g1n 10 10 1 1 10 10 1 1 n n gn g2n Multiplication Figure E17 Answers n 4 20 gn 1 1 n 10 10 gn 1 1 n 4 20 gn 1 1 n 10 10 gn 1 1 Periodic Functions 18 Find the fundamental period of each these functions a g n cos 2πn10 b g n cos πn10 cos 2πn20 c g n cos 2πn5 cos 2πn7 rob28124ch03079117indd 111 041216 119 pm C h a p t e r 3 DiscreteTime Signal Description 112 d g n e j2πn20 e j2πn20 e g n e j2πn3 e j2πn4 f g n sin 13πn8 cos 9πn6 sin 2 13πn16 cos 2 3πn4 g g n e j6πn21 cos 22πn36 sin 11πn33 Answers 20 10 20 12 252 35 16 19 If g n 15 cos 2πn12 and h n 15 cos 2πKn what are the two smallest positive values of K for which g n h n for all n Answers 1112 and 112 Signal Energy and Power 20 Identify each of these signals as either an energy signal or a power signal and find the signal energy of energy signals and the average signal power of power signals a x n u n Power Signal P x lim N 1 2N 1 nN N u n 2 lim N 1 2N 1 n0 N 1 2 lim N N 1 2N 1 1 2 b x n u n u n 10 Energy Signal E x n u n u n 10 2 n0 9 1 2 10 c x n u n 2u n 4 u n 10 Energy Signal E x n u n 2u n 4 u n 10 2 n0 3 1 2 n4 9 1 2 4 6 10 d x n 4 cos πn Power Signal P x 1 N nN 4 cos πn 2 16 N nN cos 2 πn 16 N nN 1 n 2 16 N nN 1 2 16 e x n 5 u n 3 u n 5 8 u n 1 u n 7 Energy Signal E x n 5 u n 3 u n 5 8 u n 1 u n 7 2 E x n3 6 5 u n 3 u n 5 8 u n 1 u n 7 2 E x n3 0 5 2 1 4 3 2 5 6 8 2 25 4 9 4 64 2 264 Answers 10 16 10 12 264 21 Find the signal energy of each of these signals a x n Aδ n b x n δ N 0 n c x n ramp n d x n ramp n 2ramp n 4 ramp n 8 e x n ramp n 3 u n 4 rob28124ch03079117indd 112 041216 119 pm 113 Exercises without Answers f x n δ 3 n 3 δ 6 n u n 1 u n 12 Answers 140 44 A 2 10 22 A signal x n is periodic with period N 0 6 Some selected values of x n are x 0 3 x 1 1 x 4 2 x 8 2 x 3 5 x 7 1 x 10 2 and x 3 5 What is its average signal power Answer 7333 23 Find the average signal power of a periodic signal described over one period by x n 2n 2 n 2 Answer 6 24 Find the average signal power of these signals a x n A b x n u n c x n δ N0 n d x n ramp n Answers 1 N 0 A 2 12 EXERCISES WITHOUT ANSWERS Signal Functions 25 Graph these discretetime exponential and trigonometric functions a g n 4 cos 2πn10 b g n 4 cos 22πn c g n 4 cos 18πn d g n 2 cos 2πn6 3sin 2πn6 e g n 34 n f g n 2 09 n sin 2πn4 Shifting and Scaling Functions 26 Graph these functions a g n 2u n 2 b g n u 5n c g n 2ramp n d g n 10ramp n2 e g n 7δ n 1 f g n 7δ 2 n 1 g g n 4δ 2n3 h g n 4δ 2n3 1 i g n 8 δ 4 n j g n 8 δ 4 2n 27 Graph these functions a g n u n u n b g n u n u n c g n cos 2πn12 δ 3 n d g n cos 2πn12 δ 3 n2 e g n cos 2π n 1 12 u n 1 cos 2πn 12 u n f g n m0 n cos 2πm 12 u m g g n m0 n δ 4 m δ 4 m 2 rob28124ch03079117indd 113 041216 119 pm C h a p t e r 3 DiscreteTime Signal Description 114 h g n m n δ 4 m δ 3 m u m 4 u m 5 i g n δ2 n 1 δ2 n j g n m n1 δ m m n δ m 28 Graph the magnitude and phase of each function versus k a G k 20 sin 2πk8 e jπk4 b G k δ k 8 2δ k 4 δ k 2δ k 4 δ k 8 e jπk8 29 Using MATLAB for each function below graph the original function and the shifted andor scaled function a g n 5 n 0 5 3n 0 n 4 23 n 2 4 n 8 41 n 8 g 3n vs n b g n 10 cos 2πn20 cos 2πn4 4g 2 n 1 vs n c g n 8 e j2πn16 u n g n2 vs n 30 Graph versus k in the range 10 k 10 the magnitude and phase of a X k 1 1 jk2 b X k jk 1 jk2 c X k δ 2 k e j2πk4 Differencing and Accumulation 31 Graph the accumulation of each of these discretetime functions a g n cos 2πn u n b g n cos 4πn u n 32 A discretetime function is defined by g n 3 1 n u n Another discrete time function h n is defined as the accumulation of g n from to n Find the numerical value of h 7 33 In the equation m n u m g n n 0 N w a What is name of the function g b Find the values of n 0 and N w 34 What is the numerical value of each of the following accumulations a n0 10 ramp n b n0 6 1 2 n c n u n 2 n d n10 10 δ 3 n 7 e n10 10 δ 3 2n 7 Even and Odd Functions 35 Find and graph the magnitude and phase of the even and odd parts of the discretek function G k 10 1 j4k rob28124ch03079117indd 114 041216 119 pm 115 Exercises without Answers 36 The even part of g n is g e n tri n16 and the odd part of g n is g o n u n 3 u n u n 1 u n 4 Find the value of g 3 37 Find and graph the even and odd parts of the function in Figure E37 n gn 2 6 8 6 4 2 2 4 6 6 4 2 4 8 Figure E37 38 Graph the even and odd parts of these signals a x n δ 3 n 1 b x n 15 cos 2πn9 π4 Periodic Signals 39 Find the fundamental periods if they exist of a x n δ 14 n 6 δ 8 n b x n 2 cos 3πn12 11 cos 14πn10 c x n cos n3 d x n cos 7πn3 4 sin 11πn5 e x n 9 cos 7πn15 14 sin 8πn11 f x n 22 sin 13n12 13 cos 18n23 40 What is the smallest positive value of n 0 that makes the signal 4 cos 6πn7 2 sin 15πn12 u n u n n 0 zero for all n 41 Using MATLAB graph each of these discretetime functions If a function is periodic find the period analytically and verify the period from the plot a g n sin 3πn2 sin 6πn4 b g n sin 2πn3 cos 10πn3 c g n 5cos 2πn8 Period of 8 3sin 2n5 Period of 5 d g n 10 cos n4 e g n 3 cos 2πn7 sin 2πn6 A trigonometric identity will be useful here 42 If x n sin 2πn15 and y n sin 2Aπn what is the smallest positive numerical value of A greater than 115 that makes x and y equal for all n rob28124ch03079117indd 115 041216 119 pm C h a p t e r 3 DiscreteTime Signal Description 116 Signal Energy and Power 43 Find the signal energy of each of these signals a x n 2δ n 5δ n 3 b x n u n n c x n 13 n u n d x n cos πn3 u n u n 6 e x n n 13 n u n u n 6 f x n 2 u n 2 u n 3 3tri n3 44 Find the average signal power of each of these signals a x n 1 n b x n A cos 2π F 0 n θ c x n A n 0 1 2 3 8 9 10 11 16 17 18 19 0 n 4 5 6 7 12 13 14 15 20 21 22 23 d x n e jπn2 e x n δ 3 n 3 δ 9 n 1 f x 1 n 2 cos 2πn6 45 Find the average signal power of x n 11 δ 4 2n 15 δ 5 3n 46 A periodic discretetime signal x 1 n is described over its fundamental period by n 0 1 2 x 1 n 8 3 5 A second periodic discretetime signal x 2 n is described over its fundamental period by n 0 1 2 3 x 2 n 5 2 11 4 a Find the average signal power of x 1 n b Find the average signal power of x 2 n c If y n x 1 n x 2 n find the average signal power of y n 47 A discretetime function x n is defined by Figure E47 x n 0 for n 0 and n 10 Figure E47 0 2 4 6 8 10 10 8 6 4 2 0 2 4 6 8 10 n xn rob28124ch03079117indd 116 041216 119 pm 117 Exercises without Answers A second discretetime function is defined by y n x n x n 1 a third discretetime function is defined by z n m n x m and a fourth discretetime function is defined by w n x 2n Find the following a The signal energy of x n b The signal energy of w n c The signal energy of y n d The average signal power of z n 48 A periodic discretetime signal x n is described over exactly one fundamental period by x n 5 ramp n 3 n 7 Find the average signal power of x n 49 Let two discretetime signals be defined by g 1 n 0 n 3 2n 3 n 3 0 n 3 and g 2 n 3 sin 2πn4 a Find the signal energy of g 1 n b Find the average signal power of g 2 n c Perform the transformation n 2n on g 2 n to form g 3 n and then perform the transformation n n 1 on g 3 n to form g 4 n What is the average signal power of g 4 n d Reverse the order of the two transformations of part c What is the new signal power of the new g 4 n rob28124ch03079117indd 117 041216 119 pm 118 41 INTRODUCTION AND GOALS The words signal and system were defined very generally in Chapter 1 Analysis of systems is a discipline that has been developed by engineers Engineers use mathe matical theories and tools and apply them to knowledge of the physical world to de sign things that do something useful for society The things an engineer designs are systems but as indicated in Chapter 1 the definition of a system is broader than that The term system is so broad it is difficult to define A system can be almost anything One way to define a system is as anything that performs a function Another way to define a system is as anything that responds when stimulated or excited A system can be an electrical system a mechanical system a biological system a computer system an economic system a political system and so on Systems designed by engi neers are artificial systems systems that have developed organically over a period of time through evolution and the rise of civilization are natural systems Some systems can be analyzed completely with mathematics Some systems are so complicated that mathematical analysis is extremely difficult Some systems are just not well understood because of the difficulty in measuring their characteristics In engineering the term sys tem usually refers to an artificial system that is excited by certain signals and responds with other signals Many systems were developed in earlier times by artisans who designed and im proved their systems based on their experiences and observations apparently with the use of only the simplest mathematics One of the most important distinctions between engineers and artisans is in the engineers use of higher mathematics especially calcu lus to describe and analyze systems CHAPTER GOA L S 1 To introduce nomenclature that describes important system properties 2 To illustrate the modeling of systems with differential and difference equations and block diagrams 3 To develop techniques for classifying systems according to their properties 4 C H A P T E R Description of Systems rob28124ch04118163indd 118 041216 124 pm 42 ContinuousTime Systems 119 42 CONTINUOUSTIME SYSTEMS SYSTEM MODELING One of the most important processes in signal and system analysis is the modeling of systems describing them mathematically or logically or graphically A good model is one that includes all the significant effects of a system without being so complicated that it is difficult to use Common terminology in system analysis is that if a system is excited by input signals applied at one or more inputs responses or output signals appear at one or more outputs To excite a system means to apply energy that causes it to respond One example of a system would be a boat propelled by a motor and steered by a rudder The thrust developed by the propeller the rudder position and the current of the water excite the system and the heading and speed of the boat are responses Figure 41 Notice the statement above says that the heading and speed of the boat are re sponses but it does not say that they are the responses which might imply that there are not any others Practically every system has multiple responses some significant and some insignificant In the case of the boat the heading and speed of the boat are significant but the vibration of the boat structure the sounds created by the water splashing on the sides the wake created behind the boat the rocking andor tipping of the boat and a myriad of other physical phenomena are not significant and would probably be ignored in a practical analysis of this system An automobile suspension is excited by the surface of the road as the car travels over it and the position of the chassis relative to the road is a significant response Figure 42 When we set a thermostat in a room the setting and the room temperature are input sig nals to the heating and cooling system and a response of the system is the introduction of warm or cool air to move the temperature inside the room closer to the thermostat setting Figure 41 A simplified diagram of a boat Thrust Current Rudder Figure 42 Simplified model of an automobile suspension system Automobile Chassis yt xt Spring Shock Absorber rob28124ch04118163indd 119 041216 124 pm C h a p t e r 4 Description of Systems 120 A whole class of systems measurement instruments are singleinput single output systems They are excited by the physical phenomenon being measured and the response is the instruments indication of the value of that physical phenomenon A good example is a cup anemometer The wind excites the anemometer and the angular velocity of the anemometer is the significant response Figure 43 Something that is not ordinarily thought of as a system is a highway bridge There is no obvious or deliberate input signal that produces a desired response The ideal bridge would not respond at all when excited A bridge is excited by the traffic that rolls across it the wind that blows onto it and the water currents that push on its support structure and it does move A very dramatic example showing that bridges respond when excited was the failure of the Tacoma Narrows suspension bridge in Washington state On one very windy day the bridge responded to the wind by oscillating wildly and was eventually torn apart by the forces on it This is a very dramatic example of why good analysis is important The conditions under which the bridge would respond so strongly should have been discovered in the design process so that the design could have been changed to avoid this disaster A single biological cell in a plant or animal is a system of astonishing complexity especially considering its size The human body is a system comprising a huge num ber of cells and is therefore an almost unimaginably complicated system But it can be modeled as a much simpler system in some cases to calculate an isolated effect In pharmacokinetics the human body is often modeled as a single compartment a volume containing liquid Taking a drug is an excitation and the concentration of drug in the body is the significant response Rates of infusion and excretion of the drug determine the variation of the drug concentration with time Differential Equations Below are some examples of the thinking involved in modeling systems using differen tial equations These examples were first presented in Chapter 1 ExamplE 41 Modeling a mechanical system A man 18 m tall and weighing 80 kg bungee jumps off a bridge over a river The bridge is 200 m above the water surface and the unstretched bungee cord is 30 m long The spring constant of the bungee cord is K s 11 Nm meaning that when the cord is stretched it resists the stretching with a force of 11 newtons per meter of stretch Make a mathematical model of the dynamic Wind Rotation Figure 43 Cup anemometer rob28124ch04118163indd 120 041216 124 pm 42 ContinuousTime Systems 121 position of the man as a function of time and graph the mans position versus time for the first 15 seconds When the man jumps off the bridge he goes into free fall until the bungee cord is extended to its full unstretched length This occurs when the mans feet are at 30 m below the bridge His initial velocity and position are zero using the bridge as the position reference His ac celeration is 98 ms2 until he reaches 30 m below the bridge His position is the integral of his velocity and his velocity is the integral of his acceleration So during the initial freefall time his velocity is 98t ms where t is time in seconds and his position is 49t2 m below the bridge Solving for the time of full unstretched bungeecord extension we get 247 s At that time his velocity is 2425 meters per second straight down At this point the analysis changes because the bungee cord starts having an effect There are two forces on the man 1 The downward pull of gravity mg where m is the mans mass and g is the acceleration caused by the earths gravity 2 The upward pull of the bungee cord Ksyt 30 where yt is the vertical position of the man below the bridge as a function of time Then using the principle that force equals mass times acceleration and the fact that acceleration is the second derivative of position we can write mg Ksyt 30 myʺt or myʺt Ksyt mg 30Ks This is a secondorder linear constantcoefficient inhomogeneous ordinary differential equation Its total solution is the sum of its homogeneous solution and its particular solution The homogeneous solution is a linear combination of the equations eigenfunctions The eigenfunctions are the functional forms that can satisfy this form of equation There is one eigenfunction for each eigenvalue The eigenvalues are the parameters in the eigenfunctions that make them satisfy this particular equation The eigenvalues are the solutions of the character istic equation which is mλ2 Ks 0 The solutions are λ j K s m See Web Appendix D for more on the solution of differential equations Since the eigenvalues are complex numbers it is somewhat more convenient to express the solution as a linear combination of a real sine and a real cosine instead of two complex exponentials So the homogeneous solution can be expressed in the form yht K h1 sin K s m t K h2 cos K s m t The particular solution is in the form of a linear combination of the forcing function and all its unique derivatives In this case the forcing function is a constant and all its derivatives are zero Therefore the particular solution is of the form ypt K p a constant Substituting in the form of the particular solution and solving y p t mgKs 30 The total solution is the sum of the homogeneous and particular solutions yt y h t y p t K h1 sin K s m t K h2 cos K s m t mg K s 30 K p The boundary conditions are y247 30 and ytt247 2425 Putting in numerical values for parameters applying boundary conditions and solving we get yt 1685 sin03708t 9525 cos03708t 1013 t 247 The initial variation of the mans vertical position versus time is parabolic Then at 247 s the solution becomes a sinusoid chosen to make the two solutions and the derivatives of the two solutions continuous at 247 s as is apparent in Figure 44 rob28124ch04118163indd 121 041216 124 pm C h a p t e r 4 Description of Systems 122 In Example 41 the differential equation m y t K s yt mg 30 K s describes the system This is a linear constantcoefficient inhomogeneous ordinary differential equation The right side of the equation is called its forcing function If the forcing function is zero we have a homogeneous differential equation and the solution of that equation is the homogeneous solution In signal and system analysis this solution is called the zeroinput response It is nonzero only if the initial conditions of the sys tem are nonzero meaning the system has stored energy If the system has no stored en ergy and the forcing function is not zero the response is called the zerostate response Many physical processes were ignored in the mathematical model used in Example 41 for example 1 Air resistance 2 Energy dissipation in the bungee cord 3 Horizontal components of the mans velocity 4 Rotation of the man during the fall 5 Variation of the acceleration due to gravity as a function of position 6 Variation of the water level in the river Omitting these factors kept the model mathematically simpler than it would oth erwise be System modeling is always a compromise between the accuracy and the simplicity of the model ExamplE 42 Modeling a fluidmechanical system A cylindrical water tank has crosssectional area A 1 and water level h1t and is fed by an input volumetric flow of water f1t with an orifice at height h2 whose effective crosssectional area is Figure 44 Mans vertical position versus time bridge level is zero 0 5 10 15 Time t s Elevation m Bridge Level Water Level Free Fall Bungee Stretched 200 180 160 140 120 100 80 60 40 20 0 rob28124ch04118163indd 122 041216 124 pm 42 ContinuousTime Systems 123 A 2 through which flows the output volumetric flow f 2 t Figure 45 Write a differential equa tion for the water level as a function of time and graph the water level versus time for a tank that is initially empty under different assumptions of inflow Under certain simplifying assumptions the velocity of the water flowing out of the orifice is given by Toricellis equation v 2 t 2gh1t h 2 where g is the acceleration due to earths gravity 98 ms 2 The rate of change of the volume A 1 h 1 t of water in the tank is the volumetric inflow rate minus the volumetric outflow rate d dt A 1 h1t f 1 t f 2 t and the volumetric outflow rate is the product of the effective area A 2 of the orifice and the out put flow velocity f 2 t A 2 v 2 t Combining equations we can write one differential equation for the water level A 1 d dt h 1 t A 2 2gh1t h 2 f 1 t 41 The water level in the tank is graphed in Figure 46 versus time for four constant volumetric in flows under the assumption that the tank is initially empty As the water flows in the water level increases and the increase of water level increases the water outflow The water level rises until the outflow equals the inflow and after that time the water level stays constant As first stated in Chapter 1 when the inflow is increased by a factor of two the final water level is increased by a factor of four and this fact makes the differential equation 41 nonlinear A method of finding the solution to this differential equation will be presented later in this chapter Figure 46 Water level versus time for four different volumetric inflows with the tank initially empty 0 1000 2000 3000 4000 5000 6000 7000 8000 Volumetric Inflow 0001 m3s Volumetric Inflow 0002 m3s Volumetric Inflow 0003 m3s Volumetric Inflow 0004 m3s Tank CrossSectional Area 1 m2 Orifice Area 00005 m2 Time t s Water Level h1t m 0 05 1 15 2 25 3 35 Figure 45 Tank with orifice being filled from above h1t h2 v2t f2t f1t A1 A2 rob28124ch04118163indd 123 041216 124 pm C h a p t e r 4 Description of Systems 124 Block Diagrams In system analysis it is very useful to represent systems by block diagrams A system with one input and one output would be represented as in Figure 47 The signal at the input xt is operated on by the operator to produce the signal at the output yt The operator could perform just about any operation imaginable Figure 47 A singleinput singleoutput system xt yt Figure 48 A twoinput twooutput system composed of four interconnected components 1 2 3 4 x2t x1t y2t y1t In block diagrams each input signal may go to any number of blocks and each out put signal from a block may go to any number of other blocks These signals are not af fected by being connected to any number of blocks There is no loading effect as there is in circuit analysis In an electrical analogy it would be as though the signals were voltages and the blocks all have infinite input impedance and zero output impedance In drawing block diagrams of systems some types of operations appear so often they have been assigned their own blockdiagram graphical symbols They are the amplifier the summing junction and the integrator The amplifier multiplies its input signal by a constant its gain to produce its re sponse Different symbols for amplification are used in different applications of system A system is often described and analyzed as an assembly of components A com ponent is a smaller simpler system usually one that is standard in some sense and whose properties are already known Just what is considered a component as opposed to a system depends on the situation To a circuit designer components are resistors capacitors inductors operational amplifiers and so on and systems are power amplifiers AD converters modulators filters and so forth To a communication system designer components are amplifiers modulators filters antennas and systems are microwave links fiberoptic trunk lines telephone central offices To an automobile designer com ponents are wheels engines bumpers lights seats and the system is the automobile In large complicated systems like commercial airliners telephone networks supertankers or power plants there are many levels of hierarchy of components and systems By knowing how to mathematically describe and characterize all the components in a system and how the components interact with each other an engineer can predict using mathematics how a system will work without actually building it and testing it A system made up of components is diagrammed in Figure 48 rob28124ch04118163indd 124 041216 124 pm 42 ContinuousTime Systems 125 analysis and by different authors The most common forms are shown in Figure 49 We will use Figure 49c in this text to represent an amplifier A summing junction accepts multiple input signals and responds with the sum of those signals Some of the signals may be negated before being summed so this com ponent can also produce the difference between two signals Typical graphical symbols used to represent a summing junction are illustrated in Figure 410 Figure 411 The graphical blockdiagram symbol for an integrator xt xτdτ t There are also symbols for other types of components that do special signalprocessing operations Each engineering discipline has its own preferred set of symbols for opera tions that are common in that discipline A hydraulic system diagram might have ded icated symbols for a valve a venturi a pump and a nozzle An optical system diagram might have symbols for a laser a beamsplitter a polarizer a lens and a mirror In signals and systems there are common references to two general types of sys tems openloop and closedloop An openloop system is one that simply responds directly to an input signal A closedloop system is one that responds to an input signal but also senses the output signal and feeds it back to add to or subtract from the input signal to better satisfy system requirements Any measuring instrument is an open loop system The response simply indicates what the excitation is without altering it A human driving a car is a good example of a closedloop feedback system The driver signals the car to move at a certain speed and in a certain direction by pressing the accelerator or brake and by turning the steering wheel As the car moves down a road Figure 49 Three different graphical representations of an amplifier in a system block diagram K x Kx K x Kx K x Kx a b c a b c x y xy x y xy Σ x y xy Figure 410 Three different graphical representations of a summing junction in a system block diagram We will use Figure 410c in this text to represent a summing junction If there is no plus or minus sign next to a summing junction input a plus sign is assumed An integrator when excited by any signal responds with the integral of that signal Figure 411 rob28124ch04118163indd 125 041216 124 pm C h a p t e r 4 Description of Systems 126 Figure 412 Continuoustime feedback system xt yt yt yt b a c the driver is constantly sensing the speed and position of the car relative to the road and the other cars Based on what the driver senses she modifies the input signals steering wheel accelerator andor brakes to maintain the desired direction of the car and to keep it at a safe speed and position on the road ExamplE 43 Modeling a continuoustime feedback system For the system illustrated in Figure 412 a Find its zeroinput response the response with xt 0 if the initial value of yt is y0 1 the initial rate of change of yt is y t t0 0 a 1 b 0 and c 4 b Let b 5 and find the zeroinput response for the same initial conditions as in part a c Let the system be initially at rest and let the input signal xt be a unit step Find the zerostate response for a 1 c 4 and b 1 1 5 a From the diagram we can write the differential equation for this system by realizing that the output signal from the summing junction is y t and it must equal the sum of its input signals because a 1 y t xt b y t cyt With b 0 and c 4 the response is described by the differential equation y t 4yt xt The eigenfunction is the complex exponential e st and the eigenvalues are the solutions of the characteristic equation s 2 4 0 s 12 j2 The homogeneous solution is yt K h1 e j2t K h2 e j2t Since there is no excitation this is also the total solution Applying the initial conditions y0 K h1 K h2 1 and y t t0 j2 K h1 j2 K h2 0 and solving K h1 K h2 05 The total solution is yt 05 e j2t e j2t cos2t t 0 So with b 0 the zeroinput response is a sinusoid b Now b 5 The differential equation is y t 5 y t 4yt xt the eigenvalues are s 12 14 and the solution is yt K h1 e t K h2 e 4t Applying initial conditions y0 K h1 K h2 1 and y t t0 K h1 4 K h2 0 Solving for the constants K h1 43 K h2 13 and yt 43 e t 13 e 4t t 0 This zeroinput response approaches zero for times t 0 c In this case xt is not zero and the total solution of the differential equation includes the particular solution After t 0 the input signal is a constant so the particular solution rob28124ch04118163indd 126 041216 124 pm 42 ContinuousTime Systems 127 is also a constant K p The differential equation is yt byt 4yt xt Solving for K p we get K p 025 and the total solution is yt K h1 e s 1 t K h2 e s 2 t 025 where s 12 b b 2 16 2 The response and its first derivative are both zero at t 0 Applying initial conditions and solving for the remaining two constants b s 1 s 2 K h1 K h2 1 05 j19365 05 j19365 0125 j00323 0125 j00323 1 05 j19365 05 j19365 0125 j00323 0125 j00323 5 4 1 00833 03333 The solutions are b yt 1 025 e 05t 025 cos19365t 00646 sin19365t 1 025 e 05t 025 cos19365t 00646 sin19365t 5 008333 e 4t 03333 e t 025 These zerostate responses are graphed in Figure 413 Figure 413 System responses for b 1 1 and 5 0 1 2 3 4 5 6 7 8 9 10 40 20 0 20 yt b 1 0 1 2 3 4 5 6 7 8 9 10 0 01 02 03 04 yt b 1 0 1 2 3 4 5 6 7 8 9 10 0 01 02 03 04 Time t s yt b 5 Obviously when b 1 the zerostate response grows without bound and this feedback system is unstable System dynamics are strongly influenced by feedback SYSTEM PROPERTIES Introductory Example To build an understanding of large generalized systems let us begin with examples of some simple systems that will illustrate some important system properties Circuits are rob28124ch04118163indd 127 041216 124 pm C h a p t e r 4 Description of Systems 128 R C vint voutt vint voutt it Figure 414 An RC lowpass filter a singleinput singleoutput system familiar to electrical engineers Circuits are electrical systems A very common circuit is the RC lowpass filter a singleinput singleoutput system illustrated in Figure 414 This circuit is called a lowpass filter because if the excitation is a constant amplitude sinusoid the response will be larger at low frequencies than at high frequencies So the system tends to pass low frequencies through while stop ping or blocking high frequencies Other common filter types are highpass band pass and bandstop Highpass filters pass highfrequency sinusoids and stop or block lowfrequency sinusoids Bandpass filters pass midrange frequencies and block both low and high frequencies Bandstop filters pass low and high frequencies while blocking midrange frequencies Filters will be explored in much more detail in Chapters 11 and 14 The voltage at the input of the RC lowpass filter v in t excites the system and the voltage at the output v out t is the response of the system The input voltage signal is applied to the lefthand pair of terminals and the output voltage signal appears at the righthand pair of terminals This system consists of two components familiar to elec trical engineers a resistor and a capacitor The mathematical voltagecurrent relations for resistors and capacitors are well known and are illustrated in Figure 415 Using Kirchhoffs voltage law we can write the differential equation R C v out t it v out t v in t Figure 415 Mathematical voltagecurrent relationships for a resistor and a capacitor vt vt it it vt Rit vt 1 C t iτd τ it vt R it C dvt dt rob28124ch04118163indd 128 041216 124 pm 42 ContinuousTime Systems 129 The solution of this differential equation is the sum of the homogeneous and particular solutions See Web Appendix D for more on the solution of differential equations The homogeneous solution is v outh t K h e tRC where K h is as yet unknown The particular solution depends on the functional form of v in t Let the input voltage signal v in t be a constant A volts Then since the input voltage signal is constant the par ticular solution is v outp t K p also a constant Substituting that into the differential equation and solving we get K p A and the total solution is v out t v outh t v outp t K h e tRC A The constant K h can be found by knowing the output voltage at any particular time Suppose we know the voltage across the capacitor at t 0 which is v out 0 Then v out 0 K h A K h v out 0 A and the output voltage signal can be written as v out t v out 0 e tRC A1 e tRC 42 and it is illustrated in Figure 416 This solution is written and illustrated as though it applies for all time t In prac tice that is impossible because if the solution were to apply for all time it would be unbounded as time approaches negative infinity and unbounded signals do not occur in real physical systems It is more likely in practice that the circuits initial voltage was placed on the capacitor by some means and held there until t 0 Then at t 0 the Avolt excitation was applied to the circuit and the system analysis is concerned with what happens after t 0 This solution would then apply only for that range of time and is bounded in that range of time That is v out t v out 0 e tRC A1 e tRC t 0 as illustrated in Figure 417 t voutt A RC vout0 Figure 416 RC lowpass filter response to a constant excitation t voutt A RC vout0 Figure 417 RC circuit response to an initial volt age and a constant excitation applied at time t 0 There are four determinants of the voltage response of this circuit for times t 0 the resistance R the capacitance C the initial capacitor voltage v out 0 and the applied voltage v in t The resistance and capacitance values determine the interrelationships among the voltages and currents in the system From 42 we see that if the applied voltage A is zero the response is v out t v out 0 e tRC t 0 43 and if the initial capacitor voltage v out 0 is zero the response is v out t A1 e tRC t 0 44 rob28124ch04118163indd 129 041216 124 pm C h a p t e r 4 Description of Systems 130 So the response 43 is the zeroinput response and the response 44 is the zerostate response Zerostate means no stored energy in the system and in the case of the RC lowpass filter zero state would mean the capacitor voltage is zero For this system the total response is the sum of the zeroinput and zerostate responses If the excitation is zero for all negative time then we can express it as a step of volt age v in t Aut If we assume that the circuit has been connected with this excitation between the input terminals for an infinite time since t the initial capacitor voltage at time t 0 would have to be zero Figure 418a The system would initially be in its zero state and the response would be the zerostate response Sometimes an expression like v in t Aut for the input signal is intended to represent the situation illustrated in Figure 418b In this case we are not just applying a voltage to the system we are actually changing the system by closing a switch If the initial capacitor voltage is zero in both circuits of Figure 418 the responses for times t 0 are the same R C Aut vout t R C A vout t t 0 a b Figure 418 Two ways of applying A volts to the RC lowpass filter at time t 0 R C vint voutt it iint Figure 419 RC lowpass filter with a current impulse to inject charge onto the capacitor and establish the initial capacitor voltage It is possible to include the effects of initial energy storage in a system by injecting signal energy into the system when it is in its zero state at time t 0 with a second system excitation an impulse For example in the RC lowpass filter we could put the initial voltage on the capacitor with an impulse of current from a current source in parallel with the capacitor Figure 419 When the impulse of current occurs all of its charge flows into the capacitor during the time of application of the impulse which has zero duration If the strength of the impulse is Q then the change in capacitor voltage due to the charge injected into it by the current impulse is Δ v out 1 C 0 0 i in t dt 1 C 0 0 Qδt dt Q C So choosing Q C v out 0 establishes the initial capacitor voltage as v out 0 Then the analysis of the circuit continues as though we were finding the zerostate response rob28124ch04118163indd 130 041216 124 pm 42 ContinuousTime Systems 131 to v in t and i in t instead of the zerostate response to v in t and the zeroinput response to v out 0 The total response for times t 0 is the same either way Most continuoustime systems in practice can be modeled at least approximately by differential equations in much the same way as the RC lowpass filter above was mod eled This is true of electrical mechanical chemical optical and many other kinds of sys tems So the study of signals and systems is important in a very broad array of disciplines Homogeneity If we were to double the input voltage signal of the RC lowpass filter to v in t 2 Aut the factor 2 A would carry through the analysis and the zerostate response would double to v out t 2 A1 e tRC ut Also if we were to double the initial capacitor voltage the zeroinput response would double In fact if we multiply the input voltage signal by any constant the zerostate response is also multiplied by the same constant The quality of this system that makes these statements true is called homogeneity In a homogeneous system multiplying the input signal by any constant including complex constants multiplies the zerostate response by the same constant Figure 420 illustrates in a blockdiagram sense what homogeneity means A very simple example of a system that is not homogeneous is a system character ized by the relationship yt 1 xt If x is 1 y is 2 and if x is 2 y is 3 The input signal was doubled but the output signal was not doubled What makes this system inhomogeneous is the presence of the constant 1 on the left side of the equation This system has a nonzero zeroinput response Notice that if we were to add 1 to both sides of the equation and redefine the input signal to be x new t xt 1 instead of just xt we would have yt x new t and doubling x new t would double yt The system would then be homogeneous under this new definition of the input signal Figure 420 Block diagram illustrating the concept of homogeneity for a system initially in its zero state K is any complex constant xt yt Homogeneous System xt Kyt Kxt K Multiplier ExamplE 44 Determining whether a system is homogeneous Test for homogeneity the system whose inputoutput relationship is yt expxt Let x 1 t gt Then y 1 t expgt Let x 2 t Kgt Then y 2 t expKgt expgt K K y 1 t Therefore this system is not homogeneous rob28124ch04118163indd 131 041216 124 pm C h a p t e r 4 Description of Systems 132 The analysis in Example 44 may seem like an unnecessarily formal proof for such a simple function But it is very easy to get confused in evaluating some systems even simplelooking ones unless one uses this kind of structured proof Time Invariance Suppose the system of Figure 414 were initially in its zerostate and the excitation were delayed by t 0 changing the input signal to xt Aut t 0 What would happen to the response Going through the solution process again we would find that the zerostate response is v out t A 1 e t t 0 RC ut t 0 which is exactly the original zerostate response except with t replaced by t t 0 Delaying the excitation delayed the zerostate response by the same amount without changing its functional form The quality that makes this happen is called time invariance If a system is initially in its zero state and an arbitrary input signal xt causes a response yt and an input signal xt t 0 causes a response yt t 0 for any arbitrary t 0 the system is said to be time invariant Figure 421 illustrates the concept of time invariance Figure 421 Block diagram illustrating the concept of time invariance for a system initially in its zero state yt xt TimeInvariant System xt ytt0 Delay t0 xtt0 ExamplE 46 Determining whether a system is time invariant Test for time invariance the system whose inputoutput relationship is yt xt2 Let x 1 t gt Then y 1 t gt2 Let x 2 t gt t 0 Then y 2 t gt2 t 0 y 1 t t 0 g t t 0 2 Therefore this system is not time invariant it is time variant ExamplE 45 Determining whether a system is time invariant Test for time invariance the system whose inputoutput relationship is yt expxt Let x 1 t gt Then y 1 t expgt Let x 2 t gt t 0 Then y 2 t expgt t 0 y 1 t t 0 Therefore this system is time invariant rob28124ch04118163indd 132 041216 124 pm 42 ContinuousTime Systems 133 Additivity Let the input voltage signal to the RC lowpass filter be the sum of two voltages v in t v in1 t v in2 t For a moment let v in2 t 0 and let the zerostate response for v in1 t acting alone be v out1 t The differential equation for that situation is RC v out1 t v out1 t v in1 t 45 where since we are finding the zerostate response v out1 0 0 Equation 45 and the initial condition v out1 0 0 uniquely determine the solution v out1 t Similarly if v in2 t acts alone its zerostate response obeys RC v out2 t v out2 t v in2 t 46 and v out2 t is similarly uniquely determined Adding 45 and 46 RC v out1 t v out2 t v out1 t v out2 t v in1 t v in2 t 47 The sum v in1 t v in2 t occupies the same position in 47 as v in1 t does in 45 and v out1 t v out2 t and v out1 t v out2 t occupy the same positions in 47 that v out1 t and v out1 t do in 45 Also for the zerostate response v in1 0 v in2 0 0 There fore if v in1 t produces v out1 t then v in1 t v in2 t must produce v out1 t v out2 t because both responses are uniquely determined by the same differential equation and the same initial condition This result depends on the fact that the derivative of a sum of two functions equals the sum of the derivatives of those two functions If the exci tation is the sum of two excitations the solution of this differential equation but not necessarily other differential equations is the sum of the responses to those excitations acting alone A system in which added excitations produce added zerostate responses is called additive Figure 422 If a system when excited by an arbitrary x 1 produces a zerostate response y 1 and when excited by an arbitrary x 2 produces a zerostate response y 2 and x 1 x 2 always produces the zerostate response y 1 y 2 the system is additive A very common example of a nonadditive system is a simple diode circuit Figure 423 Let the input voltage signal of the circuit V be the series connection of two constantvoltage sources V 1 and V 2 making the overall input voltage signal the sum of the two individual input voltage signals Let the overall response be the Figure 422 Block diagram illustrating the concept of additivity for a system initially in its zero state Additive System x1t x1t y1t x2t y2t x2t x1t x2t y1t y2t Summing Junction I R V1 V2 Figure 423 A DC diode circuit rob28124ch04118163indd 133 041216 124 pm C h a p t e r 4 Description of Systems 134 current I and let the individual current responses to the individual voltage sources act ing alone be I 1 and I 2 To make the result obvious let V 1 0 and let V 2 V 1 The response to V 1 acting alone is a positive current I 1 The response to V 2 acting alone is an extremely small ideally zero negative current I 2 The response I to the combined input signal V 1 V 2 is zero but the sum of the individual responses I 1 I 2 is approx imately I 1 not zero So this is not an additive system Linearity and Superposition Any system that is both homogeneous and additive is called a linear system If a linear system when excited by x 1 t produces a zerostate response y 1 t and when excited by x 2 t produces a zerostate response y 2 t then xt α x 1 t β x 2 t will produce the zerostate response yt α y 1 t β y 2 t This property of linear systems leads to an important concept called superpo sition The term superposition comes from the verb superpose The pose part of superpose means to put something into a certain position and the super part means on top of Together superpose means to place something on top of something else That is what is done when we add one input signal to another and in a linear system the overall response is one of the responses on top of added to the other The fact that superposition applies to linear systems may seem trivial and obvious but it has farreaching implications in system analysis It means that the zerostate response to any arbitrary input signal can be found by breaking the input signal down into simple pieces that add up to the original input signal finding the response to each simple piece and then adding all those responses to find the overall response to the overall input signal It also means that we can find the zerostate response and then in an independent calcu lation find the zeroinput response and then add them to find the total response This is a divideandconquer approach to solving linearsystem problems and its importance cannot be overstated Instead of solving one large complicated problem we solve multiple small simple problems And after we have solved one of the small simple problems the others are usually very easy to solve because the process is similar Linearity and superpo sition are the basis for a large and powerful set of techniques for system analysis Analysis of nonlinear systems is much more difficult than analysis of linear systems because the divideandconquer strategy usually does not work on nonlinear systems Often the only practical way to analyze a nonlinear system is with numerical as opposed to analytical methods Superposition and linearity also apply to multipleinput multipleoutput linear systems If a linear system has two inputs and we apply x1t at the first input and x 2 t at the second input and get a response yt we would get the same yt if we added the response to the first input signal acting alone y 1 t and the response to the second input signal acting alone y 2 t LTI Systems By far the most common type of system analyzed in practical system design and anal ysis is the linear timeinvariant system If a system is both linear and time invariant it is called an LTI system Analysis of LTI systems forms the overwhelming majority of the material in this text One implication of linearity that will be important later can now be proven Let an LTI system be excited by a signal x 1 t and produce a zerostate response y 1 t Also let x 2 t produce a zerostate response y 2 t Then invoking linearity α x 1 t β x 2 t will produce the zerostate response α y 1 t β y 2 t The constants α and β can be any rob28124ch04118163indd 134 041216 124 pm 42 ContinuousTime Systems 135 numbers including complex numbers Let α 1 and β j Then x 1 t j x 2 t produces the response y 1 t j y 2 t We already know that x 1 t produces y 1 t and that x 2 t produces y 2 t So we can now state the general principle When a complex excitation produces a response in an LTI system the real part of the excitation produces the real part of the response and the imaginary part of the excitation produces the imaginary part of the response This means that instead of applying a real excitation to a system to find its real response we can apply a complex excitation whose real part is the actual physical excitation find the complex response and then take its real part as the actual physical response to the actual physical excitation This is a roundabout way of solving system problems but because the eigenfunctions of real systems are complex exponentials and because of the compact notation that results when applying them in system analysis this is often a more efficient method of analysis than the direct approach This basic idea is one of the principles underlying transform methods and their applications to be presented in Chapters 6 through 9 ExamplE 47 Response of an RC lowpass filter to a square wave using superposition Use the principle of superposition to find the response of an RC lowpass filter to a square wave that is turned on at time t 0 Let the RC time constant be 1 ms let the time from one rising edge of the square wave to the next be 2 ms and let the amplitude of the square wave be 1 V Figure 424 1 2 xt t ms Figure 424 Square wave that excites an RC lowpass filter We have no formula for the response of the RC lowpass filter to a square wave but we do know how it responds to a unit step A square wave can be represented by the sum of some pos itive and negative timeshifted unit steps So xt can be expressed analytically as xt x 0 t x 1 t x 2 t x 3 t xt ut ut 0001 ut 0002 ut 0003 The RC lowpass filter is a linear timeinvariant system Therefore the response of the filter is the sum of the responses to the individual unit steps The response to one unshifted positive unit step is y 0 t 1 e 1000t ut Invoking time invariance y 1 t 1 e 1000t0001 ut 0001 y 2 t 1 e 1000t0002 ut 0002 y 3 t 1 e 1000t0003 ut 0003 rob28124ch04118163indd 135 041216 124 pm C h a p t e r 4 Description of Systems 136 Then invoking linearity and superposition yt y 0 t y 1 t y 2 t y3t yt 1 e 1000t ut 1 e 1000t0001 ut 0001 1 e 1000t0002 ut 0002 1 e 1000t0003 ut 0003 see Figure 426 Superposition is the basis of a powerful technique for finding the response of a linear system The salient characteristic of equations that describe linear systems is that the dependent variable and its integrals and derivatives appear only to the first power To illustrate this rule consider a system in which the excitation and response are related by the differential equation a y t b y 2 t xt where xt is the excitation and yt is the response If xt were changed to x new t x 1 t x 2 t the differential equation would be a y new t b y new 2 t x new t The differential equations for x 1 t and x 2 t acting alone would be a y 1 t b y 1 2 t x 1 t and a y 2 t b y 2 2 t x 2 t Figure 425 Unit steps that can be added to form a square wave 1 2 x0t t ms 1 2 x1t x2t x3t t ms 1 2 t ms 1 2 t ms t ms 8 yt 1 1 y0t y1t y3t y5t y7t y2t y4t y6t Figure 426 Response to the square wave rob28124ch04118163indd 136 041216 124 pm 42 ContinuousTime Systems 137 The sum of these two equations is a y 1 t y 2 t b y 1 2 t y 2 2 t x 1 t x 2 t x new t which is in general not equal to a y 1 t y 2 t ʺ b y 1 t y 2 t 2 x 1 t x 2 t x new t The difference is caused by the y 2 t term that is not consistent with a differential equation that describes a linear system Therefore in this system superposition does not apply A very common analysis technique in signal and system analysis is to use the methods of linear systems to analyze nonlinear systems This process is called lin earizing the system Of course the analysis is not exact because the system is not actually linear and the linearization process does not make it linear Rather linear ization replaces the exact nonlinear equations of the system by approximate lin ear equations Many nonlinear systems can be usefully analyzed by linearsystem methods if the input and output signals are small enough As an example consider a pendulum Figure 427 Assume that the mass is supported by a massless rigid rod of length L If a force xt is applied to the mass m it responds by moving The vector sum of the forces acting on the mass tangential to the direction of motion is equal to the product of the mass and the acceleration in that same direction That is xt m g sinθt mL θ t or m L θ t m g sinθt xt 48 where m is the mass at the end of the pendulum xt is a force applied to the mass tan gential to the direction of motion L is the length of the pendulum g is the acceleration due to gravity and θt is the angular position of the pendulum This system is excited by xt and responds with θt Equation 48 is nonlinear But if θt is small enough sinθt can be closely approximated by θt In that approximation m L θ t mgθt xt 49 and this is a linear equation So for small perturbations from the rest position this system can be usefully analyzed by using 49 xt θt mg sinθt L Mass Figure 427 A pendulum rob28124ch04118163indd 137 041216 124 pm C h a p t e r 4 Description of Systems 138 Stability In the RClowpassfilter example the input signal a step of voltage was bounded meaning its absolute value is less than some finite upper bound B for all time xt B for all t The response of the RC lowpass filter to this bounded input signal was a bounded output signal Any system for which the zerostate response to any arbitrary bounded excitation is also bounded is called a boundedinputboundedoutput BIBO stable system1 The most common type of system studied in signals and systems is a system whose inputoutput relationship is determined by a linear constantcoefficient ordinary differ ential equation The eigenfunction for differential equations of this type is the complex exponential So the homogeneous solution is in the form of a linear combination of com plex exponentials The behavior of each of those complex exponentials is determined by its associated eigenvalue The form of each complex exponential is e st e σt e jωt where s σ jω is the eigenvalue σ is its real part and ω is its imaginary part The fac tor e jωt has a magnitude of one for all t The factor e σt has a magnitude that gets smaller as time proceeds in the positive direction if σ is negative and gets larger if σ is positive If σ is zero the factor e σt is simply the constant one If the exponential is growing as time passes the system is unstable because a finite upper bound cannot be placed on the response If σ 0 it is possible to find a bounded input signal that makes the output signal increase without bound An input signal that is of the same functional form as the homogeneous solution of the differential equation which is bounded if the real part of the eigenvalue is zero will produce an unbounded response see Example 48 For a continuoustime LTI system described by a differential equation if the real part of any of the eigenvalues is greater than or equal to zero nonnegative the system is BIBO unstable ExamplE 48 Finding a bounded excitation that produces an unbounded response Consider an integrator for which yt t xτdτ Find the eigenvalues of the solution of this equation and find a bounded excitation that will produce an unbounded response By applying Leibnizs formula for the derivative of an integral of this type we can differentiate both sides and form the differential equation y t xt This is a very simple differential equation with one eigenvalue and the homogeneous solution is a constant because the eigenvalue is zero Therefore this system should be BIBO unstable A bounded excitation that has the same functional form as the homogeneous solution produces an unbounded re sponse In this case a constant excitation produces an unbounded response Since the response 1 The discussion of BIBO stability brings up an interesting point Is any practical system ever actually unstable by the BIBO criterion Since no practical system can ever produce an unbounded response strictly speaking all practical systems are stable The ordinary operational meaning of BIBO instability is a system described approximately by linear equations that would develop an unbounded response to a bounded excitation if the system remained linear Any practical system will become nonlinear when its response reaches some large magnitude and can never produce a truly unbounded response So a nuclear weapon is a BIBOunstable system in the ordinary sense but a BIBOstable system in the strict sense Its energy release is not unbounded even though it is extremely large compared to most other artificial systems on earth rob28124ch04118163indd 138 041216 124 pm 42 ContinuousTime Systems 139 is the integral of the excitation it should be clear that as time t passes the magnitude of the response to a constant excitation grows linearly without a finite upper bound Causality In the analysis of the systems we have considered so far we observe that each system responds only during or after the time it is excited This should seem obvious and natu ral How could a system respond before it is excited It seems obvious because we live in a physical world in which real physical systems always respond while or after they are excited But as we shall later discover in considering ideal filters in Chapter 11 some system design approaches may lead to a system that responds before it is excited Such a system cannot actually be built The fact that a real system response occurs only while or after it is excited is a re sult of the commonsense idea of cause and effect An effect has a cause and the effect occurs during or after the application of the cause Any system for which the zerostate response occurs only during or after the time in which it is excited is called a causal system All physical systems are causal because they are unable to look into the future and respond before being excited The term causal is also commonly albeit somewhat inappropriately applied to signals A causal signal is one that is zero before time t 0 This terminology comes from the fact that if an input signal that is zero before time t 0 is applied to a causal system the response is also zero before time t 0 By this definition the response would be a causal signal because it is the response of a causal system to a causal ex citation The term anticausal is sometimes used to describe signals that are zero after time t 0 In signal and system analysis we often find what is commonly referred to as the forced response of a system A very common case is one in which the input signal is periodic A periodic signal has no identifiable starting point because if a signal xt is periodic that means that xt xt nT where T is a period and n is any integer No matter how far back in time we look the signal repeats periodically So the relationship between a periodic input signal and the forced response of an LTI system which is also periodic with the same period cannot be used to determine whether a system is causal Therefore in analyzing a system for causality the system should be excited by a test signal that has an identifiable time before which it has al ways been zero A simple signal to test an LTI system for causality would be the unit impulse δt It is zero before t 0 and is zero after t 0 If the zerostate response of the system to a unit impulse occurring at t 0 is not zero before t 0 the system is not causal Chapter 5 introduces methods of determining how LTI systems respond to impulses Memory The responses of the systems we have considered so far depend on the present and past excitations In the RC lowpass filter the charge on the capacitor is determined by the current that has flowed through it in the past By this mechanism it in a sense remembers something about its past The present response of this system depends on its past excitations and that memory along with its present excitation determines its present response rob28124ch04118163indd 139 041216 124 pm C h a p t e r 4 Description of Systems 140 If any systems zerostate response at any arbitrary time depends on its exci tation at any other time the system has memory and is a dynamic system There are systems for which the present value of the response depends only on the present value of the excitation A resistive voltage divider is a good example Figure 428 If any systems response at an arbitrary time depends only on the excitation at that same time the system has no memory and is a static system The concepts of causality and memory are related All static systems are causal Also the testing for memory can be done with the same kind of test signal used to test for causality the unit impulse If the response of an LTI system to the unit impulse δt is nonzero at any time other than t 0 the system has memory Static Nonlinearity We have already seen one example of a nonlinear system one with a nonzero zeroinput response It is nonlinear because it is not homogeneous The nonlinearity is not an in trinsic result of nonlinearity of the components themselves but rather a result of the fact that the zeroinput response of the system is not zero The more common meaning of the term nonlinear system in practice is a system in which even with a zeroinput response of zero the output signal is still a nonlin ear function of the input signal This is often the result of components in the system that have static nonlinearities A statically nonlinear system is one without memory and for which the inputoutput relationship is a nonlinear function Examples of stati cally nonlinear components include diodes transistors and squarelaw detectors These components are nonlinear because if the input signal is changed by some factor the output signal can change by a different factor The difference between linear and nonlinear components of this type can be illus trated by graphing the relationship between the input and output signals For a linear resistor which is a static system the relation is determined by Ohms law vt Rit A graph of voltage versus current is linear Figure 429 Figure 428 A resistive voltage divider vit vot R1 R2 vo t R2 R1 R2 vi t rob28124ch04118163indd 140 041216 124 pm 42 ContinuousTime Systems 141 A diode is a good example of a statically nonlinear component Its voltagecurrent relationship is it I s e qv t kT 1 where I s is the reverse saturation current q is the charge on an electron k is Boltzmanns constant and T is the absolute temperature as illustrated in Figure 430 Another example of a statically nonlinear component is an analog multiplier used as a squarer An analog multiplier has two inputs and one output and the output signal is the product of the signals applied at the two inputs It is memoryless or static be cause the present output signal depends only on the present input signals Figure 431 vt vt it it Slope 1 R Resistor R R Figure 429 Voltagecurrent relationship for a resistor Diode vt it vt it Figure 430 Voltagecurrent relationship for a diode at a fixed temperature Analog Multiplier x1t yt x1tx2t x2t Squarer xt yt x2t Figure 431 An analog multiplier and a squarer The output signal yt is the product of the input signals x 1 t and x 2 t If x 1 t and x 2 t are the same signal xt then yt x 2 t This is a statically nonlinear relationship because if the excitation is multiplied by some factor A the response is multiplied by the factor A 2 making the system inhomogeneous A very common example of a static nonlinearity is the phenomenon of saturation in real as opposed to ideal operational amplifiers An operational amplifier has two inputs the inverting input and the noninverting input and one output When input rob28124ch04118163indd 141 041216 124 pm C h a p t e r 4 Description of Systems 142 voltage signals are applied to the inputs the output voltage signal of the operational amplifier is a fixed multiple of the difference between the two input voltage signals up to a point For small differences the relationship is v out t A v in t v in t But the output voltage signal is constrained by the power supply voltages and can only ap proach those voltages not exceed them Therefore if the difference between the input voltage signals is large enough that the output voltage signal calculated from v out t A v in t v in t would cause it to be outside the range V ps to V ps where ps means power supply the operational amplifier will saturate The output voltage signal will go that far and no farther When the operational amplifier is saturated the rela tionship between the input and output signals becomes statically nonlinear That is illustrated in Figure 432 Even if a system is statically nonlinear linear system analysis techniques may still be useful in analyzing it See Web Appendix C for an example of using linear system analysis to approximately analyze a nonlinear system Invertibility In the analysis of systems we usually find the zerostate response of the system given an excitation But we can often find the excitation given the zerostate response if the system is invertible A system is said to be invertible if unique excitations produce unique zerostate responses If unique excitations produce unique zerostate responses then it is possible in prin ciple at least given the zerostate response to associate it with the excitation that produced it Many practical systems are invertible Another way of describing an invertible system is to say that if a system is invert ible there exists an inverse system which when excited by the response of the first system responds with the excitation of the first system Figure 433 An example of an invertible system is any system described by a linear timeinvariant constantcoefficient differential equation of the form a k y k t a k1 y k1 t a 1 y t a 0 yt xt vintvint voutt Vps Vps Slope A Figure 432 Inputoutput signal relationship for a saturating operational amplifier rob28124ch04118163indd 142 041216 124 pm 42 ContinuousTime Systems 143 If the response yt is known then so are all its derivatives The equation indicates exactly how to calculate the excitation as a linear combination of yt and its derivatives An example of a system that is not invertible is a static system whose inputoutput functional relationship is yt sinxt 410 For any xt it is possible to determine the zerostate response yt Knowledge of the excitation uniquely determines the zerostate response However if we attempt to find the excitation given the response by rearranging the functional relationship 410 into xt sin 1 yt we encounter a problem The inverse sine function is multiplevalued Therefore knowledge of the zerostate response does not uniquely determine the exci tation This system violates the principle of invertibility because different excitations can produce the same zerostate response If at t t 0 x t 0 π4 then y t 0 2 2 But if at t t 0 x t 0 3π4 then y t 0 would have the same value 2 2 Therefore by observ ing only the zerostate response we would have no idea which excitation value caused it Another example of a system that is not invertible is one that is very familiar to electronic circuit designers the fullwave rectifier Figure 434 Assume that the transformer is an ideal 12turnsratio transformer and that the diodes are ideal so that there is no voltage drop across them in forward bias and no current through them in reverse bias Then the output voltage signal v o t and input voltage signal v i t are related by v o t v i t Suppose that at some particular time the output voltage signal is 1 V The input voltage signal at that time could be 1 V or 1 V There is no way of knowing which of these two input voltage signals is the excitation just by observing the output voltage signal Therefore we could not be assured of correctly reconstructing the excitation from the response This system is not invertible 1 x y x Figure 433 A system followed by its inverse Figure 434 A fullwave rectifier vit vot R DYNAMICS OF SECONDORDER SYSTEMS Firstorder and secondorder systems are the most common types of systems encoun tered in system design and analysis Firstorder systems are described by firstorder differential equations and secondorder systems are described by secondorder differ ential equations We have seen examples of firstorder systems As an example of a secondorder system consider the RLC circuit excited by a step in Figure 435 rob28124ch04118163indd 143 041216 124 pm C h a p t e r 4 Description of Systems 144 The sum of voltages around the loop yields LC v out t RC v out t v out t Aut 411 and the solution for the output voltage signal is v out t K 1 e R2L R2L 2 1LC t K 2 e R2L R2L 2 1LC t A and K 1 and K 2 are arbitrary constants This solution is more complicated than the solution for the RC lowpass filter was There are two exponential terms each of which has a much more complicated expo nent The exponent involves a square root of a quantity that could be negative There fore the exponent could be complexvalued For this reason the eigenfunction e st is called a complex exponential The solutions of ordinary linear differential equations with constant coefficients are always linear combinations of complex exponentials In the RLC circuit if the exponents are real the response is simply the sum of two real exponentials The more interesting case is complex exponents The exponents are complex if R2L 2 1LC 0 412 In this case the solution can be written in terms of two standard parameters of secondorder systems the natural radian frequency ω n and the damping factor α as v out t K 1 e α α 2 ω n 2 t K 2 e α α 2 ω n 2 t A 413 where ω n 2 1LC and α R2L There are two other widely used parameters of secondorder systems which are related to ω n and α the critical radian frequency ω c and the damping ratio ζ They are de fined by ζ α ω n and ω c ω n 1 ζ 2 Then we can write as v out t K 1 e α ω n ζ 2 1 t K 2 e α ω n ζ 2 1 t A When condition 412 is satisfied the system is said to be underdamped and the re sponse can be written as v out t K 1 e αj ω c t K 2 e αj ω c t A The exponents are complex conjugates of each other as they must be for v out t to be a realvalued function Assuming the circuit is initially in its zero state and applying initial conditions the output voltage signal is v out t A 1 2 1 j α ω c e αj ω c t 1 2 1 j α ω c e αj ω c t 1 R L C vint voutt Figure 435 An RLC circuit rob28124ch04118163indd 144 041216 124 pm 43 DiscreteTime Systems 145 This response appears to be a complex response of a real system with real excitation But even though the coefficients and exponents are complex the overall solution is real because using trigonometric identities the output voltage signal can be reduced to v out t A1 e αt α ω c sin ω c t cos ω c t This solution is in the form of a damped sinusoid a sinusoid multiplied by a decaying exponential The natural frequency f n ω n 2π is the frequency at which the response voltage would oscillate if the damping factor were zero The rate at which the sinu soid is damped is determined by the damping factor α Any system described by a secondorder linear differential equation could be analyzed by an analogous procedure COMPLEX SINUSOID EXCITATION An important special case of linear system analysis is an LTI system excited by a com plex sinusoid Let the input voltage signal of the RLC circuit be v in t A e j2π f 0 t It is important to realize that v in t is described exactly for all time Not only is it going to be a complex sinusoid from now on it has always been a complex sinusoid Since it began an infinite time in the past any transients that may have occurred have long since died away if the system is stable as this RLC circuit is Thus the only solution that is left at this time is the forced response The forced response is the particular solution of the describing differential equation Since all the derivatives of the complex sinusoid are also complex sinusoids the particular solution of v in t A e j2π f 0 t is simply v outp t B e j2π f 0 t where B is yet to be determined So if this LTI system is excited by a complex sinusoid the response is also a complex sinusoid at the same frequency but with a different multiplying constant in general Any LTI system excited by a complex exponential responds with a complex exponential of the same functional form except multiplied by a complex constant The forced solution can be found by the method of undetermined coefficients Substituting the form of the solution into the differential equation 411 j2π f 0 2 LCB e j2π f 0 t j2π f 0 RCB e j2π f 0 t B e j2π f 0 t A e j2π f 0 t and solving B A j2π f 0 2 LC j2π f 0 RC 1 Using the principle of superposition for LTI systems if the input signal is an ar bitrary function that is a linear combination of complex sinusoids of various frequen cies then the output signal is also a linear combination of complex sinusoids at those same frequencies This idea is the basis for the methods of Fourier series and Fourier transform analysis that will be introduced in Chapters 6 and 7 which express arbitrary signals as linear combinations of complex sinusoids 43 DISCRETETIME SYSTEMS SYSTEM MODELING Block Diagrams Just as in continuoustime systems in drawing block diagrams of discretetime sys tems there are some operations that appear so often they have been assigned their own blockdiagram graphical symbols The three essential components in a discretetime sys tem are the amplifier the summing junction and the delay The amplifier and summing rob28124ch04118163indd 145 041216 124 pm C h a p t e r 4 Description of Systems 146 junction serve the same purposes in discretetime systems as in continuoustime sys tems A delay is excited by a discretetime signal and responds with that same signal except delayed by one unit in discrete time see Figure 436 This is the most commonly used symbol but sometimes the D is replaced by an S for shift xn xn1 D Figure 436 The graphical blockdiagram symbol for a discretetime delay h1t h2 v2t f2t f1t A1 A2 Figure 437 Tank with orifice being filled from above Difference Equations Below are some examples of the thinking involved in modeling discretetime systems Three of these examples were first presented in Chapter 1 ExamplE 49 Approximate modeling of a continuoustime system using a discretetime system One use of discretetime systems is in the approximate modeling of nonlinear continuoustime systems like the fluidmechanical system in Figure 437 The fact that its differential equation A 1 d dt h 1 t A 2 2g h 1 t h 2 f 1 t Toricellis equation is nonlinear makes it harder to solve than linear differential equations One approach to finding a solution is to use a numerical method We can approximate the derivative by a finite difference d dt h 1 t h 1 n 1 T s h 1 n T s T s where T s is a finite time duration between values of h 1 at uniformly separated points in time and n indexes those points Then Toricellis equation can be approximated at those points in time by A 1 h 1 n 1 T s h 1 n T s T s A 2 2g h 1 n T s h 2 f 1 n T s rob28124ch04118163indd 146 041216 124 pm 43 DiscreteTime Systems 147 which can be rearranged into h 1 n 1 T s 1 A 1 T s f 1 n T s A 1 h 1 n T s A 2 T s 2g h 1 n T s h 2 414 which expresses the value of h 1 at the next time index n 1 in terms of the values of f 1 at the present time index n and h 1 also at the present time index We could write 414 in the simpli fied discretetime notation as h 1 n 1 1 A 1 T s f 1 n A 1 h 1 n A 2 T s 2g h 1 n h 2 or replacing n by n 1 h 1 n 1 A 1 T s f 1 n 1 A 1 h 1 n 1 A 2 T s 2g h 1 n 1 h 2 415 In 415 knowing the value of h 1 at any n we can approximately find its value at any other n The approximation is made better by making T s smaller This is an example of solving a continuoustime problem using discretetime methods Because 415 is a difference equation it defines a discretetime system Figure 438 f1n h1n Ts A1 2g h2 A1 A2Ts D D Figure 438 A system that approximately solves numerically the differential equation of fluid flow Figure 439 shows examples of the numerical solution of Toricellis equation using the discretetime system of Figure 438 for three different sampling times 100 s 500 s and 1000 s The result for T s 100 is quite accurate The result for T s 500 has the right general behavior and approaches the right final value but arrives at the final value too early The result for T s 1000 has a completely wrong shape although it does approach the correct final value The choice of a sampling time that is too large makes the solution inaccurate and in some cases can actually make a numerical algorithm unstable Below is the MATLAB code that simulates the system in Figure 438 used to solve the differential equation describing the tank with orifice g 98 Acceleration due to gravity ms2 A1 1 Area of free surface of water in tank m2 A2 00005 Effective area of orifice m2 h1 0 Height of free surface of water in tank m2 rob28124ch04118163indd 147 041216 124 pm C h a p t e r 4 Description of Systems 148 h2 0 Height of orifice m2 f1 0004 Water volumetric inflow m3s Ts 1005001000 Vector of time increments s N round8000Ts Vector of numbers of time steps for m 1lengthTs Go through the time increments h1 0 Initialize h1 to zero h h1 First entry in waterheight vector Go through the number of time increments computing the water height using the discretetime system approximation to the actual continuoustime system for n 1Nm Compute next freesurface water height h1 Tsmf1 A1h1 A2Tsmsqrt2gh1h2A1 h h h1 Append to waterheight vector end Graph the freesurface water height versus time and annotate graph subplotlengthTs1m p stemTsm0Nmhkfilled setpLineWidth2MarkerSize4 grid on if m lengthTs 0 1000 2000 3000 4000 5000 6000 7000 8000 0 1 2 3 4 h1t m h1t m h1t m Ts 100 s Ts 500 s Ts 1000 s 0 1000 2000 3000 4000 5000 6000 7000 8000 0 1 2 3 4 0 1000 2000 3000 4000 5000 6000 7000 8000 0 1 2 3 4 Time t or nTs s Figure 439 Numerical solution of Toricellis equation using the discretetime system of Figure 438 for a volumetric inflow rate of 0004 m 3 s rob28124ch04118163indd 148 041216 124 pm 43 DiscreteTime Systems 149 p xlabelTime t or itnTs s FontNameTimesFontSize18 end p ylabelh1t mFontNameTimesFontSize18 p titleitTs num2strTsm sFontNameTimesFontSize18 end ExamplE 410 Modeling a feedback system without excitation Find the output signal generated by the system illustrated in Figure 440 for times n 0 Assume the initial conditions are y0 1 and y1 0 Figure 440 A discretetime system yn yn2 yn1 197 D D The system in Figure 440 is described by the difference equation yn 197yn 1 yn 2 416 This equation along with initial conditions y0 1 and y1 0 completely determines the response yn which is the zeroinput response The zeroinput response can be found by iterat ing on 416 This yields a correct solution but it is in the form of an infinite sequence of values of the response The zeroinput response can be found in closed form by solving the difference equation see Web Appendix D Since there is no input signal exciting the system the equation is homogeneous The functional form of the homogeneous solution is the complex exponential K z n Substituting that into the difference equation we get K z n 197K z n1 K z n2 Dividing through by K z n2 we get the characteristic equation and solving it for z we get z 197 1 97 2 4 2 0985 j01726 e j01734 The fact that there are two eigenvalues means that the homogeneous solution is in the form yn K h1 z 1 n K h2 z 2 n 417 rob28124ch04118163indd 149 041216 124 pm C h a p t e r 4 Description of Systems 150 We have initial conditions y0 1 and y1 0 and we know from 417 that y0 K h1 K h2 and y1 K h1 z 1 1 K h2 z 2 1 Therefore 1 1 e j01734 e j01734 K h1 K h2 1 0 Solving for the two constants K h1 05 j2853 and K h2 05 j2853 So the complete solution is yn 05 j2853 0985 j01726 n 05 j2853 0985 j01726 n This is a correct solution but it is not in a very convenient form We can rewrite it in the form yn 05 j2853 e j01734n 05 j2853 e j01734n or yn 05 e j01734n e j01734n 2 cos 01734n j2853 e j01734n e j01734n j2 sin 01734n or yn cos01734n 5706 sin01734n The first 50 values of the signal produced by this system are illustrated in Figure 441 ExamplE 411 Modeling a simple feedback system with excitation Find the response of the system in Figure 442 if a 1 b 15 xn δn and the system is initially at rest The difference equation for this system is yn a xn byn 1 xn 15yn 1 Figure 441 Signal produced by the discretetime system in Figure 440 n 50 yn 6 6 rob28124ch04118163indd 150 041216 124 pm 43 DiscreteTime Systems 151 The solution for times n 0 is the homogeneous solution of the form K h z n Substituting and solving for z we get z 15 Therefore yn K h 15 n n 0 The constant can be found by knowing the initial value of the response which from the system diagram must be 1 Therefore y0 1 K h 15 0 K h 1 and yn 15 n n 0 This solution obviously grows without bound so the system is unstable If we chose b with a magnitude less than one the system would be stable because the solution is of the form yn b n n 0 ExamplE 412 Modeling a more complicated feedback system with excitation Find the zerostate response of the system in Figure 443 for times n 0 to xn 1 applied at time n 0 by assuming all the signals in the system are zero before time n 0 for a 1 b 15 and three different values of c 08 06 and 05 The difference equation for this system is yn axn byn 1 cyn 2 xn 15yn 1 cyn 2 418 Figure 442 A simple discretetime feedback system with a nonzero excitation xn yn b a D xn yn b a c D D Figure 443 A system with more complicated feedback The response is the total solution of the difference equation with initial conditions We can find a closedform solution by finding the total solution of the difference equation The homogeneous solution is y h n K h1 z 1 n K h2 z 2 n where z 12 075 05625 c The particular solution rob28124ch04118163indd 151 041216 124 pm C h a p t e r 4 Description of Systems 152 is in the form of a linear combination of the input signal and all its unique differences The input signal is a constant So all its differences are zero Therefore the particular solution is simply a constant K p Substituting into the difference equation K p 15 K p c K p 1 K p 1 c 05 Using 418 we can find the initial two values of yn needed to solve for the remaining two unknown constants K h1 and K h2 They are y0 1 and y1 25 In Chapter 1 three responses were illustrated for a 1 b 15 and c 08 06 and 05 Those responses are replicated in Figure 444 n 60 yn 6 a 1 b 15 c 08 n 60 yn 12 a 1 b 15 c 06 n 60 yn 140 a 1 b 15 c 05 Figure 444 System zerostate responses for three different feedback configurations The results of Example 412 demonstrate the importance of feedback in determin ing the response of a system In the first two cases the output signal is bounded But in the third case the output signal is unbounded even though the input signal is bounded Just as for continuoustime systems any time a discretetime system can exhibit an unbounded zerostate response to a bounded excitation of any kind it is classified as a BIBO unstable system So the stability of feedback systems depends on the nature of the feedback SYSTEM PROPERTIES The properties of discretetime systems are almost identical qualitatively to the prop erties of continuoustime systems In this section we explore examples illustrating some of the properties in discretetime systems Consider the system in Figure 445 The input and output signals of this system are related by the difference equation yn xn 45yn 1 The homogeneous solution is y h n K h 45 n Let xn be the unit sequence Then the particular solution is y p n 5 and the total solution is yn K h 45 n 5 See Web Appendix D for methods of solving difference equations If the system is in its zero state before time n 0 the total solution is yn 5 4 45 n n 0 0 n 0 or yn 5 445nun see Figure 446 rob28124ch04118163indd 152 041216 124 pm 43 DiscreteTime Systems 153 The similarity between the shape of the RC lowpass filters response to a unitstep excitation and the envelope of this systems response to a unitsequence is not an acci dent This system is a digital lowpass filter more on digital filters in Chapters 11 and 14 If we multiply the excitation of this system by any constant the response is multi plied by the same constant so this system is homogeneous If we delay the excitation of this system by any time n 0 we delay the response by that same time Therefore this system is also time invariant If we add any two signals to form the excitation of the sys tem the response is the sum of the responses that would have occurred by applying the two signals separately Therefore this system is an LTI discretetime system This sys tem also has a bounded response for any bounded excitation Therefore it is also stable A simple example of a system that is not time invariant would be one described by yn x2n Let x 1 n gn and let x 2 n gn 1 where gn is the signal illustrated in Figure 447 and let the response to x 1 n be y 1 n and let the response to x 2 n be y 2 n These signals are illustrated in Figure 448 Since x 2 n is the same as x 1 n except delayed by one discrete time unit for the system to be time invariant y 2 n must be the same as y 1 n except delayed by one dis cretetime unit but it is not Therefore this system is time variant xn yn 45 D Figure 445 A system 5 5 10 15 20 5 n yn Figure 446 System zerostate response to a unitsequence excitation Figure 448 Responses of the system described by yn x2n to two different excitations n 2 4 6 8 8 6 4 2 2 4 n 2 4 6 8 8 6 4 2 2 4 n 2 4 6 8 8 6 4 2 2 4 n 2 4 6 8 8 6 4 2 2 4 x1n gn y1n x12n x2 n gn1 y2n x22n y1n1 Figure 447 An excitation signal n gn 2 4 6 8 8 6 4 2 2 4 rob28124ch04118163indd 153 041216 124 pm C h a p t e r 4 Description of Systems 154 A good example of a system that is not BIBO stable is the financial system of accruing compound interest If a principle amount P of money is deposited in a fixedincome investment at an interest rate r per annum compounded annually the amount An which is the value of the investment n years later is An P 1 r n The amount An grows without bound as discretetime n passes Does that mean our banking system is unstable The amount does grow without bound and in an infinite time would approach infinity But since no one who is alive today or at any time in the future will live long enough to see this happen the fact that the system is unstable according to our definition is really of no great concern When we also consider the effects of the inevitable withdrawals from the account and monetary inflation we see that this theoretical instability is not significant The most common type of discretetime system studied in signals and systems is a system whose inputoutput relationship is determined by a linear constantcoefficient ordinary difference equation The eigenfunction is the complex exponential and the homogeneous solution is in the form of a linear combination of complex exponentials The form of each complex exponential is z n z n e j z n where z is the eigenvalue If the magnitude of z is less than one the solution form z n gets smaller in magnitude as discrete time passes and if the magnitude of z is greater than one the solution form gets larger in magnitude If the magnitude of z is exactly one it is possible to find a bounded excitation that will produce an unbounded response As was true for continuoustime systems an excitation that is of the same functional form as the homogeneous solution of the differential equation will produce an un bounded response For a discretetime system if the magnitude of any of the eigenvalues is greater than or equal to one the system is BIBO unstable ExamplE 413 Finding a bounded excitation that produces an unbounded response Consider an accumulator for which yn m n xm Find the eigenvalues of the solution of this equation and find a bounded excitation that will produce an unbounded response We can take the first backward difference of both sides of the difference equation yielding yn yn 1 xn This is a very simple difference equation with one eigenvalue and the homogeneous solution is a constant because the eigenvalue is one Therefore this system should be BIBO unstable The bounded excitation that produces an unbounded response has the same functional form as the homogeneous solution In this case a constant excitation produces an unbounded response Since the response is the accumulation of the excitation it should be clear that as discrete time n passes the magnitude of the response to a constant excitation grows lin early without an upper bound The concepts of memory causality static nonlinearity and invertibility are the same for discretetime systems as for continuoustime systems Figure 449 is an ex ample of a static system One example of a statically nonlinear system would be a twoinput OR gate in a digital logic system Suppose the logic levels are 0 V for a logical 0 and 5 V for a logical 1 If we apply 5 V to either of the two inputs with 0 V on the other the response is 5 V If we then apply 5 V to both inputs simultaneously the response is still 5 V If the Figure 449 A static system x1n x2n x3n yn rob28124ch04118163indd 154 041216 124 pm 44 Summary of Important Points 155 system were linear the response to 5 V on both inputs simultaneously would be 10 V This is also a noninvertible system If the output signal is 5 V we do not know which of three possible inputsignal combinations caused it and therefore knowledge of the output signal is insufficient to determine the input signals Even though all real physical systems must be causal in the strict sense that they cannot respond before being excited there are real signalprocessing systems that are sometimes described in a superficial sense as noncausal These are dataprocessing systems in which signals are recorded and then processed offline at a later time to produce a computed response Since the whole history of the input signals has been re corded the computed response at some designated time in the data stream can be based on values of the alreadyrecorded input signals that occurred later in time Figure 450 But since the whole data processing operation occurs after the input signals have been recorded this kind of system is still causal in the strict sense n 20 xn 2 2 n 20 yn 2 2 yn xn1 xn xn1 Figure 450 A socalled noncausal filter calculating responses from a prerecorded record of excitations 44 SUMMARY OF IMPORTANT POINTS 1 A system that is both homogeneous and additive is linear 2 A system that is both linear and time invariant is called an LTI system 3 The total response of any LTI system is the sum of its zeroinput and zerostate responses 4 Often nonlinear systems can be analyzed with linear system techniques through an approximation called linearization 5 A system is said to be BIBO stable if arbitrary bounded input signals always produce bounded output signals 6 A continuoustime LTI system is stable if all its eigenvalues have negative real parts 7 All real physical systems are causal although some may be conveniently and superficially described as noncausal rob28124ch04118163indd 155 041216 124 pm C h a p t e r 4 Description of Systems 156 8 Continuoustime systems are usually modeled by differential equations and discretetime systems are usually modeled by difference equations 9 The solution methods for difference equations are very similar to the solution methods for differential equations 10 One common use for difference equations is to approximate differential equations 11 A discretetime LTI system is stable if all its eigenvalues are less than one in magnitude EXERCISES WITH ANSWERS Answers to each exercise are in random order System Models 1 Write the differential equation for the voltage vCt in the circuit in Figure E1 for time t 0 then find an expression for the current it for time t 0 Figure E1 R1 2 Ω C 3 F Vs 10 V R2 6 Ω t 0 it ist vCt iCt Answer it 5 5 3 e t 18 t 0 2 The water tank in Figure E2 is filled by an inflow xt and is emptied by an outflow yt The outflow is controlled by a valve which offers resistance R to the flow of water out of the tank The water depth in the tank is dt and the surface area of the water is A independent of depth cylindrical tank The outflow is related to the water depth head by yt dt R The tank is 15 m high with a 1m diameter and the valve resistance is 10 s m2 a Write the differential equation for the water depth in terms of the tank dimensions and valve resistance b If the inflow is 005 m3s at what water depth will the inflow and outflow rates be equal making the water depth constant c Find an expression for the depth of water versus time after 1 m3 of water is dumped into an empty tank d If the tank is initially empty at time t 0 and the inflow is a constant 02 m3 s after time t 0 at what time will the tank start to overflow rob28124ch04118163indd 156 041216 124 pm 157 Exercises with Answers Inflow xt dt Outflow yt R Surface area A Valve Figure E2 Answers A d t dt R xt dt 05 m dt 4π e 4t10π t of 1386 10π4 1088 s 3 As derived in the text a simple pendulum is approximately described for small angles θ by the differential equation mL θ t mgθt xt where m is the mass of the pendulum L is the length of the massless rigid rod supporting the mass and θ is the angular deviation of the pendulum from vertical If the mass is 2 kg and the rod length is 05 m at what cyclic frequency will the pendulum oscillate Answer 0704 Hz 4 A block of aluminum is heated to a temperature of 100C It is then dropped into a flowing stream of water which is held at a constant temperature of 10C After 10 seconds the temperature of the block is 60C Aluminum is such a good heat conductor that its temperature is essentially uniform throughout its volume during the cooling process The rate of cooling is proportional to the temperature difference between the block and the water a Write a differential equation for this system with the temperature of the water as the excitation and the temperature of the block as the response b Compute the time constant of the system c If the same block is cooled to 0C and dropped into a flowing stream of water at 80C at time t 0 at what time will the temperature of the block reach 75C Answers T a t 90 e λt 10 τ 17 t 75 17 ln 00625 47153 Block Diagrams 5 The systems represented by these block diagrams can each be described by a differential equation of the form a N d N d t N yt a N1 d N1 d t N1 yt a 2 d 2 d t 2 yt a 1 d dt yt a 0 yt xt For each system what is the value of N For each system what are the a coefficients starting with a N and going down to a 0 In system b what range of values of A will make the system stable rob28124ch04118163indd 157 041216 124 pm C h a p t e r 4 Description of Systems 158 xt yt 2 7 6 3 xt yt 7 A a b Answers N 3 13 y t 6 y t 7 y t 2yt xt N 2 y t A y t 7yt xt For A 0 the system is stable System Properties 6 Show that a system with excitation xt and response yt described by yt u xt is nonlinear time invariant BIBO stable and noninvertible 7 Show that a system with excitation xt and response yt described by yt x t2 is linear time variant and noncausal 8 Show that the system in Figure E8 is linear time invariant BIBO unstable and dynamic xt yt 7 14 25 01 Figure E8 rob28124ch04118163indd 158 041216 124 pm 159 Exercises with Answers 9 Show that a system with excitation xn and response yn described by yn nxn is linear time variant and static 10 Show that the system described by yn x n is time variant dynamic non causal and BIBO stable 11 Show that the system described by yt sin xt is nonlinear time invariant BIBO stable static and causal 12 Show that the system described by yt x sint is linear time invariant BIBO stable and noncausal 13 Show that the system described by the equation y n e x n is nonlinear time invariant BIBO stable static causal and invertible 14 Show that a system described by yt t 2 x t 1 is time variant BIBO unstable causal dynamic and noninvertible 15 Show that a system described by yt dxt dt is not invertible and that a system described by y t t xλdλ is invertible 16 Show that a system described by y n x n2 n even 0 n odd is invertible and that a system described by y n x 2n is not invertible 17 A continuoustime system is described by the equation yt txt Show that it is linear BIBO unstable and time variant 18 Show that the system of Figure E18 is nonlinear time invariant static and invertible 5 10 yn xn Figure E18 y n 10xn 5 19 Show that the system described by yt 10 xt 2 5xt 2 xt 2 10 xt 2 is nonlinear static BIBO stable noninvertible and time invariant 20 Show that a system described by the equation yn x n 1 is nonlinear BIBO stable time invariant noncausal and noninvertible 21 Show that the system described by y n xnx n 2 x n 1 is homogeneous but not additive rob28124ch04118163indd 159 041216 124 pm C h a p t e r 4 Description of Systems 160 22 Show that the system of Figure E22 is time invariant BIBO stable and causal Figure E22 xn yn D D 1 2 025 EXERCISES WITHOUT ANSWERS System Models 23 Pharmacokinetics is the study of how drugs are absorbed into distributed through metabolized by and excreted from the human body Some drug processes can be approximately modeled by a one compartment model of the body in which V is the volume of the compartment Ct is the drug concentration in that compartment k e is a rate constant for excretion of the drug from the compartment and k 0 is the infusion rate at which the drug enters the compartment a Write a differential equation in which the infusion rate is the excitation and the drug concentration is the response b Let the parameter values be k e 04 hr 1 V 20 l and k 0 200 mghr where l is the symbol for liter If the initial drug concentration is C 0 10 mgl plot the drug concentration as a function of time in hours for the first 10 hours of infusion Find the solution as the sum of the zero excitation response and the zerostate response 24 A wellstirred vat has been fed for a long time by two streams of liquid fresh water at 02 cubic meters per second and concentrated blue dye at 01 cubic meters per second The vat contains 10 cubic meters of this mixture and the mixture is being drawn from the vat at a rate of 03 cubic meters per second to maintain a constant volume The blue dye is suddenly changed to red dye at the same flow rate At what time after the switch does the mixture drawn from the vat contain a ratio of red to blue dye of 991 25 A car rolling on a hill can be modeled as shown in Figure E25 The excitation is the force f t for which a positive value represents accelerating the car forward with the motor and a negative value represents slowing the car by braking action As it rolls the car experiences drag due to various frictional phenomena that can be approximately modeled by a coefficient k f which multiplies the cars velocity to produce a force which tends to slow the car when it moves in either direction The mass of the car is m and gravity acts on it at all times tending to make it roll down the hill in the absence of other forces Let the mass m of the car be 1000 kg let the friction coefficient k f be 5 N s m and let the angle θ be π 12 rob28124ch04118163indd 160 041216 124 pm 161 Exercises without Answers θ mg sinθ ft yt Figure E25 a Write a differential equation for this system with the force ft as the excitation and the position of the caryt as the response b If the nose of the car is initially at position y 0 0 with an initial velocity y t t0 10 m s and no applied acceleration or braking force graph the velocity of the car y t for positive time c If a constant force ft of 200 N is applied to the car what is its terminal velocity 26 At the beginning of the year 2000 the country Freedonia had a population p of 100 million people The birth rate is 4 per annum and the death rate is 2 per annum compounded daily That is the births and deaths occur every day at a uniform fraction of the current population and the next day the number of births and deaths changes because the population changed the previous day For example every day the number of people who die is the fraction 002365 of the total population at the end of the previous day neglect leapyear effects Every day 275 immigrants enter Freedonia a Write a difference equation for the population at the beginning of the nth day after January 1 2000 with the immigration rate as the excitation of the system b By finding the zeroexcitation and zerostate responses of the system determine the population of Freedonia at the beginning of the year 2050 27 In Figure E27 is a MATLAB program simulating a discretetime system a Without actually running the program find the value of x when n 10 by solving the difference equation for the system in closed form b Run the program and check the answer in part a x 1 y 3 z 0 n 0 while n 10 z y y x x 2n 09y 06z n n 1 end Figure E27 rob28124ch04118163indd 161 041216 124 pm C h a p t e r 4 Description of Systems 162 System Properties 28 A system is described by the block diagram in Figure E28 Figure E28 xt yt 1 3 025 Classify the system as to homogeneity additivity linearity time invariance BIBO stability causality memory and invertibility 29 A system is described by the differential equation t y t 8yt xt Classify the system as to linearity time invariance and BIBO stability 30 System 1 is described by yt ramp xt and system 2 is described by yt xt rampt Classify both systems as to BIBO stability linearity invertibility and time invariance 31 A system is described by y t 0 xt 0 xt x t 2 xt 0 Classify the system as to linearity time invariance memory causality and stability 32 A system is described by yt 0 t 0 xt x t 2 t 0 xt x t 2 ut Classify the system as to linearity time invariance memory causality and stability 33 A system is described by y n mnn0 nn0 xm Classify the system as to linearity time invariance and stability 34 A system is described by yn n2 1xn xn 1 Is it invertible 35 A system is described by y n 4 x n 3 x n 1 x n 4 1 x n 1 Classify the system as to linearity time invariance BIBO stability and invertibility 36 A system is described by the equation y n 9 tri n4 x n 3 Classify the system as to linearity time invariance BIBO stability memory causality and invertibility 37 A system is described by the equation yt t3 x λ dλ Classify the system as to time invariance BIBO stability and invertibility rob28124ch04118163indd 162 041216 124 pm 163 Exercises without Answers 38 A system is described by y t 1 x t 6 Classify the system as to linearity time invariance BIBO stability memory and invertibility 39 A system is described by the equation y t t3 xλdλ Classify the system as to linearity causality and invertibility 40 A system is described by y n 18y n 1 162y n 2 x n Classify this system as to linearity time invariance BIBO stability memory causality and invertibility 41 A system is described by y n m n1 x m Classify this system as to time invariance BIBO stability and invertibility 42 A system is described by yt 1 xt dxt dt 2 Classify this system as to homogeneity and additivity 43 A system is described by y n x n x n 1 Classify this system as to invertibility 44 In Figure E44 are some system descriptions in which x is the excitation and y is the response System A y t t xτdτ System B y n 2y n 1 x n 1 System C y t yt xt System D y n 8 xn 2 4xn 2 xn 2 8 xn 2 System E yt txt Figure E44 a Which systems are linear b Which systems are time invariant c Which systems are BIBO stable d Which systems are dynamic e Which systems are causal f Which systems are invertible rob28124ch04118163indd 163 041216 124 pm 164 51 INTRODUCTIO N AND GOALS The essential goal in designing systems is that they respond in the right way Therefore we must be able to calculate the response of a system to any arbitrary input signal As we will see throughout this text there are multiple approaches to doing that We have already seen how to find the response of a system described by differential or differ ence equations by finding the total solution of the equations with boundary conditions In this chapter we will develop another technique called convolution We will show that for an LTI system if we know its response to a unit impulse occurring at t 0 or n 0 that response completely characterizes the system and allows us to find the response to any input signal CHAPTER GOA L S 1 To develop techniques for finding the response of an LTI system to a unit impulse occurring at t 0 or n 0 2 To understand and apply convolution a technique for finding the response of LTI systems to arbitrary input signals for both continuoustime and discretetime systems 52 CONTINUOUS TIME IMPULSE RESPONSE We have seen techniques for finding the solutions to differential equations that describe systems The total solution is the sum of the homogeneous and particular solutions The homogeneous solution is a linear combination of eigenfunctions The particular solution depends on the form of the forcing function Although these methods work there is a more systematic way of finding how systems respond to input signals and it lends insight into important system properties It is called convolution The convolution technique for finding the response of a continuoustime LTI sys tem is based on a simple idea If we can find a way of expressing a signal as a linear combination of simple functions we can using the principles of linearity and time invariance find the response to that signal as a linear combination of the responses to those simple functions If we can find the response of an LTI system to a unitimpulse occurring at t 0 and if we can express the input signal as a linear combination of 5 C H A P T E R TimeDomain System Analysis rob28124ch05164228indd 164 041216 127 pm 52 Continuous Time 165 impulses we can find the response to it Therefore use of the convolution technique begins with the assumption that the response to a unit impulse occurring at t 0 has already been found We will call that response ht the impulse response So the first requirement in using the convolution technique to find a system response is to find the impulse response by applying a unit impulse δt occurring at t 0 The impulse in jects signal energy into the system and then goes away After the energy is injected into the system it responds with a signal determined by its dynamic character We could in principle find the impulse response experimentally by actually applying an impulse at the system input But since a true impulse cannot actually be generated this would only be an approximation Also in practice an approximation to an impulse would be a very tall pulse lasting for a very short time In reality for a real physical system a very tall pulse might actually drive it into a nonlinear mode of response and the experimentally measured impulse response would not be accurate There are other less direct but more practical ways of experimentally determining an impulse response If we have a mathematical description of the system we may be able to find the impulse response analytically The following example illustrates some methods for finding the impulse response of a system described by a differential equation ExamplE 51 Impulse response of continuoustime system 1 Find the impulse response ht of a continuoustime system characterized by the differential equation yt ayt xt 51 where xt excites the system and yt is the response We can rewrite 51 for the special case of an impulse exciting the system as ht aht δt 52 METHOD 1 Since the only excitation is the unit impulse at t 0 and the system is causal we know that the impulse response before t 0 is zero That is ht 0 t 0 The homogeneous solution for t 0 is of the form Keat This is the form of the impulse response for t 0 because in that time range the system is not being excited We now know the form of the impulse response before t 0 and after t 0 All that is left is to find out what happens at t 0 The differen tial equation 51 must be satisfied at all times That is ht aht must be a unit impulse occurring at time t 0 We can determine what happens at t 0 by integrating both sides of 52 from t 0 to t 0 infinitesimal times just before and just after zero The integral of ht is simply ht We know that at time t 0 it is zero and at time t 0 it is K h 0 K h 0 0 a 0 0 ht dt 0 0 δt dt 1 53 The homogeneous solution applies for all t 0 but at t 0 we must also consider the particular solution because the impulse is driving the system at that time The general rule for the form of the particular solution of a differential equation is a linear combination of the forcing function and all its unique derivatives The forcing function is an impulse and an impulse has infinitely many unique derivatives the doublet the triplet and so on and all of them occur exactly at t 0 Therefore until we can show a reason why an impulse andor all its derivatives cannot be in the solution we have to consider them as possibilities If ht does not have an impulse or rob28124ch05164228indd 165 041216 127 pm C h a p t e r 5 TimeDomain System Analysis 166 higherorder singularity at t 0 then 0 0 ht dt K 0 0 e at dt Ka e 0 e 0 0 0 If ht does have an impulse or higherorder singularity at t 0 then the integral may not be zero If ht has an impulse or higherorder singularity at t 0 then ht which appears on the left side of 52 must contain a doublet or higherorder singularity Since there is no doublet or higherorder singularity on the right side of 52 the equation cannot be satisfied Therefore in this example we know that there is no impulse or higherorder singularity in ht at t 0 and therefore 0 0 ht dt 0 the form of the impulse response is Keat ut and from 53 h 0 K e a 0 K 1 This is the needed initial condition to find a numerical form of the homoge neous solution that applies after t 0 The total solution is then ht eat ut Lets verify this solution by substituting it into the differential equation h t aht e at δt a e at ut a e at ut δt or using the equivalence property of the impulse e at δt e 0 δt δt Check METHOD 2 Another way to find the impulse response is to find the response of the system to a rectangular pulse of width w and height 1w beginning at t 0 and after finding the solution to let w approach zero As w approaches zero the rectangular pulse approaches a unit impulse at t 0 and the response approaches the impulse response Using the principle of linearity the response to the pulse is the sum of the responses to a step of height 1w at t 0 and the response to a step of height 1w at t w The equation for xt ut is h 1 t a h 1 t ut 54 The notation h1 t for step response follows the same logic as the coordinated notation for singularity functions The subscript indicates the number of differentiations of the impulse response In this case there is 1 differentiation or one integration in going from the unitimpulse response to the unitstep response The total response for t 0 to a unitstep is h1t Keat 1a If h1t has a discontinuity at t 0 then h1t must contain an impulse at t 0 Therefore since xt is the unit step which does not contain an impulse h1t must be continuous at t 0 otherwise 54 could not be correct Also since h1t has been zero for all negative time and it is continuous at t 0 it must also be zero at t 0 Then h 1 0 0 K e 0 1a K 1a and h1t 1a 1 eat t 0 Combining this with the fact that h1t 0 for t 0 we get the solution for all time h 1 t 1 e at a ut Using linearity and time invariance the response to a unit step occurring at t w would be h 1 t w 1 e atw a ut w Therefore the response to the rectangular pulse described above is h p t 1 e at ut 1 e atw ut w aw rob28124ch05164228indd 166 041216 127 pm 52 Continuous Time 167 Then letting w approach zero ht lim w0 h p t lim w0 1 e at ut 1 e atw ut w aw This is an indeterminate form so we must use LHôpitals rule to evaluate it lim w0 h p t lim w0 d dw 1 e at ut 1 e atw ut w d dw aw lim w0 h p t lim w0 d dw 1 e atw ut w a lim w0 h p t lim w0 1 e atw δt w a e atw ut w a lim w0 h p t 1 e at δt a e at ut a a e at ut a e at ut The impulse response is ht eat ut as before The principles used in Example 51 can be generalized to apply to finding the impulse response of a system described by a differential equation of the form a N y N t a N1 y N1 t a 1 y t a 0 yt b M x M t b M1 x M1 t b 1 x t b 0 xt 55 or k0 N a k y k t k0 M b k x k t The response ht to a unit impulse must have a functional form such that 1 When it is differentiated multiple times up to the Nth derivative all those derivatives must match a corresponding derivative of the impulse up to the Mth derivative at t 0 and 2 The linear combination of all the derivatives of ht must add to zero for any t 0 Requirement 2 is met by a solution of the form yht ut where yht is the homoge neous solution of 55 To meet requirement 1 we may need to add another function or functions to yht ut Consider three cases Case 1 M N The derivatives of yhtut provide all the singularity functions necessary to match the impulse and derivatives of the impulse on the right side and no other terms need to be added Case 2 M N We only need to add an impulse term Kδ δt Case 3 M N rob28124ch05164228indd 167 041216 127 pm C h a p t e r 5 TimeDomain System Analysis 168 The Nth derivative of the function we add to yht ut must have a term that matches the Mth derivative of the unit impulse So the function we add must be of the form K MN u MN t K MN1 u MN1 t K 0 u 0 t δt All the other derivatives of the impulse will be accounted for by differentiating the solution form yht ut multiple times Case 1 is the most common case in practice and Case 3 is rare in practice ExamplE 52 Impulse response of continuoustime system 2 Find the impulse response of a system described by yt ayt xt The impulse response must satisfy h t aht δ t 56 The highest derivative is the same for the excitation and response The form of the impulse response is ht Keat ut Kδδt and its first derivative is h t K e at δt aK e at ut K δ δ t Using the equivalence property of the impulse h t Kδt aK e at ut K δ δ t Integrating 56 from t 0 to t 0 h 0 K h 0 0 a 0 0 K e at ut K δ δtdt δ 0 0 δ 0 0 K aK 0 0 e at dt a K δ 0 0 δtdt 1 0 K aK e at a 0 0 a K δ K K e 0 e 0 0 a K δ 0 or K aKδ 0 Integrating 56 from to t and then from t 0 to t 0 we get 0 0 dt t Kδλ aK e aλ uλ K δ δ λd λ 0 0 dt t K e aλ uλ K δ δλdλ 0 0 dt t δ λdλ 0 0 Kut K e at 1ut K δ δtdt K a 0 0 1 e at utdt 0 K δ 0 0 utdt 0 0 0 dt t δ λdλ rob28124ch05164228indd 168 041216 127 pm 52 Continuous Time 169 0 0 K e at ut K δ δt dt 0 0 δtdt u0 u 0 K a 1 e at 0 0 0 K δ u 0 1 u 0 0 1 K δ 1 K a Therefore the impulse response is ht δt aeat ut Checking the solution by substituting it into 56 δ t a e at δt e 0 δt δt a 2 e at ut aδt a e at ut δ t or δt δt Check CONTINUOUSTIME CONVOLUTION Derivation Once the impulse response of a system is known we can develop a method for finding its response to a general input signal Let a system be excited by an arbitrary input sig nal xt Figure 51 How could we find the response We could find an approximate response by approximating this signal as a sequence of contiguous rectangular pulses all of the same width Tp Figure 52 Figure 51 An arbitrary signal t xt Figure 52 Contiguouspulse approximation to an arbitrary signal t xt Tp Now we can approximately find the response to the original signal as the sum of the responses to all those pulses acting individually Since all the pulses are rect angular and the same width the only differences between pulses are when they occur and how tall they are So the pulse responses all have the same form except delayed by some amount to account for time of occurrence and multiplied by a weighting con stant to account for the height We can make the approximation as good as necessary by using more pulses each of shorter duration In summary the problem of finding the response of an LTI system to an arbitrary signal becomes the problem of adding responses of a known functional form but weighted and delayed appropriately Using the rectangle function the description of the approximation to the arbitrary signal can now be written analytically The height of a pulse is the value of the signal at the time the center of the pulse occurs Then the approximation can be written as xt x T p rect t T p T p x0 rect t T p x T p rect t T p T p rob28124ch05164228indd 169 041216 127 pm C h a p t e r 5 TimeDomain System Analysis 170 Figure 53 Unit pulse response of an RC lowpass filter t Tp Tp 1 hpt Unit Pulse Figure 54 Exact and approximate xt t 1 3 xt 01 02 Exact Approximate or xt n xn T p rect t n T p T p 57 Let the response to a single pulse of width Tp and unit area centered at t 0 be a function hpt called the unitpulse response The unit pulse is 1Tp rect1Tp There fore 57 could be written in terms of shifted unit pulses as xt n T p xn T p 1 T p rect t n T p T p shifted unit pulse 58 Invoking linearity and time invariance the response to each of these pulses must be the unit pulse response hpt amplitude scaled by the factor TpxnTp and time shifted from the time origin the same amount as the pulse Then the approximation to the response is yt n T p xn T p h p t n T p 59 As an illustration let the unit pulse response hpt be that of the RC lowpass filter introduced above Figure 53 Let the input signal xt be the smooth waveform in Figure 54 which is approximated by a sequence of pulses as illustrated In Figure 55 the pulses are separated and then added to form the approximation to xt Figure 55 Approximation of xt as a sum of individual pulses xt Tp rob28124ch05164228indd 170 041216 127 pm 52 Continuous Time 171 Figure 56 Application of linearity and superposition to find the approximate system response ht ht ht ht yt yt xt ht ht ht Tp Tp Tp Since the sum of the individual pulses is the approximation of xt the approxi mate response can be found by applying the approximation of xt to the system But because the system is LTI we can alternately use the principle of superposition and apply the pulses one at a time to the system Then those responses can be added to form the approximate system response Figure 56 The system exact and approximate input signals the unitimpulse response the unitpulse response and the exact and approximate system responses are illustrated in Figure 57 based on a pulse width of 02 seconds As the pulse duration is reduced the approximation becomes better Figure 58 With a pulse width of 01 the exact and approximate responses are indistinguishable as graphed on this scale Recall from the concept of rectangularrule integration in basic calculus that a real integral of a real variable can be defined as the limit of a summation a b gx dx lim Δx0 naΔx bΔx gnΔx Δx 510 We will apply 510 to the summations of pulses and pulse responses 58 and 59 in the limit as the pulse width approaches zero As the pulse width Tp becomes smaller the exci tation and response approximations become better In the limit as Tp approaches zero the summation becomes an integral and the approximations become exact In that same limit rob28124ch05164228indd 171 041216 127 pm C h a p t e r 5 TimeDomain System Analysis 172 t 1 3 xt yt 01 02 t 1 3 ht and hpt 5 Tp 02 t 1 3 01 02 Exact Approximate UnitImpulse Response UnitPulse Response Exact Approximate Figure 57 Exact and approximate excitation unitimpulse response unitpulse response and exact and approximate system response with Tp 02 Figure 58 Exact and approximate excitation unitimpulse response unitpulse response and exact and approximate system response with Tp 01 t 1 3 xt 01 02 t 1 3 ht and hpt 5 t 1 3 yt 01 02 Tp 01 Exact Approximate UnitImpulse Response UnitPulse Response Exact Approximate the unit pulse 1Tp rect1Tp approaches a unit impulse As Tp approaches zero the points in time nTp become closer and closer together In the limit the discrete time shifts nTp merge into a continuum of timeshifts It is convenient and conventional to call that new continuous time shift τ Changing the name of the time shift amount nTp to τ and taking the limit as Tp approaches zero the width of the pulse Tp approaches a differential dτ and xt n T p dτ x n T p τ 1 T p rect t n T p T p δtτ and yt n T p dτ x n T p τ h p t n T p htτ Therefore in the limit these summations become integrals of the forms xt x τ δt τdτ 511 rob28124ch05164228indd 172 041216 127 pm 52 Continuous Time 173 and yt x τht τdτ 512 where the unitpulse response hpt approaches the unitimpulse response ht more commonly called just the impulse response of the system The integral in 511 is easily verified by application of the sampling property of the impulse The integral in 512 is called the convolution integral The convolution of two functions is conven tionally indicated by the operator 1 yt xt ht xτ ht τdτ 513 Another way of developing the convolution integral is to start with 511 which follows directly from the sampling property of the impulse The integrand of 511 is an impulse at t τ of strength xτ Since by definition ht is the response to an impulse δt and the system is homogeneous and time invariant the response to xτδt τ must be xτht τ Then invoking additivity if xt xτ δt τ dτ an integral the limit of a summation of x values then yt xτht τ dτ an integral of the ys that respond to those xs This derivation is more abstract and sophisticated and much shorter than the derivation above and is an elegant application of the properties of LTI systems and the sampling property of the impulse The impulse response of an LTI system is a very important descriptor of the way it responds because once it is determined the response to any arbitrary input signal can be found The effect of convolution can be depicted by a block diagram Figure 59 Graphical and Analytical Examples of Convolution The general mathematical form of the convolution integral is xt ht x τ ht τ dτ A graphical example of the operations implied by the convolution integral is very helpful in a conceptual understanding of convolution Let ht and xt be the functions in Figure 510 1 Do not confuse the convolution operator with the indicator for the complex conjugate of a complex number or function For example xn hn is xn convolved with hn but xn hn is the product of the complex conjugate of xn and hn Usually the difference is clear in context Figure 59 Block diagram depiction of convolution ht xt yt xtht rob28124ch05164228indd 173 041216 127 pm C h a p t e r 5 TimeDomain System Analysis 174 This impulse response ht is not typical of a practical linear system but will serve to demonstrate the process of convolution The integrand in the convolution integral is xτht τ What is ht τ It is a function of the two variables t and τ Since the variable of integration in the convolution integral is τ we should consider ht τ to be a function of τ in order to see how to do the integral We can start by graphing hτ and then hτ versus τ Figure 511 Figure 510 Two functions to be convolved t xt 2 1 1 t ht 2 1 The addition of the t in ht τ shifts the function t units to the right Figure 512 The transformation from hτ to ht τ can be described as two successive shifting scaling operations hτ ττ hτ ττt hτ t ht τ If we substitute t for τ in ht τ we have h0 From the first definition of the function ht we see that that is the point of discontinuity where ht goes from 0 to 1 That is the same point on ht τ Do the same for τ t 1 and see if it works One common confusion is to look at the integral and not understand what the process of integrating from τ to τ means Since t is not the variable of integration it is like a constant during the integration process But it is the variable in the final function that results from the convolution Think of convolution as two gen eral procedures First pick a value for t do the integration and get a result Then pick another value of t and repeat the process Each integration yields one point on the curve describing the final function Each point on the yt curve will be found by finding the total area under the product xτht τ Visualize the product xτht τ The product depends on what t is For most values of t the nonzero portions of the two functions do not overlap and the product is zero This is not typical of real impulse responses because they usually are not time limited Real impulse responses of stable systems usually begin at some time and approach zero as t approaches infinity But for some times t their nonzero portions do overlap and there is nonzero area under their product curve Consider t 5 and t 0 Figure 511 hτ and hτ graphed versus τ τ hτ 2 1 τ hτ 2 1 Figure 512 ht τ graphed versus τ τ htτ 2 t1 t rob28124ch05164228indd 174 041216 127 pm 52 Continuous Time 175 When t 5 the nonzero portions of xτ and h5 τ do not overlap and the product is zero everywhere Figure 513 Figure 513 Impulse response input signal and their product when t 5 τ 2 1 4 5 1 xτ h5τ τ 1 4 5 1 xτh5τ Figure 514 Impulse response input signal and their product when t 0 τ 2 1 1 xτ hτ τ 4 1 1 xτhτ xτhτdτ Figure 515 Product of ht τ and xτ for 1 t 0 τ 2 1 1 1t1 t 1t1 t xτ htτ 1 t 0 τ 4 htτxτ 4t For 1 t 0 the convolution of the two functions is twice the area of the h function which is 1 minus the area of a triangle of width t and height 4t Figure 515 Therefore the convolution function value over this range of t is yt 2 12t4t 21 t 2 1 t 0 For 0 t 1 the convolution of the two functions is the constant 2 For 1 t 2 the convolution of the two functions is the area of a triangle whose base width is 2 t and whose height is 8 4t or yt 122 t8 4t 22 t2 The final func tion yt is illustrated in Figure 516 When t 0 the nonzero portions of xτ and h5 τ do overlap and the product is not zero everywhere Figure 514 rob28124ch05164228indd 175 041216 127 pm C h a p t e r 5 TimeDomain System Analysis 176 Figure 516 Convolution of xt with ht t 1 t 05 t 0 t 05 t 1 t 15 t 2 4 3 2 1 1 2 3 4 2 Time t yt Figure 517 The impulse response and excitation of the RC lowpass filter τ 1 τ hτ xτ 1 RC RC As a more practical exercise let us now find the unitstep response of an RC low pass filter using convolution We already know from prior analysis that the answer is voutt 1 etRC ut First we need to find the impulse response The differential equation is RC v out t v out t v in t RC h t ht δt The form of the impulse response is ht KetRC ut Integrating once from 0 to 0 RC h 0 h 0 0 0 0 ht dt 0 u 0 1 u 0 0 h 0 1RC Then 1RC K and ht 1RCetRC ut Figure 517 Then the response vout t to a unit step vin t is vout t vin t ht or v out t v in τht τdτ uτ e tτRC RC ut τdτ We can immediately simplify the integral some by observing that the first unit step uτ makes the integrand zero for all negative τ Therefore v out t 0 e tτRC RC ut τdτ Consider the effect of the other unit step ut τ Since we are integrating over a range of τ from zero to infinity if t is negative for any τ in that range this unit step has a value rob28124ch05164228indd 176 041216 127 pm 52 Continuous Time 177 Figure 518 The relation between the two functions that form the product in the convolution integrand for t negative and t positive τ htτ xτ 1 t 0 t t τ htτ xτ 1 t 0 1 RC 1 RC of zero Figure 518 Therefore for negative t vout t 0 For positive t the unit step ut τ will be one for τ t and zero for τ t Therefore for positive t v out t 0 t e tτRC RC dτ e tτRC 0 t 1 e tRC t 0 Combining the results for negative and positive ranges of t v out t 1 e tRC ut Figure 519 and Figure 520 illustrate two more examples of convolution In each case the top row presents two functions x1t and x2t to be convolved and the flipped version of the second function x2τ which is xt τ with t 0 the flipped but notyetshifted version On the second row are the two functions in the convolution integral x1τ and x2t τ graphed versus τ for five choices of t illustrating the shifting of the second function x2t τ as t is changed On the third row are the products of Figure 519 Convolution of two rectangular pulses t 4 4 x1t 3 t 4 4 x2t 3 τ 4 4 x2τ 3 τ 4 4 x1τ and x205τ 3 τ 4 4 x1τx205τ 6 τ 4 4 x1τ and x20τ 3 τ 4 4 x1τx20τ 6 τ 4 4 x1τ and x21τ 3 τ 4 4 x1τx21τ 6 τ 4 4 x1τ and x22τ 3 τ 4 4 x1τx22τ 6 τ 4 4 x1τ and x225τ 3 τ 4 4 x1τx225τ 6 τ 4 4 x1tx2t 6 rob28124ch05164228indd 177 041216 127 pm C h a p t e r 5 TimeDomain System Analysis 178 Figure 520 Convolution of two triangular pulses t 6 6 x1t 3 t 6 6 x2t 3 τ 6 6 x2τ 3 τ 6 6 x1τx23τ 6 τ 6 6 x1τx22τ 6 τ 6 6 x1τx21τ 6 τ 6 6 x1τx20τ 6 τ 6 6 x1τ and x23τ 3 τ 6 6 x1τ and x22τ 3 τ 6 6 x1τ and x21τ 3 τ 6 6 x1τ and x20τ 3 τ 6 6 x1τ and x21τ 3 τ 6 6 x1τx21τ 6 τ 6 6 x1tx2t 6 the two functions x1τx2t τ in the convolution integral at those same times And at the bottom is a graph of the convolution of the two original functions with small dots indicating the convolution values at the five times t which are the same as the areas x 1 τx2t τdτ under the products at those times Convolution Properties An operation that appears frequently in signal and system analysis is the convolution of a signal with an impulse xt Aδt t 0 x τAδt τ t 0 dτ We can use the sampling property of the impulse to evaluate the integral The variable of integration is τ The impulse occurs in τ where t τ t 0 0 or τ t t 0 Therefore xt Aδt t 0 Axt t 0 514 This is a very important result and will show up many times in the exercises and later material Figure 521 If we define a function gt g 0 t δt then a timeshifted version gt t 0 can be expressed in either of the two alternate forms gt t 0 g 0 t t 0 δt or gt t 0 g 0 t δt t 0 rob28124ch05164228indd 178 041216 127 pm 52 Continuous Time 179 but not in the form g 0 t t 0 δt t 0 Instead g 0 t t 0 δt t 0 gt 2 t 0 This property is true not only when convolving with impulses but with any functions A shift of either of the functions being convolved but not both shifts the convolution by the same amount The commutativity associativity distributivity differentiation area and scaling prop erties of the convolution integral are proven in Web Appendix E and are summarized here Commutativity xt yt yt xt Associativity xt yt zt xt yt zt Distributivity xt yt zt xt zt yt zt If yt xt ht then Differentiation Property y t x t ht xt h t Area Property Area of y Area of x Area of h Scaling Property yat a x at hat Let the convolution of xt with ht be yt x t τhτdτ Let xt be bounded Then xt τ B for all τ where B is a finite upper bound The magnitude of the convolution integral is yt x t τhτdτ Using the principles that the magnitude of an integral of a function is less than or equal to the integral of the magnitude of the function α β gx dx α β gx dx and that the magnitude of a product of two functions is equal to the product of their magnitudes gxhx gx hx we can conclude that yt xt τ hτ dτ Figure 521 Examples of convolution with impulses t recttδt 1 2 1 t cosπtδt1 1 2 t recttδt1 1 2 3 2 1 t 2sinπt4δt 2 8 1 2 rob28124ch05164228indd 179 041216 127 pm C h a p t e r 5 TimeDomain System Analysis 180 Since xt τ is less than B in magnitude for any τ yt xt τ hτ dτ B hτ dτ or yt B hτ dτ Therefore the convolution integral converges if ht dt is bounded or in other words if ht is absolutely integrable Since convolution is commutative we can also say that if ht is bounded the condition for convergence is that xt be absolutely integrable For a convolution integral to converge the signals being convolved must both be bounded and at least one of them must be absolutely integrable ExamplE 53 Convolution of two unit rectangles Find the convolution yt of two unit rectangles xt rect t and ht rect t This convolution can be done in a direct way using the convolution integral analytically or graphically But we can exploit the differentiation property to avoid explicit integration altogether yt xt ht y t x t h t y t δt 12 δt 12 δt 12 δt 12 y t δt 1 2δt δt 1 y t ut 1 2ut ut 1 yt ramp t 1 2 ramp t ramp t 1 See Figure 522 Figure 522 Convolution of two unit rectangles t yt 1 1 1 The result of convolving two unit rectangles Example 53 is important enough to be given a name for future reference It is called the unit triangle function see Figure 523 tri t 1 t t 1 0 otherwise It is called a unit triangle because its peak height and its area are both one rob28124ch05164228indd 180 041216 127 pm 52 Continuous Time 181 System Connections Two very common connections of systems are the cascade connection and the parallel connection Figure 524 and Figure 525 Using the associativity property of convolution we can show that the cascade connection of two systems can be considered as one system whose impulse response is the convolution of the impulse responses of the two systems Using the distributivity property of convolution we can show that the parallel connection of two systems can be considered as one system whose impulse response is the sum of the impulse responses of the two systems Step Response and Impulse Response In actual system testing a system is often tested using some standard signals that are easy to generate and do not drive the system into a nonlinearity One of the most common signals of this type is the step function The response of an LTI system to a unit step is h 1 t ht ut h τut τdτ t h τdτ This proves that the response of an LTI system excited by a unit step is the integral of the impulse response Therefore we can say that just as the unit step is the integral of the impulse the unitstep response is the integral of the unitimpulse response In fact this relationship holds not just for impulse and step excitations but for any excitation If any excitation is changed to its integral the response also changes to its integral We can also turn these relationships around and say that since the first derivative is the inverse of integration if the excitation is changed to its first derivative the response is also changed to its first derivative Figure 526 Stability and Impulse Response Stability was generally defined in Chapter 4 by saying that a stable system has a bounded output signal in response to any bounded input signal We can now find a way to determine whether a system is stable by examining its impulse response We proved Figure 525 Parallel connection of two systems xt h1t h2t xth1t ytxth1txth2txth1th2t xth2t xt yt h1th2t Figure 524 Cascade connection of two systems xt h1t xth1t h2t ytxth1th2t xt h1th2t yt Figure 523 The unit triangle function t trit 1 1 1 rob28124ch05164228indd 181 041216 127 pm C h a p t e r 5 TimeDomain System Analysis 182 above that the convolution of two signals converges if both of them are bounded and at least one is absolutely integrable The response yt of a system to xt is yt xt ht Then if xt is bounded we can say that yt is bounded if ht is absolutely integrable That is if ht dt is bounded A continuoustime system is BIBO stable if its impulse response is absolutely integrable Complex Exponential Excitation and the Transfer Function Let a stable LTI system be described by a differential equation of the form k0 N a k y k t k0 M b k x k t 515 where 1 The as and bs are constants and 2 The notation xkt means the kth derivative of xt with respect to time and if k is negative that indicates integration instead of differentiation Figure 526 Relations between integrals and derivatives of excitations and responses for an LTI system t 1 6 δt 1 Impulse Excitation t 1 6 ht 1 Impulse Response t 1 6 ut 1 Step Excitation t 1 6 h1t ht 1 Step Response t 1 6 rampt 6 Ramp Excitation t 1 6 h2t ht 6 Ramp Response ht ht ht d dt d dt d dt d dt rob28124ch05164228indd 182 041216 127 pm 52 Continuous Time 183 Then let the excitation be in the form of a complex exponential xt Xest where X and s are in general complex valued This description of the excitation is valid for all time Therefore not only is the excitation a complex exponential now and in the future but it always has been a complex exponential The solution of the differential equation is the sum of the homogeneous and particular solutions The system is stable so the eigenvalues have negative real parts and the homogeneous solution approaches zero as time passes The system has been operating with this excitation for a semiinfinite time so the homo geneous solution has decayed to zero and the total solution now is the particular solution The functional form of the particular solution consists of a linear combination of the excitations functional form and all its unique derivatives Since the deriv ative of an exponential is another exponential of the same form the response yt must have the form yt Yest where Y is a complex constant Then in the differ ential equation the kth derivatives take the forms xk t sk Xest and ykt sk Yest and 515 can be written in the form k0 N a k s k Y e st k0 M b k s k X e st The equation is no longer a differential equation with real coefficients It is now an algebraic equation with complex coefficients The factors Xest and Yest can be factored out leading to Y e st k0 N a k s k X e st k0 M b k s k The ratio of the response to the excitation is then Y e st X e st Y X k0 M b k s k K0 N a k s k a ratio of two polynomials in s called a rational function This is the systems transfer function Hs k 0 M b k s k k 0 N a k s k b M s M b M1 s M1 b 2 s 2 b 1 s b 0 a N s N a N1 s N1 a 2 s 2 a 1 s a 0 516 and the response is therefore Y e st HsX e st or yt Hsxt For systems of this type the transfer function can be written directly from the differential equation If the differential equation describes the system so does the transfer function The transfer function is a fundamental concept in signals and systems and we will be using it many times in the material to follow We can also find the response to a complex exponential excitation using convolution The response yt of an LTI system with impulse response ht to xt Xest is yt ht X e st X h τ e stτ dτ Xest xt hτ e st dτ Equating the two forms of yt we get HsX e st X e st hτ e st dτ Hs hτ e st dτ rob28124ch05164228indd 183 041216 127 pm C h a p t e r 5 TimeDomain System Analysis 184 which shows how the transfer function and the impulse response are related Since both the impulse response and the transfer function completely characterize an LTI system they had to be uniquely related The integral hτ e st dτ will be revisited in Chapter 8 and will be identified as the Laplace transform of ht Frequency Response The variable s in the complex exponential est is in general complex valued Let it be of the form s σ jω where σ is the real part and ω is the imaginary part For the special case σ 0 s jω the complex exponential est becomes the complex sinusoid e jωt and the transfer function of the system Hs becomes the frequency response of the system H jω The function e jωt is called a complex sinusoid because by Eulers identity e jωt cosωt j sinωt the sum of a real cosine and an imaginary sine both of radian frequency ω From Y e st HsX e st letting s jω Y e jωt Y e jY e jωt H jωX e jωt H jω e jHjω X e jX e jωt or dividing through by e jωt Y e jY Hjω X e jH jωX Equating magnitudes we get Y H jω X and equating phases we get Y Hjω X The function H jω is called the frequency response of the system because at any radian frequency ω if we know the magnitude and phase of the excitation and the magnitude and phase of the frequency response we can find the magnitude and phase of the response In Chapter 4 we showed using principles of linearity and superposition that if a complex excitation xt is applied to a system and causes a response yt that the real part of xt causes the real part of yt and the imaginary part of xt causes the imaginary part of yt Therefore if the actual excitation of a system is xt A x cosωt θ x we can find the response of the system to an excitation x C t A x cosωt θ x j A x sinωt θ x A x e jωt θ x in the form y C t A y cosωt θ y j A y sinωt θ y A y e jωt θ y and we can take the real part yt A y cosωt θ y as the response to the real excitation xt A x cosωt θ x Using Y H jω X and Y H jω X we get A y H jω A x and θ y H jω θ x ExamplE 54 Transfer Function and Frequency Response An LTI system is described by the differential equation y t 3000 y t 2 10 6 yt 2 10 6 xt rob28124ch05164228indd 184 041216 127 pm 52 Continuous Time 185 a Find its transfer function For this differential equation of the form k0 N a k y k t k0 M b k x k t N 2 M 0 a 0 2 10 6 a 1 3000 a 2 1 and b 0 2 10 6 Therefore the transfer function is Hs 2 10 6 s 2 3000s 2 10 6 b If xt X e j400πt and yt Y e j400πt and X 3 e jπ2 find the magnitude and phase of Y The frequency response is H jω 2 10 6 jω 2 3000 jω 2 10 6 2 10 6 2 10 6 ω 2 j3000ω The radian frequency is ω 400π Therefore Hj400π 2 10 6 2 10 6 400π 2 j3000 400π 05272 e j146 Y H j400π 3 05272 3 1582 Y H j400π π2 01112 radians c If xt 8 cos200πt and yt A y cos200πt θ y find A y and θ y Using A y H j200π A x and θ y H j200π θ x A y 08078 8 64625 and θ y 08654 0 08654 radians ExamplE 55 Frequency response of a continuoustime system A continuoustime system is described by the differential equation y t 5 y t 2yt 3 x t Find and graph the magnitude and phase of its frequency response The differential equation is in the general form k0 N a k y k t k0 M b k x k t where N M 2 a 2 1 a 1 5 a 0 2 b 2 3 b 1 0 and b 0 0 The transfer function is Hs b 2 s 2 b 1 s b 0 a 2 s 2 a 1 s a 0 3 s 2 s 2 5s 2 The frequency response is replacing s by jω Hjω 3 jω 2 jω 2 j5ω 2 rob28124ch05164228indd 185 041216 127 pm C h a p t e r 5 TimeDomain System Analysis 186 See Figure 527 These graphs were generated by the following MATLAB code wmax 20 Maximum radian frequency magnitude for graph dw 01 Spacing between frequencies in graph w wmaxdwwmax Vector of radian frequencies for graph Compute the frequency response H 3jw2jw2 j5w 2 Graph and annotate the frequency response subplot211 p plotw absHk setpLineWidth2 grid on xlabelRadian frequency omegaFontSize18FontNameTimes ylabelHitjomegaFontSize18FontNameTimes subplot212 p plotwangleHk setpLineWidth2 grid on xlabelRadian frequency omegaFontSize18FontNameTimes ylabelPhase of HitjomegaFontSize18FontNameTimes 53 DISCRETE TIME IMPULSE RESPONSE Just as was true for continuoustime systems there is a convolution method for dis cretetime systems and it works in an analogous way It is based on knowing the impulse response of the system treating the input signal as a linear combination of impulses and then adding the responses to all the individual impulses Figure 527 Magnitude and phase of frequency response 20 15 10 5 0 5 10 15 20 0 05 1 15 2 25 3 Radian Frequency ω H jω 20 15 10 5 0 5 10 15 20 3 2 1 0 1 2 3 Radian Frequency ω Phase of H jω rob28124ch05164228indd 186 041216 127 pm 53 Discrete Time 187 No matter how complicated a discretetime signal is it is a sequence of impulses If we can find the response of an LTI system to a unit impulse occurring at n 0 we can find the response to any other signal Therefore use of the convolution technique begins with the assumption that the response to a unit impulse occurring at n 0 has already been found We will call the impulse response hn In finding the impulse response of a system we apply a unit impulse δn occur ring at n 0 and that is the only excitation of the system The impulse puts signal energy into the system and then goes away After the impulse energy is injected into the system it responds with a signal determined by its dynamic character In the case of continuoustime systems the actual application of an impulse to determine impulse response experimentally is problematical for practical reasons But in the case of discretetime systems this technique is much more reasonable because the discretetime impulse is a true discretetime function and a simple one at that If we have a mathematical description of the system we may be able to find the impulse response analytically Consider first a system described by a difference equa tion of the form a 0 yn a 1 yn 1 a N yn N xn 517 This is not the most general form of difference equation describing a discretetime LTI system but it is a good place to start because from the analysis of this system we can extend to finding the impulse responses of more general systems This system is causal and LTI To find the impulse response we let the excitation xn be a unit impulse at n 0 Then we can rewrite 517 for this special case as a 0 hn a 1 hn 1 a N hn N δn The system has never been excited by anything before n 0 the response hn has been zero for all negative time hn 0 n 0 and the system is in its zero state before n 0 For all times after n 0 xn is also zero and the solution of the difference equation is the homogeneous solution All we need to find the homogeneous solution after n 0 are N initial conditions we can use to evaluate the N arbitrary constants in the homogeneous solution We need an initial condition for each order of the difference equation We can always find these initial conditions by recursion The difference equation of a causal system can always be put into a recursion form in which the present response is a linear combination of the present excitation and previous responses hn δn a 1 hn 1 a N hn N a 0 Then we can find an exact homogeneous solution that is valid for n 0 That solution together with the fact that hn 0 n 0 forms the total solution the impulse response The application of an impulse to a system simply establishes some initial conditions and the system relaxes back to its former equilibrium after that if it is stable Now consider a more general system described by a difference equation of the form a 0 yn a 1 yn 1 a N yn N b 0 xn b 1 xn 1 b M xn M or k0 N a k y n k k0 M b k xn k rob28124ch05164228indd 187 041216 127 pm C h a p t e r 5 TimeDomain System Analysis 188 Since the system is LTI we can find the impulse response by first finding the impulse responses to systems described by the difference equations a 0 yn a 1 yn 1 a N yn N b 0 xn a 0 yn a 1 yn 1 a N yn N b 1 xn 1 518 a 0 yn a 1 yn 1 a N yn N a 0 yn a 1 yn 1 a N yn N b M xn M and then adding all those impulse responses Since all the equations are the same except for the strength and time of occurrence of the impulse the overall impulse response is simply the sum of a set of impulse responses of the systems in 518 weighted and delayed appropriately The impulse response of the general system must be hn b 0 h 1 n b 1 h 1 n 1 b M h 1 n M where h 1 n is the impulse response found earlier ExamplE 56 Impulse response of a system Find the impulse response hn of the system described by the difference equation 8yn 6yn 1 xn 519 If the excitation is an impulse 8hn 6hn 1 δn This equation describes a causal system so hn 0 n 0 We can find the first response to a unit impulse at n 0 from 519 n xn hn 1 hn 0 1 0 18 For n 0 the solution is the homogeneous solution of the form K h 34 n Therefore hn K h 34 n un Applying initial conditions h0 18 K h Then the impulse response of the system is hn 18 34 n un ExamplE 57 Impulse response of a system Find the impulse response hn of the system described by the difference equation 5yn 2yn 1 3yn 2 xn 520 This equation describes a causal system so hn 0 n 0 We can find the first two responses to a unit impulse at n 0 from 520 n xn hn 1 hn 0 1 0 15 1 0 15 225 rob28124ch05164228indd 188 041216 127 pm 53 Discrete Time 189 The eigenvalues are 1 and 06 So the impulse response is hn K 1 1 n K 2 06 n un Evaluating the constants h 0 K 1 K 2 15 h1 K 1 06 K 2 225 K 1 0125 K 2 0075 and the impulse response is hn 0125 1 n 0075 06 n un DISCRETETIME CONVOLUTION Derivation To demonstrate discretetime convolution suppose that an LTI system is excited by a signal xn δn δn 1 and that its impulse response is hn 07788 n un Figure 528 The excitation for any discretetime system is made up of a sequence of impulses with different strengths occurring at different times Therefore invoking linearity and time invariance the response of an LTI system will be the sum of all the individual responses to the individual impulses Since we know the response of the system to a single unit impulse occurring at n 0 we can find the responses to the individual im pulses by shifting and scaling the unitimpulse response appropriately In the example the first nonzero impulse in the excitation occurs at n 0 and its strength is one Therefore the system will respond to this with exactly its impulse re sponse The second nonzero impulse occurs at n 1 and its strength is also one The response of the system to this single impulse is the impulse response except delayed by one in discrete time So by the additivity and timeinvariance properties of LTI systems the overall system response to xn δn δn 1 is yn 07788 n un 07788 n1 un 1 Figure 528 System excitation xn system impulse response hn and system response yn n 10 30 xn 2 n 10 30 hn 2 n 10 30 yn 2 System Excitation System Impulse Response System Response rob28124ch05164228indd 189 041216 127 pm C h a p t e r 5 TimeDomain System Analysis 190 see Figure 528 Let the excitation now be xn 2δn Then since the system is LTI and the excitation is an impulse of strength two occurring at n 0 by the homogeneity property of LTI systems the system response is twice the impulse response or yn 2 07788 n un Now let the excitation be the one illustrated in Figure 529 while the impulse response remains the same The responses to the four impulses beginning at n 5 are graphed in Figure 530 10 5 5 10 15 20 1 1 xn n 10 5 5 10 15 20 1 hn n Figure 529 A sinusoid applied at time n 5 and the system impulse response Figure 530 System responses to the impulses x5 x4 x3 and x2 0 yn 1 0 yn 0 yn 10 5 5 10 15 20 1 y4n y3n y2n n 1 10 5 5 10 15 20 1 y5n n 1 10 5 5 10 15 20 1 n 1 10 5 5 10 15 20 1 n rob28124ch05164228indd 190 041216 127 pm 53 Discrete Time 191 Figure 531 illustrates the next four impulse responses When we add all the responses to all the impulses we get the total system response to the total system excitation Figure 532 Figure 531 System responses to the impulses x1 x0 x1 and x2 1 10 5 5 10 15 20 1 y1n y0n y1n y2n n 1 10 5 5 10 15 20 1 n 1 10 5 5 10 15 20 1 n 1 10 5 5 10 15 20 1 n Figure 532 The total system response 10 5 5 10 15 20 1 1 yn n Notice that there is an initial transient response but the response settles down to a sinusoid after a few discretetime units The forced response of any stable LTI system excited by a sinusoid is another sinusoid of the same frequency but generally with a different amplitude and phase We have seen graphically what happens Now it is time to see analytically what happens The total system response can be written as yn x1hn 1 x0hn x1hn 1 rob28124ch05164228indd 191 041216 127 pm C h a p t e r 5 TimeDomain System Analysis 192 or yn m xm hn m 521 The result 521 is called the convolution sum expression for the system response In words it says that the value of the response y at any discrete time n can be found by summing all the products of the excitation x at discrete times m with the impulse response h at discrete times n m for m ranging from negative to positive infinity To find a system response we only need to know the systems impulse response and we can find its response to any arbitrary excitation For an LTI system the impulse response of the system is a complete description of how it responds to any signal So we can imagine first testing a system by applying an impulse to it and recording the response Once we have it we can compute the response to any signal This is a power ful technique In system analysis we only have to solve the difference equation for the system once for the simplest possible nonzero input signal a unit impulse and then for any general signal we can find the response by convolution Compare the convolution integral for continuoustime signals with the convolu tion sum for discretetime signals yt xτht τdτ and yn m xm hn m In each case one of the two signals is timereversed and shifted and then multiplied by the other Then for continuoustime signals the product is integrated to find the total area under the product For discretetime signals the product is summed to find the total value of the product Graphical and Analytical Examples of Convolution Although the convolution operation is completely defined by 521 it is helpful to explore some graphical concepts that aid in actually performing convolution The two functions that are multiplied and then summed over m are xm and hn m To illustrate the idea of graphical convolution let the two functions xn and hn be the simple functions illustrated in Figure 533 Figure 533 Two functions xn n hn n 1 2 2 1 4 3 2 1 2 3 4 1 4 3 2 1 2 3 4 Since the summation index in 521 is m the function hn m should be considered a function of m for purposes of performing the summation in 521 With that point of view we can imagine that hn m is created by two transformations m m which changes hm to hm and then m m n which changes hm to hm n hn m The first transformation m m forms the discretetime reverse of hm rob28124ch05164228indd 192 041216 127 pm 53 Discrete Time 193 Figure 534 hm and hnm versus m m hm 1 2 1 4 3 2 1 2 3 4 m hnm 1 2 n2 4 3 2 1 n1 n n1 and the second transformation m m n shifts the alreadytimereversed function n units to the right Figure 534 Now realizing that the convolution result is yn m xmhn m the process of graphing the convolution result yn versus n is to pick a value of n and do the operation m xmhn m for that n plot that single numerical result for yn at that n and then repeat the whole process for each n Every time a new n is chosen the function hn m shifts to a new position xm stays right where it is because there is no n in xm and the summation m xmhn m is simply the sum of the products of the values of xm and hn m for that choice of n Figure 535 is an illustration of this process For all values of n not represented in Figure 535 yn 0 so we can now graph yn as illustrated in Figure 536 It is very common in engineering practice for both signals being convolved to be zero before some finite time Let x be zero before n n x and let h be zero before n n h The convolution sum is xn hn m xmhn m Since x is zero before n n x all the terms in the summation for m n x are zero and xn hn m n x xmhn m Also when n m n h the h terms are zero That puts an upper limit on m of n n h and xn hn m n x n n h xmhn m For those ns for which n n h n x the lower summation limit is greater than the upper summation limit and the convolution result is zero So it would be more complete and accurate to say that the convolution result is xn hn m n x n n h xmhn m n n h n x 0 n n h n x rob28124ch05164228indd 193 041216 127 pm C h a p t e r 5 TimeDomain System Analysis 194 ExamplE 58 Response of a movingaverage digital filter A movingaverage digital filter has an impulse response of the form hn un un NN Find the response of a movingaverage filter with N 8 to xn cos2πn16 Then change the excitation to xn cos2πn8 and find the new response n 6 4 2 1 4 3 2 1 2 3 4 yn Figure 536 Graph of yn Figure 535 yn for n 1 0 1 and 2 m h1m 1 2 3 21 4 xm m 2 1 4 3 21 2 3 4 1 2 3 4 m h0m 1 2 3 21 4 xm m 2 1 4 321 2 3 4 1 2 3 4 n 1 n 0 xmh1m m 2 1 4 3 2 1 2 3 4 y1 2 xmh0m m 4 2 1 4 3 2 1 2 3 4 y0 6 m h1m 1 2 xm m 2 1 4 321 2 3 4 43 21 1 2 3 4 m h2m 1 2 3 21 4 xm m 2 1 4 3 21 2 3 4 1 2 3 4 n 1 n 2 xmh2m m 4 1 4 3 21 2 3 4 y2 4 xmh1m m 4 2 1 4 3 21 2 3 4 y1 6 rob28124ch05164228indd 194 041216 127 pm 53 Discrete Time 195 Using convolution the response is yn xn hn cos2πn16 un un 88 Applying the definition of the convolution sum yn 1 8 m cos2πm16un m un m 8 The effect of the two unit sequence functions is to limit the summation range yn 1 8 mn7 n cos2πm16 Using the trigonometric identity cosx e jx e jx 2 yn 1 16 mn7 n e j2πm16 e j2πm16 Let q m n 7 Then yn 1 16 q0 7 e j2πqn716 e j2πqn716 yn 1 16 e j2πn716 q0 7 e j2πq16 e j2πn716 q0 7 e j2πq16 The summation of a geometric series with N terms is n0 N1 r n N r 1 1 r N 1 r r 1 This formula works for any complex value of r Therefore summing these geometric series each of length 8 yn 1 16 e jπn78 2 1 e jπ 1 e jπ8 e jπn78 2 1 e jπ 1 e jπ8 Finding a common denominator and simplifying yn 1 8 e jπn78 1 e jπ8 e jπn78 1 e jπ8 1 8 cosπn 78 cosπn 88 1 cosπ8 Then using the periodicity of the cosine cosπn 88 cosπn8 and yn 16421cosπn 78 cosπn8 Now letting xn cos2πn8 the process is essentially the same except for the period of the cosine The results are yn xn hn cos2πn8 un un 88 yn 1 16 mn7 n e j2πm8 e j2πm8 rob28124ch05164228indd 195 041216 127 pm C h a p t e r 5 TimeDomain System Analysis 196 yn 1 16 e j2πn78 q0 7 e j2πq8 e j2πn78 q0 7 e j2πq8 yn 1 16 e jπn74 1 e j2π 1 e j2π8 e jπn74 1 e j2π 1 e j2π8 0 because e j2π e j2π 1 If the averaging time of the movingaverage filter is exactly an integer number of periods of the sinusoid the response is zero because the average value of any sinusoid over any integer number of periods is zero Otherwise the response is nonzero Convolution Properties Convolution in discrete time just as in continuous time is indicated by the operator yn xn hn m xmhn m 522 The properties of convolution in discrete time are similar to the properties in contin uous time xn Aδn n 0 Axn n 0 523 yn n 0 xn h n n 0 xn n 0 hn 524 The commutativity associativity distributivity differencing and sum properties of the convolution sum are proven in Web Appendix E and are summarized below Commutativity Property xn yn yn xn Associativity Property xn yn zn xn yn zn Distributivity Property xn yn zn xn zn yn zn If yn xn hn then Differencing Property yn yn 1 xn hn hn 1 Sum Property Sum of y Sum of x Sum of h For a convolution sum to converge both signals being convolved must be bounded and at least one of them must be absolutely summable Numerical Convolution DiscreteTime Numerical Convolution MATLAB has a command conv that com putes a convolution sum The syntax is y convxh where x and h are vectors of values of discretetime signals and y is the vector containing the values of the convo lution of x with h Of course MATLAB cannot actually compute an infinite sum as indicated by 522 MATLAB can only convolve timelimited signals and the vectors x and h should contain all the nonzero values of the signals they represent They can also contain extra zero values if desired If the time of the first element in x is nx0 and the time of the first element of h is nh0 the time of the first element of y is n x0 n h0 rob28124ch05164228indd 196 041216 127 pm 53 Discrete Time 197 If the time of the last element in x is n x1 and the time of the last element in h is n h1 the time of the last element in y is n x1 n h1 The length of x is n x1 n x0 1 and the length of h is n h1 n h0 1 So the extent of y is in the range n x0 n h0 n n x1 n h1 and its length is n x1 n h1 n x0 n h0 1 n x1 n x0 1 length of x n h1 n h0 1 length of h 1 So the length of y is one less than the sum of the lengths of x and h ExamplE 59 Computing a convolution sum with MATLAB Let xn un 1 un 6 and hn trin 64 Find the convolution sum xn hn using the MATLAB conv function xn is time limited to the range 1 n 5 and hn is time limited to the range 3 n 9 Therefore any vector describing xn should be at least five elements long and any vector de scribing hn should be at least seven elements long Lets put in some extra zeros compute the convolution and graph the two signals and their convolutions using the following MATLAB code whose output is illustrated in Figure 537 Figure 537 Excitation impulse response and response of a system found using the MATLAB conv command 2 0 2 4 6 8 10 12 14 16 18 20 0 1 2 3 4 n x 2 0 2 4 6 8 10 12 14 16 18 20 0 1 2 3 4 n h 2 0 2 4 6 8 10 12 14 16 18 20 0 1 2 3 4 n y nx 28 nh 012 Set time vectors for x and h x usDn1 usDn6 Compute values of x h trinh64 Compute values of h y convxh Compute the convolution of x with h Generate a discretetime vector for y ny nx1 nh1 0lengthnx lengthnh 2 rob28124ch05164228indd 197 041216 127 pm C h a p t e r 5 TimeDomain System Analysis 198 Graph the results subplot311 stemnxxkfilled xlabeln ylabelx axis22004 subplot312 stemnhhkfilled xlabeln ylabelh axis22004 subplot313 stemnyykfilled xlabeln ylabely axis22004 ContinuousTime Numerical Convolution At this point a natural question arises Since there is no builtin MATLAB function for doing a convolution integral can we do a convolution integral using the conv function The short answer is no But if we can accept a reasonable approximation and engineers usually can the longer answer is yes approximately We can start with the convolution integral yt xt ht xτht τdτ Approximate xt and ht each as a sequence of rectangles of width Ts xt n xn T s rect t n T s T s 2 T s and ht n hn T s rect t n T s T s 2 T s The integral can be approximated at discrete points in time as yn T s m x m T s hn m T s T s This can be expressed in terms of a convolution sum as yn T s T s m xmhn m T s x n hn 525 where xn xn T s and hn hn T s and the convolution integral can be approx imated as a convolution sum under the same criteria as in the use of the conv func tion to do convolution sums For the convolution integral to converge xt or ht or both must be an energy signal Let xt be nonzero only in the time interval n x0 T s t n x1 T s and let ht be nonzero only in the time interval n h0 T s t n h1 T s Then yt is nonzero only in the time interval n x0 n h0 T s n n x1 n h1 T s and the values of T s xn hn found using the conv function cover that range To get a reasonably good approximate convolution result Ts should be chosen such that the functions xt and ht dont change much during that time interval ExamplE 510 Graphing the convolution of two continuoustime signals using the MATLAB conv function Graph the convolution yt trit trit rob28124ch05164228indd 198 041216 127 pm 53 Discrete Time 199 Although this convolution can be done analytically it is rather tedious so this is a good candidate for an approximate convolution using numerical methods namely the conv function in MATLAB The slopes of these two functions are both either plus or minus one To make a reasonably accurate approximation choose the time between samples to be 001 seconds which means that the functions change value by no more than 001 between any two adjacent samples Then from 525 y001n 001 m tri 001m tri001n m The limits on the nonzero parts of the functions are 1 t 1 which translate into limits on the corresponding discretetime signals of 100 n 100 A MATLAB program that accom plishes this approximation is below Program to do a discretetime approximation of the convolution of two unit triangles Convolution computations Ts 001 Time between samples nx 10099 nh nx Discrete time vectors for x and h x trinxTs h trinhTs Generate x and h ny nx1nh1nxendnhend Discrete time vector for y y Tsconvxh Form y by convolving x and h Graphing and annotation p plotnyTsyk setpLineWidth2 grid on xlabelTime itt sFontNameTimesFontSize18 ylabelyittFontNameTimesFontSize18 titleConvolution of Two Unshifted Unit Triangle Functions FontNameTimesFontSize18 setgcaFontNameTimesFontSize14 The graph produced is Figure 538 Figure 538 Continuoustime convolution approximation using numerical methods 2 15 1 05 0 05 1 15 2 0 01 02 03 04 05 06 07 Time t s yt Convolution of Two Unshifted Unit Triangle Functions rob28124ch05164228indd 199 041216 127 pm C h a p t e r 5 TimeDomain System Analysis 200 This graphical result agrees very closely with the analytical solution yt 16 t 2 3 ut 2 4 t 1 3 ut 1 6 t 3 ut 4 t 1 3 ut 1 t 2 3 ut 2 Stability and Impulse Response Stability was generally defined in Chapter 4 by saying that a stable system has a bounded output signal when excited by any bounded input signal We can now find a way to determine whether a system is stable by examining its impulse response The convolution of two signals converges if both of them are bounded and at least one of them is absolutely summable The response yn of a system xn is yn xn hn If xn is bounded yn is bounded if hn is absolutely summable and therefore also bounded That is if n hn is bounded A system is BIBO stable if its impulse response is absolutely summable System Connections Two very common interconnections of systems are the cascade connection and the parallel connection Figure 539 and Figure 540 Figure 540 Parallel connection of two systems xn h1n h2n h1nh2n xnh1n ynxnh1nxnh2nxnh1nh2n xnh2n xn yn Figure 539 Cascade connection of two systems xn h1n xnh1n h2n ynxnh1nh2n xn h1nh2n yn Using the associativity property of convolution we can show that the cascade con nection of two systems can be considered as one system whose impulse response is the convolution of the two individual impulse responses of the two systems Using the rob28124ch05164228indd 200 041216 127 pm 53 Discrete Time 201 distributivity property of convolution we can show that the parallel connection of two systems can be considered as one system whose impulse response is the sum of the two individual impulse responses of the two systems UnitSequence Response and Impulse Response The response of any LTI system is the convolution of the excitation with the impulse response yn xn hn m xmhn m Let the excitation be a unit sequence and let the response to a unit sequence be desig nated h 1 n Then h 1 n un hn m umhn m m0 hn m Let q n m Then h 1 n qn hq q n hq So the response of a discretetime LTI system excited by a unit sequence is the accumulation of the impulse response Just as the unit sequence is the accumula tion of the impulse response the unitsequence response is the accumulation of the unitimpulse response The subscript on h 1 n indicates the number of differences In this case there is 1 difference or one accumulation in going from the impulse response to the unitsequence response This relationship holds for any excitation If any excitation is changed to its accumulation the response also changes to its accumu lation and if the excitation is changed to its first backward difference the response is also changed to its first backward difference ExamplE 511 Finding the response of a system using convolution Find the response of the system in Figure 541 to the excitation in Figure 542 Figure 541 A system xn yn 45 D Figure 542 Excitation of the system n gn 2 4 6 8 4 2 1 rob28124ch05164228indd 201 041216 127 pm C h a p t e r 5 TimeDomain System Analysis 202 First we need the impulse response of the system We could find it directly using the meth ods presented earlier but in this case since we have already found its unitsequence response h 1 n 5 4 45 n un see Chapter 4 section on discretetime system properties pages 152 and 153 we can find the impulse response as the first backward difference of the unitsequence response hn h 1 n h 1 n 1 Combining equations hn 5 4 45 n un 5 4 45 n1 un 1 hn 5un un 1 δn 4 45 n1 45un un 1 hn 5δn 4 45 n δn δn 45 n un 1 hn 45 n un All that remains is to perform the convolution We can do that using the MATLAB program below Program to demonstrate discretetime convolution nx 515 Set a discretetime vector for the excitation x trin33 Generate the excitation vector nh 020 Set a discretetime vector for the impulse response Generate the impulse response vector h 45nhusDnh Compute the beginning and ending discrete times for the system response vector from the discretetime vectors for the excitation and the impulse response nymin nx1 nh1 nymax nxlengthnx lengthnh ny nyminnymax1 Generate the system response vector by convolving the excitation with the impulse response y convxh Graph the excitation impulse response and system response all on the same time scale for comparison Graph the excitation subplot311 p stemnxxkfilled setpLineWidth2MarkerSize4 axisnyminnymax03 xlabeln ylabelxn Graph the impulse response subplot312 p stemnhhkfilled setpLineWidth2MarkerSize4 axisnyminnymax03 xlabeln ylabelhn Graph the system response subplot313 p stemnyykfilled rob28124ch05164228indd 202 041216 127 pm 53 Discrete Time 203 setpLineWidth2MarkerSize4 axisnyminnymax03 xlabeln ylabelyn The three signals as graphed by MATLAB are illustrated in Figure 543 Complex Exponential Excitation and the Transfer Function In engineering practice the most common form of description of a discretetime system is a difference equation or a system of difference equations Consider the general form of a difference equation for a discretetime system k0 N a k y n k k0 M b k x n k 526 A complex exponential excitation causes a complex exponential response in continuoustime systems and the same is true for discretetime systems Therefore if xn X z n yn has the form yn Y z n where X and Y are complex constants Then in the difference equation xn k X z nk z k X z n and yn k z k Y z n and 526 can be written in the form k0 N a k z k Y z n k0 M b k z k X z n Figure 543 Excitation impulse response and system response 5 0 5 10 15 20 25 30 35 0 1 2 3 n xn 5 0 5 10 15 20 25 30 35 0 1 2 3 n hn 5 0 5 10 15 20 25 30 35 n yn 0 1 2 3 rob28124ch05164228indd 203 041216 127 pm C h a p t e r 5 TimeDomain System Analysis 204 The Xzn and Yzn can be factored out leading to Y z n k0 N a k z k X z n k0 M b k z k Y z n X z n Y X k0 M b k z k k0 N a k z k This ratio YX is a ratio of polynomials in z This is the transfer function for dis cretetime systems symbolized by Hz That is Hz k0 M b k z k k0 N a k z k b 0 b 1 z 1 b 2 z 2 b M z M a 0 a 1 z 1 a 2 z 2 a N z N 527 and yn Y z n HzX z n Hzxn The transfer function can then be written di rectly from the difference equation and if the difference equation describes the system so does the transfer function By multiplying numerator and denominator of 527 by zN we can express it in the alternate form Hz k0 M b k z k k0 N a k z k z NM b 0 z M b 1 z M1 b M1 z b M a 0 z N a 1 z N1 a N1 z a N 528 The two forms are equivalent but either form may be more convenient in certain situations We can also find the system response using convolution The response yn of an LTI system with impulse response hn to a complex exponential excitation xn X z n is yn hn X z n X m hm z nm X z n xn m hm z m Equating the two forms of the response HzX z n X z n m hm z m Hz m hm z m which shows the relationship between the transfer function and the impulse response of discretetime LTI systems The summation m hm z m will be identified in Chapter 9 as the z transform of hn Frequency Response The variable z in the complex exponential zn is in general complex valued Consider the special case in which z is confined to the unit circle in the complex plane such that z 1 Then z can be expressed as z e jΩ where Ω is the real variable representing radian frequency in discrete time zn becomes e jΩn a discretetime complex sinusoid e jΩ cosΩ j sinΩ and the transfer function of the system Hz becomes the frequency response of the system H e jΩ From Y z n HzX z n letting z e jΩ Y e jΩn Y e jY e jΩn H e jΩ X e jΩn H e jΩ e jH e jΩ e jΩn X e jX e jΩn rob28124ch05164228indd 204 041216 127 pm 53 Discrete Time 205 or dividing through by e jΩn Y e jY H e jΩ X e jH e jΩ X Equating magnitudes we get Y H e jΩ X and equating phases we get Y H e jΩ X The function H e jΩ is called the frequency response of the system be cause at any radian frequency Ω if we know the magnitude and phase of the excitation and the magnitude and phase of the frequency response we can find the magnitude and phase of the response As was true for continuoustime systems if a complex excitation xn is applied to a system and causes a response yn then the real part of xn causes the real part of yn and the imaginary part of xn causes the imaginary part of yn Therefore if the actual excitation of a system is xn A x cosΩn θ x we can find the response of the system to an excitation x C n A x cosΩn θ x j A x sinΩn θ x A x e jΩn θ x in the form y C n A y cosΩn θ y j A y sinΩn θ y A y e jΩn θ y and we can take the real part yn A y cosΩn θ y as the response to the real exci tation xn A x cosΩn θ x Using Y Hjω X and Y Hjω X we get A y H e jΩ A x and θ y H e jΩ θ x ExamplE 512 Transfer Function and Frequency Response An LTI system is described by the difference equation yn 075 yn 1 025 yn 2 xn a Find its transfer function For this difference equation of the form k0 N a k y n k k0 M b k x n k N 2 M 0 a 0 025 a 1 075 a 2 1 and b 0 1 Therefore the transfer function is Hz 1 z 2 075z 025 b If xn X e j05n yt Y e j05n and X 12 e jπ4 find the magnitude and phase of Y The frequency response is H e jΩ 1 e jΩ 2 075 e jΩ 025 1 e j2Ω 075 e jΩ 025 see Figure 544 rob28124ch05164228indd 205 041216 127 pm C h a p t e r 5 TimeDomain System Analysis 206 The radian frequency is Ω 05 Therefore H e jΩ 1 e j 075 e j2 025 2001 e j1303 Y H e j05 12 2001 12 24012 Y H e j05 π4 13032 π4 20886 radians c If xn 25 cos2πn5 and yn A y cos2n5 θ y find A y and θ y A y H e jπ9 A x 12489 25 312225 and θ y H e j2π5 θ x 29842 0 29842 radians dW 2pi100 Increment in discretetime radian frequency for sampling the frequency response W 2pidW2pi Discretetime radian frequency vector for graphing the frequency response Compute the frequency response H 1expj2W 075expjW 025 close all figurePosition20201200800 subplot211 ptr plotWabsHk Graph the magnitude of the frequency response grid on setptrLineWidth2 xlabelOmegaFontNameTimesFontSize36 ylabelHiteitjOmegaFontNameTimesFontSize36 titleFrequency Response MagnitudeFontNameTimesFontSize36 setgcaFontNameTimesFontSize24 axis2pi2pi025 subplot212 ptr plotWangleHk Graph the phase of the frequency response grid on setptrLineWidth2 xlabelOmegaFontNameTimesFontSize36 ylabelPhase of HiteitjOmegaFontNameTimesFontSize36 titleFrequency Response PhaseFontNameTimesFontSize36 setgcaFontNameTimesFontSize24 axis2pi2pipipi rob28124ch05164228indd 206 041216 127 pm 207 Exercises with Answers 54 SUMMARY OF IMPORTANT POINTS 1 Every LTI system is completely characterized by its impulse response 2 The response of any LTI system to an arbitrary input signal can be found by convolving the input signal with its impulse response 3 The impulse response of a cascade connection of LTI systems is the convolution of the individual impulse responses 4 The impulse response of a parallel connection of LTI systems is the sum of the individual impulse responses 5 A continuoustime LTI system is BIBO stable if its impulse response is absolutely integrable 6 A discretetime LTI system is BIBO stable if its impulse response is absolutely summable EXERCISES WITH ANSWERS Answers to each exercise are in random order Continuous Time Impulse Response 1 A continuoustime system is described by the differential equation y t 6y t x t a Write the differential equation for the special case of impulse excitation and impulse response Ω 0 05 1 15 2 25 He jΩ Frequency Response Magnitude 6 4 2 6 6 4 2 0 2 4 0 2 4 6 Ω 2 0 2 Phase of He jΩ Frequency Response Phase Figure 544 Frequency response magnitude and phase rob28124ch05164228indd 207 041216 127 pm C h a p t e r 5 TimeDomain System Analysis 208 b The impulse response is h t 6 e 6t u t δ t What is the value of the integral from t 0 to t 0 of h t c What is the value of the integral from t 0 to t 0 of h t Answers 6 h t 6h t δ t 1 2 Find the values of these integrals a 0 0 4utdt b 0 0 2e3tut 7δt dt c 0 0 t 3δλdλ dt Answers 0 0 7 3 Find the impulse responses of the systems described by these equations a y t 5y t x t b y t 6 y t 4y t x t c 2 y t 3y t x t d 4 y t 9y t 2x t x t Answers h t 116 e 9t4 u t 14 δ t h t 34 e 3t2 u t 12 δ t h t e 5t u t h t 02237 e 076t e 523t u t 4 In the system of Figure E4 a 7 and b 3 xt yt b a Figure E4 a Write the differential equation describing it b The impulse response can be written in the form h t K e λt u t Find the numerical values of K and λ Answers 13 y t x t 73 y t 7 3 5 A continuoustime system is described by the differential equation y t 6 y t 3y t x t where x is the excitation and y is the response Can the impulse response of this system contain a An impulse b A discontinuity at t 0 c A discontinuous first derivative at t 0 Answers Yes No No rob28124ch05164228indd 208 041216 127 pm 209 Exercises with Answers 6 In Figure E6 is an RC lowpass filter with excitation v in t and response v out t Let R 10 Ω and C 10 μF vout t vin t it R C Figure E6 a Write the differential equation for this circuit in terms of v in t v out t R and C b Find the impulse response of this system h t c Find the numerical value h 200 μs Answers 135335 h t e tRC RC u t v out t v out t RC v in t RC Convolution 7 If x t 2 tri t4 δ t 2 find the values of a x 1 b x 1 Answers 12 32 8 If y t 3 rect t2 rect t 3 2 what are the maximum and minimum values of y for all time Answers 0 6 9 An LTI system has an impulse response h t 2 e 3t u t a Write an expression for h t u t b Let the excitation of the system be x t u t u t 13 Write an expression for the response y t c Find the numerical value of y t at t 12 Answers 02556 y t 23 1 e 3t u t 1 e 3 t13 u t 13 h t u t 23 1 e 3t u t 10 Let y t x t h t and let x t rect t 4 rect t 1 and let h t tri t 2 tri t 6 Find the range of times over which y t is not zero Answer 35 t 85 11 Find the following values a g 3 if g t 3 e 2t u t 4δ t 1 b g 3 if g t e t u t δ t 2δ t 1 c g 1 if g t 4 sin πt 8 δ t 4 d g 1 if g t 5 rect t 4 2 δ 3t e If y t x t h t and x t 4 rect t 1 and h t 3 rect t find y 12 Answers 02209 3696 02198 53 6 rob28124ch05164228indd 209 041216 127 pm C h a p t e r 5 TimeDomain System Analysis 210 12 If x t rect t10 3 rect t 1 8 find the following numerical values a x 1 b x 5 Answers 24 15 13 If x t 5 rect t2 δt 1 δt find the values of a x 12 b x 12 c x 52 Answers 10 5 0 14 Let x t 05 rect t 2 4 and h t 3δ 2t 5δ t 1 and let y t h t x t Graph yt Answers t 10 10 05 05 1 1 15 15 2 2 25 25 15 Graph g t a g t rect t rect t2 b g t rect t 1 rect t2 c g t rect t 5 rect t 5 rect t 4 rect t 4 Answers t gt 1 1 2 5 2 3 2 1 2 1 t gt 9 11 9 t gt 1 1 2 3 2 1 2 3 2 16 Graph these functions a g t rect 4t b g t rect4t 4δ t c g t rect 4t 4δ t 2 d g t rect 4t 4δ 2t e g t rect 4t δ 1 t f g t rect 4t δ 1 t 1 g g t 12 rect 4t δ 12 t h g t 12 rect t δ 12 t Answers t 1 1 8 1 8 gt 1 1 2 t 2 1 1 8 1 8 gt 1 1 2 rob28124ch05164228indd 210 041216 127 pm 211 Exercises with Answers t gt 1 1 2 1 t 1 1 8 1 8 gt 1 1 2 t 1 1 8 1 8 gt t 2 1 8 1 8 gt t 4 gt 2 1 2 8 1 8 2 t 4 1 8 1 8 gt 17 Graph these functions a g t rect t2 δ t 2 δ t 1 b g t rect t tri t c g t e t u t e t u t d g t tri 2 t 1 2 tri 2 t 1 2 δ 2 t e g t tri 2 t 12 tri 2 t 12 δ 1 t Answers t 1 5 gt 05 t 2 2 gt 1 t 3 3 gt 1 1 t 4 1 gt 1 1 t 3 3 gt 1 1 18 A system has an impulse response h t 4 e 4t u t Find and plot the response of the system to the excitation x t rect 2 t 14 Answer t 15 yt 1 1 rob28124ch05164228indd 211 041216 127 pm C h a p t e r 5 TimeDomain System Analysis 212 19 Change the system impulse response in Exercise 18 to h t δ t 4 e 4t u t and find and plot the response to the same excitation x t rect 2 t 14 Answer t 15 yt 1 1 20 Two systems have impulse responses h 1 t u t u t a and h 2 t rect t a2 a If these two systems are connected in cascade find the response y t of the overall system to the excitation x t δ t Answer h t 4tri t a a 21 In the circuit of Figure E21 the input signal voltage is v i t and the output signal voltage is v o t L R vit vot Figure E21 a Find the impulse response in terms of R and L b If R 10 kΩ and L 100 μH graph the unitstep response Answers 004 001 t μs vot 1 h t δ t RL e RtL u t 22 Graph the responses of the systems of Exercise 1 to a unit step Answers t 5 h1t 025 t 5 h1t 05 t 5 h1t 025 rob28124ch05164228indd 212 041216 127 pm 213 Exercises with Answers t 1 h1t 02 Stability 23 A continuoustime system has an impulse response rect t δ 8 t 1 δ 8 t 5 u t Is it BIBO stable Answer No 24 Below are the impulse responses of some LTI systems In each case determine whether or not the system is BIBO stable a h t sin t u t b h t e 12t sin 30πt u t c h t rect t δ 2 t rect t 200 d h t ramp t e h t δ 1 t e t10 u t f h t δ 1 t δ 1 t 12 u t Answers 4 BIBO Unstable and 2 BIBO Stable 25 Find the impulse responses of the two systems in Figure E25 Are these systems BIBO stable xt yt xt yt a b Figure E25 Answers One BIBO Stable and one BIBO Unstable h t u t h t e t u t 26 Find the impulse response of the system in Figure E26 Is this system BIBO stable xt yt Figure E26 A doubleintegrator system Answer No rob28124ch05164228indd 213 041216 127 pm C h a p t e r 5 TimeDomain System Analysis 214 27 Find the impulse response of the system in Figure E27 and evaluate its BIBO stability xt yt 01 005 Figure E27 A twointegrator system Answers 45893e005t sin 02179t u t BIBO Unstable Frequency Response 28 A continuoustime system is described by the differential equation 4 y t 2 y t 3 y t y t 8 x t x t 4x t Its transfer function can be written in the standard form H s k0 M b k s k k0 N a k s k b M s M b M1 s M1 b 2 s 2 b 1 s b 0 a N s N a N1 s N1 a 2 s 2 a 1 s a 0 Find the values of M N and all the a and b coefficients aN a0 and bM b0 Answers 1 8 1 4 3 2 3 4 2 29 A continuoustime system is described by 2 y t 4y t x t where x is the excitation and y is the response If x t X e jωt and y t Y e jωt and H jω Y X find the numerical value of H j2 Answer 01768 e j23562 Discrete Time Impulse Response 30 Find the total numerical solution of this difference equation with initial conditions yn 01y n 1 02y n 2 5 y 0 1 y 1 4 Answer yn 62223 05 n 00794 04 n 71429 31 Find the first three numerical values starting at time n 0 of the impulse response of the discretetime system described by the difference equation 9yn 3y n 1 2y n 2 xn Answers 19 127 181 rob28124ch05164228indd 214 041216 127 pm 215 Exercises with Answers 32 Find the impulse responses of the systems described by these equations a yn xn x n 1 b 25yn 6y n 1 y n 2 xn c 4yn 5y n 1 y n 2 xn d 2yn 6y n 2 xn x n 2 Answers hn cos 2214n 0644 20 5 n hn 3 n 2 cos πn2 un 13 u n 2 hn δn δ n 1 hn 13 112 14 n un 33 A discretetime system is described by the difference equation yn 095yn 2 xn where xn is the excitation and yn is the response a Find these values h 0 h 1 h 2 h 3 h 4 b What is the numerical value of h 64 Answers 1 095 0 0 09025 0194 34 If a discretetime system is described by yn m n4 xm graph its impulse response hn Answer n hn 10 10 1 Convolution 35 Two discretetime signals xn and h n are graphed in Figure E35 If yn xn hn graph yn 10 5 0 5 10 4 3 2 1 0 1 2 3 4 n 10 5 0 5 10 4 3 2 1 0 1 2 3 4 n hn xn Figure E35 rob28124ch05164228indd 215 041216 127 pm C h a p t e r 5 TimeDomain System Analysis 216 Answer 10 5 0 5 10 20 15 10 5 0 5 10 15 20 n yn 36 For each pair of signals x 1 n and x 2 n find the numerical value of yn x 1 n x 2 n at the indicated value of n a n 10 10 x1n 2 2 10 10 2 2 10 10 2 2 10 10 2 2 10 10 2 2 10 10 2 2 x1n x1n x2n x2n x2n n n n n n n 1 b n 10 10 x1n 2 2 10 10 2 2 10 10 2 2 10 10 2 2 10 10 2 2 10 10 2 2 x1n x1n x2n x2n x2n n n n n n n 3 c n 10 10 x1n 2 2 10 10 2 2 10 10 2 2 10 10 2 2 10 10 2 2 10 10 2 2 x1n x1n x2n x2n x2n n n n n n n 2 d x 1 n 3un and x 2 n ramp n 1 n 1 Answers 2 4 0 1 37 Find the numerical values of these functions a If gn 10 cos 2πn12 δ n 8 find g 4 b If gn u n 2 u n 3 δ n 1 2δ n 2 find g2 c If gn rampn un find g 3 d If gn un u n 5 δ 2 n find g 13 e If yn xn h n and xn rampn f If gn 10 cos 2πn 12 δ n 8 find g 4 rob28124ch05164228indd 216 041216 127 pm 217 Exercises with Answers g If gn u 2n 2 u 2n 3 δ n 1 2δ n 2 find g 2 h If yn xn hn and xn rampn Answers 2 2 10 10 2 1 1 6 38 If xn 08 n un un what is the numerical value of x 3 Answer 2952 39 Graph the convolution yn xn hn where xn un u n 4 and hn δn δ n 2 Answer n 5 10 yn 1 1 40 The impulse response hn of an LTI system is illustrated in Figure E40 Find the unit sequence response h 1 n of that system over the same time range hn hn 0 n 5 n 5 5 5 5 Figure E40 Answer n 5 4 3 2 1 0 1 2 3 4 5 h 1 n 0 0 0 0 0 1 4 3 1 2 4 41 Given the excitation xn sin 2πn32 and the impulse response hn 095 n u n find a closedform expression for and plot the system response yn Answers yn 50632 sin 2πn32 1218 n 5 40 xn 1 1 Excitation 1 n 5 40 hn 1 Impulse Response n 5 40 yn 5 5 Response rob28124ch05164228indd 217 041216 127 pm C h a p t e r 5 TimeDomain System Analysis 218 42 Given the excitations xn and the impulse responses hn use MATLAB to plot the system responses yn a xn un u n 8 hn sin 2πn8 un u n 8 b xn sin 2πn8 un u n 8 hn sin 2πn8 un u n 8 Answers n 5 30 xn 1 Excitation n 5 30 hn 1 1 Impulse Response n 5 30 yn 3 3 Response n 5 30 xn 1 1 Excitation n 5 30 hn 1 1 Impulse Response n 5 30 yn 3 3 Response 43 Two systems have impulse responses h 1 n 09 n un and h 2 n δn 09 n un When these two systems are connected in parallel what is the response yn of the overall system to the excitation xn un Answer yn un 44 A discretetime system with impulse response h n 3 un u n 4 is excited by the signal xn 2 u n 2 u n 10 and the system response is yn a At what discrete time n does the first nonzero value of yn occur b At what discrete time n does the last nonzero value of yn occur c What is the maximum value of yn over all discrete time d Find the signal energy of yn Answers 2 12 24 3888 45 Find and plot the unitsequence responses of the systems in Figure E43 a xn yn 07 05 D D rob28124ch05164228indd 218 041216 127 pm 219 Exercises with Answers b 08 06 xn yn D D D Figure E43 Answers n 5 20 h1n 3 UnitSequence Response n 5 20 h1n 3 UnitSequence Response Stability 46 A discretetime system is described by yn 18y n 1 12y n 2 xn Is it BIBO stable Answer No 47 What numerical ranges of values of A and B make the system in Figure E47 BIBO stable xn yn D A B Figure E47 Answers A 1 any B rob28124ch05164228indd 219 041216 127 pm C h a p t e r 5 TimeDomain System Analysis 220 48 Below are the impulse responses of some LTI systems In each case determine whether or not the system is BIBO stable a hn 11 n un b hn un c hn tri n 4 2 d hn δ 10 nun e hn sin 2πn6 un Answers 3 BIBO Unstable and 2 BIBO Stable 49 Which of the systems in Figure E49 are BIBO stable a 09 xn yn D b 11 xn yn D c xn yn D D 05 05 d xn yn D D 15 04 Figure E49 Answers 2 BIBO Stable and 2 BIBO Unstable rob28124ch05164228indd 220 041216 127 pm 221 Exercises without Answers EXERCISES WITHOUT ANSWERS Continuous Time Impulse Response 50 Find the impulse responses of the systems described by these equations a 4 y t 2x t x t b y t 9y t 6 x t c y t 3 y t 3x t 5 x t 51 Refer to the system of Figure E51 with a 5 and b 2 xt yt a b Figure E51 a Write the differential equation for the system b The impulse response of this system can be written in the form h t K 1 e s 1 t K 2 e s 2 t u t Find the values of K 1 K 2 s 2 and s 2 c Is this system BIBO stable 52 Let y t x t h t and let x t rect t 4 rect t 1 and let h t tri t 2 tri t 6 a What is the lowest value of t for which y t is not zero b What is the highest value of t for which y t is not zero 53 The impulse response of a continuoustime system described by a differential equation consists of a linear combination of one or more functions of the form K e st where s is an eigenvalue and in some cases an impulse of the form K δ δ t How many constants of each type are needed for each system a a y t by t cx t a b and c are constants b a y t by t c x t a b and c are constants c a y t by t c x t a b and c are constants d a y t by t c x t d x t a b c and d are constants 54 A rectangular voltage pulse which begins at t 0 is 2 seconds wide and has a height of 05 V drives an RC lowpass filter in which R 10 kΩ and C 100 μF a Graph the voltage across the capacitor versus time b Change the pulse duration to 02 s and the pulse height to 5 V and repeat rob28124ch05164228indd 221 041216 127 pm C h a p t e r 5 TimeDomain System Analysis 222 c Change the pulse duration to 2 ms and the pulse height to 500 V and repeat d Change the pulse duration to 2 ms and the pulse height to 500 kV and repeat Based on these results what do you think would happen if you let the input voltage be a unit impulse Convolution 55 a If x t tri t3 δ t 2 and y t x 2t what is the numerical range of values of t for which y t is not zero b If x t tri tw δ t t 0 and y t x at what is the range of values of t in terms of w t 0 and a for which y t is not zero 56 What function convolved with 2 cos t would produce 6 sin t 57 Graph these functions a g t 3 cos 10πt 4δ t 110 b g t tri 2t δ 1 t c g t 2 tri 2t rect t 1 δ 2 t d g t 8 tri t4 δ 1 t δ 8 t e g t e 2t u t δ 4 t δ 4 t 2 58 For each graph in Figure E58 select the corresponding signal or signals from the group x 1 t x 8 t The corresponding signal may not be one of the choices A through E x 1 t δ 2 t rect t2 x 2 t 4 δ 2 t rect t2 x 3 t 14 δ 12 t rect t2 x 4 t δ 12 t rect t2 x 5 t δ 2 t rect 2t x 6 t 4 δ 2 t rect 2t x 7 t 14 δ 12 t rect 2t x 8 t δ 12 t rect 2t 4 3 2 1 1 2 3 4 1 32 1 4 2 3 4 1 32 1 4 2 3 4 1 32 1 4 2 3 4 1 32 1 4 2 3 4 1 32 1 4 2 3 4 t 4 3 2 1 1 2 3 4 t 4 3 2 1 1 2 3 4 t 4 3 2 1 1 2 3 4 t 4 3 2 1 1 2 3 4 t A B C D E Figure E58 rob28124ch05164228indd 222 041216 127 pm 223 Exercises without Answers 59 Find the average signal power of these signals a x t 4 rect t δ 4 t b x t 4 tri t δ 4 t 60 If x t u t u t 4 δ t δ t 2 what is the signal energy of x 61 A continuoustime system with impulse response h t 5 rect t is excited by x t 4 rect 2t a Find the response y t at time t 12 b Change the excitation from part a to x a t x t 1 and keep the same impulse response What is the new response y a t at time t 12 c Change the excitation from part a to x b t d dt x t and keep the same impulse response What is the new response y b t at time t 12 62 Write a functional description of the timedomain signal in Figure E62 as the convolution of two functions of t t 1 12 12 2 2 Figure E62 63 In Figure E631 are four continuouostime functions a b c and d In Figure E632 are 20 possible convolutions of pairs of these functions including convolution of a functions with itself For each convolution find the pair of signals convolved to produce it a b c d 6 4 2 0 2 4 6 4 2 0 2 4 t 6 4 2 0 2 4 6 4 2 0 2 4 t 6 42 0 2 4 6 4 2 0 2 4 t 6 4 2 0 2 4 6 4 2 0 2 4 t Figure E631 rob28124ch05164228indd 223 041216 127 pm C h a p t e r 5 TimeDomain System Analysis 224 6 4 2 0 2 4 6 8 4 0 4 8 t 6 4 2 0 2 4 6 16 8 0 8 16 t 6 4 2 0 2 4 6 8 4 0 4 8 4 t 6 4 2 0 2 4 6 4 2 0 2 t 6 4 2 0 2 4 6 16 8 0 8 16 t 64 2 0 2 4 6 32 16 0 16 32 t 6 4 2 0 2 4 6 2 1 0 1 2 t 6 4 2 0 2 4 6 16 8 0 8 16 t 6 4 2 0 2 4 6 16 8 0 8 16 t 6 4 2 0 2 4 6 4 2 0 2 4 t 6 4 2 0 2 4 6 4 2 0 2 4 4 t 6 4 2 0 2 4 6 4 2 0 2 t 6 4 2 0 2 4 6 4 2 0 2 4 2 8 t 6 4 2 0 2 4 6 2 1 0 1 t 6 4 2 0 2 4 6 8 4 0 4 t 6 4 2 0 2 4 6 8 4 0 4 8 t 6 4 2 0 2 4 6 8 4 0 4 8 t 6 4 2 0 2 4 6 8 4 0 4 8 t 6 4 2 0 2 4 6 16 8 0 8 16 t 6 4 2 0 2 4 6 2 1 0 1 2 t Figure E632 Stability 64 Write the differential equations for the systems in Figure E64 find their impulse responses and determine whether or not they are BIBO stable For each system a 05 and b 01 Then comment on the effect on system BIBO stability of redefining the response a b xt yt a b xt yt a b Figure E64 65 Find the impulse response h t of the continuoustime system described by y t βy t x t For what values of β is this system BIBO stable rob28124ch05164228indd 224 041216 127 pm 225 Exercises without Answers 66 Find the impulse response of the system in Figure E66 and evaluate its BIBO stability xt yt 2 3 1 8 Figure E66 67 For each impulse response indicate whether the LTI system it describes is BIBO stable or BIBO unstable a hn sin 2πn6 un b h t ramp t c h t δ 1 t e t10 u t d h t δ 1 t δ 1 t 12 u t Discrete Time Impulse Response 68 A discretetime system is described by the difference equation 7yn 3y n 1 y n 2 11 a The eigenvalues of this difference equation can be expressed in the polar form A e jθ where A is the magnitude and θ is the angle or phase Find the values of A and θ b The homogeneous solution approaches zero as n What value does yn approach as n 69 Find the impulse responses of the systems described by these equations a 3yn 4y n 1 y n 2 xn x n 1 b 52 yn 6y n 1 10y n 2 xn Convolution 70 Graph gn Verify with the MATLAB conv function a gn u n 1 u n 2 sin 2πn9 b gn u n 2 u n 3 sin 2πn9 c gn u n 4 u n 5 sin 2πn9 d gn u n 3 u n 4 u n 3 u n 4 δ 14 n e gn u n 3 u n 4 u n 3 u n 4 δ 7 n f gn 2 cos 2πn7 78 n un rob28124ch05164228indd 225 041216 127 pm C h a p t e r 5 TimeDomain System Analysis 226 71 If xn u n 4 u n 3 δ n 3 and yn x n 4 what is the range of values for which yn is not zero 72 Find the signal power of the discretetime function 4 sinc n δ 3 n 73 Given the function graphs 1 through 4 in Figure E731 match each convolution expression a through j to one of the functions a through h in Figure E732 if a match exists n 10 10 g1n 2 2 n 10 10 g2n 2 2 n 10 10 g3n 2 2 n 10 10 g4n 2 2 Figure E731 a g 1 n g 1 n b g 2 n g 2 n c g 3 n g 3 n d g 4 n g 4 n e g 1 n g 2 n f g 1 n g 3 n g g 1 n g 4 n h g 2 n g 3 n i g 2 n g 4 n j g 3 n g 4 n a b c d n 10 10 2 2 n 10 10 2 2 n 10 10 2 2 n 10 10 2 2 e f g h n 10 10 2 2 n 10 10 2 2 n 10 10 2 2 n 10 10 2 2 Figure E732 74 In Figure E741 are six discretetime functions All of them are zero outside the range graphed In Figure E742 are 15 candidate convolution results All of them can be formed by convolving functions AF in pairs For each convolution result identify the two functions convolved to obtain it rob28124ch05164228indd 226 041216 128 pm 227 Exercises without Answers 10 0 10 n 4 2 0 2 4 A 10 0 10 n 4 2 0 2 4 B 10 0 10 n 4 2 0 2 4 C 10 0 10 n 4 2 0 2 4 D 10 0 10 n 4 2 0 2 4 E 10 0 10 n 4 2 0 2 4 F Figure E741 10 0 10 n 4 2 0 2 4 9 10 0 10 n 4 2 0 2 4 13 10 0 10 n 4 2 0 2 4 12 10 0 10 n 4 2 0 2 4 10 10 0 10 n 4 2 0 2 4 3 10 0 10 n 4 2 0 2 4 5 10 0 10 n 4 2 0 2 4 11 10 0 10 n 4 2 0 2 4 15 10 0 10 n 4 2 0 2 4 8 10 0 10 n 4 2 0 2 4 1 10 0 10 n 4 2 0 2 4 6 10 0 10 n 4 2 0 2 4 2 10 0 10 n 4 2 0 2 4 7 10 0 10 n 4 2 0 2 4 4 10 0 10 n 4 2 0 2 4 14 Figure E742 rob28124ch05164228indd 227 041216 128 pm C h a p t e r 5 TimeDomain System Analysis 228 75 A system with an impulse response h 1 n a n un is cascade connected with a second system with impulse response h 2 n b n un If a 12 and b 23 and hn is the impulse response of the overall cascadeconnected system find h3 76 Find the impulse responses of the subsystems in Figure E75 and then convolve them to find the impulse response of the cascade connection of the two subsystems xn y1n D D y2n 08 Figure E75 Two cascaded subsystems 77 Given the excitations xn and the impulse responses hn find closedform expressions for and graph the system responses yn a xn un hn n 78 n un b xn un hn 47 δn 34 n un Stability 78 A discretetime LTI system has an impulse response of the form hn 0 n 0 a 0 a 0 a 0a 0 n 0 The pattern repeats forever in positive time For what range of values of a is the system BIBO stable 79 A system is excited by a unitramp function and the response is unbounded From these facts alone it is impossible to determine whether the system is BIBO stable or not Why 80 The impulse response of a system is zero for all negative time and for n 0 it is the alternating sequence 1 1 1 1 1 1 which continues forever Is it BIBO stable rob28124ch05164228indd 228 041216 128 pm 229 61 INTRODUCTION AND GOALS In Chapter 5 we learned how to find the response of an LTI system by expressing the excitation as a linear combination of impulses and expressing the response as a linear combination of impulse responses We called that technique convolution This type of analysis takes advantage of linearity and superposition and breaks one complicated analysis problem into multiple simpler analysis problems In this chapter we will also express an excitation as a linear combination of simple signals but now the signals will be sinusoids The response will be a linear combination of the responses to those sinusoids As we showed in Chapter 5 the re sponse of an LTI system to a sinusoid is another sinusoid of the same frequency but with a generally different amplitude and phase Expressing signals in this way leads to the frequency domain concept thinking of signals as functions of frequency instead of time Analyzing signals as linear combinations of sinusoids is not as strange as it may sound The human ear does something similar When we hear a sound what is the actual response of the brain As indicated in Chapter 1 the ear senses a time variation of air pressure Suppose this variation is a singlefrequency tone like the sound of a person whistling When we hear a whistled tone we are not aware of the very fast oscillation of air pressure with time Rather we are aware of three important charac teristics of the sound its pitch a synonym for frequency its intensity or amplitude and its duration The earbrain system effectively parameterizes the signal into three simple descriptive parameters pitch intensity and duration and does not attempt to follow the rapidly changing and very repetitive air pressure in detail In doing so the earbrain system has distilled the information in the signal down to its essence The mathematical analysis of signals as linear combinations of sinusoids does something similar but in a more mathematically precise way Looking at signals this way also lends new insight into the nature of systems and for certain types of systems greatly simplifies designing and analyzing them CH APTER GOAL S 1 To define the Fourier series as a way of representing periodic signals as linear combinations of sinusoids C H A P T E R 6 ContinuousTime Fourier Methods rob28124ch06229306indd 229 041216 130 pm C h a p t e r 6 ContinuousTime Fourier Methods 230 2 To derive using the concept of orthogonality the methods for transforming signals back and forth between time and frequency descriptions 3 To determine the types of signals that can be represented by the Fourier series 4 To develop and learn to use the properties of the Fourier series 5 To generalize the Fourier series to the Fourier transform which can represent aperiodic signals 6 To generalize the Fourier transform so it can apply to some very common useful signals 7 To develop and learn to use the properties of the Fourier transform 8 To see through examples some of the uses of the Fourier series and the Fourier transform 62 THE CONTINUOUSTIME FOURIER SERIES CONCEPTUAL BASIS A common situation in signal and system analysis is an LTI system excited by a peri odic signal A very important result from Chapter 5 is that if an LTI system is excited by a sinusoid the response is also a sinusoid with the same frequency but generally a different amplitude and phase This occurs because the complex exponential is the eigenfunction of the differential equations describing LTI systems and a sinusoid is a linear combination of complex exponentials An important result from Chapter 4 is that if an LTI system is excited by a sum of signals the overall response is the sum of the responses to each of the signals in dividually If we could find a way to express arbitrary signals as linear combinations of sinusoids we could use superposition to find the response of any LTI system to any arbitrary signal by summing the responses to the individual sinusoids The represen tation of a periodic signal by a linear combination of sinusoids is called a Fourier1 series The sinusoids can be real sinusoids of the form A cos2πtT0 θ or they can be complex sinusoids of the form A e j2πt T 0 When first introduced to the idea of expressing real signals as linear combinations of complex sinusoids students are often puzzled as to why we would want to introduce the extra and seemingly unnecessary dimension of imaginary numbers and functions Eulers identity e jx cosx j sinx illustrates the very close relationship between real and complex sinusoids It will turn out that because of the compact notation that results and because of certain mathematical simplifications that occur when using complex sinusoids they are actually more convenient and powerful in analysis than real sinusoids So the reader is encouraged to suspend disbelief for a while until the power of this method is revealed 1 Jean Baptiste Joseph Fourier was a French mathematician of the late 18th and early 19th centuries The name Fourier is commonly pronounced foreyay because of its similarity to the English word four but the proper French pronunciation is fooryay where foor rhymes with tour Fourier lived in a time of great turmoil in France the French Revolution and the reign of Napoleon Bonaparte Fourier served as secretary of the Paris Academy of Science In studying the propagation of heat in solids Fourier developed the Fourier series and the Fourier integral When he first presented his work to the great French mathematicians of the time Laplace LaGrange and LaCroix they were intrigued by his theories but they especially LaGrange thought his theories lacked mathematical rigor The publication of his paper at that time was denied Some years later Dirichlet put the theories on a firmer foundation explaining exactly what functions could and could not be expressed by a Fourier series Then Fourier published his theories in what is now a classic text Theorie analytique de la chaleur rob28124ch06229306indd 230 041216 130 pm 62 The ContinuousTime Fourier Series 231 t 4 10 Sinusoid 1 06 06 t 10 xt Exact xt 1 Approximation of xt through 1 sinusoid 4 16 t0T t0 Figure 63 Signal approximated by a constant plus a single sinusoid xt A1e j2πtT1 A2e j2πtT2 A3e j2πtT3 A1e j2πtT1 yt yt ht ht ht ht A2e j2πtT2 A3e j2πtT3 B1e j2πtT1 B2e j2πtT2 B3e j2πtT3 Figure 61 The equivalence of the response of an LTI system to an excitation signal and the sum of the systems responses to complex sinusoids whose sum is equivalent to the excitation t 4 10 Constant 06 06 t 4 10 xt 16 Exact xt 1 Approximation of xt by a constant t0T t0 Figure 62 Signal approximated by a constant If we are able to express an excitation signal as a linear combination of sinusoids we can take advantage of linearity and superposition and apply each sinusoid to the system one at a time and then add the individual responses to obtain the overall response Figure 61 Consider an arbitrary original signal xt that we would like to represent as a linear combination of sinusoids over a range of time from an initial time t 0 to a final time t 0 T as illustrated by the dashed line in Figure 62 In this illustration we will use realvalued sinusoids to make the visualization as simple as possible In Figure 62 the signal is approximated by a constant 05 which is the average value of the signal in the interval t 0 t t 0 T A constant is a special case of a si nusoid a cosine of zero frequency This is the best possible approximation of xt by a constant Best in this case means having the minimum meansquared error between xt and the approximation Of course a constant even the best one is not a very good approximation to this signal We can make the approximation better by adding to the constant a sinusoid whose fundamental period is the same as the fundamental period of xt Figure 63 This approximation is a big improvement on the previous one and is the best approximation that can be made using a constant and a single sinusoid of rob28124ch06229306indd 231 041216 130 pm C h a p t e r 6 ContinuousTime Fourier Methods 232 t 4 10 Sinusoid 2 06 06 t 4 10 xt Exact xt 1 Approximation of xt through 2 sinusoids t0T t0 Figure 64 Signal approximated by a constant plus two sinusoids Figure 65 Signal approximated by a constant plus three sinusoids t 4 10 Sinusoid 3 06 06 t 4 10 xt Exact xt 1 Approximation of xt through 3 sinusoids t0T t0 the same fundamental frequency as xt We can improve the approximation further by adding a sinusoid at a frequency of twice the fundamental frequency of xt Figure 64 If we keep adding properly chosen sinusoids at higher integer multiples of the fundamental frequency of xt we can make the approximation better and better and in the limit as the number of sinusoids approaches infinity the approximation becomes exact Figure 65 and Figure 66 The sinusoid added at three times the fundamental frequency of xt has an ampli tude of zero indicating that a sinusoid at that frequency does not improve the approxi mation After the fourth sinusoid is added the approximation is quite good being hard to distinguish in Figure 66 from the exact xt In this example the representation approaches the original signal in the represen tation time t 0 t t 0 T and also for all time because the fundamental period of the approximation is the same as the fundamental period of xt The most general rob28124ch06229306indd 232 041216 130 pm 62 The ContinuousTime Fourier Series 233 Figure 66 Signal approximated by a constant plus four sinusoids t 4 10 Sinusoid 4 06 06 t 4 10 xt Exact xt 1 Approximation of xt through 4 sinusoids t0T t0 application of Fourier series theory represents a signal in the interval t 0 t t 0 T but not necessarily outside that interval But in signal and system analysis the representa tion is almost always of a periodic signal and the fundamental period of the representa tion is almost always chosen to also be a period of the signal so that the representation is valid for all time not just in the interval t 0 t t 0 T In this example the signal and the representation have the same fundamental period but more generally the fun damental period of the representation can be chosen to be any period fundamental or not of the signal and the representation will still be valid everywhere Each of the sinusoids used in the approximations in the example above is of the form A cos2πktT θ Using the trigonometric identity cosa b cosa cosb sina sinb we can express the sinusoid in the form A cos2πktT θ A cosθ cos2πktT A sinθ sin2πktT This demonstrates that each phaseshifted cosine can be expressed also as the sum of an unshifted cosine and an unshifted sine of the same fundamental period if the am plitudes are correctly chosen The linear combination of all those sinusoids expressed as cosines and sines is called the continuoustime Fourier series CTFS and can be written in the form xt a x 0 k1 a x k cos2πktT b x k sin2πktT where a x 0 is the average value of the signal in the representation time k is the harmonic number and a x k and b x k are functions of k called harmonic functions Here we use the notation to enclose the argument k because harmonic number is always an integer The harmonic functions set the amplitudes of the sines and cosines and k determines the frequency So the higherfrequency sines and cosines have frequencies that are integer multiples of the fundamental frequency and the multiple is k The function cos2πktT is the kthharmonic cosine Its fundamental period is Tk and its fundamental cyclic frequency is kT Representing a signal this way as a linear combination of realvalued cosines and sines is called the trigonometric form of the CTFS rob28124ch06229306indd 233 041216 130 pm C h a p t e r 6 ContinuousTime Fourier Methods 234 For our purposes it is important as a prelude to later work to see the equivalence of the complex form of the CTFS Every realvalued sine and cosine can be replaced by a linear combination of complex sinusoids of the forms cos2πkt T e j2πktT e j2πktT 2 and sin2πktT e j2πktT e j2πktT j2 If we add the cosine and sine with amplitudes a x k and b x k respectively at any par ticular harmonic number k we get a x k cos2πktT b x k sin2πktT a x k e j2πktT e j2πktT 2 b x k e j2πktT e j2πktT j2 We can combine like complexsinusoid terms on the righthand side to form a x k cos2πktT b x k sin2πktT 1 2 a x k j b x k e j2πktT a x k j b x k e j2πktT Now if we define c x 0 a x 0 c x k a x k j b x k 2 k 0 and c x k c x k we can write a x k cos2πktT b x k sin2πktT c x k e j2πktT c x k e j2π k tT k 0 and we have the amplitudes c x k of the complex sinusoids e j2πktT at positive and also negative integer multiples of the fundamental cyclic frequency 1T The sum of all these complex sinusoids and the constant c x 0 is equivalent to the original function just as the sum of the sines and cosines and the constant was in the previous representation To include the constant term c x 0 in the general formulation of complex sinusoids we can let it be the zeroth k 0 harmonic of the fundamental Letting k be zero the complex sinusoid e j2πktT is just the number 1 and if we multiply it by a correctly chosen weighting factor c x 0 we can complete the complex CTFS representation It will turn out in the material to follow that the same general formula for finding c x k for any nonzero k can also be used without modification to find c x 0 and that c x 0 is simply the average value in the representation time t 0 t t 0 T of the function to be represented c x k is the complex harmonic function of xt The complex CTFS is more efficient than the trigonometric CTFS because there is only one harmonic function instead of two The CTFS representation of the function can be written compactly in the form xt k c x k e j2πktT 61 So far we have asserted that the harmonic function exists but have not indicated how it can be found That is the subject of the next section ORTHOGONALITY AND THE HARMONIC FUNCTION In the Fourier series the values of c x k determine the magnitudes and phases of com plex sinusoids that are mutually orthogonal Orthogonal means that the inner product rob28124ch06229306indd 234 041216 130 pm 62 The ContinuousTime Fourier Series 235 of the two functions of time on some time interval is zero An inner product is the integral of the product of one function and the complex conjugate of the other function over an interval in this case the time interval T For two functions x1 and x2 that are orthogonal on the interval t 0 t t 0 T x1t x 2 t inner product t 0 t 0 T x1t x 2 t dt 0 We can show that the inner product of one complex sinusoid e j2πktT and another complex sinusoid e j2πqtT on the interval t 0 t t 0 T is zero if k and q are integers and k q The inner product is e j2πktT e j2πqtT t 0 t 0 T e j2πktT e j2πqtT dt t 0 t 0 T e j2πkqtT dt Using Eulers identity e j2πktT e j2πqtT t 0 t 0 T cos 2π k q T t j sin 2π k q T t dt 62 Since k and q are both integers if k q the cosine and the sine in this integral are both being integrated over a period an integer number of fundamental periods The defi nite integral of any sinusoid of nonzero frequency over any period is zero If k q the integrand is cos0 sin0 1 and the inner product is T If k q the inner product 62 is zero So any two complex sinusoids with an integer number of fun damental periods on the interval t 0 t t 0 T are orthogonal unless they have the same number of fundamental periods Then we can conclude that functions of the form e j2πktT k constitute a countably infinite set of functions all of which are mutually orthogonal on the interval t 0 t t 0 T where t 0 is arbitrary We can now take advantage of orthogonality by multiplying the expression for the Fourier series xt k c x k e j2πktT through by e j2πqtT q an integer yielding xt e j2πqtT k c x k e j2πktT e j2πqtT k c x k e j2πkqtT If we now integrate both sides over the interval t 0 t t 0 T we get t 0 t 0 T xt e j2πqtT dt t 0 t 0 T k c x k e j2πkqtT dt Since k and t are independent variables the integral of the sum on the right side is equivalent to a sum of integrals The equation can be written as t 0 t 0 T xt e j2πqtT dt k c x k t 0 t 0 T e j2πkqtT dt and using the fact that the integral is zero unless k q the summation k c x k t 0 t 0 T e j2πkqtT dt rob28124ch06229306indd 235 041216 130 pm C h a p t e r 6 ContinuousTime Fourier Methods 236 reduces to cxqT and t 0 t 0 T xt e j2πqtT dt cxqT Solving for cxq cxq 1 T t 0 t 0 T xt e j2πqtT dt If this is a correct expression for cxq then cxk in the original Fourier series expres sion 61 must be cxk 1 T t 0 t 0 T xt e j2πktT dt 63 From this derivation we conclude that if the integral in 63 converges a periodic signal xt can be expressed as xt k cxk e j2πktT 64 where cxk 1 T T xt e j2πktT dt 65 and the notation T means the same thing as t 0 t 0 T with t 0 arbitrarily chosen Then xt and cxk form a CTFS pair which can be indicated by the notation xt ℱ𝒮 T c x k where the ℱ𝒮 means Fourier series and the T means that cxk is computed with T as the fundamental period of the CTFS representation of xt This derivation was done on the basis of using a period T of the signal as the in terval of orthogonality and also as the fundamental period of the CTFS representation T could be any period of the signal including its fundamental period T 0 In practice the most commonly used fundamental period of the representation is the fundamental period of the signal T 0 In that special case the CTFS relations become xt k cxk e j2πkt T 0 and c x k 1 T 0 T 0 xt e j2πkt T 0 dt f 0 T 0 xt e j2πk f 0 t dt where f 0 1 T 0 is the fundamental cyclic frequency of xt If the integral of a signal xt over the time interval t 0 t t 0 T diverges a CTFS cannot be found for the signal There are two other conditions on the applicability of rob28124ch06229306indd 236 041216 130 pm 62 The ContinuousTime Fourier Series 237 the CTFS which together with the condition on the convergence of the integral are called the Dirichlet conditions The Dirichlet conditions are the following 1 The signal must be absolutely integrable over the time t 0 t t 0 T That is t 0 t 0 T xt dt 2 The signal must have a finite number of maxima and minima in the time t 0 t t 0 T 3 The signal must have a finite number of discontinuities all of finite size in the time t 0 t t 0 T There are hypothetical signals for which the Dirichlet conditions are not met but they have no known engineering use THE COMPACT TRIGONOMETRIC FOURIER SERIES Consider the trigonometric Fourier series xt a x 0 k1 a x k cos2πktT b x k sin2πktT Now using A cosx B sinx A 2 B 2 cosx tan 1 BA we have xt ax0 k1 a x 2 k b x 2 k cos 2πktT tan 1 b x k a x k or xt dx0 k1 dxk cos2πktT θ x k where dx0 a x 0 d x k a x 2 k b x 2 k k 0 and θ x k tan 1 b x k a x k k 0 This is the socalled compact trigonometric Fourier series It is also expressed in purely realvalued functions and coefficients and is a little more compact than the trigonometric form but it is still not as compact or efficient as the complex form xt k cxk e j2πktT The trigonometric form is the one actually used by Jean Baptiste Joseph Fourier rob28124ch06229306indd 237 041216 130 pm C h a p t e r 6 ContinuousTime Fourier Methods 238 t sinct 1 5 4 3 2 1 2 3 4 5 1 Figure 67 The unit sinc function ExamplE 61 CTFS harmonic function of a rectangular wave Find the complex CTFS harmonic function of xt Arecttw δ T 0 t w T 0 using its fundamental period as the representation time The fundamental period is T 0 so the CTFS harmonic function is cxk 1 T 0 T 0 Arecttw δ T 0 t e j2πkt T 0 dt The integration interval can be anywhere in time as long as its length is T 0 For convenience choose to integrate over the interval T 0 2 t T 0 2 Then cxk A T 0 T 0 2 T 0 2 recttw δ T 0 t e j2πkt T 0 dt Using w T 0 and the fact that the interval contains only one rectangle function cxk A T 0 T 0 2 T 0 2 recttw e j2πkt T 0 dt A T 0 w2 w2 e j2πkt T 0 dt cxk A T 0 e j2πkt T 0 j2πk T 0 w2 w2 A e jπkw T 0 e jπkw T 0 j2πk A sinπkw T 0 πk and finally xt Arecttw δ T 0 t ℱ𝒮 T 0 cxk A sinπkw T 0 πk Even though in this example we restricted w to be less than T 0 to simplify the analysis the result is also correct for w greater than T 0 In Example 61 the harmonic function turned out to be cxk A sinπkw T 0 πk This mathematical form of the sine of a quantity divided by the quantity itself occurs often enough in Fourier analysis to deserve its own name We now define the unitsinc func tion Figure 67 as sinct sinπt πt 66 rob28124ch06229306indd 238 041216 130 pm 62 The ContinuousTime Fourier Series 239 We can now express the harmonic function from Example 61 as cxk Aw T 0 sinckw T 0 and the CTFS pair as xt A recttw δ T 0 t ℱ𝒮 T 0 cxk Aw T 0 sincwk T 0 The unitsinc function is called a unit function because its height and area are both one2 One common question when first encountering the sinc function is how to determine the value of sinc0 When the independent variable t in sinπtπt is zero both the numerator sinπt and the denominator πt evaluate to zero leaving us with an indeterminate form The solution to this problem is to use LHôpitals rule Then lim t0 sinct lim t0 sinπt πt lim t0 π cosπt π 1 So sinct is continuous at t 0 and sinc0 1 CONVERGENCE Continuous Signals In this section we will examine how the CTFS summation approaches the signal it represents as the number of terms used in the sum approaches infinity We do this by examining the partial sum x N t kN N cxk e j2πktT for successively higher values of N As a first example consider the CTFS representation of the continuous periodic signal in Figure 68 The CTFS pair is using the signals fundamental period as the fundamental period of the CTFS representation A tri2tT0 δ T 0 t ℱ𝒮 T 0 A2sinc2k2 Figure 68 A continuous signal to be represented by a CTFS t xt A T0 and the partialsum approximations x N t for N 1 3 5 and 59 are illustrated in Figure 69 2 The definition of the sinc function is generally but not universally accepted as sinct sinπtπt In some books the sinc function is defined as sinct sintt In other books this second form is called the Sa function Sat sintt How the sinc function is defined is not really critical As long as one definition is accepted and the sinc function is used in a manner consistent with that definition signal and system analysis can be done with useful results rob28124ch06229306indd 239 041216 130 pm C h a p t e r 6 ContinuousTime Fourier Methods 240 Figure 69 Successively closer approximations to a triangle wave T0 T0 T0 T0 A 2 A T0 T0 A 2 A A 2 A A 2 A t t N 3 N 1 N 5 N 59 x3t T0 T0 t t x5t x59t x1t At N 59 and probably at lower values of N it is impossible to distinguish the CTFS partialsum approximation from the original signal by observing a graph on this scale Discontinuous Signals Now consider a periodic signal with discontinuities xt Arect 2 t T 0 4 T 0 δ T 0 t Figure 610 The CTFS pair is A rect 2 t T 0 4 T 0 δ T 0 t ℱ𝒮 T 0 A2 j k sinck2 and the approximations x N t for N 1 3 5 and 59 are illustrated in Figure 611 Although the mathematical derivation indicates that the original signal and its CTFS representation are equal everywhere it is natural to wonder whether that is true after looking at Figure 611 There is an obvious overshoot and ripple near the discon tinuities that does not appear to become smaller as N increases In fact the maximum vertical overshoot near a discontinuity does not decrease with N even as N approaches infinity This overshoot is called the Gibbs phenomenon in honor of Josiah Gibbs3 3 Josiah Willard Gibbs an American physicist chemist and mathematician developed much of the theory for chemical thermodynamics and physical chemistry He invented vector analysis independently of Oliver Heaviside He earned the first American PhD in engineering from Yale in 1863 and he spent his entire career at Yale In 1901 Gibbs was awarded the Copley Medal of the Royal Society of London for being the first to apply the second law of thermodynamics to the exhaustive discussion of the relation between chemical electrical and thermal energy and capacity for external work rob28124ch06229306indd 240 041216 130 pm 62 The ContinuousTime Fourier Series 241 t t t A 2 A A 2 A A 2 A A 2 A t 2 T0 2 T0 2 T0 2 T0 2 T0 2 T0 2 T0 2 T0 x1t x3t x5t x59t N 1 N 3 N 5 N 59 Figure 611 Successively closer approximations to a square wave Figure 610 A discontinuous signal to be represented by a CTFS t xt T0 T0 A who first mathematically described it But notice also that the ripple is also confined ever more closely in the vicinity of the discontinuity as N increases In the limit as N approaches infinity the height of the overshoot is constant but its width approaches zero The error in the partialsum approximation is the difference between it and the original signal In the limit as N approaches infinity the signal power of the error ap proaches zero because the zerowidth difference at a point of discontinuity contains no signal energy Also at any particular value of t except exactly at a discontinuity rob28124ch06229306indd 241 041216 130 pm C h a p t e r 6 ContinuousTime Fourier Methods 242 the value of the CTFS representation approaches the value of the original signal as N approaches infinity At a discontinuity the functional value of the CTFS representation is always the average of the two limits of the original function approached from above and from below for any N Figure 612 is a magnified view of the CTFS representation at a discontinuity for three different values of N Since the signal energy of the difference between the two signals is zero in any finite time interval their effect on any real phys ical system is the same and they can be considered equal for any practical purpose MINIMUM ERROR OF FOURIERSERIES PARTIAL SUMS The CTFS is an infinite summation of sinusoids In general for exact equality between an arbitrary original signal and its CTFS representation infinitely many terms must be used Signals for which the equality is achieved with a finite number of terms are called bandlimited signals If a partialsum approximation x N t kN N cxk e j2πktT 67 is made to a signal xt by using only the first N harmonics of the CTFS the difference between x N t and xt is the approximation error e N t x N t xt We know that in 67 when N goes to infinity the equality is valid at every point of continuity of xt But when N is finite does the harmonic function c x k for N k N yield the best possible approximation to xt In other words could we have chosen a different har monic function c xN k that if used in place of c x k in 67 would have been a better approximation to xt The first task in answering this question is to define what is meant by best possible approximation It is usually taken to mean that the signal energy of the error e N t over Figure 612 Illustration of the Gibbs phenomenon for increasing values of N 0 A t N 19 N 59 N 199 A 2 rob28124ch06229306indd 242 041216 130 pm 62 The ContinuousTime Fourier Series 243 one period T is a minimum Lets find the harmonic function c x N k that minimizes the signal energy of the error e N t kN N c xN k e j2πktT x N t k c x k e j2πktT xt Let c y k c xN k c x k k N c x k k N Then e N t k c y k e j2πktT The signal energy of the error over one period is E e 1 T T e N t 2 dt 1 T T k c y k e j2πktT 2 dt E e 1 T T k c y k e j2πktT q c y q e j2πktT dt E e 1 T T k c y k c y k k q qk c y k c y q e j2πkqtT dt The integral of the double summation is zero for every combination of k and q for which k q because the integral over any period of e j2πkqtT is zero Therefore E e 1 T T k c y k c y k dt 1 T T k c y k 2 dt Substituting the definition of c y k we get E e 1 T T kN N c xN k c x k 2 k N c x k 2 dt E e kN N c xN k c x k 2 k N c x k 2 All the quantities being summed are nonnegative and since the second summation is fixed we want the first summation to be as small as possible It is zero if c xN k cxk proving that the harmonic function cxk gives the smallest possible meansquared error in a partial sum approximation THE FOURIER SERIES OF EVEN AND ODD PERIODIC FUNCTIONS Consider the case of representing a periodic even signal xt with fundamental period T 0 with a complex CTFS The CTFS harmonic function is cxk 1 T T xt e j2πktT dt rob28124ch06229306indd 243 041216 130 pm C h a p t e r 6 ContinuousTime Fourier Methods 244 For periodic signals this integral over a period is independent of the starting point Therefore we can rewrite the integral as c x k 1 T T2 T2 xt e j2πktT dt 1 T T2 T2 xt even cos2πktT even even dt j T2 T2 xt even sin2πktT odd odd dt Using the fact that an odd function integrated over symmetrical limits about zero is zero c x k must be real By a similar argument for a periodic odd function c x k must be imaginary For xt even and realvalued c x k is even and realvalued For xt odd and realvalued c x k is odd and purely imaginary FOURIERSERIES TABLES AND PROPERTIES The properties of the CTFS are listed in Table 61 They can all be proven using the definition of the CTFS and the harmonic function xt k cxk e j2πktT ℱ𝒮 T cxk 1T T xt e j2πktT dt In the MultiplicationConvolution Duality property the integral xt yt T xτyt τ dτ appears It looks a lot like the convolution integral we have seen earlier except that the integration range is over the fundamental period T of the CTFS representation instead of from to This operation is called periodic convolution Periodic convolution is always done with two periodic signals over a period T that is common to both of them The convolution that was introduced in Chapter 5 is aperiodic convolu tion Periodic convolution is equivalent to aperiodic convolution in the following way Any periodic signal x p t with period T can be expressed as a sum of equally spaced aperiodic signals x ap t as x p t k x ap t kT It can be shown that the periodic convolution of x p t with y p t is then x p t y p t x ap t y p t The function x ap t is not unique It can be any function that satisfies xpt k x ap t kT Table 62 shows some common CTFS pairs All but one are based on the funda mental period T of the CTFS representation being m T 0 with m being a positive integer and T 0 being the fundamental period of the signal xt k cxk e j2πktm T 0 ℱ𝒮 m T 0 cxk 1 m T 0 m T 0 xt e j2πktm T 0 dt rob28124ch06229306indd 244 041216 130 pm 62 The ContinuousTime Fourier Series 245 Table 61 CTFS properties Linearity α xt βyt ℱ𝒮 T α c x k β c y k Time Shifting xt t 0 ℱ𝒮 T e j2πk t 0 T c x k Frequency Shifting e j2π k 0 tT xt ℱ𝒮 T c x k k 0 Conjugation x t ℱ𝒮 T c x k Time Differentiation d dt xt ℱ𝒮 T j2πkT c x k Time Reversal xt ℱ𝒮 T c x k Time Integration t xτdτ ℱ𝒮 T c x k j2πkT k 0 if c x 0 0 Parsevals Theorem 1 T T xt 2 dt k c x k 2 MultiplicationConvolution Duality xtyt ℱ𝒮 T m c y m c x k m c x k c y k xtyt T xτyt τdτ ℱ𝒮 T T c x k c y k Change of Period If xt ℱ𝒮 T c x k and xt ℱ𝒮 mT c xm k c xm k c x km km an integer 0 otherwise Time Scaling If xt ℱ𝒮 T c x k and zt xmt ℱ𝒮 T c z k c z k c x km km an integer 0 otherwise Table 62 Some CTFS pairs e j2πt T 0 ℱ𝒮 m T 0 δk m cos2πk T 0 ℱ𝒮 m T 0 12δk m δk m sin2πk T 0 ℱ𝒮 m T 0 j2δk m δk m 1 ℱ𝒮 T δk T is arbitrary δ T 0 t ℱ𝒮 m T 0 1 T 0 δ m k recttw δ T 0 t ℱ𝒮 m T 0 w T 0 sincwkm T 0 δ m k tritw δ T 0 t ℱ𝒮 m T 0 w T 0 sinc 2 wkm T 0 δ m k sinctw δ T 0 t ℱ𝒮 m T 0 w T 0 rectwkm T 0 δ m k tut ut w δ T 0 t ℱ𝒮 m T 0 1 T 0 j2πkwm T 0 1 e j2πkwm T 0 1 2πkm T 0 2 δ m k rob28124ch06229306indd 245 041216 130 pm C h a p t e r 6 ContinuousTime Fourier Methods 246 ExamplE 62 Periodic excitation and response of a continuoustime system A continuoustime system is described by the differential equation y t 004 y t 158yt xt If the excitation is xt trit δ 5 t find the response yt The excitation can be expressed by a CTFS as xt k c x k e j2πkt T 0 where from Table 62 c x k w T 0 sinc 2 wkm T 0 δ m k with w 1 T 0 5 and m 1 Then xt k 15 sinc 2 k5 δ 1 k e j2πkt5 15 k sinc 2 k5 e j2πkt5 We know that the CTFS expression for the excitation is a sum of complex sinusoids and the response to each of those sinusoids will be another sinusoid of the same frequency Therefore the response can be expressed in the form yt k c y k e j2πkt5 and each complex sinusoid in yt with fundamental cyclic frequency k5 is caused by the com plex sinusoid in xt of the same frequency Substituting this form into the differential equation k j2πk5 2 c y k e j2πkt5 004 k j2πk5 c y k e j2πkt5 158 k c y k e j2πkt5 k c x k e j2πkt5 Gathering terms and simplifying k j2πk5 2 004j2πk5 158 c y k e j2πkt5 k c x k e j2πkt5 Therefore for any particular value of k the excitation and response are related by j2πk5 2 004j2πk5 158 c y k c x k and c y k c x k 1 j2πk5 2 004j2πk5 158 The quantity Hk c y k c x k is analogous to frequency response and can logically be called harmonic response The system response is yt 15 k sinc 2 k5 j2πk5 2 004j2πk5 158 e j2πkt5 rob28124ch06229306indd 246 041216 130 pm 62 The ContinuousTime Fourier Series 247 This rather intimidatinglooking expression can be easily programmed on a computer The signals their harmonic functions and the harmonic response are illustrated in Figure 613 and Figure 614 We can see from the harmonic response that the system responds strongly at harmonic number one the fundamental The fundamental period of xt is T 0 5 s So yt should have a significant response at a frequency of 02 Hz Looking at the response graph we see a signal that looks like a sinusoid and its fundamental period is 5 s so its fundamental frequency is 02 Hz The magnitudes of all the other harmonics including k 0 are almost zero That is why the Figure 613 Excitation harmonic function system harmonic response and response harmonic function k cxk 005 01 015 02 k 1 05 05 1 k 5 10 15 20 k 20 10 10 20 4 2 2 k 1 2 3 4 k 20 10 10 20 4 2 2 Hk cyk 20 20 20 20 10 10 10 10 20 20 20 20 10 10 10 10 cxk cyk Hk Figure 614 Excitation and response t t 02 04 06 08 1 10 5 5 10 10 20 30 40 50 10 20 30 40 50 yt xt rob28124ch06229306indd 247 041216 130 pm C h a p t e r 6 ContinuousTime Fourier Methods 248 average value of the response is practically zero and it looks like a sinusoid a singlefrequency signal Also notice the phase of the harmonic response at the fundamental It is 15536 radians at k 1 or almost π2 That phase shift would convert a cosine into a sine The excitation is an even function with only cosine components and the response is practically an odd function because of this phase shift We could compute xt and yt with a MATLAB program of the following form Set up a vector of ks over a wide range The ideal summation has an infinite range for k We obviously cannot do that in MATLAB but we can make the range of k so large that making it any larger would not significantly change the computational results kmax 1000 T0 5 xt and yt have a fundamental period of five dt T0100 Set the time increment for computing samples of xt and yt t T0dtT0 Set the time vector for computing samples of xt and yt x 0t y x Initialize x and y each to a vector of zeros the same length as t Compute samples of xt and yt in a for loop for k kmaxkmax Compute samples of x for one k xk sinck52expj2pikt55 Add xk to previous x x x xk Compute samples of y for one k yk xkj2pik52 004j2pik5 158 Add yk to previous y y y yk end NUMERICAL COMPUTATION OF THE FOURIER SERIES Lets consider an example of a different kind of signal for which we might want to find the CTFS Figure 615 This signal presents some problems It is not at all obvious how to describe it other than graphically It is not sinusoidal or any other obvious mathematical functional form Up to this time in our study of the CTFS in order to find a CTFS harmonic function of a signal we needed a mathematical description of it But just because we cannot describe a signal mathematically does not mean it does not have a CTFS description Most real signals that we might want to analyze in practice do not have a known exact mathematical description If we have a set of samples of the signal taken from one period we can estimate the CTFS harmonic function numerically The more samples we have the better the estimate Figure 616 The harmonic function is c x k 1 T T xt e j2πkftT dt rob28124ch06229306indd 248 041216 130 pm 62 The ContinuousTime Fourier Series 249 Since the starting point of the integral is arbitrary for convenience set it to t 0 cxk 1 T 0 T xt e j2πktT dt We dont know the function xt but if we have a set of N samples over one period starting at t 0 the time between samples is T s TN and we can approximate the integral by the sum of several integrals each covering a time of length T s cxk 1 T n0 N1 n T s n1 T s xn T s e j2πkn T s T dt 68 In Figure 616 the samples extend over one fundamental period but they could extend over any period and the analysis would still be correct If the samples are close enough together xt does not change much between samples and the integral 68 becomes a good approximation The details of the integration process are in Web Appendix F where it is shown that for harmonic numbers k N we can approximate the harmonic function as cxk 1 N n0 N1 xn T s e j2πnkN 69 The summation on the right side of 69 n0 N1 xn T s e j2πnkN is a very important operation in signal processing called the discrete Fourier transform DFT So 69 can be written as cxk 1N𝒟ℱ𝒯xn T s k N 610 where 𝒟ℱ𝒯xn T s n0 N1 xn T s e j2πnkN Figure 616 Sampling the arbitrary periodic signal to estimate its CTFS harmonic function t xt t xt T0 Ts t xt T0 Figure 615 An arbitrary periodic signal rob28124ch06229306indd 249 041216 130 pm C h a p t e r 6 ContinuousTime Fourier Methods 250 The DFT takes a set of samples representing a periodic function over one period and returns another set of numbers representing an approximation to its CTFS har monic function multiplied by the number of samples N It is a builtin function in modern highlevel programming languages like MATLAB In MATLAB the name of the function is fft which stands for fast Fourier transform The fast Fourier trans form is an efficient algorithm for computing the DFT The DFT and FFT are covered in more detail in Chapter 7 The simplest syntax of fft is X fftx where x is a vector of N samples of a function indexed by n in the range 0 n N and X is a vector of N returned numbers indexed by k in the range 0 k N The DFT n0 N1 xn T s e j2πnkN is periodic in k with period N This can be shown by finding Xk N Xk N 1 N n0 N1 xn T s e j2πnkNN 1 N n0 N1 xn T s e j2πnkN e j2πn 1 Xk The approximation 69 is for k N This includes some negative values of k But the fft function returns values of the DFT in the range 0 k N The values of the DFT for negative k are the same as the values of k in a positive range that are separated by one period So for example to find X1 find its periodic repetition XN 1 which is included in the range 0 k N This numerical technique to find the CTFS harmonic function can also be useful in cases in which the functional form of xt is known but the integral c x k 1 T T xt e j2πktT dt cannot be done analytically ExamplE 63 Using the DFT to approximate the CTFS Find the approximate CTFS harmonic function of a periodic signal xt one period of which is described by xt 1 t 2 1 t 1 The fundamental period of this signal is 2 So we can choose any integer multiple of 2 as the time over which samples are taken the representation time T Choose 128 samples over one fundamental period The following MATLAB program finds and graphs the CTFS harmonic function using the DFT Program to approximate using the DFT the CTFS of a periodic signal described over one period by xt sqrt1t2 1 t 1 N 128 Number of samples T0 2 Fundamental period T T0 Representation time Ts TN Time between samples rob28124ch06229306indd 250 041216 130 pm 62 The ContinuousTime Fourier Series 251 fs 1Ts Sampling rate n 0N 1 Time index for sampling t nTs Sampling times Compute values of xt at the sampling times x sqrt1 t2rectt2 sqrt1 t22rectt 22 sqrt1t 42rectt 42 cx fftxN DFT of samples k 0N2 1 Vector of harmonic numbers Graph the results subplot311 p plottxk setpLineWidth2 grid on axisequal axis04015 xlabelTime t s ylabelxt subplot312 p stemkabscx1N2k setpLineWidth2MarkerSize4 grid on xlabelHarmonic Number k ylabelcxk subplot313 p stemkanglecx1N2k setpLineWidth2MarkerSize4 grid on xlabelHarmonic Number k ylabelPhase of cxk Figure 617 is the graphical output of the program Figure 617 xt and cxk 0 1 2 3 4 0 05 1 15 Time t s xt 0 10 20 30 40 50 60 70 0 02 04 06 08 Harmonic Number k cxk 0 10 20 30 40 50 60 70 4 2 0 2 4 Harmonic Number k Phase of cxk rob28124ch06229306indd 251 041216 130 pm C h a p t e r 6 ContinuousTime Fourier Methods 252 Only three distinct values of phase occur in the phase graph 0 π and π The phases π and π are equivalent so they could all have been graphed as either π or π MATLAB computed the phase and because of roundoff errors in its computations it sometimes chose a number very close to π and other times chose a number very close to π The graphs of the magnitude and phase of c x k in Figure 617 are graphed only for k in the range 0 k N2 Since c x k c x k this is sufficient to define c x k in the range N2 k N2 It is often desirable to graph the harmonic function over the range N2 k N2 That can be done by realizing that the numbers returned by the DFT are exactly one period of a periodic function That being the case the second half of these numbers covering the range N2 k N is exactly the same as the set of numbers that occur in the range N2 k 0 There is a function fftshift in MATLAB that swaps the second half of the set of numbers with the first half Then the full set of N numbers covers the range N2 k N2 instead of the range 0 k N We can change the MATLAB program to analyze the signal over two fundamental periods instead of one by changing the line T T0 Representation time to T 2T0 Representation time The results are illustrated in Figure 618 Notice that now the CTFS harmonic function is zero for all odd values of k That occurred because we used two fundamental periods of xt as the representation time T The fundamental frequency of the CTFS representation is half the fundamental frequency of xt The signal power is at the fundamental frequency of xt and its harmonics which are the evennumbered harmonics of this CTFS harmonic function So only the evennumbered harmonics are nonzero The kth harmonic in the previous analysis using one fundamental period as the representation time is the same as the 2kth harmonic in this analysis Figure 618 xt and Xk using two fundamental periods as the representation time instead of one 0 1 2 3 4 0 05 1 15 Time t s xt 0 10 20 30 40 50 60 70 0 02 04 06 08 Harmonic Number k Xk 0 10 20 30 40 50 60 70 4 2 0 2 4 Harmonic Number k Phase of Xk rob28124ch06229306indd 252 041216 130 pm 62 The ContinuousTime Fourier Series 253 ExamplE 64 Total harmonic distortion computation A figure of merit for some systems is total harmonic distortion THD If the excitation signal of the system is a sinusoid the THD of the response signal is the total signal power in the response signal of all the harmonics other than the fundamental k 1 divided by the total signal power in the response signal at the fundamental k 1 An audio amplifier with a nominal gain of 100 at 4 kHz is driven by a 4 kHz sine wave with a peak amplitude of 100 mV The ideal response of the amplifier would be x i t 10 sin8000πt volts but the actual amplifier output signal xt is limited to a range of 7 volts So the actual response signal is correct for all voltages of magnitude less than 7 volts but for all ideal voltages greater than 7 in magnitude the response is clipped at 7 volts Compute the THD of the response signal The CTFS harmonic function of xt can be found analytically but it is a rather long te dious errorprone process If we are only interested in a numerical THD we can find it numeri cally using the DFT and a computer That is done in the following MATLAB program and the results are illustrated in Figure 619 f0 4000 Fundamental frequency of signal T0 1f0 Fundamental period of signal N 256 Number of samples to use in one period Ts T0N Time between samples fs 1Ts Sampling rate in samplessecond t Ts0N 1 Time vector for graphing signals Figure 619 Results of THD computation 0 005 01 015 02 025 10 5 0 5 10 Signal Power of Ideal Signal 50 Signal Power of Actual Signal 335799 0 005 01 015 02 025 10 5 0 5 10 Time t ms Time t ms Signal Power of Fundamental 329563 Total Signal Power of All Other Harmonics 062363 Total Harmonic Distortion 18923 Ideal Signal xit Actual Signal xt Error et xit xt and et x0t and Σxkt Fundamental x0t Sum of Other Harmonics xkt rob28124ch06229306indd 253 041216 130 pm C h a p t e r 6 ContinuousTime Fourier Methods 254 A 10 Ideal signal amplitude xi Asin2pif0t Ideal signal Pxi A22 Signal power of ideal signal x minxi07A Clip ideal signal at 7 volts x maxx07A Clip ideal signal at 7 volts Px meanx2 Signal power of actual signal cx fftshiftfftxN Compute harmonic function values up to k 128 k N2N2 1 Vector of harmonic numbers I0 findabsk 1 Find harmonic function values at fundamental P0 sumabscxI02 Compute signal power of fundamental Ik findabsk 1 Find harmonic function values not at fundamental Pk sumabscxIk2 Compute signal power in harmonics THD Pk100P0 Compute total harmonic distortion Compute values of fundamental component of actual signal x0 0t for kk 1lengthI0 x0 x0 cxI0kkexpj2pi kI0kkf0t end Compute values of sum of signal components not at fundamental in actual signal xk 0t for kk 1lengthIk xk xk cxIkkkexpj2pi kIkkkf0t end x0 realx0 Remove any residual imaginary parts due to roundoff xk realxk Remove any residual imaginary parts due to roundoff Graph the results and report signal powers and THD ttl Signal Power of Ideal Signal num2strPxi ttl str2matttlSignal Power of Actual Signal num2strPx subplot211 ptr plot1000txik1000txk1000txxik grid on setptrLineWidth2 xlabelTime itt msFontNameTimesFontSize24 ylabelxiitt xitt and eittFontNameTimes FontSize24 titlettlFontNameTimesFontSize24 ptr legendIdeal Signal xiittActual Signal xitt Error eitt setptrFontNameTimesFontSize18 setgcaFontSize18 subplot212 ttl Signal Power of Fundamental num2strP0 ttl str2matttlTotal Signal Power of All Other Harmonics num2strPk rob28124ch06229306indd 254 041216 130 pm 63 The ContinuousTime Fourier Transform 255 ttl str2matttlTotal Harmonic Distortion num2strTHD ptr plot1000tx0k1000txkk grid on setptrLineWidth2 xlabelTime itt msFontNameTimesFontSize24 ylabelx0itt and Sigma xitkittFontNameTimes FontSize24 titlettlFontNameTimesFontSize24 ptr legendFundamental x0ittSum of Other Harmonics xitk itt setptrFontNameTimesFontSize18 setgcaFontSize18 The THD is 18923 even with this severe 30 clipping at each positive and negative peak Therefore for good signal fidelity THD should generally be much smaller than 1 63 THE CONTINUOUSTIME FOURIER TRANSFORM The CTFS can represent any periodic signal with engineering usefulness over all time Of course some important signals are not periodic So it would be useful to somehow extend the CTFS to also be able to represent aperiodic signals over all time This can be done and the result is called the Fourier transform EXTENDING THE FOURIER SERIES TO APERIODIC SIGNALS The salient difference between a periodic signal and an aperiodic signal is that a peri odic signal repeats in a finite time T called the period It has been repeating with that period forever and will continue to repeat with that period forever An aperiodic signal does not have a finite period An aperiodic signal may repeat a pattern many times within some finite time but not over all time The transition between the Fourier series and the Fourier transform is accomplished by finding the form of the Fourier series for a periodic signal and then letting the period approach infinity Mathematically saying that a function is aperiodic and saying that a function has an infinite period are saying the same thing Consider a timedomain signal xt consisting of rectangular pulses of height A and width w with fundamental period T 0 Figure 620 This signal will illustrate the phenomena that occur in letting the fundamental period approach infinity for a general signal Representing this pulse train with a complex CTFS the harmonic function is found to be c x k Aw T 0 sinckw T 0 with T T 0 Suppose w T 0 2 meaning the waveform is at A half the time and at zero the other half a 50 duty cycle Then c x k A2sinck2 Figure 621 t xt A w T0 Figure 620 Rectangularwave signal rob28124ch06229306indd 255 041216 130 pm C h a p t e r 6 ContinuousTime Fourier Methods 256 Now let the fundamental period T 0 increase from 1 to 5 while w is unchanged Then c x 0 becomes 110 and the CTFS harmonic function is c x k 110sinck10 Figure 622 The maximum harmonic amplitude magnitude is 5 times smaller than before because the average value of the function is 5 times smaller than before As the fun damental period T 0 gets larger the harmonic amplitudes lie on a wider sinc function whose amplitude goes down as T 0 increases In the limit as T 0 approaches infinity the original timedomain waveform xt approaches a single rectangular pulse at the origin and the harmonic function approaches samples from an infinitely wide sinc function with zero amplitude If we were to multiply c x k by T 0 before graphing it the amplitude would not go to zero as T 0 approached infinity but would stay where it is and simply trace points on a widening sinc function Also graphing against k T 0 k f 0 instead of k would make the horizontal scale be frequency instead of harmonic number and the sinc function would remain the same width on that scale as T 0 increases and f 0 decreases Making those changes the last two graphs would look like Figure 623 Call this a modified harmonic function For this modified harmonic func tion T 0 c x k Awsincwk f 0 As T 0 increases without bound making the pulse train a k 50 50 05A cxk Figure 622 The magnitude of the CTFS harmonic function for a rectangularwave signal with reduced duty cycle kf0 10 10 05A T0 cxk kf0 10 10 05A T0 cxk Figure 623 Magnitudes of the modified CTFS harmonic functions for rectangularwave signals of 50 and 10 duty cycles k 50 50 05A cxk Figure 621 The magnitude of the CTFS harmonic function of a 50 dutycycle rectangularwave signal rob28124ch06229306indd 256 041216 130 pm 63 The ContinuousTime Fourier Transform 257 single pulse f 0 approaches zero and the discrete variable k f 0 approaches a continuous variable which we will call f The modified CTFS harmonic function approaches the function illustrated in Figure 624 This modified harmonic function with some notation changes becomes the continuoustime Fourier transform CTFT of that single pulse The frequency difference between adjacent CTFS harmonic amplitudes is the same as the fundamental frequency of the CTFS representation f 0 1 T 0 To empha size its relationship to a frequency differential which it will become in the limit as the fundamental period goes to infinity let this spacing be called Δf That is let Δf f 0 1 T 0 Then the complex CTFS representation of xt can be written as xt k c x k e j2πkΔft Substituting the integral expression for c x k xt k 1 T 0 t 0 t 0 T 0 xτ e j2πkΔfτ dτ e j2πkΔft The variable of integration is τ to distinguish it from the t in the function e j2πkΔft which is outside the integral Since the starting point t 0 for the integral is arbitrary let it be t 0 T 0 2 Then xt k T 0 2 T 0 2 xτ e j2πkΔfτ dτ e j2πkΔft Δf where Δf has replaced 1 T 0 In the limit as T 0 approaches infinity Δf approaches the differential df kΔf becomes a continuous variable f the integration limits approach plus and minus infinity and the summation becomes an integral xt lim T 0 k T 0 2 T 0 2 xτ e j2πkΔfτ dτ e j2πkΔft Δf xτ e j2πfτ dτ e j2πft df 611 kf0 10 10 05A T0cxk Figure 624 Limiting form of modified CTFS harmonic function for rectangularwave signal rob28124ch06229306indd 257 041216 130 pm C h a p t e r 6 ContinuousTime Fourier Methods 258 The bracketed quantity on the right side of 611 is the CTFT of xt X f xt e j2πft dt 612 and it follows that xt X f e j2πft df 613 Here we have adopted the convention that the Fourier transform of a signal is indicated by the same letter of the alphabet but uppercase instead of lowercase Notice that the Fourier transform is a function of cyclic frequency f and that the time dependence of the signal has been integrated out so the Fourier transform is not a function of time The timedomain function x and its CTFT X form a Fourier transform pair that is often indicated by the notation xt ℱ Xf Also conventional notation is Xf ℱxt and xt ℱ1Xf where ℱ is read Fourier transform of and ℱ1 is read inverse Fourier transform of Another common form of the Fourier transform is defined by making the change of variable f ω2π where ω is radian frequency Xω2π xt e jωt dt and xt 1 2π Xω2π e jωt dω 614 This is the result we obtain by simply substituting ω2π for f and dω2π for df It is much more common in engineering literature to see this form written as Xω xt e jωt dt and xt 1 2π Xω e jωt dω 615 In this second form the strict mathematical meaning of the function X has changed and that could be a source of confusion if conversion between the two forms is necessary To compound the confusion it is also quite common to see the ω form written as Xjω xt e jωt dt and xt 1 2π Xjω e jωt dω 616 again changing the meaning of the function X The reason for including a j in the functional argument is to make the Fourier transform more directly compatible with the Laplace transform Chapter 8 Suppose we have used Xf xt e j2πft dt to form the Fourier pair xt e αt ut ℱ Xf 1 j2πf α rob28124ch06229306indd 258 041216 130 pm 63 The ContinuousTime Fourier Transform 259 Ordinarily in mathematical functional notation if we were to then refer to a function X jω we would mean X f f jω Xjω 1 j2πjω α 1 2πω α But in Fourier transform literature it is very common to say that if the cyclicfrequency form of a Fourier transform is X f 1 j2πf α then the radianfrequency form is Xjω 1 j2πω2π α 1 jω α In going from X f to X jω what we have really done is to go from X f to xt using xt X f e j2πft df and then using X jω xt e jωt dt find X jω In other words X f ℱ 1 xt ℱ X jω This amounts to making the transition X f f ω2π X jω instead of X f f jω X jω In this text we will follow this traditional interpretation In any analysis it is important to choose a definition and then use it consistently In this text the forms xt X f e j2πft df ℱ X f xt e j2πft dt xt 1 2π X jω e jωt dω ℱ X jω xt e jωt dt will be used for the f and ω forms because those are the two most often encountered in engineering literature The Fourier transform as introduced here applies to continuoustime signals The CTFT is widely used in the analysis of communication systems filters and Fourier optics The ω form and the f form of the CTFT are both widely used in engineering Which one is used in any particular book or article depends on multiple factors includ ing the traditional notation conventions in a particular field and the personal preference of the author Since both forms are in common usage in this text we will use whichever form seems to be the most convenient in any individual analysis If at any time we need to change to the other form that is usually easily done by simply replacing f by ω2π or ω by 2πf In addition to the definitions presented here there are also several other alternate definitions of the Fourier transform that can be found in engineering mathematics and physics books Table 63 lists some CTFT pairs in the ω form derived directly from the definitions presented above The ω form was used here because for these functions it is a little more compact rob28124ch06229306indd 259 041216 130 pm C h a p t e r 6 ContinuousTime Fourier Methods 260 Table 63 Some CTFT pairs δ t ℱ 1 e αt ut ℱ 1jω α α 0 e αt ut ℱ 1jω α α 0 t e αt ut ℱ 1 jω α 2 α 0 t e αt ut ℱ 1 jω α 2 α 0 t n e αt ut ℱ n jω α n1 α 0 t n e αt ut ℱ n jω α n1 α 0 e αt sin ω 0 t ut ℱ ω 0 jω α 2 ω 0 2 α 0 e αt sin ω 0 tut ℱ ω 0 jω α 2 ω 0 2 α 0 e αt cos ω 0 t ut ℱ jω α jω α 2 ω 0 2 α 0 e αt cos ω 0 tut ℱ jω α jω α 2 ω 0 2 α 0 e α t ℱ 2α ω 2 α 2 α 0 THE GENERALIZED FOURIER TRANSFORM There are some important practical signals that do not have Fourier transforms in the strict sense Because these signals are so important the Fourier transform has been generalized to include them As an example of the generalized Fourier transform lets find the CTFT of a very simple function xt A a constant Using the CTFT definition xt X f e j2πft df ℱ X f xt e j2πft dt we obtain X f A e j2πft dt A e j2πft dt The integral does not converge Therefore strictly speaking the Fourier transform does not exist But we can avoid this problem by generalizing the Fourier transform with the following procedure First we will find the CTFT of x σ t A e σ t σ 0 a function that approaches the constant A as σ approaches zero Then we will let σ approach zero after finding the transform The factor e σ t is a convergence factor that allows us to evaluate the integral Figure 625 The transform is X σ f A e σ t e j2πft dt 0 A e σt e j2πft dt 0 A e σt e j2πft dt X σ f A 0 e σj2πft dt 0 e σj2πft dt A 2σ σ 2 2πf 2 Now take the limit as σ approaches zero of X σ f For f 0 lim σ0 A 2σ σ 2 2πf 2 0 Next find the area under the function X σ f as σ approaches zero Area A 2σ σ 2 2πf 2 df Using dx a 2 bx 2 1 ab tan 1 bx a rob28124ch06229306indd 260 041216 130 pm 63 The ContinuousTime Fourier Transform 261 we get Area A 2σ 2πσ tan 1 2πf σ A π π 2 π 2 A The area under the function is A and is independent of the value of σ Therefore in the limit σ 0 the Fourier transform of the constant A is a function that is zero for f 0 and has an area of A This exactly describes an impulse of strength A occurring at f 0 Therefore we can form the generalized Fouriertransform pair A ℱ Aδ f The generalization of the CTFT extends it to other useful functions including periodic functions By similar reasoning the CTFT transform pairs cos2π f 0 t ℱ 12 δ f f 0 δ f f 0 and sin2π f 0 t ℱ j2δ f f 0 δ f f 0 can be found By making the substitution f ω2π and using the scaling property of the impulse the equivalent radianfrequency forms of these transforms are found to be A ℱ 2πAδω cos ω 0 t ℱ πδω ω 0 δω ω 0 sin ω 0 t ℱ jπδω ω 0 δω ω 0 The problem that caused the need for a generalized form of the Fourier transform is that these functions constants and sinusoids are not absolutely integrable even though they are bounded The generalized Fourier transform can also be applied to Figure 625 Effect of the convergence factor e σ t t 4 4 xσt 1 f 1 1 Xσ f 4 σ decreasing σ decreasing rob28124ch06229306indd 261 041216 130 pm C h a p t e r 6 ContinuousTime Fourier Methods 262 other signals that are not absolutely integrable but are bounded for example the unit step and the signum Another way of finding the CTFT of a constant is to approach the problem from the other side by finding the inverse CTFT of an impulse X f Aδ f using the sampling property of the impulse xt Xf e j2πft df A δf e j2πft df A e 0 A This is obviously a much quicker route to finding the forward transform of a constant than the preceding development But the problem with this approach is that if we are trying to find the forward transform of a function we must first guess at the transform and then evaluate whether it is correct by finding the inverse transform ExamplE 65 CTFT of the signum and unitstep functions Find the CTFT of xt sgnt and then use that result to find the CTFT of xt ut Applying the integral formula directly we get Xf sgnt e j2πft dt 0 e j2πft dt 0 e j2πft dt and these integrals do not converge We can use a convergence factor to find the generalized CTFT Let x σ t sgnt e σ t with σ 0 Then X σ f sgnt e σ t e j2πft dt 0 e σj2πft dt 0 e σj2πft dt X σ f e σj2πft σ j2πf 0 e σj2πft σ j2πf 0 1 σ j2πf 1 σ j2πf and X f lim σ0 X σ f 1jπf or in the radianfrequency form Xjω 2jω To find the CTFT of xt ut we observe that ut 12sgnt 1 So the CTFT is U f 12sgnt 1 e j2πft dt 12 sgnt e j2πft dt ℱ sgnt 1jπf e j2πft dt ℱ1 δf U f 121jπf δ f 1j2πf 12δ f or in the radianfrequency form Ujω 1jω πδω rob28124ch06229306indd 262 041216 130 pm 63 The ContinuousTime Fourier Transform 263 ExamplE 66 Verify that the inverse CTFT of U f 1j2πf 12δ f is indeed the unit step function If we apply the inverse Fourier transform integral to U f 1j2πf 12δf we get ut 1j2πf 12δ f e j2πft df e j2πft j2πf df 12 δf e j2πft df 1 by the sampling property of the impulse ut 12 cos2πft j2πf df 0 odd integrand sin2πft 2πf even integrand df 12 2 0 sin2πft 2πf df Case 1 t 0 ut 12 2 0 0dω 12 Case 2 t 0 Let λ 2πft dλ 2πtdf ut 12 2 0 sinλ λt dλ 2πt 1 2 1 π 0 sinλ λ dλ Case 3 t 0 ut 12 2 0 sinλ λt dλ 2πt 1 2 1 π 0 sinλ λ dλ The integrals in Case 2 and Case 3 are sine integrals defined by Siz 0 z sinλ λ dλ and we can find in standard mathematical tables that lim z Siz π2 Si0 0 and Siz Siz Abramowitz and Stegun p 231 Therefore 2 0 sin2πft 2πf df 12 t 0 0 t 0 12 t 0 and ut 1 t 0 12 t 0 0 t 0 rob28124ch06229306indd 263 041216 130 pm C h a p t e r 6 ContinuousTime Fourier Methods 264 This inverse CTFT shows that for complete compatibility with Fourier transform theory the value of u0 should be defined as 12 as it was in Chapter 2 Defining the unit step this way is mathematically consistent and can occasionally have some engineering significance see Chapter 14digital filter design using the impulse invariant technique ExamplE 67 CTFT of the unitrectangle function Find the CTFT of the unitrectangle function The CTFT of the unitrectangle function is ℱrectt rectt e j2πft dt 12 12 cos2πft j sin2πftdt ℱrectt 2 0 12 cos2πftdt sinπ f πf sinc f We now have the CTFT pair rectt ℱ sincf In the ω form the pair is rectt ℱ sincω2π In this case the f form is simpler and more symmetrical than the ω form Recall the result of Example 61 A recttw δ T 0 t ℱ𝒮 T 0 Aw T 0 sincwk T 0 The CTFT of a rectangle function is a sinc function and the CTFS harmonic function of a periodically repeated rectangle function is a sampled sinc function It is sampled because k only takes on integer values This relationship between periodic repetition in time and sampling in frequency harmonic number will be important in the explora tion of sampling in Chapter 10 We can now extend the Fourier transform table to include several other func tions that often occur in Fourier analysis In Table 64 we used the cyclic frequency form of the CTFT because for these functions it is somewhat simpler and more symmetrical Table 64 More Fourier transform pairs δt ℱ 1 1 ℱ δ f sgnt ℱ 1jπf ut ℱ 12δ f 1j2πf rectt ℱ sinc f sinct ℱ rect f trit ℱ sinc 2 f sinc 2 t ℱ tri f δ T 0 t ℱ f 0 δ f 0 f f 0 1 T 0 T 0 δ T 0 t ℱ δ f 0 f T 0 1 f 0 cos2π f 0 t ℱ 12δ f f 0 δ f f 0 sin2π f 0 t ℱ j2δf f 0 δ f f 0 rob28124ch06229306indd 264 041216 130 pm 63 The ContinuousTime Fourier Transform 265 FOURIER TRANSFORM PROPERTIES Table 65 and Table 66 illustrate some properties of the CTFT derived directly from the two definitions Table 65 Fourier transform properties f form Linearity α gt β ht ℱ αG f β H f TimeShifting gt t 0 ℱ G f e j2πf t 0 Frequency Shifting e j2π f 0 t gt ℱ G f f 0 Time Scaling gat ℱ 1aG f a Frequency Scaling 1agta ℱ Gaf Time Differentiation d dt gt ℱ j2πfG f Time Integration t gλdλ ℱ G f j2πf 12G0δ f Frequency Differentiation tgt ℱ j 2π d df G f Multiplication gt ht ℱ G f H f Convolution Duality gtht ℱ G f H f Parsevals Theorem gt 2 dt G f 2 df Total Area X0 xtdt or x0 X f df Table 66 Fourier transform properties ω form Linearity α gt β ht ℱ αG jω β H jω TimeShifting gt t 0 ℱ G jω e jω t 0 Frequency Shifting e j ω 0 t gt ℱ G jω ω 0 Time Scaling gat ℱ 1aG jωa Frequency Scaling 1agta ℱ G jaω Time Differentiation d dt gt ℱ jω Gjω Time Integration t gλdλ ℱ Gjω jω πG0δω Frequency Differentiation tgt ℱ j d dω Gjω Multiplication gt ht ℱ GjωHjω Convolution Duality gtht ℱ 1 2π Gjω Hjω Parsevals Theorem gt 2 dt 1 2π Gjω 2 dω Total Area X 0 xtdt or x0 1 2π Xjωdw rob28124ch06229306indd 265 041216 130 pm C h a p t e r 6 ContinuousTime Fourier Methods 266 Any periodic signal can be expressed as a Fourier series of the form xt k c x k e j2πktT Using the frequency shifting property we can find the CTFT as X f k c x kδ f kT So the CTFT of any periodic signal consists entirely of impulses The strengths of those impulses at cyclic frequencies kT are the same as the values of the CTFS harmonic function at harmonic number k ExamplE 68 CTFS harmonic function of a periodic signal using the CTFT Use Xf k c x kδf kT to find the CTFS harmonic function of xt rect2t δ 1 t This is a convolution of two functions Therefore from the multiplicationconvolution duality property the CTFT of xt is the product of the CTFTs of the individual functions X f 12sinc f2δ1 f 12 k sinck2 δ f k and the CTFS harmonic function must therefore be c x k 12sinck2 based on T T 0 1 ExamplE 69 CTFT of a modulated sinusoid Find the CTFT of xt 24 cos100πt sin10000πt This is the product of two functions Therefore using the multiplicationconvolution duality property the CTFT will be the convolution of their individual CTFTs Using cos2π f 0 t ℱ 12δ f f 0 δ f f 0 and sin2π f 0 t ℱ j2δ f f 0 δ f f 0 we get 24 cos100πt ℱ 12δ f 50 δ f 50 rob28124ch06229306indd 266 041216 130 pm 63 The ContinuousTime Fourier Transform 267 and sin10000πt ℱ j2δ f 5000 δ f 5000 Then the overall CTFT is 24 cos100πt sin10000πt ℱ j6 δ f 4950 δ f 5050 δ f 5050 δ f 4920 The timeshifting property says that a shift in time corresponds to a phase shift in frequency As an example of why the timeshifting property makes sense let the time signal be the complex sinusoid xt e j2πt Then xt t 0 e j2πt t 0 e j2πt e j2π t 0 Figure 626 Figure 626 A complex exponential xt e j2π f 0 t and a delayed version xt 18 e j2π f 0 t18 t 2 Rext 1 Imxt 1 1 1 t 2 Rext18 1 Imxt18 1 1 1 xt ej2πt Shifting this signal in time corresponds to multiplying it by the complex number e j2π t 0 The CTFT expression xt X f e j2πft df says that any signal that is Fourier transformable can be expressed as a linear combi nation of complex sinusoids over a continuum of frequencies f and if xt is shifted by t 0 each of those complex sinusoids is multiplied by the complex number e j2πf t 0 What happens to any complex number when it is multiplied by a complex exponen tial of the form e jx where x is real The magnitude of e jx is one for any real x There fore multiplication by e jx changes the phase but not the magnitude of the complex number Changing the phase means changing its angle in the complex plane which is a simple rotation of the vector representing the number So multiplying a complex exponential function of time e j2πt by a complex constant e j2π t 0 rotates the complex exponential e j2πt with the time axis as the axis of rotation Looking at Figure 626 it is apparent that because of its unique helical shape a rotation of a complex exponential function of time and a shift along the time axis have the same net effect The frequencyshifting property can be proven by starting with a frequencyshifted version of Xf Xf f 0 and using the inverse CTFT integral The result is xt e j2π f 0 t ℱ X f f 0 rob28124ch06229306indd 267 041216 130 pm C h a p t e r 6 ContinuousTime Fourier Methods 268 Notice the similarity between the timeshifting and frequencyshifting properties They both result in multiplying by a complex sinusoid in the other domain However the sign of the exponent in the complex sinusoid is different That occurs because of the signs in the forward and inverse CTFTs X f xt e j2πft dt xt X f e j2πft df The frequencyshifting property is fundamental to understanding the effects of modu lation in communication systems One consequence of the timescaling and frequencyscaling properties is that a compression in one domain corresponds to an expansion in the other domain One interesting way of illustrating that is through the function xt e π t 2 whose CTFT is of the same functional form e π t 2 ℱ e π f 2 We can assign a characteristic width parameter w to these functions the distance between inflection points the time or fre quency between the points of maximum slope magnitude Those points occur on e π t 2 at t 1 2π so w 2π If we now timescale through the transformation t t2 for example the transform pair becomes e π t2 2 ℱ 2 e π 2f 2 Figure 627 and the width parameter of the time function becomes 2 2π while the width parameter of the frequency function becomes 2π 2 Figure 627 Time expansion and the corresponding frequency compression t 3 3 xt 2 f 3 3 X f 2 t 3 3 xt2 2 f 3 3 2X2f 2 w w w w The change of variable t t2 causes a time expansion and the corresponding effect in the frequency domain is a frequency compression accompanied by an ampli tude scale factor As the timedomain signal is expanded it falls from its maximum of one at t 0 more and more slowly as time departs from zero in either direction and in the limit as the time expansion factor approaches infinity it does not change at all and approaches the constant 1 w As the timedomain signal is expanded by some factor its CTFT is frequencycompressed and its height is multiplied by the same factor In the limit as the timedomain expansion factor approaches infinity the CTFT approaches an impulse lim a e π ta 2 1 ℱ lim a 1 a e π af 2 δ f 617 rob28124ch06229306indd 268 041216 130 pm 63 The ContinuousTime Fourier Transform 269 Figure 628 and w 0 Figure 628 Constant and impulse as limits of time and frequency scaling xt e π t 2 and its CTFT t xt 1 f X f 1 The relation between compression in one domain and expansion in the other is the basis for an idea called the uncertainty principle of Fourier analysis As a in 617 the signal energy of the timedomain function becomes less localized and the signal energy of the corresponding frequencydomain function becomes more local ized In that limit the signal energy of the signal in the frequency domain is infinitely localized to a single frequency f 0 while the time functions width becomes infinite and therefore its signal energy is infinitely unlocalized in time If we compress the time function instead it becomes an impulse at time t 0 and its signal energy occurs at one point while its CTFT becomes spread uniformly over the range f and its signal energy has no locality at all As we know the location of the signal energy of one signal better and better we lose knowledge of the location of the signal energy of its transform counterpart The name uncertainty principle comes from the principle in quantum mechanics of the same name If xt is real valued then xt x t The CTFT of xt is X f and the CTFT of x t is ℱ x t x t e j2πft dt xt e j2πft dt X f Therefore if xt x t X f X f In words if the timedomain signal is real valued its CTFT has the property that the behavior for negative frequencies is the complex conjugate of the behavior for positive frequencies This property is called Hermitian symmetry Let xt be a realvalued signal The square of the magnitude of X f is Xf 2 Xf X f Then using Xf X f we can show that the square of the magnitude of Xf is Xf 2 Xf X f X f X f X f X f X f 2 proving that the magnitude of the CTFT of a realvalued signal is an even function of frequency Using Xf X f we can also show that the phase of the CTFT of a realvalued signal can always be expressed as an odd function of frequency Since the phase of any complex function is multiplevalued there are many equally correct ways of expressing phase So we cannot say that the phase is an odd function only that it can always be expressed as an odd function Often in practical signal and system analysis the CTFT of a realvalued signal is only displayed for positive frequencies because since X f X f if we know the functional behavior for positive frequencies we also know it for negative frequencies rob28124ch06229306indd 269 041216 130 pm C h a p t e r 6 ContinuousTime Fourier Methods 270 Suppose a signal xt excites an LTI system with impulse response ht producing a response yt Then yt xt ht Using the multiplicationconvolution duality property Y jω X jωH jω In words the CTFT of xt X jω is a function of fre quency and when multiplied by H jω the CTFT of ht the result is Y jω X jω H jω the CTFT of yt X jω describes the variation of the signal xt with radian frequency and Y jω performs the same function for yt So multiplication by H jω changes the frequency description of the excitation to the frequency description of the response H jω is called the frequency response of the system This is the same frequency response first developed in Chapter 5 When two LTI systems are cascaded the impulse response of the combined system is the convolution of the impulse re sponses of the two individual systems Therefore again using the multiplicationcon volution duality property the frequency response of the cascade of two LTI systems is the product of their individual frequency responses Figure 629 Figure 629 Frequency response of a cascade of two LTI systems X jω Y jω H1 jωH2 jω X jω H1 jω H2 jω X jωH1 jω Y jωX jωH1 jωH2 jω Figure 630 xt and its integral t t 2 2 2 2 1 2 1 xt xλdλ t ExamplE 610 CTFT using the differentiation property Find the CTFT of xt rectt 12 rectt 12 using the differentiation property of the CTFT and the table entry for the CTFT of the triangle function Figure 630 The function xt is the derivative of a triangle function centered at zero with a base half width of 2 and an amplitude of 2 xt d dt 2trit2 In the table of CTFT pairs we find trit ℱ sinc 2 f Using the scaling and linearity properties 2trit2 ℱ 4 sinc 2 2f Then using the differentiation property xt ℱ j8πf sinc 2 2f If we find the CTFT of xt by using the table entry for the CTFT of a rectangle rectt ℱ sinc f and the timescaling and timeshifting properties we get xt ℱ j4sinc2f sin2πf which using the definition of the sinc function can be shown to be equivalent xt ℱ j8πf sinc 2 2f j8πf sinc2f sin2πf 2πf j4sinc2f sin2πf Parsevals theorem says that we can find the energy of a signal either in the time or frequency domain xt 2 dt X f 2 df 618 MarcAntoine Parseval des Chênes a French mathematician contemporary of Fourier of the late 18th and early 19th centuries was born April 27 1755 and died August 16 1836 The integrand X f 2 on the righthand side of 618 is called energy spectral density The name comes from the fact that its integral over all frequencies the whole rob28124ch06229306indd 270 041216 131 pm 63 The ContinuousTime Fourier Transform 271 spectrum of frequencies is the total signal energy of the signal Therefore to be consistent with the normal meaning of integration Xf 2 must be signal energy per unit cyclic frequency a signal energy density For example suppose xt represents a current in amperes A Then from the definition of signal energy the units of signal energy for this signal are A 2 s The CTFT of xt is Xf and its units are A s or AHz When we square this quantity we get the units A 2 Hz 2 A 2 s Hz signal energy cyclic frequency which confirm that the quantity X f 2 is signal energy per unit cyclic frequency ExamplE 611 Total area under a function using the CTFT Find the total area under the function xt 10sinct 47 Ordinarily we would try to directly integrate the function over all t Area xtdt 10sinc t 4 7 dt 10 sinπt 47 πt 47 dt This integral is a sine integral first mentioned in Example 66 defined by Siz 0 z sint t dt The sine integral can be found tabulated in mathematical tables books However evaluation of the sine integral is not necessary to solve this problem We can use X0 xt dt First we find the CTFT of xt which is X f 70rect7f e j8πf Then Area X0 70 ExamplE 612 CTFT of some timescaled and timeshifted sines If xt 10 sint then find a the CTFT of xt b the CTFT of x2t 1 and c the CTFT of x2t 1 a In this example the cyclic frequency of the sinusoid is 12π and the radian frequency is 1 Therefore the numbers will be simpler if we use the radianfrequency form of the CTFT Using the linearity property and looking up the transform of the general sine form sin ω 0 t ℱ jπδω ω 0 δω ω 0 sint ℱ jπδω 1 δω 1 10 sint ℱ j10πδω 1 δω 1 rob28124ch06229306indd 271 041216 131 pm C h a p t e r 6 ContinuousTime Fourier Methods 272 b From part a 10 sint ℱ j10πδω 1 δω 1 Using the time scaling property 10 sin2t ℱ j5πδω2 1 δω2 1 Then using the timeshifting property 10 sin2t 1 ℱ j5π δω2 1 δω2 1 e jω Then using the scaling property of the impulse 10 sin2t 1 ℱ j10πδω 2 δω 2 e jω or 10 sin2t 1 ℱ j10πδω 2 e j2 δω 2 e j2 c From part a 10 sint ℱ j10πδω 1 δω 1 Applying the timeshifting property first 10 sint 1 ℱ j10π δω 1 δω 1 e jω Then applying the timescaling property 10 sin2t 1 ℱ j5πδω2 1 δω2 1 e jω2 Then using the scaling property of the impulse 10 sin2t 1 ℱ j10πδω 2 δω 2 e jω2 or 10 sin2t 1 ℱ j10πδω 2 e j δω 2 e j ExamplE 613 CTFT of a scaled and shifted rectangle Find the CTFT of xt 25rectt 410 We can find the CTFT of the unit rectangle function in the table of Fourier transforms rectt ℱ sincf First apply the linearity property 25rectt ℱ 25sinc f Then apply the timescaling property 25rectt10 ℱ 250 sinc10f Then apply the timeshifting property 25 rectt 410 ℱ 250 sinc10f e j8πf ExamplE 614 CTFT of the convolution of some signals Find the CTFT of the convolution of 10 sint with 2δt 4 Method 1 Do the convolution first and find the CTFT of the result 10 sint 2δt 4 20 sint 4 rob28124ch06229306indd 272 041216 131 pm 63 The ContinuousTime Fourier Transform 273 Apply the timeshifting property 20 sint 4 ℱ j20π δω 1 δω 1 e j4ω or 20 sint 4 ℱ j10 δ f 12π δf 12π e j8πf Method 2 Do the CTFT first to avoid the convolution 10 sint 2δt 4 ℱ ℱ10 sintℱ2δt 4 2ℱ10 sintℱδt e j4ω 10 sint 2δt 4 ℱ j20πδω 1 δω 1 e j4ω or 10 sint 2δt 4 ℱ ℱ10 sintℱ2δt 4 2ℱ10 sintℱδt e j8πf 10 sint 2δt 4 ℱ j10δf 12π δf 12π e j8πf NUMERICAL COMPUTATION OF THE FOURIER TRANSFORM In cases in which the signal to be transformed is not readily describable by a mathematical function or the Fouriertransform integral cannot be done analytically we can some times find an approximation to the CTFT numerically using the discrete Fourier trans form DFT which was used to approximate the CTFS harmonic function If the signal to be transformed is a causal energy signal it can be shown Web Appendix G that we can approximate its CTFT f form at discrete frequencies by Xk f s N T s n0 N1 xn T s e j2πknN T s 𝒟ℱ𝒯xn T s k N 619 where T s 1 f s is chosen such that the signal x does not change much in that amount of time and N is chosen such that the time range 0 to NTs covers all or practically all of the signal energy of the signal x Figure 631 t xt nN n0 Ts Figure 631 A causal energy signal sampled with T s seconds between samples over a time N T s So if the signal to be transformed is a causal energy signal and we sample it over a time containing practically all of its energy and if the samples are close enough rob28124ch06229306indd 273 041216 131 pm C h a p t e r 6 ContinuousTime Fourier Methods 274 together that the signal does not change appreciably between samples the approxima tion in 619 becomes accurate for k N ExamplE 615 Using the DFT to approximate the CTFT Using the DFT find the approximate CTFT of xt t1 t 0 t 1 0 otherwise t1 trectt 12 numerically by sampling it 32 times over the time interval 0 t 2 The following MATLAB program can be used to make this approximation Program to demonstrate approximating the CTFT of t1 trectt 12 by sampling it 32 times in the time interval 0 t 2 seconds and using the DFT N 32 Sample 32 times Ts 2N Sample for two seconds and set sampling interval fs 1Ts Set sampling rate df fsN Set frequencydomain resolution n 0N 1 Vector of 32 time indices t Tsn Vector of times x t1 trectt 12 Vector of 32 xt function values X Tsfftx Vector of 32 approx Xf CTFT values k 0N2 1 Vector of 16 frequency indices Graph the results subplot311 p plottxk setpLineWidth2 grid on xlabelTime t s ylabelxt subplot312 p plotkdfabsX1N2k setpLineWidth2 grid on xlabelFrequency f Hz ylabelXf subplot313 p plotkdfangleX1N2k setpLineWidth2 grid on xlabelFrequency f Hz ylabelPhase of Xf This MATLAB program produces the graphs in Figure 632 Notice that 32 samples are taken from the timedomain signal and the DFT returns a vector of 32 numbers We only used the first 16 in these graphs The DFT is periodic and the 32 points returned represent one period Therefore the second 16 points are the same as the second 16 points occurring in the previous period and can be used to graph the DFT for negative frequencies The rob28124ch06229306indd 274 041216 131 pm 63 The ContinuousTime Fourier Transform 275 MATLAB command fftshift is provided for just that purpose Below is an example of using fftshift and graphing the approximate CTFT over equal negative and positive frequencies Program to demonstrate approximating the CTFT of t1 trectt 12 by sampling it 32 times in the time interval 0 t 2 seconds and using the DFT The frequency domain graph covers equal negative and positive frequencies N 32 Sample 32 times Ts 2N Sample for two second and set sampling interval fs 1Ts Set sampling rate df fsN Set frequencydomain resolution n 0N 1 Vector of 32 time indices t Tsn Vector of times x t1 trectt 12 Vector of 32 xt function values X fftshiftTsfftx Vector of 32 Xf approx CTFT values k N2N2 1 Vector of 32 frequency indices Graph the results subplot311 p plottxk setpLineWidth2 grid on xlabelTime t s ylabelxt subplot312 p plotkdfabsXk setpLineWidth2 grid on xlabelFrequency f Hz ylabelXf subplot313 Figure 632 A signal and its approximate CTFT found by using the DFT 0 02 04 06 08 1 12 14 16 18 2 0 01 02 03 04 Time t s xt 0 1 2 3 4 5 6 7 8 0 005 01 015 02 Frequency f Hz Xf 0 1 2 3 4 5 6 7 8 4 2 0 2 4 Frequency f Hz Phase of Xf rob28124ch06229306indd 275 041216 131 pm C h a p t e r 6 ContinuousTime Fourier Methods 276 p plotkdFangleXk setpLineWidth2 grid on xlabelFrequency f Hz ylabelPhase of Xf Figure 633 and Figure 634 show the results of this MATLAB program with 32 points and 512 points This result is a rough approximation to the CTFT because only 32 points were used If we use 512 points over a time period of 16 seconds we get an approximation with higher frequencydomain resolution and over a wider frequency range Figure 634 Approximate CTFT found by using the DFT with higher resolution 0 2 4 6 8 10 12 14 16 0 01 02 03 04 Time t s xt 20 15 10 5 0 5 10 15 20 0 005 01 015 02 Frequency f Hz Xf 20 15 10 5 0 5 10 15 20 4 2 0 2 4 Frequency f Hz Phase of Xf Figure 633 Approximate CTFT found by using the DFT graphed over equal negative and positive frequencies 0 02 04 06 08 1 12 14 16 18 2 0 01 02 03 04 Time t s xt 8 6 4 2 0 2 4 6 8 0 005 01 015 02 Frequency f Hz Xf 8 6 4 2 0 2 4 6 8 4 2 0 2 4 Frequency f Hz Phase of Xf rob28124ch06229306indd 276 041216 131 pm 63 The ContinuousTime Fourier Transform 277 ExamplE 616 System analysis using the CTFT A system described by the differential equation y t 1000yt 1000 xt is excited by xt 4rect200t Find and graph the response yt If we Fourier transform the differential equation we get j2πf Y f 1000Y f 1000X f which can be rearranged into Yf 1000X f j2πf 1000 The CTFT of the excitation is Xf 002sinc f 200 Therefore the CTFT of the response is Y f 20 sinc f 200 j2πf 1000 or using the definition of the sinc function and the exponential definition of the sine function Yf 20 sinπf200 πf200 j2πf 1000 4000 e j2πf400 e j2πf400 j2πf j2πf 1000 To find the inverse CTFT start with the CTFT pair e αt ut ℱ 1 j2πf α α 0 e 1000t ut ℱ 1 j2πf 1000 Next use the integration property t gλdλ ℱ G f j2π f 12G0δ f t e 1000λ uλdλ ℱ 1 j2πf 1 j2πf 1000 1 2000 δ f Then apply the timeshifting property gt t 0 ℱ G f e j2πf t 0 0 t1400 e 1000λ dλ ℱ 1 j2πf e j2πf400 j2πf 1000 e j2πf400 2000 δ f δ f 2000 0 t1400 e 1000λ dλ ℱ 1 j2πf e j2πf400 j2πf 1000 e j2πf400 2000 δ f δ f 2000 Subtracting the second result from the first and multiplying through by 4000 4000 t1400 e 1000λ u λ dλ 4000 t1400 e 1000λ uλdλ ℱ 4000 j2πf e j2πf400 e j2πf400 j2πf 1000 rob28124ch06229306indd 277 041216 131 pm C h a p t e r 6 ContinuousTime Fourier Methods 278 4000 t1400 e 1000λ uλdλ t1400 e 1000λ uλdλ ℱ 4000 j2πf e j2πf400 e j2πf400 j2πf 1000 The two integral expressions can be simplified as follows t1400 e 1000λ uλdλ 110001 e 1000t1400 t 1400 0 t 1400 1 1000 1 e 1000t1400 ut 1400 t1400 e 1000λ uλdλ 110001 e 1000t1400 t 1400 0 t 1400 1 1000 1 e 1000t1400 ut 1400 Then 4 1 e 1000t1400 ut 1400 1 e 1000t1400 ut 1400 ℱ 4000 j2πf e j2πf400 e j2πf400 j2πf 1000 Therefore the response is yt 41 e 1000t1400 ut 1400 1 e 1000t1400 ut 1400 Figure 635 and Figure 636 Figure 635 Magnitudes and phases of CTFTs of excitation and response and of system frequency response 02 04 06 08 002 0015 001 0005 1 002 001 0005 0015 X f f kHz f kHz f kHz f kHz f kHz f kHz H f Y f Xf Hf Yf π π 16 16 16 16 16 16 1 16 1 16 1 1 1 1 1 1 16 16 1 1 16 1 16 1 2 π 2 2 π 2 π 2 π 2 rob28124ch06229306indd 278 041216 131 pm 63 The ContinuousTime Fourier Transform 279 ExamplE 617 System analysis using the CTFT A system described by the differential equation y t 1000yt 1000xt is excited by xt 4rect200t δ 001 t Find and graph the response yt From Example 616 Y f f ω2π Y jω 1000Xjω jω 1000 The CTFT f form of the excitation is X f 2 sincf200 δ 100 f implying that X jω 2sincω400π δ 100 ω2π Using the scaling property of the periodic impulse Xjω 2sincω400π 2π δ 200π ω 4π sincω400π δ 200π ω Therefore the CTFT of the response is Yjω 4000πsincω400π δ 200π ω jω 1000 or using the definition of the periodic impulse Yjω 4000π k sincω400πδω 200πk jω 1000 Now using the equivalence property of the impulse Yjω 4000π k sinck2δω 200πk j200πk 1000 Figure 636 Rectangular pulse excitation and system response 5 1 2 3 4 1 2 3 4 10 5 10 5 10 5 10 t ms t ms xt yt rob28124ch06229306indd 279 041216 131 pm C h a p t e r 6 ContinuousTime Fourier Methods 280 and the inverse CTFT yields the response yt 2000 k sinck2 j200πk 1000 e j200πkt If we separate the k 0 term and pair each k and k this result can be written as yt 2 k1 sinck2 j01πk 05 e j200πkt sinck2 j01πk 05 e j200πkt Using the fact that the sinc function is even and combining the terms over one common denominator yt 2 k1 sinc k2 j01πk 05 e j200πkt j01πk 05 e j200πkt 01πk 2 05 2 yt 2 k1 sinc k2 cos200πkt 02πk sin200πkt 01πk 2 05 2 The response is a constant plus a linear combination of real cosines and sines at integer multi ples of 100 Hz Figure 637 and Figure 638 Figure 637 Magnitudes and phases of CTFTs of excitation and response and of system frequency response 4π 3π 2π π 4π 3π 2π π 2 02 04 06 08 1 1 1 2 1 1 2 ω ω 8π 4π 4π 8π ω ω ω ω π π π2 π2 π2 π2 8π 4π 4π 8π 8π 4π 4π 8π X jω X jω H jω Y jω H jω Y jω 103 8π 8π 8π 4π 4π 4π 4π 4π 4π 8π 8π 8π 103 103 103 103 103 2 rob28124ch06229306indd 280 041216 131 pm Exercises with Answers 281 64 SUMMARY OF IMPORTANT POINTS 1 The Fourier series is a method of representing any arbitrary signal with engineering usefulness as a linear combination of sinusoids realvalued or complex 2 The complex sinusoids used by the complex form of the Fourier series form a set of mutually orthogonal functions that can be combined in linear combinations to form any arbitrary periodic function with engineering usefulness 3 The formula for finding the harmonic function of the Fourier series can be derived using the principle of orthogonality 4 The Fourier series can be used to find the response of an LTI system to a periodic excitation 5 The Fourier series can be extended to allow the representation of aperiodic signals and the extension is called the Fourier transform 6 With a table of Fourier transform pairs and their properties the forward and inverse transforms of almost any signal of engineering significance periodic or aperiodic can be found 7 The frequency response of a stable system is the Fourier transform of its impulse response 8 The Fourier transform can be used to find the response of an LTI system to energy signals as well as to periodic signals EXERCISES WITH ANSWERS Answers to each exercise are in random order Fourier Series 1 Using MATLAB graph each sum of complex sinusoids over the time period indicated Figure 638 Excitation and response 20 10 10 20 1 2 3 4 xt t ms 20 10 10 20 1 2 3 4 yt t ms rob28124ch06229306indd 281 041216 131 pm C h a p t e r 6 ContinuousTime Fourier Methods 282 a x t 1 10 k30 30 sinc k 10 e j200πkt 15 ms t 15 ms b x t j 4 k9 9 sinc k 2 2 sinc k 2 2 e j10πkt 200 ms t 200 ms Answers t 02 02 xt 1 1 t ms 15 15 xt 1 Orthogonality 2 Show by direct analytical integration that the integral of the function g t A sin 2πt B sin 4πt is zero over the interval 12 t 12 3 A periodic signal x t with a period of 4 seconds is described over one period by x t 3 t 0 t 4 Graph the signal and find its CTFS description Then graph on the same scale approximations to the signal x N t given by x N t kN N c x k e j2πkt T 0 for N 1 2 and 3 In each case the time scale of the graph should cover at least two periods of the original signal Answers cx k 1 4 2 e j2πk 2 jπk j6πk 4 πk 2 t 8 1 3 t 8 1 3 t 8 1 3 x1t x2t x3t rob28124ch06229306indd 282 041216 131 pm Exercises with Answers 283 4 Using the CTFS table of transforms and the CTFS properties find the CTFS harmonic function of each of these periodic signals using the representation time T indicated a x t 10 sin 20πt T 110 b x t 2 cos 100π t 0005 T 150 c x t 4 cos 500πt T 150 d x t d dt e j10πt T 15 e x t rect t 4 δ 4 t T 4 f x t rect t δ 1 t T 1 g x t tri t δ 1 t T 1 Answers j5 δ k 1 δ k 1 2 δ k 5 δ k 5 sinc k4 δ k δ k j δ k 1 δ k 1 j10πδ k 1 5 Given x t ℱ𝒮 12 tri k 1 4 tri k 1 4 what is the average value of x t Answer 32 6 Given x t ℱ𝒮 8 4 u k 3 u k 4 what is the average signal power of x t Answer 112 7 The CTFS harmonic function of x t based on one fundamental period is found to be cx k 1 cos πk πk 2 a Is the signal even odd or neither b What is the average value of the signal Answers Even 12 8 A signal x t 7rect 2t 5rect 4 t 1 δ 8 t has a CTFS harmonic function cx k What is the numerical value of cx 0 It is not necessary to find a general expression for cx k to answer this question Answer 028125 9 For each function is its CTFS harmonic function purely real purely imaginary or neither a x t 8 cos 50πt 4 sin 22πt b x t 32 cos 50πt sin 22πt c x t tri t 1 4 tri t 1 4 sin 100πt d x t 100 cos 4000πt π 2 rob28124ch06229306indd 283 041216 131 pm C h a p t e r 6 ContinuousTime Fourier Methods 284 e x t 5 rect 2t 2 δ 2 t f x t 5t 3 t 3 and x t x t 6n n any integer g x t 100 sin 4000πt π 4 Answers 2 Neither 3 Purely Imaginary 2 Purely Real 10 Find the numerical values of the literal constants in these CTFS pairs a 10 sin 32πt ℱ𝒮 116 Aδ k a Bδ k b b 3 cos 44πt ℱ𝒮 111 Aδ k a Bδ k b c A rect at 1b δ 1b t ℱ𝒮 2 30 sinc 2k d d dt 2 rect 4t δ 1 t ℱ𝒮 1 Ak sinc ak T F 1 Answers π 14 32 2 2 j5 1 j5 1 152 14 12 11 The harmonic function c x k for a periodic continuoustime signal x t whose fundamental period is 4 is zero everywhere except at exactly two values of k k 8 Its value is the same at those two points c x 8 c x 8 3 What is the representation time used in calculating this harmonic function Answer 32 12 For the following signals x t the harmonic function is based on T T 0 the fundamental period of x t Find the numerical values of the literal constants a x t rect 2t 5 δ 5 t c x k A sinc kb b x t 2 sin 4πt 3 cos 12πt c x k A δ k a δ k a B δ k b δ k b Answers j 1 32 3 12 10 13 The CTFS harmonic function c x k for the signal x t rect 2 t 1 3 δ 3 t rect 2t 3 δ 3 t 1 is of the form cx k Asinc ak e jbπk Find A a and b using the fundamental period as the representation time Answers 12 16 23 14 If x t 10 tri 3t δ 2 t and d dt x t ℱ𝒮 T 0 Ak sinc 2 bk find the numerical values of A and b Answers j 5236 16 15 If x t A tri bt δ c t ℱ𝒮 8 c x k 10 sinc 2 k3 find the numerical values of A b and c Answers 30 38 8 rob28124ch06229306indd 284 041216 131 pm Exercises with Answers 285 16 In Figure E16 is a graph of one fundamental period of a periodic function x t A CTFS harmonic function c x k is found based on the representation time being the same as the fundamental period T 0 If A1 4 A 2 3 and T 0 5 what is the numerical value of c x 0 Figure E16 A1 t T0 xt A2 T0 2 If the representation period is changed to 3 T 0 what is the new numerical value of c x 0 Answers 05 05 17 Find and graph the magnitude and phase of the CTFS harmonic function of xt rect 20t δ 15 t using a representation time T 15 two ways and compare the graphs a Using the CTFS tables b Numerically with the time between points being T s 12000 Answers k 005 01 015 02 025 k 2 2 k 005 01 015 02 025 k 2 2 Analytical cxk Numerical cxk 20 10 10 20 20 10 10 20 20 10 10 20 20 10 10 20 cxk cxk rob28124ch06229306indd 285 041216 131 pm C h a p t e r 6 ContinuousTime Fourier Methods 286 18 A quantizer accepts a continuoustime input signal and responds with a continuoustime output signal that only has a finite number of equally spaced values If x in t is the input signal and x out t is the output signal and q is the difference between adjacent output levels the value of x out t at any point in time can be computed by forming the ratio x in tq rounding it off to the nearest integer and then multiplying the result by q Let the range of the signal levels accepted by the quantizer be from 10 to 10 and let the number of quantization levels be 16 Find the numerical total harmonic distortion see Example 64 in Chapter 6 of the quantizer output signal if the input signal is x in t 10 sin 2000πt Answer 02342 Forward and Inverse Fourier Transforms 19 Let a signal be defined by x t 2 cos 4πt 5 cos 15πt Find the CTFTs of x t 140 and x t 120 and identify the resultant phase shift of each sinusoid in each case Plot the phase of the CTFT and draw a straight line through the 4 phase points that result in each case What is the general relationship between the slope of that line and the time delay f 1 2 3 4 5 6 7 8 8 7 6 543 2 1 3π 4 π 10 Slope π 20 Slope X f Answers The slope of the line is 2πf times the delay 20 If xt e 5t ut and yt e 12t ut and xt ℱ X f and yt ℱ Y f and zt xt yt and zt ℱ Z f what is the numerical value of Z3 Answer 00023 e j23154 21 Find the CTFT of x t sinc t Then make the change of scale t 2t in x t and find the CTFT of the timescaled signal Answers rect f f rect f 2 f 1 1 1 2 1 2 1 1 2 1 2 rect f f rect f 2 f 1 1 1 2 1 2 1 2 rob28124ch06229306indd 286 041216 131 pm Exercises with Answers 287 22 Using the multiplicationconvolution duality of the CTFT find an expression for y t which does not use the convolution operator and graph y t a y t rect t cos πt b y t rect t cos 2πt c y t sinc t sinc t2 d y t sinc t sinc 2 t2 e y t e t u t sin 2πt Answers t 8 8 yt 1 1 t 8 8 yt 1 yt t 4 4 1 1 yt t 2 2 02 02 yt t 8 8 05 1 23 Find the following numerical values a x t 20 rect 4t X f f2 b x t 2 sinc t8 sinc t4 x 4 c x t 2 tri t4 δ t 2 x 1 and x 1 d x t 5 rect t2 δ t 1 δ t x 12 x 12 and x 52 e x t 3 rect t 1 X f f14 f x t 4 sinc 2 3t X jω ω4π g x t rect t rect 2t X f f12 h X f 10 δ f 12 δ f 12 x 1 i X jω 2 sinc ω2π 3 sinc ωπ x 0 Answers 20 49 12 3 31831 5 10 0 j270 0287 5093 24 Find the following forward or inverse Fourier transforms No final result should contain the convolution operator a ℱ 15 rect t 2 7 b ℱ1 2 tri f2 e j6πf c ℱ sin 20πt cos 200πt rob28124ch06229306indd 287 041216 131 pm C h a p t e r 6 ContinuousTime Fourier Methods 288 Answers 105 sinc 7f e j4πf 2tri f2 e j6πf j4 δ f 90 δ f 110 δ f 110 δ f 90 25 Find the signal energy of a x t 28 sinc t15 b x t 3 sinc 2 2t Answers 3 11760 26 Let y t x t h t and let x t e t u t and let h t x t What is the numerical value of y2 Answer 006765 27 Using Parsevals theorem find the signal energy of these signals a x t 4 sinc t5 b x t 2 sinc 2 3t Answers 89 80 28 What is the total area under the function g t 100 sinc t 8 30 Answer 3000 29 Let a continuoustime signal x t have a CTFT X f f f 2 0 f 2 Let y t x 4 t 2 Find the numerical values of the magnitude and phase of Y 3 where y t ℱ Y f Answer 316 0 30 Using the integration property find the CTFT of these functions and compare with the CTFT found using other properties a g t 1 t 1 2 t 1 t 2 0 elsewhere b g t 8 rect t3 Answers 3 sinc 3f sinc f 24 sinc 3f 31 Graph the magnitudes and phases of the CTFTs of these signals in the f form a x t δ t 2 b x t u t u t 1 c x t 5 rect t 2 4 d x t 25 sinc 10 t 2 e x t 6 sin 200πt f x t 2 e t u t g x t 4 e 3 t 2 rob28124ch06229306indd 288 041216 131 pm Exercises with Answers 289 32 Graph the magnitudes and phases of the CTFTs of these signals in the ω form a x t δ 2 t b x t sgn 2t c x t 10 tri t 4 20 d x t 110 sinc 2 t 1 3 e x t cos 200πt π4 4 f x t 2 e 3t u t 2 e 3t u 3t g x t 7 e 5 t Answers f 1 1 X f 20 f 1 1 π π X f f 100 100 Xf 3 f 100 100 π π X f f 5 5 Xf 1 f 5 5 π π Xf f 3 3 Xf 5 f 3 3 π π X f f 1 1 Xf 1 f 1 1 π π Xf f 2 2 Xf 1 f 2 2 π π 2 2 Xf f 10 10 X f 3 f 10 10 π π Xf Answers ω 10 10 X jω 3 ω 10 10 π π X jω ω 10 10 X jω 1 ω 10 10 15708 15708 X jω ω 2 2 X jω 200 ω 2 2 π π X jω ω 4π 4π 4π 4π X jω π ω π π X jω 4π 4π 4π 4π ω X jω 6 ω π π X jω ω 4 4 X jω 03 ω 4 4 π π X jω ω 700 700 X jω 1 ω 700 700 π π X jω rob28124ch06229306indd 289 041216 131 pm C h a p t e r 6 ContinuousTime Fourier Methods 290 33 Graph the inverse CTFTs of these functions a X f 15rect f4 b X f sinc 10f 30 c X f 18 9 f 2 d X f 1 10 jf e X f 16 δ f 3 δ f 3 f X f 8δ 5f 85 δ f g X f 3jπf Answers 10 10 t xt 0005 1 1 t xt 2 025 025t xt 20 1 1 t xt 3 3 0016667 0066667t xt 8 1 1 t xt 60 20 1 1 t xt 05 05 34 Graph the inverse CTFTs of these functions a X jω e 4 ω 2 b X jω 7 sinc 2 ωπ c X jω jπ δ ω 10π δ ω 10π d X jω π20 δ π4 ω e X jω 5πjω 10πδ ω f X jω 6 3 jω g X jω 20tri 8ω Answers 1 1 t xt 4 18 04 04 t xt 1 1 4 4 t xt 4 200 200 t xt 05 40 40 t xt 02 02 15 t xt 6 8 8 t xt 02 35 Graph the magnitudes and phases of these functions Graph the inverse CTFT of the functions also a X jω 10 3 jω 4 5 jω b X f 4 sinc f 1 2 sinc f 1 2 c X f j 10 tri f 2 8 tri f 2 8 rob28124ch06229306indd 290 041216 131 pm Exercises with Answers 291 d X f δ f 1050 δ f 950 δ f 950 δ f 1050 e X f δ f 1050 2δ f 1000 δ f 950 δ f 950 2δ f 1000 δ f 1050 Answers 1 2 6 20 20 4 20 20 π π t xt ω X jω ω X jω 1 1 16 16 10 10 5 10 10 π π t xt f X f f X f 004 004 4 4 1200 1200 1 1200 1200 π π t xt f X f f X f 004 004 8 8 1000 1000 2 1000 1000 π π t xt f X f f X f 05 05 05 05 15 15 01 15 15 π π t xt f X f f X f 36 Graph these signals versus time Graph the magnitudes and phases of the CTFTs of these signals in either the f or ω form whichever is more convenient In some cases the time graph may be conveniently done first In other cases it may be more convenient to do the time graph after the CTFT has been found by finding the inverse CTFT a x t e π t 2 sin 20πt b x t cos 400πt 1100 δ 1100 t 1100 n cos 4πn δ t n100 c x t 1 cos 400πt cos 4000πt d x t 1 rect 100t δ 150 t cos 500πt e x t rect t7 δ 1 t Answers 004 004 008 008 500 500 1 500 500 π π t xt f X f f X f 001 001 2 2 2500 2500 05 2500 2500 π π t xt f X f f X f rob28124ch06229306indd 291 041216 131 pm C h a p t e r 6 ContinuousTime Fourier Methods 292 001 001 2 2 2500 2500 05 2500 2500 π π t xt f X j2πf f X j2πf 6 6 1 4 4 7 4 4 π π t xt f X f f X f 2 2 1 1 12 12 05 12 12 π π t xt f X f f X f 37 Graph the magnitudes and phases of these functions Graph the inverse CTFTs of the functions also a X f sinc f4 δ1 f b X f sinc f 1 4 sinc f 1 4 δ1 f c X f sinc f sinc 2f Answers 2 2 1 8 8 2 8 8 π π t xt f X f f X f 2 2 4 16 16 1 16 16 π π t xt f X f f X f 2 2 05 2 2 1 2 2 π π t xt f X f f X f 38 Graph these signals versus time and the magnitudes and phases of their CTFTs a x t t sin 2πλdλ b x t t rect λdλ 0 t 12 t 12 t 12 1 t 12 c x t t 3 sinc 2λdλ rob28124ch06229306indd 292 041216 131 pm Exercises with Answers 293 Answers 1 1 1 2 2 1 2 2 π π t xt f Xf f X f 4 4 1 2 2 2 1 2 2 π π t xt f X f f X f 2 2 02 02 1 1 01 1 1 π π t xt f X f f X f Relation of CTFS to CTFT 39 The transition from the CTFS to the CTFT is illustrated by the signal x t rect tw δ T 0 t or x t n rect t n T 0 w The complex CTFS for this signal is given by c x k Aw T 0 sinc kw T 0 Graph the modified CTFS T 0 c x k Awsinc w k f 0 for w 1 and f 0 05 01 and 002 versus k f 0 for the range 8 k f 0 8 Answers 8 8 kf0 kf0 8 8 Aw π π cxk T0cxk kf0 kf0 8 8 Aw 8 8 π π cxk T0cxk kf0 kf0 8 8 8 8 Aw π π cxk T0cxk rob28124ch06229306indd 293 041216 131 pm C h a p t e r 6 ContinuousTime Fourier Methods 294 40 Find the CTFS and CTFT of these periodic functions and compare answers a x t rect t δ 2 t b x t tri 10t δ 14 t Answers X f k 12 sinc k2 δ f k2 k c x k δ f k f 0 X f 25 k sinc 2 2k5 δ f 4k k c x k δ f 4k Numerical CTFT 41 Find and graph the approximate magnitude and phase of the CTFT of x t 4 t 2 2 rect t 2 4 using the DFT to approximate the CTFT Let the time between samples of x t be 116 and sample over the time range 0 t 16 Answer f 5 5 10 15 f 2 2 X f X f 10 5 10 5 10 5 10 System Response 42 A system has a frequency response H jω 100 jω 200 a If we apply the constant signal x t 12 to this system the response is also a constant What is the numerical value of the response constant b If we apply the signal x t 3 sin 14πt to the system the response is yt A sin 14πt θ What are the numerical values of A and θ θ in radians Answers 14649 02165 6 EXERCISES WITHOUT ANSWERS Fourier Series 43 Why can the function c x k 0 k 2 k 2 k 3 0 k 3 not be the harmonic function of a real signal rob28124ch06229306indd 294 041216 131 pm 295 Exercises without Answers 44 A continuoustime signal x t consists of the periodic repetition of an even function It is represented by a CTFS using its fundamental period as the representation time a What is definitely known about its complex exponential form harmonic function b If the representation time is doubled what additional fact is known about the new complex harmonic function 45 A periodic signal x t with a fundamental period of 2 seconds is described over one period by x t sin 2πt t 12 0 12 t 1 Plot the signal and find its CTFS description Then graph on the same scale approximations to the signal x N t given by x N t kN N X k e j2πkt T 0 for N 1 2 and 3 In each case the time scale of the graph should cover at least two periods of the original signal 46 Using MATLAB graph the following signals over the time range 3 t 3 a x 0 t 1 b x1 t x 0 t 2 cos 2πt c x2 t x1 t 2 cos 4πt d x 20 t x19 t 2 cos 40πt For each part a through d numerically evaluate the area of the signal over the time range 12 t 12 47 Identify which of these functions has a complex CTFS cg k for which 1 Re cg k 0 for all k or 2 Im c g k 0 for all k or 3 neither of these conditions applies a g t 18 cos 200πt 22 cos 240πt b g t 4 sin 10πt sin 2000πt c g t tri t 1 4 δ 10 t 48 A continuoustime signal x t with fundamental period T 0 has a CTFS harmonic function c x k tri k10 using the representation time T T 0 If z t x 2t and c z k is its CTFS harmonic function using the same representation time T T 0 find the numerical values of c z 1 and c z 2 49 Each signal in Figure E49 is graphed over a range of exactly one fundamental period Which of the signals have harmonic functions c x k that have a purely real value for every value of k Which have a purely imaginary value for every value of k rob28124ch06229306indd 295 041216 131 pm C h a p t e r 6 ContinuousTime Fourier Methods 296 a b 0 05 1 2 1 0 1 2 t 1 0 1 0 05 1 15 t xt xt Figure E49 50 Find the CTFS harmonic function c x k for the continuoustime function x t sinc 8t 12 δ 12 t using its fundamental period as the representation time 51 If x t 4sinc t2 δ 9 t and the representation time is one fundamental period use Parsevals theorem to find the numerical value of the signal power of x t 52 In Figure E52 is graphed exactly one period of a periodic function x t Its harmonic function c x k with T T 0 can be written as c x k Ag bk e jck t 10 10 10 10 xt Figure E52 What is the name of the function g What are the numerical values of A b and c 53 Find the numerical values of the literal constants a 8 cos 30πt ℱ𝒮 115 A δ k m δ k m Find A and m b 5 cos 10πt ℱ𝒮 1 A δ k m δ k m Find A and m c 3 rect t6 δ 18 t ℱ𝒮 18 A sinc bk Find A and b d 9 rect 3t δ 2 t ℱ𝒮 4 A sinc bk δ m k Find A b and m rob28124ch06229306indd 296 041216 131 pm 297 Exercises without Answers 54 If x t ℱ𝒮 200 ms 4 δ k 1 3δ k δ k 1 a What is average value of x t b Is x t an even or an odd function or neither c If y t d dt x t what is the value of c y 1 55 If c x k 3 δ k 1 δ k 1 and c y k j2 δ k 2 δ k 2 and both are based on the same representation time and z t x t y t c z k can be written in the form c z k A δ k a δ k b δ k c δ k d Find the numerical values of A a b c and d 56 Let x t ℱ𝒮 10 4 sinc k5 and let y t d dt x t and let y t ℱ𝒮 10 c y k a What is the numerical value of c y 3 b Is x t an even function an odd function neither or impossible to know c Is y t an even function an odd function neither or impossible to know 57 If x t ℱ𝒮 T 0 c x k u k 3 u k 4 find the average signal power P x of x t 58 In Figure E58 is a graph of one fundamental period of the product of an unshifted cosine x1 t B cos 2π f x t and an unshifted sine x2 t B sin 2π f x t of equal amplitude and frequency If A 6 and T 0 10 find B and f x Figure E58 A t T0 x1 tx2 t 59 Four continuoustime signals have the following descriptions c x1 k 3 sinc k2 δ k T T 0 1 x 2 t is periodic and one period of x 2 t t δ 1 t rect t5 25 t 25 and T T 0 5 x 3 t is periodic and one period of x 3 t t 1 2 t 2 and T T 0 4 rob28124ch06229306indd 297 041216 131 pm C h a p t e r 6 ContinuousTime Fourier Methods 298 10 0 10 0 2 4 6 k 10 0 10 2 0 2 k cx4 k cx4 k c x4 k 0 k 10 T T 0 10 Answer the following questions 1 Which continuoustime signals are even functions 2 Which continuoustime signals are not even but can be made even by adding or subtracting a constant 3 Which continuoustime signals are odd functions 4 Which continuoustime signals are not odd but can be made odd by adding or subtracting a constant 5 Which continuoustime signals have an average value of zero 6 Which of the continuoustime signals are square waves 7 What is the average signal power of x1 t 8 What is the average signal power of x4 t 60 In some types of communication systems binary data are transmitted using a technique called binary phaseshift keying BPSK in which a 1 is represented by a burst of a continuoustime sine wave and a 0 is represented by a burst which is the exact negative of the burst that represents a 1 Let the sine frequency be 1 MHz and let the burst width be 10 periods of the sine wave Find and graph the CTFS harmonic function for a periodic binary signal consisting of alternating 1s and 0s using its fundamental period as the representation time 61 Match the CTFS magnitude graphs to the time functions In all cases T 4 rob28124ch06229306indd 298 041216 131 pm 299 Exercises without Answers 0 2 4 t 0 2 4 t 0 2 4 t 0 2 4 3 2 1 0 1 2 3 3 2 1 0 1 2 3 3 2 1 0 1 2 3 3 2 1 0 1 2 3 t 0 2 4 3 2 1 0 1 2 3 t xt xt xt xt xt a b c d e f g h i j 10 0 10 0 02 04 06 08 1 0 02 04 06 08 1 0 02 04 06 08 1 0 02 04 06 08 1 0 02 04 06 08 1 k 10 0 10 k 10 0 10 k 10 0 10 k 10 0 10 k 10 0 10 k 10 0 10 k 10 0 10 k 10 0 10 0 02 04 06 08 1 0 02 04 06 08 1 0 02 04 06 08 1 0 02 04 06 08 1 0 02 04 06 08 1 k Xk Xk Xk Xk Xk Xk Xk Xk Xk Xk 10 0 10 k 62 A continuoustime system is described by the differential equation a 2 y t a 1 y t a 0 y t b 2 x t b 1 x t b 0 and the system is excited by x t rect tw δ T 0 t a Let a 2 1 a 1 20 a 0 250100 b 2 1 b 1 0 and b 0 250000 Also let T 0 3 2π b 0 and let w T 0 2 Graph the response yt over the time range 0 t 2 T 0 At what harmonic number is the magnitude of the harmonic response a minimum What cyclic frequency does that correspond to Can you see in y t the effect of this minimum magnitude response b Change T 0 to 2π b 0 and repeat part a rob28124ch06229306indd 299 041216 131 pm C h a p t e r 6 ContinuousTime Fourier Methods 300 Forward and Inverse Fourier Transforms 63 A system is excited by a signal x t 4rect t2 and its response is yt 10 1 e t1 u t 1 1 e t1 u t 1 What is its impulse response 64 Graph the magnitudes and phases of the CTFTs of the following functions a gt 5δ 4t b g t 4 δ 4 t 1 δ 4 t 3 c g t u 2t u t 1 d gt sgnt sgn t e gt rect t 1 2 rect t 1 2 f gt rect t4 g gt 5 tri t5 2 tri t2 h gt 32 rect t8 rect t2 65 Graph the magnitudes and phases of the CTFTs of the following functions a rect 4t b rect4t 4δt c rect4t 4δ t 2 d rect 4t 4δ 2t e rect 4t δ 1 t f rect 4t δ 1 t 1 g 12 rect 4t δ 12 t h 12 rectt δ 12 t 66 Given that yt x t 2 x t 2 and that Y f 3 sinc 2 2f find xt 67 An LTI system has a frequency response H jω 1 jω j6 1 jω j6 a Find an expression for its impulse response h t which does not contain the square root of minus one j b Is this system stable Explain how you know 68 A periodic signal has a fundamental period of 4 seconds a What is the lowest positive frequency at which its CTFT could be nonzero b What is the nextlowest positive frequency at which its CTFT could be nonzero 69 A signal xt has a CTFT X f j2πf 3 jf10 a What is the total net area under the signal xt b Let yt be the integral of xt yt t x λ dλ What is the total net area under yt c What is the numerical value of X f in the limit as f 70 Answer the following questions a A signal x 1 t has a CTFT X 1 f If x 2 t x 1 t 4 what is the relationship between X 1 f and X 2 f rob28124ch06229306indd 300 041216 131 pm 301 Exercises without Answers b A signal x 1 t has a CTFT X 1 f If x 2 t x 1 t5 what is the relationship between the maximum value of X 1 f and the maximum value of X 2 f c A CTFT has the value e jπ4 at a frequency f 20 What is the value of that same CTFT at a frequency of f 20 71 If xt δ T 0 t rect t4 and x t ℱ X f find three different numerical values of T 0 for which X f Aδ f and the corresponding numerical impulse strengths A 72 If yt ℱ Y f and d dt y t ℱ 1 e jπf2 find and graph yt 73 Let a signal xt have a CTFT X f f f 2 0 f 2 Let yt x 4 t 2 Find the values of the magnitude and phase of Y 3 where y t ℱ Y f 74 Graph the magnitude and phase of the CTFT of each of the signals in Figure E74 ω form a t xt 01 20 20 b t xt 3 10 c t xt 7 5 5 10 10 d t xt 7 2 3 7 8 13 Figure E74 75 Graph the inverse CTFTs of the functions in Figure E75 a 20 4 4 4 4 f X f f X f b f X f 20 4 4 f 4 4 π 2 π 2 X f rob28124ch06229306indd 301 041216 131 pm C h a p t e r 6 ContinuousTime Fourier Methods 302 c f X f 5 5 f 5 5 2 π π X f d f X f 5 5 f 5 5 5 8 X f Figure E75 76 If a CTFT in the f form is 24 δ 3 f and in the ω form it is A δ b cω what are the numerical values of A b and c 77 Find the numerical values of the literal constants in the following CTFT pairs a A e at u t ℱ 8 jω 2 b A e at u t ℱ 1 j8ω 2 c 3 δ 4 t ℱ A δ a f d 4 δ 3 5t ℱ A δ a bf e 4 δ 3 5t ℱ A δ a f Hint Use the definition of the continuoustime periodic impulse and continuoustime impulse properties f A sin 200πt π3 ℱ j8 δ f a δ f a e bf g Arect ta rect tb ℱ 40 sinc 2f sinc 4f h d dt 3tri 5t ℱ A f sinc 2 af i 10 sin 20t ℱ A δ ω b δ ω b j 4 cos 6t π2 ℱ A δ f b δ f b e cf k δ t 2 δ t 2 ℱ A sin 2πbf l u t 5 u t 5 ℱ A sinc bω m u t 3 u t 6 ℱ A sinc bf e cf n A sinc bt cos ct ℱ rect f 50 4 rect f 50 4 o A cos b t c ℱ δ f 5 e jπ4 δ f 5 e jπ4 p A rect bt δ c t ℱ 9 sinc f10 δ 1 f rob28124ch06229306indd 302 041216 131 pm 303 Exercises without Answers 78 Below are two lists one of timedomain functions and one of frequency domain functions By writing the number of a timedomain function match the frequencydomain functions to their inverse CTFTs in the list of timedomain functions a Time Domain Frequency Domain 1 12 δ 18 t 5 δ f 200 δ f 200 2 5 sinc 2 t 2 52 rect f2 e j4πf 3 3δ 3t 9 180 sinc 20f e j8πf 4 7 sinc 2 t12 84 tri 12f 5 5 sinc 2 t 2 96 sinc 4f e j2πf 6 5 cos 200πt 4 δ 8 f 7 2 tri t 5 10 e j6πf 8 3δ t 3 10 sinc 2 5f e j10πf 9 24 u t 1 u t 3 10 2 δ 14 t 11 9 rect t 4 20 12 2 tri t 10 5 13 24 u t 3 u t 1 14 10 cos 400πt b Time Domain Frequency Domain 1 3δ t 3 4 δ 8 f 2 3 sinc 8t 7 0375 rect ω16π e j7ω 3 rect t 3 6 e j3ω 4 12 u t 3 u t 5 12 tri 3f e j2πf 5 4 sinc 2 t 1 3 0375 rect f8 e j7πf4 6 10 sin 5πt j10π δ ω 10π δ ω 10π 7 12 δ 18 t 125 sinc 2 f4 e j4πf 8 3 sinc 8 t 7 3 e j3ω 9 3δ 3t 9 96 sinc 4ωπ e jω 10 12 u t 3 u t 5 6 sinc 6f e j6πf rob28124ch06229306indd 303 041216 131 pm C h a p t e r 6 ContinuousTime Fourier Methods 304 11 18 tri 6 t 5 3 sinc 2 3ωπ e j5ω 12 5 tri 4 t 2 13 2 δ 4 t 14 5 sin 10πt 79 Find the inverse CTFT of the real frequencydomain function in Figure E79 and graph it Let A 1 f 1 95 kHz and f 2 105 kHz A f1 f1 f2 f2 f X f Figure E79 80 In Figure E801 are graphs of some continuoustime energy signals In Figure E802 are graphs of some magnitude CTFTs in random order Find the CTFT magnitude that matches each energy signal 10 8 6 4 2 0 2 4 6 8 10 2 1 0 1 2 t 10 8 6 4 2 0 2 4 6 8 10 2 1 0 1 2 t xt xt 10 8 6 4 2 0 2 4 6 8 10 2 1 0 1 2 t 10 8 6 4 2 0 2 4 6 8 10 2 1 0 1 2 t xt xt 1 2 3 4 Figure E801 rob28124ch06229306indd 304 041216 131 pm 305 Exercises without Answers Figure E802 4 2 0 2 4 0 1 2 3 4 5 6 7 8 f 4 2 0 2 4 0 1 2 3 4 5 6 7 8 f X f Xf X f Xf X f X f 4 2 0 2 4 0 1 2 3 4 5 6 7 8 f 4 2 0 2 4 0 1 2 3 4 5 6 7 8 f 4 2 0 2 4 0 1 2 3 4 5 6 7 8 f 4 2 0 2 4 0 1 2 3 4 5 6 7 8 f b a d c f e 81 In many communication systems a device called a mixer is used In its simplest form a mixer is simply an analog multiplier That is its response signal yt is the product of its two excitation signals If the two excitation signals are x1t 10sinc 20t and x2t 5 cos 2000πt graph the magnitude of the CTFT of yt Y f and compare it to the magnitude of the CTFT of x1t In simple terms what does a mixer do 82 One major problem in real instrumentation systems is electromagnetic interference caused by the 60 Hz power lines A system with an impulse response of the form ht Aut ut t0 can reject 60 Hz and all its harmonics Find the numerical value of t 0 that makes this happen 83 In electronics one of the first circuits studied is the rectifier There are two forms the halfwave rectifier and the fullwave rectifier The halfwave rectifier cuts off half of an excitation sinusoid and leaves the other half intact The fullwave rectifier reverses the polarity of half of the excitation sinusoid and leaves the other half intact Let the excitation sinusoid be a typical household voltage 120 Vrms at 60 Hz and let both types of rectifiers alter the negative half of the sinusoid while leaving the positive half unchanged Find and graph the magnitudes of the CTFTs of the responses of both types of rectifiers either form System Response 84 An LTI continuoustime system is described by the differential equation 2 y t 5yt xt rob28124ch06229306indd 305 041216 131 pm C h a p t e r 6 ContinuousTime Fourier Methods 306 Which is the correct description of this systems frequency response a The system attenuates low frequencies more than high frequencies b The system attenuates high frequencies more than low frequencies c The system has the same effect on all frequencies Relation of CTFS to CTFT 85 An aperiodic signal xt has a CTFT X f 0 f 4 4 f 2 f 4 2 f 2 4tri f 4 2tri f 2 Let x p t x t δ 2 t k x t 2k and let X p k be its CTFS harmonic function a Draw the magnitude of X f vs f b Find an expression for c x p k c Find the numerical values of c x p 3 c x p 5 c x p 10 rob28124ch06229306indd 306 041216 131 pm 307 71 INTRODUCTION AND GOALS In Chapter 6 we developed the continuoustime Fourier series as a method of repre senting periodic continuoustime signals and finding the response of a continuoustime LTI system to a periodic excitation Then we extended the Fourier series to the Fourier transform by letting the period of the periodic signal approach infinity In this chapter we take a similar path applied to discretetime systems Most of the basic concepts are the same but there are a few important differences CH APTER GOAL S 1 To develop methods of expressing discretetime signals as linear combinations of sinusoids real or complex 2 To explore the general properties of these ways of expressing discretetime signals 3 To generalize the discretetime Fourier series to include aperiodic signals by defining the discretetime Fourier transform 4 To establish the types of signals that can or cannot be described by a discrete time Fourier transform 5 To derive and demonstrate the properties of the discretetime Fourier transform 6 To illustrate the interrelationships among the Fourier methods 72 THE DISCRETETIME FOURIER SERIES AND THE DISCRETE FOURIER TRANSFORM LINEARITY AND COMPLEXEXPONENTIAL EXCITATION As was true in continuoustime if a discretetime LTI system is excited by a sinusoid the response is also a sinusoid with the same frequency but generally a different mag nitude and phase If an LTI system is excited by a sum of signals the overall response is the sum of the responses to each of the signals individually The discretetime Fourier series DTFS expresses arbitrary periodic signals as linear combinations of sinusoids realvalued or complex so we can use superposition to find the response of any LTI system to any arbitrary signal by summing the responses to the individual sinusoids Figure 71 C H A P T E R 7 DiscreteTime Fourier Methods rob28124ch07307353indd 307 041216 140 pm C h a p t e r 7 DiscreteTime Fourier Methods 308 The sinusoids can be realvalued or complex Realvalued sinusoids and complex sinusoids are related by cos x e jx e jx 2 and sin x e jx e jx j2 and this relationship is illustrated in Figure 72 Consider an arbitrary periodic signal xn that we would like to represent as a linear combination of sinusoids as illustrated by the center graph in Figure 73 Here we use realvalued sinusoids to simplify the visualization Figure 71 The equivalence of the response of an LTI system to a signal and the sum of its responses to complex sinusoids whose sum is equivalent to the signal xn A1e j2πnN1A2e j2πnN2A3e j2πnN3 A1e j2πnN1 yn yn hn hn A2e j2πnN2 hn A3e j2πnN3 B1e j2πnN1 B2e j2πnN2 B3e j2πnN3 hn Figure 72 Addition and subtraction of e j2πn16 and e j2πn16 to form 2 cos 2πn16 and j2 sin 2πn16 Ree j2πn16 Ime j2πn16 n Reej2πn16 Imej2πn16 n Ree j2πn16ej2πn16 Ime j2πn16ej2πn16 Ime j2πn16ej2πn16 n n Ree j2πn16ej2πn16 rob28124ch07307353indd 308 041216 140 pm 72 The DiscreteTime Fourier Series and the Discrete Fourier Transform 309 In Figure 73 the signal is approximated by a constant 02197 which is the average value of the signal A constant is a special case of a sinusoid in this case 02197cos2πknN with k 0 This is the best possible approximation of xn by a constant because the mean squared error between xn and the approximation is a mini mum We can make this poor approximation better by adding to the constant a sinusoid whose fundamental period N is the fundamental period of xn Figure 74 This is the best approximation that can be made using a constant and a single sinusoid of the same fundamental period as xn We can improve the approximation further by adding a sinusoid at a frequency of twice the fundamental frequency of xn Figure 75 Figure 73 Signal approximated by a constant n 10 30 Constant 06 06 n 10 30 xn 1 Approximation by a Constant n 10 30 Exact xn 1 n0N n0 Figure 74 Signal approximated by a constant plus a single sinusoid n 10 30 Sinusoid 1 06 06 n 10 30 Exact xn 1 n 10 30 xn 1 Approximation Through 1 Sinusoid n0N n0 n 10 30 Sinusoid 2 06 06 n 10 30 Exact xn 1 n 10 30 xn 1 Approximation Through 2 Sinusoids n0N n0 Figure 75 Signal approximated by a constant plus two sinusoids rob28124ch07307353indd 309 041216 140 pm C h a p t e r 7 DiscreteTime Fourier Methods 310 If we keep adding properly chosen sinusoids at higher integer multiples of the fun damental frequency of xn we can make the approximation better and better Unlike the general case in continuous time with a finite number of sinusoids the representation becomes exact Figure 76 This illustrates one significant difference between continuoustime and dis cretetime Fourier series representations In discrete time exact representation of a periodic signal is always achieved with a finite number of sinusoids Just as in the CTFS k is called the harmonic number and all the sinusoids have frequencies that are k times the fundamental cyclic frequency which for the DTFS is 1N The DTFS represents a discretetime periodic signal of fundamental period N0 as a linear combination of complex sinusoids of the form xn kN c x k e j2πknN where N m N 0 m an integer and cxk is the DTFS harmonic function The notation kN is equivalent to k n 0 n 0 N1 where n0 is arbitrary in other words a summation over any set of N consecutive values of k Although the most commonly used value for N is the fundamental period of the signal N0 m 1 N does not have to be N0 N can be any period of the signal In discretetime signal and system analysis there is a very similar form of represen tation of discretetime periodic signals using the discrete Fourier transform DFT first mentioned in Chapter 6 It also represents periodic discretetime signals as a linear combination of complex sinusoids The inverse DFT is usually written in the form xn 1 N k0 N1 Xk e j2πknN where Xk is the DFT harmonic function of xn and Xk N cxk The name ends with transform instead of series but since it is a linear combination of sinusoids at a discrete set of frequencies for terminological consistency it probably should have Figure 76 Signal represented by a constant plus six sinusoids n 10 30 Sinusoid 6 06 06 n 10 30 Exact xn 1 n 10 30 xn 1 Exact Representation by 6 Sinusoids n0N n0 rob28124ch07307353indd 310 041216 140 pm 72 The DiscreteTime Fourier Series and the Discrete Fourier Transform 311 been called a series The name transform probably came out of its use in digital signal processing in which it is often used to find a numerical approximation to the CTFT The DFT is so widely used and so similar to the DTFS that in this text we will concentrate on the DFT knowing that conversion to the DTFS is very simple The formula xn 1 N k0 N1 Xk e j2πknN is the inverse DFT It forms the time domain function as a linear combination of complex sinusoids The forward DFT is Xk n0 N1 xn e j2πknN where N is any period of xn It forms the harmonic function from the timedomain function As shown in Chapter 6 one important property of the DFT is that Xk is periodic Xk Xk N k any integer So now it should be clear why the summation in the inverse DFT is over a finite range of k values The harmonic function Xk is periodic with period N and therefore has only N unique values The summation needs only N terms to utilize all the unique values of Xk The formula for the inverse DFT is most commonly written as xn 1 N k0 N1 Xk e j2πknN but since Xk is periodic with period N it can be written more generally as xn 1 N k N Xk e j2πknN ORTHOGONALITY AND THE HARMONIC FUNCTION We can find the forward DFT Xk of xn by a process analogous to the one used for the CTFS To streamline the notation in the equations to follow let W N e j2πN 71 Since the beginning point of the summation k N Xk e j2πknN is arbitrary let it be k 0 Then if we write e j2πknN for each n in n 0 n n 0 N using 71 we can write the matrix equation x n 0 x n 0 1 x n 0 N 1 x 1 N W N 0 W N n 0 W N n 0 N1 W N 0 W N n 0 1 W N n 0 1 N1 W N 0 W N n 0 N1 W N n 0 N1 N1 W X0 X1 XN 1 X 72 or in the compact form Nx WX If W is nonsingular we can directly find X as X W 1 Nx Equation 72 can also be written in the form rob28124ch07307353indd 311 041216 140 pm C h a p t e r 7 DiscreteTime Fourier Methods 312 N x n 0 x n 0 1 x n 0 N 1 1 1 1 k0 X0 W N n 0 W N n 0 1 W N n 0 N1 k1 X1 W N n 0 N1 W N n 0 1 N1 W N n 0 N1 N1 kN1 XN 1 73 or Nx w 0 X0 w 1 X1 w N1 XN 1 74 where W w 0 w 1 w N1 The elements of the first column vector w0 are all the constant one and can be thought of as the function values in a unitamplitude complex sinusoid of zero frequency The second column vector w1 consists of the function val ues from one cycle of a unitamplitude complex sinusoid over the time period n 0 n n 0 N Each succeeding column vector consists of the function values from k cycles of a unitamplitude complex sinusoid at the next higher harmonic number over the time period n 0 n n 0 N Figure 77 illustrates these complex sinusoids for the case of N 8 and n0 0 Notice that the sequence of complex sinusoid values versus n for k 7 looks just like the sequence for k 1 except rotating in the opposite direction In fact the sequence for k 7 is the same as the sequence for k 1 This had to happen because of the periodicity of the DFT Figure 77 Illustration of a complete set of orthogonal basis vectors for N 8 and n0 0 k 0 n Ime j2πkn8 Ree j2πkn8 k 1 n Ime j2πkn8 Ree j2πkn8 k 2 n Ime j2πkn8 Ree j2πkn8 k 3 n Ime j2πkn8 Ree j2πkn8 k 4 n Ime j2πkn8 Ree j2πkn8 k 5 n Ime j2πkn8 Ree j2πkn8 k 6 n Ime j2πkn8 Ree j2πkn8 k 7 n Ime j2πkn8 Ree j2πkn8 These vectors form a family of orthogonal basis vectors Recall from basic linear algebra or vector analysis that the projection p of a real vector x in the direction of another real vector y is p x T y y T y y 75 and when that projection is zero x and y are said to be orthogonal That happens when the dot product or scalar product or inner product of x and y x T y is zero If the vec tors are complexvalued the theory is practically the same except the dot product is x H y and the projection is p x H y y H y y 76 rob28124ch07307353indd 312 041216 140 pm 72 The DiscreteTime Fourier Series and the Discrete Fourier Transform 313 where the notation x H means the complexconjugate of the transpose of x This is such a common operation on complexvalued matrices that the transpose of a complexvalued matrix is often defined as including the complexconjugation oper ation This is true for transposing matrices in MATLAB A set of correctly chosen orthogonal vectors can form a basis An orthogonal vector basis is a set of vectors that can be combined in linear combinations to form any arbitrary vector of the same dimension The dot product of the first two basis vectors in 74 is w 0 H w 1 1 1 1 W N n 0 W N n 0 1 W N n 0 N1 W N n 0 1 W N W N N1 77 The sum of a finitelength geometric series is n0 N1 r n N r 1 1 r N 1 r r 1 Summing the geometric series in 77 w 0 H w 1 W n 0 1 W N N 1 W N W N n 0 1 e j2π 1 e j2πN 0 proving that they are indeed orthogonal if N 1 In general the dot product of the k1harmonic vector and the k2harmonic vector is w k 1 H w k 2 W N n 0 k 1 W N n 0 1 k 1 W N n 0 N1 k 1 W N n 0 k 2 W N n 0 1 k 2 W N n 0 N1 k 2 w k 1 H w k 2 W N n 0 k 2 k 1 1 W N k 2 k 1 W N N1 k 2 k 1 w k 1 H w k 2 W N n 0 k 2 k 1 1 W N k 2 k 1 N 1 W N k 2 k 1 W N n 0 k 2 k 1 1 e j2π k 2 k 1 1 e j2π k 2 k 1 N w k 1 H w k 2 0 k 1 k 2 N k 1 k 2 Nδ k 1 k 2 This result is zero for k 1 k 2 because the numerator is zero and the denominator is not The numerator is zero because both k1 and k2 are integers and therefore e j2π k 2 k 1 is one The denominator is not zero because both k1 and k2 are in the range 0 k 1 k 2 N and the ratio k 2 k 1 N cannot be an integer if k 1 k 2 and N 1 So all the vectors in 74 are mutually orthogonal rob28124ch07307353indd 313 041216 140 pm C h a p t e r 7 DiscreteTime Fourier Methods 314 The fact that the columns of W are orthogonal leads to an interesting interpretation of how X can be calculated If we premultiply all the terms in 74 by w 0 H we get w 0 H Nx w 0 H w 0 N X0 w 0 H w 1 0 X1 w 0 H w N1 0 XN 1 NX0 Then we can solve for X0 as X0 w 0 H Nx w 0 H w 0 N w 0 H x The vector X0w0 is the projection of the vector Nx in the direction of the basis vec tor w0 Similarly each Xkwk is the projection of the vector Nx in the direction of the basis vector wk The value of the harmonic function Xk can be found at each harmonic number as Xk w k H x and we can summarize the entire process of finding the harmonic function as X w 0 H w 1 H w N1 H x W H x 78 Because of the orthogonality of the vectors w k 1 and w k 2 k 1 k 2 the product of W and its complexconjugate transpose W H is W W H w 0 w 1 w N1 w 0 H w 1 H w N1 H N 0 0 0 N 0 0 0 N NI Dividing both sides by N W W H N 1 0 0 0 1 0 0 0 1 I Therefore the inverse of W is W 1 W H N and from X W 1 Nx we can solve for X as X W H x 79 rob28124ch07307353indd 314 041216 140 pm 72 The DiscreteTime Fourier Series and the Discrete Fourier Transform 315 which is the same as 78 Equations 78 and 79 can be written in a summation form Xk n n 0 n 0 N1 xn e j2πknN Now we have the forward and inverse DFT formulas as Xk n n 0 n 0 N1 xn e j2πknN xn 1 N k N 0 Xk e j2πknN 710 If the timedomain function xn is bounded on the time n 0 n n 0 N the harmonic function can always be found and is itself bounded because it is a finite summation of bounded terms In most of the literature concerning the DFT the transform pair is written in this form xk n0 N1 xn e j2πknN xn 1 N k0 N1 Xk e j2πknN 711 Here the beginning point for xn is taken as n0 0 and the beginning point for Xk is taken as k 0 This is the form of the DFT that is implemented in practically all computer languages So in using the DFT on a computer the user should be aware that the result returned by the computer is based on the assumption that the first element in the vector of N values of x sent to the DFT for processing is x0 If the first element is x n 0 n 0 0 then the DFT result will have an extra linear phase shift of e j2πk n 0 N This can be compensated for by multiplying the DFT result by e j2πk n 0 N Similarly if the first value of Xk is not at k 0 the inverse DFT result will be multiplied by a complex sinusoid DISCRETE FOURIER TRANSFORM PROPERTIES In all the properties listed in Table 71 xn 𝒟ℱ𝒯 N Xk and yn 𝒟ℱ𝒯 N Yk If a signal xn is even and periodic with period N then its harmonic function is Xk n0 N1 xn e j2πknN If N is an even number Xk x0 n1 N21 xn e j2πknN xN2 e jπk nN21 N1 xn e j2πknN Xk x0 n1 N21 xn e j2πknN nN1 N21 xn e j2πknN 1 k xN2 Knowing that x is periodic with period N we can subtract N from n in the second summation yielding Xk x0 n1 N21 xn e j2πknN n1 N21 xn e j2πk nN N 1 k xN2 Xk x0 n1 N21 xn e j2πknN e j2πk 1 n1 N21 xn e j2πknN 1 k xN2 rob28124ch07307353indd 315 041216 140 pm C h a p t e r 7 DiscreteTime Fourier Methods 316 Table 71 DFT properties Linearity αxn βyn 𝒟ℱ𝒯 N αXk βYk Time Shifting xn n 0 𝒟ℱ𝒯 N Xk e j2πk n 0 N Frequency Shifting xn e j2π k 0 nN 𝒟ℱ𝒯 N Xk k 0 Time Reversal xn xN n 𝒟ℱ𝒯 N Xk XN k Conjugation x n 𝒟ℱ𝒯 N X k X N k x n x N n 𝒟ℱ𝒯 N X k Time Scaling zn xnm nm an integer 0 otherwise N mN Zk Xk Change of Period N qN q a positive integer X q k qXkq kq an integer 0 otherwise MultiplicationConvolution Duality xnyn 𝒟ℱ𝒯 N 1N YkXk xnyn 𝒟ℱ𝒯 N YkXk where xnyn m N xmyn m Parsevals Theorem 1 N n N xn 2 1 N2 k N Xk 2 X k x 0 n1 N21 xn e j2πknN xn e j2πknN 1 k xN2 Now since xn xn Xk x0 2 n1 N21 xn cos 2πkN 1 k xN2 All these terms are realvalued therefore Xk is also A similar analysis shows that if N is an odd number the result is the same Xk has all real values Also if xn is an odd periodic function with period N all the values of Xk are purely imaginary ExamplE 71 DFT of a periodically repeated rectangular pulse 1 Find the DFT of xn un un n x δ N 0 n 0 n x N 0 using N 0 as the representation time un un n x δ N 0 n 𝒟ℱ𝒯 N 0 n0 n x 1 e j2πkn N 0 rob28124ch07307353indd 316 041216 140 pm 72 The DiscreteTime Fourier Series and the Discrete Fourier Transform 317 Summing the finitelength geometric series un un n x δ N 0 n 𝒟ℱ𝒯 N 0 1 e j2πk n x N 0 1 e j2πkn N 0 e jπk n x N 0 e jπk N 0 e jπk n x N 0 e jπk n x N 0 e jπk N 0 e jπk N 0 un un n x δ N 0 n 𝒟ℱ𝒯 N 0 e jπk n x 1 N 0 sinπk n x N 0 sinπk N 0 0 n x N 0 ExamplE 72 DFT of a periodically repeated rectangular pulse 2 Find the DFT of xn un n 0 un n 1 δ N 0 n 0 n 1 n 0 N 0 From Example 71 we already know the DFT pair un un n x δ N 0 n 𝒟ℱ𝒯 N 0 e jπk n x 1 N 0 sin πk n x N 0 sin πk N 0 0 n x N 0 If we apply the timeshifting property xn n y 𝒟ℱ𝒯 N Xk e j2πk n y N to this result we have un n y un n y n x δ N 0 n 𝒟ℱ𝒯 N 0 e jπk n x 1 N 0 e j2πk n y N 0 sin πk n x N 0 sin πk N 0 0 n x N 0 un n y un n y n x δ N 0 n 𝒟ℱ𝒯 N 0 e jπk n x 2 n y 1 N 0 sin πk n x N 0 sin πk N 0 0 n x N 0 Now let n0 ny and let n 1 n y n x un n 0 un n 1 δ N 0 n 𝒟ℱ𝒯 N 0 e jπk n 0 n 1 1 N sin πk n 1 n 0 N 0 sin πk N 0 0 n 1 n 0 N 0 Consider the special case in which n 0 n 1 1 Then un n 0 un n 1 δ N 0 n 𝒟ℱ𝒯 N 0 sin πk n 1 n 0 N 0 sin πk N 0 n 0 n 1 1 This is the case of a rectangular pulse of width n 1 n 0 2 n 1 1 centered at n 0 This is analogous to a continuoustime periodically repeated pulse of the form T 0 recttw δ T 0 t Compare their harmonic functions T 0 recttw δ T 0 t ℱ𝒮 T 0 w sincwk T 0 sin πwk T 0 πk T 0 un n 0 un n 1 δ N 0 n 𝒟ℱ𝒯 N 0 sin πk n 1 n 0 N 0 sin πk N 0 n 0 n 1 1 rob28124ch07307353indd 317 041216 140 pm C h a p t e r 7 DiscreteTime Fourier Methods 318 The harmonic function of T 0 recttw δ T 0 t is a sinc function Although it may not yet be obvious the harmonic function of un n 0 un n 1 δ N 0 n is a periodically repeated sinc function The DFT harmonic function of a periodically repeated rectangular pulse can be found using this result It can also be found numerically using the fft function in MATLAB This MATLAB program computes the harmonic function both ways and graphs the results for com parison The phase graphs are not identical but they only differ when the phase is π radians and these two phases are equivalent See Figure 78 So the two methods of computing the har monic function yield the same result N 16 Set fundamental period to 16 n0 2 Turn on rectangular pulse at n2 n1 7 Turn off rectangular pulse at n7 n 0N1 Discretetime vector for computing xn over one fundamental period Compute values of xn over one fundamental period x usDnn0 usDnn1 usD is a userwritten unit sequence function X fftx Compute the DFT harmonic function Xk of xn using fft k 0N1 Harmonic number vector for graphing Xk Compute harmonic function Xk analytically Xa expjpikn1n0Nn1n0drclkNn1n0expjpikN close all figurePosition20201200800 subplot221 ptr stemnabsXkfilled grid on setptrLineWidth2MarkerSize4 xlabelitkFontNameTimesFontSize36 ylabelXitkFontNameTimesFontSize36 titlefft ResultFontNameTimesFontSize36 setgcaFontNameTimesFontSize24 subplot223 ptr stemnangleXkfilled grid on setptrLineWidth2MarkerSize4 xlabelitkFontNameTimesFontSize36 ylabelPhase of XitkFontNameTimesFontSize36 setgcaFontNameTimesFontSize24 subplot222 ptr stemnabsXakfilled grid on setptrLineWidth2MarkerSize4 xlabelitkFontNameTimesFontSize36 ylabelXitkFontNameTimesFontSize36 titleAnalytical ResultFontNameTimesFontSize36 setgcaFontNameTimesFontSize24 subplot224 ptr stemnangleXakfilled grid on setptrLineWidth2MarkerSize4 rob28124ch07307353indd 318 041216 140 pm 72 The DiscreteTime Fourier Series and the Discrete Fourier Transform 319 xlabelitkFontNameTimesFontSize36 ylabelPhase of XitkFontNameTimesFontSize36 setgcaFontNameTimesFontSize24 Figure 78 A comparison of the numerical and analytical DFT of a periodicallyrepeated discretetime rectangular pulse k 0 1 2 3 4 5 fft Result k 4 2 0 2 4 k 0 1 2 3 4 5 Xk Xk Analytical Result 0 5 10 15 0 5 10 15 0 5 10 15 0 5 10 15 k 4 2 0 2 4 Phase of Xk Phase of Xk The functional form sin πNx N sin πx see Example 72 appears commonly enough in the analysis of signals and systems to be given the name Dirichlet function Figure 79 drcl t N sin πNt N sin πt 712 Figure 79 The Dirichlet function for N 4 5 7 and 13 t 2 2 drclt4 1 1 t 2 2 drclt5 1 1 t 2 2 drclt7 1 1 t 2 2 drclt13 1 1 rob28124ch07307353indd 319 041216 140 pm C h a p t e r 7 DiscreteTime Fourier Methods 320 For N odd the similarity to a sinc function is obvious the Dirichlet function is an infinite sum of uniformly spaced sinc functions The numerator sin Nπt is zero when t is any integer multiple of 1N Therefore the Dirichlet function is zero at those points unless the denominator is also zero The denominator N sin πt is zero for every integer value of t Therefore we must use LHôpitals rule to evaluate the Dirichlet function at integer values of t lim tm drclt N lim tm sinNπt N sinπt lim tm Nπ cosNπt Nπ cosπt 1 m an integer If N is even the extrema of the Dirichlet function alternate between 1 and 1 If N is odd the extrema are all 1 A version of the Dirichlet function is a part of the MATLAB signal toolbox with the function name diric It is defined as diricx N sinNx2 N sinx2 Therefore drclt N diric2πt N Function to compute values of the Dirichlet function Works for vectors or scalars equally well x sinNpitNsinpit function x drcltN x diric2pitN Function to implement the Dirichlet function without using the MATLAB diric function Works for vectors or scalars equally well x sinNpitNsinpit function x drcltN num sinNpit den Nsinpit I findabsden 10eps numI cosNpitI denI cospitI x numden Using the definition of the dirichlet function the DFT pair from Example 72 can be written as un n 0 un n 1 δ N n 𝒟ℱ𝒯 N e jπk n 1 n 0 N e jπkN n 1 n 0 drclkN n 1 n 0 rob28124ch07307353indd 320 041216 140 pm 72 The DiscreteTime Fourier Series and the Discrete Fourier Transform 321 Table 72 shows several common DFT pairs Table 72 DFT pairs For each pair m is a positive integer e j2πnN 𝒟ℱ𝒯 mN mN δ mN k m cos 2πqnN 𝒟ℱ𝒯 mN mN2 δ mN k mq δ mN k mq sin 2πqnN 𝒟ℱ𝒯 mN jmN2 δ mN k mq δ mN k mq δ N n 𝒟ℱ𝒯 mN m δ mN k 1 𝒟ℱ𝒯 N N δ N k un n 0 un n 1 δ N n 𝒟ℱ𝒯 N e jπk n 1 n 0 N e jπkN n 1 n 0 drclkN n 1 n 0 trin N w δ N n 𝒟ℱ𝒯 N N w drcl 2 kN N w N w an integer sincnw δ N n 𝒟ℱ𝒯 N wrectwkN δ N k THE FAST FOURIER TRANSFORM The forward DFT is defined by Xk n0 N1 xn e j2πnkN A straightforward way of computing the DFT would be by the following algorithm written in MATLAB which directly implements the operations indicated above Acquire the input data in an array x with N elements Initialize the DFT array to a column vector of zeros X zerosN1 Compute the Xks in a nested double for loop for k 0N1 for n 0N1 Xk1 Xk1 xn1expj2pinkN end end rob28124ch07307353indd 321 041216 140 pm C h a p t e r 7 DiscreteTime Fourier Methods 322 One should never actually write this program in MATLAB because the DFT is al ready built in to MATLAB as an intrinsic function called fft The computation of a DFT using this algorithm requires N 2 complex multi plyadd operations Therefore the number of computations increases as the square of the number of elements in the input vector that is being transformed In 1965 James Cooley1 and John Tukey2 popularized an algorithm that is much more effi cient in computing time for large input arrays whose length is an integer power of 2 This algorithm for computing the DFT is the socalled fast Fourier transform or just FFT The reduction in calculation time for the fast Fourier transform algorithm versus the doubleforloop approach presented above is illustrated in Table 73 in which A is the number of complexnumber additions required and M is the number of complexnumber multiplications required the subscript DFT indicates using the straightforward doubleforloop approach and FFT indicates the FFT algorithm As the number of points N in the transformation process is increased the speed advantage of the FFT grows very quickly But these speed improvement factors do not apply if N is not an integer power of two For this reason practically all actual DFT analysis is done with the FFT using a data vector length that is an integer power of 2 In MATLAB if the input vector is an integer power of 2 in length the algorithm used in the MATLAB function fft is the FFT algorithm just discussed If it is not an inte ger power of 2 in length the DFT is still computed but the speed suffers because a less efficient algorithm must be used Table 73 Numbers of additions and multiplications and ratios for several Ns γ N 2γ ADFT MDFT AFFT MFFT ADFT AFFT MDFT MFFT 1 2 2 4 2 1 1 4 2 4 12 16 8 4 15 4 3 8 56 64 24 12 233 533 4 16 240 256 64 32 375 8 5 32 992 1024 160 80 62 128 6 64 4032 4096 384 192 105 213 7 128 16256 16384 896 448 181 366 8 256 65280 65536 2048 1024 319 64 9 512 261632 262144 4608 2304 568 1138 10 1024 1047552 1048576 10240 5120 1023 2048 1 James Cooley received his PhD in applied mathematics from Columbia University in 1961 Cooley was a pioneer in the digital signal processing field having developed with John Tukey the fast Fourier transform He developed the FFT through mathematical theory and applications and has helped make it more widely available by devising algorithms for scientific and engineering applications 2 John Tukey received his PhD from Princeton in mathematics in 1939 He worked at Bell Labs from 1945 to 1970 He developed new techniques in data analysis and graphing and plotting methods that now appear in standard statistics texts He wrote many publications on time series analysis and other aspects of digital signal processing that are now very important in engineering and science He developed along with James Cooley the fast Fourier transform algorithm He is credited with having coined as a contraction of binary digit the word bit the smallest unit of information used by a computer rob28124ch07307353indd 322 041216 140 pm 73 The DiscreteTime Fourier Transform 323 73 THE DISCRETETIME FOURIER TRANSFORM EXTENDING THE DISCRETE FOURIER TRANSFORM TO APERIODIC SIGNALS Consider a discretetime rectangularwave signal Figure 710 Figure 710 A general discretetime rectangularwave signal xn n N0 N0 Nw Nw 1 The DFT harmonic function based on one fundamental period N N 0 is Xk 2 N w 1 drclk N 0 2 N w 1 a sampled Dirichlet function with maxima of 2 N w 1 and a period of N 0 To illustrate the effects of different fundamental periods N 0 let N w 5 and graph the magnitude of Xk versus k for N0 22 44 and 88 Figure 711 The effect on the DFT harmonic function of increasing the fundamental period of xn is to spread it out as a function of harmonic number k So in the limit as N 0 ap proaches infinity the period of the DFT harmonic function also approaches infinity If the period of a function is infinite it is no longer periodic We can normalize by graph ing the DFT harmonic function versus discretetime cyclic frequency k N 0 instead of harmonic number k Then the fundamental period of the DFT harmonic function as graphed is always one rather than N 0 Figure 712 As N 0 approaches infinity the separation between points of Xk approaches zero and the discrete frequency graph becomes a continuous frequency graph Figure 713 Figure 712 Magnitude of the DF T harmonic function of a rectangularwave signal graphed versus kN0 instead of k 1 1 Xk 12 Nw 5 N0 22 1 1 12 1 1 12 k N0 k N0 k N0 Xk Nw 5 N0 44 Xk Nw 5 N0 88 Figure 711 Effect of the fundamental period N 0 on the magnitude of the DFT harmonic function of a rectangularwave signal k 88 88 Xk 11 Nw 5 N0 22 Xk Nw 5 N0 44 k 88 88 11 k 88 88 11 Xk Nw 5 N0 88 rob28124ch07307353indd 323 041216 140 pm C h a p t e r 7 DiscreteTime Fourier Methods 324 Figure 713 Limiting DFT harmonic function of a rectangularwave signal 1 1 Xk 11 k N0 DERIVATION AND DEFINITION To analytically extend the DFT to aperiodic signals let ΔF 1 N 0 a finite increment in discretetime cyclic frequency F Then xn can be written as the inverse DFT of Xk xn 1 N 0 k N 0 Xk e j2πkn N 0 ΔF k N 0 Xk e j2πkΔFn Substituting the summation expression for Xk in the DFT definition xn ΔF k N 0 m0 N 0 1 xm e j2πkΔFm e j2kπΔFn The index of summation n in the expression for Xk has been changed to m to avoid confusion with the n in the expression for xn since they are independent variables Since xn is periodic with fundamental period N0 the inner summation can be over any period and the previous equation can be written as xn k N 0 m N 0 xm e j2πkΔFm e j2πkΔFn ΔF Let the range of the inner summation be N 0 2 m N 0 2 for N0 even or N 0 12 m N 0 12 for N 0 odd The outer summation is over any arbitrary range of k of width N0 so let its range be k 0 k k 0 N 0 Then xn k k 0 k 0 N 0 1 m N 0 2 N 0 21 xm e j2πkΔFm e j2πkΔFn ΔF N 0 even 713 or xn k k 0 k 0 N 0 1 m N 0 1 2 N 0 1 2 xm e j2πkΔFm e j2πkΔFn ΔF N 0 odd 714 Now let the fundamental period N0 of the DFT approach infinity In that limit the following things happen 1 ΔF approaches the differential discretetime frequency dF 2 kΔF becomes discretetime frequency F a continuous independent variable because ΔF is approaching dF rob28124ch07307353indd 324 041216 140 pm 73 The DiscreteTime Fourier Transform 325 3 The outer summation approaches an integral in F kΔF The summation covers a range of k 0 k k 0 N 0 The equivalent range of limits on the integral it approaches can be found using the relationships F kdF k N 0 Dividing the harmonicnumber range k 0 k k 0 N 0 by N 0 translates it to the discretetime frequency range F 0 F F 0 1 where F 0 is arbitrary because k 0 is arbitrary The inner summation covers an infinite range because N 0 is approaching infinity Then in the limit 713 and 714 both become xn 1 m xm e j2πFm ℱ xm e j2πFn dF The equivalent radianfrequency form is xn 1 2π 2π m xm e jΩm e jΩn dΩ in which Ω 2πF and dF dΩ2π These results define the discretetime Fourier transform DTFT as xn 1 XF e j2πFn dF ℱ XF n xn e j2πFn or xn 12π 2π X e jΩ e jΩn dΩ ℱ X e jΩ n xn e jΩn Table 74 has some DTFT pairs for some typical simple signals Table 74 Some DTFT pairs derived directly from the definition δ n ℱ 1 α n un ℱ e jΩ e jΩ α 1 1 α e jΩ α 1 α n un 1 ℱ e jΩ e jΩ α 1 1 α e jΩ α 1 n α n un ℱ α e jΩ e jΩ α 2 α e jΩ 1 α e jΩ 2 α 1 n α n un 1 ℱ α e jΩ e jΩ α 2 α e jΩ 1 α e jΩ 2 α 1 α n sin Ω 0 n un ℱ e jΩ α sin Ω 0 e j2Ω 2α e jΩ cos Ω 0 α 2 α 1 α n sin Ω 0 n un 1 ℱ e jΩ α sin Ω 0 e j2Ω 2α e jΩ cos Ω 0 α 2 α 1 α n cos Ω 0 n un ℱ e jΩ e jΩ α cos Ω 0 e j2Ω 2α e jΩ cos Ω 0 α 2 α 1 α n cos Ω 0 n un 1 ℱ e jΩ e jΩ α cos Ω 0 e j2Ω 2α e jΩ cos Ω 0 α 2 α 1 α n ℱ e jΩ e jΩ α e jΩ e jΩ 1α α 1 rob28124ch07307353indd 325 041216 140 pm C h a p t e r 7 DiscreteTime Fourier Methods 326 Here we are faced with the same notational decision we encountered in deriving the CTFT in Chapter 6 XF is defined by XF n xn e j2πFn and X e jΩ is de fined by X e jΩ n xn e jΩn but the two Xs are actually mathematically different functions because X e jΩ X F F e jΩ The decision here will be similar to the one reached in Chapter 6 We will use the forms XF and X e jΩ for the same reasons The use of X e jΩ instead of the simpler form XΩ is motivated by the desire to maintain consistency of functional definition between the DTFT and the z transform to be pre sented in Chapter 9 THE GENERALIZED DTFT Just as we saw in continuous time in discrete time there are some important practical signals that do not have a DTFT in the strict sense Because these signals are so import ant the DTFT has been generalized to include them Consider the DTFT of xn A a constant XF n A e j2πFn A n e j2πFn The series does not converge Therefore strictly speaking the DTFT does not exist We faced a similar situation with the CTFT and found that the generalized CTFT of a constant is an impulse at f 0 or ω 0 Because of the close relationship between the CTFT and DTFT we might expect a similar result for the DTFT of a constant But all DTFTs must be periodic So a periodic impulse is the logical choice Let a signal xn have a DTFT of the form A δ 1 F Then xn can be found by finding the inverse DTFT of A δ 1 F xn 1 A δ 1 F e j2πFn dF A 12 12 δF e j2πFn dF A This establishes the DTFT pairs A ℱ A δ 1 F or A ℱ 2πA δ 2π Ω If we now generalize to the form A δ 1 F F 0 12 F 0 12 we get xn 1 A δ 1 F F 0 e j2πFn dF A 12 12 δF F 0 e j2πFn dF A e j2π F 0 n Then if xn A cos 2π F 0 n A2 e j2π F 0 n e j2π F 0 n we get the DTFT pairs A cos 2π F 0 n ℱ A2 δ 1 F F 0 δ 1 F F 0 or A cos Ω 0 n ℱ πA δ 1 Ω Ω 0 δ 1 Ω Ω 0 By a similar process we can also derive the DTFT pairs A sin 2π F 0 n ℱ jA2 δ 1 F F 0 δ 1 F F 0 rob28124ch07307353indd 326 041216 140 pm 73 The DiscreteTime Fourier Transform 327 or A sin Ω 0 n ℱ jπA δ 1 Ω Ω 0 δ 1 Ω Ω 0 Now we can extend the table of DTFT pairs to include more useful functions Table 75 Table 75 More DTFT pairs δ n ℱ 1 un ℱ 1 1 e j2πF 12 δ 1 F un ℱ 1 1 e jΩ π δ 1 Ω sincnw ℱ w rectwF δ 1 F sincnw ℱ w rectwΩ2π δ 2π Ω trinw ℱ w drcl 2 F w trinw ℱ w drcl 2 Ω2π w 1 ℱ δ 1 F 1 ℱ 2π δ 2π Ω δ N 0 n ℱ 1 N 0 δ 1 N 0 F δ N 0 n ℱ 2π N 0 δ 2π N 0 Ω cos 2π F 0 n ℱ 12 δ 1 F F 0 δ 1 F F 0 cos Ω 0 n ℱ π δ 2π Ω Ω 0 δ 2π Ω Ω 0 sin 2π F 0 n ℱ j2 δ 1 F F 0 δ 1 F F 0 sin Ω 0 n ℱ jπ δ 2π Ω Ω 0 δ 2π Ω Ω 0 un n 0 un n 1 𝒵 e j2πF e j2πF 1 e j2π n 0 F e j2π n 1 F e jπF n 0 n 1 e jπF n 1 n 0 drclF n 1 n 0 un n 0 un n 1 𝒵 e jΩ e jΩ 1 e j n 0 Ω e j n 1 Ω e jΩ n 0 n 1 2 e jΩ2 n 1 n 0 drclΩ2π n 1 n 0 CONVERGENCE OF THE DISCRETETIME FOURIER TRANSFORM The condition for convergence of the DTFT is simply that the summation in XF n xn e j2πFn or X e jΩ n xn e jΩn 715 actually converges It will converge if n xn 716 If the DTFT function is bounded the inverse transform xn 1 XF e j2πFn dF or xn 1 2π 2π X e jΩ e jΩn dΩ 717 will always converge because the integration interval is finite DTFT PROPERTIES Let xn and yn be two signals whose DTFTs are XF and YF or X e jΩ and Y e jΩ Then the properties in Table 76 apply rob28124ch07307353indd 327 041216 140 pm C h a p t e r 7 DiscreteTime Fourier Methods 328 Table 76 DTFT properties α xn β yn ℱ α XF β Y F α xn β yn ℱ αX e jΩ β Y e jΩ xn n 0 ℱ e j2πF n 0 XF xn n 0 ℱ e jΩ n 0 X e jΩ e j2π F 0 n xn ℱ XF F 0 e j Ω 0 n xn ℱ X e jΩ Ω 0 If zn x nm nm an integer 0 otherwise then zn ℱ XmF or zn ℱ X e jmΩ x n ℱ X F x n ℱ X e jΩ xn xn 1 ℱ 1 e j2πF XF xn xn 1 ℱ 1 e jΩ X e jΩ m n xm ℱ XF 1 e j2πF 1 2 X 0 δ 1 F m n xm ℱ X e jΩ 1 e jΩ πX e j0 1 δ 2π Ω x n ℱ XF xn ℱ X e jΩ xn yn ℱ XFYF xn yn ℱ X e jΩ Y e jΩ xnyn ℱ XF YF xnyn ℱ 12πX e jΩ Y e jΩ n e j2πFn δ 1 F n e jΩn 2π δ 2π Ω n xn 2 1 XF 2 dF n xn 2 12π 2π X e jΩ 2 dΩ In the property xnyn ℱ 12πX e jΩ Y e jΩ the operator indicates periodic convolution which was first introduced in Chapter 6 In this case X e jΩ Y e jΩ 2π X e jΦ Y e jΩΦ dΦ ExamplE 73 Inverse DTFT of two periodic shifted rectangles Find and graph the inverse DTFT of XF rect50F 14 rect50F 14 δ 1 F Figure 714 Figure 714 Magnitude of XF XF F 1 1 1 rob28124ch07307353indd 328 041216 140 pm 73 The DiscreteTime Fourier Transform 329 We can start with the table entry sincnw ℱ wrectwF δ 1 F or in this case 150 sincn50 ℱ rect50F δ 1 F Now apply the frequencyshifting property e j2π F 0 n xn ℱ XF F 0 e jπn2 150sincn50 ℱ rect50F 14 δ 1 F 718 and e jπn2 150sincn50 ℱ rect50F 14 δ 1 F 719 Remember when two functions are convolved a shift of either one of them but not both shifts the convolution by the same amount Finally combining 718 and 719 and simplifying 125 sincn50 cosπn2 ℱ rect50F 14 rect50F 14 δ 1 F Time scaling in the DTFT is quite different from time scaling in the CTFT because of the differences between discrete time and continuous time Let zn xan If a is not an integer some values of zn are undefined and a DTFT cannot be found for it If a is an inte ger greater than one some values of xn will not appear in zn because of decimation and there cannot be a unique relationship between the DTFTs of xn and zn Figure 715 In Figure 715 the two signals x 1 n and x 2 n are different signals but have the same values at even values of n Each of them when decimated by a factor of 2 yields the same decimated signal zn Therefore the DTFT of a signal and the DTFT of a decimated version of that signal are not uniquely related and no timescaling property can be found for that kind of time scaling However if zn is a timeexpanded version of xn formed by inserting zeros between values of xn there is a unique relationship between the DTFTs of xn and zn Let zn xnm nm an integer 0 otherwise where m is an integer Then Z e jΩ X e jmΩ and the timescaling property of the DTFT is If zn xnm nm an integer 0 otherwise then zn 𝒯 XmF zn ℱ X e jmΩ 720 40 n 1 1 n 40 x1n 1 1 40 n 1 1 x2n zn Figure 715 Two different signals which when decimated by a factor of 2 yield the same signal rob28124ch07307353indd 329 041216 140 pm C h a p t e r 7 DiscreteTime Fourier Methods 330 These results can also be interpreted as a frequencyscaling property Given a DTFT X e jΩ if we scale Ω to mΩ where m 1 the effect in the time domain is to insert m 1 zeros between the points in xn The only scaling that can be done in the frequency domain is compression and only by a factor that is an integer This is necessary because all DTFTs must have a period not necessarily a fundamental period of 2π in Ω ExamplE 74 General expression for the DTFT of a periodic impulse Given the DTFT pair 1 ℱ 2π δ 2π Ω use the timescaling property to find a general expres sion for the DTFT of δ N 0 n The constant 1 can be expressed as δ 1 n The periodic impulse δ N 0 n is a timescaled version of δ 1 n scaled by the integer N 0 That is δ N 0 n δ 1 n N 0 n N 0 an integer 0 otherwise Therefore from 720 δ N 0 n ℱ 2π δ 2π N 0 Ω 2π N 0 δ 2π N 0 Ω The implications of multiplicationconvolution duality for signal and system analysis are the same for discretetime signals and systems as for continuoustime signals and systems The response of a system is the convolution of the excitation with the impulse response The equivalent statement in the frequency domain is that the DTFT of the response of a system is the product of the DTFT of the excitation and the frequency response which is the DTFT of the impulse response Figure 716 The implications for cascade connections of systems are also the same Figure 717 If the excitation is a sinusoid of the form xn A cos 2πn N 0 θ then X e jΩ πA δ 2π Ω Ω 0 δ 2π Ω Ω 0 e jθΩ Ω 0 Figure 716 Equivalence of convolution in the time domain and multiplication in the frequency domain xn xnhn hn Xe jΩ Xe jΩHe jΩ He jΩ Figure 717 Cascade connection of systems Xe jΩ Xe jΩH1e jΩ Ye jΩ Xe jΩH1e jΩH2e jΩ H1e jΩ H2e jΩ Xe jΩ Ye jΩ H1e jΩH2e jΩ rob28124ch07307353indd 330 041216 140 pm 73 The DiscreteTime Fourier Transform 331 where Ω 0 2π N 0 Then Y e jΩ X e jΩ H e jΩ H e jΩ πA δ 2π Ω Ω 0 δ 2π Ω Ω 0 e jθΩ Ω 0 Using the equivalence property of the impulse the periodicity of the DTFT and the conjugation property of the CTFT Y e jΩ πA H e j Ω 0 δ 2π Ω Ω 0 H e j Ω 0 H e j Ω 0 δ 2π Ω Ω 0 e jθΩ Ω 0 Y e jΩ πA ReH e j Ω 0 δ 2π Ω Ω 0 δ 2π Ω Ω 0 j Im H e j Ω 0 δ 2π Ω Ω 0 δ 2π Ω Ω 0 e jθΩ Ω 0 yn AReH e j Ω 0 cos 2πn N 0 θ Im H e j Ω 0 sin2πn N 0 θ yn A H e j2π N 0 cos 2πn N 0 θ H e j2π N 0 ExamplE 75 Frequency response of a system Graph the magnitude and phase of the frequency response of the system in Figure 718 If the sys tem is excited by a signal xn sin Ω 0 n find and graph the response yn for Ω0 π4 π2 3π4 The difference equation describing the system is yn 07yn 1 xn and the impulse response is hn 07 n un The frequency response is the Fourier transform of the impulse response We can use the DTFT pair α n un ℱ 1 1 α e jΩ Figure 718 A discretetime system xn yn 07 D to get hn 07 n un ℱ H e jΩ 1 1 07 e jΩ Since the frequency response is periodic in Ω with period 2π a range π Ω π will show all the frequencyresponse behavior At Ω 0 the frequency response is H e j0 05882 At Ω π the frequency response is H e jπ 3333 The response at Ω Ω0 is yn H e j Ω 0 sin Ω 0 n H e j Ω 0 rob28124ch07307353indd 331 041216 140 pm C h a p t e r 7 DiscreteTime Fourier Methods 332 Figure 719 Figure 719 Frequency response and three sinusoidal signals and responses to them 3 2 1 0 1 2 3 0 1 2 3 4 He jΩ 3 2 1 0 1 2 3 15 1 05 0 05 1 15 Ω Ω 0 10 20 30 40 50 60 1 1 2 2 n 0 10 20 30 40 50 60 n 0 10 20 30 40 50 60 n 0 10 20 30 40 50 60 n 0 10 20 30 40 50 60 n 0 10 20 30 40 50 60 n xn yn xn yn xn yn 1 1 1 1 2 2 2 2 Ω0 π4 He jΩ Ω0 π2 Ω0 3π4 ExamplE 76 Signal energy of a sinc signal Find the signal energy of xn 15sincn100 The signal energy of a signal is defined as E x n xn 2 But we can avoid doing a complicated infinite summation by using Parsevals theorem The DTFT of xn can be found by starting with the Fourier pair sincnw ℱ wrectwF δ 1 F and applying the linearity property to form 15sincn100 ℱ 20rect100F δ 1 F rob28124ch07307353indd 332 041216 140 pm 73 The DiscreteTime Fourier Transform 333 Parsevals theorem is n xn 2 1 XF 2 dF So the signal energy is E x 1 20rect100F δ 1 F 2 dF 20rect100F 2 dF or E x 400 1200 1200 dF 4 ExamplE 77 Inverse DTFT of a periodically repeated rectangle Find the inverse DTFT of XF rectwF δ 1 F w 1 using the definition of the DTFT xn 1 XF e j2πFn dF 1 rectwF δ 1 F e j2πFn dF Since we can choose to integrate over any interval in F of width one lets choose the simplest one xn 12 12 rectwF δ 1 F e j2πFn dF In this integration interval there is exactly one rectangle function of width 1w because w 1 and xn 12w 12w e j2πFn dF 2 0 12w cos 2πFndF sin πnw πn 1 w sinc n w 721 From this result we can also establish the handy DTFT pair which appears in the table of DTFT pairs sincnw ℱ wrectwF δ 1 F w 1 or sincnw ℱ w k rectwF k w 1 or in radianfrequency form using the convolution property yt xt ht yat a xat hat we get sincnw ℱ wrectwΩ2π δ 2π Ω w 1 or sincnw ℱ w k rectwΩ 2πk2π w 1 rob28124ch07307353indd 333 041216 140 pm C h a p t e r 7 DiscreteTime Fourier Methods 334 Although these Fourier pairs we derived under the condition w 1 to make the inversion integral 721 simpler they are actually also correct for w 1 NUMERICAL COMPUTATION OF THE DISCRETETIME FOURIER TRANSFORM The DTFT is defined by XF n xn e j2πFn and the DFT is defined by Xk n0 N1 xn e j2πknN If the signal xn is causal and time limited the summation in the DTFT is over a finite range of n values beginning with n 0 We can set the value of N by letting N 1 be the last value of n needed to cover that finite range Then XF n0 N1 xn e j2πFn If we now make the change of variable F kN we get X F FkN XkN n0 N1 xn e j2πknN Xk or in the radianfrequency form X e jΩ Ω2πkN X e j2πkN n0 N1 xn e j2πknN Xk So the DTFT of xn can be found from the DFT of xn at a discrete set of frequencies F kN or equivalently Ω 2πkN k being any integer If it is desired to increase the resolution of this set of discrete frequencies we can just make N larger The extra val ues of xn corresponding to the larger value of N will all be zero This technique for increasing the frequencydomain resolution is called zero padding The inverse DTFT is defined by xn 1 XF e j2πFn dF and the inverse DFT is defined by xn 1 N k0 N1 Xk e j2πknN We can approximate the inverse DTFT by the sum of N integrals that together approx imate the inverse DTFT integral xn k0 N1 kN k1N XkN e j2πFn dF k0 N1 XkN kN k1N e j2πFn dF xn k0 N1 XkN e j2πk1nN e j2πkN j2πn e j2πnN 1 j2πn k0 N1 XkN e j2πknN xn e jπnN j2 sin πnN j2πn k0 N1 XkN e j2πknN e jπnN sincnN 1 N k0 N1 XkN e j2πknN rob28124ch07307353indd 334 041216 140 pm 73 The DiscreteTime Fourier Transform 335 For n N xn 1 N k0 N1 XkN e j2πknN or in the radianfrequency form xn 1 N k0 N1 X e j2πkN e j2πknN This is the inverse DFT with Xk X F FkN XkN or Xk X e jΩ Ω2πkN X e j2πkN ExamplE 78 Inverse DTFT using the DFT Find the approximate inverse DTFT of XF rect 50F 14 rect 50F 14 δ 1 F using the DFT N 512 Number of pts to approximate XF k 0N1 Harmonic numbers Compute samples from XF between 0 and 1 assuming periodic repetition with period 1 X rect50kN 14 rect50kN 34 Compute the approximate inverse DTFT and center the function on n 0 xa realfftshiftifftX n N2N21 Vector of discrete times for plotting Compute exact xn from exact inverse DTFT xe sincn50cospin225 Graph the exact inverse DTFT subplot211 p stemnxekfilled setpLineWidth1 MarkerSize2 axisN2N2005005 grid on xlabelitnFontNameTimesFontSize18 ylabelxitnFontNameTimesFontSize18 titleExactFontNameTimesFontSize24 rob28124ch07307353indd 335 041216 140 pm C h a p t e r 7 DiscreteTime Fourier Methods 336 Graph the approximate inverse DTFT subplot212 p stemnxakfilled setpLineWidth1 MarkerSize2 axisN2N2005005 grid on xlabelitnFontNameTimesFontSize18 ylabelxitnFontNameTimesFontSize18 titleApproximation Using the DFTFontNameTimesFontSize24 The exact and approximate inverse DTFT results are illustrated in Figure 720 Notice that the exact and approximate xn are practically the same near n 0 but are noticeably differ ent near n 256 This occurs because the approximate result is periodic and the overlap of the periodically repeated sinc functions causes these errors near plus or minus half a period Figure 720 Exact and approximate inverse DTFT of XF 250 200 150 100 50 0 50 100 150 200 250 005 0 005 n xn Exact 250 200 150 100 50 0 50 100 150 200 250 005 0 005 n xn Approximation Using the DFT Example 79 illustrates a common analysis problem and a different kind of solution ExamplE 79 System response using the DTFT and the DFT A system with frequency response H e jΩ e jΩ e jΩ 07 is excited by xn trin 88 Find the system response The DTFT of the excitation is X e jΩ 8 drcl 2 Ω2π8 e j8Ω So the DTFT of the response is Y e jΩ e jΩ e jΩ 07 8 drcl 2 Ω2π8 e j8Ω rob28124ch07307353indd 336 041216 140 pm 73 The DiscreteTime Fourier Transform 337 Here we have a problem How do we find the inverse DTFT of Y e jΩ For an analytical solution it would probably be easier in this case to do the convolution in the time domain than to use transforms But there is another way We could use the inverse DFT to approximate the inverse DTFT and find Y e jΩ numerically When we compute the inverse DFT the number of values of yn will be the same as the num ber of values of Y e j2πkN we use N To make this a good approximation we need a value of N large enough to cover the time range over which we expect yn to have values significantly different from zero The triangle signal has a full base width of 16 and the impulse response of the system is 07 n un This is a decaying exponential which approaches but never reaches zero If we use the width at which its value goes below 1 of its initial value we get a width of about 13 Since the convolu tion will be the sum of those two widths minus one we need an N of at least 28 Also remember that the approximation relies on the inequality n N for a good approximation So lets use N 128 in doing the computations and then use only the first 30 values Below is a MATLAB program to find this inverse DTFT Following that are the three graphs produced by the program Figure 721 Figure 721 Excitation impulse response and system response 0 5 10 15 20 25 30 0 05 1 Excitation xn 0 5 10 15 20 25 30 0 05 1 Impulse Response hn 0 5 10 15 20 25 30 5 0 5 n n n System Response yn Program to find an inverse DTFT using the inverse DFT N 128 Number of points to use k 0N1 Vector of harmonic numbers n k Vector of discrete times x trin88 Vector of excitation signal values Compute the DTFT of the excitation X 8drclkN82expj16pikN Compute the frequency response of the system H expj2pikNexpj2pikN 07 rob28124ch07307353indd 337 041216 140 pm C h a p t e r 7 DiscreteTime Fourier Methods 338 h 07nuDn Vector of impulse response values Y HX Compute the DTFT of the response y realifftY n k Vector of system response values Graph the excitation impulse response and response n n130 x x130 h h130 y y130 subplot311 ptr stemnxkfilled grid on setptrLineWidth2MarkerSize4 xlabelitnFontSize24FontNameTimes ylabelxitnFontSize24FontNameTimes titleExcitationFontSize24FontNameTimes setgcaFontSize18FontNameTimes subplot312 ptr stemnhkfilled grid on setptrLineWidth2MarkerSize4 xlabelitnFontSize24FontNameTimes ylabelhitnFontSize24FontNameTimes titleImpulse ResponseFontSize24FontNameTimes setgcaFontSize18FontNameTimes subplot313 ptr stemnykfilled grid on setptrLineWidth2MarkerSize4 xlabelitnFontSize24FontNameTimes ylabelyitnFontSize24FontNameTimes titleSystem ResponseFontSize24FontNameTimes setgcaFontSize18FontNameTimes ExamplE 710 Using the DFT to find a system response A set of samples n 0 1 2 3 4 5 6 7 8 9 10 x n 9 8 6 4 4 9 9 1 2 5 6 is taken from an experiment and processed by a smoothing filter whose impulse response is hn n 07 n un Find the filter response yn We can find a DTFT of hn in the table But xn is not an identifiable functional form We could find the transform of xn by using the direct formula X e jΩ z0 10 xn e jΩn rob28124ch07307353indd 338 041216 140 pm 73 The DiscreteTime Fourier Transform 339 But this is pretty tedious and time consuming If the nonzero portion of xn were much longer this would become quite impractical Instead we can find the solution numerically using the relation derived above for approximating a DTFT with the DFT X e j2πkN n0 N1 xn e j2πknN This problem could also be solved in the time domain using numerical convolution But there are two reasons why using the DFT might be preferable First if the number of points used is an integer power of two the fft algorithm that is used to implement the DFT on computers is very efficient and may have a significant advantage in a shorter computing time than timedomain convolution Second using the DFT method the time scale for the excitation impulse response and system response are all the same That is not true when using numerical timedomain convolution The following MATLAB program solves this problem numerically using the DFT Figure 722 shows the graphs of the excitation impulse response and system response Figure 722 Excitation impulse response and system response 0 5 10 15 20 25 30 35 10 0 10 n n n 0 5 10 15 20 25 30 35 0 1 2 0 5 10 15 20 25 30 35 20 0 20 xn hn yn Program to find a discretetime system response using the DFT N 32 Use 32 points n 0N1 Time vector Set excitation values x 98644991256zeros121 h n07nuDn Compute impulse response X fftx DFT of excitation H ffth DFT of impulse response Y XH DFT of system response y realifftY System response Graph the excitation impulse response and system response subplot311 rob28124ch07307353indd 339 041216 140 pm C h a p t e r 7 DiscreteTime Fourier Methods 340 ptr stemnxkfilled setptrLineWidth2MarkerSize4 grid on xlabelitnFontNameTimesFontSize24 ylabelxitnFontNameTimesFontSize24 setgcaFontNameTimesFontSize18 subplot312 ptr stemnhkfilled setptrLineWidth2MarkerSize4 grid on xlabelitnFontNameTimesFontSize24 ylabelhitnFontNameTimesFontSize24 setgcaFontNameTimesFontSize18 subplot313 ptr stemnykfilled setptrLineWidth2MarkerSize4 grid on xlabelitnFontNameTimesFontSize24 ylabelyitnFontNameTimesFontSize24 setgcaFontNameTimesFontSize18 74 FOURIER METHOD COMPARISONS The DTFT completes the four Fourier analysis methods These four methods form a matrix of methods for the four combinations of continuous and discrete time and continuous and discrete frequency expressed as harmonic number Figure 723 Figure 723 Fourier methods matrix CTFT DTFT CTFS DFT Continuous Time Discrete Time Continuous Frequency Discrete Frequency In Figure 724 are four rectangles or periodically repeated rectangles in both continuous and discrete time along with their Fourier transforms or harmonic functions The CTFT of a single continuoustime rectangle is a single continu ousfrequency sinc function If that continuoustime rectangle is sampled to pro duce a single discretetime rectangle its DTFT is similar to the CTFT except that it is now periodically repeated If the continuoustime rectangle is periodically repeated its CTFS harmonic function is similar to the CTFT except that it has been sampled in frequency harmonic number If the original continuoustime rectangle is both periodically repeated and sampled its DFT is also both periodi cally repeated and sampled So in general periodic repetition in one domain time or frequency corresponds to sampling in the other domain frequency or time and rob28124ch07307353indd 340 041216 140 pm 75 Summary of Important Points 341 sampling in one domain time or frequency corresponds to periodic repetition in the other domain frequency or time These relationships will be important in Chapter 10 on sampling 75 SUMMARY OF IMPORTANT POINTS 1 Any discretetime signal of engineering significance can be represented by a discretetime Fourier series or inverse discrete Fourier transform DFT and the number of harmonics needed in the representation is the same as the fundamental period of the representation 2 The complex sinusoids used in the DFT constitute a set of orthogonal basis functions 3 The fast Fourier transform FFT is an efficient computer algorithm for computing the DFT if the representation time is an integer power of two 4 The DFT can be extended to a discretetime Fourier transform DTFT for aperiodic signals by letting the representation time approach infinity 5 By allowing impulses in the transforms the DTFT can be generalized to apply to some important signals 6 The DFT and inverse DFT can be used to numerically approximate the DTFT and inverse DTFT under certain conditions 7 With a table of discretetime Fourier transform pairs and their properties the forward and inverse transforms of almost any signal of engineering significance can be found 8 The CTFS CTFT DFT and DTFT are closely related analysis methods for periodic or aperiodic continuoustime or discretetime signals Figure 724 Fourier transform comparison for four related signals t xt T0 w w1 n xn 1 N0 Nw k Xk k π π N0 N0 k cxk 1 k π π w T0 t 1 1 1 1 2 2 xt 1 f 4 4 X f 1 f 4 4 π π n xn 1 Ω Ω 4π 4π 2Nw1 2π π π X e jΩ CTFS DFT CTFT DTFT Nw Nw 2Nw1 X e jΩ X f Xk 2Nw1 cxk 4π 4π rob28124ch07307353indd 341 041216 140 pm C h a p t e r 7 DiscreteTime Fourier Methods 342 EXERCISES WITH ANSWERS Answers to each exercise are in random order Orthogonality 1 Without using a calculator or computer find the dot products of a w 1 and w 1 b w 1 and w 2 c w 11 and w 37 where w k W 4 0 W 4 k W 4 2k W 4 3k and W N e j2πN to show that they are orthogonal Answers 0 0 0 2 Find the projection p of the vector x 11 4 in the direction of the vector y 2 1 Answer 18 25 15 3 Find the projection p of the vector x 2 3 1 5 in the direction of the vector y 1 j 1 j Then find the DFT of x and compare this result with X 3 y4 Answer 14 j2 2 j4 14 j2 2 j4 Discrete Fourier Transform 4 A periodic discretetime signal with fundamental period N 3 has the values x 1 7 x 2 3 x 3 1 If x n 𝒟ℱ𝒯 3 X k find the magnitude and angle in radians of X 1 Answer 87178 16858 5 Using the direct summation formula find the DFT harmonic function of δ 10 n with N 10 and compare it with the DFT given in the table Answer δ 1 k 6 Without using a computer find the forward DFT of the following sequence of data and then find the inverse DFT of that sequence and verify that you get back the original sequence x 0 x 1 x 2 x 3 3 4 1 2 Answer Forward DFT is 6 2 j6 2 2 j6 rob28124ch07307353indd 342 041216 140 pm 343 Exercises with Answers 7 A signal x is sampled eight times The samples are x 0 x 7 a b c d e f g h These samples are sent to a DFT algorithm and the output from that algorithm is X a set of eight numbers X 0 X 7 a In terms of a b c d e f g and h what is X 0 b In terms of a b c d e f g and h what is X 4 c If X 3 2 j5 what is the numerical value of X 3 d If X 5 3 e jπ3 what is the numerical value of X 3 e If X 5 9 e j3π4 what is the numerical value of X 3 Answers 9 e j3π4 a b c d e f g h 3 e jπ3 a b c d e f g h 2 j5 8 A discretetime periodic signal with fundamental period N 0 6 has the values x 4 3 x 9 2 x 1 1 x 14 5 x 24 3 x 7 9 Also x n 𝒟ℱ𝒯 6 X k a Find x 5 b Find x 322 c Find X 2 Answers 9 3 149332 e 27862 9 Find the numerical values of the literal constants in a 8 u n 3 u n 2 δ 12 n 𝒟ℱ𝒯 12 A e bk drcl ck D b 5 δ 8 n 2 𝒟ℱ𝒯 8 A e jakπ c δ 4 n 1 δ 4 n 1 𝒟ℱ𝒯 4 jA δ 4 k a δ 4 k a Answers A 5 a 12 A 2 a 1 A 40 b jπ6 c 112 D 5 10 The signal x n 1 has a fundamental period N 0 1 a Find its DFT harmonic function using that fundamental period as the representation time b Now let z n xn4 n4 an integer 0 otherwise Find the DFT harmonic function for z n using its fundamental period as the representation time c Verify that z 0 1 and that z 1 0 by using the DFT representation of z n Answers N δ N k 0 N δ N k 11 If x n 5 cos 2πn5 𝒟ℱ𝒯 15 X k find the numerical values of X 11 X 33 X 9 X 12 X 24 X 48 and X 75 Answers 752 752 752 0 0 0 0 rob28124ch07307353indd 343 041216 140 pm C h a p t e r 7 DiscreteTime Fourier Methods 344 12 Find the DFT harmonic function of x n u n u n 20 δ 20 n using its fundamental period as the representation time There are at least two ways of computing X k and one of them is much easier than the other Find the easy way Answer 20 δ 20 k 13 For each of these signals find the DFT over one fundamental period and show that X N 0 2 is real a x n u n 2 u n 3 δ 12 n b x n u n 3 u n 2 δ 12 n c x n cos 14πn16 cos 2πn16 d x n cos 12πn14 cos 2π n 3 14 Answers 1 1 4 7 DiscreteTime Fourier Transform Definition 14 From the summation definition find the DTFT of x n 10 u n 4 u n 5 and compare with the DTFT table Answer 90 drcl Ω2π 9 15 From the definition derive a general expression for the Ω form of the DTFT of functions of the form x n α n sin Ω 0 n u n α 1 Compare with the DTFT table Answer α e jΩ sin Ω 0 e j2Ω 2α e jΩ cos Ω 0 α 2 α 1 16 Given the DTFT pairs below convert them from the radianfrequency form to the cyclic frequency form using Ω 2π F without doing any inverse DTFTs a α n cos Ω 0 n u n 𝒵 z z α cos Ω 0 z 2 2αz cos Ω 0 α 2 z α b cos Ω 0 n ℱ π δ 2π Ω Ω 0 δ 2π Ω Ω 0 Answers 12 δ 1 F F 0 δ 1 F F 0 1 α cos 2π F 0 e j2πF 1 2α cos 2π F 0 e j2πF α 2 e j4πF α 1 17 If x n n 2 u n u n 3 and x n ℱ X e jΩ what is the value of X e jΩ Ω0 Answer 5 rob28124ch07307353indd 344 041216 140 pm 345 Exercises with Answers Forward and Inverse DiscreteTime Fourier Transforms 18 A discretetime signal is defined by x n sin πn6 Graph the magnitude and phase of the DTFT of x n 3 and x n 12 Answers n 12 12 xn 1 1 F 1 1 05 F 1 1 π π XF XF n 12 12 xn 1 1 F 1 1 XF 05 F 1 1 π π XF 19 If X F 3 δ 1 F 14 δ 1 F 14 j4 δ 1 F 19 δ F 19 and x n ℱ X F what is the fundamental period of x n Answer 36 20 If X F δ 1 F 110 δ 1 F 110 δ 116 F and x n ℱ X F what is the fundamental period of x n Answer 80 21 Graph the magnitude and phase of the DTFT of x n u n 4 u n 5 cos 2πn6 Then graph x n Answers n 12 12 xn 2 2 F 1 1 XF 1 F 1 1 Phase of X F π π rob28124ch07307353indd 345 041216 140 pm C h a p t e r 7 DiscreteTime Fourier Methods 346 22 Graph the inverse DTFT of X F 12 rect 4F δ 1 F δ 12 F Answer xn n 025 01 16 16 23 Let X e jΩ 4π j6π sin 2Ω Its inverse DTFT is x n Find the numerical values of x n for 3 n 3 Answers 0 0 32 32 0 2 24 A signal x n has a DTFT X F 5 drcl F5 What is its signal energy Answer 20 25 Find the numerical values of the literal constants a A u n W u n W 1 e jBπn ℱ 10 sin 5π F 1 sin π F 1 b 2 δ 15 n 3 u n 3 u n 4 ℱ A e jBΩ c 23 n u n 2 ℱ A e jBΩ 1 α e jΩ d 4 sinc n10 ℱ A rect BF δ 1 F e 10 cos 5πn 14 ℱ A δ 1 F a δ 1 F a f 4 δ n 3 δ n 3 ℱ A sin aF g 8 u n 3 u n 2 ℱ A e bΩ drcl cΩ D h 7 u n 3 u n 4 4 sin 2πn12 ℱ A drcl aΩ b δ 2π Ω c δ 2π Ω c i j42 drcl F 5 δ 1 F 116 δ 1 F 116 e j4πF A δ 1 F 116 δ 1 F 116 j δ 1 F 116 δ 1 F 116 j A cos 2π n n 0 N 0 ℱ j 36 2 1 j δ 1 F 116 1 j δ 1 F 116 Answers A 10 W 2 B 2 A 40 B 10 A 5 a 528 01786 A j253148 A 2 B 3 A 94 B 2 α 23 A j8 a 6π 1885 A j196π a 12π b 7 c π6 A 72 N 0 16 n 0 2 A 40 b j c 12π D 5 26 Given the DTFT pair x n ℱ 10 1 06 e jΩ and y n xn2 n2 an integer 0 otherwise rob28124ch07307353indd 346 041216 140 pm 347 Exercises with Answers find the magnitude and phase of Y e jΩ Ωπ4 Answer 8575 05404 radians 27 Let x n ℱ X F 8 tri 2F e j2πF δ 1 F a phaseshifted triangle in the range 12 F 12 that repeats that pattern periodically with fundamental period one Also let y n xn3 n3 an integer 0 if n3 is not an integer and let y n ℱ Y F a Find the magnitude and angle in radians of X 03 b Find the magnitude and angle in radians of X 22 c What is the fundamental period of Y F d Find the magnitude and angle in radians of Y 055 Answers 13 32 1885 23984 21997 48 12566 28 A signal x n has a DTFT X F Some of the values of x n are n 2 1 0 1 2 3 4 5 6 x n 8 2 1 5 7 9 8 2 3 Let Y F X 2F with y n ℱ Y F Find the numerical values of y n for 2 n 4 Answers 1 5 0 2 0 0 29 Using the differencing property of the DTFT and the transform pair tri n2 ℱ 1 cos Ω find the DTFT of 12 δ n 1 δ n δ n 1 δ n 2 Compare it with Fourier transform found using the table Answer 12 e jΩ 1 e jΩ e j2Ω 30 A signal is described by x n lnn 1 0 n 10 lnn 1 10 n 0 0 otherwise Graph the magnitude and phase of its DTFT over the range π Ω π rob28124ch07307353indd 347 041216 140 pm C h a p t e r 7 DiscreteTime Fourier Methods 348 Answer Ω 10 8 6 4 2 0 2 4 6 8 10 4 2 0 2 4 n xn 4 3 2 1 0 1 2 3 4 0 10 20 30 Ω Xe jΩ 4 3 2 1 0 1 2 3 4 2 1 0 1 2 Phase of Xe jΩ EXERCISES WITHOUT ANSWERS Discrete Fourier Transform 31 If x 3 j 3 j 7 and y 9 4 j 5 2 j7 find x H y 32 Fill in the blanks with correct numbers for this DFT harmonic function of a real valued signal with N 8 k 0 1 2 3 4 5 6 7 X k 5 2 j7 4 j2 3 9 j4 k 11 9 26 47 X k 33 A discretetime signal x n is periodic with period 8 One period of its DFT harmonic function is X 0 X 7 34 j5 4 j31 j5 41 j5 4 j34 j5 a What is the average value of x n b What is the signal power of x n c Is x n even odd or neither 34 A set of samples x n from a signal is converted to a set of numbers X k by using the DFT rob28124ch07307353indd 348 041216 140 pm 349 Exercises without Answers a If x 0 x 1 x 2 x 3 x 4 2 8 3 1 9 find the numerical value of X 0 b If x n consists of 24 samples taken from exactly three periods of a sinusoid at a sampling rate which is exactly eight times the frequency of the sinusoid two values of X k in the range 0 k 24 are not zero Which ones are they c If x n consists of seven samples all of which are the same 5 in the range 0 k 7 which X k values are zero d If x 0 x 1 x 2 x 3 a b 0 b and X 0 X 1 X 2 X 3 A B 0 B how are a and b related to each other Express B in terms of a and b 35 If x 1 n 10 cos 2πn8 𝒟ℱ𝒯 8 X 1 k and x 2 n 𝒟ℱ𝒯 32 X 1 k find the numerical values of x 2 2 x 2 4 x 2 8 and x 2 204 36 A periodic discretetime signal x n is exactly described for all discrete time by its DFT X k 8 δ 8 k 1 δ 8 k 1 j2 δ 8 k 2 j16 δ 8 k 2 e jπk4 a Write a correct analytical expression for x n in which 1 j does not appear b What is the numerical value of x n at n 10 37 A discretetime signal x n with fundamental period N 0 4 has a DFT X k k 0 1 2 3 X k 4 2 j3 1 2 j3 a Find X 5 b Find X 22 c Find x 0 d Find x 3 38 A signal x t is sampled four times and the samples are x 0 x 1 x 2 x 3 Its DFT harmonic function is X 0 X 1 X 2 X 3 X 3 can be written as X 3 ax 0 bx 1 cx 2 dx 3 What are the numerical values of a b c and d 39 The DFT harmonic function X k of x n 5 cos πn 3 sin πn2 using N 4 can be written in the form X k A δ 4 k a δ 4 k a δ 4 k b δ 4 k b Find the numerical values of A a and b 40 Four data points x 0 x 1 x 2 x 3 are converted by the DFT into four corresponding data points X 0 X 1 X 2 X 3 If x 0 2 x 1 3 x 3 7 X 1 3 j10 X 2 3 find the numerical values of the missing data x 2 X 0 and X 3 Be careful to observe which symbols are lower and upper case 41 Demonstrate with a counterexample that the fundamental period of a discrete time function and the fundamental period of its DFT harmonic function are not necessarily the same 42 In Figure E42 is graphed exactly one period of a periodic function x n Its harmonic function X k with N N 0 can be written in the form X k A B e jbk e jck D e jdk rob28124ch07307353indd 349 041216 140 pm C h a p t e r 7 DiscreteTime Fourier Methods 350 Find the numerical values of the constants 10 8 6 4 2 0 2 4 6 8 10 10 8 6 4 2 0 2 4 6 8 10 n xn Figure E42 43 Associate each discretetime signal in Figure E431 with its corresponding DFT magnitude in Figure E432 0 5 10 15 0 05 1 15 2 n xn xn xn xn xn xn 0 5 10 15 0 1 2 3 4 n 0 5 10 15 2 1 0 1 2 n 0 5 10 15 0 1 2 3 n 0 5 10 15 4 2 0 2 4 n 0 5 10 15 0 05 1 15 2 n 1 3 2 4 6 5 Figure E431 0 5 10 15 0 5 10 15 k Xk Xk Xk Xk Xk Xk Xk Xk Xk Xk Xk Xk A 0 5 10 15 0 10 20 30 k B 0 5 10 15 0 10 20 30 k C 0 5 10 15 0 2 4 6 8 k D 0 5 10 15 0 10 20 30 k E 0 5 10 15 0 5 10 15 k F 0 5 10 15 0 2 4 6 8 k G 0 5 10 15 0 1 2 k H 0 5 10 15 0 10 20 k I 0 5 10 15 0 20 40 k J 0 5 10 15 0 2 4 6 8 k K 0 5 10 15 0 10 20 30 k L rob28124ch07307353indd 350 041216 140 pm 351 Exercises without Answers 0 5 10 15 0 5 10 15 k Xk Xk Xk Xk Xk Xk Xk Xk Xk Xk Xk Xk A 0 5 10 15 0 10 20 30 k B 0 5 10 15 0 10 20 30 k C 0 5 10 15 0 2 4 6 8 k D 0 5 10 15 0 10 20 30 k E 0 5 10 15 0 5 10 15 k F 0 5 10 15 0 2 4 6 8 k G 0 5 10 15 0 1 2 k H 0 5 10 15 0 10 20 k I 0 5 10 15 0 20 40 k J 0 5 10 15 0 2 4 6 8 k K 0 5 10 15 0 10 20 30 k L Figure E432 44 Associate each discretetime signal in Figure E441 with its corresponding DFT magnitude in Figure E442 0 5 10 15 2 0 2 n xn xn xn xn xn xn xn xn xn xn 0 5 10 15 4 2 0 2 4 n 0 5 10 15 2 0 2 n 0 5 10 15 5 0 5 n 0 5 10 15 2 0 2 n 0 5 10 15 0 05 1 n 0 5 10 15 0 05 1 n 0 5 10 15 0 05 1 n 0 5 10 15 1 0 1 n 0 5 10 15 1 0 1 2 n 1 6 7 8 9 10 2 3 4 5 Figure E441 0 5 10 15 0 5 10 15 k B 0 5 10 15 0 10 20 30 k Xk Xk Xk Xk Xk Xk Xk Xk Xk Xk Xk Xk Xk Xk Xk A 0 5 10 15 0 10 20 30 k I 0 5 10 15 0 20 40 60 k M 0 5 10 15 0 5 10 15 k O 0 5 10 15 0 05 1 k D 0 5 10 15 0 1 2 k J 0 5 10 15 0 2 4 k G 0 5 10 15 0 2 4 k C 0 5 10 15 0 1 2 3 k H 0 5 10 15 0 10 20 k E 0 5 10 15 0 10 20 k N 0 5 10 15 0 5 10 15 k F 0 5 10 15 0 2 4 6 8 k L 0 5 10 15 0 5 10 k K Figure E442 rob28124ch07307353indd 351 041216 140 pm C h a p t e r 7 DiscreteTime Fourier Methods 352 Forward and Inverse DiscreteTime Fourier Transforms 45 Given the DTFT pairs below convert them from the cyclicfrequency form to the radianfrequency form using Ω 2πF without doing any inverse DTFTs a δ N n ℱ 1N δ 1N F b sin 2π F 0 n ℱ j2 δ 1 F F 0 δ 1 F F 0 c sinc nw ℱ wrect wF δ 1 F 46 Find the DTFT of each of these signals a x n 13 n u n 1 b x n sin πn4 14 n u n 2 c x n sinc 2πn8 sinc 2π n 4 8 d x n sinc 2 2πn8 47 Graph the inverse DTFTs of these functions a X F δ 1 F δ 1 F 12 b X e jΩ j2π δ 2π Ω π4 δ 2π Ω π4 c X e jΩ 2π δ Ω π2 δ Ω 3π8 δ Ω 5π8 δ 2π 2Ω 48 A signal x n has a DTFT X e jΩ 10 drcl Ω2π5 What is its signal energy 49 Given this DTFT pair x 1 n ℱ X 1 e jΩ e jΩ 2 e j2Ω 3 e j3Ω and the related pair x 2 n 3 x 1 n 2 ℱ X 2 e jΩ find the value of X 2 e jΩ Ωπ4 50 A signal x n has a DTFT X e jΩ 2π δ 2π Ω π2 δ 2π Ω π2 j δ 2π Ω 2π3 j δ 2π Ω 2π3 What is the fundamental period N0 of x n 51 Let x n tri n4 δ 6 n and let x n ℱ X e jΩ a What is x 21 b What is the lowest positive frequency Ω at which X e jΩ 0 52 Find an expression for the inverse DTFT x n of X F rect 10F δ 1 F 12 δ 1 F 14 δ 1 F 14 and evaluate it at the discrete time n 2 53 If X F tri 4F δ 1 F δ 1 F 14 δ 1 F 14 then what is the numerical value of X 118 54 Find the numerical values of the literal constants a x n 2δ n 3 3δ n 3 ℱ X e jΩ A sin bΩ C e dΩ b 2 cos 2πn 24 cos 2πn 4 ℱ A δ 1 F a δ 1 F a δ 1 F b δ 1 F b rob28124ch07307353indd 352 041216 140 pm 353 Exercises without Answers c 4 u n 4 u n 5 δ 9 n ℱ A δ b Ω d 5 cos 2πn14 ℱ A δ 2π Ω b δ 2π Ω b e 2 sin 2π n 3 9 ℱ A δ 1 F b δ 1 F b e cF f 7sinc n20 ℱ Arect bF δ 1c cF g A B cos nc ℱ 4π δ π Ω h Asinc nb sin nc ℱ 12 rect 5 F 15 rect 5 F 15 δ 1 F i A δ b n δ b n c ℱ π 1 e j2Ω 3 δ π3 Ω 55 Let x n be a signal and let y n m n xm If Y e jΩ cos 2Ω x n consists of exactly four discretetime impulses What are their strengths and locations rob28124ch07307353indd 353 041216 140 pm 354 81 INTRODUCTION AND GOALS The continuoustime Fourier transform CTFT is a powerful tool for signal and system analysis but it has its limitations There are some useful signals that do not have a CTFT even in the generalized sense which allows for impulses in the CTFT of a signal The CTFT expresses signals as linear combinations of complex sinusoids In this chapter we extend the CTFT to the Laplace transform which expresses signals as linear combinations of complex exponentials the eigenfunctions of the differential equations that describe continuoustime LTI systems Complex sinusoids are a special case of complex exponentials Some signals that do not have a CTFT do have a Laplace transform The impulse responses of LTI systems completely characterize them Because the Laplace transform describes the impulse responses of LTI systems as linear combina tions of the eigenfunctions of LTI systems it directly encapsulates the characteristics of a system in a very useful way Many system analysis and design techniques are based on the Laplace transform CHAPTER GOA L S 1 To develop the Laplace transform which is applicable to some signals that do not have a CTFT 2 To define the range of signals to which the Laplace transform applies 3 To develop a technique for realizing a system directly from its transfer function 4 To learn how to find forward and inverse Laplace transforms 5 To derive and illustrate the properties of the Laplace transform especially those that do not have a direct counterpart in the Fourier transform 6 To define the unilateral Laplace transform and explore its unique features 7 To learn how to solve differential equations with initial conditions using the unilateral Laplace transform 8 To relate the pole and zero locations of a transfer function of a system directly to the frequency response of the system 9 To learn how MATLAB represents the transfer functions of systems 8 C H A P T E R The Laplace Transform rob28124ch08354405indd 354 041216 141 pm 82 Development of the Laplace Transform 355 82 DEVELOPMENT OF THE LAPLACE TRANSFORM When we extended the Fourier series to the Fourier transform we let the fundamental period of a periodic signal increase to infinity making the discrete frequencies k f 0 in the CTFS merge into the continuum of frequencies f in the CTFT This led to the two alternate definitions of the Fourier transform X jω xt e jωt dt xt 12π X jω e jωt dω and X f xt e j2πft dt xt X f e j2πft df There are two common approaches to introducing the Laplace transform One ap proach is to conceive the Laplace transform as a generalization of the Fourier trans form by expressing functions as linear combinations of complex exponentials instead of as linear combinations of the more restricted class of functions complex sinusoids used in the Fourier transform The other approach is to exploit the unique nature of the complex exponential as the eigenfunction of the differential equations that describe linear systems and to realize that an LTI system excited by a complex exponential responds with another complex exponential The relation between the excitation and response complex exponentials of an LTI system is the Laplace transform We will consider both approaches GENERALIZING THE FOURIER TRANSFORM If we simply generalize the forward Fourier transform by replacing complex sinusoids of the form e jωt ω a real variable with complex exponentials e st s a complex variable we get xt Xs xt e st dt which defines a forward Laplace1 transform where the notation means Laplace transform of Being a complex variable s can have values anywhere in the complex plane It has a real part called σ and an imaginary part called ω so s σ jω Then for the special case in which σ is zero and the Fourier transform of the function xt exists in the strict sense the forward Laplace transform is equivalent to a forward Fourier transform X jω X s sjω This relationship between the Fourier and Laplace transforms is the reason for choos ing in Chapter 6 the functional notation for the CTFT as Xjω instead of Xω This choice preserves the strict mathematical meaning of the function X 1 Pierre Simon Laplace attended a Benedictine priory school until the age of 16 when he entered Caen University intending to study theology But he soon realized that his real talent and love were in mathematics He quit the university and went to Paris where he was befriended by dAlambert who secured for him a teaching position in a military school He produced in the next few years a sequence of many papers on various topics all of high quality He was elected to the Paris Academy in 1773 at the age of 23 He spent most of his career working in the areas of probability and celestial mechanics rob28124ch08354405indd 355 041216 141 pm C h a p t e r 8 The Laplace Transform 356 Using s σ jω in the forward Laplace transform we get Xs xt e σjωt dt xt e σt e jωt dt ℱ xt e σt So one way of conceptualizing the Laplace transform is that it is equivalent to a Fourier transform of the product of the function xt and a real exponential convergence factor of the form e σt as illustrated in Figure 81 eσt xt t xteσt t Figure 81 The effect of the decayingexponential convergence factor on the original function The convergence factor allows us in some cases to find transforms for which the Fourier transform cannot be found As mentioned in an earlier chapter the Fourier transforms of some functions do not strictly speaking exist For example the function gt A ut would have the Fourier transform G jω A ut e jωt dt A 0 e jωt dt This integral does not converge The technique used in Chapter 6 to make the Fourier transform converge was to multiply the signal by a convergence factor e σ t where σ is positive real Then the Fourier transform of the modified signal can be found and the limit taken as σ approaches zero The Fourier transform found by this technique was called a generalized Fourier transform in which the impulse was allowed as a part of the transform Notice that for time t 0 this convergence factor is the same in the Laplace transform and the generalized Fourier transform but in the Laplace transform the limit as σ approaches zero is not taken As we will soon see there are other useful functions that do not have even a generalized Fourier transform Now to formally derive the forward and inverse Laplace transforms from the Fourier transform we take the Fourier transform of g σ t gt e σt instead of the orig inal function gt That integral would then be ℱ g σ t G σ jω g σ t e jωt dt gt e σjωt dt rob28124ch08354405indd 356 041216 141 pm 82 Development of the Laplace Transform 357 This integral may or may not converge depending on the nature of the function gt and the choice of the value of σ We will soon explore the conditions under which the integral converges Using the notation s σ jω ℱ g σ t gt G s gt e st dt This is the Laplace transform of gt if the integral converges The inverse Fourier transform would be ℱ 1 G σ jω g σ t 1 2π G σ jω e jωt dω 1 2π G s e jωt dω Using s σ jω and ds jdω we get g σ t 1 j2π σj σj G s e sσt ds e σt j2π σj σj G s e st ds or dividing both sides by e σt gt 1 j2π σj σj G s e st ds This defines an inverse Laplace transform When we are dealing only with Laplace transforms the subscript will not be needed to avoid confusion with Fourier trans forms and the forward and inverse transforms can be written as Xs xt e st dt and xt 1 j2π σj σj Xs e st ds 81 This result shows that a function can be expressed as a linear combination of complex exponentials a generalization of the Fourier transform in which a function is expressed as a linear combination of complex sinusoids A common notational convention is xt Xs indicating that xt and Xs form a Laplacetransform pair COMPLEX EXPONENTIAL EXCITATION AND RESPONSE Another approach to the Laplace transform is to consider the response of an LTI sys tem to a complex exponential excitation of the form xt K e st where s σ jω and σ ω and K are all realvalued Using convolution the response yt of an LTI system with impulse response ht to xt is yt ht K e st K hτ e stτ dτ K e st x t hτ e sτ dτ The response of an LTI system to a complexexponential excitation is that same excitation multiplied by the quantity hτ e sτ dτ if this integral converges This is the integral of rob28124ch08354405indd 357 041216 141 pm C h a p t e r 8 The Laplace Transform 358 the product of the impulse response hτ and a complex exponential e sτ over all τ and the result of this operation is a function of s only This result is usually written as Hs ht e st dt 82 and Hs is called the Laplace transform of ht The name of the variable of integra tion was changed from τ to t but that does not change the result Hs For an LTI system knowledge of ht is enough to completely characterize the system Hs also contains enough information to completely characterize the system but the information is in a different form The fact that this form is different can lend insight into the systems operation that is more difficult to see by examining ht alone In the chapters to follow we will see many examples of the advantage of viewing system properties and performance through Hs in addition to ht 83 THE TRANSFER FUNCTION Now lets find the Laplace transform Ys of the response yt of an LTI system with impulse response ht to an excitation xt Ys yt e st dt ht xt e st dt hτ xt τdτ e st dt Separating the two integrals Ys hτdτ xt τ e st dt Let λ t τ dλ dt Then Ys hτdτ xλ e s λτ dλ hτ e sτ dτ Hs xλ e sλ dλ Xs The Laplace transform Ys of the response yt is Ys HsXs 83 the product of the Laplace transforms of the excitation and impulse response if all the transforms exist Hs is called the transfer function of the system because it de scribes in the s domain how the system transfers the excitation to the response This is a fundamental result in system analysis In this new s domain timeconvolution becomes sdomain multiplication just as it did using the Fourier transform yt xt ht Ys XsHs 84 CASCADECONNECTED SYSTEMS If the response of one system is the excitation of another system they are said to be cascade connected Figure 82 The Laplace transform of the overall system response is then Ys H 2 s H 1 sXs H 1 s H 2 s Xs rob28124ch08354405indd 358 041216 141 pm 85 Direct Form II Realization 359 and the cascadeconnected systems are equivalent to a single system whose transfer function is Hs H1sH2s 85 DIRECT FORM II REALIZATION System realization is the process of putting together system components to form an overall system with a desired transfer function In Chapter 5 we found that if a system is described by a linear differential equation of the form k0 N a k y k t k0 N b k x k t its transfer function is a ratio of polynomials in s and the coefficients of the powers of s are the same as the coefficients of derivatives of x and y in the differential equation Hs Ys Xs k0 N b k s k k0 N a k s k b N s N b N1 s N1 b 1 s b 0 a N s N a N1 s N1 a 1 s a 0 84 Here the nominal orders of the numerator and denominator are both assumed to be N If the numerator order is actually less than N some of the higherorder b coefficients will be zero The denominator order must be N and a N cannot be zero if this is an Nthorder system One standard form of system realization is called Direct Form II The transfer function can be thought of as the product of two transfer functions H1s Y1s Xs 1 a N s N a N1 s N1 a 1 s a 0 85 and H2s Ys Y1s b N s N b N1 s N1 b 1 s b 0 Figure 83 where the output signal of the first system Y1s is the input signal of the second system Figure 83 A system conceived as two cascaded systems Xs Y1s H1s H2s bNsN bN1sN1b1sb0 aNsN aN1sN1 a1sa0 1 Ys Figure 82 Cascade connection of systems H1s H2s XsH1s Xs Xs YsXsH1sH2s Ys H1sH2s We can draw a block diagram of H 1 s by rewriting 85 as Xs a N s N a N1 s N1 a 1 s a 0 Y 1 s rob28124ch08354405indd 359 041216 141 pm C h a p t e r 8 The Laplace Transform 360 or Xs a N s N Y 1 s a N1 s N1 Y 1 s a 1 s Y 1 s a 0 Y 1 s or s N Y 1 s 1 a N Xs a N1 s N1 Y 1 s a 1 s Y 1 s a 0 Y 1 s Figure 84 Xs sNY1s sN1Y1s sN2Y1s sY1s Y1s aN1 aN2 a1 a0 1aN s1 s1 s1 Figure 84 Realization of H 1 s Figure 85 Overall Direct Form II system realization aN1 aN2 bN2 bN1 a1 a0 b0 b1 bN Xs Ys 1aN s1 s1 s1 Now we can immediately synthesize the overall response Ys as a linear combination of the various powers of s multiplying Y 1 s Figure 85 86 THE INVERSE LAPLACE TRANSFORM In the practical application of the Laplace transform we need a way to convert Ys to yt an inverse Laplace transform It was shown in 81 that yt 1 j2π σj σj Ys e st ds where σ is the real part of s This is a contour integral in the complex s plane and is beyond the scope of this text The inversion integral is rarely used in practical problem solving because the Laplace transforms of most useful signals have already been found and tabulated 87 EXISTENCE OF THE LAPLACE TRANSFORM We should now explore under what conditions the Laplace transform Xs xt e st dt actually exists It exists if the integral converges and whether or not the integral con verges depend on xt and s rob28124ch08354405indd 360 041216 141 pm 87 Existence of the Laplace Transform 361 TIMELIMITED SIGNALS If xt 0 for t t 0 and t t 1 with t 0 and t 1 finite it is called a timelimited signal If xt is also finite for all t the Laplacetransform integral converges for any value of s and the Laplace transform of xt exists Figure 86 RIGHT AND LEFTSIDED SIGNALS If xt 0 for t t 0 it is called a rightsided signal and the Laplace transform becomes Xs t 0 xt e st dt Figure 87a Consider the Laplace transform Xs of the rightsided signal xt e αt ut t 0 α ℝ Xs t 0 e αt e st dt t 0 e ασt e jωt dt Figure 88a t t0 xt t xt t0 a b Figure 87 a A rightsided signal b a leftsided signal a b xt t t0 xt t t0 Figure 88 a xt e αt ut t 0 α ℝ b xt e βt u t 0 t β ℝ If σ α the integral converges The inequality σ α defines a region in the s plane called the region of convergence ROC Figure 89a Figure 89 Regions of convergence for a the rightsided signal xt e αt ut t 0 α ℝ and b the leftsided signal xt e βt u t 0 t β ℝ a b σ ω s α ROC σ ω s β Figure 86 A finite timelimited signal t t0 xt t1 rob28124ch08354405indd 361 041216 141 pm C h a p t e r 8 The Laplace Transform 362 If xt 0 for t t 0 it is called a leftsided signal Figure 87b The Laplace transform becomes Xs t 0 xt e st dt If xt e βt u t 0 t β ℝ Xs t 0 e βt e st dt t 0 e βσt e jωt dt and the integral converges for any σ β Figure 88b and Figure 89b Any signal can be expressed as the sum of a rightsided signal and a leftsided signal Figure 810 If xt x r t x l t where x r t is the rightsided part and x l t is the leftsided part and if x r t K r e αt and x l t K l e βt where K r and K l are constants then the Laplacetransform integral converges and the Laplace transform exists for α σ β This implies that if α β a Laplace transform can be found and the ROC in the s plane is the region α σ β If α β the Laplace transform does not exist For rightsided signals the ROC is always the region of the s plane to the right of α For leftsided signals the ROC is always the region of the s plane to the left of β 88 LAPLACETRANSFORM PAIRS We can build a table of Laplacetransform pairs starting with signals described by δt and e αt cos ω 0 t ut Using the definition δt δt e st dt 1 All s e αt cos ω 0 t ut e αt cos ω 0 tut e st dt 0 e j ω 0 t e j ω 0 t 2 e sαt dt σ α e αt cos ω 0 t ut 12 0 e sj ω 0 αt e sj ω 0 αt dt σ α Figure 810 A signal divided into a leftsided part a and a rightsided part b a b xlt t0 t0 xrt t xt t t rob28124ch08354405indd 362 041216 141 pm 88 LaplaceTransform Pairs 363 e αt cos ω 0 t ut 12 1 s j ω 0 α 1 s j ω 0 α σ α e αt cos ω 0 t ut s α s α 2 ω 0 2 σ α If α 0 cos ω 0 t ut s s 2 ω 0 2 σ 0 If ω 0 0 e αt ut 1 s α σ α If α ω 0 0 ut 1s σ 0 Using similar methods we can build a table of the most often used Laplacetransform pairs Table 81 To illustrate the importance of specifying not only the algebraic form of the Laplace transform but also its ROC consider the Laplace transforms of e αt ut and e αt u t e αt ut 1 s α σ α and e αt ut 1 s α σ α Table 81 Some common Laplacetransform pairs δt 1 All σ ut 1s σ 0 ut 1s σ 0 rampt t ut 1 s 2 σ 0 rampt t ut 1 s 2 σ 0 e αt ut 1s α σ α e αt ut 1s α σ α t n ut n s n1 σ 0 t n ut n s n1 σ 0 t e αt ut 1s α2 σ α t e αt ut 1s α2 σ α t n e αt ut n s α n1 σ α t n e αt ut n s α n1 σ α sin ω 0 t ut ω 0 s 2 ω 0 2 σ 0 sin ω 0 t ut ω 0 s 2 ω 0 2 σ 0 cos ω 0 t ut s s 2 ω 0 2 σ 0 cos ω 0 t ut s s 2 ω 0 2 σ 0 e αt sin ω 0 t ut ω 0 s α 2 ω 0 2 σ α e αt sin ω 0 t ut ω 0 s α 2 ω 0 2 σ α e αt cos ω 0 t ut s α s α 2 ω 0 2 σ α e αt cos ω 0 t ut s α s α 2 ω 0 2 σ α e α t 1 s α 1 s α 2α s 2 α 2 α σ α rob28124ch08354405indd 363 041216 141 pm C h a p t e r 8 The Laplace Transform 364 The algebraic expression for the Laplace transform is the same in each case but the ROCs are totally different in fact mutually exclusive That means that the Laplace transform of a linear combination of these two functions cannot be found because we cannot find a region in the s plane that is common to the ROCs of both e αt ut and e αt ut An observant reader may have noticed that some very common signal functions do not appear in Table 81 for example a constant The function xt ut appears but xt 1 does not The Laplace transform of xt 1 would be Xs e st dt 0 e σt e jωt dt ROC σ0 0 e σt e jωt dt ROC σ0 There is no ROC common to both of these integrals therefore the Laplace transform does not exist For the same reason cos ω 0 t sin ω 0 t sgnt and δ T 0 t do not appear in the table although cos ω 0 tut and sin ω 0 tut do appear The Laplace transform 1s α is finite at every point in the s plane except the point s α This unique point is called a pole of 1s α In general a pole of a Laplace transform is a value of s at which the transform tends to infinity The opposite concept is a zero of a Laplace transform a value of s at which the transform is zero For 1s α there is a single zero at infinity The Laplace transform cos ω 0 tut s s 2 ω 0 2 has poles at s j ω 0 a zero at s 0 and a zero at infinity A useful tool in signal and system analysis is the polezero diagram in which an x marks a pole and an o marks a zero in the s plane Figure 811 2 2 2 2 4 6 8 10 4 6 8 10 4 6 10 8 σ ω 2 2 2 4 6 8 10 4 6 8 10 4 10 8 6 σ ω s2 4s 20 s2 8s 32 2 2 2 4 6 8 10 4 6 8 10 4 10 8 6 σ ω s2 s 2s 6 s 8s 4 s 10s 6s 4 Figure 811 Example polezero diagrams The small 2 next to the zero in the rightmost polezero diagram in Figure 811 indi cates that there is a double zero at s 0 As we will see in later material the poles and zeros of the Laplace transform of a function contain much valuable information about the nature of the function rob28124ch08354405indd 364 041216 141 pm 88 LaplaceTransform Pairs 365 ExamplE 82 Inverse Laplace transforms Find the inverse Laplace transforms of a Xs 4 s 3 10 s 6 3 σ 6 ExamplE 81 Laplace transform of a noncausal exponential signal Find the Laplace transform of xt e t ut e 2t ut The Laplace transform of this sum is the sum of the Laplace transforms of the individual terms e t ut and e 2t ut The ROC of the sum is the region in the s plane that is common to the two ROCs From Table 81 e t ut 1 s 1 σ 1 and e 2t ut 1 s 2 σ 2 In this case the region in the s plane that is common to both ROCs is 1 σ 2 and e t ut e 2t ut 1 s 1 1 s 2 1 σ 2 Figure 812 This Laplace transform has poles at s 1 and s 2 and two zeros at infinity σ ω s 2 s 1 ROC s Figure 812 ROC for the Laplace transform of xt e t ut e 2t ut rob28124ch08354405indd 365 041216 141 pm C h a p t e r 8 The Laplace Transform 366 b In this case the ROC is to the right of both poles and both timedomain signals must be right sided and using eαt ut 1 s α σ α xt 4 e 3t ut 10 e 6t ut Figure 813b c In this case the ROC is to the left of both poles and both timedomain signals must be left sided and using e αt u t 1 s α σ α xt 4 e 3t ut 10 e 6t ut Figure 813c b Xs 4 s 3 10 s 6 σ 6 c Xs 4 s 3 10 s 6 σ 3 a Xs is the sum of two sdomain functions and the inverse Laplace transform must be the sum of two timedomain functions Xs has two poles one at s 3 and one at s 6 We know that for rightsided signals the ROC is always to the right of the pole and for leftsided signals the ROC is always to the left of the pole Therefore 4 s 3 must inverse transform into a rightsided signal and 10 s 6 must inverse transform into a leftsided signal Then using e αt ut 1 s α σ α and e αt ut 1 s α σ α we get xt 4 e 3t ut 10 e 6t ut Figure 813a Figure 813 Three inverse Laplace transforms t 30 03 03 03 xt 10 t 30 xt 60 t 30 xt 10 6 a b c rob28124ch08354405indd 366 041216 141 pm 89 PartialFraction Expansion 367 89 PARTIALFRACTION EXPANSION In Example 82 each sdomain expression was in the form of two terms each of which can be found directly in Table 81 But what do we do when the Laplacetransform expression is in a more complicated form For example how do we find the inverse Laplace transform of Xs s s 2 4s 3 s s 3s 1 σ 1 This form does not appear in Table 81 In a case like this a technique called partialfraction expansion becomes very useful Using that technique it is possible to write Xs as Xs 32 s 3 12 s 1 1 2 3 s 3 1 s 1 σ 1 Then the inverse transform can be found as xt 123 e 3t e t ut The most common type of problem in signal and system analysis using Laplace methods is to find the inverse transform of a rational function in s of the form Gs b M s M b M1 s M1 b 1 s b 0 s N a N1 s N1 a 1 s a 0 where the numerator and denominator coefficients a and b are constants Since the orders of the numerator and denominator are arbitrary this function does not appear in standard tables of Laplace transforms But using partialfraction expansion it can be expressed as a sum of functions that do appear in standard tables of Laplace transforms It is always possible numerically if not analytically to factor the denominator polynomial and to express the function in the form Gs b M s M b M1 s M1 b 1 s b 0 s p 1 s p 2 s p N where the ps are the finite poles of Gs Lets consider for now the simplest case that there are no repeated finite poles and that N M making the fraction proper in s Once the poles have been identified we should be able to write the function in the partialfraction form Gs K 1 s p 1 K 2 s p 2 K N s p N if we can find the correct values of the Ks For this form of the function to be correct the identity b M s M b M1 s M1 b 1 s b 0 s p 1 s p 2 s p N K 1 s p 1 K 2 s p 2 K N s p N 86 must be satisfied for any arbitrary value of s The Ks can be found by putting the right side into the form of a single fraction with a common denominator that is the same as the leftside denominator and then setting the coefficients of each power of s in the rob28124ch08354405indd 367 041216 141 pm C h a p t e r 8 The Laplace Transform 368 numerators equal and solving those equations for the Ks But there is another way that is often easier Multiply both sides of 86 by s p 1 s p 1 b M s M b M1 s M1 b 1 s b 0 s p 1 s p 2 s p N s p 1 K 1 s p 1 s p 1 K 2 s p 2 s p 1 K N s p N or b M s M b M1 s M1 b 1 s b 0 s p 2 s p N K 1 s p 1 K 2 s p 2 s p 1 K N s p N 87 Since 86 must be satisfied for any arbitrary value of s let s p 1 All the factors s p 1 on the right side become zero 87 becomes K 1 b M p 1 M b M1 p 1 M1 b 1 p 1 b 0 p 1 p 2 p 1 p N and we immediately have the value of K 1 We can use the same technique to find all the other Ks Then using the Laplacetransform pairs e αt ut 1 s α σ α and e αt ut 1 s α σ α we can find the inverse Laplace transform ExamplE 83 Inverse Laplace transform using partialfraction expansion Find the inverse Laplace transform of Gs 10s s 3s 1 σ 1 We can expand this expression in partial fractions yielding Gs 10s s 1 s3 s 3 10s s 3 s1 s 1 σ 1 Gs 15 s 3 5 s 1 σ 1 Then using e at ut 1 s a σ α we get gt 53 e 3t e t ut The most common situation in practice is that there are no repeated poles but lets see what happens if we have two poles that are identical Gs b M s M b M1 s M1 b 1 s b 0 s p 1 2 s p 3 s p N rob28124ch08354405indd 368 041216 141 pm 89 PartialFraction Expansion 369 If we try the same technique to find the partialfraction form we get Gs K 11 s p 1 K 12 s p 1 K 3 s p 3 K N s p N But this can be written as Gs K 11 K 12 s p 1 K 3 s p 3 K N s p N K 1 s p 1 K 3 s p 3 K N s p N and we see that the sum of two arbitrary constants K 11 K 12 is really only a single arbitrary constant There are really only N 1 Ks instead of N Ks and when we form the common denominator of the partialfraction sum it is not the same as the denominator of the original function We could change the form of the partialfraction expansion to Gs K 1 s p 1 2 K3 s p3 KN s pN Then if we tried to solve the equation by finding a common denominator and equating equal powers of s we would find that we have N equations in N 1 unknowns and there is no unique solution The solution to this problem is to find a partialfraction expansion in the form Gs K 12 s p 1 2 K11 s p1 K3 s p3 KN s pN We can find K 12 by multiplying both sides of b M s M b M1 s M1 b 1 s b 0 s p 1 2 s p 3 s p N K 12 s p 1 2 K11 s p1 K3 s p3 KN s pN 88 by s p 1 2 yielding b M s M b M1 s M1 b 1 s b 0 s p 3 s p N K 12 s p 1 K 11 s p 1 2 K 3 s p 3 s p 1 2 K N s p N and then letting s p 1 yielding K 12 b M p 1 M b M1 p 1 M1 b 1 p 1 b 0 p1 p3p1 p N But when we try to find K 11 by the usual technique we encounter another problem s p 1 b M s M b M1 s M1 b 1 s b 0 s p 1 2 s p 3 s p N s p 1 K 12 s p 1 2 s p 1 K 11 s p 1 s p 1 K 3 s p 3 s p 1 K N s p N or b M s M b M1 s M1 b 1 s b 0 s p 1 s p 3 s p N K 12 s p 1 K11 rob28124ch08354405indd 369 041216 141 pm C h a p t e r 8 The Laplace Transform 370 Now if we set s p 1 we get division by zero on both sides of the equation and we cannot directly solve it for K 11 But we can avoid this problem by multiplying 88 through by s p 1 2 yielding b M s M b M1 s M1 b 1 s b 0 s p 3 s p N K 12 s p 1 K 11 s p 1 2 K 3 s p 3 s p 1 2 K N s p N differentiating with respect to s yielding d ds b M s M b M1 s M1 b 1 s b 0 s p 3 s p N K 11 s p 3 2s p 1 s p 1 2 s p 3 2 K 3 s p q 2s p 1 s p 1 2 s p N 2 K N and then setting s p 1 and solving for K 11 K 11 d ds b M s M b M1 s M1 b 1 s b 0 s p 3 s p N s p 1 d ds s p 1 2 Gs s p 1 If there were a higherorder repeated pole such as a triple quadruple and so on very unusual in practice we could find the coefficients by extending this differentiation idea to multiple derivatives In general if Hs is of the form Hs b M s M b M1 s M1 b 1 s b 0 s p 1 s p 2 s p N1 s p N m with N 1 distinct finite poles and a repeated Nth pole of order m it can be written as Hs K 1 s p 1 K 2 s p 2 K N1 s p N1 K Nm s p N m K Nm1 s p N m1 K N1 s p N where the Ks for the distinct poles are found as before and where the K for a repeated pole p q of order m for the denominator of the form s p q mk is K qk 1 m k d mk d s mk s p q m Hs s p q k 12 m 89 and it is understood that 0 1 ExamplE 84 Inverse Laplace transform using partialfraction expansion Find the inverse Laplace transform of Gs s 5 s 2 s 2 σ 0 This function has a repeated pole at s 0 Therefore the form of the partialfraction ex pansion must be Gs K 12 s 2 K 11 s K 3 s 2 σ 0 rob28124ch08354405indd 370 041216 141 pm 89 PartialFraction Expansion 371 We find K 12 by multiplying Gs by s 2 and setting s to zero in the remaining expression yielding K 12 s 2 Gs s0 52 We find K 11 by multiplying Gs by s 2 differentiating with respect to s and setting s to zero in the remaining expression yielding K 11 d ds s 2 Gs s0 d ds s 5 s 2 s0 s 2 s 5 s 2 2 s0 3 4 We find K 3 by the usual method to be 34 So Gs 5 2 s 2 3 4s 3 4s 2 σ 0 and the inverse transform is gt 5 2 t 3 4 3 4 e 2t ut 10t 31 e 2t 4 ut Lets now examine the effect of a violation of one of the assumptions in the origi nal explanation of the partialfraction expansion method the assumption that Gs b M s M b M1 s M1 b 1 s b 0 s p 1 s p 2 s p N is a proper fraction in s If M N we cannot expand in partial fractions because the partialfraction expression is in the form Gs K 1 s p 1 K 2 s p 2 K N s p N Combining these terms over a common denominator Gs K 1 k1 k1 kN s p k K 2 k1 k2 kN s p k K 2 k1 kN kN s p k s p 1 s p 2 s p N The highest power of s in the numerator is N 1 Therefore any ratio of polynomials in s that is to be expanded in partial fractions must have a numerator degree in s no greater than N 1 making it proper in s This is not really much of a restriction be cause if the fraction is improper in s we can always synthetically divide the numerator by the denominator until we have a remainder that is of lower order than the denomina tor Then we will have an expression consisting of the sum of terms with nonnegative integer powers of s plus a proper fraction in s The terms with nonnegative powers of s have inverse Laplace transforms that are impulses and higherorder singularities ExamplE 85 Inverse Laplace transform using partialfraction expansion Find the inverse Laplace transform of Gs 10 s 2 s 1s 3 σ 0 rob28124ch08354405indd 371 041216 141 pm C h a p t e r 8 The Laplace Transform 372 This rational function is an improper fraction in s Synthetically dividing the numerator by the denominator we get s 2 4s 3 10 s 2 10 s 2 40s 30 40s 30 10 10 s 2 s 1s 3 10 40s 30 s 2 4s 3 Therefore Gs 10 40s 30 s 1s 3 σ 0 Expanding the proper fraction in s in partial fractions Gs 10 5 9 s 3 1 s 1 σ 0 Then using e at ut 1 s a and δt 1 we get gt 10δt 59 e 3t e t ut Figure 814 Figure 814 Inverse Laplace transform of Gs 10 s 2 s 1s 3 1 10 40 2 3 4 5 t gt ExamplE 86 Inverse Laplace transform using partialfraction expansion Find the inverse Laplace transform of Gs s s 3 s 2 4s 5 σ 2 If we take the usual route of finding a partialfraction expansion we must first factor the denominator Gs s s 3s 2 js 2 j σ 2 rob28124ch08354405indd 372 041216 141 pm 89 PartialFraction Expansion 373 and find that we have a pair of complexconjugate poles The partialfraction method still works with complex poles Expanding in partial fractions Gs 32 s 3 3 j4 s 2 j 3 j4 s 2 j σ 2 With complex poles like this we have a choice We can either 1 Continue as though they were real poles find a timedomain expression and then simplify it or 2 Combine the last two fractions into one fraction with all real coefficients and find its inverse Laplace transform by looking up that form in a table Method 1 gs 3 2 e 3t 3 j 4 e 2jt 3 j 4 e 2jt u t This is a correct expression for gt but it is not in the most convenient form We can manipulate it into an expression containing only realvalued functions Finding a common denominator and recognizing trigonometric functions gt 3 2 e 3t 3 e 2jt 3 e 2jt j e 2jt j e 2jt 4 ut gt 3 2 e 3t e 2t 3 e jt e jt j e jt e jt 4 ut gt 32 e 2t cost 13 sin t e 3t ut Method 2 Gs 32 s 3 1 4 3 js 2 j 3 js 2 j s 2 4s 5 σ 2 When we simplify the numerator we have a firstdegree polynomial in s divided by a second degree polynomial in s Gs 32 s 3 1 4 6s 10 s 4s 5 32 s 3 6 4 s 53 s 2 2 1 σ 2 In the table of transforms we find e αt cos ω 0 t ut s α s α 2 ω 0 2 σ α and e αt sin ω 0 t ut ω 0 s α 2 ω 0 2 σ α Our denominator form matches these denominators but the numerator form does not But we can add and subtract numerator forms to form two rational functions whose numerator forms do appear in the table Gs 32 s 3 3 2 s 2 s 2 2 1 13 1 s 2 2 1 σ 2 810 Now we can directly find the inverse transform gt 32 e 2t cos t 13 sin t e 3t ut rob28124ch08354405indd 373 041216 141 pm C h a p t e r 8 The Laplace Transform 374 Realizing that there are two complex conjugate roots we could have combined the two terms with the complex roots into one with a common denominator of the form Gs A s 3 K 2 s p 2 K 3 s p 3 A s 3 s K 2 K 3 K 3 p 2 K 2 p 3 s 2 4s 5 or since K 2 and K 3 are arbitrary constants Gs A s 3 Bs C s 2 4s 5 Both B and C will be real numbers because K 2 and K 3 are complex conjugates and so are p2 and p3 Then we can find the partialfraction expansion in this form A is found exactly as before to be 32 Since Gs and its partialfraction expansion must be equal for any arbitrary value of s and Gs s s 3 s 2 4s 5 we can write s s 3 s 2 4s 5 s0 32 s 3 Bs C s 2 4s 5 s0 or 0 12 C5 C 52 Then s s 3 s 2 4s 5 32 s 3 Bs 52 s 2 4s 5 and we can find B by letting s be any convenient number for example one Then 1 4 3 4 B 52 2 B 3 2 and Gs 32 s 3 3 2 s 53 s 2 4s 5 This result is identical to 810 and the rest of the solution is therefore the same MATLAB has a function residue that can be used in finding partialfraction expansions The syntax is rpk residueba where b is a vector of coefficients of descending powers of s in the numerator of the expression and a is a vector of coefficients of descending powers of s in the denominator of the expression r is a vector of residues p is a vector of finite pole locations and k is a vector of socalled direct terms which result when the degree of the numerator is equal to or greater than the degree of the denominator The vectors a and b must always include all powers of s down through zero The term residue comes from theories of closedcontour integration in the complex plane a topic that is beyond the scope of this text For our purposes residues are simply the numerators in the partialfraction expansion rob28124ch08354405indd 374 041216 141 pm 89 PartialFraction Expansion 375 ExamplE 87 Partialfraction expansion using MATLABs residue function Expand the expression Hs s2 3s 1 s4 5s3 2s2 7s 3 in partial fractions In MATLAB b 1 3 1 a 1 5 2 7 3 rpk residueba r r 00856 00496 02369i 00496 02369i 00135 p p 48587 01441 11902i 01441 11902i 04295 k k There are four poles at 48587 01441 j11902 01441 j11902 and 04295 and the residues at those poles are 00856 00496 j02369 00496 j02369 and 00135 respec tively There are no direct terms because Hs is a proper fraction in s Now we can write Hs as Hs 00496 j02369 s 01441 j11902 00496 j02369 s 01441 j11902 00856 s 48587 00135 s 04295 or combining the two terms with complex poles and residues into one term with all real coefficients Hs 00991s 05495 s2 02883s 1437 00856 s 048587 00135 s 04295 ExamplE 88 Response of an LTI system Find the response yt of an LTI system a With impulse response ht 5 e 4t ut if excited by xt ut b With impulse response ht 5 e 4t ut if excited by xt ut c With impulse response ht 5 e 4t ut if excited by xt ut rob28124ch08354405indd 375 041216 141 pm C h a p t e r 8 The Laplace Transform 376 d With impulse response ht 5 e 4t ut if excited by xt ut a ht 5 e 4t ut Hs 5 s 4 σ 4 xt ut Xs 1s σ 0 Therefore Ys HsXs 5 ss 4 σ 0 Ys can be expressed in the partialfraction form Ys 54 s 54 s 4 σ 0 yt 541 e 4t ut Ys 54 s 54 s 4 σ 0 Figure 815 Figure 815 The four system responses t 15 15 yt 15 15 t 15 15 yt 15 15 t 15 15 yt 15 15 t 15 15 yt 15 15 ht 5e4t ut xt ut ht 5e4t ut xt ut ht 5e4t ut xt ut ht 5e4t ut xt ut b xt ut Xs 1s σ 0 Ys HsXs 5 ss 4 4 σ 0 Ys 54 s 54 s 4 4 σ 0 yt 54e4t ut ut Ys 54 s 54 s 4 4 σ 0 Figure 815 rob28124ch08354405indd 376 041216 141 pm 810 LaplaceTransform Properties 377 c ht 5 e 4t ut Hs 5 s 4 σ 4 Ys HsXs 5 ss 4 0 σ 4 Ys 54 s 54 s 4 0 σ 4 yt 54 ut e 4t ut Ys 54 s 54 s 4 0 σ 4 Figure 815 d Ys HsXs 5 ss 4 σ 0 Ys 54 s 54 s 4 σ 0 yt 54ut e 4t ut Ys 54 s 54 s 4 σ 4 Figure 815 810 LAPLACETRANSFORM PROPERTIES Let gt and ht have Laplace transforms Gs and Hs with regions of convergence ROC G and ROC H respectively Then it can be shown that the following properties apply Table 82 Table 82 Laplacetransform properties Linearity α gt βht t 0 αGs βHs ROC ROC G ROC H TimeShifting gt t 0 Gs e s t 0 ROC ROC G sDomain Shift e s 0 t gt Gs s 0 ROC ROC G shifted by s 0 s is in ROC if s s 0 is in ROC G Time Scaling gat 1aGsa ROC ROC G scaled by a s is in ROC if sa is in ROC G Time Differentiation d dt gt sGs ROC ROC G sDomain Differentiation tgτ d ds Gs ROC ROC G Time Integration t gτdτ Gs s ROC ROC G σ 0 Convolution in Time gt ht Gs Hs ROC ROC G ROC H If gt 0 t 0 and there are no impulses or higherorder singularities at t 0 then Initial Value Theorem g0 lim s sGs Final Value Theorem lim t gt lim s0 sGs if lim t gt exists The final value theorem applies if the limit lim t g t exists The limit lim s0 s G s may exist even if the limit lim t g t does not exist For example if X s s s 2 4 then lim s0 s G s lim s0 s 2 s 2 4 0 But x t cos 4t u t and lim t g t lim t cos 4t u t does not exist Therefore the conclusion that the final value is zero is wrong It can be shown that For the final value theorem to apply to a function G s all the finite poles of the func tion sG s must lie in the open left half of the s plane rob28124ch08354405indd 377 041216 141 pm C h a p t e r 8 The Laplace Transform 378 ExamplE 89 Use of the sdomain shifting property If X1s 1 s 5 σ 5 and X2s X1s j4 X1s j4 σ 5 find x2t e5t ut 1 s 5 σ 5 Using the sdomain shifting property e 5j4t ut 1 s j4 5 σ 5 and e 5j4t ut 1 s j4 5 σ 5 Therefore x2t e 5j4t ut e 5j4t ut e 5t e j4t e j4t ut 2 e 5t cos 4t ut The effect of shifting equal amounts in opposite directions parallel to the ω axis in the s domain and adding corresponds to multiplication by a causal cosine in the time domain The overall effect is doublesideband suppressed carrier modulation which will be discussed in Web Chapter 15 ExamplE 810 Laplace transforms of two timescaled rectangular pulses Find the Laplace transforms of xt ut ut a and x2t u2t u2t a We have already found the Laplace transform of ut which is 1s σ 0 Using the linear ity and timeshifting properties ut ut a 1 e as s all σ Now using the timescaling property u2t u2t a 1 2 1 e as2 s2 1 e as2 s all σ This result is sensible when we consider that u2t ut and u2t a u2t a2 ut a2 ExamplE 811 Using sdomain differentiation to derive a transform pair Using sdomain differentiation and the basic Laplace transform ut 1s σ 0 find the inverse Laplace transform of 1 s 2 σ 0 ut 1s σ 0 Using tgt d ds Gs t ut 1s2 σ 0 rob28124ch08354405indd 378 041216 141 pm 811 The Unilateral Laplace Transform 379 Therefore rampt t ut 1s2 σ 0 By induction we can extend this to the general case d ds 1 s 1 s 2 d 2 d s 2 1 s 2 s 3 d 3 d s 3 1 s 6 s 4 d 4 d s 4 1 s 24 s 5 d n d s n 1 s 1 n n s n1 The corresponding transform pairs are t ut 1 s 2 σ 0 t 2 2 ut 1 s 3 σ 0 t3 6 ut 1 s 4 σ 0 t n n ut 1 s n1 σ 0 ExamplE 812 Using the time integration property to derive a transform pair In Example 811 we used complexfrequency differentiation to derive the Laplacetransform pair t ut 1 s 2 σ 0 Derive the same pair from ut 1s σ 0 using the time integration property instead t uτdτ 0 t dτ t t 0 0 t 0 t ut Therefore t ut 1 s 1 s 1 s 2 σ 0 Successive integrations of ut yield t ut t 2 2 ut t 3 6 ut and these can be used to derive the general form t n n ut 1 s n1 σ 0 811 THE UNILATERAL LAPLACE TRANSFORM DEFINITION In the introduction to the Laplace transform it was apparent that if we consider the full range of possible signals to transform sometimes a region of convergence can be rob28124ch08354405indd 379 041216 141 pm C h a p t e r 8 The Laplace Transform 380 found and sometimes it cannot be found If we leave out some pathological functions like t t or e t 2 which grow faster than an exponential and have no known engineering usefulness and restrict ourselves to functions that are zero before or after time t 0 the Laplace transform and its ROC become considerably simpler The quality that made the functions g1t A e αt ut α 0 and g2t A e αt ut α 0 Laplace transformable was that each of them was restricted by the unitstep function to be zero over a semiinfinite range of time Even a function as benign as gt A which is bounded for all t causes problems because a single convergence factor that makes the Laplace transform converge for all time cannot be found But the function gt Aut is Laplace transformable The presence of the unit step allows the choice of a convergence factor for positive time that makes the Laplacetransform integral converge For this reason and other reasons a modification of the Laplace transform that avoids many convergence issues is usually used in practical analysis Let us now redefine the Laplace transform as Gs 0 gt e st dt Only the lower limit of integration has changed The Laplace transform defined by Gs gt e st dt is conventionally called the twosided or bilateral Laplace transform The Laplace transform defined by Gs 0 gt e st dt is conventionally called the onesided or unilateral Laplace transform The unilateral Laplace transform is restrictive in the sense that it excludes the negativetime behavior of functions But since in the analy sis of any real system a time origin can be chosen to make all signals zero before that time this is not really a practical problem and actually has some advantages Since the lower limit of integration is t 0 any functional behavior of gt before time t 0 is irrelevant to the transform This means that any other function that has the same behavior at or after time t 0 will have the same transform Therefore for the trans form to be unique to one timedomain function it should only be applied to functions that are zero before time t 02 The inverse unilateral Laplace transform is exactly the same as derived above for the bilateral Laplace transform gt 1 j2π σj σj Gs e st ds It is common to see the Laplacetransform pair defined by gt Gs 0 gt e st dt 1Gs gt 1 j2π σj σj Gs e st ds 811 The unilateral Laplace transform has a simple ROC It is always the region of the s plane to the right of all the finite poles of the transform Figure 816 2Even for times t 0 the transform is not actually unique to a single timedomain function As mentioned in Chapter 2 in the discussion of the definition of the unitstep function all the definitions have exactly the signal energy over any finite time range and yet their values are different at the discontinuity time t 0 This is a mathematical point without any real engineering significance Their effects on any real system will be identical because there is no signal energy in a signal at a point unless there is an impulse at the point and real systems respond to the energy of input signals Also if two functions differ in value at a finite number of points the Laplace transform integral will yield the same transform for the two functions because the area under a point is zero rob28124ch08354405indd 380 041216 141 pm 811 The Unilateral Laplace Transform 381 PROPERTIES UNIQUE TO THE UNILATERAL LAPLACE TRANSFORM Most of the properties of the unilateral Laplace transform are the same as the properties of the bilateral Laplace transform but there are a few differences If gt 0 for t 0 and ht 0 for t 0 and gt Gs and ht Hs then the properties in Table 83 that are different for the unilateral Laplace transform can be shown to apply The timeshifting property is now only valid for time shifts to the right time de lays because only for delayed signals is the entire nonzero part of the signal still guar anteed to be included in the integral from 0 to infinity If a signal were shifted to the left advanced in time some of it might occur before time t 0 and not be included within the limits of the Laplacetransform integral That would destroy the unique relation between the transform of the signal and the transform of its shifted version making it impossible to relate them in any general way Figure 817 σ ω ROC s Figure 816 ROC for a unilateral Laplace transform Table 83 Unilateral Laplacetransform properties that differ from bilateral Laplace transform properties Time Shifting gt t0 Gse st 0 t0 0 Time Scaling gat 1aGsa a 0 First Time Derivative d dt gt sGs g0 Nth Time Derivative dN dt N gt sNGs n1 N s Nn d n1 dt n1 gt t0 Time Integration 0 t gτdτ Gss rob28124ch08354405indd 381 041216 141 pm C h a p t e r 8 The Laplace Transform 382 Similarly in the timescaling and frequencyscaling properties the constant a can not be negative because that would turn a causal signal into a noncausal signal and the unilateral Laplace transform is only valid for causal signals The time derivative properties are important properties of the unilateral Laplace transform These are the properties that make the solution of differential equations with initial conditions systematic When using the differentiation properties in solving differential equations the initial conditions are automatically called for in the proper form as an inherent part of the transform process Table 84 has several commonly used unilateral Laplace transforms Figure 817 Shifts of a causal function t gt t gtt0 t gtt0 t0 0 t0 0 Table 84 Common unilateral Laplacetransform pairs δt 1 All s ut 1s σ 0 u n t ut ut n1 convolutions 1 s n σ 0 rampt tut 1 s 2 σ 0 e αt ut 1 s α σ α t n ut n s n1 σ 0 t e αt ut 1 s α 2 σ α t n e αt ut n s α n1 σ α sin ω 0 t ut ω 0 s 2 ω 0 2 σ 0 cos ω 0 t ut s s 2 ω 0 2 σ 0 e αt sin ω 0 t ut ω 0 s α 2 ω 0 2 σ α e αt cos ω 0 t ut s α s α 2 ω 0 2 σ α rob28124ch08354405indd 382 041216 141 pm 811 The Unilateral Laplace Transform 383 SOLUTION OF DIFFERENTIAL EQUATIONS WITH INITIAL CONDITIONS The power of the Laplace transform lies in its use in the analysis of linear system dynamics This comes about because linear continuoustime systems are described by linear differential equations and after Laplace transformation differentiation is rep resented by multiplication by s Therefore the solution of the differential equation is transformed into the solution of an algebraic equation The unilateral Laplace trans form is especially convenient for transient analysis of systems whose excitation begins at an initial time which can be identified as t 0 and of unstable systems or systems driven by forcing functions that are unbounded as time increases ExamplE 813 Solution of a differential equation with initial conditions using the unilateral Laplace transform Solve the differential equation x t 7 x t 12xt 0 for times t 0 subject to the initial conditions x 0 2 and d dt xt t 0 4 First Laplace transform both sides of the equation s 2 Xs sx 0 d dt xt t 0 7sXs x 0 12Xs 0 Then solve for Xs Xs sx 0 7x 0 d dt xt t 0 s2 7s 12 or Xs 2s 10 s 2 7s 12 Expanding Xs in partial fractions Xs 4 s 3 2 s 4 From the Laplacetransform table e αt ut 1 s α Inverse Laplace transforming xt 4e3t 2e4t ut Substituting this result into the original differential equation for times t 0 d 2 d t 2 4e3t 2e4t 7 d dt 4e3t 2e4t 124e3t 2e4t 0 36e3t 32e4t 84e3t 56e4t 48e3t 24e4t 0 0 0 rob28124ch08354405indd 383 041216 141 pm C h a p t e r 8 The Laplace Transform 384 proving that the xt found actually solves the differential equation Also x 0 4 2 2 and d dt xt t 0 12 8 4 which verifies that the solution also satisfies the stated initial conditions ExamplE 814 Response of a bridgedT network In Figure 818 the excitation voltage is vi t 10 ut volts Find the zerostate response v R L t Figure 818 BridgedT network R1 10 kΩ RL 1 kΩ R2 10 kΩ C1 1 μF C2 1 μF vx t vi t vRL t We can write nodal equations C 1 d dt vxt vit C 2 d dt v x t v RL t G 1 v x t 0 C 2 d dt v RL t v x t G L v RL t G 2 v RL t v i t 0 where G 1 1 R 1 10 4 S G 2 1 R 2 10 4 S and G L 10 3 S Laplace transforming the equations C 1 s V x s v x 0 s V i s v i 0 C 2 s V x s v x 0 s V RL s v RL 0 G 1 V x s 0 C 2 s V RL s v RL 0 s V x s v x 0 G L V RL s G 2 V RL s V i s 0 Since we seek the zerostate response all the initial conditions are zero and the equations simplify to s C 1 V x s V i s s C 2 V x s V RL s G 1 V x s 0 s C 2 V RL s V x s G L V RL s G 2 V RL s V i s 0 The Laplace transform of the excitation is V i s 10s Then s C 1 C 2 G 1 s C 2 s C 2 s C 2 G L G 2 V x s V RL s 10 C 1 10 G 2 s rob28124ch08354405indd 384 041216 141 pm 812 PoleZero Diagrams and Frequency Response 385 The determinant of the 2 by 2 matrix is Δ s C 1 C 2 G 1 s C 2 G L G 2 s 2 C 2 2 s 2 C 1 C 2 s G 1 C 2 G L G 2 C 1 C 2 G 1 G L G 2 and by Cramers rule the solution for the Laplace transform of the response is V RL s s C 1 C 2 G 1 10 C 1 s C 2 10 G 2 s s 2 C 1 C 2 s G 1 C 2 G L G 2 C 1 C 2 G 1 G L G 2 V RL s 10 s 2 C 1 C 2 s G 2 C 1 C 2 G 1 G 2 s s 2 C 1 C 2 s G 1 C 2 G L G 2 C 1 C 2 G 1 G L G 2 or V RL s 10 s 2 s G 2 C 1 C 2 C 1 C 2 G 1 G 2 C 1 C 2 s s 2 s G 1 C 1 G L G 2 C 1 C 2 C 1 C 2 G 1 G L G 2 C 1 C 2 Using the component numerical values V RL s 10 s 2 200s 10000 s s 2 2300s 110000 Expanding in partial fractions V RL s 09091 s 0243 s 4886 9334 s 2251 Inverse Laplace transforming v RL t 09091 0243 e 4886t 9334 e 2251t ut As a partial check on the correctness of this solution the response approaches 09091 as t This is exactly the voltage found using voltage division between the two resistors considering the capacitors to be open circuits So the final value looks right The initial response at time t 0 is 10 V The capacitors are initially uncharged so at time t 0 their voltages are both zero and the excitation and response voltages must be the same So the initial value also looks right These two checks on the solution do not guarantee that it is correct for all time but they are very good checks on the reasonableness of the solution and will often detect an error 812 POLEZERO DIAGRAMS AND FREQUENCY RESPONSE In practice the most common kind of transfer function is one that can be expressed as a ratio of polynomials in s Hs Ns Ds This type of transfer function can be factored into the form Hs A s z 1 s z 2 s z M s p 1 s p 2 s p N rob28124ch08354405indd 385 041216 141 pm C h a p t e r 8 The Laplace Transform 386 Then the frequency response of the system is H jω A jω z 1 jω z 2 jω z M jω p 1 jω p 2 jω p N To illustrate a graphical interpretation of this result with an example let the transfer function be Hs 3s s 3 This transfer function has a zero at s 0 and a pole at s 3 Figure 819 Converting the transfer function to a frequency response H jω 3 jω jω 3 The frequency response is three times the ratio of jω to jω 3 The numerator and denominator can be conceived as vectors in the s plane as illustrated in Figure 820 for an arbitrary choice of ω σ ω s 3 s 0 s Figure 819 Polezero plot for Hs 3ss 3 Figure 820 Diagram showing the vectors jω and jω 3 σ ω s 3 s 0 s ω jω jω 3 As the frequency ω is changed the vectors change also The magnitude of the frequency response at any particular frequency is three times the magnitude of the numerator vector divided by the magnitude of the denominator vector H jω 3 jω jω 3 The phase of the frequency response at any particular frequency is the phase of the constant 3 which is zero plus the phase of the numerator jω a constant π2 radians for positive frequencies and a constant π2 radians for negative frequencies minus the phase of the denominator jω 3 H jω 3 0 jω jω 3 At frequencies approaching zero from the positive side the numerator vector length approaches zero and the denominator vector length approaches a minimum rob28124ch08354405indd 386 041216 141 pm 812 PoleZero Diagrams and Frequency Response 387 value of 3 making the overall frequency response magnitude approach zero In that same limit the phase of jω is π2 radians and the phase of jω 3 approaches zero so that the overall frequency response phase approaches π2 radians lim ω 0 H jω lim ω 0 3 jω jω 3 0 and lim ω 0 Hjω lim ω 0 jω lim ω 0 jω 3 π2 0 π2 At frequencies approaching zero from the negative side the numerator vector length approaches zero and the denominator vector length approaches a minimum value of 3 making the overall frequency response magnitude approach zero as before In that same limit the phase of jω is π2 radians and the phase of jω 3 approaches zero so that the overall frequency response phase approaches π2 radians lim ω 0 H jω lim ω 0 3 jω jω 3 0 and lim ω 0 H jω lim ω 0 jω lim ω 0 jω 3 π2 0 π2 At frequencies approaching positive infinity the two vector lengths approach the same value and the overall frequency response magnitude approaches 3 In that same limit the phase of jω is π2 radians and the phase of jω 3 approach π2 radians so that the overall frequencyresponse phase approaches zero lim ω H jω lim ω 3 jω jω 3 3 and lim ω Hjω lim ω jω lim ω jω 3 π2 π2 0 At frequencies approaching negative infinity the two vector lengths approach the same value and the overall frequency response magnitude approaches 3 as before In that same limit the phase of jω is π2 radians and the phase of jω 3 approach π2 radians so that the overall frequencyresponse phase approaches zero lim ω Hjω lim ω 3 jω jω 3 3 and lim ω H jω lim ω jω lim ω jω 3 π2 π2 0 These attributes of the frequency response inferred from the polezero plot are borne out by a graph of the magnitude and phase frequency response Figure 821 This system attenuates low frequencies relative to higher frequencies A system with this type of frequency response is often called a highpass filter because it generally lets high frequencies pass through and generally blocks low frequencies rob28124ch08354405indd 387 041216 141 pm C h a p t e r 8 The Laplace Transform 388 ExamplE 815 Frequency response of a system from its polezero diagram Find the magnitude and phase frequency response of a system whose transfer function is Hs s 2 2s 17 s 2 4s 104 This can be factored into Hs s 1 j4s 1 j4 s 2 j10s 2 j10 So the poles and zeros of this transfer function are z 1 1 j4 z 2 1 j4 p 1 2 j10 and p 2 2 j10 as illustrated in Figure 822 Converting the transfer function to a frequency response Hjω jω 1 j4 jω 1 j4 jω 2 j10 jω 2 j10 The magnitude of the frequency response at any particular frequency is the product of the numeratorvector magnitudes divided by the product of the denominatorvector magnitudes Hjω jω 1 j4 jω 1 j4 jω 2 j10 jω 2 j10 The phase of the frequency response at any particular frequency is the sum of the numeratorvector angles minus the sum of the denominatorvector angles H jω jω 1 j4 jω 1 j4 jω 2 j10 jω 2 j10 This transfer function has no poles or zeros on the ω axis Therefore its frequency response is neither zero nor infinite at any real frequency But the finite poles and finite zeros are near the real axis and because of that proximity will strongly influence the frequency response for real frequencies near those poles and zeros For a real frequency ω near the pole p 1 the denominator factor jω 2 j10 becomes very small and that makes the overall frequency response mag nitude become very large Conversely for a real frequency ω near the zero z 1 the numerator Figure 821 Magnitude and phase frequency response of a system whose transfer function is Hs 3ss 3 ω 20 20 H jω 3 ω 20 20 π 2 π 4 π 2 σ ω s 3 s 0 s ω 3 ω 3 3 2 π 4 3 2 H jω Figure 822 Polezero plot of Hs s 2 2s 17 s 2 4s 104 σ ω 2 2 2 2 4 6 8 10 4 6 8 10 8 6 4 10 z1 z2 p1 p2 rob28124ch08354405indd 388 041216 141 pm 812 PoleZero Diagrams and Frequency Response 389 factor jω 1 j4 becomes very small and that makes the overall frequency response magnitude become very small So not only does the frequency response magnitude go to zero at zeros and to infinity at poles it also becomes small near zeros and it becomes large near poles The frequency response magnitude and phase are illustrated in Figure 823 Figure 823 Magnitude and phase frequency response of a system whose transfer function is Hs s 2 2s 17 s 2 4s 104 ω 40 40 10 10 4 10 4 4 10 4 H jω 22536 ω 40 40 π π H jω Figure 824 Onefinitepole lowpass filter σ ω s 5 f 10 10 H f 02 t 02 12 h1t 02 Frequency response can be graphed using the MATLAB control toolbox command bode and polezero diagrams can be plotted using the MATLAB control toolbox command pzmap By using graphical concepts to interpret polezero plots one can with practice perceive approximately how the frequency response looks There is one aspect of the transfer function that is not evident in the polezero plot The frequencyindependent gain A has no effect on the polezero plot and therefore cannot be determined by observing it But all the dynamic behavior of the system is determinable from the polezero plot to within a gain constant Below is a sequence of illustrations of how frequency response and step response change as the number and locations of the finite poles and zeros of a system are changed In Figure 824 is a polezero diagram of a system with one finite pole and no finite zeros Its frequency response emphasizes low frequencies relative to high frequencies rob28124ch08354405indd 389 041216 141 pm C h a p t e r 8 The Laplace Transform 390 making it a lowpass filter and its step response reflects that fact by not jumping dis continuously at time t 0 and approaching a nonzero final value The continuity of the step response at time t 0 is a consequence of the fact that the highfrequency content of the unit step has been attenuated so that the response cannot change discontinuously In Figure 825 a zero at zero has been added to the system in Figure 824 This changes the frequency response to that of a highpass filter This is reflected in the step response in the fact that it jumps discontinuously at time t 0 and approaches a final value of zero The final value of the step response must be zero because the filter completely blocks the zerofrequency content of the input signal The jump at t 0 is discontinuous because the highfrequency content of the unit step has been retained In Figure 826 is a lowpass filter with two real finite poles and no finite zeros The step response does not jump discontinuously at time t 0 and approaches a non zero final value The response is similar to that in Figure 824 but the attenuation of highfrequency content is stronger as can be seen in the fact that the frequency re sponse falls faster with increasing frequency than the response in Figure 824 The step response is also slightly different starting at time t 0 with a zero slope instead of the nonzero slope in Figure 824 Figure 825 Onefinitepole onefinitezero highpass filter σ ω s 5 f 10 10 H f 1 t 02 12 h1t 1 Figure 826 Twofinite pole system σ ω s 5 2 f 10 10 H f 01 t 05 3 h1t 01 In Figure 827 a zero at zero has been added to the system of Figure 826 The step response does not jump discontinuously at time t 0 and approaches a final value of zero because the system attenuates both the highfrequency content and the lowfrequency content relative to the midrange frequencies A system with this general form of frequency response is called a bandpass filter Attenuating the highfrequency content makes the step response continuous and attenuating the lowfrequency content makes the final value of the step response zero rob28124ch08354405indd 390 041216 141 pm 812 PoleZero Diagrams and Frequency Response 391 In Figure 828 another zero at zero has been added to the filter of Figure 827 making it a highpass filter The step response jumps discontinuously at time t 0 and the response approaches a final value of zero The lowfrequency attenuation is stronger than the system of Figure 825 and that also affects the step response making it undershoot zero before settling to zero Figure 827 Twofinitepole onefinitezero bandpass filter σ ω s 5 2 f 10 10 H f 016 t 05 35 h1t 012 Figure 828 Twofinitepole twofinitezero highpass filter 2 σ ω s 5 2 f 10 10 H f 1 t 05 25 h1t 02 1 In Figure 829 is another twofinitepole lowpass filter but with a frequency re sponse that is noticeably different from the system in Figure 826 because the poles are now complex conjugates instead of real The frequency response increases and reaches a peak at frequencies near the two poles before it falls at high frequencies A system with this general form of frequency response is said to be underdamped In an under damped system the step response overshoots its final value and rings before settling Figure 829 Twofinitepole underdamped lowpass filter σ ω s 1 5 5 f 10 10 H f 01 t 6 1 h1t 006 rob28124ch08354405indd 391 041216 141 pm C h a p t e r 8 The Laplace Transform 392 The step response is still continuous everywhere and still approaches a nonzero final value but in a different way than in Figure 826 In Figure 830 a zero at zero has been added to the system of Figure 829 This changes it from lowpass to bandpass but now because of the complexconjugate pole lo cations the response is underdamped as is seen in the peaking in the frequency response and the ringing in the step response as compared with the system in Figure 827 Figure 830 Twofinitepole onefinitezero underdamped bandpass filter σ ω s 1 5 5 f 10 10 H f 05 t 6 1 h1t 01 015 Figure 831 Twofinitepole twofinitezero underdamped highpass filter 2 σ ω s 1 5 5 f 10 10 H f 3 t 6 1 h1t 06 1 We see in these examples that moving the poles nearer to the ω axis decreases the damping makes the step response ring for a longer time and makes the frequency response peak to a higher value What would happen if we put the poles on the ω axis Having two finite poles on the ω axis and no finite zeros means that there are poles at s j ω 0 the transfer function is of the form Hs K ω 0 s 2 ω 0 2 and the impulse response is of the form ht K sin ω 0 t ut The response to an impulse is equal to a sinusoid after t 0 and oscillates with stable amplitude forever thereafter The frequency response is H jω K ω 0 jω 2 ω 0 2 So if the system is excited by a sinusoid xt A sin ω 0 t the response is infinite an unbounded response to a bounded excitation If the system were excited by a sinusoid applied at t 0 xt Asin ω 0 t ut the response would be yt KA 2 sin ω 0 t ω 0 t cos ω 0 t ut In Figure 831 another zero at zero has been added to the system of Figure 830 making it a highpass filter It is still underdamped as is evident in the peaking of the frequency response and the ringing in the step response rob28124ch08354405indd 392 041216 141 pm 813 MATLAB System Objects 393 This contains a sinusoid starting at time t 0 and growing in amplitude linearly for ever in positive time Again this is an unbounded response to a bounded excitation indicating an unstable system Undamped resonance is never achieved in a real passive system but it can be achieved in a system with active components that can compensate for energy losses and drive the damping ratio to zero 813 MATLAB SYSTEM OBJECTS The MATLAB control toolbox contains many helpful commands for the analysis of systems They are based on the idea of a system object a special type of variable in MATLAB for the description of systems One way of creating a system description in MATLAB is through the use of the tf transfer function command The syntax for creating a system object with tf is sys tfnumden This command creates a system object sys from two vectors num and den The two vectors are all the coefficients of s including any zeros in descending order in the nu merator and denominator of a transfer function For example let the transfer function be H 1 s s 2 4 s 5 4 s 4 7 s 3 15 s 2 31s 75 In MATLAB we can form H1s with num 1 0 4 den 1 4 7 15 31 75 H1 tfnumden H1 Transfer function s2 4 s5 4 s4 7 s3 15 s2 31 s 75 Alternately we can form a system description by specifying the finite zeros finite poles and a gain constant of a system using the zpk command The syntax is sys zpkzpk where z is a vector of finite zeros of the system p is a vector of finite poles of the system and k is the gain constant For example suppose we know that a system has a transfer function H 2 s 20 s 4 s 3s 10 We can form the system description with z 4 p 3 10 k 20 H2 zpkzpk H2 Zeropolegain 20 s4 s3 s10 rob28124ch08354405indd 393 041216 141 pm C h a p t e r 8 The Laplace Transform 394 Another way of forming a system object in MATLAB is to first define s as the independent variable of the Laplace transform with the command s tfs Then we can simply write a transfer function like H3s ss 3 s2 2s 8 in the same way we would on paper H3 ss3s22s8 Transfer function s2 3 s s2 2 s 8 We can convert one type of system description to the other type tfH2 Transfer function 20 s 80 s2 13 s 30 zpkH1 Zeropolegain s2 4 s3081 s2 2901s 545 s2 1982s 4467 We can get information about systems from their descriptions using the two com mands tfdata and zpkdata For example numden tfdataH2v num num 0 20 80 den den 1 13 30 zpk zpkdataH1v z z 0 20000i 0 20000i p p 30807 14505 18291i 14505 18291i 09909 18669i 09909 18669i k k 1 rob28124ch08354405indd 394 041216 141 pm Exercises with Answers 395 The v argument in these commands indicates that the answers should be returned in vector form This last result indicates that the transfer function H1s has zeros at j2 and poles at 30807 14505 j1829 and 09909 j18669 MATLAB has some handy functions for doing frequencyresponse analysis in the control toolbox The command H freqsnumdenw accepts the two vectors num and den and interprets them as the coefficients of the pow ers of s in the numerator and denominator of the transfer function Hs starting with the highest power and going all the way to the zero power not skipping any It returns in H the complex frequency response at the radian frequencies in the vector w 814 SUMMARY OF IMPORTANT POINTS 1 The Laplace transform can be used to determine the transfer function of an LTI system and the transfer function can be used to find the response of an LTI system to an arbitrary excitation 2 The Laplace transform exists for signals whose magnitudes do not grow any faster than an exponential in either positive or negative time 3 The region of convergence of the Laplace transform of a signal depends on whether the signal is right or left sided 4 Systems described by ordinary linear constantcoefficient differential equations have transfer functions in the form of a ratio of polynomials in s 5 Polezero diagrams of a systems transfer function encapsulate most of its properties and can be used to determine its frequency response to within a gain constant 6 MATLAB has an object defined to represent a system transfer function and many functions to operate on objects of this type 7 With a table of Laplace transform pairs and properties the forward and inverse transforms of almost any signal of engineering significance can be found 8 The unilateral Laplace transform is commonly used in practical problem solving because it does not require any involved consideration of the region of convergence and is therefore simpler than the bilateral form EXERCISES WITH ANSWERS Answers to each exercise are in random order LaplaceTransform Definition 1 Starting with the definition of the Laplace transform ℒ gt G s 0 gt e st dt find the Laplace transforms of these signals a xt e t ut b x t e 2t cos 200πt u t c xt rampt d x t t e t ut Answers 1 s 1 σ 1 1 s 1 2 σ 1 s 2 s 2 2 200π 2 σ 2 1 s 2 σ 0 rob28124ch08354405indd 395 041216 141 pm C h a p t e r 8 The Laplace Transform 396 Direct Form II System Realization 2 Draw Direct Form II system diagrams of the systems with these transfer functions a Hs 1 s 1 b Hs 4 s 3 s 10 Answers Xs Ys s1 Xs Ys 4 12 10 s1 Forward and Inverse Laplace Transforms 3 Using the complexfrequencyshifting property find and graph the inverse Laplace transform of Xs 1 s j4 3 1 s j4 3 σ 3 Answer t 01 2 xt 2 2 4 Using the timescaling property find the Laplace transforms of these signals a xt δ4t b xt u4t Answers 1 s σ 0 14 All s 5 Using the convolution in time property find the Laplace transforms of these signals and graph the signals versus time a xt e t ut u t b x t e t sin 20πt ut u t c x t 8 cos πt2 ut ut u t 1 d xt 8 cos 2πt u t u t u t 1 Answers t 1 4 xt 2 2 t 1 5 xt 1 5 t 1 xt 002 002 t 1 8 xt 10 10 rob28124ch08354405indd 396 041216 141 pm Exercises with Answers 397 6 A system impulse response ht has a unilateral Laplace transform H s s s 4 s 3 s 2 a What is the region of convergence for Hs b Could the CTFT of ht be found directly from Hs If not why not Answers σ 2 No 7 Find the inverse Laplace transforms of these functions a Xs 24 s s 8 σ 0 b Xs 20 s 2 4s 3 σ 3 c Xs 5 s 2 6s 73 σ 3 d Xs 10 s s 2 6s 73 σ 0 e Xs 4 s 2 s 2 6s 73 σ 0 f Xs 2s s 2 2s 13 σ 1 g Xs s s 3 σ 3 h Xs s s 2 4s 4 σ 2 i Xs s 2 s 2 4s 4 σ 2 j Xs 10s s 4 4 s 2 4 σ 2 Answers 5 2 t sin 2 t ut δ t 4 e 2t t 1 u t 01371 e3t cos8t 0375 sin8t u t 1 73 2 292t 24 24 e 3t cos8t 55 48 sin8t ut 58 e3t sin 8t ut e2t1 2t ut 10e3t et ut 31 e8t ut δt 3 e 3t ut 2et 1 12 sin 12t cos 12t u t 8 Using the initial and final value theorems find the initial and final values if possible of the signals whose Laplace transforms are these functions a Xs 10 s 8 σ 8 b Xs s 3 s 3 2 4 σ 3 c Xs s s 2 4 σ 0 d Xs 10s s 2 10s 300 σ 5 e Xs 8 s s 20 σ 0 f Xs 8 s 2 s 20 σ 0 g Xs s 3 s s 5 h Xs s 2 7 s 2 4 Answers 0 Does not apply 0 25 10 0 Does not apply Does not apply 1 Does not apply 1 0 1 06 Does not apply Does not apply rob28124ch08354405indd 397 041216 141 pm C h a p t e r 8 The Laplace Transform 398 9 A system has a transfer function H s s 2 2s 3 s 2 s 1 Its impulse response is h t and its unit step response is h 1 t Find the final values of each of them Answers 0 3 10 Find the numerical values of the literal constants in the form b 2 s 2 b 1 s b 0 a 2 s 2 a 1 s a 0 e s t 0 of the bilateral Laplace transforms of these functions a 3 e 8 t1 u t 1 b 4 cos 32πt u t c 4 e t2 sin 32π t 2 u t 2 Answers 0 0 3 1 0 8 1 0 4 0 1 0 32π2 0 0 0 128π 1 2 32π 2 2 11 Let the function x t be defined by x t ℒ s s 5 s 2 16 σ 0 x t can be written as the sum of three functions two of which are causal sinusoids a What is the third function b What is the cyclic frequency of the causal sinusoids Answers Impulse 0637 Hz 12 A system has a transfer function H s s s 1 s 2 s a which can be expanded in partial fractions in the form H s A B s 2 C s a If a 2 and B 3 2 find the values of a A and C Answers 6 1 105 13 Find the numerical values of the literal constants a 4 e 5t cos 25πt u t ℒ A s a s 2 bs c b 6 s 4 s a 2 s 4 b s a c A sin at B cos at u t ℒ 3 3s 4 s 2 9 d A e at sin bt B cos bt u t ℒ 35s 325 s 2 18s 85 e Aδt Bt C ut ℒ 3 s 1 s 2 s 2 f At B 1 e bt ut ℒ s 1 s 2 s 3 g 3 e 2t cos 8t 24 u t 3 ℒ A s a s 2 bs c e ds h 4 e at u t A e t2 u t ℒ 36 s 2 bs 3 σ 12 σ a Answers 9 6 65 3 6 9 4 5 10 61935 4 9 3 5 9 2 00074 2 4 68 3 7 2 13 49 3 rob28124ch08354405indd 398 041216 141 pm Exercises with Answers 399 Unilateral LaplaceTransform Integral 14 Starting with the definition of the unilateral Laplace transform ℒ gt G s 0 g t e st dt find the Laplace transforms of these signals a xt e t ut b xt e 2t cos 200πt u t c xt u t 4 d xt u t 4 Answers 1 s 1 σ 1 1 s σ 0 e 4s s σ 0 s 2 s 2 2 200π 2 σ 2 Solving Differential Equations 15 Using the differentiation properties of the unilateral Laplace transform write the Laplace transform of the differential equation x t 2 x t 4x t u t Answer s2 Xs sx 0 d dt x t t 0 2 sXs x 0 4Xs 1 s 16 Using the unilateral Laplace transform solve these differential equations for t 0 a x t 10xt ut x 0 1 b x t 2xt 4x t u t x 0 0 d dt x t t 0 4 c xt 2x t sin 2πt ut x 0 4 Answers 14 1 e t cos 3 t 17 3 e t sin 3 t ut 2π e 2t 2π cos 2πt 2 sin 2πt 4 2π 2 4 e 2t ut 1 9 e 10t 10 u t 17 Write the differential equations describing the systems in Figure E17 and find and graph the indicated responses a xt ut yt is the response y0 0 xt yt 4 b v0 10 vt is the response R 1 kΩ C 1 μF vt Figure E17 rob28124ch08354405indd 399 041216 141 pm C h a p t e r 8 The Laplace Transform 400 Answers t 0004 vt 10 t 1 yt 025 EXERCISES WITHOUT ANSWERS Region of Convergence 18 Find the regions of convergence of the Laplace transforms of the following functions a 3 e 2t ut b 10 e t2 ut c 10 e 2t ut d 5δ 3t 1 e 12 32e2t ut f 12 ut 32 e 2t ut g 5 e 2t ut 7et ut h δ t 2 e 2t ut Existence of the Laplace Transform 19 Graph the polezero plot and region of convergence if it exists for these signals a xt et ut e4t ut b xt e2t ut et ut Direct Form II System Realization 20 Draw Direct Form II system diagrams of the systems with these transfer functions a Hs 10 s2 8 s3 3s2 7s 22 b Hs 10 s 20 s 4 s 8 s 14 21 Fill in the blanks in the block diagram in Figure E21 with numbers for a system whose transfer function is Hs 7 s 2 s s 4 Figure E21 Xs Ys s1 s1 rob28124ch08354405indd 400 041216 141 pm 401 Exercises without Answers Forward and Inverse Laplace Transforms 22 Using a table of Laplace transforms and the properties find the Laplace transforms of the following functions a g t 5 sin 2π t 1 u t 1 b g t 5 sin 2πt u t 1 c g t 2 cos 10πt cos 100πt u t d g t d dt u t 2 e g t 0 t u τ dτ f g t d dt 5 e tτ 2 u t τ τ 0 g g t 2 e 5t cos 10πt u t h x t 5 sin πt π8 ut 23 Given gt ℒ s 1 s s 4 σ 0 find the Laplace transforms of a g2t b d dt gt c g t 4 d gt gt 24 Find the timedomain functions that are the inverse Laplace transforms of these functions Then using the initial and final value theorems verify that they agree with the timedomain functions a Gs 4s s 3 s 8 σ 3 b Gs 4 s 3 s 8 σ 3 c Gs s s 2 2s 2 σ 1 d Gs e2s s2 2s 2 σ 1 25 Given e4t ut ℒ Gs find the inverse Laplace transforms of a Gs3 σ 4 b Gs 2 Gs 2 σ 4 c Gss σ 4 26 Given the Laplacetransform pair gt ℒ 3ss 5 s 2s2 2s 8 and the fact that gt is continuous at t 0 complete the following Laplacetransform pairs a d dt g t ℒ b g t 3 ℒ 27 For each timedomain function in the column on the left find its Laplace transform in the column on the right It may or may not be there 1 e3t1 ut 1 A e1 s 3 2 e3t1 ut 1 B e2s s 3 rob28124ch08354405indd 401 041216 141 pm C h a p t e r 8 The Laplace Transform 402 3 et31 ut 1 C e3s s 3 4 e3t ut 1 D e4s s 13 5 et3 ut 1 E es13 s 13 6 e3t1 ut F es s 3 28 A causal signal x t has a Laplace transform X s s s 4 16 If y t 5x 3t Ys can be expressed in the form Y s As s 4 a 4 What are the numerical values of A and a 29 Find the numerical values of the literal constants a Aδt Beat ebt ut ℒ 3s2 s2 5s 4 b Aeat1 bt ut ℒ 5s s 1 2 c 2 sin10πt ut ℒ A s2 as b σ 0 d 7 cos 4t 1 u t 14 ℒ As s 2 as b e cs σ 0 e 7 e 3t sin 2πt u t ℒ A s 2 as b σ c f A rampat B Cect ut ℒ 12 s 1 s 2 s 3 σ 0 g Aδt B e bt C e ct ut ℒ 3s s 3 s 2 4s 3 h At B Cect ut ℒ 1 s2s 8 i A Bramp t t b C D e d t t d u t t e ℒ 10e3s s2s 1 j K 1 t e at K 2 e bt K 3 e ct ut ℒ 2 s 3 s 2 s 5 k A e at sin 3t ut ℒ 12 s 2 6s b l 3e2t cos 8t 24 u t 3 ℒ A s a s2 bs c eds 30 Find the numerical values of the constants K 0 K 1 K 2 p 1 and p 2 s2 3 3s2 s 9 K 0 K 1 s p 1 K 2 s p 2 31 Given that gt ut ℒ 8 s 2 s s 4 find the function gt 32 Given that gt ut 5 s 12 find the function gt rob28124ch08354405indd 402 041216 141 pm 403 Exercises without Answers Solution of Differential Equations 33 Write the differential equations describing the systems in Figure E33 and find and graph the indicated responses a xt ut yt is the response y0 5 d dt yt t0 10 xt yt 2 10 b is t ut vt is the response No initial energy storage vt it R2 1 kΩ R1 2 kΩ C2 1 μF C1 3 μF ist c is t cos2000πt vt is the response No initial energy storage vt it R2 1 kΩ R1 2 kΩ C2 1 μF C1 3 μF ist Figure E33 PoleZero Diagrams and Frequency Response 34 Draw polezero diagrams of these transfer functions a H s s 3 s 1 s s 2 s 6 b H s s s 2 s 1 c H s s s 10 s 2 11s 10 d H s 1 s 1 s 2 1618s 1 s 2 0618s 1 rob28124ch08354405indd 403 041216 141 pm C h a p t e r 8 The Laplace Transform 404 35 In Figure E35 are some polezero plots of transfer functions of systems of the general form H s A s z1 s zN s p1 s pD in which A 1 the zs are the zeros and the ps are the poles Answer the following questions 6 4 2 0 2 4 6 6 4 2 0 2 4 6 a b c d e f σ ω s 6 4 2 0 2 4 6 6 4 2 0 2 4 6 σ ω s 6 4 2 0 2 4 6 6 4 2 0 2 4 6 σ ω s 6 4 2 0 2 4 6 6 4 2 0 2 4 6 σ ω s 6 4 2 0 2 4 6 6 4 2 0 2 4 6 σ ω s 6 4 2 0 2 4 6 6 4 2 0 2 4 6 σ ω s Figure E35 a Which ones have a magnitude frequency response that is nonzero at ω 0 b Which ones have a magnitude frequency response that is nonzero as ω c There are two which have a bandpass frequency response zero at zero and zero at infinity Which one is more underdamped higher Q d Which one has a magnitude frequency response whose shape is closest to being a bandstop filter e Which ones have a magnitude frequency response that approaches K ω 6 at very high frequencies K is a constant f Which one has a magnitude frequency response that is constant g Which one has a magnitude frequency response whose shape is closest to an ideal lowpass filter h Which ones have a phase frequency response that is discontinuous at ω 0 rob28124ch08354405indd 404 041216 141 pm 405 Exercises without Answers 36 For each of the polezero plots in Figure E36 determine whether the frequency response is that of a practical lowpass bandpass highpass or bandstop filter ω σ s a ω σ s b ω σ s c ω σ s d 3 Figure E36 rob28124ch08354405indd 405 041216 141 pm 406 9 C H A P T E R 91 INTRODUCTION AND GOALS Every analysis method used in continuous time has a corresponding method in discrete time The counterpart to the Laplace transform is the z transform which expresses signals as linear combinations of discretetime complex exponentials Although the transform methods in discrete time are very similar to those used in continuous time there are a few important differences This material is important because in modern system designs digital signal processing is being used more and more An understanding of discretetime concepts is needed to grasp the analysis and design of systems that process both continuoustime and discretetime signals and convert back and forth between them with sampling and interpolation CHAPTER GOA L S The chapter goals in this chapter parallel those of Chapter 8 but as applied to discretetime signals and systems 1 To develop the z transform as a more general analysis technique for systems than the discretetime Fourier transform and as a natural result of the convolution process when a discretetime system is excited by its eigenfunction 2 To define the z transform and its inverse and to determine for what signals it exists 3 To define the transfer function of discretetime systems and learn a way of realizing a discretetime system directly from a transfer function 4 To build tables of z transform pairs and properties and learn how to use them with partialfraction expansion to find inverse z transforms 5 To define a unilateral z transform 6 To solve difference equations with initial conditions using the unilateral z transform 7 To relate the pole and zero locations of a transfer function of a system directly to the frequency response of the system 8 To learn how MATLAB represents the transfer functions of systems 9 To compare the usefulness and efficiency of different transform methods in some typical problems The z Transform rob28124ch09406445indd 406 041216 144 pm 92 Generalizing the DiscreteTime Fourier Transform 407 92 GENERALIZING THE DISCRETETIME FOURIER TRANSFORM The Laplace transform is a generalization of the continuoustime Fourier transform CTFT which allows consideration of signals and impulse responses that do not have a CTFT In Chapter 8 we saw how this generalization allowed analysis of signals and systems that could not be analyzed with the Fourier transform and also how it gives insight into system performance through analysis of the locations of the poles and zeros of the transfer function in the splane The z transform is a generalization of the discretetime Fourier transform DTFT with similar advantages The z transform is to discretetime signal and system analysis what the Laplace transform is to continu oustime signal and system analysis There are two approaches to deriving the z transform that are analogous to the two approaches taken to the derivation of the Laplace transform generalizing the DTFT and exploiting the unique properties of complex exponentials as the eigenfunctions of LTI systems The DTFT is defined by xn 1 2π 2π X e jΩ e jΩn d Ω ℱ X e jΩ n xn e jΩn or xn 1 XF e j2πFn dF ℱ XF n xn e j2πFn The Laplace transform generalizes the CTFT by changing complex sinusoids of the form ejωt where ω is a real variable to complex exponentials of the form est where s is a complex variable The independent variable in the DTFT is discretetime radian frequency Ω The exponential function e jΩn appears in both the forward and inverse transforms as ejΩn 1e jΩn in the forward transform For real Ω e jΩn is a discrete time complex sinusoid and has a magnitude of one for any value of discrete time n which is real By analogy with the Laplace transform we could generalize the DTFT by replacing the real variable Ω with a complex variable S and thereby replace e jΩn with eSn a complex exponential For complex S eS can lie anywhere in the complex plane We can simplify the notation by letting z eS and expressing discretetime signals as linear combinations of zn instead of eSn Replacing e jΩn with zn in the DTFT leads directly to the conventional definition of a forward z transform Xz n xn z n 91 and xn and Xz are said to form a ztransform pair xn 𝒵 Xz The fact that z can range anywhere in the complex plane means that we can use discretetime complex exponentials instead of just discretetime complex sinusoids in representing a discretetime signal Some signals cannot be represented by linear combinations of complex sinusoids but can be represented by a linear combination of complex exponentials rob28124ch09406445indd 407 041216 144 pm C h a p t e r 9 The z Transform 408 93 COMPLEX EXPONENTIAL EXCITATION AND RESPONSE Let the excitation of a discretetime LTI system be a complex exponential of the form K zn where z is in general complex and K is any constant Using convolution the response yn of an LTI system with impulse response hn to a complex exponential excitation xn K zn is yn hn K z n K m hm z nm K z n xn m hm z m So the response to a complex exponential is that same complex exponential multiplied by m hm z m if the series converges This is identical to 91 94 THE TRANSFER FUNCTION If an LTI system with impulse response hn is excited by a signal xn the z transform Yz of the response yn is Yz n yn z n n hn xn z n n m hm xn m z n Separating the two summations Yz m hm n xn m z n Let q n m Then Yz m hm q xq z qm m hm z m Hz q xq z q Xz So in a manner similar to the Laplace transform Yz Hz Xz and Hz is called the transfer function of the discretetime system just as first introduced in Chapter 5 95 CASCADECONNECTED SYSTEMS The transfer functions of components in the cascade connection of discretetime sys tems combine in the same way they do in continuoustime systems Figure 91 The overall transfer function of two systems in cascade connection is the product of their individual transfer functions Figure 91 Cascade connection of systems H1z H2z XzH1z Xz Xz YzXzH1zH2z Yz H1zH2z rob28124ch09406445indd 408 041216 144 pm 96 Direct Form II System Realization 409 96 DIRECT FORM II SYSTEM REALIZATION In engineering practice the most common form of description of a discretetime system is a difference equation or a system of difference equations We showed in Chapter 5 that for a discretetime system described by a difference equation of the form k0 N a k yn k k0 M b k xn k 92 the transfer function is Hz k0 M b k z k k0 N a k z k b 0 b 1 z 1 b 2 z 2 b M z M a 0 a 1 z 1 a 2 z 2 a N z N 93 or alternately Hz k0 M b k z k k0 N a k z k z NM b 0 z M b 1 z M1 b M1 z b M a 0 z N a 1 z N1 a N1 z a N 94 The Direct Form II realization of discretetime systems is directly analogous to Direct Form II in continuous time The transfer function Hz Yz Xz b 0 b 1 z 1 b N z N a 0 a 1 z 1 a N z N b 0 z N b 1 z N1 b N a 0 z N a 1 z N1 a N can be separated into the cascade of two subsystem transfer functions H 1 z Y 1 z Xz 1 a 0 z N a 1 z N1 a N 95 and H 2 z Yz Y 1 z b 0 z N b 1 z N1 b N Here the order of the numerator and denominator are both indicated as N If the order of the numerator is actually less than N some of the bs will be zero But a 0 must not be zero From 95 z N Y 1 z 1 a 0 Xz a 1 z N1 Y 1 z a N Y 1 z Figure 92 All the terms of the form z k Y 1 z that are needed to form H 2 z are available in the realization of H 1 z Combining them in a linear combination using the b coefficients we get the Direct Form II realization of the overall system Figure 93 rob28124ch09406445indd 409 041216 144 pm C h a p t e r 9 The z Transform 410 Figure 93 Overall Direct Form II canonical system realization b1 b0 b2 bN1 bN a1 a2 aN1 aN Xz Yz 1a0 z1 z1 z1 Figure 94 A timelimited discretetime signal n xn 97 THE INVERSE z TRANSFORM The conversion from Hz to hn is the inverse z transform and can be done by the direct formula hn 1 j2π C Hz z n1 dz This is a contour integral around a circle in the complex z plane and is beyond the scope of this text Most practical inverse z transforms are done using a table of ztransform pairs and the properties of the z transform 98 EXISTENCE OF THE z TRANSFORM TIMELIMITED SIGNALS The conditions for existence of the z transform are analogous to the conditions for ex istence of the Laplace transform If a discretetime signal is time limited and bounded the ztransform summation is finite and its z transform exists for any finite nonzero value of z Figure 94 Figure 92 Direct Form II canonical realization of H 1 z Xz zNY1z a1 1a0 a2 aN1 aN zN1Y1z z1 z1 z1 zN2Y1z zY1z Y1z rob28124ch09406445indd 410 041216 144 pm 411 98 Existence of the z Transform An impulse δn is a very simple bounded timelimited signal and its z transform is n δn z n 1 This z transform has no zeros or poles For any nonzero value of z the transform of this impulse exists If we shift the impulse in time in either direction we get a slightly different result δn 1 𝒵 z 1 pole at zero δn 1 𝒵 z pole at infinity So the z transform of δn 1 exists for every nonzero value of z and the z transform of δn 1 exists for every finite value of z RIGHT AND LEFTSIDED SIGNALS A rightsided signal x r n is one for which x r n 0 for any n n 0 and a leftsided signal x l n is one for which x l n 0 for any n n 0 Figure 95 Figure 96 a xn α n un n 0 α ℂ b xn β n u n 0 n β ℂ n0 n0 n n xn xn a b Figure 95 a Rightsided discretetime signal b leftsided discretetime signal n xrn n xln a b Consider the rightsided signal xn α n un n 0 α ℂ Figure 96a Its z transform is Xz n α n un n 0 z n n n 0 α z 1 n rob28124ch09406445indd 411 041216 144 pm C h a p t e r 9 The z Transform 412 Figure 98 Some noncausal signals and their ROCs if they exist n 12 12 10 xn 12n un3n un1 xn 085ncos2πn6un09ncos2πn6un1 xn 11ncos2πn6un105ncos2πn6un1 xn 095nun09nun1 n 12 12 4 n 12 12 4 4 n 12 12 4 4 ROC is 12 z 3 No ROC No ROC ROC is 085 z 09 if the series converges and the series converges if αz 1 or z α This region of the z plane is called the region of convergence ROC Figure 97a If xn 0 for n n 0 it is called a leftsided signal Figure 96b If xn β n u n 0 n β ℂ Xz n n 0 β n z n n n 0 β z 1 n n n 0 β 1 z n and the summation converges for β 1 z 1 or z β Figure 97b Figure 97 Region of convergence for a the rightsided signal xn α n un n 0 α ℂ and b the leftsided signal xn β n u n 0 n β ℂ z ROC ROC z β α a b rob28124ch09406445indd 412 041216 144 pm 413 99 zTransform Pairs Figure 99 a xn K α n α 1 b xn K α n α 1 xn xn n n Just as in continuous time any discretetime signal can be expressed as a sum of a rightsided signal and a leftsided signal If xn x r n x l n and if x r n K r α n and x l n K l β n where K r and K l are constants then the summation converges and the z transform exists for α z β This implies that if α β a z transform can be found and the ROC in the z plane is the region α z β If α β the z transform does not exist Figure 98 ExamplE 91 z transform of a noncausal signal Find the z transform of xn K α n α ℝ Its variation with n depends on the value of α Figure 99 It can be written as xn K α n un α n un 1 If α 1 then α α 1 no ROC can be found and it does not have a z transform If α 1 then α α 1 the ROC is α z α 1 and the z transform is K α n 𝒵 K n α n z n K n0 α z 1 n n 0 α 1 z 1 n 1 α z α 1 K α n 𝒵 K n0 α z 1 n n0 αz n 1 α z α 1 This consists of two summations plus a constant Each summation is a geometric series of the form n0 r n and the series converges to 11 r if r 1 K α n 𝒵 K 1 1 α z 1 1 1 αz 1 K z z α z z α 1 α z α 1 99 zTRANSFORM PAIRS We can start a useful table of z transforms with the impulse δn and the damped cosine α n cos Ω 0 n un As we have already seen δn 𝒵 1 The z transform of the damped cosine is α n cos Ω 0 n un 𝒵 n α n cos Ω 0 n un z n rob28124ch09406445indd 413 041216 144 pm C h a p t e r 9 The z Transform 414 α n cos Ω 0 n un 𝒵 n0 α n e j Ω 0 n e j Ω 0 n 2 z n α n cos Ω 0 n un 𝒵 12 n0 α e j Ω 0 z 1 n α e j Ω 0 z 1 n For convergence of this z transform z α and α n cos Ω 0 n un 𝒵 12 1 1 α e j Ω 0 z 1 1 1 α e j Ω 0 z 1 z α This can be simplified to either of the two alternate forms α n cos Ω 0 n un 𝒵 1 α cos Ω 0 z 1 1 2α cos Ω 0 z 1 α 2 z 2 z α or α n cos Ω 0 n un 𝒵 zz α cos Ω 0 z 2 2α cos Ω 0 z α 2 z α If α 1 then cos Ω 0 n un 𝒵 zz cos Ω 0 z 2 2 cos Ω 0 z 1 1 cos Ω 0 z 1 1 2 cos Ω 0 z 1 z 2 z 1 If Ω 0 0 then α n un 𝒵 z z α 1 1 α z 1 z α If α 1 and Ω 0 0 then un 𝒵 z z 1 1 1 z 1 z 1 rob28124ch09406445indd 414 041216 144 pm 415 99 zTransform Pairs ExamplE 92 Inverse z transforms Find the inverse z transforms of a Xz z z 05 z z 2 05 z 2 b Xz z z 05 z z 2 z 2 c Xz z z 05 z z 2 z 05 a Rightsided signals have ROCs that are outside a circle and leftsided signals have ROCs that are inside a circle Therefore using α n un 𝒵 z z α 1 1 α z 1 z α Table 91 shows the ztransform pairs for several commonly used functions Table 91 Some ztransform pairs δn 𝒵 1 All z un 𝒵 z z 1 1 1 z 1 z 1 un 1 𝒵 z z 1 z 1 α n un 𝒵 z z α 1 1 α z 1 z α α n un 1 𝒵 z z α 1 1 α z 1 z α n un 𝒵 z z 1 2 z 1 1 z 1 2 z 1 nun 1 𝒵 z z 1 2 z 1 1 z 1 2 z 1 n 2 un 𝒵 z z 1 z 1 3 1 z 1 z 1 z 1 3 z 1 n 2 un 1 𝒵 zz 1 z 1 3 1 z 1 z 1 z 1 3 z 1 n α n un 𝒵 αz z α 2 α z 1 1 α z 1 2 z α n α n un 1 𝒵 αz z α 2 α z 1 1 α z 1 2 z α sin Ω 0 n un 𝒵 z sin Ω 0 z 2 2z cos Ω 0 1 z 1 sin Ω 0 n un 1 𝒵 z sin Ω 0 z 2 2z cos Ω 0 1 z 1 cos Ω 0 n un 𝒵 zz cos Ω 0 z 2 2z cos Ω 0 1 z 1 cos Ω 0 n un 1 𝒵 zz cos Ω 0 z 2 2z cos Ω 0 1 z 1 α n sin Ω 0 n un 𝒵 zα sin Ω 0 z 2 2αz cos Ω 0 α 2 z α α n sin Ω 0 n un 1 𝒵 zα sin Ω 0 z 2 2αz cos Ω 0 α 2 z α α n cos Ω 0 n un 𝒵 zz α cos Ω 0 z 2 2αz cos Ω 0 α 2 z α α n cos Ω 0 n un 1 𝒵 z z α cos Ω 0 z 2 2αz cos Ω 0 α 2 z α α n 𝒵 z z α z z α 1 α z α 1 un n 0 un n 1 𝒵 z z 1 z n 0 z n 1 z n 1 n 0 1 z n 1 n 0 2 z 1 z n 1 1 z 0 rob28124ch09406445indd 415 041216 144 pm C h a p t e r 9 The z Transform 416 and α n un 1 𝒵 z z α 1 1 α z 1 z α we get 05 n un 2 n un 1 𝒵 Xz z z 05 z z 2 05 z 2 or 05 n un 2 n un 1 𝒵 Xz z z 05 z z 2 05 z 2 b In this case both signals are right sided 05 n 2 n un 𝒵 Xz z z 05 z z 2 z 2 c In this case both signals are left sided 05 n 2 n u n 1 𝒵 Xz z z 05 z z 2 z 05 910 zTRANSFORM PROPERTIES Given the ztransform pairs gn 𝒵 Gz and hn 𝒵 Hz with ROCs of ROCG and ROCH respectively the properties of the z transform are listed in Table 92 Table 92 ztransform properties Linearity α gn β hn 𝒵 α Gz β Hz ROC ROC G ROC H Time Shifting gn n 0 𝒵 z n 0 Gz ROC ROC G except perhaps z 0 or z Change of Scale in z α n gn 𝒵 Gzα ROC α ROC G Time Reversal gn 𝒵 G z 1 ROC 1 ROC G Time Expansion gnk nk an integer 0 otherwise 𝒵 G z k ROC ROC G 1k Conjugation g n 𝒵 G z ROC ROC G zDomain Differentiation n gn 𝒵 z d dz Gz ROC ROC G Convolution gn hn 𝒵 HzGz First Backward Difference gn gn 1 𝒵 1 z 1 Gz ROC ROC G z 0 Accumulation m n gm 𝒵 z z 1 Gz ROC ROC G z 1 Initial Value Theorem If gn 0 n 0 then g0 lim z Gz Final Value Theorem If gn 0 n 0 lim n gn lim z1 z 1Gz if lim n gn exists rob28124ch09406445indd 416 041216 144 pm 417 911 Inverse zTransform Methods 911 INVERSE zTRANSFORM METHODS SYNTHETIC DIVISION For rational functions of z of the form Hz b M z M b M1 z M1 b 1 z b 0 a N z N a N1 z N1 a 1 z a 0 we can always synthetically divide the numerator by the denominator and get a sequence of powers of z For example if the function is Hz z 12 z 07 z 04 z 02 z 08 z 05 z 08 or Hz z 3 01 z 2 104z 0336 z 3 05 z 2 034z 008 z 08 the synthetic division process produces z 3 05 z 2 034z 008 z 3 01 z 2 104z 0336 z 3 05 z 2 034z 008 04 z 2 07z 0256 04 z 2 02z 0136 0032 z 1 05z 012 0032 z 1 1 04 z 1 05 z 2 Then the inverse z transform is hn δn 04δn 1 05δn 2 There is an alternate form of synthetic division 008 034z 05 z 2 z 3 0336 104z 01 z 2 z 3 0336 1428z 21 z 2 42 z 3 2468z 22 z 2 52 z 3 2468z 10489 z 2 15425 z 3 3085 z 4 12689 z 2 10225 z 3 3085 z 4 42 3085z 158613 z 2 From this result we might conclude that the inverse z transform would be 42δn 3085δn 1 158613δn 2 It is natural at this point to wonder why these two results are different and which one is correct The key to knowing which one is correct is the ROC z 08 This implies a rightsided inverse transform and the first synthetic division result is of that form That series converges for z 08 The second series converges for z 02 and would be the correct answer if the ROC were z 02 rob28124ch09406445indd 417 041216 144 pm C h a p t e r 9 The z Transform 418 Synthetic division always works for a rational function but the answer is usu ally in the form of an infinite series In most practical analyses a closed form is more useful PARTIALFRACTION EXPANSION The technique of partialfraction expansion to find the inverse z transform is alge braically identical to the method used to find inverse Laplace transforms with the variable s replaced by the variable z But there is a situation in inverse z transforms that deserves mention It is very common to have zdomain functions in which the number of finite zeros equals the number of finite poles making the expression improper in z with at least one zero at z 0 Hz z NM z z 1 z z 2 z z M z p 1 z p 2 z p N N M We cannot immediately expand Hz in partial fractions because it is an improper rational function of z In a case like this it is convenient to divide both sides of the equation by z Hz z z NM1 z z 1 z z 2 z z M z p 1 z p 2 z p N Hzz is a proper fraction in z and can be expanded in partial fractions Hz z K 1 z p 1 K 2 z p 2 K N z p N Then both sides can be multiplied by z and the inverse transform can be found Hz z K 1 z p 1 z K 2 z p 2 z K N z p N hn K 1 p 1 n un K 2 p 2 n un K N p N n un Just as we did in finding inverse Laplace transforms we could have solved this problem using synthetic division to obtain a proper remainder But this new technique is often simpler EXAMPLES OF FORWARD AND INVERSE z TRANSFORMS The timeshifting property is very important in converting zdomain transferfunction expressions into actual systems and other than the linearity property is probably the most oftenused property of the z transform ExamplE 93 System block diagram from a transfer function using the timeshifting property A system has a transfer function Hz Yz Xz z 12 z 2 z 29 z 23 Draw a system block diagram using delays amplifiers and summing junctions rob28124ch09406445indd 418 041216 144 pm 419 911 Inverse zTransform Methods We can rearrange the transferfunction equation into Yz z 2 z 29 Xz z 12 or z 2 Yz zXz 12 Xz zYz 29 Yz Multiplying this equation through by z 2 we get Yz z 1 Xz 12 z 2 Xz z 1 Yz 29 z 2 Yz Now using the timeshifting property if xn 𝒵 Xz and yn 𝒵 Yz then the inverse z transform is yn xn 1 12xn 2 yn 1 29yn 2 This is called a recursion relationship between xn and yn expressing the value of yn at discrete time n as a linear combination of the values of both xn and yn at discrete times n n 1 n 2 From it we can directly synthesize a block diagram of the system Figure 910 This system realization uses four delays two amplifiers and two summing junctions This block diagram was drawn in a natural way by directly implementing the recursion relation in the diagram Realized in Direct Form II the realization uses two delays three amplifiers and three sum ming junctions Figure 911 There are multiple other ways of realizing the system see Chapter 13 A special case of the changeofscaleinz property α n gn 𝒵 Gz α is of particular interest Let the constant α be e j Ω 0 with Ω 0 real Then e j Ω 0 n gn 𝒵 Gz e j Ω 0 Every value of z is changed to z e j Ω 0 This accomplishes a counterclockwise rotation of the transform Gz in the z plane by the angle Ω 0 because e j Ω 0 has a magnitude of one and a phase of Ω 0 This effect is a little hard to see in the abstract An example will illustrate it better Let Gz z 1 z 08 e jπ4 z 08 e jπ4 Figure 910 Timedomain system block diagram for the transfer function Hz z 12 z 2 z 29 xn yn D D D D 12 29 29yn2 12xn2 xn1 yn1 yn2 xn2 Figure 911 Direct Form II realization of Hz z 12 z 2 z 29 xn D D yn 05 1 022222 rob28124ch09406445indd 419 041216 144 pm C h a p t e r 9 The z Transform 420 and let Ω 0 π8 Then Gz e j Ω 0 Gz e jπ8 z e jπ8 1 z e jπ8 08 e jπ4 z e jπ8 08 e jπ4 or Gz e jπ8 e jπ8 z e jπ8 e jπ8 z 08 e jπ8 e jπ8 z 08 e j3π8 e jπ8 z e jπ8 z 08 e jπ8 z 08 e j3π8 The original function has finite poles at z 08 e jπ4 and a zero at z 1 The trans formed function Gz e jπ8 has finite poles at z 08 e jπ8 and 08 e j3π8 and a zero at z e jπ8 So the finite pole and zero locations have been rotated counterclockwise by π8 radians Figure 912 Figure 912 Illustration of the frequencyscaling property of the z transform for the special case of a scaling by e j Ω 0 z PoleZero Plot of Gz z Ω0 PoleZero Plot of Gze jΩ0 A multiplication by a complex sinusoid of the form e j Ω 0 n in the time domain corresponds to a rotation of its z transform ExamplE 94 z transforms of a causal exponential and a causal exponentially damped sinusoid Find the z transforms of xn e n40 un and x m n e n40 sin2πn8un and draw polezero diagrams for Xz and X m z Using α n un 𝒵 z z α 1 1 α z 1 z α we get e n40 un 𝒵 z z e 140 z e 140 Therefore Xz z z e 140 z e 140 rob28124ch09406445indd 420 041216 144 pm 421 911 Inverse zTransform Methods We can rewrite x m n as x m n e n40 e j2πn8 e j2πn8 j2 un or x m n j 2 e n40 e j2πn8 e n40 e j2πn8 un Then starting with e n40 un 𝒵 z z e 140 z e 140 and using the changeofscale property α n gn 𝒵 Gzα we get e j2πn8 e n40 un 𝒵 z e j2π8 z e j2π8 e 140 z e 140 and e j2πn8 e n40 un 𝒵 z e j2π8 z e j2π8 e 140 z e 140 and j 2 e n40 e j2πn8 e n40 e j2πn8 un 𝒵 j 2 z e j2π8 z e j2π8 e 140 z e j2π8 z e j2π8 e 140 z e 140 or X m z j 2 z e j2π8 z e j2π8 e 140 z e j2π8 z e j2π8 e 140 z e 140 sin 2π8 z 2 2z e 140 cos 2π8 e 120 z e 140 or X m z 06896z z 2 13793z 09512 06896z z 06896 j06896 z 06896 j06896 z e 140 Figure 913 z PoleZero Plot of Xz PoleZero Plot of Xmz 09753 09753 Unit Circle z Unit Circle π 4 Figure 913 Polezero plots of Xz and X m z rob28124ch09406445indd 421 041216 144 pm C h a p t e r 9 The z Transform 422 ExamplE 95 z transform using the differentiation property Using the zdomain differentiation property show that the z transform of n un is z z 1 2 z 1 Start with un 𝒵 z z 1 z 1 Then using the zdomain differentiation property n un 𝒵 z d dz z z 1 z z 1 2 z 1 or n un 𝒵 z z 1 2 z 1 ExamplE 96 z transform using the accumulation property Using the accumulation property show that the z transform of n un is z z 1 2 z 1 First express n un as an accumulation n un m0 n um 1 Then using the timeshifting property find the z transform of un 1 un 1 𝒵 z 1 z z 1 1 z 1 z 1 Then applying the accumulation property n un m0 n um 1 𝒵 z z 1 1 z 1 z z 1 2 z 1 As was true for the Laplace transform the final value theorem applies if the limit lim n gn exists The limit lim z1 z 1Gz may exist even if the limit lim n gn does not For example if Xz z z 2 z 2 then lim z1 z 1Xz lim z1 z 1 z z 2 0 But xn 2 n un and the limit lim n xn does not exist Therefore the conclusion that the final value is zero is wrong In a manner similar to the analogous proof for Laplace transforms the following can be shown For the final value theorem to apply to a function Gz all the finite poles of the function z 1Gz must lie in the open interior of the unit circle of the z plane rob28124ch09406445indd 422 041216 144 pm 423 912 The Unilateral z Transform ExamplE 97 z transform of an anticausal signal Find the z transform of xn 4 03 n un Using α n un 1 𝒵 z z α 1 1 α z 1 z α Identify α as 0 3 1 Then 0 3 1 n un 1 𝒵 z z 0 3 1 z 0 3 1 103 n u n 1 𝒵 z z 103 z 103 Use the timeshifting property 103 n1 u n 1 1 𝒵 z 1 z z 103 1 z 103 z 103 310 103 n un 𝒵 1 z 103 z 103 310 103 n un 𝒵 1 z 103 z 103 Using the linearity property multiply both sides by 4310 or 403 4 03 n un 𝒵 403 z 103 40 3z 10 z 103 912 THE UNILATERAL z TRANSFORM The unilateral Laplace transform proved convenient for continuoustime functions and the unilateral z transform is convenient for discretetime functions for the same rea sons We can define a unilateral z transform which is only valid for functions that are zero before discrete time n 0 and avoid in most practical problems any complicated consideration of the region of convergence The unilateral z transform is defined by Xz n0 xn z n 96 The region of convergence of the unilateral z transform is always the open exterior of a cir cle centered at the origin of the z plane whose radius is the largest finite pole magnitude PROPERTIES UNIQUE TO THE UNILATERAL z TRANSFORM The properties of the unilateral z transform are very similar to the properties of the bilateral z transform The timeshifting property is a little different Let gn 0 n 0 Then for the unilateral z transform gn n 0 𝒵 z n 0 Gz n 0 0 z n 0 Gz m0 n 0 1 gm z m n 0 0 rob28124ch09406445indd 423 041216 144 pm C h a p t e r 9 The z Transform 424 This property must be different for shifts to the left because when a causal function is shifted to the left some nonzero values may no longer lie in the summation range of the unilateral z transform which begins at n 0 The extra terms m0 n 0 1 gm z m account for any function values that are shifted into the n 0 range The accumulation property for the unilateral z transform is m0 n gm 𝒵 z z 1 Gz Only the lower summation limit has changed Actually the bilateral form m n gm 𝒵 z z 1 Gz would still work because for a causal signal gn m n gm m0 n gm The unilateral z transform of any causal signal is exactly the same as the bilateral z transform of that signal So the table of bilateral z transforms can be used as a table of unilateral z transforms SOLUTION OF DIFFERENCE EQUATIONS One way of looking at the z transform is that it bears a relationship to difference equa tions analogous to the relationship of the Laplace transform to differential equations A linear difference equation with initial conditions can be converted by the z transform into an algebraic equation Then it is solved and the solution in the time domain is found by an inverse z transform ExamplE 98 Solution of a difference equation with initial conditions using the z transform Solve the difference equation yn 2 32yn 1 12yn 14 n for n 0 with initial conditions y0 10 and y1 4 Initial conditions for a secondorder differential equation usually consist of a specification of the initial value of the function and its first derivative Initial conditions for a secondorder difference equation usually consist of the specification of the initial two values of the function in this case y0 and y1 Taking the z transform of both sides of the difference equation using the timeshifting property of the z transform z 2 Yz y0 z 1 y1 32zYz y0 12Yz z z 14 Solving for Yz Yz z z 14 z 2 y0 z y1 32z y0 z 2 32z 12 rob28124ch09406445indd 424 041216 144 pm 913 PoleZero Diagrams and Frequency Response 425 Yz z z 2 y0 z7y04 y1 y14 3 y08 1 z 14 z 2 32z 12 Substituting in the numerical values of the initial conditions Yz z 10 z 2 272z 154 z 14z 12z 1 Dividing both sides by z Yz z 10 z 2 272z 154 z 14z 12z 1 This is a proper fraction in z and can therefore be expanded in partial fractions as Yz z 163 z 14 4 z 12 23 z 1 Yz 16z3 z 14 4z z 12 2z3 z 1 Then using α n un 𝒵 z z α and taking the inverse z transform y n 5333 025 n 4 05 n 0667un Evaluating this expression for n 0 and n 1 yields y0 5333 025 0 4 05 0 0667 10 y1 5333 025 1 4 05 1 0667 1333 2 0667 4 which agree with the initial conditions Substituting the solution into the difference equation 5333 025 n2 4 05 n2 0667 155333 025 n1 4 05 n1 0667 055333 025 n 4 05 n 0667 025 n for n 0 or 0333 025 n 05 n 0667 2 025 n 3 05 n 1 2667 025 n 2 05 n 0333 025 n for n 0 or 025 n 025 n for n 0 which proves that the solution does indeed solve the difference equation 913 POLEZERO DIAGRAMS AND FREQUENCY RESPONSE To examine the frequency response of discretetime systems we can specialize the z transform to the DTFT through the transformation z e jΩ with Ω being a real variable representing discretetime radian frequency The fact that Ω is real means that in determining frequency response the only values of z that we are now considering are those on the unit circle in the z plane because e jΩ 1 for any real Ω This is directly analogous to determining the frequency response of a continuous time system by examining the behavior of its sdomain transfer function as s moves along the ω axis in the splane and a similar graphical technique can be used Suppose the transfer function of a system is Hz z z 2 z2 516 z z p 1 z p 2 rob28124ch09406445indd 425 041216 144 pm C h a p t e r 9 The z Transform 426 where p 1 1 j2 4 and p 2 1 j2 4 The transfer function has a zero at zero and two complexconjugate finite poles Figure 914 The frequency response of the system at any particular radian frequency Ω 0 is determined to within a multiplicative constant by the vectors from the finite poles and finite zeros of the transfer function to the zplane point z 0 e j Ω 0 The magnitude of the frequency response is the product of the magnitudes of the zero vectors divided by the product of the magnitudes of the pole vectors In this case H e jΩ e jΩ e jΩ p 1 e jΩ p 2 97 It is apparent that as e jΩ approaches a pole p 1 for example the magnitude of the difference e jΩ p 1 becomes small making the magnitude of the denominator small and tending to make the magnitude of the transfer function larger The opposite effect occurs when e jΩ approaches a zero The phase of the frequency response is the sum of the angles of the zero vectors minus the sum of the angles of the pole vectors In this case H e jΩ e jΩ e jΩ p 1 e jΩ p 2 Figure 915 The maximum magnitude frequency response occurs at approximately z e j111 which are the points on the unit circle at the same angle as the finite poles of the transfer function and therefore the points on the unit circle at which the denominator factors e j Ω 0 p 1 and e j Ω 0 p 2 in 97 reach their minimum magnitudes An important difference between the frequency response of continuoustime and discretetime systems is that for discretetime systems the frequency response is always periodic with period 2π in Ω That difference can be seen directly in this graphical technique because as Ω moves from zero in a positive direction it traverses the entire unit circle in a counterclockwise direction and then on its second traversal of the unit circle retraces its previous positions repeating the same frequency responses found on the first traversal z z0ejΩ0 z0p1 z0 z0p2 Figure 914 zdomain polezero diagram of a system transfer function rob28124ch09406445indd 426 041216 144 pm 913 PoleZero Diagrams and Frequency Response 427 ExamplE 99 Polezero plot and frequency response from a transfer function 1 Draw the polezero plot and graph the frequency response for the system whose transfer function is Hz z 2 096z 09028 z 2 156z 08109 The transfer function can be factored into Hz z 048 j082z 048 j082 z 078 j045z 078 j045 The polezero diagram is in Figure 916 The magnitude and phase frequency responses of the system are illustrated in Figure 917 Figure 916 Polezero diagram of the transfer function Hz z 2 096z 09028 z 2 156z 08109 Rez Imz z Figure 917 Magnitude and phase frequency response of the system whose transfer function is Hz z 2 096z 09028 z 2 156z 08109 Ω Ω 2π 2π 2π 2π He jΩ 10 π π He jΩ Figure 915 Magnitude and phase frequency response of the system whose transfer function is Hz z z 2 z2 516 HejΩ 2 Ω Ω 2π 2π 2π 2π π π Closest Approach to a Pole Closest Approach to a Pole HejΩ rob28124ch09406445indd 427 041216 144 pm C h a p t e r 9 The z Transform 428 ExamplE 910 Polezero plot and frequency response from a transfer function 2 Draw the polezero plot and graph the frequency response for the system whose transfer function is Hz 00686 z 2 1087z 03132 z 2 1315z 06182 This transfer function can be factored into Hz 00686 z 05435 j01333 z 05435 j01333 z 06575 j04312 z 06575 j04312 The polezero diagram is illustrated in Figure 918 The magnitude and phase frequency responses of the system are illustrated in Figure 919 914 MATLAB SYSTEM OBJECTS Discretetime system objects can be created and used in much the same way as continuoustime system objects The syntax for creating a system object with tf is almost the same sys tfnumdenTs but with the extra argument Ts the time between samples on the assumption that the discretetime signals were created by sampling continuoustime signals For example let the transfer function be H 1 z z 2 z 08 z 03 z 2 14z 02 z 3 08 z 2 z 3 11 z 2 022z 006 Figure 918 Polezero diagram for the transfer function Hz 00686 z 2 1087z 03132 z 2 1315z 06182 Rez Imz z Ω HejΩ 1 Ω 2π 2π 2π 2π π π HejΩ Figure 919 Magnitude and phase frequency response of the system whose transfer function is Hz 00686 z 2 1087z 03132 z 2 1315z 06182 rob28124ch09406445indd 428 041216 144 pm 914 MATLAB System Objects 429 IN MATLAB num 1 08 0 0 den 1 11 022 006 Ts 0008 H1 tfnumdenTs H1 Transfer function z3 08 z2 z3 11 z2 022 z 006 Sampling time 0008 We can also use zpk z 04 p 07 06 k 3 H2 zpkzpkTs H2 Zeropolegain 3 z04 z07 z06 Sampling time 0008 We can also define z as the independent variable of the z transform with the command z tfzTs H3 7zz202z08 H3 Transfer function 7 z z2 02 z 08 Sampling time 0008 We are not required to specify the sampling time z tfz H3 7zz202z08 H3 Transfer function 7 z z2 02 z 08 Sampling time unspecified The command H freqznumdenW rob28124ch09406445indd 429 041216 144 pm C h a p t e r 9 The z Transform 430 accepts the two vectors num and den and interprets them as the coefficients of the powers of z in the numerator and denominator of the transfer function Hz It returns in H the complex frequency response at the discretetime radian frequencies in the vector W 915 TRANSFORM METHOD COMPARISONS Each type of transform has a niche in signal and system analysis where it is particularly convenient If we want to find the total response of a discretetime system to a causal or anticausal excitation we would probably use the z transform If we are interested in the frequency response of a system the DTFT is convenient If we want to find the forced response of a system to a periodic excitation we might use the DTFT or the discrete Fourier transform depending on the type of analysis needed and the form in which the excitation is known analytically or numerically ExamplE 911 Total system response using the z transform and the DTFT A system with transfer function Hz z z 03z 08 z 08 is excited by a unit sequence Find the total response The z transform of the response is Yz Hz Xz z z 03 z 08 z z 1 z 1 Expanding in partial fractions Yz z 2 z 03z 08z 1 01169 z 03 03232 z 08 07937 z 1 z 1 Therefore the total response is yn 01169 03 n1 03232 08 n1 07937 un 1 This problem can also be analyzed using the DTFT but the notation is significantly clumsier mainly because the DTFT of a unit sequence is 1 1 e jΩ π δ 2π Ω The system frequency response is H e jΩ e jΩ e jΩ 03 e jΩ 08 The DTFT of the system response is Y e jΩ H e jΩ X e jΩ e jΩ e jΩ 03 e jΩ 08 1 1 e jΩ π δ 2π Ω or Y e jΩ e j2Ω e jΩ 03 e jΩ 08 e jΩ 1 π e jΩ e jΩ 03 e jΩ 08 δ 2π Ω Expanding in partial fractions Y e jΩ 01169 e jΩ 03 03232 e jΩ 08 07937 e jΩ 1 π 1 031 08 δ 2π Ω rob28124ch09406445indd 430 041216 144 pm 915 Transform Method Comparisons 431 Using the equivalence property of the impulse and the periodicity of both δ 2π Ω and e jΩ Y e jΩ 01169 e jΩ 1 03 e jΩ 03232 e jΩ 1 08 e jΩ 07937 e jΩ 1 e jΩ 24933 δ 2π Ω Then manipulating this expression into a form for which the inverse DTFT is direct Y e jΩ 01169 e jΩ 1 03 e jΩ 03232 e jΩ 1 08 e jΩ 07937 e jΩ 1 e jΩ π δ 2π Ω 07937π δ 2π Ω 24933 δ 2π Ω 0 Y e jΩ 01169 e jΩ 1 03 e jΩ 03232 e jΩ 1 08 e jΩ 07937 e jΩ 1 e jΩ π δ 2π Ω And finally taking the inverse DTFT yn 01169 03 n1 03232 08 n1 07937 un 1 The result is the same but the effort and the probability of error are considerably greater ExamplE 912 System response to a sinusoid A system with transfer function Hz z z 09 z 09 is excited by the sinusoid xn cos2πn12 Find the response The excitation is the pure sinusoid xn cos2πn12 not the causal sinusoid xn cos2πn12 un Pure sinusoids do not appear in the table of z transforms Since the excitation is a pure sinusoid we are finding the forced response of the system and we can use the DTFT pairs cos Ω 0 n ℱ π δ 2π Ω Ω 0 δ 2π Ω Ω 0 and δ N 0 n ℱ 2π N 0 δ 2π N 0 Ω and the duality of multiplication and convolution xn yn ℱ X e jΩ Y e jΩ The DTFT of the response of the system is Y e jΩ e jΩ e jΩ 09 π δ 2π Ω π6 δ 2π Ω π6 Y e jΩ π e jΩ δ 2π Ω π6 e jΩ 09 e jΩ δ 2π Ω π6 e jΩ 09 Using the equivalence property of the impulse and the fact that both e jΩ and δ 2π Ω have a fundamental period of 2π Y e jΩ π e jπ6 δ 2π Ω π6 e jπ6 09 e jπ6 δ 2π Ω π6 e jπ6 09 rob28124ch09406445indd 431 041216 144 pm C h a p t e r 9 The z Transform 432 Finding a common denominator and simplifying Y e jΩ π δ 2π Ω π61 09 e jπ6 δ 2π Ω π61 09 e jπ6 181 18 cosπ6 Y e jΩ π 02206 δ 2π Ω π6 δ 2π Ω π6 j045 δ 2π Ω π6 δ 2π Ω π6 02512 Y e jΩ 27589 δ 2π Ω π6 δ 2π Ω π6 j56278 δ 2π Ω π6 δ 2π Ω π6 Recognizing the DTFTs of a cosine and a sine yn 08782 cos2πn12 17914 sin2πn12 Using A cosx B sinx A 2 B 2 cosx tan 1 BA yn 1995 cos2πn12 1115 We did not use the z transform because there is no entry in the table of ztransform pairs for a sinusoid But there is an entry for a sinusoid multiplied by a unit sequence cos Ω 0 n un 𝒵 zz cos Ω 0 z 2 2z cos Ω 0 1 z 1 It is instructive to find the response of the system to this different but similar excitation The transfer function is Hz z z 09 z 09 The z transform of the response is Yz z z 09 zz cosπ6 z 2 2z cosπ6 1 z 1 Expanding in partial fractions Yz 01217z z 09 08783 z 2 01353z z 2 1732z 1 z 1 To find the inverse z transform we need to manipulate the expressions into forms similar to the table entries The first fraction form appears directly in the table The second fraction has a denominator of the same form as the z transforms of cos Ω 0 n un and sin Ω 0 n un but the numerator is not in exactly the right form But by adding and subtracting the right amounts in the numerator we can express Yz in the form Yz 01217 z 09 08783 zz 0866 z 2 1732z 1 204 05z z 2 1732z 1 z 1 yn 01217 09 n un 08783cos2πn12 204 sin2πn12 un yn 01217 09 n un 1995 cos2πn12 1115 un Notice that the response consists of two parts a transient response 01217 09 n un and a forced response 1995 cos2πn12 1115 un that except for the unit sequence factor is exactly the same as the forced response we found using the DTFT So even though we do not have a z transform of a sinusoid in the z transform table we can use the z transforms of cos Ω 0 n un and sin Ω 0 n un to find the forced response to a sinusoid rob28124ch09406445indd 432 041216 144 pm 915 Transform Method Comparisons 433 The analysis in Example 912 a system excited by a sinusoid is very common in some types of signal and system analysis It is important enough to generalize the process If the transfer function of the system is Hz Nz Dz the response of the system to cos Ω 0 n un is Yz Nz Dz zz cos Ω 0 z 2 2z cos Ω 0 1 The poles of this response are the poles of the transfer function plus the roots of z 2 2z cos Ω 0 1 0 which are the complex conjugate pair p 1 e j Ω 0 and p 2 e j Ω 0 Therefore p 1 p 2 p 1 p 2 2 cos Ω 0 p 1 p 2 j2 sin Ω 0 and p 1 p 2 1 Then if Ω 0 mπ m an integer and if there is no polezero cancellation these poles are dis tinct The response can be written in partialfraction form as Yz z N 1 z Dz 1 p 1 p 2 H p 1 p 1 cos Ω 0 z p 1 1 p 2 p 1 H p 2 p 2 cos Ω 0 z p 2 or after simplification Yz z N 1 z Dz H r p 1 z p 1r H i p 1 p 1i z 2 z2 p 1r 1 where p 1 p 1r j p 1i and H p 1 H r p 1 j H i p 1 This can be written in terms of the original parameters as Yz z N 1 z Dz ReHcos Ω 0 j sin Ω 0 z 2 z cos Ω 0 z 2 z2 cos Ω 0 1 ImHcos Ω 0 j sin Ω 0 z sin Ω 0 z 2 z2 cos Ω 0 1 The inverse z transform is yn Z 1 z N 1 z Dz ReHcos Ω 0 j sin Ω 0 cos Ω 0 n ImHcos Ω 0 j sin Ω 0 sin Ω 0 n un or using ReA cos Ω 0 n Im A sin Ω 0 n A cos Ω 0 n A yn Z 1 z N 1 z D z Hcos Ω 0 j sin Ω 0 cos Ω 0 n Hcos Ω 0 j sin Ω 0 un or finally yn Z 1 z N 1 z Dz H p 1 cos Ω 0 n H p 1 un 98 rob28124ch09406445indd 433 041216 144 pm C h a p t e r 9 The z Transform 434 If the system is stable the term Z 1 z N 1 z Dz the natural or transient response decays to zero with discrete time and the term H p 1 cos Ω 0 n H p 1 un is equal to a sinusoid after discrete time n 0 and persists forever Using this result we could now solve the problem in Example 912 much more quickly The response to xn cos2πn12 un is yn Z 1 z N 1 z Dz H p 1 cos Ω 0 n H p 1 un and the response to xn cos2πn12 is y f n H p 1 cos Ω 0 n H p 1 where Hz z z 09 and p 1 e jπ6 Therefore H e jπ6 e jπ6 e jπ6 09 08783 j17917 1995 1115 and y f n 1995 cos Ω 0 n 0115 916 SUMMARY OF IMPORTANT POINTS 1 The z transform can be used to determine the transfer function of a discrete time LTI system and the transfer function can be used to find the response of a discretetime LTI system to an arbitrary excitation 2 The z transform exists for discretetime signals whose magnitudes do not grow any faster than an exponential in either positive or negative time 3 The region of convergence of the z transform of a signal depends on whether the signal is right or left sided 4 Systems described by ordinary linear constantcoefficient difference equations have transfer functions in the form of a ratio of polynomials in z and the systems can be realized directly from the transfer function 5 With a table of z transform pairs and ztransform properties the forward and inverse transforms of almost any signal of engineering significance can be found 6 The unilateral z transform is commonly used in practical problem solving because it does not require any involved consideration of the region of convergence and is therefore simpler than the bilateral form 7 Polezero diagrams of a systems transfer function encapsulate most of its properties and can be used to determine its frequency response 8 MATLAB has an object defined to represent a discretetime system transfer function and many functions to operate on objects of this type rob28124ch09406445indd 434 041216 144 pm 435 Exercises with Answers EXERCISES WITH ANSWERS DirectForm II System Realization 1 Draw a Direct Form II block diagram for each of these system transfer functions a H z z z 1 z 2 15z 08 b H z z 2 2z 4 z 12 2 z 2 z 1 Answers Xz Yz 15 08 1 z1 z1 Xz Yz z1 z1 z1 14 14 2 1 12 Existence of the z Transform 2 Find the region of convergence if it exists in the z plane of the z transform of these signals a x n u n u n b x n u n u n 10 c x n 4n u n 1 Hint Express the timedomain function as the sum of a causal function and an anticausal function combine the ztransform results over a common denominator and simplify d x n 4n u n 1 e x n 12 085 n cos 2πn10 u n 1 3 04 n2 u n 2 Answers z 1 z 0 z 1 Does Not Exist 04 z 085 Forward and Inverse z Transforms 3 Using the timeshifting property find the z transforms of these signals a x n u n 5 b x n u n 2 c x n 23 n u n 2 Answers z 3 z 1 z 1 9 4 z 3 z 23 z 23 z 4 z 1 z 1 4 Using the changeofscale property find the z transform of x n sin 2πn32 cos 2πn8 u n Answer sin 2πn32 cos 2πn8 u n 𝒵 z 01379 z 2 03827z 01379 z 4 27741 z 3 38478 z 2 27741z 1 z 1 rob28124ch09406445indd 435 041216 144 pm C h a p t e r 9 The z Transform 436 5 Using the zdomaindifferentiation property find the z transform of x n n 58 n u n Answer n 58 n u n 𝒵 5z8 z 58 2 z 58 6 Using the convolution property find the z transforms of these signals a x n 09 n u n u n b x n 09 n u n 06 n u n Answers z 2 z 2 19z 09 z 1 z 2 z 2 15z 054 z 09 7 Using the differencing property and the z transform of the unit sequence find the z transform of the unit impulse and verify your result by checking the ztransform table 8 Find the z transform of x n u n u n 10 and using that result and the differencing property find the z transform of x n δ n δ n 10 Compare this result with the z transform found directly by applying the time shifting property to an impulse Answer 1 z 10 All z 9 Using the accumulation property find the z transforms of these signals a x n ramp n b x n m n u m 5 u m Answers z 2 z 5 1 z 1 2 z 1 z z 1 2 z 1 10 A discretetime signal y n is related to another discretetime signal x n by y n m0 n x m If y n 𝒵 1 z 1 2 what are the values of x 1 x 0 x 1 and x 2 Answers 1 0 0 0 11 Using the finalvalue theorem find the final value of functions that are the inverse z transforms of these functions if the theorem applies a X z z z 1 z 1 b X z z 2z 74 z 2 74z 34 z 1 c X z z 3 2 z 2 3z 7 z 1 z 2 18z 09 z 1 Answers 1 1 70 rob28124ch09406445indd 436 041216 144 pm 437 Exercises with Answers 12 A discretetime signal x n has a z transform X z 5 z 2 z 3 2 z 2 1 2 z 1 4 What is the numerical value of x 0 Answer 25 13 Find the inverse z transforms of these functions in series form by synthetic division a X z z z 12 z 12 b X z z 1 z 2 2z 1 z 1 c X z z z 12 z 12 d X z z 2 4 z 2 2z 3 z 3 2 Answers δ n 1 δ n 2 δ n k δ n 12 δ n 1 1 2 k δ n k 0667δ n 0778δ n 1 03704δ n 2 2δ n 1 4δ n 2 8δ n 3 2 k δ n k 14 Find the inverse z transforms of these functions in closed form using partial fraction expansions a ztransform table and the properties of the z transform a X z 1 z z 12 z 12 b X z z 2 z 12 z 34 z 12 c X z z 2 z 2 18z 082 z 09055 d X z z 1 3 z 2 2z 2 z 08165 e X z 2 z 2 01488z z 2 175z 1 Answers 2 12 n 3 34 n u n 1 2 12 n1 δ n 1 u n 1 09055 n cos 3031n 903 sin 3031n u n 04472 08165 n 12247 sin 11503 n 1 u n sin 11503n u n 1 2 cos 05054n 3 sin 05054n u n 15 A discretetime system has a transfer function the z transform of its impulse response H z z z 2 z 024 If a unit sequence u n is applied as an excitation to this system what are the numerical values of the responses y 0 y 1 and y 2 Answers 0 0 1 rob28124ch09406445indd 437 041216 144 pm C h a p t e r 9 The z Transform 438 16 The z transform of a discretetime signal x n is X z z 4 z 4 z 2 1 What are the numerical values of x 0 x 1 and x 2 Answer 1 0 1 17 If H z z 2 z 12 z 13 z 12 then by finding the partialfraction expansion of this improper fraction in z two different ways its inverse z transform h n can be written in two different forms h n A 12 n B 13 n u n and h n δ n C 12 n1 D 13 n1 u n 1 Find the numerical values of A B C and D Answers 06 04 03 01333 Unilateral zTransform Properties 18 Using the timeshifting property find the unilateral z transforms of these signals a x n u n 5 b x n u n 2 c x n 23 n u n 2 Answers z 4 z 1 z 1 z z 23 z 23 z z 1 z 1 19 If the unilateral z transform of x n is X z z z 1 what are the z transforms of x n 1 and x n 1 Answers 1 z 1 z z 1 20 The unilateral z transform of x n 5 07 n1 u n 1 can be written in the form X z A z z a Find the numerical values of A and a Answers 35 07 Solution of Difference Equations 21 Using the z transform find the total solutions to these difference equations with initial conditions for discrete time n 0 a 2y n 1 y n sin 2πn16 u n y 0 1 b 5y n 2 3y n 1 y n 08 n u n y 0 1 y 1 10 Answers y n 02934 12 n1 u n 1 12 n u n 02934 cos π8 n 1 2812 sin π8 n 1 u n 1 rob28124ch09406445indd 438 041216 144 pm 439 Exercises with Answers y n 04444 08 n u n δ n 95556 04472 n1 cos 08355 n 1 09325 sin 08355 n 1 u n 1 22 For each block diagram in Figure E22 write the difference equation and find and graph the response y n of the system for discrete times n 0 assuming no initial energy storage in the system and impulse excitation x n δ n a D xn yn b D xn yn 08 c D D xn yn 09 05 Figure E22 Answers n 5 20 yn 1 1 n 5 20 yn 1 1 n 5 20 yn 1 1 PoleZero Diagrams and Frequency Response 23 Sketch the magnitude frequency response of the systems in Figure E23 from their polezero diagrams rob28124ch09406445indd 439 041216 144 pm C h a p t e r 9 The z Transform 440 a Rez Imz z 05 b Rez Imz z 1 05 c Rez Imz z 05 05 05 Figure E23 Answers Ω π He jΩ 2 Ω π He jΩ 3 Ω π π He jΩ 2 π π 24 Where are the finite poles and zeros of H z z 2 z 2 6 z 3 4 z 2 3z Answers 2 0333 j06236 0 rob28124ch09406445indd 440 041216 144 pm 441 Exercises without Answers EXERCISES WITHOUT ANSWERS Direct Form II System Realization 25 Draw a Direct Form II block diagram for each of these system transfer functions a H z z 2 2 z 4 12 z 3 106 z 2 008z 002 b H z z 2 z 2 08z 02 2 z 2 2z 1 z 2 12z 05 Existence of the z Transform 26 Graph the region of convergence if it exists in the z plane of the z transform of these discretetime signals a x n 12 n u n b x n 54 n u n 107 n u n Forward and Inverse zTransforms 27 Find the inverse z transform of H z z 2 z 12 Is a system with this transfer function causal Why or why not 28 Find the forward z transforms of these discretetime functions a 4 cos 2πn4 u n b 4 cos 2π n 4 4 u n 4 c 4 cos 2πn4 u n 4 d 4 cos 2π n 4 4 u n 29 Using the timeshifting property find the z transforms of these signals a x n 23 n1 u n 1 b x n 23 n u n 1 c x n sin 2π n 1 4 u n 1 30 If the z transform of x n is X z 1 z 34 z 34 and Y z j X e jπ6 z X e jπ6 z what is y n 31 Using the convolution property find the z transforms of these signals a x n sin 2πn8 u n u n b x n sin 2πn8 u n u n u n 8 32 A digital filter has an impulse response h n δ n δ n 1 δ n 2 10 a How many finite poles and finite zeros are there in its transfer function and what are their numerical locations b If the excitation x n of this system is a unit sequence what is the final numerical value of the response lim n y n rob28124ch09406445indd 441 061216 242 pm C h a p t e r 9 The z Transform 442 33 The forward z transform h n 45 n u n u n can be expressed in the general form H z b 2 z 2 b 1 z b 0 a 2 z 2 a 1 z a 0 Find the numerical values of b 2 b 1 b 0 a 2 a 1 and a 0 34 Find the inverse z transforms of these functions in closed form using partialfraction expansions a ztransform table and the properties of the z transform a X z z 1 z 2 18z 082 z 09055 b X z z 1 z z 2 18z 082 z 09055 c X z z 2 z 2 z 14 z 05 d X z z 03 z 2 08z 016 z 03 z 04 2 01 z 04 2 1 z 04 z 04 e X z z 2 08z 03 z 3 z 1 08 z 2 03 z 3 z 0 f X z 2z z 2 z 074 z 086 g X z z 3 z 1 2 z 1 2 z 1 35 The z transform of a signal x n is X z z 4 z 4 z 2 1 z 1 What are the numerical values of x 2 x 1 x 0 x 1 x 2 x 3 and x 4 36 Find the numerical values of the literal constants a 10 04 n sin πn8 u n 𝒵 b 2 z 2 b 1 z b 0 z 2 a 1 z a 0 b A a n sin bn u n 𝒵 12z z 2 064 c A a n B b n u n 𝒵 z z 04 z 2 15z 03 d Atri n n 0 b 𝒵 z 2 2z 1 z 2 e δ n 2δ n 2 07 n u n 𝒵 A z B z 1 z C f A a n cos bn B sin bn u n 𝒵 z 2 z 2 z 08 g 4u n 1 𝒵 Az z B rob28124ch09406445indd 442 041216 144 pm 443 Exercises without Answers h 4 5 n u n u n 𝒵 b 2 z 2 b 1 z b 0 a 2 z 2 a 1 z a 0 i 1 e j 2πn 8 u n 𝒵 b 2 z 2 b 1 z b 0 a 2 z 2 a 1 z a 0 j A α n cos Ω 0 n B sin Ω 0 n u n 𝒵 6 z z 12 z 2 49 k 23 n u n 2 𝒵 A z b z c l 6 08 n1 cos π n 1 12 u n 1 𝒵 b 2 z 2 b 1 z b 0 a 2 z 2 a 1 z a 0 m A b n u n 1 C d n u n 𝒵 z 2 z 05 z 02 02 z 05 n 03 n u n 8 𝒵 A z b z a z c o A a nc cos 2πb n c B sin 2πb n c u n c 𝒵 z 1 15625z 1 z 2 064 z 08 p 3 u n 1 u n 1 𝒵 A z a B z b PoleZero Diagrams and Frequency Response 37 A digital filter has a transfer function H z 09525 z 2 z 1 z 2 095z 09025 a What are the numerical locations of its poles and zeros b Find the numerical frequency response magnitude at the radian frequencies Ω 0 and Ω π3 38 A filter has an impulse response h n δ n δ n 1 2 A sinusoid x n is created by sampling at f s 10 Hz a continuoustime sinusoid with cyclic frequency f 0 What is the minimum positive numerical value of f 0 for which the forced filter response is zero 39 Find the magnitude of the transfer function of the systems with the polezero plots in Figure E39 at the specified frequencies In each case assume the transfer function is of the general form H z K z z 1 z z 2 z z N z p 1 z p 2 z p D where the zs are the zeros and the ps are the poles and let K 1 a Ω 0 b Ω π 1 0 1 1 05 0 05 1 Rez Imz z 08 1 0 1 1 05 0 05 1 Rez Imz z 04 07 07 Figure E39 rob28124ch09406445indd 443 041216 144 pm C h a p t e r 9 The z Transform 444 40 For each of the systems with the polezero plots in Figure E40 find the discrete time radian frequencies Ω max and Ω min in the range π Ω π for which the transfer function magnitude is a maximum and a minimum If there is more than one value of Ω max or Ω min find all such values a b 1 0 1 1 05 0 05 1 Rez Rez Imz z 05 08 08 1 0 1 1 05 0 05 1 Imz z 08 Figure E40 41 Sketch the magnitude frequency response of the systems in Figure E41 from their polezero diagrams a Rez Imz z 05 05 0866 02 b Rez Imz z 1 045 045 0866 Figure E41 42 Match the polezero plots in Figure E42 to the corresponding magnitude frequency responses 1 0 a b c d e 1 1 0 1 1 0 1 1 0 1 1 0 1 1 0 1 1 0 1 1 0 1 1 0 1 1 0 1 z z z z 2 0 2 2 0 2 2 0 2 2 0 2 2 0 2 0 05 1 15 H H H H H 0 2 4 6 8 0 05 1 15 0 2 4 0 05 1 15 z Ω Ω Ω Ω Ω Figure E42 rob28124ch09406445indd 444 041216 144 pm 445 Exercises without Answers 43 Using the following definitions of lowpass highpass bandpass and bandstop classify the systems whose transfer functions have the polezero diagrams in Figure E43 Some may not be classifiable In each case the transfer function is H z LP H 1 0 and H 1 0 HP H 1 0 and H 1 0 BP H 1 0 and H 1 0 and H z 0 for some range of z 1 BS H 1 0 and H 1 0 and H z 0 for at least one z 1 z Unit Circle z z z z Unit Circle Unit Circle Unit Circle Unit Circle Figure E43 44 For each magnitude frequency response and each unit sequence response in Figure E44 find the corresponding polezero diagram 1 0 1 1 05 0 05 1 Rez Imz 1 0 1 1 05 0 05 1 Rez Imz 1 0 1 1 05 0 05 1 Rez Imz 1 0 1 1 05 0 05 1 Rez Imz 1 0 1 1 05 0 05 1 Rez Imz PoleZero Diagram z PoleZero Diagram z PoleZero Diagram z PoleZero Diagram z PoleZero Diagram z 2 0 2 0 2 4 6 8 Ω Ω Ω Ω Magnitude Frequency Response Magnitude Frequency Response Magnitude Frequency Response Magnitude Frequency Response Magnitude Frequency Response 2 0 2 0 05 1 2 0 2 0 05 1 2 0 2 0 05 1 2 0 2 0 05 1 15 2 0 10 20 30 0 05 1 15 n h1n h1n h1n h1n h1n Unit Sequence Response 0 10 20 30 0 05 1 n Unit Sequence Response 0 10 20 30 1 05 0 05 1 n Unit Sequence Response 0 10 20 30 0 05 1 15 n Unit Sequence Response 0 10 20 30 0 05 1 n Unit Sequence Response He jΩ He jΩ He jΩ He jΩ HejΩ Figure E44 rob28124ch09406445indd 445 041216 144 pm 446 101 INTRODUCTION AND GOALS In the application of signal processing to real signals in real systems we often do not have a mathematical description of the signals We must measure and analyze them to discover their characteristics If the signal is unknown the process of analysis be gins with the acquisition of the signals measuring and recording the signals over a period of time This could be done with a tape recorder or other analog recording device but the most common technique of acquiring signals today is by sampling The term analog refers to continuoustime signals and systems Sampling converts a continuoustime signal into a discretetime signal In previous chapters we have ex plored ways of analyzing continuoustime signals and discretetime signals In this chapter we investigate the relationships between them Much signal processing and analysis today is done using digital signal processing DSP A DSP system can acquire store and perform mathematical calculations on numbers A computer can be used as a DSP system Since the memory and mass storage capacity of any DSP system are finite it can only handle a finite number of numbers Therefore if a DSP system is to be used to analyze a signal it can only be sampled for a finite time The salient question addressed in this chapter is To what extent do the samples accurately describe the signal from which they are taken We will see that whether and how much information is lost by sampling depends on the way the samples are taken We will find that under certain circumstances practically all of the signal information can be stored in a finite number of numerical samples Many filtering operations that were once done with analog filters now use digital filters which operate on samples from a signal instead of the original continuoustime signal Modern cellular telephone systems use DSP to improve voice quality separate channels and switch users between cells Longdistance telephone communication systems use DSP to efficiently use long trunk lines and microwave links Television sets use DSP to improve picture quality Robotic vision is based on signals from cam eras that digitize sample an image and then analyze it with computation techniques to recognize features Modern control systems in automobiles manufacturing plants and scientific instrumentation usually have embedded processors that analyze signals and make decisions using DSP CHAPTER GOA L S 1 To determine how a continuoustime signal must be sampled to retain most or all of its information 10 C H A P T E R Sampling and Signal Processing rob28124ch10446508indd 446 041216 145 pm 102 ContinuousTime Sampling 447 2 To learn how to reconstruct a continuoustime signal from its samples 3 To apply sampling techniques to discretetime signals and to see the similarities with continuoustime sampling 102 CONTINUOUSTIME SAMPLING SAMPLING METHODS Sampling of electrical signals occasionally currents but usually voltages is most com monly done with two devices the sampleandhold SH and the analogtodigital converter ADC The excitation of the SH is the analog voltage at its input When the SH is clocked it responds with that voltage at its output and holds that voltage until it is clocked to acquire another voltage Figure 101 t vint t voutt t ct Aperture Time Figure 101 Operation of a sampleandhold In Figure 101 the signal ct is the clock signal The acquisition of the input volt age signal of the SH occurs during the aperture time which is the width of a clock pulse During the clock pulse the output voltage signal very quickly moves from its previous value to track the excitation At the end of the clock pulse the output voltage signal is held at a fixed value until the next clock pulse occurs An ADC accepts an analog voltage at its input and responds with a set of binary bits often called a code The ADC response can be serial or a parallel If the ADC has a serial response it produces on one output pin a single output voltage signal that is a timed sequence of high and low voltages representing the 1s and 0s of the set of binary bits If the ADC has a parallel response there is a response voltage for each bit and each bit appears simultaneously on a dedicated output pin of the ADC as a high rob28124ch10446508indd 447 041216 145 pm C h a p t e r 10 Sampling and Signal Processing 448 or low voltage representing a 1 or a 0 in the set of binary bits Figure 102 An ADC may be preceded by a SH to keep its excitation constant during the conversion time The excitation of the ADC is a continuoustime signal and the response is a discretetime signal Not only is the response of the ADC discretetime but is also quantized and encoded The number of binary bits produced by the ADC is finite Therefore the number of unique bit patterns it can produce is also finite If the num ber of bits the ADC produces is n the number of unique bit patterns it can produce is 2 n Quantization is the effect of converting a continuum of infinitely many excitation values into a finite number of response values Since the response has an error due to quantization it is as though the signal has noise on it and this noise is called quantization noise If the number of bits used to represent the response is large enough quantization noise is often negligible in comparison with other noise sources After quantization the ADC encodes the signal also Encoding is the conversion from an analog voltage to a binary bit pattern The relation between the excitation and re sponse of an ADC whose input voltage range is V 0 vint V 0 is illustrated in Figure 103 for a 3bit ADC A 3bit ADC is rarely if ever actually used but it does illustrate the quantization effect nicely because the number of unique bit patterns is small and the quantization noise is large Figure 103 ADC excitationresponse relationship Excitation Voltage Response Code 100 101 110 111 000 001 010 011 V0 V0 Serial ADC Parallel ADC Figure 102 Serial and parallel ADC operation The effects of quantization are easy to see in a sinusoid quantized by a 3bit ADC Figure 104 When the signal is quantized to 8 bits the quantization error is much smaller Figure 105 The opposite of analogtodigital conversion is obviously digitaltoanalog con version done by a digitaltoanalog converter DAC A DAC accepts binary bit rob28124ch10446508indd 448 041216 145 pm 102 ContinuousTime Sampling 449 Figure 104 Sinusoid quantized to 3 bits Original Sinusoid t 3bit Quantized Approximation V0 V0 xt 8bit Quantization t V0 V0 Figure 105 Sinusoid quantized to 8 bits Figure 106 DAC excitationresponse relationship Excitation Code Response Voltage 100 101 110 111 000 001 010 011 V0 V0 patterns as its excitation and produces an analog voltage as its response Since the number of unique bit patterns it can accept is finite the DAC response signal is an an alog voltage that is quantized The relation between excitation and response for a 3bit DAC is shown in Figure 106 In the material to follow the effects of quantization will not be considered The model for analyzing the effects of sampling will be that the sampler is ideal in the sense that the response signals quantization noise is zero THE SAMPLING THEOREM Qualitative Concepts If we are to use samples from a continuoustime signal instead of the signal itself the most important question to answer is how to sample the signal so as to retain the in formation it carries If the signal can be exactly reconstructed from the samples then the samples contain all the information in the signal We must decide how fast to sample the signal and how long to sample it Consider the signal xt Figure 107a Suppose this signal is sampled at the sampling rate illustrated in Figure 107b Most people would probably intuitively say that there are enough samples here to describe the signal adequately by drawing a smooth curve through the points How about the sampling rate in Figure 107c Is this sampling rate adequate How about the rate in Figure 107d Most people would probably agree that the sampling rate in Figure 107d is inadequate rob28124ch10446508indd 449 041216 145 pm C h a p t e r 10 Sampling and Signal Processing 450 A naturally drawn smooth curve through the last sample set would not look very much like the original curve Although the last sampling rate was inadequate for this signal it might be just fine for another signal Figure 108 It seems adequate for the signal of Figure 108 because it is much smoother and more slowly varying The minimum rate at which samples can be taken while retaining the information in the signal depends on how fast the signal varies with time the frequency content of the signal The question of how fast samples have to be taken to describe a signal was answered definitively by the sampling theorem Claude Shannon1 of Bell Labs was a major contributor to theories of sampling 1 Claude Shannon arrived as a graduate student at the Massachusetts Institute of Technology in 1936 In 1937 he wrote a thesis on the use of electrical circuits to make decisions based on Boolean logic In 1948 while working at Bell Labs he wrote A Mathematical Theory of Communication which outlined what we now call information theory This work has been called the Magna Carta of the information age He was appointed a professor of communication sciences and mathematics at MIT in 1957 but remained a consultant to Bell Labs He was often seen in the corridors of MIT on a unicycle sometimes juggling at the same time He also devised one of the first chessplaying programs t n n n xt xn xn xn a b c d Figure 107 a A continuoustime signal bd discretetime signals formed by sampling the continuoustime signal at different rates Figure 108 A discretetime signal formed by sampling a slowly varying signal n xn rob28124ch10446508indd 450 041216 145 pm 102 ContinuousTime Sampling 451 Figure 109 Pulse train t pt w 1 Ts xt yt Multiplier Sampling Theorem Derivation Let the process of sampling a continuoustime signal xt be to multiply it by a periodic pulse train pt Let the amplitude of each pulse be one let the width of each pulse be w and let the fundamental period of the pulse train be Ts Figure 109 The pulse train can be mathematically described by pt recttw δ T s t The output signal is yt xtpt xtrecttw δ T s t The average of the signal yt over the width of the pulse centered at t n T s can be considered an approximate sample of xt at time t n T s The continuoustime Fourier transform CTFT of yt is Y f X f wsincwf f s δ f s f where f s 1 T s is the pulse repetition rate pulse train fundamental frequency and Y f X f w f s k sincwk f s δ f k f s Y f w f s k sincwk f s X f k f s The CTFT Y f of the response is a set of replicas of the CTFT of the input signal xt repeated periodically at integer multiples of the pulse repetition rate f s and also multi plied by the value of a sinc function whose width is determined by the pulse width w Figure 1010 Replicas of the spectrum of the input signal occur multiple times in the spectrum of the output signal each centered at an integer multiple of the pulse repeti tion rate and multiplied by a different constant As we make each pulse shorter its average value approaches the exact value of the signal at its center The approximation of ideal sampling improves as w approaches zero In the limit as w approaches zero yt lim w0 n xt rectt nTsw In that limit the signal power of yt approaches zero But if we now modify the sam pling process to compensate for that effect by making the area of each sampling pulse rob28124ch10446508indd 451 041216 145 pm C h a p t e r 10 Sampling and Signal Processing 452 one instead of the height we get the new pulse train pt 1wrecttw δ T s t and now yt is yt n xt1wrectt n T s w Let the response in this limit as w approaches zero be designated x δ t In that limit the rectangular pulses 1wrectt n T s w approach unit impulses and x δ t lim w0 yt n xtδt n T s xt δ T s t This operation is called impulse sampling or sometimes impulse modulation Of course as a practical matter this kind of sampling is impossible because we cannot generate impulses But the analysis of this hypothetical type of sampling is still useful because it leads to relationships between the values of a signal at discrete points and the values of the signal at all other times Notice that in this model of sampling the response of the sampler is still a continuoustime signal but one whose value is zero except at the sampling instants It is revealing to examine the CTFT of the newly defined response x δ t It is X δ f X f 1 T s δ 1 T s f f s X f δ f s f This is the sum of equalsize replicas of the CTFT X f of the original signal xt each shifted by a different integer multiple of the sampling frequency f s and multiplied by f s Figure 1011 These replicas are called aliases In Figure 1011 the dashed lines rep resent the aliases of the original signals CTFT magnitude and the solid line represents the magnitude of the sum of those aliases Obviously the shape of the original signals CTFT magnitude is lost in the overlapping process But if X f is zero for all f f m and if f s 2 f m the aliases do not overlap Figure 1012 Figure 1010 Magnitude CTFT of input and output signals Xf fm fm f Xpf fm fm fs fs f Sinc Function rob28124ch10446508indd 452 041216 145 pm 102 ContinuousTime Sampling 453 Figure 1011 CTFT of an impulsesampled signal Xδf f fs fs Xf f A Afs Figure 1012 CTFT of a bandlimited signal impulsesampled above twice its bandlimit f fs fs Xf f A Afs fm fm fm fm Xδf Signals for which X f is zero for all f f m are called strictly bandlimited or more often just bandlimited signals If the aliases do not overlap then at least in principle the original signal could be recovered from the impulsesampled signal by filtering out the aliases centered at f f s 2 f s 3 f s with a lowpass filter whose frequency response is H f T s f f c 0 otherwise Ts rect f 2 f c where fm fc fs fm an ideal lowpass filter This fact forms the basis for what is commonly known as the sampling theorem If a continuoustime signal is sampled for all time at a rate f s that is more than twice the bandlimit f m of the signal the original continuoustime signal can be recovered exactly from the samples If the highest frequency present in a signal is f m the sampling rate must be above 2 f m and the frequency 2 f m is called the Nyquist2 rate The words rate and frequency both describe something that happens periodically In this text the word frequency will refer to the frequencies present in a signal and the word rate will refer to the way a signal is sampled A signal sampled at greater than its Nyquist rate is said to be oversampled and a 2 Harry Nyquist received his PhD from Yale in 1917 From 1917 to 1934 he was employed by Bell Labs where he worked on transmitting pictures using telegraphy and on voice transmission He was the first to quantitatively explain thermal noise He invented the vestigial sideband transmission technique still widely used in the transmission of television signals He invented the Nyquist diagram for determining the stability of feedback systems rob28124ch10446508indd 453 041216 145 pm C h a p t e r 10 Sampling and Signal Processing 454 signal sampled at less than its Nyquist rate is said to be undersampled When a signal is sampled at a rate f s the frequency f s 2 is called the Nyquist frequency Therefore if a signal has any signal power at or above the Nyquist frequency the aliases will overlap Another sampling model that we have used in previous chapters is the creation of a discretetime signal xn from a continuoustime signal xt through xn xn T s where T s is the time between consecutive samples This may look like a more realistic model of practical sampling and in some ways it is but instantaneous sampling at a point in time is also not possible practically We will refer to this sampling model as simply sampling instead of impulse sampling Recall that the DTFT of any discretetime signal is always periodic The CTFT of an impulsesampled signal is also periodic The CTFT of an impulsesampled continuoustime signal x δ t and the DTFT of a discretetime signal x s n formed by sampling that same continuoustime signal are similar Figure 1013 The s subscript on x s n is there to help avoid confusion between the different transforms that follow The waveshapes are the same The only difference is that the DTFT is based on nor malized frequency F or Ω and the CTFT on actual frequency f or ω The sampling theorem can be derived using the DTFT instead of the CTFT and the result is the same Figure 1013 Comparison between the CTFT of an impulsesampled signal and the DTFT of a sampled signal f fs fs Xf Xse jΩ f A Afs fm Ωm fm fm fm Ω fs fs 2π 2π Afs Xδf F 1 1 Afs fm XsF fs fm Ωm fs ALIASING The phenomenon of aliasing overlapping of aliases is not an exotic mathemati cal concept that is outside the experience of ordinary people Almost everyone has observed aliasing but probably without knowing what to call it A very common expe rience that illustrates aliasing sometimes occurs while watching television Suppose you are watching a Western movie on television and there is a picture of a horsedrawn wagon with spoked wheels If the wheels on the wagon gradually rotate faster and faster a point is reached at which the wheels appear to stop rotating forward and begin to appear to rotate backward even though the wagon is obviously moving forward If the speed of rotation were increased further the wheels would eventually appear to stop and then rotate forward again This is an example of the phenomenon of aliasing rob28124ch10446508indd 454 041216 145 pm 102 ContinuousTime Sampling 455 Figure 1014 Wagon wheel angular positions at four sampling times t 0 t Ts t 2Ts t 3Ts Slow Fast Although it is not apparent to the human eye the image on a television screen is flashed upon the screen 30 times per second under the NTSC video standard That is the image is effectively sampled at a rate of 30 samplessecond Figure 1014 shows the positions of a spoked wheel at four sampling instants for several different rotational velocities starting with a lower rotational velocity at the top and progressing toward a higher rotational velocity at the bottom A small index dot has been added to the wheel to help in seeing the actual rotation of the wheel as opposed to the apparent rotation This wheel has eight spokes so upon rotation by oneeighth of a complete revolution the wheel looks exactly the same as in its initial position Therefore the image of the wheel has an angular period of π4 radians or 45 the angular spacing between spokes If the rotational velocity of the wheel is f 0 revolutionssecond Hz the image fundamental frequency is 8 f 0 Hz The image repeats exactly eight times in one complete wheel rotation Let the image be sampled at 30 Hz T s 130 s On the top row the wheel is ro tating clockwise at 5 T s 150s or 0416 revs so that in the top row the spokes have rotated by 0 5 10 and 15 clockwise The eye and brain of the observer inter pret the succession of images to mean that the wheel is rotating clockwise because of the progression of angles at the sampling instants In this case the wheel appears to be and is rotating at an image rotational frequency of 150s In the second row the rotational speed is four times faster than in the top row and the angles of rotation at the sampling instants are 0 20 40 and 60 clockwise The wheel still correctly appears to be rotating clockwise at its actual rotational frequency of 600s In the third row the rotational speed is 675s Now the ambiguity caused by sampling begins If the index dot were not there it would be impossible to determine whether the wheel is rotating 225 per sample or 225 per sample because the image samples are identical for those two cases It is impossible by simply looking at the sam ple images to determine whether the rotation is clockwise or counterclockwise In the fourth row the wheel is rotating at 1200s Now ignoring the index dot the wheel definitely appears to be rotating at 5 per sample instead of the actual rotational fre quency of 40 per sample The perception of the human brain would be that the wheel is rotating 5 counterclockwise per sample instead of 40 clockwise In the bottom row rob28124ch10446508indd 455 041216 145 pm C h a p t e r 10 Sampling and Signal Processing 456 the wheel rotation is 1350s or clockwise 45 per sample Now the wheel appears to be standing still even though it is rotating clockwise Its angular velocity seems to be zero because it is being sampled at a rate exactly equal to the image fundamental frequency ExamplE 101 Finding Nyquist rates of signals Find the Nyquist rate for each of the following signals a xt 25 cos500πt X f 125δ f 250 δ f 250 The highest frequency and the only frequency present in this signal is f m 250 Hz The Nyquist rate is 500 samplessecond b xt 15rectt2 X f 30sinc2 f Since the sinc function never goes to zero and stays there at a finite frequency the highest frequency in the signal is infinite and the Nyquist rate is also infinite The rectangle function is not bandlimited c xt 10sinc5t X f 2rect f 5 The highest frequency present in xt is the value of f at which the rect function has its discontinuous transition from one to zero f m 25 Hz Therefore the Nyquist rate is 5 samplessecond d xt 2sinc5000t sin500000πt X f 1 2500 rect f 5000 j 2 δ f 250000 δ f 250000 X f j 5000 rect f 250000 5000 rect f 250000 5000 The highest frequency in xt is f m 2525 kHz Therefore the Nyquist rate is 505000 samplessecond ExamplE 102 Analysis of an RC filter as an antialiasing filter Suppose a signal that is to be acquired by a data acquisition system is known to have an ampli tude spectrum that is flat out to 100 kHz and drops suddenly there to zero Suppose further that the fastest rate at which our data acquisition system can sample the signal is 60 kHz Design an RC lowpass antialiasing filter that will reduce the signals amplitude spectrum at 30 kHz to less than 1 of its value at very low frequencies so that aliasing will be minimized The frequency response of a unitygain RC lowpass filter is H f 1 j2π f RC 1 rob28124ch10446508indd 456 041216 145 pm 102 ContinuousTime Sampling 457 The squared magnitude of the frequency response is H f 2 1 2π f RC 2 1 and its value at very low frequencies approaches one Set the RC time constant so that at 30 kHz the squared magnitude of H f is 001 2 H30000 2 1 2π 30000 RC 2 1 001 2 Solving for RC RC 05305 ms The corner frequency 3 dB frequency of this RC lowpass filter is 300 Hz which is 100 times lower than the Nyquist frequency of 30 kHz Figure 1015 It must be set this low to meet the specification using a singlepole filter because its frequency response rolls off so slowly For this reason most practical antialiasing filters are designed as higherorder filters with much faster transitions from the pass band to the stop band Figure 1015b shows the frequency response of a 6thorder Butterworth lowpass filter Butterworth filters are covered in Chapter 14 The higherorder filter preserves much more of the signal than the RC filter TIMELIMITED AND BANDLIMITED SIGNALS Recall that the original mathematical statement of the way a signal is sampled is x s n xn T s This equation holds true for any integer value of n and that implies that the signal xt is sampled for all time Therefore infinitely many samples are needed to describe xt exactly from the information in x s n The sampling theorem is predicated on sampling this way So even though the Nyquist rate has been found and may be finite one must in general still take infinitely many samples to exactly reconstruct the original signal from its samples even if it is bandlimited and we oversample It is tempting to think that if a signal is time limited having nonzero values only over a finite time one could then sample only over that time knowing all the other samples are zero and have all the information in the signal The problem with that idea Figure 1015 a Magnitude frequency response of the antialiasing RC lowpass filter b magnitude frequency response of a 6thorder Butterworth antialiasing lowpass filter 1 30 10 20 30 10 20 02 04 06 08 1 02 04 06 08 Hf Hf fkHz fkHz a b rob28124ch10446508indd 457 041216 145 pm C h a p t e r 10 Sampling and Signal Processing 458 is that no timelimited signal can also be bandlimited and therefore no finite sampling rate is adequate The fact that a signal cannot be simultaneously time limited and bandlimited is a fundamental law of Fourier analysis The validity of this law can be demonstrated by the following argument Let a signal xt have no nonzero values outside the time range t 1 t t 2 Let its CTFT be X f If xt is time limited to the time range t 1 t t 2 it can be multiplied by a rectangle function whose nonzero portion covers this same time range without changing the signal That is xt xtrect t t 0 Δt 101 where t 0 t 1 t 2 2 and Δt t 2 t 1 Figure 1016 Finding the CTFT of both sides of 101 we obtain X f X f Δt sincΔtf e j2πf t 0 This last equation says that X f is unaffected by being convolved with a sinc function Since sincΔtf has an infinite nonzero extent in f if it is convolved with an X f that has a finite nonzero extent in f the convolution of the two will have an infinite nonzero extent in f Therefore the last equation cannot be satisfied by any X f that has a finite nonzero extent in f proving that if a signal is time limited it cannot be bandlimited The converse that a bandlimited signal cannot be time limited can be proven by a similar argument A signal can be unlimited in both time and frequency but it cannot be limited in both time and frequency INTERPOLATION Ideal Interpolation The description given above on how to recover the original signal indicated that we could filter the impulsesampled signal to remove all the aliases except the one cen tered at zero frequency If that filter were an ideal lowpass filter with a constant gain of T s 1 f s in its passband and bandwidth f c where f m f c f s f m that operation in the frequency domain would be described by X f T s rect f 2 f c X δ f T s rect f 2 f c f s X f δ f s f If we inverse transform this expression we get xt T s f s 1 2 f c sinc2 f c t xt1 f s δ T s t 1 f s n xn T s δt n T s Figure 1016 A timelimited function and a rectangle timelimited to the same time t1 t2 t xt 1 rob28124ch10446508indd 458 041216 145 pm 102 ContinuousTime Sampling 459 or xt 2 f c f s sinc2 f c t n xn T s δt n T s xt 2 f c f s n xn T s sinc2 f c t n T s 102 By pursuing an admittedly impractical idea impulse sampling we have arrived at a result that allows us to fill in the values of a signal for all time given its values at equally spaced points in time There are no impulses in 102 only the sample values which are the strengths of the impulses that would have been created by im pulse sampling The process of filling in the missing values between the samples is called interpolation Consider the special case f c f s 2 In this case the interpolation process is de scribed by the simpler expression xt n xn T s sinct n T s T s Now interpolation consists simply of multiplying each sinc function by its correspond ing sample value and then adding all the scaled and shifted sinc functions as illustrated in Figure 1017 Referring to Figure 1017 notice that each sinc function peaks at its sample time and is zero at every other sample time So the interpolation is obviously correct at the sample times The derivation above shows that it is also correct at all the points between sample times Practical Interpolation The interpolation method in the previous section reconstructs the signal exactly but it is based on an assumption that is never justified in practice the availability of in finitely many samples The interpolated value at any point is the sum of contributions from infinitely many weighted sinc functions each of infinite time extent But since as a practical matter we cannot acquire infinitely many samples much less process them we must approximately reconstruct the signal using a finite number of samples Many techniques can be used The selection of the one to be used in any given situa tion depends on what accuracy of reconstruction is required and how oversampled the signal is Figure 1017 Interpolation process for an ideal lowpass filter corner frequency set to half the sampling rate t xt Ts rob28124ch10446508indd 459 041216 145 pm C h a p t e r 10 Sampling and Signal Processing 460 ZeroOrder Hold Probably the simplest approximate reconstruction idea is to simply let the reconstruction always be the value of the most recent sample Figure 1018 This is a simple technique because the samples in the form of numerical codes can be the input signal to a DAC that is clocked to produce a new output signal with every clock pulse The signal produced by this technique has a stairstep shape that follows the original signal This type of signal reconstruction can be modeled by impulse sampling the signal and letting the impulsesampled signal excite a system called a zeroorder hold whose impulse response is ht 1 0 t T s 0 otherwise rect t T s 2 T s Figure 1019 One popular way of further reducing the effects of the aliases is to follow the zeroorder hold with a practical lowpass filter that smooths out the steps caused by the zeroorder hold The zeroorder hold inevitably causes a delay relative to the original signal because it is causal and any practical lowpass smoothing filter will add still more delay FirstOrder Hold Another natural idea is to interpolate between samples with straight lines Figure 1020 This is obviously a better approximation to the original signal but it is a little harder to implement As drawn in Figure 1020 the value of the interpolated signal at any time depends on the value of the previous sample and the value of the next sample This cannot be done in real time because the value of the next sample is not known in real time But if we are willing to delay the reconstructed signal by one sample time T s we can make the reconstruction process occur in real time The reconstructed signal would appear as shown in Figure 1021 Figure 1018 Zeroorder hold signal reconstruction xt t Figure 1019 Impulse response of a zeroorder hold ht t Ts 1 Figure 1020 Signal reconstruction by straightline interpolation xt t Figure 1021 Straightline signal reconstruction delayed by one sample time xt t rob28124ch10446508indd 460 041216 146 pm 102 ContinuousTime Sampling 461 This interpolation can be accomplished by following the zeroorder hold by an identical zeroorder hold This means that the impulse response of such an interpo lation system would be the convolution of the zeroorder hold impulse response with itself ht rect t Ts 2 Ts rect t Ts 2 Ts tri t Ts Ts Figure 1022 This type of interpolation system is called a firstorder hold Figure 1022 Impulse response of a firstorder hold ht t 2Ts 1 One very familiar example of the use of sampling and signal reconstruction is the playback of an audio compact disk CD A CD stores samples of a musical signal that have been taken at a rate of 44100 samplessecond Half of that sampling rate is 2205 kHz The frequency response of a young healthy human ear is conventionally taken to span from about 20 Hz to about 20 kHz with some variability in that range So the sampling rate is a little more than twice the highest frequency the human ear can detect SAMPLING BANDPASS SIGNALS The sampling theorem as stated above was based on a simple idea If we sample fast enough the aliases do not overlap and the original signal can be recovered by an ideal lowpass filter We found that if we sample faster than twice the highest frequency in the signal we can recover the signal from the samples That is true for all signals but for some signals the minimum sampling rate can be reduced In making the argument that we must sample at a rate greater than twice the highest frequency in the signal we were implicitly assuming that if we sampled at any lower rate the aliases would overlap In the spectra used above to illustrate the ideas the aliases would overlap But that is not true of all signals For example let a continuoustime signal have a narrow bandpass spectrum that is nonzero only for 15 kHz f 20 kHz Then the bandwidth of this signal is 5 kHz Figure 1023 Figure 1023 A narrowbandpass signal spectrum 20 15 f kHz 15 20 x f rob28124ch10446508indd 461 041216 146 pm C h a p t e r 10 Sampling and Signal Processing 462 If we impulse sample this signal at 20 kHz we would get the aliases illustrated in Figure 1024 These aliases do not overlap Therefore it must be possible with knowl edge of the original signals spectrum and the right kind of filtering to recover the sig nal from the samples We could even sample at 10 kHz half the highest frequency get the aliases in Figure 1025 and still recover the original signal theoretically with that same filter But if we sampled at any lower rate the aliases would definitely overlap and we could not recover the original signal Notice that this minimum sampling rate is not twice the highest frequency in the signal but rather twice the bandwidth of the signal Figure 1024 The spectrum of a bandpass signal impulsesampled at 20 kHz Xδ f 40 20 20 40 f kHz Figure 1026 Magnitude spectrum of a general bandpass signal fH fH fL fL f X f Figure 1025 The spectrum of a bandpass signal impulsesampled at 10 kHz Xδ f 40 30 20 10 10 20 30 40 f kHz The aliases occur at shifts of integer multiples of the sampling rate Let the integer k index the aliases Then the k 1 th alias must lie wholly below f L and the kth alias must lie wholly above f H That is k 1 f s f L f L k 1 f s 2 f L and k f s f H f H k f s 2 f H Rearranging these two inequalities we get k 1 f s 2 f H B In this example the ratio of the highest frequency to the bandwidth of the signal was an integer When that ratio is not an integer it becomes more difficult to find the minimum sampling rate that avoids aliasing Figure 1026 rob28124ch10446508indd 462 041216 146 pm 102 ContinuousTime Sampling 463 where B is the bandwidth f H f L and 1 f s k 2 f H Now set the product of the left sides of these inequalities less than the product of the right sides of these inequalities k 1 f H B k f H k f H B Since k must be an integer that means that the real limit on k is k max f H B the greatest integer in f H B So the two conditions k max f H B and k max 2 f H f smin or the single condition f smin 2 f H f H B determine the minimum sampling rate for which aliasing does not occur ExamplE 103 Minimum sampling rate to avoid aliasing Let a signal have no nonzero spectral components outside the range 34 kHz f 47 kHz What is the minimum sampling rate that avoids aliasing f smin 2 f H f H B 94 kHz 47 kHz 13 kHz 31333 samplessecond ExamplE 104 Minimum sampling rate to avoid aliasing Let a signal have no nonzero spectral components outside the range 0 f 580 kHz What is the minimum sampling rate that avoids aliasing f smin 2 f H f H B 1160 kHz 580 kHz 580 kHz 1160000 samplessecond This is a lowpass signal and the minimum sampling rate is twice the highest frequency as orig inally determined in the sampling theorem In most real engineering design situations choosing the sampling rate to be more than twice the highest frequency in the signal is the practical solution As we will soon see that rate is usually considerably above the Nyquist rate in order to simplify some of the other signal processing operations rob28124ch10446508indd 463 041216 146 pm C h a p t e r 10 Sampling and Signal Processing 464 SAMPLING A SINUSOID The whole point of Fourier analysis is that any signal can be decomposed into sinusoids real or complex Therefore lets explore sampling by looking at some real sinusoids sampled above below and at the Nyquist rate In each example a sample occurs at time t 0 This sets a definite phase relationship between an exactly described mathe matical signal and the way it is sampled This is arbitrary but there must always be a samplingtime reference and when we get to sampling for finite times the first sample will always be at time t 0 unless otherwise stated Also in the usual use of the DFT in DSP the first sample is normally assumed to occur at time t 0 Case 1 A cosine sampled at a rate that is four times its frequency or at twice its Nyquist rate Figure 1027 Figure 1028 Cosine sampled at its Nyquist rate n xt xn Figure 1029 Sinusoid with same samples as a cosine sampled at its Nyquist rate n xn Figure 1027 Cosine sampled at twice its Nyquist rate n xt xn It is clear here that the sample values and the knowledge that the signal is sampled fast enough are adequate to uniquely describe this sinusoid No other sinusoid of this or any other frequency below the Nyquist frequency could pass exactly through all the samples in the full time range n In fact no other signal of any kind that is bandlimited to below the Nyquist frequency could pass exactly through all the samples Case 2 A cosine sampled at twice its frequency or at its Nyquist rate Figure 1028 Is this sampling adequate to uniquely determine the signal No Consider the sinusoi dal signal in Figure 1029 which is of the same frequency and passes exactly through the same samples This is a special case that illustrates the subtlety mentioned earlier in the sampling theorem To be sure of exactly reconstructing any general signal from its samples the sampling rate must be more than the Nyquist rate instead of at least the Nyquist rate In earlier examples it did not matter because the signal power at exactly the Nyquist frequency was zero no impulse in the amplitude spectrum there If there is a sinusoid rob28124ch10446508indd 464 041216 146 pm 102 ContinuousTime Sampling 465 in a signal exactly at its bandlimit the sampling must exceed the Nyquist rate for exact reconstruction in general Notice that there is no ambiguity about the frequency of the signal But there is ambiguity about the amplitude and phase as illustrated above If the sincfunctioninterpolation procedure derived earlier were applied to the samples in Figure 1029 the cosine in Figure 1028 that was sampled at its peaks would result Any sinusoid at some frequency can be expressed as the sum of an unshifted cosine of some amplitude at the same frequency and an unshifted sine of some amplitude at the same frequency The amplitudes of the unshifted sine and cosine depend on the phase of the original sinusoid Using a trigonometric identity A cos2π f 0 t θ A cos2π f 0 t cosθ A sin2π f 0 t sinθ A cos2π f 0 t θ A cosθ A c cos2π f 0 t A sinθ A s sin2π f 0 t A cos2π f 0 t θ A c cos2π f 0 t A s sin2π f 0 t When a sinusoid is sampled at exactly the Nyquist rate the sincfunction interpolation always yields the cosine part and drops the sine part an effect of aliasing The cosine part of a general sinusoid is often called the inphase part and the sine part is often called the quadrature part The dropping of the quadrature part of a sinusoid can easily be seen in the time domain by sampling an unshifted sine function at exactly the Nyquist rate All the samples are zero Figure 1030 Figure 1030 Sine sampled at its Nyquist rate n xt xn If we were to add a sine function of any amplitude at exactly this frequency to any signal and then sample the new signal the samples would be the same as if the sine function were not there because its value is zero at each sample time Figure 1031 Therefore the quadrature or sine part of a signal that is at exactly the Nyquist frequency is lost when the signal is sampled Case 3 A sinusoid sampled at slightly above the Nyquist rate Figure 1032 Now because the sampling rate is higher than the Nyquist rate the samples do not all occur at zero crossings and there is enough information in the samples to reconstruct the signal There is only one sinusoid whose frequency is less than the Nyquist fre quency of a unique amplitude phase and frequency that passes exactly through all these samples Case 4 Two sinusoids of different frequencies sampled at the same rate with the same sample values Figure 1033 In this case the lowerfrequency sinusoid is oversampled and the higherfrequency sinusoid is undersampled This illustrates the ambiguity caused by undersampling If we only had access to the samples from the higherfrequency sinusoid and we believed that the signal had been properly sampled according to the sampling theorem we would interpret them as having come from the lowerfrequency sinusoid rob28124ch10446508indd 465 041216 146 pm C h a p t e r 10 Sampling and Signal Processing 466 Figure 1032 Sine sampled at slightly above its Nyquist rate n xt xn Figure 1031 Effect on samples of adding a sine at the Nyquist frequency 2 1 1 2 n 2 1 1 2 n 2 1 1 2 xnAsinπn n Asinπn xt xn xnTs Asin2π fs2t xtAsin2πfs2t Figure 1033 Two sinusoids of different frequencies that have the same sample values n xn If a sinusoid x 1 t A cos2π f 0 t θ is sampled at a rate f s the samples will be the same as the samples from another sinusoid x 2 t A cos2π f 0 k f s t θ where k is any integer including negative integers This can be shown by expanding the argu ment of x 2 t A cos2π f 0 t 2πk f s t θ The samples occur at times n T s where n is an integer Therefore the nth sample values of the two sinusoids are x1n T s A cos2π f 0 n T s θ and x2n T s A cos2π f 0 n T s 2πk f s n T s θ rob28124ch10446508indd 466 041216 146 pm 102 ContinuousTime Sampling 467 and since f s T s 1 the second equation simplifies to x 2 n T s A cos2π f 0 n T s 2kπn θ Since kn is the product of integers and therefore also an integer and since adding an integer multiple of 2π to the argument of a sinusoid does not change its value x 2 n T s A cos2π f 0 n T s 2kπn θ A cos2π f 0 n T s θ x 1 n T s BANDLIMITED PERIODIC SIGNALS In a previous section we saw what the requirements were for adequately sampling a sig nal We also learned that in general for perfect reconstruction of the signal infinitely many samples are required Since any DSP system has a finite storage capability it is important to explore methods of signal analysis using a finite number of samples There is one type of signal that can be completely described by a finite number of samples a bandlimited periodic signal Knowledge of what happens in one period is sufficient to describe all periods and one period is finite in duration Figure 1034 Figure 1034 A bandlimited periodic continuoustime signal and a discretetime signal formed by sampling it eight times per fundamental period n xn n N0 t xt t T0 Therefore a finite number of samples over one period of a bandlimited periodic signal taken at a rate that is above the Nyquist rate and is also an integer multiple of the fundamental frequency is a complete description of the signal Making the sampling rate an integer multiple of the fundamental frequency ensures that the sam ples from any fundamental period are exactly the same as the samples from any other fundamental period Let the signal formed by sampling a bandlimited periodic signal xt above its Nyquist rate be the periodic signal x s n and let an impulsesampled version of xt sampled at the same rate be x δ t Figure 1035 Only one fundamental period of samples is shown in Figure 1035 to emphasize that one fundamental period of samples is enough to completely describe the bandlim ited periodic signal We can find the appropriate Fourier transforms of these signals Figure 1036 The CTFT of xt consists only of impulses because it is periodic and it consists of a finite number of impulses because it is bandlimited So a finite number of numbers completely characterizes the signal in both the time and frequency domains If we mul tiply the impulse strengths in X f by the sampling rate f s we get the impulse strengths in the same frequency range of X δ f rob28124ch10446508indd 467 041216 146 pm C h a p t e r 10 Sampling and Signal Processing 468 ExamplE 105 Finding a CTFS harmonic function from a DFT harmonic function Find the CTFS harmonic function for the signal xt 4 2 cos20πt 3 sin40πt by sampling above the Nyquist rate at an integer multiple of the fundamental frequency over one fundamental period and finding the DFT harmonic function of the samples There are exactly three frequencies present in the signal 0 Hz 10 Hz and 20 Hz Therefore the highest frequency present in the signal is f m 20 Hz and the Nyquist rate is 40 samplessecond The fundamental frequency is the greatest common divisor of 10 Hz and 20 Hz which is 10 Hz So we must sample for 110 second If we were to sample at the Nyquist rate for exactly one funda mental period we would get four samples If we are to sample above the Nyquist rate at an integer multiple of the fundamental frequency we must take five or more samples in one fundamental period To keep the calculations simple we will sample eight times in one fundamental period a sampling rate of 80 samplessecond Then beginning the sampling at time t 0 the samples are x0 x1 x7 6 1 2 4 7 2 2 1 2 4 7 2 Using the formula for finding the DFT harmonic function of a discretetime function Xk nN0 xn e j2πkn N 0 Figure 1035 A bandlimited periodic continuoustime signal a discretetime signal and a continuoustime impulse signal created by sampling it above its Nyquist rate t 02 05 05 T0 n 26 05 05 N0 t 02 05 05 T0 xt xδt xδn Figure 1036 Magnitudes of the Fourier transforms of the three timedomain signals of Figure 1035 f 390 390 Xf 016 F 3 3 XδF 016 f 390 390 Xδ f 208 CTFT DTFT CTFT rob28124ch10446508indd 468 041216 146 pm 102 ContinuousTime Sampling 469 we get X0 X1 X7 32 8 j12 0 0 0 j12 8 The righthand side of this equation is one fundamental period of the DFT harmonic function Xk of the function xn Finding the CTFS harmonic function of xt 4 2 cos20πt 3 sin40πt directly using c x k 1 T 0 T 0 xt e j2πkt T 0 dt we get c x 4 c x 3 c x 4 0 0 j32 1 4 1 j32 0 0 From the two results 1N times the values X0 X1 X2 X3 X4 in the DFT harmonic function and the CTFS harmonic values c x 0 c x 1 c x 2 c x 3 c x 4 are the same and using the fact that Xk is periodic with fundamental period 8 18X4 X3 X2 X 1 and cx4 c x 3 c x 2 c x 1 are the same also Now lets violate the sampling theorem by sampling at the Nyquist rate In this case there are four samples in one fundamental period x0 x1 x2x3 6 4 2 4 and one period of the DFT harmonic function is X0 X1 X2 X3 16 4 0 4 The nonzero values of the CTFS harmonic function are the set c x 2 c x 1 c x 2 j32 1 4 1 j32 The j32 for c x 2 is missing from the DFT harmonic function because X2 0 This is the amplitude of the sine function at 40 Hz This is a demonstration that when we sample a sine function at exactly the Nyquist rate we dont see it in the samples because we sample it exactly at its zero crossings A thoughtful reader may have noticed that the description of a signal based on samples in the time domain from one fundamental period consists of a finite set of numbers x s n n 0 n n 0 N which contains N independent real numbers and the corresponding DFT harmonicfunction description of the signal in the frequency domain consists of the finite set of numbers X s k k 0 k k 0 N which contains N complex numbers and therefore 2N real numbers two real numbers for each complex number the real and imaginary parts So it might seem that the description in the time domain is more efficient than in the frequency domain since it is accomplished with fewer real numbers But how can this be when the set X s k k 0 k k 0 N is calcu lated directly from the set x s n n 0 n n 0 N with no extra information A closer examination of the relationship between the two sets of numbers will reveal that this apparent difference is an illusion As first discussed in Chapter 7 X s 0 is always real It can be computed by the DFT formula as Xs0 nN xsn rob28124ch10446508indd 469 041216 146 pm C h a p t e r 10 Sampling and Signal Processing 470 Since all the x s n s are real X s 0 must also be real because it is simply the sum of all the x s n s So this number never has a nonzero imaginary part There are two cases to consider next N even and N odd Case 1 N even For simplicity and without loss of generality in X s k nN x s n e jπknN n k 0 k 0 N1 x s n e jπknN let k 0 N2 Then X s k 0 X s N2 nN x s n e jπn nN x s n 1 n and X s N2 is guaranteed to be real All the DFT harmonic function values in one period other than X s 0 and X s N2 occur in pairs X s k and X s k Next recall that for any real x s n X s k X s k That is once we know X s k we also know X s k So even though each X s k contains two real numbers and each X s k does also X s k does not add any information since we already know that X s k X s k X s k is not independent of X s k So now we have as independent numbers X s 0 X s N2 and X s k for 1 k N2 All the X s ks from k 1 to k N2 1 yield a total of 2N2 1 N 2 independent real numbers Add the two guaranteedreal values X s 0 and X s N2 and we finally have a total of N independent real numbers in the frequencydomain description of this signal Case 2 N odd For simplicity and without loss of generality let k 0 N 12 In this case we simply have X s 0 plus N 12 complex conjugate pairs X s k and X s k We have already seen that X s k X s k So we have the real number X s 0 and two indepen dent real numbers per complex conjugate pair or N 1 independent real numbers for a total of N independent real numbers The information content in the form of independent real numbers is conserved in the process of converting from the time to the frequency domain SIGNAL PROCESSING USING THE DFT CTFTDFT Relationship In the following development of the relationship between the CTFT and the DFT all the processing steps from the CTFT of the original function to the DFT will be illus trated by an example signal Then several uses of the DFT are developed for signal processing operations We will use the F form of the DTFT because the transform relationships are a little more symmetrical than in the Ω form Let a signal xt be sampled and let the total number of samples taken be N where N T f s T is the total sampling time and f s is the sampling frequency Then the time between samples is T s 1 f s Below is an example of an original signal in both the time and frequency domains Figure 1037 The first processing step in converting from the CTFT to the DFT is to sample the signal xt to form a signal x s n xn T s The frequencydomain counterpart of the discretetime function is its DTFT In the next section we will look at the relation between these two transforms rob28124ch10446508indd 470 041216 146 pm 102 ContinuousTime Sampling 471 CTFTDTFT Relationship The CTFT is the Fourier transform of a continuoustime function and the DTFT is the Fourier transform of a discretetime function If we mul tiply a continuoustime function xt by a periodic impulse of period T s we create the continuoustime impulse function x δ t xt δ T s t n xn T s δt n T s 103 If we now form a function x s n whose values are the values of the original continuous time function xt at integer multiples of T s and are therefore also the strengths of the impulses in the continuoustime impulse function x δ t we get the relationship x s n xn T s The two functions x s n and x δ t are described by the same set of numbers the impulse strengths and contain the same information If we now find the CTFT of 103 we get X δ f X f f s δ f s f n xn T s e j2πfn T s where f s 1 T s and xt ℱ X f or X δ f f s k X f k f s n x s n e j2πfn f s If we make the change of variable f f s F we get X δ f s F f s k X f s F k n x s n e j2πnF X s F The last expression is exactly the definition of the DTFT of x s n which is X s F Summarizing if xsn xnTs and xδ t n x s nδt n T s then X s F X δ f s F 104 Figure 1037 An original signal and its CTFT xt 1 1 t 28 X f 36329 Original Signal f 0381 X f π π f rob28124ch10446508indd 471 041216 146 pm C h a p t e r 10 Sampling and Signal Processing 472 or X δ f X s f f s 105 Also X s F f s k X f s F k 106 Figure 1038 Now we can write the DTFT of x s n which is X s F in terms of the CTFT of xt which is X f It is X s F f s X f s F δ 1 F f s k X f s F k a frequencyscaled and periodically repeated version of X f Figure 1039 Next we must limit the number of samples to those occurring in the total discretetime sampling time N Let the time of the first sample be n 0 This is the default assumption in the DFT Other time references could be used but the effect of a different time reference is simply a phase shift that varies linearly with frequency This can be accomplished by multiplying x s n by a window function wn 1 0 n N 0 otherwise Figure 1038 Fourier spectra of original signal impulsesampled signal and sampled signal n xn 1 5 5 3 1 A 3 Ts t xδt A Ts t xt A 1 F XF 1 2A DTFT fs fs f Xδ f 2A CTFT fs fs f X f 2Ts A CTFT ℱ ℱ ℱ rob28124ch10446508indd 472 041216 146 pm 102 ContinuousTime Sampling 473 as illustrated in Figure 1040 This window function has exactly N nonzero values the first one being at discrete time n 0 Call the sampledandwindowed signal x sw n Then x sw n wn x s n x s n 0 n N 0 otherwise The process of limiting a signal to the finite range N in discrete time is called windowing because we are considering only that part of the sampled signal that can be seen through a window of finite length The window function need not be a rect angle Other window shapes are often used in practice to minimize an effect called leakage described below in the frequency domain The DTFT of x sw n is the peri odic convolution of the DTFT of the signal x s n and the DTFT of the window function wn which is X sw F WF X s F The DTFT of the window function is WF e jπFN1 N drclF N Then X sw F e jπFN1 N drclF N f s k X f s F k or using the fact that periodic convolution with a periodic signal is equivalent to ape riodic convolution with any aperiodic signal that can be periodically repeated to form the periodic signal X sw F f s e jπFN1 N drclF N X f s F 107 So the effect in the frequency domain of windowing in discretetime is that the Fourier transform of the timesampled signal has been periodically convolved with WF e jπFN1 N drclF N Figure 1041 Figure 1040 Original signal timesampled and windowed to form a discretetime signal and the DTFT of that discretetime signal xswn 1 1 n 63 XswF 29154 F 2 2 π π F 2 2 s XswF Figure 1039 Original signal timesampled to form a discretetime signal and the DTFT of the discretetime signal xsn 1 1 n 63 XsF 83039 F 2 2 π π F 2 2 XsF rob28124ch10446508indd 473 041216 146 pm C h a p t e r 10 Sampling and Signal Processing 474 The convolution process will tend to spread X s F in the frequency domain which causes the power of X s F at any frequency to leak over into nearby frequencies in X sw F This is where the term leakage comes from The use of a different window function whose DTFT is more confined in the frequency domain reduces but can never completely eliminate leakage As can be seen in Figure 1041 as the number of samples N increases the width of the main lobe of each fundamental period of this function decreases reducing leakage So another way to reduce leakage is to use a larger set of samples At this point in the process we have a finite sequence of numbers from the sampledandwindowed signal but the DTFT of the windowed signal is a periodic function in continuous frequency F and therefore not appropriate for computer storage and manipulation The fact that the timedomain function has become time limited by the windowing process and the fact that the frequencydomain function is periodic allow us to sample now in the frequency domain over one fundamental period to com pletely describe the frequencydomain function It is natural at this point to wonder how a frequencydomain function must be sampled to be able to reconstruct it from its samples The answer is almost identical to the answer for sampling timedomain signals except that time and frequency have exchanged roles The relations between the time and frequency domains are almost identical because of the duality of the forward and inverse Fourier transforms Sampling and PeriodicRepetition Relationship The inverse DFT of a periodic function xn with fundamental period N is defined by xn 1 N kN Xk e j2πknN 108 Taking the DTFT of both sides using the DTFT pair e j2π F 0 n ℱ δ1F F 0 we can find the DTFT of xn yielding XF 1 N kN Xk δ1F kN 109 WF 32 N 8 F WF 32 N 16 F 32 N 32 F 1 1 1 WF Figure 1041 Magnitude of the DTFT of the rectangular window function wn 1 0 n N 0 otherwise for three different window widths rob28124ch10446508indd 474 041216 146 pm 102 ContinuousTime Sampling 475 Then XF 1 N kN Xk q δF kN q 1 N k XkδF kN 1010 This shows that for periodic functions the DFT is simply a scaled special case of the DTFT If a function xn is periodic its DTFT consists only of impulses occurring at kN with strengths XkN Figure 1042 Figure 1042 Harmonic function and DTFT of xn A21 cos2πn4 xn N0 k Xk 1 1 3 7 5 3 5 7 F n XF A DFT DTFT 1 1 AN0 2 A 2 ℱ풮 ℱ Summarizing for a periodic function xn with fundamental period N XF 1 N k XkδF kN 1011 Let xn be an aperiodic function with DTFT XF Let x p n be a periodic exten sion of xn with fundamental period N p such that x p n m xn m N p xn δ N p n 1012 Figure 1043 Using the multiplicationconvolution duality of the DTFT and finding the DTFT of 1012 X p F XF1 N p δ 1 N p F 1 N p k Xk N p δF k N p 1013 Using 1011 and 1013 X p k Xk N p 1014 rob28124ch10446508indd 475 041216 146 pm C h a p t e r 10 Sampling and Signal Processing 476 where X p k is the DFT of x p n If an aperiodic signal xn is periodically repeated with fundamental period N p to form a periodic signal x p n the values of its DFT har monic function X p k can be found from XF which is the DTFT of xn evaluated at the discrete frequencies k N p This forms a correspondence between sampling in the frequency domain and periodic repetition in the time domain If we now form a periodic repetition of x sw n x swp n m x sw n mN with fundamental period N its DFT is X swp k X sw kN k an integer or from 107 X swp k f s e jπFN1 N drclF N X f s F FkN The effect of the last operation sampling in the frequency domain is sometimes called picket fencing Figure 1044 Since the nonzero length of x sw n is exactly N x swp n is a periodic repetition of x sw n with a fundamental period equal to its length so the multiple replicas of x sw n do not overlap but instead just touch Therefore x sw n can be recovered from x swp n by simply isolating one fundamental period of x swp n in the discretetime range 0 n N Figure 1043 A signal and its DTFT and the periodic repetition of the signal and its DFT harmonic function xn xpn 1 1 8 π π π π Signal xn n 64 XF 8 F F XF 2 2 Periodically Repeated Signal xpn n 64 Xpk k 32 32 2 2 Xpk k 32 32 rob28124ch10446508indd 476 041216 146 pm 102 ContinuousTime Sampling 477 The result X swp k f s e jπFN1 N drclF N X f s F FkN is the DFT of a periodic extension of the discretetime signal formed by sampling the original signal over a finite time In summary in moving from the CTFT of a continuoustime signal to the DFT of samples of the continuoustime signal taken over a finite time we do the following In the time domain 1 Sample the continuous time signal 2 Window the samples by multiplying them by a window function 3 Periodically repeat the nonzero samples from step 2 In the frequency domain 1 Find the DTFT of the sampled signal which is a scaled and periodically repeated version of the CTFT of the original signal 2 Periodically convolve the DTFT of the sampled signal with the DTFT of the window function 3 Sample in frequency the result of step 2 The DFT and inverse DFT being strictly numerical operations form an exact correspondence between a set of N real numbers and a set of N complex numbers If the set of real numbers is a set of N signal values over exactly one period of a periodic discretetime signal xn then the set of N complex numbers is a set of complex ampli tudes over one period of the DFT Xk of that discretetime signal These are the com plex amplitudes of complex discretetime sinusoids which when added will produce the periodic discretetime signal N xn Figure 1044 Original signal timesampled windowed and periodically repeated to form a periodic discretetime signal and the DFT of that signal xswpn 1 1 n 63 Xswpk 29154 Sampled Windowed and Periodically Repeated Signal k 32 32 π π k 32 32 Xswpk rob28124ch10446508indd 477 041216 146 pm C h a p t e r 10 Sampling and Signal Processing 478 If the set of N real numbers is a set of samples from one period of a bandlimited periodic continuoustime signal sampled above its Nyquist rate and at a rate that is an integer multiple of its fundamental frequency the numbers returned by the DFT can be scaled and interpreted as complex amplitudes of continuoustime complex sinusoids which when added will recreate the periodic continuoustime signal So when using the DFT in the analysis of periodic discretetime signals or bandlimited periodic continuoustime signals we can obtain results that can be used to exactly compute the DTFS or DTFT or CTFS or CTFT of the periodic signal When we use the DFT in the analysis of aperiodic signals we are inherently making an approximation because the DFT and inverse DFT are only exact for periodic signals If the set of N real numbers represents all or practically all the nonzero values of an aperiodic discretetime energy signal we can find an approximation to the DTFT of that signal at a set of discrete frequencies using the results returned by the DFT If the set of N real numbers represents samples from all or practically all the nonzero range of an aperiodic continuoustime signal we can find an approximation to the CTFT of that continuoustime signal at a set of discrete frequencies using the results returned by the DFT Computing the CTFS Harmonic Function with the DFT It can be shown that if a signal xt is periodic with fundamental frequency f 0 and if it is sampled at a rate f s that is above the Nyquist rate and if the ratio of the sampling rate to the fundamental frequency f s f 0 is an integer that the DFT of the samples Xk is related to the CTFS harmonic function of the signal c x k by Xk N c x k δ N k In this special case the relationship is exact Approximating the CTFT with the DFT Forward CTFT In cases in which the signal to be transformed is not readily de scribable by a mathematical function or the Fouriertransform integral cannot be done analytically we can sometimes find an approximation to the CTFT numerically using the DFT If the signal to be transformed is a causal energy signal it can be shown that we can approximate its CTFT at discrete frequencies k f s N by Xk f s N T s n0 N1 xn T s e j2πknN T s 𝒟ℱ𝒯xn T s k N 1015 where T s 1 f s and N is chosen such that the time range 0 to N T s covers all or practically all of the signal energy of the signal x Figure 1045 So if the signal to be transformed Figure 1045 A causal energy signal sampled with T s seconds between samples over a time N T s t xt n N n 0 Ts rob28124ch10446508indd 478 041216 146 pm 102 ContinuousTime Sampling 479 is a causal energy signal and we sample it over a time containing practically all of its energy the approximation in 1015 becomes accurate for k N Inverse CTFT The inverse CTFT is defined by xt X f e j2πft df If we know Xk f s N in the range N k max k k max N and if the magnitude of Xk f s N is negligible outside that range then it can be shown that for n N xn T s f s 𝒟ℱ𝒯1 X ext k f s N where X ext k f s N X k f s N k max k k max 0 k max k N2 and X ext k f s N X ext k mN f s N Approximating the DTFT with the DFT The numerical approximation of the DTFT using the DFT was derived in Chapter 7 The DTFT of xn computed at frequencies F kN or Ω 2πkN is Xk N 𝒟ℱ𝒯xn 1016 Approximating ContinuousTime Convolution with the DFT Aperiodic Convolution Another common use of the DFT is to approximate the con volution of two continuoustime signals using samples from them Suppose we want to convolve two aperiodic energy signals xt and ht to form yt It can be shown that for n N yn T s T s 𝒟ℱ𝒯1𝒟ℱ𝒯xn T s 𝒟ℱ𝒯hn T s 1017 Periodic Convolution Let xt and ht be two periodic continuoustime signals with a common period T and sample them over exactly that time at a rate f s above the Nyquist rate taking N samples of each signal Let yt be the periodic convolution of xt with ht Then it can be shown that yn T s T s 𝒟ℱ𝒯1𝒟ℱ𝒯xn T s 𝒟ℱ𝒯hn T s 1018 DiscreteTime Convolution with the DFT Aperiodic Convolution If xn and hn are energy signals and most or all of their energy occurs in the time range 0 n N then it can be shown that for n N yn 𝒟ℱ𝒯1𝒟ℱ𝒯xn 𝒟ℱ𝒯hn 1019 Periodic Convolution Let xn and hn be two periodic signals with a common period N Let yn be the periodic convolution of xn with hn Then it can be shown that yn 𝒟ℱ𝒯1𝒟ℱ𝒯xn 𝒟ℱ𝒯hn 1020 rob28124ch10446508indd 479 041216 146 pm C h a p t e r 10 Sampling and Signal Processing 480 Summary of signal processing using the DFT CTFS cxk ejπkN sinck N N Xk k N CTFS Xk N cxk δN k if fs fNyq and fs f0 is an integer CTFT Xkfs N Ts 𝒟ℱ𝒯xnTs DTFT XkN 𝒟ℱ𝒯xn ContinuousTime xt httnTs Ts 𝒟ℱ𝒯1𝒟ℱ𝒯xnTs 𝒟ℱ𝒯hnTs Aperiodic Convolution DiscreteTime xn hn 𝒟ℱ𝒯1𝒟ℱ𝒯xn 𝒟ℱ𝒯hn Aperiodic Convolution ContinuousTime xt httnTs Ts 𝒟ℱ𝒯1𝒟ℱ𝒯xnTs 𝒟ℱ𝒯hnTs Periodic Convolution DiscreteTime xn hn 𝒟ℱ𝒯1𝒟ℱ𝒯xn 𝒟ℱ𝒯hn Periodic Convolution Figure 1046 Sixteen samples taken from a continuoustime signal xn 1496 13356 n 15 What do we know so far We know the value of xt at 16 points in time over a time span of 16 ms We dont know what signal values preceded or followed xt We also dont know what values occurred between the samples we acquired So to draw any reasonable conclusions about xt and its CTFT we will need more information Suppose we know that xt is bandlimited to less than 500 Hz If it is bandlimited it cannot be time limited so we know that outside the time over which we acquired the data the signal values were not all zero In fact they cannot be any constant because if they were we could subtract that constant from the signal creating a timelimited signal which cannot be bandlimited The signal values outside the 16 ms time range could vary in many different ways or could repeat in a periodic pattern If they repeat in a periodic pattern with this set of 16 values as the fundamental period then xt is a bandlimited periodic signal and is unique It is the only bandlimited signal with that fundamental period that could have produced the samples The samples and the DFT of the samples form a DFT pair xn 𝒟ℱ𝒯 16 Xk The CTFS harmonic function c x k can be found from the DFT through Xk N c x k δ N k if f s f Nyq and f s f 0 is an integer A typical use of the DFT is to estimate the CTFT of a continuoustime signal using only a finite set of samples taken from it Suppose we sample a continuoustime signal xt 16 times at a 1000 samplessecond rate and acquire the samples xn illustrated in Figure 1046 rob28124ch10446508indd 480 041216 146 pm 103 DiscreteTime Sampling 481 and xt can therefore be recovered exactly Also the CTFT is a set of impulses spaced apart by the signals fundamental frequency whose strengths are the same as the values of the CTFS harmonic function Now lets make a different assumption about what happened outside the 16ms time of the sample set Suppose we know that xt is zero outside the 16ms range over which we sampled Then it is time limited and cannot be bandlimited so we can not exactly satisfy the sampling theorem But if the signal is smooth enough and we sample fast enough it is possible that the amount of signal energy in the CTFT above the Nyquist frequency is negligible and we can compute good approximations of the CTFT of xt at a discrete set of frequencies using Xk f s N T s 𝒟ℱ𝒯xn T S 103 DISCRETETIME SAMPLING PERIODICIMPULSE SAMPLING In the previous sections all the signals that were sampled were continuoustime sig nals Discretetime signals can also be sampled Just as in sampling continuoustime signals the main concern in sampling discretetime signals is whether the information in the signal is preserved by the sampling process There are two complementary pro cesses used in discretetime signal processing to change the sampling rate of a signal decimation and interpolation Decimation is a process of reducing the number of samples and interpolation is a process of increasing the number of samples We will consider decimation first We impulsesampled a continuoustime signal by multiplying it by a continuoustime periodic impulse Analogously we can sample a discretetime signal by multiplying it by a discretetime periodic impulse Let the discretetime signal to be sampled be xn Then the sampled signal would be x s n xn δ N s n where N s is the discrete time between samples Figure 1047 The DTFT of the sampled signal is X s F XF F s δ F s F F s 1 N s Figure 1048 The similarity of discretetime sampling to continuoustime sampling is obvious In both cases if the aliases do not overlap the original signal can be recovered from the samples and there is a minimum sampling rate for recovery of the signals The sampling rate must satisfy the inequality F s 2 F m where F m is the discretetime cyclic frequency above which the DTFT of the original discretetime signal is zero in the base fundamental period F 12 That is for F m F 1 F m the DTFT of the original signal is zero A discretetime signal that satisfies this requirement is bandlim ited in the discretetime sense Just as with continuoustime sampling if a signal is properly sampled we can re construct it from the samples using interpolation The process of recovering the original signal is described in the discretetimefrequency domain as a lowpass filtering operation XF X s F 1 F s rectF2 F c δ 1 F rob28124ch10446508indd 481 041216 146 pm C h a p t e r 10 Sampling and Signal Processing 482 where F c is the cutoff discretetime frequency of the ideal lowpass discretetime filter The equivalent operation in the discretetime domain is a discretetime convolution xn x s n 2 F c F s sinc2 F c n In the practical application of sampling discretetime signals it does not make much sense to retain all those zero values between the sampling points because we al ready know they are zero Therefore it is common to create a new signal x d n which has only the values of the discretetime signal x s n at integer multiples of the sampling interval N s The process of forming this new signal is called decimation Decimation was briefly discussed in Chapter 3 The relations between the signals are given by x d n x s N s n x N s n This operation is discretetime time scaling which for N s 1 causes discretetime time compression and the corresponding effect in the discretetimefrequency domain is discretetime frequency expansion The DTFT of x d n is X d F n x d n e j2πFn n x s N s n e j2πFn We can make a change of variable m N s n yielding X d F m minteger multiple of N s x s m e j2πFm N s Now taking advantage of the fact that all the values of x s n between the allowed values m integer multiple of N s are zero we can include the zeros in the summation yielding X d F m x s m e j2πF N s m X s F N s Figure 1047 An example of discretetime sampling n xn n xsn n δ4n 1 Figure 1048 DTFT of discretetime signal and a sampled version of it F XF Fm 1 1 A F XsF Fs 1 1 AFs F Fs Fs 1 1 Fs δFsF rob28124ch10446508indd 482 041216 146 pm 103 DiscreteTime Sampling 483 So the DTFT of the decimated signal is a discretetimefrequencyscaled version of the DTFT of the sampled signal Figure 1049 Observe carefully that the DTFT of the decimated signal is not a discretetimefrequency scaled version of the DTFT of the original signal but rather a discretetimefrequency scaled version of the DTFT of the discretetimesampled original signal X d F X s F N s XF N s The term downsampling is sometimes used instead of decimation This term comes from the idea that the discretetime signal was produced by sampling a continuoustime signal If the continuoustime signal was oversampled by some factor then the discretetime signal can be decimated by the same factor without losing information about the original continuoustime signal thus reducing the effective sampling rate or downsampling INTERPOLATION The opposite of decimation is interpolation or upsampling The process is simply the reverse of decimation First extra zeros are placed between samples then the signal so cre ated is filtered by an ideal discretetime lowpass filter Let the original discretetime signal be xn and let the signal created by adding N s 1 zeros between samples be x s n Then x s n xn N s n N s an integer 0 otherwise This discretetime expansion of xn to form x s n is the exact opposite of the discretetime compression of x s n to form x d n in decimation so we should expect the effect in the discretetimefrequency domain to be the opposite also A discretetime expansion by a factor of N s creates a discretetimefrequency compression by the same factor X s F X N s F Figure 1049 Comparison of the discretetimedomain and discretetimefrequency domain effects of sampling and decimation n xn n xsn n xdn F XF Fm 1 1 A F XsF Fs 1 1 F XdF Fs 1 1 AFs AFs rob28124ch10446508indd 483 041216 146 pm C h a p t e r 10 Sampling and Signal Processing 484 Figure 1050 The signal x s n can be lowpass filtered to interpolate between the nonzero values If we use an ideal unitygain lowpass filter with a transfer function HF rect N s F δ 1 F we get an interpolated signal X i F X s F rect N s F δ 1 F and the equivalent in the discretetime domain is x i n x s n 1 N s sincn N s Figure 1051 Figure 1050 Effects in both the discretetime and discretetimefrequency domains of inserting N s 1 zeros between samples Ns 1 n xsn n xn F XF 1 1 A F XsF 1 1 A Figure 1051 Comparison of the discretetimedomain and discretetimefrequency domain effects of expansion and interpolation Ns 1 n xsn n xin n xn F XF 1 1 A F XsF 1 1 A F XiF 1 1 A rob28124ch10446508indd 484 041216 146 pm 103 DiscreteTime Sampling 485 Notice that the interpolation using the unitygain ideal lowpass filter introduced a gain factor of 1 N s reducing the amplitude of the interpolated signal x i n relative to the original signal xn This can be compensated for by using an ideal lowpass filter with a gain of N s HF N s rect N s F δ 1 F instead of unity gain ExamplE 106 Sample the signal xt 5 sin2000πt cos20000πt at 80 kHz over one fundamental period to form a discretetime signal xn Take every fourth sample of xn to form x s n and decimate x s n to form x d n Then upsample x d n by a factor of eight to form x i n Figure 1052 and Figure 1053 Figure 1052 Original sampled and decimated discretetime signals and their DTFTs xn 5 5 n 96 xdn 5 5 n 96 xsn 5 5 n 96 XF F 1 1 5 4 XdF F 1 1 XsF F 1 1 5 16 5 4 rob28124ch10446508indd 485 041216 146 pm C h a p t e r 10 Sampling and Signal Processing 486 104 SUMMARY OF IMPORTANT POINTS 1 A sampled or impulsesampled signal has a Fourier spectrum that is a periodically repeated version of the spectrum of the signal sampled Each repetition is called an alias 2 If the aliases in the spectrum of the sampled signal do not overlap the original signal can be recovered from the samples 3 If the signal is sampled at a rate more than twice its highest frequency the aliases will not overlap 4 A signal cannot be simultaneously time limited and bandlimited 5 The ideal interpolating function is the sinc function but since it is noncausal other methods must be used in practice 6 A bandlimited periodic signal can be completely described by a finite set of numbers 7 The CTFT of a signal and the DFT of samples from it are related through the operations sampling in time windowing and sampling in frequency 8 The DFT can be used to approximate the CTFT the CTFS and other common signalprocessing operations and as the sampling rate andor number of samples are increased the approximation gets better 9 The techniques used in sampling a continuoustime signal can be used in almost the same way in sampling discretetime signals There are analogous concepts of bandwidth minimum sampling rate aliasing and so on Figure 1053 Original upsampled and discretetimelowpassfiltered discretetime signals xdn 5 5 n 192 xin n xsn n 192 192 5 5 5 5 rob28124ch10446508indd 486 041216 146 pm Exercises with Answers 487 EXERCISES WITH ANSWERS Answers to each exercise are in random order Pulse Amplitude Modulation 1 Sample the signal xt 10 sinc 500t by multiplying it by the pulse train pt rect104t δ1mst to form the signal xpt Graph the magnitude of the CTFT Xp f of xpt Answer f 20000 20000 Xf 0002 2 Let x t 10 sinc500t and form a signal xpt xtδ1mst rect104t Graph the magnitude of the CTFT Xp f of xpt Answer f 20000 20000 Xf 0002 Sampling 3 A signal x t 25 sin 200πt is sampled at 300 samplessecond with the first sample being taken at time t 0 What is the value of the fifth sample Answer 21651 4 The signal x t 30 cos 2000πt sin 50πt is sampled at a rate f s 104 samples second with the first sample occurring at time t 0 What is the value of the third sample Answer 02912 5 A continuoustime signal x1t 20 sin 100πt is undersampled at f s 40 samplessecond to form a discretetime signal x n If the samples in x n had rob28124ch10446508indd 487 041216 146 pm C h a p t e r 10 Sampling and Signal Processing 488 been instead taken from a continuoustime signal x2 t at a rate more than twice the highest frequency in x2 t what would be a correct mathematical description of x2 t Answer x2 t 20 sin 20πt 6 A continuoustime signal x t 10 sinc 25t is sampled at f s 20 samples second to form x n and x n ℱ X e jΩ a Find an expression for X e jΩ b What is the maximum numerical magnitude of Xe jΩ Answers 8 rect Ω25π δ2πΩ 16 7 Given a signal x t tri 100t form a signal x n by sampling x t at a rate f s 800 and form an informationequivalent impulse signal x δ t by multiplying x t by a periodic sequence of unit impulses whose fundamental frequency is the same f 0 f s 800 Graph the magnitude of the DTFT of xn and the CTFT of x δ t Change the sampling rate to f s 5000 and repeat Answers f 1600 1600 Xδ f 8 fs 800 Xδ f fs 5000 2 2 2 2 F 10000 10000 XF XF 8 f 50 50 F 8 Given a bandlimited signal x t sinc t4 cos 2πt form a signal x n by sampling x t at a rate f s 4 and form an informationequivalent impulse signal x δ t by multiplying x t by a periodic sequence of unit impulses whose fundamental frequency is the same f 0 f s 4 Graph the magnitude of the DTFT of x n and the CTFT of x δ t Change the sampling rate to f s 2 and repeat Answers f 8 8 Xδ f 8 fs 4 F 2 2 XF 8 f 4 4 Xδ f 8 fs 2 F 2 2 XF 8 rob28124ch10446508indd 488 041216 146 pm Exercises with Answers 489 Impulse Sampling 9 Let x δ t K tri t4 δ 4 t t 0 and let x δ t ℱ X δ f For t 0 0 X δ f X δ0 f and for t 0 2 X δ f G f X δ0 f What is the function G f Answer cos 4πf 10 Let x t 8 rect t5 let an impulsesampled version be x δ t x t δ T s t and let x δ t ℱ X δ f The functional behavior of X δ f generally depends on T s but for this signal for all values of T s above some minimum value T smin X δ f is the same What is the numerical value of T smin Answer 25 11 For each signal x t impulse sample it at the rate specified by multiplying it by a periodic impulse δ T s t T s 1 f s and graph the impulsesampled signal x δ t over the time range specified and the magnitude and phase of its CTFT X δ f over the frequency range specified a x t rect 100t fs 1100 20 ms t 20 ms 3 kHz f 3 kHz b x t rect 100t fs 110 Ts 1110 9091 ms 20 ms t 20 ms 3 kHz f 3 kHz c x t tri 45t fs 180 100 ms t 100 ms 400 f 400 Answers Xδt t 002 002 f 3000 3000 3000 f 3000 1 12 Xδ f Xδ f π π t 01 01 f 400 400 Xδt 1 Xδ f f 400 400 4 1 1 Xδ f 1 1 Xδ f t 002 002 xδt 1 f 3000 3000 2 f 3000 3000 Xδ f rob28124ch10446508indd 489 041216 146 pm C h a p t e r 10 Sampling and Signal Processing 490 12 Given a signal x t tri 200t δ 005 t impulse sample it at the rate f s specified by multiplying it by a periodic impulse of the form δ T s t T s 1 f s Then filter the impulsesampled signal xδt with an ideal lowpass filter whose gain is Ts in its passband and whose corner frequency is the Nyquist frequency Graph the signal x t and the response of the lowpass filter x f t over the time range 60 ms t 60 ms a fs 1000 b fs 200 c fs 100 Answers t 005 005 Xδt xft t 1 08 06 04 02 025 006 005 005 006 05 t xδt xδt t xft xft 1 08 06 04 02 1 08 06 04 02 005 005 005 005 t 005 t 005 006 005 006 005 025 05 099 13 Given a signal x t 8 cos 24πt 6 cos 104πt impulse sample it at the rate specified by multiplying it by a periodic impulse of the form δ T s t T s 1 f s Then filter the impulsesampled signal with an ideal lowpass filter whose gain is Ts in its passband and whose corner frequency is the Nyquist frequency Graph the signal x t and the response of the lowpass filter x i t over two fundamental periods of x i t a f s 100 b f s 50 c f s 40 Answers 15 15 02 t xt 15 15 02 t xit rob28124ch10446508indd 490 041216 146 pm Exercises with Answers 491 15 15 1 t xt 15 15 1 t xit 15 15 02 t 15 15 02 t xt xit Nyquist Rates 14 Find the Nyquist rates for these signals a x t sinc 20t b x t 4 sinc 2 100t c x t 8 sin 50πt d x t 4 sin 30πt 3 cos 70πt e x t rect 300t f x t 10 sin 40πt cos 300πt g x t sinc t2 δ 10 t h x t sinc t2 δ 01 t i x t 8tri t 4 12 j x t 13 e 20t cos 70πt u t k x t u t u t 5 Answers 70 Infinite 200 20 Infinite 04 340 Infinite Infinite Infinite 50 15 Let x t 10 cos 4πt a Is x t bandlimited Explain your answer If it is bandlimited what is its Nyquist rate b If we multiply x t by rect t to form y t is y t bandlimited Explain your answer If it is bandlimited what is its Nyquist rate c If we multiply x t by sinc t to form y t is y t bandlimited Explain your answer If it is bandlimited what is its Nyquist rate d If we multiply x t by e π t 2 to form y t is y t bandlimited Explain your answer If it is bandlimited what is its Nyquist rate Answers No No Yes 4 Hz Yes 5 Hz 16 Two sinusoids one at 40 Hz and the other at 150 Hz are combined to form a single signal x t a If they are added what is the Nyquist rate for x t b If they are multiplied what is the Nyquist rate for x t Answers 380 Hz 300 Hz rob28124ch10446508indd 491 041216 146 pm C h a p t e r 10 Sampling and Signal Processing 492 TimeLimited and Bandlimited Signals 17 A continuoustime signal x t is described by x t 4 cos 2πt sin 20πt If x t is filtered by a unitygain ideal lowpass filter with a bandwidth of 10 Hz the response is a sinusoid What are the amplitude and frequency of that sinusoid Answers 2 9 Hz 18 Graph these timelimited signals and find and graph the magnitude of their CTFTs and confirm that they are not bandlimited a x t 5 rect t100 b x t 10 tri 5t c x t rect t 1 cos 2πt d x t rect t 1 cos 2πt cos 16πt Answers t 04 04 10 xt f 20 20 2 Xf t 1 1 2 xt 2 f 12 12 1 Xf t 1 1 2 xt f 5 5 1 Xf t 200 200 5 xt f 004 004 500 Xf 19 Graph the magnitudes of these bandlimitedsignal CTFTs and find and graph their inverse CTFTs and confirm that they are not time limited a X f rect f e j4πf b X f tri 100f e jπf c X f δ f 4 δ f 4 d X f j δ f 4 δ f 4 rect 8f Answers f 6 6 1 Xf t 16 16 025 xt 025 f 002 002 1 X f t 400 400 001 xt 0005 f 4 4 1 Xf t 1 1 2 xt 2 f 1 1 1 X f t 6 6 1 xt 05 rob28124ch10446508indd 492 041216 146 pm Exercises with Answers 493 Interpolation 20 Sample the signal x t sin 2πt at a sampling rate f s Then using MATLAB graph the interpolation between samples in the time range 1 t 1 using the approximation x t 2 f c f s nN N x n T s sinc 2 f c t n T s with these combinations of f s f c and N a f s 4 f c 2 N 1 b f s 4 f c 2 N 2 c f s 8 f c 4 N 4 d f s 8 f c 2 N 4 e f s 16 f c 8 N 8 f f s 16 f c 8 N 16 Answers t 1 1 xt 1 1 t 1 1 xt 1 1 t 1 1 xt 1 1 t 1 1 xt 1 1 t 1 1 xt 1 1 t 1 1 xt 1 1 21 For each signal and specified sampling rate graph the original signal and an interpolation between samples of the signal using a zeroorder hold over the time range 1 t 1 The MATLAB function stairs could be useful here a xt sin2πt f s 8 b xt sin2πt f s 32 c xt rect t f s 8 d xt trit f s 8 Answers t 1 1 xt 1 1 t 1 1 xt 1 1 t 1 1 xt 1 t 1 1 xt 1 1 1 22 Repeat Exercise 21 except use a firstorder hold instead of a zeroorder hold Answers t 1 15 xt 1 1 t 1 15 t 1 15 xt 1 1 t 1 15 xt 1 1 xt 1 1 rob28124ch10446508indd 493 041216 146 pm C h a p t e r 10 Sampling and Signal Processing 494 23 Sample each signal x t N times at the rate f s creating the signal x n Graph x t versus t and x n versus n T s over the time range 0 t N T s Find the DFT X k of the N samples Then graph the magnitude and phase of X f versus f and T s X k versus kΔf over the frequency range f s 2 f f s 2 where Δ f f s N Graph T s X k as a continuous function of kΔf using the MATLAB plot command a xt 5 rect 2 t 2 f s 16 N 64 b xt 3 sinc t 20 5 f s 1 N 40 c xt 2 rect t 2 sin 8πt f s 32 N 128 d xt 10 tri t 2 2 tri t 6 2 f s 8 N 64 e xt 5 cos 2πt cos 16πt f s 64 N 128 Answers t 4 xt 2 2 nTs 4 xn 2 2 f 16 16 X f 1 f 16 16 π π kfsNF 16 16 TsXk 1 kfsNF 16 16 π π X f TsXk t 4 xt 5 nTs 4 xn 5 f 8 8 X f 25 f 8 8 π π kfsNF 8 8 TsXk 25 kfsNF 8 8 π π X f TsXk t 2 xt 5 5 nTs 2 xn 5 5 f 32 32 X f 125 f 32 32 π π kfsNF 32 32 125 kfsNF 32 32 π π X f TsXk TsXk t 40 xt 1 3 nTs 40 xn 1 3 f 05 05 X f 15 f 05 05 π π kfsNF 05 05 TsXk 15 kfsNF 05 05 π π X f TsXk t 8 xt 10 10 nTs 8 xn 10 10 f 4 4 X f 40 f 4 4 π π kfsNF 4 4 TsXk 40 kfsNF 4 4 π π X f TsXk rob28124ch10446508indd 494 041216 146 pm Exercises with Answers 495 Aliasing 24 For each pair of signals below sample at the specified rate and find the DTFT of the sampled signals In each case explain by examining the DTFTs of the two signals why the samples are the same a x1 t 4 cos 16πt and x2 t 4 cos 76πt f s 30 b x1 t 6 sinc 8t and x 2 t 6 sinc 8t cos 400πt f s 100 c x1 t 9 cos 14πt and x2 t 9 cos 98πt f s 56 25 For each sinusoid find the two other sinusoids whose frequencies are nearest to the frequency of the given sinusoid and which when sampled at the specified rate have exactly the same samples a x t 4 cos 8πt f s 20 b x t 4 sin 8πt f s 20 c x t 2 sin 20πt f s 50 d x t 2 cos 20πt f s 50 e x t 5 cos 30πt π4 f s 50 Answers 4 cos 48πt and 4 cos 32πt 2 sin 80πt and 2 sin 120πt 5 cos 130πt π4 and 5 cos 70πt π4 2 cos 80πt and 2 cos 120πt 4 sin 48πt and 4 sin 32πt Bandlimited Periodic Signals 26 Sample the following signals x t to form signals x n Sample at the Nyquist rate and then at the next higher rate for which f s f 0 is an integer which implies that the total sampling time divided by the time between samples is also an integer Graph the signals and the magnitudes of the CTFTs of the continuoustime signals and the DTFTs of the discretetime signals a x t 2 sin 30πt 5 cos 18πt b x t 6 sin 6πt cos 24πt Answers t 025 025 xt 8 8 n 8 8 xNyqn 5 5 n 8 8 x11n 5 5 f 15 15 X f 3 F 1 1 XNyqF 3 F 1 1 X11F 3 t 025 025 xt 8 8 n 8 8 xNyqn 8 8 n 8 8 x11n 8 8 f 15 15 X f 3 F 1 1 XNyqF 15 F 1 1 X11F 15 CTFTCTFSDFT Relationships 27 Start with a signal x t 8 cos 30πt and sample window and periodically repeat it using a sampling rate of f s 60 and a window width of N 32 For each signal in the process graph the signal and its transform either CTFT or DTFT rob28124ch10446508indd 495 041216 146 pm C h a p t e r 10 Sampling and Signal Processing 496 Answers n 16 48 xsn 8 8 F 1 1 Xs f 4 F 1 1 π π Xs f n 16 48 xswn 8 8 F 1 1 XswF 128 F 1 1 π π XswF n 16 48 xswsn 8 8 F 1 1 XswsF 4 F 1 1 π π XswsF t 03 08 xt 8 8 f 15 15 X f 4 f 15 15 π π X f 28 Sample each signal x t N times at the rate f s creating the signal x n Graph x t versus t and x n versus n T s over the time range 0 t N T s Find the DFT X k of the N samples Then graph the magnitude and phase of X f versus f and X k N versus kΔf over the frequency range f s 2 f f s 2 where Δf f s N Graph X k N as an impulse function of kΔf using the MATLAB stem command to represent the impulses a x t 4 cos 200πt f s 800 N 32 b x t 6 rect 2t δ 1 t f s 16 N 128 c x t 6 sinc 4t δ 1 t f s 16 N 128 d x t 5 cos 2πt cos 16πt f s 64 N 128 Answers t 8 xt 6 nTs 8 xn 6 f 8 8 X f 3 f 8 8 π π kfsNF 8 8 XkNF 3 kfsNF 8 8 π π X f XkNF t 004 xt 4 4 nTs 004 xn 4 4 f 400 400 X f 2 f 400 400 π π kfsNF 400 400 XkNF 2 kfsNF 400 400 π π X f XkNF t 8 xt 2 6 nTs 8 xn 2 6 f 8 8 X f 15 f 8 8 π π kfsNF 8 8 XkNF 15 kfsNF 8 8 π π X f XkNF t 2 xt 5 5 nTs 2 xn 5 5 f 32 32 X f 125 f 32 32 X f π π kfsNF 32 32 125 kfsNF 32 32 π π XkNF XkNF rob28124ch10446508indd 496 041216 146 pm Exercises with Answers 497 Windows 29 Sometimes window shapes other than a rectangle are used Using MATLAB find and graph the magnitudes of the DFTs of these window functions with N 32 a von Hann or Hanning wn 1 2 1 cos 2πn N 1 0 n N b Bartlett wn 2n N 1 0 n N 1 2 2 2n N 1 N 1 2 n N c Hamming wn 054 046 cos 2πn N 1 0 n N d Blackman wn 042 05 cos 2πn N 1 008 cos 4πn N 1 0 n N Answers n 31 wn 1 k 32 32 Wk 16 n 31 wn 1 k 32 32 Wk 16 n 31 wn 1 k 32 32 Wk 16 n 31 wn 1 k 32 32 Wk 16 DFT 30 A signal x t is sampled four times to produce the signal x n and the sample values are x 0 x 1 x 2 x 3 7 3 4 a This set of four numbers is the set of input data to the DFT which returns the set X 0 X 1 X 2 X 3 a What numerical value of a makes X 1 a purely real number b Let a 9 What is the numerical value of X 29 c If X15 9 j2 what is the numerical value of X 1 Answers 3 11 j6 9 j2 rob28124ch10446508indd 497 041216 146 pm C h a p t e r 10 Sampling and Signal Processing 498 31 Sample the following signals at the specified rates for the specified times and graph the magnitudes and phases of the DFTs versus harmonic number in the range N2 k N2 1 a x t tri t 1 f s 2 N 16 b x t tri t 1 f s 8 N 16 c x t tri t 1 f s 16 N 256 d x t tri t tri t 4 f s 2 N 8 e x t tri t tri t 4 f s 8 N 32 f x t tri t tri t 4 f s 64 N 256 Answers k 4 3 Xk 2 fs 2 N 8 k 4 3 π π Xk k 16 15 Xk 8 fs 8 N 32 k 16 15 π π Xk k 128 127 Xk 16 fs 16 N 256 k 128 127 π π Xk k 8 7 Xk 2 fs 2 N 16 k 8 7 π π Xk k 8 7 Xk 8 fs 8 N 16 k 8 7 π π Xk k 128 127 Xk 64 fs 64 N 256 k 128 127 Xk π π rob28124ch10446508indd 498 041216 146 pm Exercises with Answers 499 32 For each signal graph the original signal and the decimated signal for the specified sampling interval Also graph the magnitudes of the DTFTs of both signals a xn tri n 10 Ns 2 xd n tri n 5 b xn 095n sin 2πn 10 un Ns 2 c xn cos 2πn 8 Ns 7 Answers n 20 20 1 xn 1 n 20 20 1 1 xdn F 1 1 05 XF F 1 1 05 XdF n 20 20 1 xn n 20 20 1 xdn F 1 1 10 XF F 1 1 10 XdF n 5 40 1 xn 1 n 5 40 1 xdn 1 F 1 1 100 XF F 1 1 100 XdF 33 For each signal in Exercise 32 insert the specified number of zeros between samples lowpass discretetime filter the signals with the specified cutoff frequency and graph the resulting signal and the magnitude of its DTFT a Insert 1 zero between points Cutoff frequency is Fc 01 b Insert 4 zeros between points Cutoff frequency is Fc 02 c Insert 4 zeros between points Cutoff frequency is Fc 002 rob28124ch10446508indd 499 041216 146 pm C h a p t e r 10 Sampling and Signal Processing 500 Answers n 5 40 Ω 2π 2π Xie jΩ 05 05 xin 200 n 20 20 05 F 1 1 XiF xin 5 No graph needed EXERCISES WITHOUT ANSWERS Sampling 34 The theoretically perfect interpolation function is the sinc function But we cannot actually use it in practice Why not 35 A continuoustime signal with a fundamental period of 2 seconds is sampled at a rate of 6 samplessecond Some selected values of the discretetime signal that result are x 0 3 x 13 1 x 7 7 x 33 0 x 17 3 Find the following numerical values if it is possible to do so If it is impossible explain why a x24 b x18 c x21 d x103 36 The signal x t 5trit 1 δ 2 t is sampled at a rate of 8 samplessecond with the first sample sample number 1 occurring at time t 0 a What is the numerical value of sample number 6 b What is the numerical value of sample number 63 37 Using MATLAB or an equivalent mathematical computer tool graph the signal x t 3 cos 20πt 2 sin 30πt over a time range of 0 t 400 ms Also graph the signal formed by sampling this function at the following sampling intervals a T s 1120 s b T s 160 s c T s 130 s and d T s 115 s Based on what you observe what can you say about how fast this signal should be sampled so that it could be reconstructed from the samples 38 A signal x t 20 cos 1000πt is impulse sampled at a sampling rate of 2000 samplessecond Graph two periods of the impulsesampled signal x δ t Let the one sample be at time t 0 Then graph four periods centered at 0 Hz of the CTFT X δ f of the impulsesampled signal x δ t Change the sampling rate to 500 samplessecond and repeat 39 A signal x t 10 rect t4 is impulse sampled at a sampling rate of 2 samples second Graph the impulsesampled signal x δ t on the interval 4 t 4 Then rob28124ch10446508indd 500 041216 146 pm 501 Exercises without Answers graph three periods centered at f 0 of the CTFT X δ f of the impulsesampled signal x δ t Change the sampling rate to 12 samplessecond and repeat 40 A signal x t 4 sinc 10t is impulse sampled at a sampling rate of 20 samples second Graph the impulsesampled signal x δ t on the interval 05 t 05 Then graph three periods centered at f 0 of the CTFT X δ f of the impulse sampled signal x δ t Change the sampling rate to 4 samplessecond and repeat 41 A signal x n is formed by sampling a signal x t 20 cos 8πt at a sampling rate of 20 samplessecond Graph x n over 10 periods versus discrete time Then do the same for sampling frequencies of 8 samplessecond and 6 samplessecond 42 A signal x n is formed by sampling a signal x t 4 sin 200πt at a sampling rate of 400 samplessecond Graph x n over 10 periods versus discrete time Then do the same for sampling frequencies of 200 samplessecond and 60 samplessecond 43 A signal x t is sampled above its Nyquist rate to form a signal x n and is also impulse sampled at the same rate to form an impulse signal x δ t The DTFT of x n is X F 10 rect 5F δ 1 F or X e jΩ 10 rect 5Ω2π δ 2π Ω a If the sampling rate is 100 samplessecond what is the highest frequency at which the CTFT of x t is nonzero b What is the lowest positive frequency greater than the highest frequency in x t at which the CTFT of x δ t is nonzero c If the original signal x t is to be recovered from the impulse sampled signal x δ t by using an ideal lowpass filter with impulse response h t A sinc wt what is the maximum possible value of w 44 Below is a graph of some samples taken from a sinusoid 0 05 1 15 2 15 1 05 0 05 1 15 2 Time t s xnTs a What is the sampling rate f s b If these samples have been taken properly according to Shannons sampling theorem what is the fundamental frequency f 0 of the sinusoid c The sinusoid from which the samples came can be expressed in the form A cos 2π f 0 t What is the numerical value of A d Specify the fundamental frequencies f 01 and f 02 of two other cosines of the same amplitude which when sampled at the same rate would yield the same set of samples 45 A bandlimited periodic continuoustime signal is sampled at twice its Nyquist rate over exactly one fundamental period at f s 220 Hz with the first sample occurring at t 0 and the samples are 1 2 4 6 3 5 97 a What is the maximum frequency at which the continuoustime signal could have any signal power rob28124ch10446508indd 501 041216 146 pm C h a p t e r 10 Sampling and Signal Processing 502 b What is the fundamental period T 0 of the continuoustime signal c If sample 1 occurs at time t 0 and the sampling continued indefinitely what would be the numerical value of sample 317 and at what time would it occur 46 A wagon wheel has eight spokes It is rotating at a constant angular velocity Four snapshots of the wheel are taken at the four times illustrated in Figure E46 Let the sampling interval time between snapshots be 10 ms a In Case 1 what are the three lowest positive angular velocities in revolutions per second rps at which it could be rotating b In Case 2 what are the three lowest positive angular velocities in revolutions per second rps at which it could be rotating t 3Ts t 2Ts t Ts t 0 Case 1 Case 2 Figure E46 Impulse Sampling 47 A signal x t is impulsesampled at 100 samplessecond which is above its Nyquist rate to form the impulse signal x δ t The CTFTs of the two signals are X f and X δ f What is the numerical ratio X δ 0 X 0 48 A sinusoidal signal of frequency f 0 is impulse sampled at a rate of f s The impulsesampled signal is the excitation of an ideal lowpass filter with corner cutoff frequency of f c For each set of parameters below what frequencies are present in the response of the filter List only the nonnegative frequencies including zero if present a f 0 20 f s 50 f c 210 b f 0 20 f s 50 f c 40 c f 0 20 f s 15 f c 35 d f 0 20 f s 5 f c 22 e f 0 20 f s 20 f c 50 49 Each signal x below is sampled to form x s by being multiplied by a periodic impulse of the form δ T s t for continuoustime signals or δ N s n for discretetime signals and f s 1 T s and N s 1 F s a x t 4 cos 20πt f s 40 What is the first positive frequency above 10 Hz at which X s f is not zero b x n 3 sin 2πn 10 N s 3 If the sampled signal is filtered by an ideal lowpass discretetime filter what is the maximum corner frequency that rob28124ch10446508indd 502 041216 146 pm 503 Exercises without Answers would produce a pure sinusoidal response from the filter What is the maximum corner frequency that would produce no response at all from the filter c x t 10 tri t f s 4 If the sampled signal is interpolated by simply always holding the last sample value what would be the value of the interpolated signal at time t 09 d x n 20 N s 4 The DTFT of the sampled signal consists entirely of impulses all of the same strength 50 A continuoustime signal x t is sampled above its Nyquist rate to form a discretetime signal x n and is also impulse sampled at the same rate to form a continuous time impulse signal x δ t The DTFT of x n is X F 10 rect 5F δ 1 F or X e jΩ 10 rect 5Ω 2π δ 2π Ω a If the sampling rate is 100 samplessecond what is the highest frequency in x t b What is the lowest positive frequency greater than the highest frequency of x t at which the CTFT of x δ t is nonzero c If the original signal x t is to be recovered from the impulse sampled signal x δ t by using an ideal lowpass filter with impulse response h t A sinc wt what is the maximum possible value of w 51 For each signal x t impulse sample it at the rate specified by multiplying it by a periodic impulse δ T s t T s 1 f s and graph the impulsesampled signal x δ t over the time range specified and the magnitude and phase of its CTFT X δ f over the frequency range specified a x t 5 1 cos 200πt rect 100t f s 1600 20 ms t 20 ms 2000 f 2000 b x t e t 2 2 f s 5 5 t 5 15 f 15 c x t 10 e t20 u t f s 1 10 t 100 3 f 3 52 Given a signal x t rect 20t δ 01 t and an ideal lowpass filter whose frequency response is T s rect f f s process x t in two different ways Process 1 Filter the signal and multiply it by f s Process 2 Impulse sample the signal at the rate specified then filter the impulsesampled signal For each sampling rate graph the original signal x t and the processed signal y t over the time range 05 t 05 In each case by examining the CTFTs of the signals explain why the two signals do or do not look the same a f s 1000 b f s 200 c f s 50 d f s 20 e f s 10 f f s 4 g f s 2 rob28124ch10446508indd 503 041216 146 pm C h a p t e r 10 Sampling and Signal Processing 504 53 Sample the signal x t 4 sin 20πt 02 t 02 0 otherwise 4 sin 20πt rect t04 over the time range 05 t 05 at the specified sampling rates and approximately reconstruct the signal by using the sincfunction technique xt 2 f c f s n xn T s sinc2 f c t n T s except with a finite set of samples and with the specified filter cutoff frequency That is use xt 2 f c f s nN N xn T s sinc2 f c t n T s where N 05Ts Graph the reconstructed signal in each case a fs 20 fc 10 fs 20 Ts 005 N 10 b fs 40 fc 10 fs 40 Ts 0025 N 20 c fs 40 fc 20 fs 40 Ts 0025 N 20 d fs 100 fc 10 fs 100 Ts 001 N 50 e fs 100 fc 20 fs 100 Ts 001 N 50 f fs 100 fc 50 fs 100 Ts 001 N 50 Nyquist Rates 54 Find the Nyquist rates for these signals a xt 15 rect300t cos104πt b xt 7 sinc40t cos150πt c xt 15rect500t δ1100 t cos104πt d xt 4sinc500t δ1200t e xt 2 sinc500t δ1200t cos104πt f xt t t 10 0 t 10 g xt 8 sinc101t 4 cos200πt h xt 32 sinc101t cos200πt i xt 7 sinc99t δ1t j xt 6 tri100t cos20000πt k xt 2 cos2πt t 12 1 t 12 l xt 4 sinc20t δ3ttrit 10 rob28124ch10446508indd 504 041216 146 pm 505 Exercises without Answers 55 A signal x1t 5 sin20πt is sampled at four times its Nyquist rate Another signal x2t 5 sin2πf0t is sampled at the same rate What is the smallest value of f0 which is greater than 10 and for which the samples of x2t are exactly the same as the samples of x1t 56 A signal xt 4 cos200πt 7 sin200πt is sampled at its Nyquist rate with one of the samples occurring at time t 0 If an attempt is made to reconstruct this signal from these samples by ideal sincfunction interpolation what signal will actually be created by this interpolation process Aliasing 57 Graph the signal xn formed by sampling the signal xt 10 sin8πt at twice the Nyquist rate and xt itself Then on the same graph at least two other continuoustime sinusoids which would yield exactly the same samples if sampled at the same times 58 A cosine xt and a sine signal yt of the same frequency are added to form a composite signal zt The signal zt is then sampled at exactly its Nyquist rate with the usual assumption that a sample occurs at time t 0 Which of the two signals xt or yt would if sampled by itself produce exactly the same set of samples 59 Each signal x below is impulse sampled to form xs by being multiplied by a periodic impulse function of the form δTst and fs 1Ts a xt 4 cos20πt fs 40 What is the first positive frequency above 10 Hz at which Xs f is not zero b xt 10 trit fs 4 If the sampled signal is interpolated by simply always holding the last sample value what would be the value of the interpolated signal at time t 09 Practical Sampling 60 Graph the magnitude of the CTFT of xt 25 sinc2 t 6 What is the minimum sampling rate required to exactly reconstruct xt from its samples Infinitely many samples would be required to exactly reconstruct xt from its samples If one were to make a practical compromise in which one sampled over the minimum possible time which could contain 99 of the energy of this waveform how many samples would be required 61 Graph the magnitude of the CTFT of xt 8 rect3t This signal is not bandlimited so it cannot be sampled adequately to exactly reconstruct the signal from the samples As a practical compromise assume that a bandwidth which contains 99 of the energy of xt is great enough to practically reconstruct xt from its samples What is the minimum required sampling rate in this case X f 83 sinc f 3 Bandlimited Periodic Signals 62 A discretetime signal xn is formed by sampling a continuoustime sinusoid sinusoid xt A cos 2π fot π 3 θ at exactly its Nyquist rate with one of the samples occurring exactly at time t 0 a What value of θmax in the range π 2 θmax π 2 will maximize the signal power of xn and in terms of A what is that maximum signal power rob28124ch10446508indd 505 041216 146 pm C h a p t e r 10 Sampling and Signal Processing 506 b What value of θmin in the range π 2 θmin π 2 will minimize the signal power of xn and in terms of A what is that minimum signal power 63 How many sample values are required to yield enough information to exactly describe these bandlimited periodic signals a xt 8 3 cos8πt 9 sin4πt fm 4 fNyq 8 b xt 8 3 cos7πt 9 sin4πt fm 35 fNyq 7 64 Sample the signal xt 15 sinc5t δ2t sin32πt to form the signal xn Sample at the Nyquist rate and then at the next higher rate for which the number of samples per cycle is an integer Graph the signals and the magnitude of the CTFT of the continuoustime signal and the DTFT of the discretetime signal 65 A signal xt is periodic and one period of the signal is described by xt 3t 0 t 55 0 55 t 8 Find the samples of this signal over one period sampled at a rate of 1 samples second beginning at time t 0 Then graph on the same scale two periods of the original signal and two periods of a periodic signal which is bandlimited to 05 Hz or less that would have these same samples DFT 66 A signal xt is sampled four times and the samples are x0 x1 x2 x3 Its DFT is X0 X1 X2 X3 X3 can be written as X3 ax0 bx 1 cx2 d x3 What are the numerical values of a b c and d 67 Sample the bandlimited periodic signal xt 8 cos50πt 12 sin80πt at exactly its Nyquist rate over exactly one period of xt Find the DFT of those samples From the DFT find the CTFS Graph the CTFS representation of the signal that results and compare it with xt Explain any differences Repeat for a sampling rate of twice the Nyquist rate 68 An arbitrary realvalued signal is sampled 32 times to form a set of numbers x0 x1 x31 These are fed into an fft algorithm on a computer and it returns the set of numbers X0 X1 X31 Which of these returned numbers are guaranteed to be purely real 69 A bandlimited periodic signal xt whose highest frequency is 25 Hz is sampled at 100 samplessecond over exactly one fundamental period to form the signal xn The samples are x0 x1 x2 x3 a b c d a Let one period of the DFT of those samples be X0 X1 X2 X3 What is the value of X1 in terms of a b c and d b What is the average value of xt in terms of a b c and d rob28124ch10446508indd 506 041216 146 pm 507 Exercises without Answers c One of the numbers X0 X1 X2 X3 must be zero Which one is it and why d Two of the numbers X0 X1 X2 X3 must be real numbers Which ones are they and why e If X1 2 j3 what is the numerical value of X3 and why 70 Using MATLAB a Generate a pseudorandom sequence of 256 data points in a vector x using the randn function which is built in to MATLAB b Find the DFT of that sequence of data and put it in a vector X c Set a vector Xlpf equal to X d Change all the values in Xlpf to zero except the first 8 points and the last 8 points e Take the real part of the inverse DFT of Xlpf and put it in a vector xlpf f Generate a set of 256 sample times t which begin with 0 and are uniformly separated by 1 g Graph x and xlpf versus t on the same scale and compare What kind of effect does this operation have on a set of data Why is the output array called xlpf 71 In Figure E71 match functions to their DFT magnitudes n 16 xn 1 2 A n 16 xn 1 2 B n 16 xn 1 2 C n 16 xn 1 2 D n 16 xn 1 2 E k 16 Xk 8 k 16 Xk 8 k 16 Xk 8 k 16 Xk 8 k 16 Xk 8 1 2 3 4 5 Figure E71 rob28124ch10446508indd 507 041216 146 pm C h a p t e r 10 Sampling and Signal Processing 508 72 For each xn in ah in Figure E72 find the DFT magnitude Xk corresponding to it n 16 xn 6 6 a n 16 xn 6 6 b n 16 xn 6 6 c n 16 xn 6 6 d n 16 xn 6 6 e n 16 xn 6 6 f n 16 xn 6 6 g n 16 xn 6 6 h k 16 Xk 50 8 k 16 Xk 8 4 k 16 Xk 2 12 k 16 Xk 35 6 k 16 Xk 50 11 k 16 Xk 35 10 k 16 Xk 6 3 k 16 Xk 20 2 k 16 Xk 16 1 k 16 Xk 16 7 k 16 Xk 4 9 k 16 Xk 16 5 Figure E72 DiscreteTime Sampling 73 A discretetime signal xn sinc 2n 19 is sampled by multiplying it by δNsn What is the maximum value of Ns for which the original signal can be theoretically reconstructed exactly from the samples 74 A discretetime signal is passed through a discretetime lowpass filter with frequency response HF rect85F δ1 F If every Nth point of the filters output signal is sampled what is the maximum numerical value of N for which all the information in the original discretetime signal is preserved rob28124ch10446508indd 508 041216 146 pm 509 111 INTRODUCTION AND GOALS Up to this point in this text the material has been highly mathematical and abstract We have seen some occasional examples of the use of these signal and system analysis techniques but no really indepth exploration of their use We are now at the point at which we have enough analytical tools to attack some important types of signals and systems and demonstrate why frequencydomain methods are so popular and powerful in analysis of many systems Once we have developed a real facility and familiarity with frequencydomain methods we will understand why many professional engineers spend practically their whole careers in the frequency domain creating designing and analyzing systems with transform methods Every linear timeinvariant LTI system has an impulse response and through the Fourier transform a frequency response and through the Laplace transform a transfer function We will analyze systems called filters that are designed to have a certain fre quency response We will define the term ideal filter and we will see ways of approximat ing the ideal filter Since frequency response is so important in the analysis of systems we will develop efficient methods of finding the frequency responses of complicated systems CH APTER GOAL S 1 To demonstrate the use of transform methods in the analysis of some systems with practical engineering applications 2 To develop an appreciation of the power of signal and system analysis done directly in the frequency domain 112 FREQUENCY RESPONSE Probably the most familiar example of frequency response in everyday life is the re sponse of the human ear to sounds Figure 111 illustrates the variation of the percep tion by the average healthy human ear of the loudness of a single sinusoidal frequency of a constant midlevel intensity as a function of frequency from 20 Hz to 20 kHz This range of frequencies is commonly called the audio range This frequency response is a result of the structure of the human ear A system de signed with the ears response in mind is a homeentertainment audio system This is an C H A P T E R 11 Frequency Response Analysis rob28124ch11509591indd 509 061216 218 pm C h a p t e r 11 Frequency Response Analysis 510 example of a system that is designed without knowing exactly what signals it will process or exactly how they should be processed But it is known that the signals will lie in the audio frequency range Since different people have different tastes in music and how it should sound such a system should have some flexibility An audio system typically has an amplifier that is capable of adjusting the relative loudness of one frequency versus another through tone controls like bass adjustment treble adjustment loudness compen sation or a graphic equalizer These controls allow any individual user of the system to adjust its frequency response for the most pleasing sound with any kind of music Audioamplifier controls are good examples of systems designed in the frequency domain Their purpose is to shape the frequency response of the amplifier The term filter is commonly used for systems whose main purpose is to shape a frequency re sponse We have already seen a few examples of filters characterized as lowpass high pass bandpass or bandstop What does the word filter mean in general It is a device for separating something desirable from something undesirable A coffee filter sepa rates the desirable coffee from the undesirable coffee grounds An oil filter removes undesirable particulates In signal and system analysis a filter separates the desirable part of a signal from the undesirable part A filter is conventionally defined in signal and system analysis as a device that emphasizes the power of a signal in one frequency range while deemphasizing the power in another frequency range 113 CONTINUOUSTIME FILTERS EXAMPLES OF FILTERS Filters have passbands and stopbands A passband is a frequency range in which the filter allows the signal power to pass through relatively unaffected A stopband is a fre quency range in which the filter significantly attenuates the signal power allowing very Figure 111 Average human ears perception of the loudness of a constantamplitude audio tone as a function of frequency 0 2 4 6 8 10 12 14 16 18 20 0 01 02 03 04 05 06 07 08 09 1 Frequency f kHz Perceived Loudness HumanEar Perception of Loudness vs Frequency Normalized to 4 kHz rob28124ch11509591indd 510 061216 218 pm 113 ContinuousTime Filters 511 little to pass through The four basic types of filters are lowpass highpass bandpass and bandstop filters In a lowpass filter the passband is a region of low frequency and the stopband is a region of high frequency In a highpass filter those bands are reversed Low frequencies are attenuated and high frequencies are not A bandpass filter has a passband in a midrange of frequencies and stopbands at both low and high frequen cies A bandstop filter reverses the pass and stop bands of the bandpass filter Simple adjustments of the bass and treble low and high frequencies volume in an audio amplifier could be accomplished by using lowpass and highpass filters with adjustable corner frequencies We have seen a circuit realization of a lowpass filter We can also make a lowpass filter using standard continuoustime system building blocks integrators amplifiers and summing junctions Figure 112a Figure 112 Simple filters a lowpass b highpass xt yt ωc ωc a xt yt ωc b The system in Figure 112a is a lowpass filter with a corner frequency of ω c in radianssecond and a frequency response magnitude that approaches one at low frequencies This is a very simple Direct Form II system The transfer function is Hs ω c s ω c therefore the frequency response is H jω H s sjω ω c jω ω c or H f H s sj2πf 2π f c j2π f 2π f c f c j f f c where ω c 2π f c The system in Figure 112b is a highpass filter with a corner fre quency of ω c Its transfer function and frequency response are Hs s s ω c Hjω jω jω ω c H f jf jf f c In either filter if ω c can be varied the relative power of the signals at low and high frequencies can be adjusted These two systems can be cascade connected to form a bandpass filter Figure 113 The transfer function and frequency response of the bandpass filter are Hs s s ωca ωcb s ωcb ω cb s s 2 ω ca ω cb s ω ca ω cb H jω jω ω cb jω 2 jω ω ca ω cb ω ca ω cb H f jf f cb jf 2 jf f ca f cb f ca f cb f ca ω ca 2π f cb ω cb 2π rob28124ch11509591indd 511 061216 218 pm C h a p t e r 11 Frequency Response Analysis 512 As an example let ω ca 100 and let ω cb 50000 Then the frequency responses of the lowpass highpass and bandpass filters are as illustrated in Figure 114 A bandstop filter can be made by parallel connecting a lowpass and highpass filter if the corner frequency of the lowpass filter is lower than the corner frequency of the highpass filter Figure 115 Figure 114 High low and bandpass filter frequency responses 100 104 105 106 102 103 101 100 104 105 106 101 102 103 100 104 105 106 101 102 103 100 104 105 106 102 103 101 104 105 106 102 103 101 100 104 105 106 101 102 103 0 02 04 06 08 1 H jω Highpass Filter 15 1 05 0 05 1 15 15 1 05 0 05 1 15 15 1 05 0 05 1 15 H jω Lowpass Filter ω ω ω ω ω ω 100 104 105 106 101 102 103 0 02 04 06 08 1 H jω Bandpass Filter H jω H jω H jω 0 02 04 06 08 1 Figure 115 A bandstop filter formed by parallel connecting a lowpass filter and a highpass filter xt ωcb ωcb yt ωca ωca ωcb Figure 113 A bandpass filter formed by cascading a highpass filter and a lowpass filter xt ωcb ωcb yt ωca rob28124ch11509591indd 512 061216 218 pm 113 ContinuousTime Filters 513 The transfer function and frequency response of the bandstop filter are Hs s 2 2 ω cb s ω ca ω cb s 2 ω ca ω cb s ω ca ω cb H jω jω 2 j2ω ω cb ω ca ω cb jω 2 jω ω ca ω cb ω ca ω cb H f jf 2 j2f f cb f ca f cb jf 2 jf f ca f cb f ca f cb f ca ω ca 2π f cb ω cb 2π Figure 116 Bandstop filter frequency response 100 101 102 103 104 105 106 0 02 04 06 08 1 H jω Bandstop Filter 100 101 102 103 104 105 106 15 1 05 0 05 1 15 ω H jω Figure 117 A biquadratic system xt yt ω0 2 10β 2ω0 2ω010β ω0 2 If for example ω ca 50000 and ω cb 100 the frequency response would look like Figure 116 A graphic equalizer is a little more complicated than a simple lowpass highpass or bandpass filter It has several cascaded filters each of which can increase or decrease the frequency response of the amplifier in a narrow range of frequencies Consider the system in Figure 117 Its transfer function and frequency response are Hs s 2 2 ω 0 s 10 β ω 0 2 s 2 2 ω 0 s 10 β ω 0 2 H jω jω 2 j2 ω 0 ω 10 β ω 0 2 jω 2 j2 ω 0 ω 10 β ω 0 2 This transfer function is biquadratic in s a ratio of two quadratic polynomials If we graph the frequency response magnitude with ω 0 1 for several values of the param eter β we can see how this system could be used as one filter in a graphic equalizer Figure 118 It is apparent that with proper selection of the parameter β this filter can either emphasize or deemphasize signals near its center frequency ω 0 and has a frequency rob28124ch11509591indd 513 061216 218 pm C h a p t e r 11 Frequency Response Analysis 514 response approaching one for frequencies far from its center frequency A set of cas caded filters of this type each with a different center frequency can be used to empha size or deemphasize multiple bands of frequencies and thereby to tailor the frequency response to almost any shape a listener might desire Figure 119 Figure 119 Conceptual block diagram of a graphic equalizer yt xt f0 10 f0 20 f0 40 f0 20480 Figure 118 Frequency response magnitude for H jω jω 2 j2ω 10 β 1 jω 2 j2ω 10 β 1 101 100 101 0 05 1 15 2 25 3 β 02 β 015 β 01 β 005 β 005 β 01 β 015 β 02 Radian Frequency ω H jω β 0 With all the filters set to emphasize their frequency range the magnitude frequency responses of the subsystems could look like Figure 1110 The center frequencies of these filters are 20 Hz 40 Hz 80 Hz 20480 Hz The filters are spaced at octave intervals in frequency An octave is a factoroftwo change in frequency That makes the individualfilter center frequencies be uniformly spaced on a logarithmic scale and the bandwidths of the filters are also uniform on a logarithmic scale Another example of a system designed to handle unknown signals would be an instrumentation system measuring pressure temperature flow and so on in an indus trial process We do not know exactly how these process parameters vary But they normally lie within some known range and can vary no faster than some maximum rate because of the physical limitations of the process Again this knowledge allows us to design a signal processing system appropriate for these types of signals Even though a signals exact characteristics may be unknown we usually know something about it We often know its approximate power spectrum That is we have an approximate description of the signal power of the signal in the frequency domain If we could not mathematically calculate the power spectrum we could estimate it based on knowledge of the physics of the system that created it or we could measure it One way to measure it would be through the use of filters rob28124ch11509591indd 514 061216 218 pm 113 ContinuousTime Filters 515 IDEAL FILTERS Distortion An ideal lowpass filter would pass all signal power at frequencies below some maxi mum without distorting the signal at all in that frequency range and completely stop or block all signal power at frequencies above that maximum It is important here to define precisely what is meant by distortion Distortion is commonly construed in signal and system analysis to mean changing the shape of a signal This does not mean that if we change the signal we necessarily distort it Multiplication of the signal by a constant or a time shift of the signal are changes that are not considered to be distortion Suppose a signal xt has the shape illustrated at the top of Figure 1111a Then the signal at the bottom of Figure 1111a is an undistorted version of that signal Figure 1111b illustrates one type of distortion Figure 1110 Frequency response magnitudes for 11 filters spanning the audio range 102 103 104 08 09 1 11 12 13 14 15 16 17 18 Cyclic Frequency f H f Figure 1111 a An original signal and a changed but undistorted version of it b an original signal and a distorted version of it t 1 xt 1 1 Original Signal t 02 08 xt 1 a 1 TimeShifted Signal t 1 xt 1 1 Original Signal t 1 xt 1 b 1 Clipped Signal rob28124ch11509591indd 515 061216 218 pm C h a p t e r 11 Frequency Response Analysis 516 The response of any LTI system is the convolution of its excitation with its impulse response Any signal convolved with a unit impulse at the origin is unchanged xt δt xt If the impulse has a strength other than one the signal is multiplied by that strength but the shape is still unchanged xt Aδt Axt If the impulse is time shifted the convolution is time shifted also but without changing the shape xt Aδt t 0 Axt t 0 Therefore the impulse response of a filter that does not distort would be an impulse possibly with strength other than one and possibly shifted in time The most general form of impulse response of a distortionless system would be ht Aδt t 0 The corresponding frequency response would be the CTFT of the impulse response H f A e j2πf t 0 The frequency response can be charac terized by its magnitude and phase H f A and H f 2πf t 0 Therefore a distortionless system has a frequency response magnitude that is constant with frequency and a phase that is linear with frequency Figure 1112 It should be noted here that a distortionless impulse response or frequency re sponse is a concept that cannot actually be realized in any real physical system No real system can have a frequency response that is constant all the way to an infinite fre quency Therefore the frequency responses of all real physical systems must approach zero as frequency approaches infinity Filter Classifications Since the purpose of a filter is to remove the undesirable part of a signal and leave the rest no filter not even an ideal one is distortionless because its magnitude is not constant with frequency But an ideal filter is distortionless within its passband Its fre quency response magnitude is constant within the passband and its frequencyresponse phase is linear within the passband We can now define the four basic types of ideal filter In the following descriptions f m f L and f H are all positive and finite An ideal lowpass filter passes signal power for frequencies 0 f f m without distortion and stops signal power at other frequencies An ideal highpass filter stops signal power for frequencies 0 f f m and passes signal power at other frequencies without distortion An ideal bandpass filter passes signal power for frequencies f L f f H without distortion and stops signal power at other frequencies An ideal bandstop filter stops signal power for frequencies f L f f H and passes signal power at other frequencies without distortion Ideal Filter Frequency Responses Figure 1113 and Figure 1114 illustrate typical magnitude and phase frequency responses of the four basic types of ideal filters It is appropriate here to define a word that is very commonly used in signal and sys tem analysis bandwidth The term bandwidth is applied to both signals and systems It generally means a range of frequencies This could be the range of frequencies present in a signal or the range of frequencies a system passes or stops For historical reasons it is usually construed to mean a range of frequencies in positive frequency space For example an ideal lowpass filter with corner frequencies of f m as illustrated in Figure 1113 is said to have a bandwidth of f m even though the width of the filters nonzero magnitude frequency response is obviously 2 f m The ideal bandpass filter has a bandwidth of f H f L which is the width of the passband in positive frequency space Figure 1112 Magnitude and phase of a distortionless system H f f f A f 1 t0 2π H f rob28124ch11509591indd 516 061216 218 pm 113 ContinuousTime Filters 517 Figure 1115 Examples of bandwidth definitions Null Bandwidth f H f HalfPower Bandwidth f H f 2 A A 2 Absolute Bandwidth f H f Figure 1113 Magnitude and phase frequency responses of ideal lowpass and highpass filters Figure 1114 Magnitude and phase frequency responses of ideal bandpass and bandstop filters f f fL fH fL fH f f fL fH fL fH Ideal Bandpass Filter H f Ideal Bandstop Filter H f H f H f f f Ideal Lowpass Filter H f f f fm fm fm fm Ideal Highpass Filter H f H f H f There are many different kinds of definitions of bandwidth absolute bandwidth halfpower bandwidth null bandwidth and so on Figure 1115 Each of them is a range of frequencies but defined in different ways For example if a signal has no signal power at all below some minimum positive frequency and above some maximum posi tive frequency its absolute bandwidth is the difference between those two frequencies If a signal has a finite absolute bandwidth it is said to be strictly bandlimited or more commonly just bandlimited Most real signals are not known to be bandlimited so other definitions of bandwidth are needed Impulse Responses and Causality The impulse responses of ideal filters are the inverse transforms of their frequency responses The impulse and frequency responses of the four basic types of ideal filter are summarized in Figure 1116 These descriptions are general in the sense that they involve an arbitrary gain constant A and an arbitrary time delay t 0 Notice that the ideal highpass filter and the ideal bandstop filter have frequency responses extending all the way to infinity This is rob28124ch11509591indd 517 061216 218 pm C h a p t e r 11 Frequency Response Analysis 518 impossible in any real physical system Therefore practical approximations to the ideal highpass and bandstop filter allow higherfrequency signals to pass but only up to some high not infinite frequency High is a relative term and as a practical matter usually means beyond the frequencies of any signals actually expected to occur in the system In Figure 1117 are some typical shapes of impulse responses for the four basic types of ideal filter Figure 1116 Frequency responses and impulse responses of the four basic types of ideal filter Ideal Filter Type Frequency Response Lowpass H f Arect f 2 f m e j2πf t 0 Highpass H f A1 rect f 2 f m e j2πf t 0 Bandpass H f Arect f f 0 Δf rect f f 0 Δf e j2πf t 0 Bandstop H f A1 rect f f 0 Δf rect f f 0 Δf e j2πf t 0 Ideal Filter Type Impulse Response Lowpass ht 2A f m sinc2 f m t t 0 Highpass ht Aδ t t 0 2A f m sinc 2 f m t t 0 Bandpass ht 2AΔf sincΔf t t 0 cos2π f 0 t t 0 Bandstop ht Aδt t 0 2AΔf sincΔf t t 0 cos2π f 0 t t 0 Δ f f H f L f 0 f H f L 2 Figure 1117 Typical impulse responses of ideal lowpass highpass bandpass and bandstop filters t Ideal Lowpass ht t Ideal Highpass ht t Ideal Bandpass ht t Ideal Bandstop ht As mentioned above one reason ideal filters are called ideal is that they cannot physically exist The reason is not simply that perfect system components with ideal characteristics do not exist although that would be sufficient It is more fundamen tal than that Consider the impulse responses depicted in Figure 1117 They are the responses of the filters to a unit impulse applied at time t 0 Notice that all of the impulse responses of these ideal filters have a nonzero response before the impulse is applied at time t 0 In fact all of these particular impulse responses begin at an infinite time before time t 0 It should be intuitively obvious that a real system cannot look into the future and anticipate the application of the excitation and start responding before it occurs All ideal filters are noncausal rob28124ch11509591indd 518 061216 218 pm 113 ContinuousTime Filters 519 Although ideal filters cannot be built useful approximations to them can be built In Figure 1118 and Figure 1119 are some examples of the impulse responses frequency responses and responses to square waves of some nonideal causal filters that approximate the four common types of ideal filters Figure 1118 Impulse responses frequency responses and responses to square waves of causal lowpass and bandpass filters t 05 2 ht 05 3 Causal Lowpass ht t 2 xt 1 Excitation t 2 yt 07 Response f 4 4 H f 1 f 4 4 4 4 Hf t 05 2 ht 3 3 Causal Bandpass ht t 2 xt 1 Excitation t 2 yt 06 06 Response ω 30 30 H jω 1 ω 30 30 4 4 H jω Figure 1119 Impulse responses frequency responses and responses to square waves of causal highpass and bandstop filters t 05 2 ht 10 2 Causal Highpass ht t 2 xt 1 Excitation t 2 yt 15 15 Response f 4 4 H f 1 f 4 4 4 4 H f t 05 2 ht 10 4 Causal Bandstop ht t 2 xt 1 Excitation t 2 yt 02 12 Response ω 30 30 H jω 1 ω 30 30 4 4 H jω The lowpass filter smooths the square wave by removing highfrequency signal power from it but leaves the lowfrequency signal power including zero frequency making the average values of the input and output signals the same because the fre quency response at zero frequency is one The bandpass filter removes highfrequency signal power smoothing the signal and removes lowfrequency power including zero frequency making the average value of the response zero The highpass filter removes lowfrequency signal power from the square wave making the average value of the response zero But the highfrequency signal power that defines the sharp discontinuities in the square wave is preserved The bandstop filter removes signal power in a small range of frequencies and leaves the very lowfrequency and very highfrequency signal power So the discontinuities and the average value of the square wave are both preserved but some of the midfrequency signal power is removed rob28124ch11509591indd 519 061216 218 pm C h a p t e r 11 Frequency Response Analysis 520 Figure 1120 A system to measure the power spectrum of a signal H f f 2Δf Squarer x2 x2 x2 x2 x Time Averager H f f x Squarer Time Averager x Squarer Time Averager x Squarer Time Averager Px0 Δf H f f f Δf f2 f2 fN1 fN1 f1 f1 H f Δf xt Px f1 Px f2 Px fN1 The Power Spectrum One purpose of launching into filter analysis was to explain one way of determining the power spectrum of a signal by measuring it That could be accomplished by the system illustrated in Figure 1120 The signal is routed to multiple bandpass filters each with the same bandwidth but a unique center frequency Each filters response is that part of the signal lying in the frequency range of the filter Then the output signal from each filter is the input signal of a squarer and its output signal is the input signal of a time averager A squarer simply takes the square of the signal This is not a linear operation so this is not a linear system The output signal from any squarer is that part of the instantaneous signal power of the original xt that lies in the passband of the band pass filter Then the time averager forms the timeaverage signal power Each output response P x f n is a measure of the signal power of the original xt in a narrow band of frequencies centered at f n Taken together the Ps are an indication of the variation of the signal power with frequency the power spectrum It is unlikely that an engineer today would actually build a system like this to mea sure the power spectrum of a signal A better way to measure it is to use an instrument called a spectrum analyzer But this illustration is useful because it reinforces the concept of what a filter does and what the term power spectrum means Noise Removal Every useful signal always has another undesirable signal called noise added to it One very important use of filters is in removing noise from a signal The sources of noise are many and varied By careful design noise can often be greatly reduced but can never be completely eliminated As an example of filtering suppose the signal power is confined to a range of low frequencies and the noise power is spread over a much rob28124ch11509591indd 520 061216 218 pm 113 ContinuousTime Filters 521 wider range of frequencies a very common situation We can filter the signal plus noise with a lowpass filter and reduce the noise power without having much effect on the signal power Figure 1121 The ratio of the signal power of the desired signal to the signal power of the noise is called the signaltonoise ratio often abbreviated SNR Probably the most funda mental consideration in communication system design is to maximize the SNR and filtering is a very important technique in maximizing SNR BODE DIAGRAMS The Decibel In graphing frequency response the magnitude of the frequency response is often con verted to a logarithmic scale using a unit called the decibel dB If the frequency response magnitude is H jω Y jω X jω then that magnitude expressed in decibels is H jω dB 20 log 10 H jω 20 log 10 Y jω X jω Y jω dB X jω dB 111 The name decibel comes from the original unit defined by Bell Telephone engineers the bel B named in honor of Alexander Graham Bell1 the inventor of the tele phone The bel is defined as the common logarithm base 10 of a power ratio For example if the response signal power of a system is 100 and the input signal power 1 Alexander Graham Bell was born in Scotland in a family specializing in elocution In 1864 he became a resident master in Elgins Weston House Academy in Scotland where he studied sound and first thought of transmitting speech with electricity He moved to Canada in 1870 to recuperate from tuberculosis and later settled in Boston There he continued working on transmitting sound over wires and on March 7 1876 he was granted a patent for the telephone arguably the most valuable patent ever issued He became independently wealthy as a result of the income derived from this patent In 1898 he became president of the National Geographic Society Figure 1121 Partial removal of noise by a lowpass filter X f f X f N f f N f f Y f f H f f xt yt nt ht LPF rob28124ch11509591indd 521 061216 218 pm C h a p t e r 11 Frequency Response Analysis 522 expressed in the same units is 20 the signalpower gain of the system expressed in bels would be log 10 P Y P X log 10 10020 0699 B Since the prefix deci is the international standard for onetenth a decibel is onetenth of a bel and that same power ratio expressed in dB would be 699 dB So the power gain expressed in dB would be 10 log 10 P Y P X Since signal power is proportional to the square of the signal itself the ratio of powers expressed directly in terms of the signals would be 10 log 10 P Y P X 10 log 10 Y 2 X 2 10 log 10 YX 2 20 log 10 YX In a system in which multiple subsystems are cascaded the overall frequency response is the product of the individual frequency responses but the overall frequency re sponse expressed in dB is the sum of the individual frequency responses expressed in dB because of the logarithmic definition of the dB Also use of decibels may reveal frequency response behavior that is difficult to see on a linear graph Before considering the frequency responses of practical filters it is useful to be come familiar with a very helpful and common way of displaying frequency response Often linear graphs of frequency response although accurate do not reveal important system behavior As an example consider the graphs of the two quite differentlooking frequency responses H 1 jω 1 jω 1 and H 2 jω 30 30 ω 2 j31ω Figure 1122 Figure 1123 Logmagnitude graphs of the two frequency responses ω 20π 20π H1 jωdB H2 jωdB ω 20π 20π Figure 1122 Comparison of the magnitudes of two apparently different frequency responses ω 80 80 H1 jω H2 jω 1 ω 80 80 1 Graphed this way the two magnitude frequency response graphs look identical yet we know the frequency responses are different One way of seeing small differ ences between frequency responses is to graph them in dB The decibel is defined logarithmically A logarithmic graph deemphasizes large values and emphasizes small values Then small differences between frequency responses can be more easily seen Figure 1123 rob28124ch11509591indd 522 061216 218 pm 113 ContinuousTime Filters 523 In the linear graphs the behavior of the magnitude frequency response looked identical because at very small values the two graphs look the same In a dB graph the difference between the two magnitude frequency responses at very small values can be seen Although this type of graph is used sometimes a more common way of displaying frequency response is the Bode2 diagram or Bode plot Like the logmagnitude graph the Bode diagram reveals small differences between frequency responses but it is also a systematic way of quickly sketching or estimating the overall frequency response of a system that may contain multiple cascaded frequency responses A logmagnitude graph is logarithmic in one dimension A magnitude Bode diagram is logarithmic in both dimensions A magnitude frequency response Bode diagram is a graph of the frequency response magnitude in dB against a logarithmic frequency scale Since the frequency scale is now logarithmic only positive frequencies can be used in a graph That is not a loss of information since for frequency responses of real systems the value of the frequency response at any negative frequency is the complex conjugate of the value at the corresponding positive frequency Returning now to the two different system frequency responses H 1 jω 1 jω 1 and H 2 jω 30 30 ω 2 j31ω if we make a Bode diagram of each of them their difference becomes more evident Figure 1124 The dB scale makes the behavior of the two magnitude frequency responses at the higher frequencies distinguishable 2 Hendrik Bode received a BA in 1924 and an MA in 1926 from Ohio State University In 1926 he started work at Bell Telephone Laboratories and worked with electronic filters and equalizers While employed at Bell Labs he went to Columbia University Graduate School and received his PhD in 1935 In 1938 Bode used the magnitude and phase frequency response plots of a complex function He investigated closed loop stability using the notions of gain and phase margin These Bode plots are used extensively with many electronic systems He published Network Analysis and Feedback Amplifier Design considered to be a very important book in this field Bode retired in October 1967 and was elected Gordon Mckay Professor of Systems Engineering at Harvard University Figure 1124 Bode diagrams of the two example of frequency responses 100 101 100 101 100 101 100 101 40 30 20 10 0 Radian Frequency ω 3 25 2 15 1 05 0 Radian Frequency ω 40 30 20 10 0 Radian Frequency ω 3 25 2 15 1 05 0 Radian Frequency ω H1 jω H1 jωdB H2 jωdB H2 jω rob28124ch11509591indd 523 061216 218 pm C h a p t e r 11 Frequency Response Analysis 524 Although the fact that differences between low levels of magnitude frequency response can be better seen with a Bode diagram is a good reason to use it it is by no means the only reason It is not even the main reason The fact that system gains in dB add instead of multiplying when systems are cascaded makes the quick graphical estimation of overall system gain characteristics easier using Bode diagrams than using linear graphs Most LTI systems are described by linear differential equations with constant coefficients The most general form of such an equation is k0 N a k d k d t k yt k0 M b k d k d t k xt 112 where xt is the excitation and yt is the response From Chapter 5 we know that the transfer function is Hs b M s M b M1 s M1 b 1 s b 0 a N s N a N1 s N1 a 1 s b 0 The numerator and denominator polynomials can be factored putting the transfer function into the form Hs A 1 s z 1 1 s z 2 1 s z M 1 s p 1 1 s p 2 1 s p N where the zs and ps are the zeros and poles For real systems the coefficients a and b in 112 are all real and all the finite ps and zs in the factored forms must either be real or must occur in complex conjugate pairs so that when the factored numerator and denominator are multiplied out to ob tain the ratioofpolynomials form all the coefficients of the powers of s are real From the factored form the system transfer function can be considered as being the cascade of a frequencyindependent gain A and multiple subsystems each having a transfer function with one finite pole or one finite zero If we now convert the transfer function to a frequency response through s jω we can think of the overall frequency response as resulting from the cascade of multiple components each with a simple frequency response Figure 1125 Figure 1125 A system represented as a cascade of simpler systems X jω Y jω H jω A z1 1 jω z2 1 jω zM jω 1 p1 jω 1 1 p2 jω 1 1 pN jω 1 1 Each component system will have a Bode diagram and because the magnitude Bode diagrams are graphed in dB the overall magnitude Bode diagram is the sum of the individual magnitude Bode diagrams Phase is graphed linearly as before against a logarithmic frequency scale and the overall phase Bode diagram is the sum of all the phases contributed by the components rob28124ch11509591indd 524 061216 218 pm 113 ContinuousTime Filters 525 The OneRealPole System Consider the frequency response of a subsystem with a single real pole at s p k and no finite zeros Hs 1 1 s p k H jω 1 1 jω p k 113 Before proceeding first consider the inverse CTFT of Hjω We can use the CTFT pair e at ut ℱ 1 a jω Rea 0 and rewrite 113 as H jω p k jω p k Then it follows that p k e p k t ut ℱ p k jω p k p k 0 114 This shows that the pole must have a negative real value for the frequency response to have meaning If it is positive we cannot do the inverse CTFT to find the correspond ing time function If p k is negative the exponential in 114 decays to zero in positive time If it were positive that would indicate a growing exponential in positive time and the system would be unstable The Fourier transform of a growing exponential does not exist Also frequency response has no practical meaning for an unstable system because it could never actually be tested The magnitudes and phases of H jω 11 jω p k versus frequency are graphed in Figure 1126 For frequencies ω p k the frequency response approaches H jω 1 the magnitude response is approximately zero dB and the phase response is approxi mately zero radians For frequencies ω p k the frequency response is approximately H jω p k jω the magnitude frequency response approaches a linear slope of 6 dB per octave or 20 dB per decade and the phase response approaches a constant π2 radians An octave is a factorof2 change in frequency and a decade is a factorof10 change in frequency These limiting behaviors for extreme frequencies define magni tude and phase asymptotes The intersection of the two magnitude asymptotes occurs at ω p k which is called the corner frequency At the corner frequency ω p k the frequency response is H jω 1 1 j p k p k 1 1 j p k 0 and its magnitude is 1 2 0707 We can convert this to decibels 0707 dB 20 log 10 0707 3 dB At that point the actual Bode diagram is 3 dB below the corner formed by the asymp totes This is the point of largest deviation of this magnitude Bode diagram from its as ymptotes The phase Bode diagram goes through π4 radians at the corner frequency and approaches zero radians below and π2 radians above the corner frequency rob28124ch11509591indd 525 061216 218 pm C h a p t e r 11 Frequency Response Analysis 526 ExamplE 111 Bode diagram of frequency response of an RC lowpass filter Draw magnitude and phase Bode diagrams for an RC lowpass filter frequency response with a time constant of 50 μs The form of the RC lowpass filter transfer function is Hs 1 sRC 1 The time constant is RC Therefore Hs 1 50 10 6 s 1 1 s20000 1 Setting the denominator equal to zero and solving for the pole location we get a pole at s 20000 Then we can write the frequency response in the standard onenegativerealpole form H jω Hssjω 1 1 jω20000 The corresponding corner frequency on the Bode diagram is at ω 20000 Figure 1127 The OneRealZero System An analysis similar to the onerealpole system anal ysis yields the magnitude and phase Bode diagrams for a subsystem with a single negativereal zero and no finite poles Hs 1 s z k H jω 1 jω z k z k 0 Figure 1126 The magnitude and phase frequency response of a singlenegativerealpole subsystem 3 dB ω ω Slope of 6 dBoctave or 20 dBdecade 4 π 2 π pk pk 10pk 01 pk pk 10pk 01 Asymptotes Asymptote pk jω 1 1 pk jω 1 1 Figure 1127 Magnitude and phase Bode diagram for the RC lowpass filter frequency response 20000 20000 3 dB ω ω 20 dBdecade 4 π 2 π jω 1 1 20000 2000 200000 20000 2000 200000 jω 1 1 rob28124ch11509591indd 526 061216 218 pm 113 ContinuousTime Filters 527 The diagrams are very similar to those for the singlenegativereal pole except that the magnitude asymptote above the corner frequency has a slope of 6 dB per octave or 20 dB per decade and the phase approaches π2 radians instead of π2 radians They are basically the singlenegativerealpole Bode diagrams turned upside down For a subsystem with a singlepositivereal zero and no finite poles of the form H jω 1 jω z k z k 0 the magnitude graph is the same as in Figure 1128 but the phase approaches π2 instead of π2 at frequencies above the corner frequency Integrators and Differentiators We must also consider a pole or a zero at zero frequency Figure 1129 and Figure 1130 A system component with a single pole at s 0 is called an integrator because its transfer function is Hs 1s and division by s corresponds to integration in the time domain A system component with a single zero at s 0 is called a differentiator because its transfer function is Hs s and multiplication by s corresponds to differentiation in the time domain FrequencyIndependent Gain The only remaining type of simple system compo nent is a frequencyindependent gain Figure 1131 In Figure 1131 the gain constant A is assumed to be positive That is why the phase is zero If A is negative the phase is π radians The asymptotes are helpful in drawing the actual Bode diagram and they are espe cially helpful in sketching the overall Bode diagram for a more complicated system The asymptotes can be quickly sketched from knowledge of a few simple rules and added together Then the magnitude Bode diagram can be sketched approximately by drawing a smooth curve that approaches the asymptotes and deviates at the corners by 3 dB Figure 1129 The magnitude and phase frequency response of a single pole at s 0 ω ω jω 1 jω 1 Slope of 6 dBoctave or 20 dBdecade 2 π 1 10 01 1 10 01 Figure 1128 The magnitude and phase frequency response of a singlenegativerealzero subsystem ω 3 dB Slope of 6 dBoctave or 20 dBdecade ω 4 π 2 π zk 10zk 01 Asymptotes Asymptote zk jω 1 zk jω 1 zk 10zk 01 Figure 1128 rob28124ch11509591indd 527 061216 218 pm C h a p t e r 11 Frequency Response Analysis 528 ExamplE 112 Bode diagram of the frequency response of an RC circuit Graph the Bode diagram for the voltage frequency response of the circuit in Figure 1132 where C 1 1 F C 2 2 F R s 4 Ω R 1 2 Ω R 2 3 Ω Figure 1130 The magnitude and phase frequency response of a single zero at s 0 ω Slope of 6 dBoctave or 20 dBdecade jω ω 2 π 10 1 01 10 1 01 jω Figure 1131 The magnitude and phase frequency response of a frequencyindependent gain A ω ω A 20log10A A Figure 1132 An RC Circuit vit vot R1 Rs C1 R2 C2 The transfer function is Hs 1 R s C 2 s 1 R 1 C 1 s 2 C 1 C 2 R s C 1 C 2 R 1 C 1 R 2 C 2 R 1 R 2 C 1 C 2 s R 1 R 2 R s R 1 R 2 R s C 1 C 2 Substituting s jω and using numerical values for the components the frequency response is H jω 3 j2ω 1 48 jω 2 j50ω 9 0125 jω 05 jω 02316 jω 08104 H jω 0333 1 jω 05 1 jω 02316 1 jω 08104 A 1 jω z 1 1 jω p 1 1 jω p 2 rob28124ch11509591indd 528 061216 218 pm 113 ContinuousTime Filters 529 The following MATLAB program demonstrates some techniques for drawing Bode diagrams Set up a logarithmic vector of radian frequencies for graphing the Bode diagram from 001 to 10 radsec w logspace21200 Set the gain zero and pole values A 03333 z1 05 p1 02316 p2 08104 Compute the complex frequency response H A1jwz11jwp11jwp2 Graph the magnitude Bode diagram subplot211 p semilogxw20log10absHk setpLineWidth2 grid on xlabelomegaFontSize18FontNameTimes Figure 1133 Individual asymptotic and overall asymptotic and exact Bode magnitude and phase diagrams for the circuit voltage frequency response 40 20 20 40 2 2 1 decade 1 decade 1 decade 01 1 40 20 20 40 ω 2 2 01 1 ω 01 1 ω 2 2 01 1 ω 1 decade 1 decade 2 2 01 1 ω 2 2 1 decade 1 decade 01 1 ω 40 20 20 40 1 decade 01 1 ω 40 20 20 40 1 decade 01 1 ω 01 1 ω 30 20 10 HdB jω 01 1 ω A Zero dB Radians Pole 1 Pole 2 H jω where A 0333 z 1 05 p 1 02316 p 2 08104 So this frequency response has two finite poles one finite zero and one frequency independent gain We can quickly construct an overall asymptotic Bode diagram by adding the asymptotic Bode diagrams for the four individual components of the overall frequency response Figure 1133 rob28124ch11509591indd 529 061216 218 pm C h a p t e r 11 Frequency Response Analysis 530 ylabelHitjomegadBFontSize18FontNameTimes titleMagnitudeFontSize24FontNameTimes setgcaFontSize14FontNameTimes Graph the phase Bode diagram subplot212 p semilogxwangleHk setpLineWidth2 grid on xlabelomegaFontSize18FontNameTimes ylabelPhase of HitjomegaFontSize18FontNameTimes titlePhaseFontSize24FontNameTimes setgcaFontSize14FontNameTimes The resulting magnitude and phase Bode diagrams are illustrated in Figure 1134 Complex Pole and Zero Pairs Now consider the more complicated case of complex poles and zeros For real system functions they always occur in complex conjugate pairs So a complex conjugate pair of poles with no finite zeros would form a subsys tem transfer function Hs 1 1 sp11 sp2 1 1 1 p 1 1 p 1 s s 2 p 1 p 1 and frequency response H jω 1 1 jωp11 jωp2 1 1 jω1 p 1 1 p 1 jω 2 p 1 p 1 Figure 1134 Magnitude and phase Bode diagrams of the frequency response of the filter 102 101 100 101 40 30 20 10 0 ω H jωdB Magnitude 102 101 100 101 2 15 1 05 0 ω Phase Phase of H jω rob28124ch11509591indd 530 061216 218 pm 113 ContinuousTime Filters 531 or H jω 1 1 jω 2Re p 1 p 1 2 jω 2 p 1 2 From the table of Fourier pairs we find the pair e ω n ζt sin ω n 1 ζ 2 t u t ℱ ω n 1 ζ 2 jω 2 jω2ζ ω n ω n 2 in the ω domain which can be expressed in the form ω n e ω n ζt sin ω n 1 ζ 2 t 1 ζ 2 ut ℱ 1 1 jω 2ζ ω n ω n 2 jω ω n 2 2 whose right side is of the same functional form as H jω 1 1 jω 2 Re p 1 p 1 2 jω 2 p 1 2 This is a standard form of a secondorder underdamped system response where the natural radian frequency is ω n and the damping ratio is ζ Therefore for this type of subsystem ω n 2 p 1 2 p 1 p 2 and ζ Re p 1 ω n p1 p2 2 p1 p2 The Bode diagram for this subsystem is illustrated in Figure 1135 Figure 1135 Magnitude and phase Bode diagram for a secondorder complex pole pair 50 40 30 20 10 0 10 20 0 ζ 005 ζ 01 ζ 02 ζ 05 ζ 1 ζ 005 ζ 01 ζ 02 ζ 05 ζ 1 ωn 01ωn 10ωn 01ωn 10ωn ωn ω ω π 2 π H jωdB H jω rob28124ch11509591indd 531 061216 218 pm C h a p t e r 11 Frequency Response Analysis 532 A complex pair of zeros would form a subsystem frequency response of the form H jω 1 jω z1 1 jω z2 1 jω 1 z1 1 z 1 jω 2 z 1 z 1 1 jω 2 Re z 1 z 1 2 jω 2 z 1 2 In this type of subsystem we can identify the natural radian frequency and the damping ratio as ω n 2 z 1 2 z 1 z 2 and ζ Re z 1 ω n z 1 z 2 2 z 1 z 2 The Bode diagram for this subsystem is illustrated in Figure 1136 Figure 1136 Magnitude and phase Bode diagram for a secondorder complex zero pair 30 20 10 0 10 20 30 40 0 ζ 1 ζ 05 ζ 02 ζ 01 ζ 005 ζ 1 ζ 05 ζ 02 ζ 01 ζ 005 ωn 01ωn 10ωn 01ωn 10ωn ωn ω ω π 2 π H jωdB H jω PRACTICAL FILTERS Passive Filters The Lowpass Filter Approximations to the ideal lowpass and bandpass filters can be made with certain types of circuits The simplest approximation to the ideal lowpass filter is the one we have already analyzed more than once the socalled RC lowpass filter Figure 1137 We have found its response to a step and to a sinusoid Let us now analyze it directly in the frequency domain The differential equation describing this circuit is RC v out t v out t v in t Laplace transforming both sides assuming no initial charge on the capacitor sRC V out s V out s V in s We can now solve directly for the transfer function Hs V out s V in s 1 sRC 1 rob28124ch11509591indd 532 061216 218 pm 113 ContinuousTime Filters 533 The method commonly used in elementary circuit analysis to solve for the frequency response is based on the phasor and impedance concepts Impedance is a generalization of the idea of resistance to apply to inductors and capacitors Recall the voltagecurrent relationships for resistors capacitors and inductors Figure 1138 If we Laplace transform these relationships we get Vs R Is Vs sL Is and Is sC Vs The impedance concept comes from the similarity of the inductor and capacitor equa tions to Ohms law for resistors If we form the ratios of voltage to current we get Vs Is R Vs Is sL and Vs Is 1 sC For resistors this ratio is called resistance In the generalization this ratio is called impedance Impedance is conventionally symbolized by Z Using that symbol Z R s R Z L s sL and Z C s 1sC This allows us to apply many of the techniques of resistive circuit analysis to circuits that contain inductors and capacitors and are analyzed in the frequency domain In the case of the RC lowpass filter we can view it as a voltage divider Figure 1139 Figure 1137 Practical RC lowpass filter R C vint voutt Figure 1138 Defining equations for resistors capacitors and inductors vt it vt vt it it vt Rit vt L i t it C v t Figure 1139 Impedance voltage divider representation of the RC lowpass filter ZR jω ZC jω Vin jω Vout jω Then we can directly write the transfer function in the frequency domain Hs V out s V in s Z c s Z c s Z f s 1sC 1sC R 1 sRC 1 and the frequency response as H jω 1 jωRC 1 or H f 1 j2π f RC 1 rob28124ch11509591indd 533 061216 218 pm C h a p t e r 11 Frequency Response Analysis 534 arriving at the same result as before without a direct reference to the time domain The magnitude and phase of the RC lowpass filter frequency response are illustrated in Figure 1140 The impulse response of the RC lowpass filter is the inverse CTFT of its frequency response ht e tRC RC ut as illustrated in Figure 1141 For this physically realizable filter the impulse response is zero before time t 0 The filter is causal At very low frequencies approaching zero the capacitors impedance is much greater in magnitude than the resistors impedance the voltage division ratio ap proaches one and the output voltage signal and input voltage signal are about the same At very high frequencies the capacitors impedance becomes much smaller in magnitude than the resistors impedance and the voltage division ratio approaches zero Thus we can say approximately that low frequencies pass through and high frequencies get stopped This qualitative analysis of the circuit agrees with the math ematical form of the frequency response H jω 1 jωRC 1 At low frequencies lim ω0 H jω 1 and at high frequencies lim ω H jω 0 The RC lowpass filter is lowpass only because the excitation is defined as the volt age at the input and the response is defined as the voltage at the output If the response Figure 1140 Magnitude and phase frequency responses of an RC lowpass filter 2 1 RC 1 RC 1 1 ω ω H jω H jω 45 90 45 90 Figure 1141 Impulse response of an RC lowpass filter t ht RC 1 RC rob28124ch11509591indd 534 061216 218 pm 113 ContinuousTime Filters 535 The Bandpass Filter One of the simplest forms of a practical bandpass filter is illustrated in Figure 1143 Hs V out s V in s sRC s 2 sRC 1LC H jω jωRC jω 2 jωRC 1LC At very low frequencies the capacitor is an open circuit and the inductor is a perfect conductor Therefore at very low frequencies the output voltage signal is practically zero At very high frequencies the inductor is an open circuit and the capacitor is a perfect conductor again making the output voltage signal zero The impedance of the parallel inductorcapacitor combination is Z LC s sLsC sL 1sC sL s 2 LC 1 For s 2 LC 1 0 s j 1LC ω 1 LC the impedance is infinite This frequency is called the resonant frequency So at the resonant frequency of the parallelLC circuit the impedance of that parallel combination of inductor and ca pacitor goes to infinity and the output voltage signal is the same as the input voltage signal The overall behavior of the circuit is to approximately pass frequencies near the resonant frequency and block other frequencies hence it is a practical bandpass filter Figure 1143 An RLC practical bandpass filter R C L vint voutt Figure 1142 Alternate form of a practical lowpass filter L R vint voutt had been defined as the current the nature of the filtering process would change com pletely In that case the frequency response would become H jω I jω V in jω 1 Z R jω Z c jω 1 1jωC R jωC jωRC 1 With this definition at low frequencies the capacitor impedance is very large blocking current flow so the response approaches zero At high frequencies the capacitor im pedance approaches zero so the circuit responds as though the capacitor were a perfect conductor and the current flow is determined by the resistance R Mathematically the response approaches zero at low frequencies and approaches the constant 1R at high frequencies This defines a highpass filter lim ω0 H jω 0 and lim ω H jω 1R Another much less common form of lowpass filter is illustrated in Figure 1142 Hs V out s V in s R sL R H jω R jωL R Using the impedance and voltage divider ideas can you explain in words why this circuit is a lowpass filter rob28124ch11509591indd 535 061216 218 pm C h a p t e r 11 Frequency Response Analysis 536 A graph of the magnitude and phase of the frequency response Figure 1144 for a particular choice of component values reveals its bandpass nature The impulse response of the RLC bandpass filter is ht 2ζ ω n e ζ ω n t cos ω c t ζ 1 ζ 2 sin ω c t ut where 2ζ ω n 1RC ω n 2 1LC and ω c ω n 1 ζ 2 Figure 1145 Notice that the impulse response of this physically realizable filter is causal All physical systems are filters in the sense that each of them has a response that has a characteristic variation with frequency This is what gives a musical instrument and each human voice its characteristic sound To see how important this is try playing just the mouthpiece of any wind instrument The sound is very unpleasant until the in strument is attached then it becomes very pleasant when played by a good musician The sun periodically heats the earth as it rotates and the earth acts like a lowpass filter smoothing out the daily variations and responding with a lagging seasonal variation of temperature In prehistoric times people tended to live in caves because the thermal mass of the rock around them smoothed out the seasonal variation of temperature and allowed them to be cooler in the summer and warmer in the winter another exam ple of lowpass filtering Industrial foamrubber ear plugs are designed to allow lower frequencies through so that people wearing them can converse but to block intense highfrequency sounds that may damage the ear The list of examples of systems that we are familiar with in daily life that perform filtering operations is endless Active Filters All the practical filters we have examined so far have been passive filters The term passive means they contained no devices with the capability of producing an output signal with more actual power not signal power than the input signal Many modern filters are active filters They contain active devices like transistors andor operational amplifiers and require an external source of power to operate properly With the use of active devices the actual output power can be greater than the actual input power Figure 1144 Magnitude and phase frequency responses of a practical RLC bandpass filter ω ω H jω 1 1 LC H jω 1 LC 90 90 Figure 1145 Impulse response of a practical RLC bandpass filter t ht 1 RC 2π ωn 1ζ2 rob28124ch11509591indd 536 061216 218 pm 113 ContinuousTime Filters 537 The subject of active filters is a large one and only the simplest forms of active filters will be introduced here3 Operational Amplifiers There are two commonly used forms of operational am plifier circuits the inverting amplifier form and the noninverting amplifier form Figure 1146 The analysis here will use the simplest possible model for the opera tional amplifier the ideal operational amplifier An ideal operational amplifier has infinite input impedance zero output impedance infinite gain and infinite bandwidth For each type of amplifier there are two impedances Z i s and Z f s that control the transfer function The transfer function of the inverting amplifier can be derived by observing that since the operational amplifier input impedance is infinite the current flowing into either input terminal is zero and therefore I f s I i s 115 Also since the output voltage is finite and the operational amplifier gain is infinite the voltage difference between the two input terminals must be zero Therefore I i s V i s Z i s 116 and I f s V f s Z f s 117 Equating 116 and 117 according to 115 and solving for the transfer function Hs V o s V i s Z f s Z i s 118 Similarly it can be shown that the noninverting amplifier transfer function is Hs V o s V i s Z f s Z i s Z i s 1 Z f s Z i s 119 3 In some passive circuits there is voltage gain at some frequencies The output voltage signal can be larger than the input voltage signal Therefore the output signal power as defined previously would be greater than the input signal power But this is not actual power gain because that higher output voltage signal is across a higher impedance Figure 1146 Two common forms of amplifiers utilizing operational amplifiers Inverting Amplifier Zis Zfs Vis Iis Vos Ifs Vis Vos Zf s Zis Noninverting Amplifier rob28124ch11509591indd 537 061216 218 pm C h a p t e r 11 Frequency Response Analysis 538 The Integrator Probably the most common and simplest form of active filter is the active integrator Figure 1147 Using the inverting amplifier gain formula 118 for the transfer function Hs Z f s Z i s 1sC R 1 sRC H f 1 j2π f RC The action of the integrator is easier to see if the frequency response is rearranged to the form V o f 1 RC V i f j2π f or Vo jω 1 RC V i jω jω The integrator integrates the signal but at the same time multiplies it by 1RC Notice that we did not introduce a practical passive integrator The passive RC lowpass filter acts much like an integrator for frequencies well above its corner frequency but at a low enough frequency its response is not like an integrator So the active device the operational amplifier in this case has given the filter designer another degree of freedom in design Figure 1148 An active RC lowpass filter vit vot vxt Ri C iit ift Rf Figure 1147 An active integrator vit vot vxt R C iit ift The Lowpass Filter The integrator is easily changed to a lowpass filter by the addi tion of a single resistor Figure 1148 For this circuit Hs V 0 s V i s R f R i 1 sC R f 1 H jω V 0 jω V i jω R f R i 1 jωC R f 1 This frequency response has the same functional form as the passive RC lowpass filter except for the factor R f R i So this is a filter with gain It filters and amplifies the sig nal simultaneously In this case the voltage gain is negative ExamplE 113 Bode diagram of the frequency response of a twostage active filter Graph the Bode magnitude and phase diagrams for the twostage active filter in Figure 1149 The transfer function of the first stage is H 1 s Z f 1 s Z i1 s R f 1 R i1 1 1 s C f 1 R f 1 rob28124ch11509591indd 538 061216 218 pm 113 ContinuousTime Filters 539 The transfer function of the second stage is H 2 s Z f 1 s Z i1 s s R f 2 C i2 1 s R f 2 C f 2 Since the output impedance of an ideal operational amplifier is zero the second stage does not load the first stage and the overall transfer function is simply the product of the two transfer functions Hs R f 1 R i1 s R f 2 C i2 1 s C f 1 R f 1 1 s R f 2 C f 2 Substituting in parameter values and letting s j2π f we get the frequency response H f j1000 f 1000 jf 101000 jf Figure 1150 This is a practical bandpass filter Figure 1149 A twostage active filter Ri1 160 Ω Rf1 160 Ω Ci2 1 μF Cf 2 1 μF Rf 2 160 Ω Cf1 01 μF vi f vo f Figure 1150 Bode diagram of the frequency response of the twostage active filter f H f dB 40 30 20 10 f 102 103 104 105 106 102 103 104 105 106 π 2 π 2 Asymptotes H f rob28124ch11509591indd 539 061216 218 pm C h a p t e r 11 Frequency Response Analysis 540 ExamplE 114 Design of an active highpass filter Design an active filter that attenuates signals at 60 Hz and below by more than 40 dB and amplifies signals at 10 kHz and above with a positive gain that deviates from 20 dB by no more than 2 dB This specifies a highpass filter The gain must be positive A positive gain and some high pass filtering can be accomplished by one noninverting amplifier However looking at the trans fer function and frequency response for the noninverting amplifier Hs V o s V i s Z f s Z i s Z i s H jω Z f jω Z i jω Z i jω we see that if the two impedances consist of only resistors and capacitors its gain is never less than one and we need attenuation a gain less than one at low frequencies If we were to use both inductors and capacitors we could make the magnitude of the sum Z f jω Z i jω be less than the magnitude of Z i jω at some frequencies and achieve a gain less than one But we could not make that occur for all frequencies below 60 Hz and the use of inductors is generally avoided in practical design unless absolutely necessary There are other practical difficulties with this idea also using real as opposed to ideal operational amplifiers If we use one inverting amplifier we have a negative gain But we could follow it with an other inverting amplifier making the overall gain positive Gain is the opposite of attenuation If the attenuation is 60 dB the gain is 60 dB If the gain at 60 Hz is 40 dB and the response is that of a singlepole highpass filter the Bode diagram asymptote on the magnitude frequency response would pass through 20 dB of gain at 600 Hz 0 dB of gain at 6 kHz and 20 dB of gain at 60 kHz But we need 20 dB of gain at 10 kHz so a singlepole filter is inadequate to meet the specifications We need a twopole highpass filter We can achieve that with a cascade of two singlepole highpass filters meeting the requirements for attenuation and positive gain simultaneously Now we must choose Z f jω and Z i jω to make the inverting amplifier a highpass filter Figure 1148 illustrates an active lowpass filter That filter is lowpass because the gain is Z f jω Z i jω Z i jω is constant and Z f jω has a larger magnitude at low frequencies than at high frequencies There is more than one way to make a highpass filter using the same inverting amplifier configuration We could make the magnitude of Z f jω be small at low fre quencies and larger at high frequencies That requires the use of an inductor but again for prac tical reasons inductors should be avoided unless really needed We could make Z f jω constant and make the magnitude of Z i jω large at low frequencies and small at high frequencies That general goal can be accomplished by either a parallel or series combination of a resistor and a capacitor Figure 1151 Figure 1151 Two ideas for a highpass filter using only capacitors and resistors Rf Ri Ci Rf Ri Ci a b rob28124ch11509591indd 540 061216 218 pm 113 ContinuousTime Filters 541 If we just think about the limiting behavior of these two design ideas at very low and very high frequencies we immediately see that only one of them meets the specifications of this de sign Design a has a finite gain at very low frequencies and a gain that rises with frequency at higher frequencies never approaching a constant Design b has a gain that falls with frequency at low frequencies approaching zero at zero frequency and approaching a constant gain at high frequencies Design b can be used to meet our specification So now the design is a cascade of two inverting amplifiers Figure 1152 Figure 1152 Cascade of two inverting highpass active filters Rf2 Ri2 Ci2 Rf1 Ri1 Ci1 At this point we must select the resistor and capacitor values to meet the attenuation and gain requirements There are many ways of doing that The design is not unique We can begin by selecting the resistors to meet the highfrequency gain requirement of 20 dB That is an overall highfrequency gain of 10 which we can apportion any way we want between the two amplifiers Lets let the two stage gains be approximately the same Then the resistor ratios in each stage should be about 316 We should choose resistors large enough not to load the outputs of the operational amplifiers but small enough that stray capacitances dont cause problems Resistors in the range of 500 Ω to 50 kΩ are usually good choices But unless we are willing to pay a lot we cannot arbitrarily choose a resistor value Resistors come in standard values typically in a sequence of 1 12 15 18 22 27 33 39 47 56 68 82 10 n where n sets the decade of the resistance value Some ratios that are very near 316 are 39 12 325 47 15 313 56 18 311 68 22 309 82 27 303 To set the overall gain very near 10 we can choose the firststage ratio to be 3912 325 and the secondstage ratio to be 6822 309 and achieve an overall highfrequency gain of 10043 So we set R f 1 39 kΩ R i1 12 kΩ R f 2 68 kΩ R i 2 22 kΩ Now we must choose the capacitor values to achieve the attenuation at 60 Hz and below and the gain at 10 kHz and above To simplify the design lets set the two corner frequencies of the two stages at the same or very nearly the same value With a twopole lowfrequency rolloff of 40 dB per decade and a highfrequency gain of approximately 20 dB we get a 60dB difference between the frequency response magnitude at 60 Hz and 10 kHz If we were to set the gain at 60 Hz to be exactly 40 dB then at 600 Hz we would have approximately 0 dB gain and at 6 kHz we would have a gain of 40 dB and it would be higher at 10 kHz This does not meet the specification We can start at the highfrequency end and set the gain at 10 kHz to be approximately 10 meaning that the corner for the lowfrequency rolloff should be well below 10 kHz If we put it rob28124ch11509591indd 541 061216 218 pm C h a p t e r 11 Frequency Response Analysis 542 at 1 kHz the approximate gain at 100 Hz based on asymptotic approximations will be 20 dB and at 10 Hz it will be 60 dB We need 40 dB at 60 Hz But we only get about 29 dB at 60 Hz So we need to put the corner frequency a little higher say 3 kHz If we put the corner frequency at 3 kHz the calculated capacitor values will be C i1 46 nF and C i2 24 nF Again we cannot arbitrarily choose a capacitor value Standard capacitor values are typically arrayed at the same intervals as standard resistor values 1 12 15 18 22 27 33 39 47 56 68 82 10 n There is some leeway in the location of the corner frequency so we probably dont need a really precise value of capacitance We can choose C i1 047 nF and C i2 22 nF making one a little high and one a little low This will separate the poles slightly but will still create the desired 40 dB per decade lowfrequency rolloff This looks like a good design but we need to verify its performance by drawing a Bode diagram Figure 1153 Figure 1153 Bode diagram for twostage active highpass filter design f H f dB 80 40 20 f π π 10 kHz 60 Hz H f It is apparent from the diagram that the attenuation at 60 Hz is adequate Calculation of the gain at 10 kHz yields about 192 dB which also meets specifications These results are based on exact values of resistors and capacitors In reality all resistors and capacitors are typically chosen based on their nominal values but their actual values may differ from the nominal by a few percent So any good design should have some tolerance in the specifications to allow for small deviations of component values from the design values ExamplE 115 SallenKey bandpass filter A popular filter design that can be found in many books on electronics or filters is the twopole singlestage SallenKey or constantK bandpass filter Figure 1154 The triangle symbol with the K inside represents an ideal noninverting amplifier with a finite voltage gain K an infinite input impedance a zero output impedance and infinite rob28124ch11509591indd 542 061216 218 pm 113 ContinuousTime Filters 543 Figure 1154 SallenKey or constantK bandpass filter vit vot K R2 R1 C2 C1 bandwidth not an operational amplifier The overall bandpassfilter transfer function and frequency response are Hs V o s V i s s K 1 K 1 R 1 C 2 s 2 1 R 1 C 1 1 R 2 C 2 1 R 1 C 2 1 K s 1 R 1 R 2 C 1 C 2 and H jω V o jω V i jω jω K 1 K 1 R 1 C 2 jω 2 jω 1 R 1 C 1 1 R 2 C 2 1 R 1 C 2 1 K 1 R 1 R 2 C 1 C 2 The frequency response is of the form H jω H 0 j2ζ ω 0 2 jω 2 2ζ ω 0 jω ω 0 2 jωA jω 2 2ζ ω 0 jω ω 0 2 where A K 1 K 1 R 1 C 2 ω 0 2 1 R 1 R 2 C 1 C 2 ζ R 1 C 1 R 2 C 2 R 2 C 1 1 K 2 R 1 R 2 C 1 C 2 Q 1 2ζ R 1 R 2 C 1 C 2 R 1 C 1 R 2 C 2 R 2 C 1 1 K and H 0 K 1 1 K C 2 C 1 R 1 R 2 The recommended design procedure is to choose the Q and the resonant frequency f 0 ω 0 2π choose C 1 C 2 C as some convenient value and then calculate R 1 R 2 1 2π f 0 C and K 3Q 1 2Q 1 and H 0 3Q 1 rob28124ch11509591indd 543 061216 218 pm C h a p t e r 11 Frequency Response Analysis 544 Also it is recommended that Q should be less than 10 for this design Design a filter of this type with a Q of 5 and a center frequency of 50 kHz We can pick convenient values of capacitance so let C 1 C 2 C 10 nF Then R 1 R 2 318 Ω and K 1556 and H 0 14 That makes the frequency response H jω jω8792 10 5 jω 2 64 10 4 jω 986 10 10 or written as a function of cyclic frequency H f j2π f 8792 10 5 j2πf 2 64 10 4 j2π f 986 10 10 Figure 1155 Figure 1155 Bode diagram of the SallenKey bandpass filter frequency response f H f 50 20 f π 50 kHz 103 104 105 106 107 226 dB 2 π 2 H f 103 104 105 106 107 As in the previous example we cannot choose the component values to be exactly those calculated but we can come close We would probably have to use nominal 330 Ω resistors and that would alter the frequency response slightly depending on their actual values and the actual values of the capacitors ExamplE 116 Biquadratic RLC active filter The biquadratic filter introduced in Section 112 can be realized as an active filter Figure 1156 Under the assumption of an ideal operational amplifier the transfer function can be found using standard circuit analysis techniques It is Hs V o s V i s s 2 R R 1 R 2 R 1 R f R 2 L R 1 R 2 s 1 LC s 2 R R 1 R 2 R 2 R s R 1 L R 1 R 2 s 1 LC rob28124ch11509591indd 544 061216 218 pm 113 ContinuousTime Filters 545 Consider the two cases R 1 0 R 2 0 and R 1 0 R 2 0 If R 1 0 R 2 0 then the frequency response is H jω jω 2 jωR R f L 1LC jω 2 jωRL 1LC The natural radian frequency is ω n 1 LC There are poles at jω R2L R2L 2 1LC and zeros at jω R R f 2L R R f 2L 2 1 LC and at low and high frequencies and at resonance lim ω0 Hjω 1 lim ω H jω 1 H j ω n R R f R 1 If R 2 L C and R R f 2 L C the poles are complex and the zeros are real and the dominant effect near ω n is an increase in the frequency response magnitude Notice that in this case the frequency response does not depend on R 1 This condition is just like having the RLC resonant circuit in the feedback with the potentiometer removed If R 1 0 R 2 0 then H jω jω 2 jω R L 1 LC jω 2 jω R R s L 1 LC The natural radian frequency is ω n 1 LC There are zeros at jω R 2L R 2L 2 1 LC and poles at jω R R s 2L R R s 2L 2 1 LC Figure 1156 Active RLC realization of a biquadratic filter Rf vo vx vy R L C Rs R2 R1 vi rob28124ch11509591indd 545 061216 218 pm C h a p t e r 11 Frequency Response Analysis 546 and at low and high frequencies and at resonance lim ω0 H jω 1 lim ω H jω 1 H j ω n R R R s 1 If R 2 L C and R R s 2 L C the zeros are complex and the poles are real and the dominant effect near ω n is a decrease in the frequency response magnitude Notice that in this case the frequency response does not depend on R 2 This condition is just like having the RLC resonant circuit on the input of the amplifier with the potentiometer removed If R 1 R 2 and R f R s the frequency response is H jω 1 and the output signal is the same as the input signal So one potentiometer can determine whether the frequency response magnitude is in creased or decreased near a resonant frequency The graphic equalizer of Section 112 could be realized with a cascade connection of 9 to 11 such biquadratic filters with their resonant frequencies spaced apart by octaves But it can also be realized with only one operational am plifier as illustrated in Figure 1157 Because of the interaction of the passive RLC networks the operation of this circuit is not identical to that of multiple cascadeconnected biquadratic filters but it accomplishes the same goal with fewer parts Figure 1157 A circuit realization of a graphic equalizer with only one operational amplifier Rf vo vx vy vi Ra La Ca Rs R2a R1a vx Rb Lb Cb R2b R1b vx Rc Lc Cc R2c R1c vx Rk Lk Ck R2k R1k 114 DISCRETETIME FILTERS NOTATION The discretetime Fourier transform DTFT was derived from the z transform by mak ing the change of variable z e j2πF or z e jΩ where F and Ω are both real variables representing frequency cyclic and radian In the literature on discretetime digital rob28124ch11509591indd 546 061216 218 pm 114 DiscreteTime Filters 547 systems the most commonly used variable for frequency is radian frequency Ω So in the following sections on discretetime filters we will also use Ω predominantly4 IDEAL FILTERS The analysis and design of discretetime filters have many parallels with the analysis and design of continuoustime filters In this and the next section we will explore the characteristics of discretetime filters using many of the techniques and much of the terminology developed for continuoustime filters Distortion The term distortion means the same thing for discretetime filters as it does for continuoustime filters changing the shape of a signal Suppose a signal xn has the shape illustrated at the top of Figure 1158a Then the signal at the bottom of Figure 1158a is an undistorted version of that signal Figure 1158b illustrates one type of distortion Just as was true for continuoustime filters the impulse response of a filter that does not distort is an impulse possibly with a strength other than one and possibly shifted in time The most general form of an impulse response of a distortionless sys tem is hn Aδn n 0 The corresponding frequency response is the DTFT of the impulse response H e jΩ A e jΩ n 0 The frequency response can be characterized by its magnitude and phase H e jΩ A and H e jΩ Ω n 0 Therefore a distortionless 4 The reader should be aware that notation varies widely among books and papers in this area The DTFT of a discretetime function xn might be written in any of the forms X e j2πf X e jΩ XΩ X e jω Xω Some authors use the same symbol ω for radian frequency in both continuous and discrete time Some authors use ω and f in discrete time and Ω and F in continuous time Some authors preserve the meaning of X as the z transform of x by replacing z by e jΩ or e jω Other authors redefine the function X and the DTFT by using Ω or ω as the independent variable All notation forms have advantages and disadvantages Figure 1158 a An original signal and a changed but undistorted version of it b an original signal and a distorted version of it n 32 xn 1 1 Original Signal n 32 xn 1 1 Attenuated Signal a n 32 xn 4 Original Signal n 32 xn 1 LogAmplified Signal b rob28124ch11509591indd 547 061216 218 pm C h a p t e r 11 Frequency Response Analysis 548 system has a frequency response magnitude that is constant with frequency and a phase that is linear with frequency Figure 1159 The magnitude frequency response of a distortionless system is constant and the phase frequency response is linear over the range π Ω π and repeats periodically outside that range Since n 0 is an integer the magnitude and phase of a distortionless filter are guaranteed to repeat every time Ω changes by 2π Filter Classifications The terms passband and stopband have the same significance for discretetime filters as they do for continuoustime filters The descriptions of ideal discretetime filters are similar in concept but have to be modified slightly because of the fact that all discretetime systems have periodic frequency responses They are periodic because in the signal A cos Ω 0 n if Ω 0 is changed by adding 2πm m an integer the signal be comes A cos Ω 0 2π mn and the signal is unchanged because A cos Ω 0 n A cos Ω 0 2πmn A cos Ω 0 n 2πmn m an integer Therefore a discretetime filter is classified by its frequency response over the base period π Ω π An ideal lowpass filter passes signal power for frequencies 0 Ω Ω m π without distortion and stops signal power at other frequencies in the range π Ω π An ideal highpass filter stops signal power for frequencies 0 Ω Ω m π and passes signal power at other frequencies in the range π Ω π without distortion An ideal bandpass filter passes signal power for frequencies 0 Ω L Ω Ω H π without distortion and stops signal power at other frequencies in the range π Ω π An ideal bandstop filter stops signal power for frequencies 0 Ω L Ω Ω H π and passes signal power at other frequencies in the range π Ω π without distortion Frequency Responses In Figure 1160 and Figure 1161 are the magnitude and phase frequency responses of the four basic types of ideal filters Impulse Responses and Causality The impulse responses of ideal filters are the inverse transforms of their frequency responses The impulse and frequency responses of the four basic types of ideal filter are summarized in Figure 1162 These descriptions are general in the sense that they involve an arbitrary gain constant A and an arbitrary time delay n 0 In Figure 1163 are some typical shapes of impulse responses for the four basic types of ideal filter The consideration of causality is the same for discretetime filters as for continuoustime filters Like ideal continuoustime filters ideal discretetime filters have noncausal impulse responses and are therefore physically impossible to build Figure 1159 Magnitude and phase of a distortionless system HejΩ Ω Ω A Ω 2πn0 2π 2π 2π HejΩ rob28124ch11509591indd 548 061216 218 pm 114 DiscreteTime Filters 549 Figure 1162 Frequency responses and impulse responses of the four basic types of ideal filter Filter Type Frequency Response Lowpass H e jΩ A rectΩ2 Ω m e jΩ n 0 δ 2π Ω Highpass H e jΩ A e jΩ n 0 1 rectΩ2 Ω m δ 2π Ω Bandpass H e jΩ A rect Ω Ω 0 ΔΩ rect Ω Ω 0 ΔΩ e jΩ n 0 δ 2π Ω Bandstop H e jΩ A e jΩ n 0 1 rect Ω Ω 0 ΔΩ rect Ω Ω 0 ΔΩ δ 2π Ω Filter Type Impulse Response Lowpass hn A Ω m π sinc Ω m n n 0 π Highpass hn Aδn n 0 A Ω m π sinc Ω m n n 0 π Bandpass hn 2AΔf sincΔf t t 0 cos 2π f 0 t t 0 Bandstop hn Aδn n 0 AΔΩπsincΔΩn n 0 2π cos Ω 0 n n 0 ΔΩ Ω H Ω L Ω 0 Ω H Ω L 2 Figure 1160 Magnitude and phase frequency responses of ideal lowpass and highpass filters Ideal Lowpass Filter Ideal Highpass Filter Ω He jΩ Ω Ωm Ωm He jΩ Ω He jΩ Ω Ωm Ωm HejΩ 2π 2π 2π 2π 2π 2π 2π 2π Figure 1161 Magnitude and phase frequency responses of ideal bandpass and bandstop filters Ω HejΩ Ω ΩL ΩH ΩL ΩH Ideal Bandstop Filter Ω HejΩ Ω Ideal Bandpass Filter ΩL ΩH ΩL ΩH 2π 2π 2π 2π 2π 2π 2π 2π HejΩ HejΩ In Figure 1164 and Figure 1165 are some examples of the impulse responses frequency responses and responses to rectangular waves of some nonideal causal fil ters that approximate the four common types of ideal filters In each case the frequency response is graphed only over the base period π Ω π The effects of these practical filters on the rectangular waves are similar to those shown for the corresponding continuoustime filters Filtering Images One interesting way to demonstrate what filters do is to filter an image An image is a twodimensional signal Images can be acquired in various ways A film camera exposes lightsensitive film to a scene through a lens system which puts an optical rob28124ch11509591indd 549 061216 218 pm C h a p t e r 11 Frequency Response Analysis 550 Figure 1164 Impulse responses frequency responses and responses to rectangular waves of causal lowpass and bandpass filters n 5 25 hn 005 035 Causal Lowpass hn n 5 25 xn 1 Excitation n 5 25 yn 02 12 Response Ω π π HejΩ 1 Ω π π 4 4 HejΩ n 5 25 hn 02 025 Causal Bandpass hn n 5 25 xn 1 Excitation n 5 25 yn 1 08 Response π π 1 π π 4 4 He jΩ He jΩ Ω Ω Figure 1165 Impulse responses frequency responses and responses to rectangular waves of causal highpass and bandstop filters n 5 25 hn 06 04 Causal Highpass hn 5 25 xn 1 Excitation n 5 25 yn 03 03 Response Ω π π He jΩ 1 Ω π π 4 4 He jΩ n n 5 25 hn 06 08 Causal Bandstop hn n 5 25 xn 1 Excitation n 5 25 yn 02 12 Response π π 1 π π 4 4 He jΩ Ω Ω He jΩ Figure 1163 Typical impulse responses of ideal lowpass highpass bandpass and bandstop filters n hn Ideal Lowpass n hn Ideal Highpass n hn Ideal Bandpass n hn Ideal Bandstop rob28124ch11509591indd 550 061216 218 pm 114 DiscreteTime Filters 551 Figure 1166 A white cross on a black background Figure 1167 Brightness of the top row of pixels in the whitecross image nx 99 1 Brightness of the Top Row of the Image bnx image of the scene on the film The photograph could be a color photograph or a blackandwhite monochrome photograph This discussion will be confined to mono chrome images A digital camera acquires an image by imaging the scene on a usu ally rectangular array of detectors which convert light energy to electrical charge Each detector in effect sees a very tiny part of the image called a pixel short for picture element The image acquired by the digital camera then consists of an array of numbers one for each pixel indicating the light intensity at that point again assuming a monochrome image A photograph is a continuousspace function of two spatial coordinates conven tionally called x and y An acquired digital image is a discretespace function of two discretespace coordinates n x and n y In principle a photograph could be directly fil tered In fact there are optical techniques that do just that But by far the most common type of image filtering is done digitally meaning that an acquired digital image is filtered by a computer using numerical methods The techniques used to filter images are very similar to the techniques used to filter time signals except that they are done in two dimensions Consider the very simple example image in Figure 1166 One technique for filtering an image is to treat one row of pixels as a onedimensional signal and filter it just like a discretetime signal Figure 1167 is a graph of the brightness of the pixels in the top row of the image versus horizontal discretespace n x Figure 1168 Brightness of the top row of pixels after being lowpass filtered by a causal lowpass filter 99 1 CausallyFiltered Brightness nx bnx If the signal were actually a function of discretetime and we were filtering in real time meaning we would not have future values available during the filtering process the lowpassfiltered signal might look like Figure 1168 rob28124ch11509591indd 551 061216 218 pm C h a p t e r 11 Frequency Response Analysis 552 After lowpass filtering all the rows in the image would look smeared or smoothed in the horizontal direction and unaltered in the vertical direction Figure 1169 If we had filtered the columns instead of the rows the effect would have been as illustrated in Figure 1170 One nice thing about image filtering is that usually causality is not relevant to the filtering process Usually the entire image is acquired and then processed Following the analogy between time and space during horizontal filtering past signal values would lie to the left and future values to the right In realtime filtering of time sig nals we cannot use future values because we dont yet know what they are In image fil tering we have the entire image before we begin filtering and therefore future values are available If we horizontally filtered the top row of the image with a noncausal lowpass filter the effect might look as illustrated in Figure 1171 Figure 1171 Brightness of top row of pixels after being lowpass filtered by a noncausal lowpass filter 99 1 Noncausally Filtered Brightness nx bnx Figure 1169 Whitecross image after all rows have been lowpass filtered by a causal lowpass filter Figure 1170 Whitecross image after all columns have been lowpass filtered by a causal lowpass filter If we horizontally lowpass filtered the entire image with a noncausal lowpass filter the result would look like Figure 1172 The overall effect of this type of filtering can be seen in Figure 1173 where both the rows and columns of the image have been filtered by a lowpass filter Of course the filter referred to above as noncausal is actually causal because all the image data are acquired before the filtering process begins Knowledge of the future is never required It is only called noncausal because if a space coordinate were instead time and we were doing realtime filtering the filtering would be noncausal Figure 1174 illustrates some other images and other filtering operations rob28124ch11509591indd 552 061216 218 pm 114 DiscreteTime Filters 553 In each image in Figure 1174 the pixel values range from black to white with gray lev els in between To grasp the filtering effects think of a black pixel as having a value of 0 and a white pixel as having a value of 1 Then medium gray would have a pixel value of 05 Image a is a checkerboard pattern filtered by a highpass filter in both dimensions The effect of the highpass filter is to emphasize the edges and to deemphasize the constant values between the edges The edges contain the highspatialfrequency information in the image So the highpassfiltered image has an average value of 05 medium gray and the black and white squares which were very different in the original image look about the same in the filtered image The checkerboard in b is filtered by a bandpass filter This type of filter smooths edges because it has little response at high frequencies Figure 1173 Whitecross image filtered by a lowpass filter a causal b noncausal a b Figure 1172 Whitecross image after all rows have been lowpass filtered by a noncausal lowpass filter Figure 1174 Examples of different types of image filtering d M J Roberts Noncausal Highpass Noncausal Bandpass Causal Lowpass Noncausal Highpass a b c d rob28124ch11509591indd 553 061216 218 pm C h a p t e r 11 Frequency Response Analysis 554 It also attenuates the average values because it also has little response at very low fre quencies including zero Image c is a random dot pattern filtered by a causal lowpass filter We can see that it is a causal filter because the smoothing of the dots always occurs to the right and below the dots which would be later times if the signals were time signals The response of a filter to a very small point of light in an image is called its point spread function The point spread function is analogous to the impulse response in timedomain systems A small dot of light approximates a twodimensional impulse and the point spread function is the approximate twodimensional impulse response The last image d is of the face of a dog It is highpass filtered The effect is to form an image that looks like an outline of the original image because it emphasizes sudden changes edges and deemphasizes the slowly varying parts of the image PRACTICAL FILTERS Comparison with ContinuousTime Filters Figure 1175 is an example of an LTI lowpass filter Its unitsequence response is 5 4 08 n un Figure 1176 Figure 1175 A lowpass filter xn yn 4 5 D Figure 1176 Unitsequence response of the lowpass filter 5 5 10 15 20 5 n yn Figure 1177 Impulse response of the lowpass filter n 5 20 hn 1 The impulse response of any discretetime system is the first backward difference of its unitsequence response In this case that is hn 5 4 45 n un 5 4 45 n1 un 1 which reduces to hn 08 n un Figure 1177 The transfer function and frequency response are Hz z z 08 H e jΩ e jΩ e jΩ 08 rob28124ch11509591indd 554 061216 218 pm 114 DiscreteTime Filters 555 Figure 1178 Figure 1178 Frequency response of the lowpass filter Ω π π HejΩ 5 Ω π π π π HejΩ Figure 1179 A comparison of the impulse responses of a discretetime and an RC lowpass filter n 5 20 hn 1 t ht RC 1 RC Figure 1180 Frequency responses of discretetime and continuoustime lowpass filters Ω π π HejΩ 5 Ω π π He jΩ π π 2 1 2πRC 1 2πRC 1 1 f f H f 45 45 90 90 H f It is instructive to compare the impulse and frequency responses of this lowpass filter and the impulse and frequency responses of the RC lowpass filter The impulse response of the discretetime lowpass filter looks like a sampled version of the impulse response of the RC lowpass filter Figure 1179 Their frequency responses also have some similarities Figure 1180 rob28124ch11509591indd 555 061216 218 pm C h a p t e r 11 Frequency Response Analysis 556 Figure 1181 A highpass filter xn α 1 1 yn D 1 α 0 Figure 1182 A bandpass filter xn α 1 1 yn D β D 1 α 0 1 β 0 Figure 1183 A bandstop filter xn α 1 1 yn D β D 1 β α 0 If we compare the shapes of the magnitudes and phases of these frequency re sponses over the frequency range π Ω π they look very much alike magnitudes more than phases But a discretetime frequency response is always periodic and can never be lowpass in the same sense as the frequency response of the RC lowpass filter The name lowpass applies accurately to the behavior of the frequency response in the range π Ω π and that is the only sense in which the designation lowpass is cor rectly used for discretetime systems Highpass Bandpass and Bandstop Filters Of course we can have highpass and bandpass discretetime filters also Figure 1181 through Figure 1183 The transfer functions and frequency responses of these filters are Hz z 1 z α H e jΩ e jΩ 1 e jΩ α rob28124ch11509591indd 556 061216 218 pm 114 DiscreteTime Filters 557 for the highpass filter H e jΩ zz 1 z 2 α βz αβ H e jΩ e jΩ e jΩ 1 e j2Ω α β e jΩ αβ for the bandpass filter and H e jΩ 2 z 2 1 β αz β z 2 α βz αβ H e jΩ 2 e j2Ω 1 β α e jΩ β e j2Ω α β e jΩ αβ 1 β α 0 for the bandstop filter ExamplE 117 Response of a highpass filter to a sinusoid A sinusoidal signal xn 5 sin 2πn18 excites a highpass filter with transfer function Hz z 1 z 07 Graph the response yn The filters frequency response is H e jΩ e jΩ 1 e jΩ 07 The DTFT of the excitation is X e jΩ j5π δ 2π Ω π9 δ 2π Ω π9 The DTFT of the response is the product of these two Y e jΩ e jΩ 1 e jΩ 07 j 5π δ 2π Ω π9 δ 2π Ω π9 Using the equivalence property of the impulse and the fact that both are periodic with period 2π Y e jΩ j5π δ 2π Ω π9 e jπ9 1 e jπ9 07 δ 2π Ω π9 e jπ9 1 e jπ9 07 Y e jΩ j5π e jπ9 1 e jπ9 07 δ 2π Ω π9 e jπ9 1 e jπ9 07 δ 2π Ω π9 e jπ9 07 e jπ9 07 Y e jΩ j5π 17 e jπ9 07 e jπ9 δ 2π Ω π9 17 07 e jπ9 e jπ9 δ 2π Ω π9 149 14 cos π9 Y e jΩ j2867π 17 δ 2π Ω π9 δ 2π Ω π9 07 e jπ9 δ 2π Ω π9 e jπ9 δ 2π Ω π9 e jπ9 δ 2π Ω π9 07 e jπ9 δ 2π Ω π9 Y e jΩ j2867π 17 δ 2π Ω π9 δ 2π Ω π9 07 cos π9 j07 sin π9 δ 2π Ω π9 cos π9 j sin π9 δ 2π Ω π9 cos π9 j sin π9 δ 2π Ω π9 07 cos π9 j07 sin π9 δ 2π Ω π9 Y e jΩ j2867π 171 cos π9 δ 2π Ω π9 δ 2π Ω π9 j03 sin π9 δ 2π Ω π9 δ 2π Ω π9 Inverse transforming yn 2867 171 cosπ9 sin2πn18 2867 03 sinπ9 cos2πn18 yn 2939 sin2πn18 29412 cos2πn18 4158 sin2πn18 0786 rob28124ch11509591indd 557 061216 218 pm C h a p t e r 11 Frequency Response Analysis 558 Figure 1184 Excitation and response of a highpass filter 0 5 10 15 20 25 30 35 5 0 5 n xn 0 5 10 15 20 25 30 35 5 0 5 n yn Figure 1184 shows the excitation and response of the filter ExamplE 118 Effects of filters on example signals Test the filter in Figure 1185 with a unit impulse a unit sequence and a random signal to show the filtering effects at all three outputs H LP e jΩ Y LP e jΩ X e jΩ 01 1 09 e jΩ H HP e jΩ Y HP e jΩ X e jΩ 095 1 e jΩ 1 09 e jΩ H BP e jΩ Y BP e jΩ X e jΩ 02 1 e jΩ 1 18 e jΩ 081 e j2Ω Figure 1185 Filter with lowpass highpass and bandpass outputs xn yLPn 09 1 1 D 09 D yHPn yBPn 01 095 02 rob28124ch11509591indd 558 061216 218 pm 114 DiscreteTime Filters 559 Notice in Figure 1186 that sums of the highpass and bandpass impulse responses are zero because the frequency response is zero at Ω 0 Figure 1187 Unitsequence responses at the three outputs 0 10 20 30 40 50 60 0 10 20 30 40 50 60 0 05 1 h1n h1n h1n Lowpass Unit Sequence Response 0 05 1 Highpass Unit Sequence Response 0 05 1 n Bandpass Unit Sequence Response 0 10 20 30 40 50 60 Figure 1186 Impulse responses at the three outputs 0 10 20 30 40 50 60 0 005 01 hn Lowpass Unit Impulse Response 0 10 20 30 40 50 60 05 0 05 1 hn Highpass Unit Impulse Response 0 10 20 30 40 50 60 02 0 02 n hn Bandpass Unit Impulse Response rob28124ch11509591indd 559 061216 218 pm C h a p t e r 11 Frequency Response Analysis 560 The lowpass filters response to a unit sequence Figure 1187 approaches a nonzero final value because the filter passes the average value of the unit sequence The unitsequence responses of the highpass and bandpass filters both approach zero Also the unitsequence response of the highpass filter jumps suddenly at the application of the unit sequence but the lowpass and bandpass filters both respond much more slowly indicating that they do not allow highfrequency signals to pass through The lowpassfilter output signal Figure 1188 is a smoothed version of the input signal The rapidly changing highfrequency content has been removed by the filter The highpassfilter response has an average value of zero and all the rapid changes in the input signal appear as rapid changes in the output signal The bandpass filter removes the average value of the signal and also smooths it to some extent because it removes both the very low and very high frequencies The Moving Average Filter A very common type of lowpass filter that will illustrate some principles of dis cretetime filter design and analysis is the movingaverage filter Figure 1189 The difference equation describing this filter is yn xn xn 1 xn 2 xn N 1 N and its impulse response is hn un un N N Figure 1188 Responses at the three outputs to a random signal 0 10 20 30 40 50 60 2 0 2 xn Random Excitation Signal 0 10 20 30 40 50 60 2 0 2 yLPn Lowpass Filter Response 0 10 20 30 40 50 60 2 0 2 yHPn Highpass Filter Response 0 10 20 30 40 50 60 2 0 2 n yBPn Bandpass Filter Response rob28124ch11509591indd 560 061216 219 pm 114 DiscreteTime Filters 561 Figure 1190 Its frequency response is H e jΩ e jN1Ω2 N sin NΩ2 sin Ω2 e jN1Ω2 drclΩ2π N Figure 1191 Figure 1189 A movingaverage filter Figure 1190 Impulse response of a movingaverage filter n hn N 1 N xn yn 1 2 N1 1 N D D D This filter is usually described as a smoothing filter because it generally attenu ates higher frequencies That designation would be consistent with being a lowpass fil ter However observing the nulls in the frequency response magnitude one might be tempted to call it a multiple bandstop filter This illustrates that classification of a filter as lowpass highpass bandpass or bandstop is not always clear However because of the traditional use of this filter to smooth a set of data it is usually classified as lowpass Figure 1191 Frequency response of a movingaverage filter for two different averaging times Ω 2π 2π 2π 2π He jΩ 1 N 4 Ω 2π 4 4 Ω 2π 2π He jΩ 1 N 9 Ω 2π 4 4 π π π π He jΩ He jΩ rob28124ch11509591indd 561 061216 219 pm C h a p t e r 11 Frequency Response Analysis 562 ExamplE 119 Filtering a pulse with a movingaverage filter Filter the signal xn un un 9 a with a movingaverage filter with N 6 b with the bandpass filter in Figure 1182 with α 08 and β 05 Using MATLAB graph the zerostate response yn from each filter The zerostate response is the convolution of the impulse response with the excitation The impulse response for the movingaverage filter is hn 16un un 6 The frequency response of the bandpass filter is H e jΩ Y e jΩ X e jΩ 1 e jΩ 1 13 e jΩ 04 e j2Ω 1 1 08 e jΩ 1 e jΩ 1 05 e jΩ Therefore its impulse response is hn 08 n un 05 n un 05 n1 un 1 The MATLAB program has a main script file It calls a function convD to do the discretetime convolutions Program to graph the response of a moving average filter and a discretetime bandpass filter to a rectangular pulse close all Close all open figure windows figurePosition2020800600 Open a new figure window n 530 Set up a time vector for the responses x uD n uDn 9 Excitation vector Moving average filter response h uD n uD n 6 Moving average filter impulse response yn convDTxnhnn Response of moving average filter Graph the response subplot211 p stemnykfilled setpLineWidth2MarkerSize4 grid on xlabelitnFontNameTimesFontSize18 ylabelyitnFontNameTimesFontSize18 titleMovingAverage FilterFontNameTimesFontSize24 Bandpass filter response Find bandpass filter impulse response rob28124ch11509591indd 562 061216 219 pm 114 DiscreteTime Filters 563 h1 08nuD n h2 05nuD n 05n1uD n1 hn convD h1nh2nn yn convD xnhnn Response of bandpass filter Graph the response subplot212 p stemnykfilled setpLineWidth2 MarkerSize4 grid on xlabelitnFontNameTimesFontSize18 ylabelyitnFontNameTimesFontSize18 titleBandpass FilterFontNameTimesFontSize24 Function to perform a discretetime convolution on two signals and return their convolution at specified discrete times The two signals are in column vectors x1 and x2 and their times are in column vectors n1 and n2 The discrete times at which the convolution is desired are in the column n12 The returned convolution is in column vector x12 and its time is in column vector n12 If n12 is not included in the function call it is generated in the function as the total time determined by the individual time vectors x12n12 convD x1n1x2n2n12 function x12n12 convD x1n1x2n2n12 Convolve the two vectors using the MATLAB conv command xtmp convx1x2 Set a temporary vector of times for the convolution based on the input time vectors ntmp n11 n21 0lengthn1lengthn22 Set the first and last times in temporary vector nmin ntmp1 nmax ntmplengthntmp if nargin 5 If no input time vector is specified use ntmp x12 xtmp n12 ntmp else If an input time vector is specified compute the convolution at those times x12 0n12 Initialize output convolution to zero Find the indices of the desired times which are between the minimum and maximum of the temporary time vector I12intmp findn12 nmin n12 nmax rob28124ch11509591indd 563 061216 219 pm C h a p t e r 11 Frequency Response Analysis 564 Translate them to the indices in the temporary time vector Itmp n12I12intmp nmin 1 Replace the convolution values for those times in the desired time vector x12 I12intmp xtmpItmp end The graphs created are in Figure 1192 Figure 1193 Ideal discretetime lowpass filter impulse response n hn The Almost Ideal Lowpass Filter If we want to approach the frequencydomain performance of the ideal lowpass filter we must design a discretetime filter with an impulse response that closely approaches the inverse DTFT of the ideal frequency response We have previously shown that the ideal lowpass filter is noncausal and cannot be physically realized However we can closely approach it The ideal lowpassfilter impulse response is illustrated in Figure 1193 Figure 1192 Two filter responses 5 0 5 10 15 20 25 30 0 1 2 3 4 5 6 n yn MovingAverage Filter 5 0 5 10 15 20 25 30 15 1 05 0 05 1 15 n yn Bandpass Filter rob28124ch11509591indd 564 061216 219 pm 114 DiscreteTime Filters 565 Figure 1194 Almostideal discretetime lowpass filter impulse response n 64 hn 025 The problem in realizing this filter physically is the part of the impulse response that occurs before time n 0 If we arrange to delay the impulse response by a large amount then the signal energy of the impulse response that occurs before time n 0 will become very small and we can chop it off and closely approach the ideal frequency response Figure 1194 and Figure 1195 Figure 1195 Almostideal discretetime lowpass filter frequency response Ω 2π 2π He jΩ 1 Ω 2π 2π π π He jΩ Figure 1196 Almostideal discretetime lowpass filter frequency response plotted on a dB scale Ω 2π 2π He jΩdB 100 The magnitude response in the stopband is so small that we cannot see its shape when it is plotted on a linear scale as in Figure 1195 In cases like this a logmagnitude plot helps see what the real attenuation is in the stopband Figure 1196 rob28124ch11509591indd 565 061216 219 pm C h a p t e r 11 Frequency Response Analysis 566 This filter has a very nice lowpassfilter magnitude response but it comes at a price We must wait for it to respond The closer a filter approaches the ideal the greater time delay there is in the impulse response This is apparent in the time delay of the impulse response and the phase shift of the frequency response The fact that a long delay is required for filters that approach the ideal is also true of highpass bandpass and bandstop filters and it is true for both continuoustime and discretetime filters It is a general principle of filter design that any filter designed to be able to discriminate between two closely spaced frequencies and pass one while stopping the other must in a sense observe them for a long time to be able to distinguish one from the other The closer they are in frequency the longer the filter must observe them to be able to make the distinction That is the basic reason for the requirement for a long time delay in the response of a filter that approaches an ideal filter Advantages Compared to ContinuousTime Filters One might wonder why we would want to use a discretetime filter instead of a continuoustime filter There are several reasons Discretetime filters are built with three basic elements a delay device a multiplier and an adder These can be implemented with digital devices As long as we stay within their intended ranges of operation these de vices always do exactly the same thing That cannot be said of devices such as resistors capacitors and operational amplifiers which make up continuoustime filters A resistor of a certain nominal resistance is never exactly that value even under ideal conditions And even if it were at some time temperature effects or other environmental effects would change it The same thing can be said of capacitors inductors transistors and so on So discretetime filters are more stable and reproducible than continuoustime filters It is often difficult to implement a continuoustime filter at very low frequencies because the component sizes become unwieldy for example very large capacitor val ues may be needed Also at very low frequencies thermal drift effects on components become a big problem because they are indistinguishable from signal changes in the same frequency range Discretetime filters do not have these problems Discretetime filters are often implemented with programmable digital hardware That means that this type of discretetime filter can be reprogrammed to perform a dif ferent function without changing the hardware Continuoustime filters do not have this flexibility Also some types of discretetime filters are so computationally sophisticated that they would be practically impossible to implement as continuoustime filters Discretetime signals can be reliably stored for very long times without any signif icant degradation on magnetic disk or tape or CDROM Continuoustime signals can be stored on analog magnetic tape but over time the values can degrade By timemultiplexing discretetime signals one filter can accommodate multiple sig nals in a way that seems to be and effectively is simultaneous Continuoustime filters can not do that because to operate correctly they require that the input signal always be present 115 SUMMARY OF IMPORTANT POINTS 1 Frequency response and impulse response of LTI systems are related through the Fourier transform 2 Characterization of systems in the frequency domain allows generalized design procedures for systems that process certain types of signals 3 An ideal filter is distortionless within its passband 4 All ideal filters are noncausal and therefore cannot be built 5 Filtering techniques can be applied to images as well as signals 6 Practical discretetime filters can be implemented as discretetime systems using only amplifiers summing junctions and delays rob28124ch11509591indd 566 061216 219 pm Exercises with Answers 567 7 All the ideas that apply to continuoustime filters apply in a similar way to discretetime filters 8 Discretetime filters have several advantages over continuoustime filters EXERCISES WITH ANSWERS Answers to each exercise are in random order ContinuousTime Frequency Response 1 A system has an impulse response hLP t 3e10t ut and another system has an impulse response hHP t δt 3e10t ut a Sketch the magnitude and phase of the frequency response of these two systems in a parallel connection b Sketch the magnitude and phase of the frequency response of these two systems in a cascade connection Answers ω 40 40 HC jω 025 ω 40 40 π π HC jω ω 40 40 HP jω 1 ω 40 40 HP jω π π ContinuousTime Ideal Filters 2 A system has an impulse response ht 10 rect t 001 002 What is its null bandwidth Answer 50 Hz ContinuousTime Causality 3 Determine whether or not the systems with these frequency responses are causal a H f sinc f b H f sinc f ejπf c H jω rectω d H jω rectωejω e H f A f H f Ae j2πf g H jω 1 ejω h H f rect f 20ej40πf i He jΩ e jΩ e jΩ 09 j He jΩ e j2Ω e jΩ 13 Answers 4 Causal and 6 Noncausal rob28124ch11509591indd 567 061216 219 pm C h a p t e r 11 Frequency Response Analysis 568 Logarithmic Graphs Bode Diagrams and Decibels 4 A system is excited by a sinusoid whose signal power is 001 and the response is a sinusoid of the same frequency with a signal power of 4 What is the magnitude of the transfer function of the system at the frequency of the sinusoid expressed in decibels dB Answer 26 dB 5 A system is excited by a sinusoid whose amplitude is 1 µV and the response is a sinusoid of the same frequency with an amplitude of 5 V What is the magnitude of the transfer function of the system at the frequency of the sinusoid expressed in decibels dB Answer 134 dB 6 A system has a transfer function H s 10 s 2 s 2 11s 10 a Find the values of its frequency response magnitude in dB and its frequency response angle in radians at these frequencies ω 001 ω 1 ω 10 ω 1000 b Draw the overall asymptotic magnitude Bode diagram for this frequency response in the radian frequency range 10 2 ω 10 3 Answers 20 dB and 0011 radians 169465 dB and 08851 radians 80 dB and 31306 radians 30535 dB and 22565 radians 103 102 101 100 101 102 100 80 60 40 20 0 20 40 ω H jωdB 7 In an inverting opamp amplifier the feedback component is a 1000 Ω resistor and the component between the input voltage terminal and the operational amplifiers inverting input is a 10 µF capacitor If the voltage transfer function is H f what are the magnitude and phase of H 200 Answer 1257 and 157 radians 8 An active opamp integrator has a frequency response magnitude Bode diagram that goes through 40 dB at ω 500 At what numerical value of ω is the system frequency response magnitude 100 times smaller than it is at ω 500 Answer 50000 rob28124ch11509591indd 568 061216 219 pm Exercises with Answers 569 9 Graph the magnitude frequency responses both on a linearmagnitude and on a logmagnitude scale of the systems with these frequency responses over the frequency range specified a H f 20 20 4π 2 f 2 j42πf 100 f 100 b H jω 2 10 5 100 jω 1700 ω 2 j20ω 500 ω 500 Answers f 100 100 H f 1 f 100 100 lnH f 10 ω 500 500 H jω 2 ω 500 500 lnH jω 10 10 Draw asymptotic and exact magnitude and phase Bode diagrams for the frequency responses of the following circuits and systems a An RC lowpass filter with R 1 MΩ and C 01 µF b The circuit of Figure E10b R 10 Ω C 1 μF L 1 mH vit vLt Figure E10b Answers 101 100 101 102 103 50 40 30 20 10 0 ω 101 100 101 102 103 ω 2 15 1 05 0 H jωdB H jω 100 80 60 40 20 0 20 ω ω H jωdB 102 103 104 105 106 107 102 103 104 105 106 107 0 05 1 15 2 25 3 35 H jω rob28124ch11509591indd 569 061216 219 pm C h a p t e r 11 Frequency Response Analysis 570 ContinuousTime Practical Passive Filters 11 Find and graph the frequency response of each of the circuits in Figure E11 given the indicated excitation and response a Excitationv i t Response v Lt R 10 Ω C 1 μF L 1 mH vit vLt b Excitationv i t Response iC t R 1 kΩ C 1 μF vit iCt c Excitation vi t Response vR t R 1 kΩ C 1 μF L 1 mH vit vRt d Excitation vi t Response vR t R 100 Ω C 1 μF L 1 mH iit vRt Figure E11 rob28124ch11509591indd 570 061216 219 pm Exercises with Answers 571 Answers ω 150000 150000 H jω 3 ω 150000 150000 π π H jω ω 1500 1500 H jω 0001 ω 1500 1500 π π H jω ω 1000000 1000000 H jω 100 ω 1000000 1000000 π π H jω ω 50000 50000 H jω 1 ω 50000 50000 π π H jω 12 With reference to the circuit schematic diagram in Figure E12 Figure E12 L R C voutt vint a Find a general expression for the frequency response H jω b Is this circuit a practical lowpass highpass bandpass or bandstop filter c What is the slope in dBdecade of the magnitude Bode diagram for the frequency response of this filter at very low frequencies d What is the slope in dBdecade of the magnitude Bode diagram for the frequency response of this filter at very high frequencies Answers Lowpass 40 dBdecade H jω 1 LC 1 jω 2 jωRC 1LC 0 dBdecade 13 In Figure E13 is a practical passive continuoustime filter Let C 16 µF and R 1000 Ω C R vit vot Figure E13 a Find its transfer function H f in terms of R C and f as variables b At what frequency f is its transfer function magnitude a minimum and what are the transfer function magnitude and phase at that frequency rob28124ch11509591indd 571 061216 219 pm C h a p t e r 11 Frequency Response Analysis 572 c At what frequency f is its transfer function magnitude a maximum and what are the transfer function magnitude and phase at that frequency d What are the magnitude and phase of the transfer function at a frequency of 10 Hz e If you keep R 1000 Ω and choose a new capacitor value C to make the magnitude of the transfer function at 100 Hz less than 30 of the maximum transfer function magnitude what is the largest numerical value of C you could use Answers 0709 07828 05005 µF H f R R 1j2π f C j2π f RC j2π f RC 1 1 0 0 0 undefined 14 For the practical passive filter in Figure E14 with transfer function Hs Vout sVins R C L voutt vint Figure E14 a What is the slope in dB per decade of a magnitude Bode diagram of its frequency response at frequencies approaching zero and at frequencies approaching infinity b Find the frequency at which the magnitude of the transfer function is a maximum c Find the nonzero frequency at which the phase of the transfer function is zero d Find the phase shift of the transfer function just above and just below f 0 e Classify this filter as a practical approximation to the ideal lowpass highpass bandpass or bandstop filter Answers 20 and 20 dBdecade 7118 π2 and π2 Bandpass 7118 15 For each circuit in Figure E15 the frequency response is the ratio H f V o f V i f Which circuits have a Zero frequency response at f 0 b Zero frequency response at f c Transfer function magnitude of one at f 0 d Transfer function magnitude of one at f e Transfer function magnitude nonzero and phase of zero at some frequency 0 f at a finite nonzero frequency rob28124ch11509591indd 572 061216 219 pm Exercises with Answers 573 R C L vit vot iit R C L vit vot iit vit iit vot C R R C vit vot iit R L vit vot iit R L vit vot iit R C L vit vot iit R C L vit vot iit a b c d e f g h Figure E15 Answers adfg bceh bh bdfh aceg 16 The causal square wave voltage signal illustrated in Figure E16 is the excitation for five practical passive filters numbered 15 also in Figure E16 The voltage responses of the five filters are illustrated below them Match the responses to the filters a b c d e Excitation 2 0 2 4 6 8 10 1 0 1 vit Time t s C 01F vint voutt R 1 Ω vit C 01 F R 1 Ω vot vot vit R 1 Ω C 1 F vint C 1F R 1 Ω L 1H voutt vint R 1 Ω C 1F voutt 2 0 2 4 6 8 10 1 0 1 Time t s vot 2 0 2 4 6 8 10 1 0 1 Time t s vot 2 0 2 4 6 8 10 1 0 1 Time t s vot 2 0 2 4 6 8 10 1 0 1 Time t s vot 2 0 2 4 6 8 10 1 0 1 Time t s vot 1 2 3 4 5 Figure E16 Answers 1E 2D 3B 4A 5C 17 Classify each of these frequency responses as lowpass highpass bandpass or bandstop a H f 1 1 jf b H f jf 1 jf c H jω j10ω 100 ω 2 j10ω Answers Bandpass Highpass Lowpass rob28124ch11509591indd 573 061216 219 pm C h a p t e r 11 Frequency Response Analysis 574 18 In the circuit in Figure E18 let R 10 Ω L 10 mH and C 100 µF and let H jω V o jω V i jω Figure E18 R L C vit iit vot a H jω can be expressed in the form A jω 2 jBω C Find the numerical values of A B and C b Find the numerical value of H 0 c Find the numerical value of lim ω H jω For parts d e and f redefine the frequency response as H jω I i jω V i jω d H jω can now be expressed in the form jωA jω 2 jBω C Find the numerical values of A B and C e Find the numerical value of H 0 f Find the numerical value of lim ω H jω Answers 100 1000 1000000 1000000 1000 1000000 1 0 0 0 19 A passive circuit consists of a resistor R and an inductor L in parallel Define the input signal to the system as the voltage v t across both the resistor and inductor and define the response signal as the total current i t that flows into the parallel combination of the resistor and inductor a Find an expression for the frequency response H f as a ratio of two polynomials in f b If R 1 Ω and L 01 H find the numerical value of H 10 Answers 10126e j01578 R j2πfL j2πfRL ContinuousTime Practical Active Filters 20 Match each circuit in Figure E20 to the magnitude asymptotic Bode diagram of its frequency response H jω V o jω V i jω rob28124ch11509591indd 574 061216 219 pm Exercises with Answers 575 vit vot vxt Rf Ri Ci iit ift vit vot vxt Rf Ci iit ift vit vot vxt R C iit ift vit vot vxt Ri C iit ift Rf H jωdB 10k 10k1 10k2 10k3 ω 20 20 40 60 40 H jωdB 10k 10k1 10k2 10k3 ω 20 20 40 60 40 H jωdB 10k 10k1 10k2 10k3 ω 20 20 40 60 40 20 40 H jωdB 10k 10k1 10k2 10k3 ω 20 40 60 A B C D 1 2 3 4 Figure E20 Answers A3 B1 C2 D4 21 The transfer function for the system in Figure E20 can be written in the form Hs b2s2 b1s b0 s2 a1s a0 ωca 400 ω cb 600 Figure E20 ωcb ωcb ωca yt xt a Find numerical values for the as and bs b Find the systems numerical frequency response magnitude in dB and phase in radians at 150 Hz Answers 61193 dB 060251 radians 0 600 0 1000 240000 DiscreteTime Frequency Response 22 A system has an impulse response h n 7 8 n u n What is its halfpower discretetimefrequency bandwidth Answer 01337 radians 23 Match each polezero diagram in Figure E23 to its magnitude frequency response Assume that the transfer functions are of the form H s A s z 1 s z 2 s z M s p 1 s p 2 s p N rob28124ch11509591indd 575 061216 219 pm C h a p t e r 11 Frequency Response Analysis 576 and that A 1 In each case all finite poles and zeros are shown Figure E23 4 2 0 5 0 5 s 20 0 20 0 05 1 ω H H 4 2 0 5 0 5 s 20 0 20 0 01 02 ω H F 4 2 0 5 0 5 s 20 0 20 0 05 1 H D 4 2 0 5 0 5 20 0 20 0 05 1 ω H I 4 2 0 5 0 5 20 0 20 0 005 01 ω H G 4 2 0 5 0 5 20 0 20 0 5 H E 4 2 0 5 0 5 20 0 20 0 1 2 H A 4 2 0 5 0 5 20 0 20 0 1 2 H B 4 2 0 5 0 5 20 0 20 0 02 H C 4 2 0 5 0 5 20 0 20 0 10 ω H J 8 6 4 9 7 5 1 2 3 10 s s s s s s s Answers 1H 2F 3D 4I 5G 6E 7A 8B 9C 10J DiscreteTime Ideal Filters 24 Classify each of the frequency responses in Figure E24 as lowpass highpass bandpass or bandstop a b c d F 2 2 HF 1 Ω 4π 4π 1 HejΩ Ω 4π 4π 1 He jΩ F 2 2 HF 1 Figure E24 Answers Highpass Bandpass Lowpass Bandstop DiscreteTime Causality 25 For the system frequency response HejΩ e jAΩ 1 08e jΩ what numerical range of integer values of A will produce a causal system Answer A 0 rob28124ch11509591indd 576 061216 219 pm Exercises with Answers 577 DiscreteTime Practical Filters 26 Find the frequency response H e jΩ Y e jΩ X e jΩ and graph the frequency response of each of the filters in Figure E26 over the range 2π Ω 2π a xn yn D b xn yn D 09 c xn yn 09 D D d D D xn yn 06 Figure E26 Answers Ω 2π 2π He jΩ 2 Ω 2π 2π π π He jΩ He jΩ 20 Ω 2π 2π Ω 2π π π He jΩ Ω 2π 2π 10 Ω 2π 2π π π He jΩ jΩ He jΩ Ω 2π 2π He jΩ 2 Ω 2π 2π π π He jΩ 27 Find the minimum stop band attenuation of a movingaverage filter with N 3 Define the stop band as the frequency region Ω C Ω π where Ω c is the discretetime frequency of the first null in the frequency response Answer 954 dB rob28124ch11509591indd 577 061216 219 pm C h a p t e r 11 Frequency Response Analysis 578 28 In Figure E28 match each polezero diagram to its magnitude frequency response graph The gain constants are not all one Figure E28 1 0 1 1 0 1 2 0 2 0 2 4 Ω H 1 0 1 1 0 1 2 0 2 0 1 2 Ω H 1 0 1 1 0 1 2 0 2 0 2 4 Ω H 1 0 1 1 0 1 2 0 2 0 2 4 Ω H A B C D E 1 0 1 1 0 1 z z z z z 2 0 2 0 2 4 Ω H 1 2 3 4 5 Answers 1E 2D 3B 4A 5C 29 In Figure E29 match each polezero diagram to its magnitude frequency response and unit sequence response Figure E29 1 2 3 4 5 1 0 1 1 05 0 05 1 z 1 0 1 1 05 0 05 1 z 1 0 1 1 05 0 05 1 1 0 1 1 05 0 05 1 1 0 1 1 05 0 05 1 2 0 2 0 2 4 6 8 Ω 2 0 2 0 05 1 Ω 2 0 2 0 05 1 Ω 2 0 2 0 05 1 Ω 2 0 2 0 05 1 15 2 Ω 0 10 20 30 0 05 1 15 n 0 10 20 30 0 05 1 n 0 10 20 30 1 05 0 05 1 n 0 10 20 30 0 05 1 15 n 0 10 20 30 0 05 1 n h 1n h 1n h 1n h 1n h 1n H H H H H Af Bf Cf Df Ef As Bs Cs Ds Es z z z Answers 1BfBs 2CfEs 3EfAs 4DfDs 5AfCs rob28124ch11509591indd 578 061216 219 pm Exercises without Answers 579 EXERCISES WITHOUT ANSWERS ContinuousTime Frequency Response 30 Why is it impossible to actually make a distortionless system 31 Why is it impossible to actually make an ideal filter 32 One problem with causal filters is that the response of the filter always lags the excitation This problem cannot be eliminated if the filtering is done in real time but if the signal is recorded for later offline filtering one simple way of eliminating the lag effect is to filter the signal record the response and then filter that recorded response with the same filter but playing the signal back through the system backward Suppose the filter is a singlepole filter with a frequency response of the form H jω 1 1 jω ωc where ω c is the cutoff frequency halfpower frequency of the filter a What is the effective frequency response of the entire process of filtering the signal forward then backward b What is the effective impulse response ContinuousTime Ideal Filters 33 A signal xt is described by xt rect 1000t δ 0002 t a If xt is the excitation of an ideal lowpass filter with a cutoff frequency of 3 kHz graph the excitation xt and the response yt on the same scale and compare b If xt is the excitation of an ideal bandpass filter with a low cutoff frequency of 1 kHz and a high cutoff frequency of 5 kHz graph the excitation xt and the response yt on the same scale and compare ContinuousTime Causality 34 Determine whether or not the systems with these frequency responses are causal a H jω 2 jω b H jω 10 6 j4ω c H jω 4 25 ω 2 j6ω 4 jω 3 2 16 d H jω 4 25 ω 2 j6ω e jω 4 jω 3 2 16 e jω e H jω 4 25 ω 2 j6ω e jω 4 jω 3 2 16 e jω f H jω jω 9 45 ω 2 j6ω jω 3 jω 3 2 36 6 jω 3 2 36 g H jω 49 49 ω 2 rob28124ch11509591indd 579 061216 219 pm C h a p t e r 11 Frequency Response Analysis 580 Bode Diagrams 35 Draw asymptotic and exact magnitude and phase Bode diagrams for the frequency responses of the circuits and systems in Figure E35 a R1 1 kΩ C1 1 μF vit R2 10 kΩ C2 01 μF vC2t b 10 jω10 jω jω10 X jω Y jω c A system whose frequency response is H jω j20ω 10000 ω 2 j20ω Figure E35 36 A highpass filter is made with one resistor and one practical inductor The practical inductor can be modeled as an ideal inductor in series with a small resistance A Bode diagram of the actual filters frequency response deviates from the Bode diagram of the frequency response of a highpass filter made with an ideal inductor Draw a sketch of the ideal Bode diagram and the actual Bode diagram to illustrate the difference 37 A system has a transfer function Hs 3 s 2 7s s 2 8s 4 a In a magnitude Bode diagram of its frequency response what are the values of all the corner frequencies in radianssecond b What is the slope in dBdecade of the magnitude Bode diagram at very low frequencies approaching zero c What is the slope in dBdecade of the magnitude Bode diagram at very high frequencies approaching infinity ContinuousTime Practical Passive Filters 38 Find and graph the frequency response of each of the circuits in Figure E38 given the indicated excitation and response a Excitation vi t Response vC2 t R1 1 kΩ C1 1 μF vit R2 10 kΩ C2 01 μF vC2t rob28124ch11509591indd 580 061216 219 pm Exercises without Answers 581 b Excitation vi t Response iC1 t R1 1 kΩ C1 1 μF vit iC1t R2 10 kΩ C2 01 μF c Excitation vi t Response vR2 t C1 1 μF vit vR2t R2 10 kΩ R1 10 kΩ C2 1 μF d Excitation ii t Response vR1 t C1 1 μF iit vR1t R2 10 kΩ R1 10 kΩ C2 1 μF e Excitation vi t Response vRL t vit vRLt R1 10 kΩ RL 1 kΩ R2 10 kΩ C1 1 μF C2 1 μF Figure E38 rob28124ch11509591indd 581 061216 219 pm C h a p t e r 11 Frequency Response Analysis 582 39 Find and graph versus frequency the magnitude and phase of the input impedance Z in jω V i jω I i jω and frequency response H jω V o jω V i jω for each of the filters in Figure E39 1 μF vit iit vot 1 kΩ 10 nF vit vot 100 Ω 50 mH iit a b Figure E39 40 The signal x t in Exercise 33 is the excitation of an RC lowpass filter with R 1 kΩ and C 03 µF Sketch the excitation and response voltages versus time on the same scale ContinuousTime Filters 41 In Figure E41 are some descriptions of filters in the form of an impulse response a frequency response magnitude and a circuit diagram For each of these to the extent possible classify the filters as ideal or practical causal or noncausal lowpass highpass bandpass or bandstop a b c d e f g t ht R C L vint 1 ω H jω t t 05 2 12 L R C R voutt voutt vot vit vint iit ht ht Figure E41 ContinuousTime Practical Active Filters 42 Design an active highpass filter using an ideal operational amplifier two resistors and one capacitor and derive its frequency response to verify that it is highpass 43 Find the frequency response H jω V o jω V i jω of the active filter in Figure E43 with R i 1000 Ω C i 1μF and R f 5000 Ω rob28124ch11509591indd 582 061216 219 pm Exercises without Answers 583 Rf Ri Ci iit vit vot vxt ift Figure E43 a Find all the corner frequencies in radians per second in a magnitude Bode diagram of this frequency response b At very low and very high frequencies what is the slope of the magnitude Bode diagram in dBdecade 44 In the active filters in Figure E44 all resistors are 1 ohm and all capacitors are 1 farad For each filter the frequency response is H jω V 0 jω V i jω Identify the frequency response magnitude Bode diagram for each circuit vit vot vxt Ci Rf iit if t vit vot vxt Ci Cf iit ift Rf vit vot vxt Ri Cf iit ift vit vot vxt Ri C iit ift Rf vit vot vxt Rf iit ift Cf Ri vit vot vxt ift iit Rf Cf Ri vit vot vxt ift iit Rf Cf Ri vit vot vxt if t iit Rf Ri Ci A B C D E F G H 101 100 101 20 10 0 10 20 ω H jωdB 101 100 101 20 10 0 10 20 ω H jωdB 101 100 101 20 10 0 0 10 20 ω H jωdB 101 100 101 20 10 0 10 20 ω H jωdB 101 100 101 20 10 0 10 20 ω H jωdB 101 100 101 20 10 0 10 20 ω H jωdB 101 100 101 20 10 10 20 ω H jωdB 101 100 101 20 10 0 10 20 ω H jωdB 1 2 3 4 5 6 7 8 Figure E44 rob28124ch11509591indd 583 061216 219 pm C h a p t e r 11 Frequency Response Analysis 584 45 You have available two resistors of 100 Ω and 1000 Ω two capacitors of 10 µF and 100 µF and one ideal operational amplifier with its noninverting input grounded from which to design some active filters a Draw the circuit for an active lowpass filter with a corner frequency of 159 Hz Be sure to label the values of the resistors and capacitors b Draw the circuit for an active differentiator with the largest possible frequency response magnitude at 1 Hz Be sure to label the values of the resistors and capacitors c Draw the circuit for an active bandpass filter Make the low corner frequency and the high corner frequency as far apart as possible Be sure to label the values of the resistors and capacitors 46 Using only resistors and capacitors put single components into the circuit diagram in Figure E46 in the numbered positions that will make the frequency response of this filter H jω V 0 jω V i jω bandpass in nature with two poles You need not put values on the components just indicate whether they are capacitors or resistors The triangle with a K in it is a voltage amplifier of gain K not an operational amplifier There is more than one correct answer Figure E46 Vo Vi K 1 2 3 4 47 Classify the following transfer functions of filters as lowpass bandpass highpass or bandstop according to these definitions Lowpass H0 0 H jω ω 0 Highpass H0 0 H jω ω 0 Bandpass H0 0 H jω ω 0 H jω0 0 0 ω0 Bandstop H0 0 H jω ω 0 H jω0 0 for 0 ω 0 a H s has finite poles at s 2 and s 7 and a finite zero at s 20 b H s has finite poles at s 2 and s 7 and a finite zero at s 0 c H s has three finite poles in the left halfplane and no finite zeros d H s has finite poles at s 2 and s 7 and a double finite zero at s 0 48 There are eight polezero graphs of system transfer functions in Figure E48 Answer the following questions about their frequency responses impulse responses and step responses Frequency response a Which have a phase approaching zero at very high frequencies rob28124ch11509591indd 584 061216 219 pm Exercises without Answers 585 b Which have a phase that is discontinuous at zero frequency c Which have a magnitude approaching zero at high frequencies Step response a Which have a step response that is nonzero in the limit t b Which have a step response that is discontinuous at t 0 Impulse response a Which have an impulse response that contains an impulse b Which have an impulse response approaching zero in the limit t D E F A B C G H 6 4 2 0 2 4 6 6 4 2 0 2 4 6 σ ω s 6 4 2 0 2 4 6 6 4 2 0 2 4 6 σ ω s 6 4 2 0 2 4 6 6 4 2 0 2 4 6 σ ω s 6 4 2 0 2 4 6 6 4 2 0 2 4 6 σ ω s 6 4 2 0 2 4 6 6 4 2 0 2 4 6 σ ω s 6 4 2 0 2 4 6 6 4 2 0 2 4 6 σ ω s 6 4 2 0 2 4 6 6 4 2 0 2 4 6 σ ω s 6 4 2 0 2 4 6 6 4 2 0 2 4 6 σ ω s Figure E48 rob28124ch11509591indd 585 061216 219 pm C h a p t e r 11 Frequency Response Analysis 586 49 In Figure E49 are some polezero diagrams and some magnitude frequency responses Match the frequency responses to the polezero diagrams 4 2 0 2 5 0 5 20 0 20 0 01 02 H A 4 2 0 2 5 0 5 20 0 20 0 05 1 H B 4 2 0 2 5 0 5 20 0 20 0 05 1 H C 4 2 0 2 5 0 5 s s s s s 20 0 20 0 02 04 H D 1 2 3 4 5 6 7 8 4 2 0 2 5 0 5 20 0 20 0 005 01 H E 4 2 0 2 5 0 5 20 0 20 0 1 2 H F 4 2 0 2 5 0 5 20 0 20 0 02 04 H G 4 2 0 2 5 0 5 20 0 20 0 02 04 H H 20 0 20 0 005 01 H I ω 20 0 20 0 1 2 H J ω 20 0 20 0 05 1 15 H K ω 20 0 20 0 05 1 H L ω s s s Figure E49 DiscreteTime Causality 50 Determine whether or not the systems with these frequency responses are causal a He jΩ rect 5Ωπ δ2πΩej10Ω hn 1 10 sinc n 10 10 rob28124ch11509591indd 586 061216 219 pm Exercises without Answers 587 b He jΩ j sin Ω e jΩ e Ω 2 hn 12 δn 1 δn 1 c He jΩ 1 ej2Ω hn δn δn 2 d He jΩ 8e jΩ 8 5e jΩ e jΩ 1 58e jΩ hn 5 8 n1 u n 1 DiscreteTime Filters 51 In Figure E51 are pairs of excitations x and responses y For each pair identify the type of filtering that was done lowpass highpass bandpass or bandstop a b n 10 10 60 xn 20 n 60 yn 4 8 n 10 60 xn 1 4 n 10 60 yn 16 c d n 10 60 xn 16 n 10 60 yn 3 3 n 10 60 xn 20 n 10 60 yn 15 5 rob28124ch11509591indd 587 061216 219 pm C h a p t e r 11 Frequency Response Analysis 588 e f n 10 60 xn 1 n 10 60 yn 07 n 10 60 xn 1 n 10 60 yn 1 1 Figure E51 52 Classify each of these frequency responses as lowpass highpass bandpass or bandstop a He jΩ sin 3Ω 2 sin Ω 2 b He jΩ jsinΩ sin2Ω 53 For each of the systems with the polezero diagrams in Figure E53 find the discretetime radian frequencies Ω max and Ω min in the range π Ω π for which the transfer function magnitude is a maximum and a minimum If there is more than one value of Ω max or Ω min find all such values a b 1 0 1 1 05 0 05 1 Rez Imz 05 08 08 z 1 0 1 1 05 0 05 1 Rez Imz z 08 Figure E53 54 Find the frequency response He jΩ Y e jΩ X e jΩ and graph it for each of the filters in Figure E54 rob28124ch11509591indd 588 061216 219 pm Exercises without Answers 589 a Xz Yz 1143 0135 0413 00675 00675 z1 z1 b Xz Yz 1257 00914 0467 00593 00593 z1 z1 c Xz Yz 05747 0403 10216 0403 05747 0528 1083 03254 04228 z1 z1 z1 z1 Figure E54 55 Referring to the system block diagram in Figure E55 D D yn vn wn y1n xn β 1 α Figure E55 a Write the difference equations i Relating w n to x n without reference to v n and rob28124ch11509591indd 589 061216 219 pm C h a p t e r 11 Frequency Response Analysis 590 1 0 1 1 0 1 z A 1 0 1 1 0 1 B 1 0 1 1 0 1 C 1 0 1 1 0 1 D 0 10 20 30 2 0 2 0 10 20 30 2 0 2 0 10 20 30 2 0 2 0 10 20 30 2 0 2 h1n h1n h1n h1n 0 2 0 5 Ω 0 2 0 5 Ω 0 2 0 5 Ω 0 2 0 5 Ω 1s 2s 3s 4s 1f 2f 3f 4f H H H H z z z 57 In Figure E57 match the polezero diagrams to the corresponding magnitude frequency responses by filling in each blank with the correct letter 1 2 3 4 5 A B C D E 1 0 1 1 0 1 z z z z z 1 0 1 1 0 1 1 0 1 1 0 1 1 0 1 1 0 1 1 0 1 1 0 1 2 0 2 0 05 1 15 Ω H 2 0 2 0 2 4 6 8 Ω H 0 05 1 15 2 0 2 Ω H H 0 2 4 2 0 2 Ω H 0 05 1 15 2 0 2 Ω H ii Relating v n to x n without reference to w n b Then z transform the equations from part a and find the transfer function c Then find the transfer function Yz Y1z d The transfer function of this entire filter can be expressed in the form H z b0 z2 b1z b2 z2 a1z a2 If α 08 and β 05 find the numerical values of the constants 56 In Figure E56 match each system polezero diagram to its step response and frequency response rob28124ch11509591indd 590 061216 219 pm Exercises without Answers 591 1 2 3 4 5 A B C D E 1 0 1 1 0 1 z z z z z 1 0 1 1 0 1 1 0 1 1 0 1 1 0 1 1 0 1 1 0 1 1 0 1 2 0 2 0 05 1 15 Ω H 2 0 2 0 2 4 6 8 Ω H 0 05 1 15 2 0 2 Ω H H 0 2 4 2 0 2 Ω H 0 05 1 15 2 0 2 Ω H Figure E56 58 For each system transfer function below which type of ideal filter does it most closely approximate lowpass highpass bandpass or bandstop a H z z 1 z 09 b H z z 2 1 z 2 08 rob28124ch11509591indd 591 061216 219 pm 592 121 INTRODUCTION AND GOALS Pierre Laplace invented the Laplace transform as a method of solving linear constant coefficient differential equations Most continuoustime LTI systems are described at least approximately by differential equations of that type The Laplace transform describes the impulse responses of LTI systems as linear combinations of the eigenfunctions of the differential equations that describe them Because of this Laplace transform directly encapsulates the characteristics of a system in a powerful way Many system analysis and design techniques are based on the use of the Laplace transform without ever directly referring to the differential equations that describe them In this chapter we will explore some of the most common applications of the Laplace transform in system analysis CHAPTER GOA L S 1 To apply the Laplace transform to the generalized analysis of LTI systems including feedback systems for stability timedomain response to standard signals and frequency response 2 To develop techniques for realizing systems in different forms 122 SYSTEM REPRESENTATIONS The discipline of system analysis includes systems of many kinds electrical hydraulic pneumatic chemical and so on LTI systems can be described by differential equations or block diagrams Differential equations can be transformed into algebraic equations by the Laplace transform and these transformed equations form an alternate type of system description Electrical systems can be described by circuit diagrams Circuit analysis can be done in the time domain but it is often done in the frequency domain because of the power of linear algebra in expressing system interrelationships in terms of algebraic instead of differential equations Circuits are interconnections of circuit elements such as resistors capacitors inductors transistors diodes transformers voltage sources current sources and so forth To the extent that these elements can be characterized by linear frequencydomain relationships the circuit can be analyzed 12 C H A P T E R Laplace System Analysis rob28124ch12592649indd 592 041216 154 pm 122 System Representations 593 Figure 121 Timedomain circuit diagram of an RLC circuit R1 vgt vCt iLt R2 L C by frequencydomain techniques Nonlinear elements such as transistors diodes and transformers can often be modeled approximately over small signal ranges as linear devices These models consist of linear resistors capacitors and inductors plus dependent voltage and current sources all of which can be characterized by LTI system transfer functions As an example of circuit analysis using Laplace methods consider the circuit of Figure 121 which illustrates a circuit description in the time domain This circuit can be described by two coupled differential equations v g t R 1 i L t C d dt v C t L d dt i L t 0 L d dt i L t v C t R 2 C d dt v C t 0 If we Laplace transform both equations we get V g s R 1 I L s Cs V C s v c 0 sL I L s i L 0 0 sL I L s i L 0 V C s R 2 C s V C s v c 0 0 If there is initially no energy stored in the circuit it is in its zero state these equations simplify to V g s R 1 I L s s R 1 C V C s sL I L s 0 sL I L s V C s s R 2 C V C s 0 It is common to rewrite the equations in the form R 1 sL s R 1 C sL 1 s R 2 C I L s V C s V g s 0 or Z R 1 s Z L s Z R 1 s Z C s Z L s 1 Z R 2 s Z C s I L s V C s V g s 0 where Z R 1 s R 1 Z R 2 s R 2 Z L s sL Z C s 1sC The equations are written this way to emphasize the impedance concept of circuit analysis The terms sL and 1sC are the impedances of the inductor and capacitor rob28124ch12592649indd 593 041216 154 pm C h a p t e r 12 Laplace System Analysis 594 respectively Impedance is a generalization of the concept of resistance Using this concept equations can be written directly from circuit diagrams using relations similar to Ohms law for resistors V R s Z R Is RIs V L s Z L Is sLIs V C s Z C Is 1sCIs Now the circuit of Figure 121 can be conceived as the circuit of Figure 122 The circuit equations can now be written directly from Figure 122 as two equa tions in the complex frequency domain without ever writing the timedomain equa tions again if there is initially no stored energy in the circuit V g s R 1 I L s sC V C s sL I L s 0 sL I L s V C s s R 2 C V C s 0 These circuit equations can be interpreted in a system sense as differentiation andor multiplication by a constant and summation of signals in this case ILs and VC s R 1 I L s multiplication by a constant s R 1 C V C s differentiation and multiplication by a constant sL I L s differentiation and multiplication by a constant summation V g s sL I L s differentiation and multiplication by a constant V C s s R 2 C V C s differentiation and multiplication by a constant summation 0 A block diagram could be drawn for this system using integrators amplifiers and sum ming junctions Other kinds of systems can also be modeled by interconnections of integrators amplifiers and summing junctions These elements may represent various physical systems that have the same mathematical relationships between an excitation and a response As a very simple example suppose a mass m is acted upon by a force an excitation ft It responds by moving The response could be the position pt of the mass in some appropriate coordinate system According to classical Newtonian me chanics the acceleration of a body in any coordinate direction is the force applied to the body in that direction divided by the mass of the body d 2 d t 2 pt ft m Figure 122 Frequencydomain circuit diagram of an RLC circuit R1 Vgs VCs ILs R2 sL sC 1 rob28124ch12592649indd 594 041216 154 pm 122 System Representations 595 This can be directly stated in the Laplace domain assuming the initial position and velocity are zero as s 2 Ps Fs m So this very simple system could be modeled by a multiplication by a constant and two integrations Figure 123 ft pt Fs Ps 1m 1m s1 s1 Figure 123 Block diagrams of d2 ptdt ftm and s2 Ps Fsm Figure 124 A mechanical system System at Rest ft is the system excitation signal System in Motion m1 m1 m2 Ks1 Ks2 Kd ft m2 Ks1 Ks2 Kd x1 x2 We can also represent with block diagrams more complicated systems like Figure 124 In Figure 124 the positions x1t and x2t are the distances from the rest positions of masses m1 and m2 respectively Summing forces on mass m1 ft K d x 1 t K s1 x 1 t x 2 t m 1 x 1 t Summing forces on mass m2 K s1 x 1 t x 2 t K s2 x 2 t m 2 x 2 t rob28124ch12592649indd 595 041216 154 pm C h a p t e r 12 Laplace System Analysis 596 Laplace transforming both equations Fs K d s X 1 s K s1 X 1 s X 2 s m 1 s 2 X 1 s K s1 X 1 s X 2 s K s2 X 2 s m 2 s 2 X 2 s We can also model the mechanical system with a block diagram Figure 125 Kdm1 Ks2m2 Ks2m2 Kdm1 Ks1 ft x1t x2t Ks1 Fs X1s X2s 1m1 1m1 1m1 1m2 1m1 1m2 s1 s1 s1 s1 Figure 125 Timedomain and frequencydomain block diagrams of the mechanical system of Figure 124 123 SYSTEM STABILITY A very important consideration in system analysis is system stability As shown in Chapter 5 a continuoustime system is boundedinputboundedoutput BIBO stable if its impulse response is absolutely integrable The Laplace transform of the impulse response is the transfer function For systems that can be described by differential equations of the form k0 N a k d k d t k yt k0 M b k d k d t k xt where aN 1 without loss of generality the transfer function is of the form Hs Ys Xs k0 M b k s k k0 N a k s k b M s M b M1 s M1 b 1 s b 0 s N a N1 s N1 a 1 s a 0 The denominator can always be factored numerically if necessary so the transfer function can also be written in the form Hs Ys Xs b M s M b M1 s M1 b 1 s b 0 s p 1 s p 2 s p N If there are any polezero pairs that lie at exactly the same location in the splane they cancel in the transfer function and should be removed before examining the transfer function for stability If M N and none of the poles is repeated then the transfer function can be expressed in partialfraction form as Hs K 1 s p 1 K 2 s p 2 K N s p N rob28124ch12592649indd 596 041216 154 pm 123 System Stability 597 and the impulse response is then of the form ht K 1 e p 1 t K 2 e p 2 t K N e p N t ut where the ps are the poles of the transfer function For ht to be absolutely integrable each of the terms must be individually absolutely integrable The integral of the magnitude of a typical term is I K e pt ut dt K 0 e Rept e jImpt dt I K 0 e Rept e jImpt 1 dt K 0 e Rept dt In the last integral eRept is nonnegative over the range of integration Therefore I K 0 e Rept dt For this integral to converge the real part of the pole p must be negative For BIBO stability of a linear timeinvariant LTI system all the poles of its transfer function must lie in the open left halfplane LHP The term open left halfplane means the left halfplane not including the ω axis If there are simple nonrepeated poles on the ω axis and no poles are in the right halfplane RHP the system is called marginally stable because even though the impulse response does not decay with time it does not grow either Marginal stability is a special case of BIBO instability because in these cases it is possible to find a bounded input signal that will produce an unbounded output signal Even though it sounds strange a marginally stable system is also a BIBO unstable system If there is a repeated pole of order n in the transfer function the impulse response will have terms of the general form t n1ept ut where p is the location of the repeated pole If the real part of p is not negative terms of this form grow without bound in positive time indicating there is an unbounded response to a bounded excitation and that the system is BIBO unstable Therefore if a systems transfer function has repeated poles the rule is unchanged The poles must all be in the open LHP for system stability However there is one small difference from the case of simple poles If there are repeated poles on the ω axis and no poles in the right halfplane RHP the system is not marginally stable it is simply unstable These conditions are summarized in Table 121 Table 121 Conditions for system stability marginal stability or instability which includes marginal stability as a special case Stability Marginal Stability Instability All poles in the open LHP One or more simple poles on One or more poles in the the ω axis but no repeated poles open RHP or on the ω axis on the ω axis and no poles in includes marginal stability the open RHP rob28124ch12592649indd 597 041216 154 pm C h a p t e r 12 Laplace System Analysis 598 If we excite the system in Figure 126 a by applying an impulse of horizontal force to the sphere it responds by moving and then rolling back and forth If there is even the slightest bit of rolling friction or any other loss mechanism like air re sistance the sphere eventually returns to its initial equilibrium position This is an example of a stable system If there is no friction or any other loss mechanism the sphere will oscillate back and forth forever but will remain confined near the relative lowpoint of the surface Its response does not grow with time but it does not decay either In this case the system is marginally stable If we excite the sphere in Figure 126 b even the slightest bit the sphere rolls down the hill and never returns If the hill is infinitely high the spheres speed will approach infinity an unbounded response to a bounded excitation This is an unstable system In Figure 126 c if we excite the sphere with an impulse of horizontal force it responds by rolling If there is any loss mechanism the sphere eventually comes to rest but not at its original point This is a bounded response to a bounded excitation and the system is stable If there is no loss mechanism the sphere will roll forever without accelerating This is marginal stability again ExamplE 121 Repeated pole on the ω axis The simplest form of a system with a repeated pole on the ω axis is the double integrator with transfer function Hs A s 2 where A is a constant Find its impulse response Using t n ut n s n1 we find the transform pair At ut A s 2 a ramp function that grows without bound in positive time indicating that the system is unstable and not mar ginally stable Stable Equilibrium Stable Equilibrium Unstable Equilibrium Unstable Equilibrium Marginally Stable Equilibrium Without Rolling Friction With Rolling Friction Marginally Stable Equilibrium a b c Figure 126 Illustrations of three types of stability An analogy that is sometimes helpful in remembering the different descriptions of system stability or instability is to consider the movement of a sphere placed on different kinds of surfaces Figure 126 rob28124ch12592649indd 598 041216 154 pm 124 System Connections 599 124 SYSTEM CONNECTIONS CASCADE AND PARALLEL CONNECTIONS Earlier we found the impulse response and frequency responses of cascade and parallel connections of systems The results for these types of systems are the same for transfer functions as they were for frequency responses Figure 127 and Figure 128 Figure 127 Cascade connection of systems H1sH2s Xs Ys H1s H2s Xs XsH1s Ys XsH1sH2s Figure 128 Parallel connection of systems H1s H2s Xs Ys H1s H2s Xs Ys XsH1s XsH2s XsH1s H2s XsH1s XsH2s Figure 129 Feedback connection of systems H1s H2s Xs Ys Es THE FEEDBACK CONNECTION Terminology and Basic Relationships Another type of connection that is very important in system analysis is the feedback connection Figure 129 The transfer function H1s is in the forward path and the transfer function H2s is in the feedback path In the controlsystem literature it is common to call the forwardpath transfer function H1s the plant because it is usually an established system designed to produce something and the feedbackpath transfer function H2s the sensor because it is usually a system added to the plant to help control it or stabilize it by sensing the plant response and feeding it back to the summing junction at the plant input The excitation of the plant is called the error signal and is given by Es Xs H2s Ys and the response of H1s which is Ys H1sEs is the excitation for the sensor H2s Combining equations and solving for the overall transfer function Hs Ys Xs H 1 s 1 H 1 sH 2 s 121 In the block diagram illustrating feedback in Figure 129 the feedback signal is subtracted from the input signal This is a very common convention in feedback system analysis and stems from the history of feedback used as negative feedback to stabilize a system The basic idea behind the term negative is that if the plant output signal rob28124ch12592649indd 599 041216 154 pm C h a p t e r 12 Laplace System Analysis 600 goes too far in some direction the sensor will feed back a signal proportional to the plant output signal which is subtracted from the input signal and therefore tends to move the plant output signal in the opposite direction moderating it This of course assumes that the signal fed back by the sensor really does have the quality of stabilizing the system Whether the sensor signal actually does stabilize the system depends on its dynamic response and the dynamic response of the plant It is conventional in system analysis to give the product of the forward and feedbackpath transfer functions the special name loop transfer function Ts H1s H2s because it shows up so much in feedback system analysis In electronic feedback amplifier design this is sometimes called the loop transmission It is given the name loop transfer function or loop transmission because it represents what happens to a signal as it goes from any point in the loop around the loop exactly one time and back to the starting point except for the effect of the minus sign on the summing junction So the transfer function of the feedback system is the forwardpath transfer function H1s divided by one plus the loop transfer function or Hs H 1 s 1 Ts Notice that when H2s goes to zero meaning there is no feedback that Ts does also and the system transfer function Hs becomes the same as the forwardpath transfer function H1s Feedback Effects on Stability It is important to realize that feedback can have a very dramatic effect on system re sponse changing it from slow to fast fast to slow stable to unstable or unstable to stable The simplest type of feedback is to feed back a signal directly proportional to the output signal That means that H2s K a constant In that case the overall system transfer function becomes Hs H 1 s 1 K H 1 s Suppose the forwardpath system is an integrator with transfer function H1s 1s which is marginally stable Then Hs 1s 1 Ks 1 s K The forwardpath transfer function H1s has a pole at s 0 but Hs has a pole at s K If K is positive the overall feedback system is stable having one pole in the open LHP If K is negative the overall feedback system is unstable with a pole in the RHP As K is made a larger positive value the pole moves farther from the origin of the splane and the system responds more quickly to an input signal This is a simple demonstration of an effect of feedback There is much more to learn about feedback and usually a full semester of feedback control theory is needed for a real appreciation of the effects of feedback on system dynamics Feeding the forwardpath output signal back to alter its own input signal is often called closing the loop for obvious reasons and if there is no feedback path the system is said to be operating openloop Politicians business executives and other wouldbe movers and shakers in our society want to be in the loop This terminology probably came from feedback loop concepts because one who is in the loop has the chance of affecting the system performance and therefore has power in the political economic or social system rob28124ch12592649indd 600 041216 154 pm 124 System Connections 601 If K is large enough then at least for some values of s KH2s 1 and Hs 1H2s and the overall transfer function of the feedback system performs the approximate inverse of the operation of the feedback path That means that if we were to cascade connect a system with transfer function H2s to this feedback system the overall system transfer function would be approximately one Figure 1211 over some range of values of s It is natural to wonder at this point what has been accomplished because the system of Figure 1211 seems to have no net effect There are real situations in which a signal has been changed by some kind of unavoidable system effect and we desire to restore the original signal This is very common in communication systems in which a signal has been sent over a channel that ideally would not change the signal but actually does for reasons beyond the control of the designer An equalization filter can be used to restore the original signal It is designed to have the inverse of the effect of the channel on the signal as nearly as possible Some systems designed to measure physical phenomena use sensors that have inherently lowpass transfer functions usually because of some unavoidable mechanical or thermal inertia The measurement system can be made to respond more quickly by cascading the sensor with an electronic signalprocessing system whose transfer function is the approximate inverse of the sensors transfer function Another beneficial effect of feedback is to reduce the sensitivity of a system to parameter changes A very common example of this benefit is the use of feedback in an operational amplifier configured as in Figure 1212 Figure 1210 A feedback system K H2s Xs Ys Es Figure 1211 A system cascaded with another system designed to be its approximate inverse H2s Xs Ys H2s K Figure 1212 An inverting voltage amplifier using an operational amplifier with feedback Zis Zf s Ves Vos Vis Beneficial Effects of Feedback Feedback is used for many different purposes One interesting effect of feedback can be seen in a system like Figure 1210 The overall transfer function is Hs K 1 K H 2 s A typical approximate expression for the gain of an operational amplifier with the noninverting input grounded H1s in the feedback block diagram is H 1 s V o s V e s A 0 1 sp rob28124ch12592649indd 601 041216 154 pm C h a p t e r 12 Laplace System Analysis 602 where A0 is the magnitude of the operational amplifier voltage gain at low frequencies and p is a single pole on the negative real axis of the splane The overall transfer func tion can be found using standard circuit analysis techniques But it can also be found by using feedback concepts The error voltage Ves is a function of Vis and Vos Since the input impedance of the operational amplifier is typically very large compared with the two external impedances Zi s and Zf s the error voltage is V e s V o s V i s V o s Z f s Z i s Z f s or V e s V o s Z i s Z i s Z f s V i s Z f s Z i s Z f s So we can model the system using the block diagram in Figure 1213 According to the general feedbacksystem transfer function Hs Ys Xs H 1 s 1 H 1 sH 2 s the amplifier transfer function should be V o s V i s Z f s Z i s Z f s A 0 1 sp 1 A 0 1 sp Z i s Z i s Z f s Simplifying and forming the ratio of Vo s to Vi s as the desired overall transfer function V o s V i s A 0 Z f s 1 sp A 0 Z i s 1 spZ f s If the lowfrequency gain magnitude A0 is very large which it usually is then we can approximate this transfer function at low frequencies as V o s V i s Z f s Z i s Figure 1213 Block diagram of an inverting voltage amplifier using feedback on an operational amplifier Zf s Zis Zf s Zis Zis Zf s Vis Ves Vos A0 1 sp rob28124ch12592649indd 602 041216 154 pm 124 System Connections 603 This is the wellknown idealoperationalamplifier formula for the gain of an inverting voltage amplifier In this case being large means that A0 is large enough that the de nominator of the transfer function is approximately A0 Zi s which means that A 0 1 s p and A 0 1 s p Z f s Z i s Its exact value is not important as long as it is very large and that fact represents the reduction in the systems sensitivity to changes in parameter values that affect A0 or p To illustrate the effects of feedback on amplifier performance let A0 107 and p 100 Also let Zf s be a resistor of 10 kΩ and let Zi s be a resistor of 1 kΩ Then the overall system transfer function is V o s V i s 10 8 11 1 s100 10 7 The numerical value of the transfer function at a real radian frequency of ω 100 a cyclic frequency of f 1002π 159 Hz is V o s V i s 10 8 11 j11 10 7 9999989 j0000011 Now let the operational amplifiers lowfrequency gain be reduced by a factor of 10 to A0 106 When we recalculate the transfer function at 159 Hz we get V o s V i s 10 7 11 j11 10 6 999989 j000011 which represents a change of approximately 0001 in the magnitude of the transfer function So a change in the forwardpath transfer function by a factor of 10 produced a change in the overall system transfer function magnitude of about 0001 The feed back connection made the overall transfer function very insensitive to changes in the operational amplifier gain even very large changes In amplifier design this is a very beneficial result because resistors and especially resistor ratios can be made very insensitive to environmental factors and can hold the system transfer function almost constant even if components in the operational amplifier change by large percentages from their nominal values Another consequence of the relative insensitivity of the system transfer function to the gain A0 of the operational amplifier is that if A0 is a function of signal level mak ing the operational amplifier gain nonlinear as long as A0 is large the system transfer function is still very accurate Figure 1214 and practically linear Another beneficial effect of feedback can be seen by calculating the bandwidth of the operational amplifier itself and comparing that to the bandwidth of the inverting amplifier with feedback The corner frequency of the operational amplifier itself in this example is 159 Hz The corner frequency of the inverting amplifier with feed back is the frequency at which the real and imaginary parts of the denominator of the overall transfer function are equal in magnitude That occurs at a cyclic frequency of f 145 MHz This is an increase in bandwidth by a factor of approximately 910000 It is hard to overstate the importance of feedback principles in improving the performance of many systems in many ways rob28124ch12592649indd 603 041216 154 pm C h a p t e r 12 Laplace System Analysis 604 The transfer function of the operational amplifier is a very large number at low frequencies So the operational amplifier has a large voltage gain at low frequencies The voltage gain of the feedback amplifier is typically much smaller So in using feedback we have lost voltage gain but gained gain stability and bandwidth among other things In effect we have traded gain for improvements in other amplifier characteristics Feedback can be used to stabilize an otherwise unstable system The F117 stealth fighter is inherently aerodynamically unstable It can fly under a pilots control only with the help of a computercontrolled feedback system that senses the aircrafts position speed and attitude and constantly compensates when it starts to go unstable A very simple example of stabilization of an unstable system using feedback would be a system whose transfer function is H 1 s 1 s p p 0 With a pole in the RHP this system is unstable If we use a feedbackpath transfer func tion that is a constant gain K we get the overall system transfer function Hs 1 s p 1 K s p 1 s p K For any value of K satisfying K p the feedback system is stable Instability Caused by Feedback Although feedback can have many very beneficial effects there is another effect of feedback in systems that is also very important and can be a problem rather than a benefit The addition of feedback to a stable system can cause it to become unstable The overall feedbacksystem transfer function is Hs Ys X s H 1 s 1 H 1 sH 2 s Even though all the poles of H1s and H2s may lie in the open LHP the poles of Hs may not Consider the forward and feedback transfer functions H 1 s K s 3s 5 and H 2 s 1 s 4 Figure 1214 Linear and nonlinear operational amplifier gain Ideal Gain Actual Gain vt vt vot rob28124ch12592649indd 604 041216 154 pm 124 System Connections 605 H1s and H2s are both BIBO stable But if we put them into a feedback system like Figure 129 the overall system gain is then Hs Ks 4 s 3s 4s 5 K Ks 4 s 3 12 s 2 47s 60 K Whether or not this feedback system is stable now depends on the value of K If K is 5 the poles lie at 5904 and 3048 j1311 They are all in the open LHP and the feedback system is stable But if K is 700 the poles lie at 12917 and 04583 j7657 Two poles are in the RHP and the system is unstable Almost everyone has experienced a system made unstable by feedback Often when large crowds gather to hear someone speak a publicaddress system is used The speaker speaks into a microphone His voice is amplified and fed to one or more speakers so everyone in the audience can hear his voice Of course the sound emanating from the speakers is also detected and amplified by the microphone and amplifier This is an example of feedback because the output signal of the public address system is fed back to the input of the system Anyone who has ever heard it will never forget the sound of the public address system when it goes unstable usually a very loud tone And we probably know the usual solution turn down the amplifier gain This tone can occur even when no one is speaking into the microphone Why does the system go unstable with no apparent input signal and why does turning down the amplifier gain not just reduce the volume of the tone but eliminate it entirely Albert Einstein was famous for the Gedankenversuch thought experiment We can understand the feedback phenomenon through a thought experiment Imagine that we have a microphone amplifier and speaker in the middle of a desert with no one around and no wind or other acoustic disturbance and that the amplifier gain is initially turned down to zero If we tap on the microphone we hear only the direct sound of tapping and nothing from the speakers Then we turn the amplifier gain up a little Now when we tap on the microphone we hear the tap directly but also some sound from the speakers slightly delayed because of the distance the sound has to travel from the speakers to our ears assuming the speakers are farther away from our ears than the microphone As we turn the gain up more and more increasing the loop transfer function T the tapping sound from the speakers rises in volume Figure 1215 In Figure 1215 pt is acous tic pressure as a function of time As we increase the magnitude of the loop transfer function T by turning up the amplifier gain when we tap on the microphone we gradually notice a change not just in the volume but also in the nature of the sound from the speakers We hear not only the tap but also what is commonly called reverberation multiple echoes of the tap These multiple echoes are caused by the sound of the tap coming from the speaker to the microphone being amplified and going to the speaker again and returning to the microphone again multiple times As the gain is increased this phenomenon becomes more obvious and at some gain level a loud tone begins and continues without any tapping or any other acoustic input to the microphone until we turn the gain back down At some level of amplifier gain any signal from the microphone no matter how weak is amplified fed to the speaker returns to the microphone and causes a new sig nal in the microphone which is the same strength as the original signal At this gain the signal never dies it just keeps on circulating If the gain is made slightly higher the signal grows every time it makes the round trip from microphone to speaker and back If the public address system were truly linear that signal would increase without bound But no real public address system is truly linear and at some volume level the amplifier is driving the speaker as hard as it can but the sound level does not increase any more rob28124ch12592649indd 605 041216 154 pm C h a p t e r 12 Laplace System Analysis 606 It is natural to wonder how this process begins without any acoustic input to the microphone First as a practical matter it is impossible to arrange to have absolutely no ambient sound strike the microphone Second even if that were possible the ampli fier has inherent random noise processes that cause an acoustic signal from the speaker and that is enough to start the feedback process Now carry the experiment a little further With the amplifier gain high enough to cause the tone we move the speaker farther from the microphone As we move the speaker away the pitch of the loud tone changes and at some distance the tone stops The pitch changes because the frequency of the tone depends on the time sound takes to propagate from the speaker to the microphone The loud tone stops at some distance because the sound intensity from the speaker is reduced as it is moved farther away and the return signal due to feedback is less than the original signal and the signal dies away instead of increasing in power Now we will mathematically model the public address system with the tools we have learned and see exactly how feedback instability occurs Figure 1216 To keep the model simple yet illustrative we will let the transfer functions of the microphone amplifier and speaker be the constants Km KA and Ks Then we model the propagation of sound from the speaker to the microphone as a simple delay with a gain that is in versely proportional to the square of the distance d from the speaker to the microphone p m t K p s t dv d 2 122 where Pst is the sound acoustic pressure from the speaker Pmt is the sound ar riving at the microphone v is the speed of sound in air and K is a constant Laplace transforming both sides of 122 P m s K d 2 P s s e dsv Figure 1215 Public address system sound from tapping on the microphone for three different system loop transfer functions t 06 1 1 T 09 pt t 06 1 1 pt t 06 1 1 T 03 T 06 pt Echos Figure 1216 A public address system Amplifier rob28124ch12592649indd 606 041216 154 pm 124 System Connections 607 Then we can model the public address system as a feedback system with a forwardpath transfer function H1s KmKAKs and a feedbackpath transfer function H 2 s K d 2 e dsv Figure 1217 The overall transfer function is Hs K m K A K s 1 K m K A K s K d 2 e dsv The poles p of this system transfer function lie at the zeros of 1Km KA Ks Kd2edpv Solving 1 K m K A K s K d 2 e dpv 0 123 or e dpv d 2 K m K A K s K Any value of p that solves this equation is a pole location If we take the logarithm of both sides and solve for p we get p v d ln d 2 K m K A K s K So this is a solution of 123 But it is not the only solution It is just the only realvalued solution If we add any integer multiple of j2πvd to p we get another solution because e dpj2nπvdv e dpv e j2nπ 1 e dpv where n is any integer That means that there are infinitely many poles all with the same real part v d ln d 2 K m K A K s K Figure 1218 This system is a little different from the systems we have been analyzing because this system has infinitely many poles one for each integer n But that is not a problem in this analysis because we are only trying to establish the conditions under which the Figure 1217 Block diagram of a public address system smt sst KmKAKs K e d2 d s v rob28124ch12592649indd 607 041216 154 pm C h a p t e r 12 Laplace System Analysis 608 system is stable As we have already seen stability requires that all poles lie in the open LHP That means in this case that v d ln d 2 K m K A K s K 0 or ln d 2 K m K A K s K 0 or K m K A K s K d 2 1 124 In words the product of all the transferfunction magnitudes around the feedback loop must be less than one This makes common sense because if the product of all the transferfunction magnitudes around the loop exceeds one that means that when a signal makes a complete roundtrip through the feedback loop it is bigger when it comes back than when it left and that causes it to grow without bound So when we turn down the amplifier gain KA to stop the loud tone caused by feedback we are satisfying 124 Suppose we increase the loop transfer function magnitude KmKAKsKd2 by turning up the amplifier gain KA The poles move to the right parallel to the σ axis and at some gain value they reach the ω axis Now suppose instead we increase the loop transfer function magnitude by moving the microphone and speaker closer together This moves the poles to the right but also away from the σ axis so that when we reach marginal stability the poles are all at higher radian frequencies A system that obeys this simple model can oscillate at multiple frequencies simultaneously In reality that is unlikely A real public address system microphone amplifier and speaker would have transfer functions that are functions of frequency and would therefore change the pole locations so that only one pair of poles would lie on the ω axis at marginal stability If the gain is turned up above the gain for marginal stability the system is driven into a nonlinear mode of operation and linear system analysis methods fail to predict exactly how it will oscillate But linear system methods do predict accurately that it will oscillate and that is very important Stable Oscillation Using Feedback The oscillation of the public address system in the last section was an undesirable system response But some systems are designed to oscillate Examples are labora tory function generators computer clocks local oscillators in radio receivers quartz Figure 1218 Polezero diagram of the public address system 2 ω σ πv d πv d πv d πv d 3 2 4 v d ln KmKAKs K d2 rob28124ch12592649indd 608 041216 154 pm 124 System Connections 609 crystals in wristwatches a pendulum on a grandfather clock and so on Some systems are designed to oscillate in a nonlinear mode in which they simply alternate between two or more unstable states and their response signals are not necessarily sinusoidal Freerunning computer clocks are a good example But some systems are designed to operate as an LTI system in a marginally stable mode with a true sinusoidal oscilla tion Since marginal stability requires that the system have poles on the ω axis of the splane this mode of operation is very exacting The slightest movement of the system poles due to any parameter variation will cause the oscillation either to grow or decay with time So systems that operate in this mode must have some mechanism for keep ing the poles on the ω axis The prototype feedback diagram in Figure 1219 has an excitation and a response A system designed to oscillate does not have an apparent excitation that is Xs 0 Figure 1220 The sign is changed on H2s to make the system in Figure 1220 be just like the system in Figure 1219 with Xs 0 How can we have a response if we have no excitation The short answer is we cannot However it is important to realize that every system is constantly being excited whether we intend it or not Every system has random noise processes that cause signal fluctuations The system responds to these noise fluctuations just as it would to an intentional excitation Figure 1219 Prototype feedback system Xs Es Ys H2s H1s Figure 1220 Oscillator feedback system H2s Ys H1s The key to having a stable oscillation is having a transfer function with poles on the ω axis of the form Hs A s 2 ω 0 2 Then the system gain at the radian frequency ω0 s jω0 is infinite implying that the response is infinitely greater than the excitation That could mean either that a finite excitation produces an infinite response or that a zero excitation produces a finite response Therefore a system with poles on the ω axis can produce a stable nonzero response with no excitation1 One very interesting and important example of a system designed to oscillate in a marginally stable mode is a laser The acronym LASER stands for Light Amplifi cation by Stimulated Emission of Radiation A laser is not actually a light amplifier although internally light amplification does occur it is a light oscillator But the acronym for Light Oscillation by Stimulated Emission of Radiation LOSER de scribed itself and did not catch on 1 It is important here to distinguish between two uses of the word stable A BIBO stable system is one that has a bounded response to any arbitrary bounded excitation A stable oscillation in the context of this section is an oscillating output signal that maintains a constant amplitude neither growing nor decaying If an LTI system has an impulse response that is a stable oscillation the system is marginally stable a special case of BIBO unstable That is for such a system there exists a bounded excitation that would produce an unbounded response If we were to actually excite any real system with such an excitation its response would grow for a while but then at some signal level would change from an LTI system to a nonlinear system or would start to reveal that it was never actually an LTI system in the first place and the response signal would remain bounded rob28124ch12592649indd 609 041216 154 pm C h a p t e r 12 Laplace System Analysis 610 Even though the laser is an oscillator light amplification is an inherent process in its operation A laser is filled with a medium that has been pumped by an external power source in such a way that light of the right wavelength propagating through the pumped medium experiences an increase in power as it propagates Figure 1221 The device illustrated in Figure 1221 is a onepass travellingwave light amplifier not a laser The oscillation of light in a laser is caused by introducing into the onepass travellingwave light amplifier mirrors on each end that reflect some or all of the light striking them At each mirror some or all of the light is fed back into the pumped laser medium for further amplification Figure 1222 Pumped Laser Medium Mirror Mirror Figure 1222 A laser Pumped Laser Medium Light In Light Out Figure 1221 A onepass travellingwave light amplifier It would be possible in principle to introduce light at one end of this device through a partial mirror and amplify it Such a device is called a regenerative travellingwave light amplifier But it is much more common to make the mirror at one end as reflective as possible essentially reflecting all the light that strikes it and make the mirror at the other end a partial mirror reflecting some of the light that strikes it and transmitting the rest A laser operates without any external light signal as an excitation The light that it emits begins in the pumped laser medium itself A phenomenon called spontaneous emission causes light to be generated at random times and in random directions in the pumped medium Any such light that happens to propagate perpendicular to a mirror gets amplified on its way to the mirror then reflected and further amplified as it bounces between the mirrors The closer the propagation is to perpendicular to the mirrors the longer the beam bounces and the more it is amplified by the multiple passes through the laser medium In steadystate operation the light that is perpendic ular to the mirrors has the highest power of all the light propagation inside the laser cavity because it has the greatest gain advantage One mirror is always a partial mirror so some light transmits at each bounce off that mirror This light constitutes the output light beam of the laser Figure 1223 In order for light oscillation to be sustained the loop transfer function of the sys tem must be the real number 1 under the assumed negative feedback sign on the prototype feedback system of Figure 1219 or it must be the real number 1 under the assumption of the oscillator system of Figure 1220 Under either assumption for stable oscillation the light as it travels from a starting point to one mirror back to the other mirror and then back to the starting point must experience an overall gain mag nitude of one and phase shift of an integer multiple of 2π radians This simply means that the wavelength of the light must be such that it fits into the laser cavity with exactly an integer number of waves in one roundtrip path It is important to realize here that the wavelength of light in lasers is typically somewhere in the range from 100 nm to many microns ultraviolet to far infrared and typical lengths of laser cavities are in the range of a 100 µm for a laser diode to more than a meter in some cases Therefore as light propagates between the mirrors it may experience more than a million radians of phase shift and even in the shortest cavities rob28124ch12592649indd 610 041216 154 pm 124 System Connections 611 the phase shift is usually a large multiple of 2π radians So in a laser the exact wave length of oscillation is determined by which optical wavelength fits into the roundtrip path with exactly an integer number of waves There are infinitely many wavelengths that satisfy this criterion the wave that fits into the round trip exactly once plus all its harmonics Figure 1224 Although all these wavelengths of light could theoretically oscillate there are other mechanisms atomic or molecular resonances wavelengthselective mirrors etc that limit the actual oscillation to a small number of these wavelengths that experience enough gain to oscillate A laser can be modeled by a block diagram with a forward path and a feedback path Figure 1225 The constants KF and KR represent the magnitude of the gain ex perienced by the electric field of the light as it propagates from one mirror to the other 100 Mirror Partial Mirror 100 Mirror Partial Mirror 100 Mirror Partial Mirror 100 Mirror Partial Mirror 100 Mirror Partial Mirror 100 Mirror Partial Mirror Figure 1223 Multiple light reflections at different initial angles Figure 1224 Illustrations of wavelengths that fit into the laser cavity an integer number of times Es Kro Kto Kr KFe L c s KRe L c s Figure 1225 Laser block diagram along the forward and reverse paths respectively The factors eLcs account for the phase shift due to propagation time where L is the distance between the mirrors and c is the speed of light in the laser cavity The constant Kto is the electric field transmission coefficient for light exiting the laser cavity through the output partial mirror and the constant Kro is the electric field reflection coefficient for light reflected at the output partial mirror back into the laser cavity The constant Kr is the electric field reflection coefficient for light reflected at the 100 mirror back into the laser cavity Kto Kro and Kr are in general complex indicating that there is a phase shift of the electric field rob28124ch12592649indd 611 041216 154 pm C h a p t e r 12 Laplace System Analysis 612 during reflection and transmission The loop transfer function is using the definition developed based on the sign convention in Figure 1219 Ts K F K ro K R K r e 2Lcs Its value is 1 when K F K ro K R K r 1 and e 2Lcs 1 or equivalently s j2πn c 2L j πc L n n any integer where the quantity c2L is the roundtrip travel time for the propagating light wave These are values of s on the ω axis at harmonics of a fundamental radian frequency πcL Since this is the fundamental frequency it is also the spacing between frequen cies which is conventionally called the axial mode spacing Δωax When a laser is first turned on the medium is pumped and a light beam starts through spontaneous emission It grows in intensity because at first the magnitude of the roundtrip gain is greater than one K F K ro K R K r 1 But as it grows it extracts energy from the pumped medium and that reduces the gains KF and KR An equilibrium is reached when the beam strength is exactly the right magnitude to keep the loop transfer function magnitude K F K ro K R K r at exactly one The pumping and lightamplification mechanisms in a laser together form a selflimiting process that stabilizes at a loop transfer function magnitude of one So as long as there is enough pumping power and the mirrors are reflective enough to achieve a loop transfer function magnitude of one at some very low output power the laser will oscillate stably The RootLocus Method A very common situation in feedback system analysis is a system for which the forwardpath gain H1s contains a gain constant K that can be adjusted That is H 1 s K P 1 s Q 1 s The adjustable gain parameter K conventionally taken to be nonnegative has a strong effect on the systems dynamics The overall system transfer function is Hs H 1 s 1 H 1 sH 2 s and the loop transfer function is Ts H1sH2s The poles of Hs are the zeros of 1 Ts The loop transfer function can be written in the form of K times a numerator divided by a denominator Ts K P 1 s Q 1 s P 2 s Q 2 s K Ps Qs 125 rob28124ch12592649indd 612 041216 154 pm 124 System Connections 613 so the poles of Hs occur where 1 K Ps Qs 0 which can be expressed in the two alternate forms Qs K Ps 0 126 and Qs K Ps 0 127 From 125 we see that if Ts is proper Qs is of higher order than Ps the zeros of Qs constitute all the poles of Ts and the zeros of Ps are all finite zeros of Ts but because the order of Ps is less than the order of Qs there are also one or more zeros of Ts at infinity The full range of possible adjustment of K is from zero to infinity First let K approach zero In that limit from 126 the zeros of 1 Ts which are the poles of Hs are the zeros of Qs and the poles of Hs are therefore the poles of Ts because Ts K PsQs Now consider the opposite case K approaching infinity In that limit from 127 the zeros of 1 Ts are the zeros of Ps and the poles of Hs are the zeros of Ts including any zeros at infinity So the loop transfer function poles and zeros are very important in the analysis of the feedback system As the gain factor K moves from zero to infinity the poles of the feedback system move from the poles of the loop transfer function to the zeros of the loop transfer function some of which may be at infinity A rootlocus plot is a plot of the locations of the feedbacksystem poles as the gain factor K is varied from zero to infinity The name root locus comes from the location locus of a root of 1 Ts as the gain factor K is varied We will first examine two simple examples of the rootlocus method and then establish some general rules for drawing the root locus of any system Consider first a system whose forwardpath gain is H 1 s K s 1s 2 and whose feedbackpath gain is H2s 1 Then Ts K s 1s 2 and the rootlocus plot begins at s 1 and s 2 the poles of Ts All the zeros of Ts are at infinity and those are the zeros that the root locus approaches as the gain factor K is increased Figure 1226 The roots of 1 Ts are the roots of s 1s 2 K s 2 3s 2 K 0 and using the quadratic formula the roots are at 3 1 4K 2 For K 0 we get roots at s 1 and s 2 the poles of Ts For K 14 we get a repeated root at 32 For K 14 we get two complexconjugate roots whose imaginary parts go to plus and rob28124ch12592649indd 613 041216 154 pm C h a p t e r 12 Laplace System Analysis 614 minus infinity as K increases but whose real parts stay at 32 Since this root locus ex tends to infinity in the imaginary dimension with a real part that always places the roots in the LHP this feedback system is stable for any value of K Now add one pole to the forwardpath transfer function making it H 1 s K s 1s 2s 3 The new root locus is the locus of solutions to the equation s3 6s2 11s 6 K 0 Figure 1227 At or above the value of K for which two branches of the root locus cross the ω axis this system is unstable So this system which is openloop stable can be made unstable by using feedback The poles are at the roots of s3 6s2 11s 6 K 0 It is possible to find a general solution for a cubic equation of this type but it is very tedious It is much easier to generate multiple values for K and solve for the roots nu merically to find the value of K that causes the poles of Hs to move into the RHP In Figure 1228 we can see that K 60 puts two poles exactly on the ω axis So any value of K greater than or equal to 60 will cause this feedback system to be unstable Figure 1228 Roots of s3 6s2 11s 6 K 0 for several values of K Figure 1227 Root locus of 1 Ts 1 K s 1s 2s 3 s K60 K10 ω σ 1 3 2 K1 Figure 1226 Root locus of 1 Ts 1 K s 1s 2 s ω σ 2 1 K Roots 0 3 2 1 025 311 173 116 05 319 14 j025 14 j025 1 332 134 j056 134 j056 2 352 124 j086 124 j086 10 431 085 j173 085 j173 30 521 039 j260 039 j260 60 600 000 j332 000 j332 100 671 036 j396 036 j396 rob28124ch12592649indd 614 041216 154 pm 124 System Connections 615 Figure 1229 illustrates some rootlocus plots for different numbers and different locations of the poles and zeros of Ts There are several rules for plotting a root locus These rules come from rules of algebra derived by mathematicians about the locations of the roots of polynomial equations 1 The number of branches in a root locus is the greater of the degree of the numerator polynomial and the degree of the denominator polynomial of Ts 2 Each rootlocus branch begins on a pole of Ts and terminates on a zero of Ts 3 Any region of the real axis for which the sum of the number of real poles andor real zeros lying to its right on the real axis is odd is a part of the root locus and all other regions of the real axis are not part of the root locus The regions that are part of the root locus are called allowed regions 4 The root locus is symmetrical about the real axis 5 If the number of finite poles of Ts exceeds the number of finite zeros of Ts by an integer m then m branches of the root locus terminate on zeros of Ts that lie at infinity Each of these branches approaches a straightline asymptote and the angles of these asymptotes are 2k 1πm k 0 1 m 1 with respect to the positive real axis These asymptotes intersect on the real axis at the location σ 1 m finite poles finite zeros called the centroid of the root locus These are sums of all finite poles and all finite zeros not just the ones on the real axis 6 The breakaway or breakin points where the rootlocus branches intersect occur where d ds 1 Ts 0 Figure 1229 Example rootlocus plots ω σ ω σ ω σ ω σ ω σ ω σ ω σ ω σ rob28124ch12592649indd 615 041216 154 pm C h a p t e r 12 Laplace System Analysis 616 ExamplE 122 Root locus 1 Draw a root locus for a system whose loop transfer function is Ts s 4s 5 s 1s 2s 3 The thinking steps in figuring out where the rootlocus branches go are the following 1 Ts has poles at σ 1 σ 2 and σ 3 and zeros at σ 4 σ 5 and s 2 The number of rootlocus branches is 3 Rule 1 3 The allowed regions on the real axis are in the ranges 2 σ 1 4 σ 3 and σ 5 Figure 1230 Rule 3 Figure 1231 Initial stage of drawing a root locus s 10 5 5 5 ω σ Figure 1230 Allowed regions on the real axis s ω σ 10 5 5 5 Allowed Regions on Real Axis 4 The rootlocus branches must begin at σ 1 σ 2 and σ 3 Rule 2 5 Two rootlocus branches must terminate on σ 4 and σ 5 and the third branch must terminate on the zero at infinity Rule 2 6 The two rootlocus branches beginning at σ 1 and σ 2 initially move toward each other because they must stay in an allowed region Rule 3 When they intersect they must both become complex and must be complex conjugates of each other Rule 4 7 The third rootlocus branch beginning at σ 3 must move to the left toward the zero at σ 4 Rule 3 This branch cannot go anywhere else and at the same time preserve the symmetry about the real axis So this branch simply terminates on the zero at σ 4 Rule 2 Figure 1231 8 Now we know that the two other rootlocus branches must terminate on the zero at σ 5 and the zero at s They are already complex Therefore they have to move to the left and back down to the σ axis and then one must go to the right to terminate on the zero at σ 5 while the other one moves to the left on the real axis approaching negative infinity Rule 2 9 There are three finite poles and two finite zeros That means there is only one root locus branch going to a zero at infinity as we have already seen The angle that branch approaches should be π radians the negative real axis Rule 5 This agrees with the previous conclusion number 8 rob28124ch12592649indd 616 041216 154 pm 124 System Connections 617 10 The point at which the two branches break out of the real axis and the point at which the two branches break back into the real axis must both occur where dds1Ts 0 Rule 6 d ds 1 Ts d ds s 1s 2s 3 s 4s 5 0 Differentiating and equating to zero we get s4 18s3 103s2 228s 166 0 The roots are at s 947 s 434 s 269 and s 150 So the breakout point is at σ 150 and the breakin point is at σ 947 The root locus never moves into the RHP so this system is stable for any nonnegative value of the gain factor K Figure 1232 The other two solutions of s4 18s3 103s2 228s 166 0 s 434 and s 269 are the breakout and breakin points for the socalled complementary root locus The comple mentary root locus is the locus of the poles of Hs as K goes from zero to negative infinity ExamplE 123 Root locus 2 Draw a root locus for a system whose forward path plant is the system of Figure 1233 with a2 1 a1 2 a0 2 b2 0 b1 1 and b0 0 and whose feedback path sensor is the system of Figure 1233 with a2 1 a1 2 a0 0 b2 1 b1 1 b0 0 and K 1 Figure 1232 Completed root locus s 10 5 5 5 ω σ Figure 1233 A secondorder system with a gain factor K xt yt 1a2 a1 b0 a0 b1 b2 K rob28124ch12592649indd 617 041216 154 pm C h a p t e r 12 Laplace System Analysis 618 The forwardpath transfer function H1s and the feedbackpath transfer function H2s are H 1 s Ks s 2 2s 2 and H 2 s s 2 s s 2 2s s 1 s 2 The loop transfer function is Ts H 1 s H 2 s Ks s 1 s 2 2s 2s 2 The poles of T are at s 1 j and s 2 The zeros are at s 0 s 1 and s Since H1s has poles in the RHP the forwardpath system is unstable 1 The root locus has three branches Rule 1 2 The allowed regions on the real axis are 1 σ 0 and σ 2 Rule 3 3 The root locus begins on the poles of Ts So the branch that begins at s 2 can only go to the left and remain in an allowed region on the real axis It can never leave the real axis because of symmetry requirements Rule 4 Therefore this branch terminates on the zero at infinity 4 The other two branches begin on complex conjugate poles at s 1 j They must terminate on the remaining two zeros at s 0 and s 2 To reach these zeros and at the same time preserve symmetry about the real axis Rule 4 they must migrate to the left and down into the allowed region 1 σ 0 5 The breakin point can be found be setting dds1 Ts 0 The solution gives us a breakin point at s 04652 Figure 1234 In this example the overall feedback system starts out unstable at a low K value but as K is increased the poles that were initially in the RHP migrate into the LHP So if K is large enough the overall feedback system becomes stable even though the forwardpath system is unstable Tracking Errors in UnityGain Feedback Systems A very common type of feedback system is one in which the purpose of the system is to make the output signal track the input signal using unitygain feedback H2s 1 Figure 1235 Figure 1235 A unitygain feedback system H1s Xs Ys Es Figure 1234 Complete root locus s ω σ 3 3 3 This type of system is called unitygain because the output signal is always com pared directly with the input signal and if there is any difference error signal that is amplified by the forwardpath gain of the system in an attempt to bring the output signal closer to the input signal If the forwardpath gain of the system is large that forces the error signal to be small making the output and input signals closer together Whether or not the error signal can be forced to zero depends on the forwardpath transfer function H1s and the type of excitation It is natural to wonder at this point what the purpose is of a system whose out put signal equals its input signal What have we gained If the system is an electronic rob28124ch12592649indd 618 041216 154 pm 124 System Connections 619 amplifier and the signals are voltages we have a voltage gain of one but the input im pedance could be very high and the response voltage could drive a very low impedance so that the actual power in watts delivered by the output signal is much greater than the actual power supplied by the input signal In other systems the input signal could be a voltage set by a lowpower amplifier or a potentiometer and the output signal could be a voltage indicating the position of some large mechanical device like a crane an artillery piece an astronomical telescope and so on Now we will mathematically determine the nature of the steadystate error The term steadystate means the behavior as time approaches infinity The error signal is Es Xs Ys Xs H 1 sEs Solving for Es Es Xs 1 H 1 s We can find the steadystate value of the error signal using the finalvalue theorem lim t et lim s0 s Es lim s0 s Xs 1 H 1 s If the input signal is a step of the form xt Aut then Xs As and lim t et lim s0 A 1 H 1 s and the steadystate error is zero if lim s0 1 1 H 1 s is zero If H1s is in the familiar form of a ratio of polynomials in s H 1 s b N s N b N1 s N1 b 2 s 2 b 1 s b 0 a D s D a D1 s D1 a 2 s 2 a 1 s a 0 128 then lim t et lim s0 1 1 b N s N b N1 s N1 b 2 s 2 b 1 s b 0 a D s D a D1 s D1 a 2 s 2 a 1 s a 0 a 0 a 0 b 0 and if a0 0 and b0 0 the steadystate error is zero If a0 0 then H1s can be expressed in the form H 1 s b N s N b N1 s N1 b 2 s 2 b 1 s b 0 s a D s D1 a D1 s D2 a 2 s a 1 and it is immediately apparent that H1s has a pole at s 0 So we can summarize by saying that if a stable unitygain feedback system has a forwardpath transfer function with a pole at s 0 the steadystate error for a step excitation is zero If there is no pole at s 0 the steadystate error is a0 a0 b0 and the larger b0 is in comparison with a0 the smaller the steadystate error This makes sense from another point of view because if the forwardpath gain is of the form 128 the feedbacksystem lowfrequency gain is b0 a0 b0 which approaches one for b0 a0 indicating that the input and output signals approach the same value rob28124ch12592649indd 619 041216 154 pm C h a p t e r 12 Laplace System Analysis 620 A unitygain feedback system with a forwardpath transfer function H1s that has no poles at s 0 is called a type 0 system If it has one pole at s 0 the system is a type 1 system In general any unitygain feedback system is a type n system where n is the number of poles at s 0 in H1s So summarizing using the new terminology 1 A stable type 0 system has a finite steadystate error for step excitation 2 A stable type n system n 1 has a zero steadystate error for step excitation Figure 1236 illustrates typical steadystate responses to step excitation for stable type 0 and type 1 systems Figure 1236 Type 0 and type 1 system responses to a step Type 0 System yt xt t h1t Type 1 System t yt xt h1t Now we will consider a ramp excitation xt A rampt At ut whose Laplace transform is Xs As2 The steadystate error is lim t et lim s0 A s1 H 1 s Again if H1s is a ratio of polynomials in s lim t et lim s0 1 s 1 1 b N s N b N1 s N1 b 2 s 2 b 1 s b 0 a D s D a D1 s D1 a 2 s 2 a 1 s a 0 or lim t et lim s0 a D s D a D1 s D1 a 2 s 2 a 1 s a 0 s a D s D a D1 s D1 a 2 s 2 a 1 s a 0 b N s N b N1 s N1 b 2 s 2 b 1 s b 0 This limit depends on the values of the as and bs If a0 0 the steadystate error is infinite If a0 0 and b0 0 the limit is a1b0 indicating that the steadystate error is a nonzero constant If a0 0 and a1 0 and b0 0 the steadystate error is zero The condition a0 0 and a1 0 means there is a repeated pole at s 0 in the for wardpath transfer function So for a stable type 2 system the steadystate error under ramp excitation is zero Summarizing 1 A stable type 0 system has an infinite steadystate error for ramp excitation 2 A stable type 1 system has a finite steadystate error for ramp excitation 3 A stable type n system n 2 has a zero steadystate error for ramp excitation Figure 1237 illustrates typical steadystate responses to ramp excitation for stable type 0 type 1 and type 2 systems These results can be extrapolated to higherorder excitations At 2 ut At 3 ut etc When the highest power of s in the denominator of the transform rob28124ch12592649indd 620 041216 154 pm 125 System Analysis Using MATLAB 621 of the excitation is the same as or lower than the type number 0 1 2 etc of the system and the system is stable the steadystate error is zero This result was illustrated with forwardpath transfer functions in the form of a ratio of polynomials but the result can be shown to be true for any form of transfer function based only on the number of poles at s 0 It may seem that more poles in the forwardpath transfer function at s 0 are generally desirable because they reduce the steadystate error in the overall feedback system But generally speaking the more poles in the forwardpath transfer function the harder it is to make a feedback system stable So we may trade one problem for another by putting poles at s 0 in the forwardpath transfer function ExamplE 124 Instability caused by adding a pole at zero in the forward transfer function Let the forward transfer function of a unitygain feedback system be H 1 s 100 ss 4 Then the overall transfer function is Hs 100 s 2 4s 100 with poles at s 2 j9798 Both poles are in the LHP so the system is stable Now add a pole at zero to H1s and reevaluate the stability of the system The new H1s is H 1 s 100 s 2 s 4 and the new overall transfer function is Hs 100 s 3 4 s 2 100 with poles at s 64235 and s 1212 j3755 Two of the poles are in the RHP and the overall system is unstable 125 SYSTEM ANALYSIS USING MATLAB The MATLAB system object was introduced in Chapter 6 The syntax for creating a system object with tf is sys tfnumden Figure 1237 Type 0 1 and 2 system responses to a ramp Type 0 System yt xt t h2t Type 1 System yt xt h2t t Type 2 System yt xt h2t t rob28124ch12592649indd 621 041216 154 pm C h a p t e r 12 Laplace System Analysis 622 The syntax for creating a system object with zpk is sys zpkzpk The real power of the controlsystem toolbox comes in interconnecting systems Suppose we want the overall transfer function Hs H1sH2s of the two systems H 1 s s 2 4 s 5 4 s 4 7 s 3 15 s 2 31s 75 and H 2 s 20 s 4 s 3s 10 in a cascade connection In MATLAB num 1 0 4 den 1 4 7 15 31 75 H1 tfnumden z 4 p 3 10 k 20 H2 zpkzpk Hc H1H2 Hc Zeropolegain 20 s4 s2 4 s3081 s3 s10 s2 2901s 545 s2 1982s 4467 tfHc Transfer function 20 s3 80 s2 80 s 320 s7 17 s6 89 s5 226 s4 436 s3 928 s2 1905 s 2250 If we want to know what the transfer function of these two systems in parallel would be Hp H1 H2 Hp Zeropolegain 20 s4023 s3077 s2 2881s 5486 s2 1982s 4505 s3081 s3 s10 s2 2901s 545 s2 1982s 4467 tfHp rob28124ch12592649indd 622 041216 154 pm 126 System Responses to Standard Signals 623 Transfer function 20 s6 160 s5 461 s4 873 s3 1854 s2 4032 s 6120 s7 17 s6 89 s5 226 s4 436 s3 928 s2 1905 s 2250 There is also a command feedback for forming the overall transfer function of a feed back system Hf feedbackH1H2 Hf Zeropolegain s3 s10 s2 4 s9973 s2 6465s 1069 s2 2587s 5163 s2 2025s 4669 Sometimes when manipulating system objects the result will not be in the ideal form It may have a pole and zero at the same location Although there is nothing mathematically wrong with this it is generally better to cancel that pole and zero to simplify the transfer function This can be done using the command minreal for minimum realization Once we have a system described we can graph its step response with step its im pulse response with impulse and a Bode diagram of its frequency response with bode We can also draw its polezero diagram using the MATLAB command pzmap MATLAB has a function called freqresp that does frequency response graphs The syntax is H freqrespsysw where sys is the MATLABsystem object w is a vector of radian frequencies ω and H is the frequency response of the system at those radian frequencies The MATLAB control toolbox also has a command for plotting the root locus of a system loop transfer function The syntax is rlocussys where sys is a MATLAB system object There are many other useful commands in the control toolbox which can be examined by typing help control 126 SYSTEM RESPONSES TO STANDARD SIGNALS We have seen in previous signal and system analysis that an LTI system is completely characterized by its impulse response In testing real systems the application of an im pulse to find the systems impulse response is not practical First a true impulse cannot be generated and second even if we could generate a true impulse since it has an un bounded amplitude it would inevitably drive a real system into a nonlinear mode of oper ation We could generate an approximation to the true unit impulse in the form of a very short duration and very tall pulse of unit area Its time duration should be so small that making it any smaller would not significantly change any signals in the system Although this type of test is possible a very tall pulse may drive a system into nonlinearity It is much easier to generate a good approximation to a step than to an impulse and the step amplitude can be small enough so as to not cause the system to go nonlinear Sinusoids are also easy to generate and are confined to varying between finite bounds that can be small enough that the sinusoid will not overdrive the system and rob28124ch12592649indd 623 041216 154 pm C h a p t e r 12 Laplace System Analysis 624 force it into nonlinearity The frequency of the sinusoid can be varied to determine the frequency response of the system Since sinusoids are very closely related to complex exponentials this type of testing can directly yield information about the system characteristics UNITSTEP RESPONSE Let the transfer function of an LTI system be of the form Hs N H s D H s where NHs is of a lower degree in s than DHs Then the Laplace transform of the zerostate response Ys to Xs is Ys N H s D H s X s Let xt be a unit step Then the Laplace transform of the zerostate response is Ys H 1 s N H s s D H s Using the partial fraction expansion technique this can be separated into two terms Ys N H1 s D H s H0 s If the system is BIBO stable the roots of DH s are all in the open LHP and the inverse Laplace transform of NH1sDH s is called the natural response or the transient response because it decays to zero as time t approaches infinity The forced response of the system to a unit step is the inverse Laplace transform of H0s which is H0 ut The expression Ys N H1 s D H s H0 s has two terms The first term has poles that are identical to the system poles and the second term has a pole at the same location as the Laplace transform of the unit step This result can be generalized to an arbitrary excitation If the Laplace transform of the excitation is Xs N x s D x s then the Laplace transform of the system response is Ys N H s D H s Xs N H s D H s N x s D x s N H1 s D H s same poles as system N x1 s D x s same poles as excitation Now lets examine the unitstep response of some simple systems The simplest dynamic system is a firstorder system whose transfer function is of the form Hs A 1 sp Ap s p rob28124ch12592649indd 624 041216 154 pm 126 System Responses to Standard Signals 625 where A is the lowfrequency transfer function of the system and p is the pole location in the splane The Laplace transform of the step response is Ys H 1 s A 1 sps Ap 1 sp A s A s A s p Inverse Laplace transforming yt A1 eptut If p is positive the system is unstable and the magnitude of the response to a unit step increases exponentially with time Figure 1238 Unstable Systems Stable Systems σ ω 1 4 3 2 1 2 3 4 s σ ω 1 4 3 2 1 2 3 4 s t p 1 p 2 p 4 yt h1t t A 0632A p 1 p 2 p 4 4 1 2 1 1 yt h1t Figure 1238 Responses of a firstorder system to a unit step and the corresponding polezero diagrams The speed of the exponential increase depends on the magnitude of p being greater for a larger magnitude of p If p is negative the system is stable and the response approaches a constant A with time The speed of the approach to A depends on the magni tude of p being greater for a larger magnitude of p The negative reciprocal of p is called the time constant τ of the system τ 1p and for a stable system the response to a unit step moves 632 of the distance to the final value in a time equal to one time constant Now consider a secondorder system whose transfer function is of the form Ηs A ω n 2 s 2 2ζ ω n s ω n 2 ω n 0 This form of a secondorder system transfer function has three parameters the lowfrequency gain A the damping ratio ζ and the natural radian frequency ωn The form of the unitstep response depends on these parameter values The Laplace transform of the system unitstep response is H 1 s A ω n 2 s s 2 2ζ ω n s ω n 2 A ω n 2 ss ω n ζ ζ 2 1 s ω n ζ ζ 2 1 This can be expanded in partial fractions if ζ 1 as H 1 s A 1 s 1 2 ζ 2 1 ζ ζ 2 1 s ω n ζ ζ 2 1 1 2 ζ 2 1 ζ ζ 2 1 s ω n ζ ζ 2 1 rob28124ch12592649indd 625 041216 154 pm C h a p t e r 12 Laplace System Analysis 626 and the timedomain response is then h 1 t A e ω n ζ ζ 2 1 t 2 ζ 2 1 ζ ζ 2 1 e ω n ζ ζ 2 1 t 2 ζ 2 1 ζ ζ 2 1 1 ut For the special case of ζ 1 the system unitstep response is H 1 s A ω n 2 s ω n 2 s the two poles are identical the partial fraction expansion is H 1 s A 1 s ω n s ω n 2 1 s ω n and the timedomain response is h 1 t A1 1 ω n t e ω n t ut Aut 1 1 ω n t e ω n t ζ 1 1 1 ω n t e ω n t ζ 1 It is difficult just by examining the mathematical functional form of the unitstep response to immediately determine what it will look like for an arbitrary choice of parameters To explore the effect of the parameters lets first set A and ωn constant and examine the effect of the damping ratio ζ Let A 1 and let ωn 1 Then the unitstep response and the corresponding polezero diagrams are as illustrated in Figure 1239 for six choices of ζ Figure 1239 Secondorder system responses to a unit step and the corresponding polezero diagrams 1 t t ζ 5 ζ 5 ζ 5 ζ 5 ζ 1 ζ 5 ζ 5 ζ 5 ζ 5 ζ 1 ζ 1 ζ 02 ζ 02 ζ 02 ζ 02 ζ 02 ζ 02 ζ 02 ζ 02 σ ω 1 01 99 s σ ω 1 01 99 s yt h1t yt h1t rob28124ch12592649indd 626 041216 154 pm 126 System Responses to Standard Signals 627 We can see why these different types of behavior occur if we examine the unitstep response h 1 t A e ω n ζ ζ 2 1 t 2 ζ 2 1 ζ ζ 2 1 e ω n ζ ζ 2 1 t 2 ζ 2 1 ζ ζ 2 1 1 ut 129 and in particular the exponents of e ω n ζ ζ 2 1 t The signs of the real parts of these exponents determine whether the response grows or decays with time t 0 For times t 0 the response is zero because of the unit step ut Case 1 ζ 0 If ζ 0 then the exponent of e in both terms in 129 has a positive real part for positive time and the step response therefore grows with time and the system is unstable The exact form of the unitstep response depends on the value of ζ It is a simple increasing exponential for ζ 1 and an exponentially growing oscillation for 1 ζ 0 But either way the system is unstable Case 2 ζ 0 If ζ 0 then the exponent of e in both terms in 129 has a negative real part for positive time and the step response therefore decays with time and the system is stable Case 2a ζ 1 If ζ 1 then ζ2 1 0 and the coefficients of t in 129 ω n ζ ζ 2 1 t are both negative real numbers and the unitstep response is in the form of a constant plus the sum of two decaying exponentials This case ζ 1 is called the overdamped case Case 2b 0 ζ 1 If 0 ζ 1 then ζ2 1 0 and the coefficients of t in 129 ω n ζ ζ 2 1 t are both complex numbers in a complexconjugate pair with negative real parts and the unitstep response is in the form of a constant plus the sum of two sinusoids multiplied by a decaying exponential Even though the response overshoots its final value it still settles to a constant value and is therefore the response of a stable system This case 0 ζ 1 is called the underdamped case The dividing line between the overdamped and underdamped cases is the case ζ 1 This condition is called critical damping Now lets examine the effect of changing ωn while holding the other parameters constant Let A 1 and ζ 05 The step response is illustrated in Figure 1240 for 3 values of ωn Since ωn is the natural radian frequency it is logical that it would affect the ringing rate of the step response The response of any LTI system to a unit step can be found using the MATLAB control toolbox command step SINUSOID RESPONSE Now lets examine the response of a system to a causal sinusoid one applied to the system at time t 0 Again let the system transfer function be of the form Hs N H s D H s rob28124ch12592649indd 627 041216 154 pm C h a p t e r 12 Laplace System Analysis 628 Then the Laplace transform of the zerostate response to xt cosω0t ut would be Ys N H s D H s s s 2 ω 0 2 This can be separated into partial fractions in the form Ys N H1 s D H s 1 2 H j ω 0 s j ω 0 1 2 H j ω 0 s j ω 0 N H1 s D H s 1 2 H j ω 0 s j ω 0 1 2 H j ω 0 s j ω 0 or Ys N H1 s D H s 1 2 H j ω 0 s j ω 0 H j ω 0 s j ω 0 s 2 ω 0 2 Ys N H1 s D H s 1 2 s s 2 ω 0 2 H j ω 0 H j ω 0 j ω 0 s 2 ω 0 2 H j ω 0 H j ω 0 Ys N H1 s D H s ReH j ω 0 s s 2 ω 0 2 ImH j ω 0 ω 0 s 2 ω 0 2 The inverse Laplace transform of the term ReHjω0ss2 ω02 is the product of a unit step and a cosine at ω0 with an amplitude of ReHjω0 and the inverse Laplace transform of the term ImHjω0ω0s2 ω02 is the product of a unit step and a sine at ω0 with an amplitude of ImHjω0 That is yt 1 N H1 s D H s ReH j ω 0 cos ω 0 t ImH j ω 0 sin ω 0 tut or using ReAcosω0t ImAsinω0t Acosω0t A yt 1 N H1 s D H s H j ω 0 cos ω 0 t H j ω 0 ut If the system is stable the roots of DHs are all in the open LHP and the inverse Laplace transform of NH1sDHs the transient response decays to zero as time t approaches infinity Therefore the forced response that persists after the transient Figure 1240 Secondorder system response for three different values of ωn and the corresponding polezero plots 1 t ωn 1 ωn 1 ωn 05 ωn 05 ωn 02 ωn 02 σ ω s h1t rob28124ch12592649indd 628 041216 154 pm 126 System Responses to Standard Signals 629 response has died away is a causal sinusoid of the same frequency as the excitation and with an amplitude and phase determined by the transfer function evaluated at s jω0 The forced response is exactly the same as the response obtained by using Fourier methods because the Fourier methods assume that the excitation is a true sinusoid applied at time t not a causal sinusoid and therefore there is no transient response in the solution ExamplE 125 Zerostate response of a system to a causal cosine Find the total zerostate response of a system characterized by the transfer function Hs 10 s 10 to a unitamplitude causal cosine at a frequency of 2 Hz The radian frequency ω0 of the cosine is 4π Therefore the Laplace transform of the response is Ys 10 s 10 s s 2 4π 2 Ys 0388 s 10 ReH j4π s s 2 4π 2 ImH j4π ω 0 s 2 4π 2 and the timedomain response is yt 1 0388 s 10 Hj4π cos4πt H j4π ut or yt 0388 e 10t 10 j4π 10 cos4πt j4π 10 ut or yt 0388 e 10t 0623 cos4πt 0899 ut The excitation and response are illustrated in Figure 1241 Looking at the graph we see that the response appears to reach a stable amplitude in less than one second This is reasonable Figure 1241 Excitation and response of a firstorder system excited by a cosine applied at time t 0 t Excitation Response 1 2 1 1 rob28124ch12592649indd 629 041216 154 pm C h a p t e r 12 Laplace System Analysis 630 given that the transient response has a time constant of 110 of a second After the response stabilizes its amplitude is about 62 of the excitation amplitude and its phase is shifted so that it lags the excitation by about a 0899 radian phase shift which is equivalent to about a 72ms time delay If we solve for the response of the system using Fourier methods we write the frequency response from the transfer function as H jω 10 jω 10 If we make the excitation of the system a true cosine it is xt cos4πt and its continuoustime Fourier transform CTFT is Xjω πδω 4π δω 4π Then the system response is Y jω πδω 4π δω 4π 10 jω 10 10π δω 4π j4π 10 δω 4π j4π 10 or Y jω 10π 10δω 4π δω 4π j4πδω 4π δω 4π 16 π 2 100 Inverse Fourier transforming yt 0388 cos4πt 0487 sin4πt or using ReA cos ω 0 t ImA sin ω 0 t A cos ω 0 t A yt 0623cos4πt 0899 This is exactly the same except for the unit step as the forced response of the previous solution which was found using Laplace transforms 127 STANDARD REALIZATIONS OF SYSTEMS The process of system design as opposed to system analysis is to develop a desired transfer function for a class of excitations that yields a desired response or responses Once we have found the desired transfer function the next logical step is to actually build or perhaps simulate the system The usual first step in building or simulating a system is to form a block diagram that describes the interactions among all the signals in the system This step is called realization arising from the concept of making a real system instead of just a set of equations that describe its behavior There are several standard types of system realization We have already seen Direct Form II in Chapter 8 We will explore two more here CASCADE REALIZATION The second standard system realization is the cascade form The numerator and denominator of the general transfer function form Hs Ys Xs k0 M b k s k k0 N a k s k b M s M b M1 s M1 b 1 s b 0 s N a N1 s N1 a 1 s a 0 a N 1 1210 rob28124ch12592649indd 630 041216 154 pm 127 Standard Realizations of Systems 631 where M N can be factored yielding a transfer function expression of the form Hs A s z 1 s p 1 s z 2 s p 2 s z M s p M 1 s p M1 1 s p M2 1 s p N Any of the component fractions Y k s X k s s z k s p k or Y k s X k s 1 s p k represents a subsystem that can be realized by writing the relationship as H k s 1 s p k H k1 s s z k H k2 s or H k s 1 s p k and realizing it as a Direct Form II system Figure 1242 Then the entire original system can be realized in cascade form Figure 1243 H k s s z k s p k H k s 1 s p k Figure 1242 Direct Form II realization of a single subsystem in the cascade realization zk pk pk Xks Xks Yks Yks s1 s1 Figure 1243 Overall cascade system realization z1 z2 p1 p2 pN1 pN Ys Xs s1 s1 s1 s1 A problem sometimes arises with this type of cascade realization Sometimes the firstorder subsystems have complex poles This necessitates multiplication by complex numbers and that usually cannot be done in a system realization In such cases two subsystems with complex conjugate poles should be combined into one secondorder subsystem of the form H k s s b 0 s 2 a 1 s a 0 which can always be realized with real coefficients Figure 1244 rob28124ch12592649indd 631 041216 154 pm C h a p t e r 12 Laplace System Analysis 632 PARALLEL REALIZATION The last standard realization of a system is the parallel realization This can be accom plished by expanding the standard transfer function form 1210 in partial fractions of the form Hs K 1 s p 1 K 2 s p 2 K N s p N Figure 1245 128 SUMMARY OF IMPORTANT POINTS 1 Continuoustime systems can be described by differential equations block diagrams or circuit diagrams in the time or frequency domain 2 A continuoustime LTI system is stable if all the poles of its transfer function lie in the open left halfplane 3 Marginally stable systems form a subset of unstable systems 4 The three most important types of system interconnections are the cascade connection the parallel connection and the feedback connection 5 The unit step and the sinusoid are important practical signals for testing system characteristics 6 The Direct Form II cascade and parallel realizations are important standard ways of realizing systems Figure 1245 Overall parallel system realization Xs Ys p1 p2 K1 K1 pN KN s1 s1 s1 Figure 1244 A standardform secondorder subsystem b0 a1 a0 Xs Ys s1 s1 rob28124ch12592649indd 632 041216 154 pm 633 Exercises with Answers EXERCISES WITH ANSWERS Answers to each exercise are in random order Transfer Functions 1 For each circuit in Figure E1 write the transfer function between the indicated excitation and indicated response Express each transfer function in the standard form H s A s M b N1 s M1 b 2 s 2 b 1 s b 0 s N a D1 s N1 a 2 s 2 a 1 s a 0 a Excitation v s t Response v o t R1 L C R2 vst vot b Excitation i s t Response v o t R1 R2 vot ist C1 C2 c Excitation i s t Response i 1 t R1 C2 C1 R2 vst i1t Figure E1 rob28124ch12592649indd 633 041216 154 pm C h a p t e r 12 Laplace System Analysis 634 Answers 1 R 1 s2 s R2C2 s2 s 1 R2C2 1 R2C1 1 R1C1 1 R1R2C1C2 1 R 1 C 1 C 2 1 s2 s 1 R2C2 1 R1C1 1 R1R2C1C2 R 2 R 1 LC 1 s2 s 1 R1C R2 L R2 R1 R1LC 2 For each block diagram in Figure E2 write the transfer function between the excitation xt and the response yt a xt yt 8 2 b xt yt 4 10 1 Figure E2 Answers s 1 s 3 4 s 2 10s 1 s 3 8 s 2 2s Stability 3 Evaluate the stability of each of these system transfer functions a H s 100 s 200 b H s 80 s 4 rob28124ch12592649indd 634 041216 154 pm 635 c H s 6 s s 1 d H s 15s s 2 4s 4 e H s 3 s 10 s 2 4s 29 f H s 3 s 2 4 s 2 4s 29 g H s 1 s 2 64 h H s 10 s 3 4 s 2 29s i H s 1 s j H s s k H s s 2 8 s 2 6s 9 l H s s 3 s 2 9 m H s s 3 s 2 9 n H s s 2 8 s 2 6s 9 o H s 1 s s 3 p H s 1 s 2 Answers 4 Stable 6 Marginally Stable therefore unstable 6 Unstable and not marginally stable 4 In the feedback system in Figure E5 H 1 s s 3 s 2 and H 2 s K What range of real values of K makes this system stable H2s H1s Ys Xs Es Figure E5 Answer 1 K 2 3 5 If H 1 s K s 2 9 and H 2 s s 2 4 for what range of K is this system unstable For what range of K is this system marginally stable Answers All K K 9 4 Parallel Cascade and Feedback Connections 6 Find the overall transfer functions of the systems in Figure E6 in the form of a single ratio of polynomials in s a 10 s2 3s 2 s2 s2 3s 2 Xs Ys Exercises with Answers rob28124ch12592649indd 635 041216 154 pm C h a p t e r 12 Laplace System Analysis 636 b s 1 s2 2s 13 1 s 10 Xs Ys c s s2 s 5 Xs Ys d 20s s2 200s 290000 1 s 400 Xs Ys Figure E6 Answers 10 s 2 s 4 6 s 3 13 s 2 12s 4 2 s 2 13 2 s 23 2 s 3 12 s 2 33s 130 20 s 2 400s s 3 600 s 2 370020s 116 10 8 s s2 2s 5 7 In the feedback system in Figure E7 find the overall system transfer function for the given values of forwardpath gain K a K 10 6 b K 10 5 c K 10 d K 1 e K 1 f K 10 K 01 Xs Ys Figure E7 Answers 10 111 0909 10 5 8 Which of the systems with these descriptions are stable a H s 13 s s 4 b Unitygain feedback system with H 1 s 5 s 2 3 c System with impulse response h t 7 cos 22πt u t d System with zeros at s 2 and s 05 and poles at s 08 and s 18 Answers 1 Stable 3 Unstable rob28124ch12592649indd 636 041216 154 pm 637 9 For what range of values of K is the system in Figure E9 stable Graph the step responses for K 0 K 4 and K 8 Ks Xs Ys 1 s2 4s 4 Figure E9 Answers t 1 4 h1t h1t h1t 3000 K 0 t 1 4 05 05 K 4 t 1 4 025 K 8 The system is unstable if K 4 and stable if K 4 10 Graph the impulse response and the polezero diagram for the forward path and the overall system in Figure E10 xt yt 100 s22s26 10 s20 Figure E10 Answers t 05 4 h1t 20 20 Forward Path t 05 8 ht 30 30 Overall System s σ ω 1 5 s σ ω 5 00612 829 829 2212 20 H1s Hs Root Locus 11 The loop transfer function of a feedback system has 4 finite poles and 1 finite zero What is the angle in radians between the asymptotes of the system root locus Answer 2π 3 radians Exercises with Answers rob28124ch12592649indd 637 041216 154 pm C h a p t e r 12 Laplace System Analysis 638 12 For each H 1 s in Figure E12 is there a finite positive value of K for which this system is unstable H1s K Ys Xs Figure E12 a H 1 s 1000 s 4 s 10 b H 1 s 25 s2 2s 1 c H 1 s 10 s 3 s 8 s 22 Answers Yes No Yes 13 Draw the root locus for each of the systems that have these loop transfer functions and identify the transfer functions that are stable for all finite positive real values of K a T s K s 3 s 8 b T s Ks s 3 s 8 c T s K s 2 s 3 s 8 d T s K s 1 s 2 4s 8 Answers σ ω 1 Unstable σ 8 3 ω Stable σ 8 3 ω Stable σ 3 ω Stable 14 Sketch a root locus for each of the diagrams in Figure E14 The diagrams show the poles and zeros of the loop transfer function T s of a feedback system rob28124ch12592649indd 638 041216 154 pm 639 8 6 4 2 0 2 4 5 0 5 s s s s s s s s s A 8 6 4 2 0 2 4 5 0 5 B 8 6 4 2 0 2 4 5 0 5 D 8 6 4 2 0 2 4 5 0 5 F 8 6 4 2 0 2 4 5 0 5 C 8 6 4 2 0 2 4 5 0 5 E 8 6 4 2 0 2 4 5 0 5 I 8 6 4 2 0 2 4 5 0 5 G 8 6 4 2 0 2 4 5 0 5 H Figure E14 For each system indicate whether it will be unstable at some finite positive value of the gain constant K Answers 5 Stable and 4 Unstable Tracking Errors in UnityGain Feedback Systems 15 In Figure E15 is a block diagram of a continuoustime feedback system Answer the questions H1s H2s Ys Xs Es Figure E15 a If H 1 s K s 4 and H 2 s 3 s 10 is this system stable for all positive values of K Exercises with Answers rob28124ch12592649indd 639 041216 154 pm C h a p t e r 12 Laplace System Analysis 640 b If H 1 s 4 s 5 and H 2 s 1 what is the steadystate error when the excitation signal is a unit step Steadystate error is defined as lim t e t where e t ℒ E s c If H 1 s 4 s s 5 and H 2 s 1 what is the steadystate error when the excitation signal is a unit step d If H 1 s K s 3 and H 2 s 1 sketch a root locus of the system Answers 0 59 σ ω s 3 Yes 16 A continuoustime unitygain tracking feedback system has a forwardpath transfer function with exactly one pole at the origin of the splane a Describe in as much detail as possible the steadystate error signal in response to a step excitation b Describe in as much detail as possible the steadystate error signal in response to a ramp excitation Answer Zero Nonzero Finite 17 The transfer function of a continuoustime system has one finite pole in the open left halfplane at s s 0 and no finite zeros If you wanted to make its response to a step excitation approach its final value faster how would you change s 0 Answer Make s 0 more negative System Responses to Standard Signals 18 Using the Laplace transform find and graph the timedomain response y t of the systems with these transfer functions to the sinusoidal excitation x t cos 10πt u t a H s 1 s 1 b H s s 2 s 2 2 16 Answers t 1 yt 0033333 0033333 rob28124ch12592649indd 640 041216 154 pm 641 t 1 5 yt 10 5 19 Find the responses of the systems with these transfer functions to a unit step and a unit amplitude 1 Hz cosine applied at time t 0 Also find the responses to a true unit amplitude 1 Hz cosine using the CTFT and compare to the steadystate part of the total solution found using the Laplace transform Use the results of the chapter in which if a cosine is applied at time t 0 to a system whose transfer function is H s N s D s the response is y t ℒ 1 N 1 s D s H j ω 0 cos ω 0 t H j ω 0 u t where N 1 s D s is the partial fraction involving the transfer function poles a H s 1 s b H s s s 1 c H s s s 2 2s 40 d H s s 2 2s 40 s 2 Answers 8 π 2 cos 2πt 3277 sin 2πt 15819 y t sin 2πt 2π 03184 sin 2πt 00132 cos 2πt 2π 2 cos 2πt 2π sin 2πt 1 2π 2 System Realization 20 Draw cascade system diagrams of the systems with these transfer functions a H s s s 1 b H s s 4 s 2 s 12 c H s 20 s s 2 5s 10 Answers Ys Xs 2 12 4 s1 s1 Ys Xs s1 Ys s1 s1 s1 Xs 20 10 5 Exercises with Answers rob28124ch12592649indd 641 041216 154 pm C h a p t e r 12 Laplace System Analysis 642 EXERCISES WITHOUT ANSWERS Stability 21 A continuoustime system with a bounded impulse response is excited by x t cos 200πt u t Find and report the numerical locations for a pair of poles in the system transfer function that would cause the response to x t to grow in positive time t 0 without bound Transfer Functions 22 In the feedback system in Figure E22 graph the response of the system to a unit step for the time period 0 t 10 then write the expression for the overall system transfer function and draw a polezero diagram for the given values of K a K 20 b K 10 c K 1 d K 1 e K 10 f K 20 K 01 es Xs Ys OneSecond Time Delay Figure E22 23 Find the sdomain transfer functions for the circuits in Figure E23 and then draw block diagrams for them as systems with Vis as the excitation and Vos as the response R 10 kΩ L 5 mH R 10 kΩ R 10 kΩ C 1 μF C 1 μF C 1 μF vit vit vot vot R 10 kΩ L 5 mH C 1 μF vit vot a b d R 10 kΩ L 5 mH C 1 μF vit vot c Figure E23 rob28124ch12592649indd 642 041216 154 pm 643 Exercises without Answers Stability 24 Determine whether the systems with these transfer functions are stable marginally stable or unstable and not marginally stable a H s s s 2 s 2 8 b H s s s 2 s 2 8 c H s s 2 s 2 4s 8 d H s s 2 s 2 4s 8 e H s s s 3 4 s 2 8s Parallel Cascade and Feedback Connections 25 Find the expression for the overall system transfer function of the system in Figure E25 For what positive values of K is the system stable xt yt K s1s2 Figure E25 26 A laser operates on the fundamental principle that a pumped medium amplifies a travelling light beam as it propagates through the medium Without mirrors a laser becomes a singlepass travelling wave amplifier Figure E261 Medium Light In Light Out Pumped Laser Figure E261 This is a system without feedback If we now place mirrors at each end of the pumped medium we introduce feedback into the system Medium Mirror Mirror Pumped Laser Figure E262 When the gain of the medium becomes large enough the system oscillates creating a coherent output light beam That is laser operation If the gain of the medium is less that that required to sustain oscillation the system is known as a regenerative travellingwave amplifier RTWA rob28124ch12592649indd 643 041216 154 pm C h a p t e r 12 Laplace System Analysis 644 Let the electric field of a light beam incident on the RTWA from the left be the excitation of the system Einc s and let the electric fields of the reflected light Erefl s and the transmitted light Etrans s be the responses of the system Figure E263 jti grp gfp jti jto ri ri ro Etranss Eincs Ecircs Erefls Figure E263 Electric field reflectivity of the input mirror r i 099 Electric field transmissivity of the input mirror t i 1 r i 2 Electric field reflectivity of the output mirror r o 098 Electric field transmissivity of the output mirror t o 1 r o 2 Forward and reverse path electric field gains g fp s g rp s 101 e 10 9 s Find an expression for the frequency response E trans f E inc f and plot its magnitude over the frequency range 3 10 14 5 10 8 Hz 27 A classical example of the use of feedback is the phaselocked loop used to demodulate frequencymodulated signals Figure E27 Phase Detector Loop Filter HLFs Voltage Controlled Oscillator xt xLFt yt yVCOt Figure E27 The excitation x t is a frequencymodulated sinusoid The phase detector detects the phase difference between the excitation and the signal produced by the voltagecontrolled oscillator The response of the phase detector is a voltage proportional to phase difference The loop filter filters that voltage Then the loop filter response controls the frequency of the voltagecontrolled oscillator When the excitation is at a constant frequency and the loop is locked the phase difference between the two phasedetector excitation signals is zero In an actual phase detector the phase difference is near 90 at lock But that is not significant in this analysis since that only causes a near 90 phase shift and has no impact on system performance or stability As the frequency of the excitation x t varies rob28124ch12592649indd 644 041216 154 pm 645 the loop detects the accompanying phase variation and tracks it The overall response signal y t is a signal proportional to the frequency of the excitation The actual excitation in a system sense of this system is not x t but rather the phase of x t ϕ x t because the phase detector detects differences in phase not voltage Let the frequency of x t be f x t The relation between phase and frequency can be seen by examining a sinusoid Let x t A cos 2π f 0 t The phase of this cosine is 2π f 0 t and for a simple sinusoid f 0 constant it increases linearly with time The frequency is f0 the derivative of the phase Therefore the relation between phase and frequency for a frequencymodulated signal is f x t 1 2π d dt ϕ x t Let the frequency of the excitation be 100 MHz Let the transfer function of the voltagecontrolled oscillator be 10 8 Hz V Let the transfer function of the loop filter be H LF s 1 s 12 10 5 Let the transfer function of the phase detector be 1 V radian If the frequency of the excitation signal suddenly changes to 100001 MHz graph the change in the output signal Δy t 28 The circuit in Figure E28 is a simple approximate model of an operational amplifier with the inverting input grounded Ri Ro Rx Cx Output A0vit vxt vit vxt Figure E28 Ri 1 MΩ Rx 1 kΩ Cx 8 μF Ro 10 Ω Ao 106 a Define the excitation of the circuit as the current of a current source applied to the noninverting input and define the response as the voltage developed between the noninverting input and ground Find the transfer function and graph its frequency response This transfer function is the input impedance b Define the excitation of the circuit as the current of a current source applied to the output and define the response as the voltage developed between the output and ground with the noninverting input grounded Find the transfer function and graph its frequency response This transfer function is the output impedance c Define the excitation of the circuit as the voltage of a voltage source applied to the noninverting input and define the response as the voltage developed between the output and ground Find the transfer function and graph its frequency response This transfer function is the voltage gain Exercises without Answers rob28124ch12592649indd 645 041216 154 pm C h a p t e r 12 Laplace System Analysis 646 Root Locus 30 A feedback system has a forwardpath transfer function H 1 s K s 3 s 6 and a feedbackpath transfer function H 2 s s 10 s 2 2s 4 Is there a finite positive value of K that makes this system unstable Explain how you know 31 Each polezero diagram in Figure E31 represents the poles and zeros of the loop transfer function T s of a feedback system Sketch a root locus on each diagram Two of these systems will become unstable at a finite nonzero value of the gain factor K in the loop transfer function Which ones Three of these systems are marginally stable for an infinite K Which ones Ri Rf Rs Ro Rx Cx Output A0vit vxt vxt vit Figure E29 4 2 0 5 0 5 G 4 2 0 5 0 5 D 4 2 0 5 0 5 F 4 2 0 5 0 5 H 4 2 0 5 0 5 C 4 2 0 5 0 5 A 4 2 0 5 0 5 B 4 2 0 5 0 5 E 4 2 0 5 0 5 s s s s s s s s s I 29 Change the circuit of Exercise 28 to the circuit in Figure E29 This is a feedback circuit which establishes a positive closedloop voltage gain of the overall amplifier Repeat steps a b and c of Exercise 28 for the feedback circuit and compare the results What are the important effects of feedback for this circuit Ri 1 MΩ Rx 1 kΩ Cx 8 μF Ro 10 Ω A0 106 Rf 10 kΩ Rs 5 kΩ rob28124ch12592649indd 646 041216 154 pm 647 Figure E31 4 2 0 5 0 5 G 4 2 0 5 0 5 D 4 2 0 5 0 5 F 4 2 0 5 0 5 H 4 2 0 5 0 5 C 4 2 0 5 0 5 A 4 2 0 5 0 5 B 4 2 0 5 0 5 E 4 2 0 5 0 5 s s s s s s s s s I Tracking Errors in UnityGain Feedback Systems 32 A unitygain feedback system with overall transfer function Hs has a forward path transfer function H1s For each of the following forwardpath transfer functions assuming the overall system is stable determine whether the responses to a unitstep and unitramp excitation of the overall unitygain feedback system have zero finite or infinite steadystate error a H 1 s 10s s 10 b H 1 s 7s s 4 s 12 c H 1 s s 5 s 8 s 2 s 1 s 25 d H 1 s 1 s s 11 s 32 Response to Standard Signals 33 A system with one finite pole in the left half of the splane and no finite zeros has a step response which approaches its final value on a 20ms time constant The pole is then shifted to a more negative real value on the σ axis What happens to the time constant 34 Given an LTI system transfer function Hs find the timedomain response yt to the excitation x t a x t sin 2πt u t H s 1 s 1 b x t u t H s 3 s 2 Exercises without Answers rob28124ch12592649indd 647 041216 154 pm C h a p t e r 12 Laplace System Analysis 648 c x t u t H s 3s s 2 d x t u t H s 5s s 2 2s 2 e x t sin 2πt u t H s 5s s 2 2s 2 35 Two systems A and B in Figure E35 have the two polezero diagrams below Which of them responds more quickly to a unitstep excitation approaches the final value at a faster rate Explain your answer A B σ ω 1 4321 2 3 4 s σ ω 1 4321 2 3 4 s Figure E35 36 Two systems A and B in Figure E36 have the two polezero diagrams below Which of them has a unitstep response that overshoots the final value before settling to the final value Explain your answer A B σ ω 1 s σ ω 1 s Figure E36 37 A secondorder system is excited by a unit step and the response is as illustrated in Figure E37 Write an expression for the transfer function of the system Time sec Amplitude Step Response 0 10 20 30 40 50 60 0 002 004 006 008 01 012 014 016 018 02 Figure E37 rob28124ch12592649indd 648 041216 154 pm 649 38 The excitation signal x t 20 cos 40πt u t is applied to a system whose transfer function is H s 5 s 150 The response contains a transient term and a steadystate term After the transient term has died away what is the amplitude A response of the response and what is the phase difference between the excitation and response signals Δθ θ excitation θ response System Realization 39 Draw cascade system diagrams of the systems with these transfer functions a H s 50 s 2 s 3 8 s 2 13s 40 b H s s 3 s 3 18 s 2 92s 120 40 Draw parallel system diagrams of the systems with these transfer functions a H s 10 s 3 s 3 4 s 2 9s 3 b H s 5 6 s 3 77 s 2 228s 189 Exercises without Answers rob28124ch12592649indd 649 041216 154 pm 650 13 C H A P T E R 131 INTRODUCTION AND GOALS This chapter follows a path similar to that of Chapter 12 on system analysis using the Laplace transform except as applied to discretetime signals and systems instead of continuoustime signals and systems CHAPTER GOA L S 1 To appreciate the relationship between the z and Laplace transforms 2 To apply the z transform to the generalized analysis of LTI systems including feedback systems for stability and timedomain response to standard signals 3 To develop techniques for realizing discretetime systems in different forms 132 SYSTEM MODELS DIFFERENCE EQUATIONS The real power of the Laplace transform is in the analysis of the dynamic behavior of continuoustime systems In an analogous manner the real power of the z transform is in the analysis of the dynamic behavior of discretetime systems Most continu oustime systems analyzed by engineers are described by differential equations and most discretetime systems are described by difference equations The general form of a difference equation describing a discretetime system with an excitation x n and a response y n is k0 N a k yn k k0 M b k xn k If both x n and y n are causal and we z transform both sides we get k0 N a k z k Yz k0 M b k z k Xz zTransform System Analysis rob28124ch13650679indd 650 041216 201 pm 133 System Stability 651 The transfer function Hz is the ratio of Yz to Xz Hz Yz Xz k0 M b k z k k0 N a k z k b 0 b 1 z 1 b M z M a 0 a 1 z 1 a N z N or Hz z NM b 0 z M b 1 z M1 b M1 z b M a 0 z N a 1 z N1 a N1 z a N So the transfer function of a discretetime system described by a difference equation is a ratio of polynomials in z just as the transfer function of a continuoustime system described by a differential equation is a ratio of polynomials in s BLOCK DIAGRAMS Discretetime systems are conveniently modeled by block diagrams just as continuoustime systems are and transfer functions can be written directly from block diagrams Consider the system in Figure 131 The describing difference equation is yn 2xn xn 1 12yn 1 We can redraw the block diagram to make it a zdomain block diagram instead of a timedomain block diagram Figure 132 In the z domain the describing equation is Yz 2 Xz z 1 Xz 12 z 1 Yz and the transfer function is Hz Yz Xz 2 z 1 1 12 z 1 2z 1 z 12 Figure 131 Timedomain block diagram of a system D D xn yn 2 1 2 Figure 132 zdomain block diagram of a system Xz Yz 2 1 2 z1 z1 133 SYSTEM STABILITY A causal discretetime system is BIBO stable if its impulse response is absolutely sum mable that is if the sum of the magnitudes of the impulses in the impulse response is finite For a system whose transfer function is a ratio of polynomials in z of the form Hz b 0 z M b 1 z M1 b M a 0 z N a 1 z N1 a N with M N and all distinct poles the transfer function can be written in the partial fraction form Hz K 1 z p 1 K 2 z p 2 K N z p N rob28124ch13650679indd 651 041216 201 pm C h a p t e r 13 zTransform System Analysis 652 and the impulse response is then of the form hn K 1 p 1 n1 K 2 p 2 n1 K N p N n1 un 1 some of the ps may be complex For the system to be stable each term must be abso lutely summable The summation of the absolute value of a typical term is n K p n1 un 1 K n1 p n1 K n0 p n e jp n K n0 p n e jnp 1 n K p n1 un 1 K n0 p n Convergence of this last summation requires that p 1 Therefore for stability all the poles must satisfy the condition p k 1 In a discretetime system all the poles of the transfer function must lie in the open interior of the unit circle in the z plane for system stability This is directly analogous to the requirement in continuoustime systems that all the poles lie in the open left half of the splane for system stability This analysis was done for the most common case in which all the poles are distinct If there are repeated poles it can be shown that the requirement that all the poles lie in the open interior of the unit circle for system stability is unchanged 134 SYSTEM CONNECTIONS The transfer functions of components in the cascade parallel and feedback connec tions of discretetime systems combine in the same way they do in continuoustime systems Figure 133 through Figure 135 We find the overall transfer function of a feedback system by the same technique used for continuoustime systems and the result is Hz Yz Xz H 1 z 1 H 1 zH 2 z H 1 z 1 Tz 131 where Tz H 1 zH 2 z is the loop transfer function Figure 133 Cascade connection of systems H1z XzH1z Xz Yz XzH1zH2z H2z Yz Xz H1zH2z Figure 134 Parallel connection of systems H1z H2z Xz Yz Xz Yz XzH1z XzH2z XzH1z H2z XzH2z XzH1z H1z H2z Figure 135 Feedback connection of systems H1z H2z Xz Yz Ez rob28124ch13650679indd 652 041216 201 pm 134 System Connections 653 Just as was true for continuoustime feedback systems a root locus can be drawn for a discretetime feedback system for which H 1 z K P 1 z Q 1 z and H 2 z P 2 z Q 2 z The procedure for drawing the root locus is exactly the same as for continuoustime systems except that the loop transfer function Tz H 1 zH 2 z is a function of z instead of s However the interpretation of the root locus after it is drawn is a little different For continuoustime systems the forwardpath gain K at which the root locus crosses into the right halfplane is the value at which the system becomes unstable For discretetime systems the statement is the same except that right halfplane is replaced with exterior of the unit circle ExamplE 131 Discretetime system stability analysis using root locus Draw a root locus for the discretetime system whose forwardpath transfer function is H 1 z K z 1 z 12 and whose feedbackpath transfer function is H 2 z z 23 z 13 The loop transfer function is Tz K z 1 z 12 z 23 z 13 There are two zeros at z 23 and z 1 and two poles at z 12 and z 13 It is apparent from the root locus Figure 136 that this system is unconditionally stable for any finite positive K Figure 136 Root locus of Tz K z 1 z 12 z 23 z 13 1 1 1 1 Imz Rez rob28124ch13650679indd 653 041216 201 pm C h a p t e r 13 zTransform System Analysis 654 135 SYSTEM RESPONSES TO STANDARD SIGNALS As indicated in Chapter 12 it is impractical to find the impulse response of a continuoustime system by actually applying an impulse to the system In contrast the discretetime impulse is a simple wellbehaved function and can be applied in a practical situation with no real problems In addition to finding impulse response finding the responses of systems to the unit sequence and to a sinusoid applied to the system at time n 0 are also good ways of testing system dynamic performance UNITSEQUENCE RESPONSE Let the transfer function of a system be Hz N H z D H z Then the unitsequence response of the system in the z domain is Yz z z 1 N H z D H z The unitsequence response can be written in the partialfraction form Yz z N H1 z D H z H1 z 1 z N H1 z D H z H1 z z 1 If the system is stable and causal the inverse z transform of the term z N H1 z D H z is a signal that decays with time the transient response and the inverse z transform of the term H1 z z 1 is a unit sequence multiplied by the value of the transfer function at z 1 the forced response ExamplE 132 Unitsequence response using the z transform A system has a transfer function Hz 100z z 12 Find and graph the unitsequence response In the z domain the unitsequence response is H 1 z z z 1 100z z 12 z 100 z 12 200 z 1 100 2z z 1 z z 12 The timedomain unitsequence response is the inverse z transform which is h 1 n 1002 12 n un Figure 137 The final value that the unitsequence response approaches is 200 and that is the same as H1 rob28124ch13650679indd 654 041216 201 pm 135 System Responses to Standard Signals 655 In signal and system analysis the two most commonly encountered systems are onepole and twopole systems The typical transfer function of a onepole system is of the form Hz Kz z p where p is the location of a real pole in the z plane Its zdomain response to a unitsequence is H 1 z z z 1 Kz z p K 1 p z z 1 pz z p and its timedomain response is h 1 n K 1 p 1 p n1 un To simplify this expression and isolate effects let the gain constant K be 1 p Then h 1 n 1 p n1 un The forced response is u n and the transient response is p n1 un This is the discretetime counterpart of the classic unitstep response of a onepole continuoustime system and the speed of the response is determined by the pole loca tion For 0 p 1 the system is stable and the closer p is to 1 the slower the response is Figure 138 For p 1 the system is unstable A typical transfer function for a secondorder system is of the form Hz K z 2 z 2 2 r 0 cos Ω 0 z r 0 2 The poles of Hz lie at p 12 r 0 e j Ω 0 If r 0 1 both poles lie inside the unit circle and the system is stable The z transform of the unitsequence response is H 1 z K z z 1 z 2 z 2 2 r 0 cos Ω 0 z r 0 2 Figure 137 Unitsequence response 5 10 15 20 200 h1n n rob28124ch13650679indd 655 041216 201 pm C h a p t e r 13 zTransform System Analysis 656 For Ω 0 mπ m an integer the partial fraction expansion of H 1 zKz is H 1 z Kz 1 1 2 r 0 cos Ω 0 r 0 2 1 z 1 r 0 2 2 r 0 cos Ω 0 z r 0 2 z 2 2 r 0 cos Ω 0 z r 0 2 Then H 1 z Kz 1 2 r 0 cos Ω 0 r 0 2 1 z 1 r 0 2 2 r 0 cos Ω 0 z r 0 2 z 2 2 r 0 cos Ω 0 z r 0 2 or H 1 z H1 z z 1 z r 0 2 2 r 0 cos Ω 0 z r 0 2 z 2 2 r 0 cos Ω 0 z r 0 2 which can be written as H 1 z H1 z z 1 r 0 r 0 2 cos Ω 0 z 2 r 0 cos Ω 0 z z 2 2 r 0 cos Ω 0 z r 0 2 1 r 0 2 cos Ω 0 cos Ω 0 sin Ω 0 z r 0 sin Ω 0 z 2 2 r 0 cos Ω 0 z r 0 2 The inverse z transform is h 1 n H1 1 r 0 r 0 2 cos Ω 0 r 0 n cosn Ω 0 1 r 0 2 cos Ω 0 cos Ω 0 sin Ω 0 r 0 n sin n Ω 0 un This is the general solution for the unitsequence response of this kind of second order system If we let K 1 2 r 0 cos Ω 0 r 0 2 then the system has unity gain H1 1 ExamplE 133 Polezero diagrams and unitsequence response using the z transform A system has a transfer function of the form Hz K z 2 z 2 2 r 0 cos Ω 0 z r 0 2 with K 1 2 r 0 cos Ω 0 r 0 2 Figure 138 Response of a onepole system to a unit sequence as the pole location changes Re Im z Re Im z Re Im z Re Im z n 5 15 h1n 1 n 5 15 h1n 1 n 5 15 h1n 1 n 5 15 h1n 1 rob28124ch13650679indd 656 041216 201 pm 135 System Responses to Standard Signals 657 Plot the polezero diagrams and graph the unitsequence responses for a r 0 12 Ω 0 π6 b r 0 12 Ω 0 π3 c r 0 34 Ω 0 π6 and d r 0 34 Ω 0 π3 Figure 139 shows the polezero diagrams and unitsequence responses for the values of r 0 and Ω 0 given above Figure 139 Polezero diagrams and unitsequence responses of a unitygain secondorder system for four combinations of r 0 and Ω 0 Re Im z n 5 15 h1n 1 a Re Im z n 5 15 h1n 1 b Re Im z n 5 15 h1n 1 c Re Im z n 5 15 h1n 1 d As r 0 is increased the response becomes more underdamped ringing for a longer time As Ω 0 is increased the speed of the ringing is increased So we can generalize by saying that poles near the unit circle cause a more underdamped response than poles farther away from and inside the unit circle We can also say that the rate of ringing of the response depends on the angle of the poles being greater for a greater angle RESPONSE TO A CAUSAL SINUSOID The response of a system to a unitamplitude cosine of radian frequency Ω 0 applied to the system at time n 0 is Yz N H z D H z zz cos Ω 0 z 2 2z cos Ω 0 1 The poles of this response are the poles of the transfer function plus the roots of z 2 2z cos Ω 0 1 0 which are the complex conjugate pair p 1 e j Ω 0 and p 2 e j Ω 0 Therefore p 1 p 2 p 1 p 2 2 cos Ω 0 p 1 p 2 j2 sin Ω 0 and p 1 p 2 1 Then if Ω 0 mπ m an integer and if there is no polezero cancellation these poles are distinct and the response can be written in partialfraction form as Yz z N H1 z D H z 1 p 1 p 2 H p 1 p 1 cos Ω 0 z p 1 1 p 2 p 1 H p 2 p 2 cos Ω 0 z p 2 or after simplification Yz z N H1 z D H z H r p 1 z p 1r H i p 1 p 1i z 2 z2 p 1r 1 rob28124ch13650679indd 657 041216 201 pm C h a p t e r 13 zTransform System Analysis 658 where p 1 p 1r j p 1i and H p 1 H r p 1 j H i p 1 This can be written in terms of the original parameters as Yz z N H1 z D H z ReHcos Ω 0 j sin Ω 0 z 2 z cos Ω 0 z 2 z2 cos Ω 0 1 Im Hcos Ω 0 j sin Ω 0 z sin Ω 0 z 2 z2 cos Ω 0 1 The inverse z transform is yn Z 1 z N H1 z D H z ReHcos Ω 0 j sin Ω 0 cos Ω 0 n Im Hcos Ω 0 j sin Ω 0 sin Ω 0 n un or using ReA cos Ω 0 n Im A sin Ω 0 n A cos Ω 0 n A yn Z 1 z N H1 z D H z Hcos Ω 0 j sin Ω 0 cos Ω 0 n Hcos Ω 0 j sin Ω 0 un or yn Z 1 z N H1 z D H z H p 1 cos Ω 0 n H p 1 un 132 If the system is stable the term Z 1 z N H1 z D H z the transient response decays to zero with discrete time and the term H p 1 cos Ω 0 n H p 1 un the forced response is equal to a sinusoid after dis crete time n 0 and persists forever ExamplE 134 System response to a causal cosine using the z transform The system of Example 132 has a transfer function Hz 100z z 12 Find and graph the response to x n cos Ω 0 nu n with Ω 0 π4 In the z domain the response is of the form Yz Kz z p zz cos Ω 0 z 2 2z cos Ω 0 1 Kz z p zz cos Ω 0 z e j Ω 0 z e j Ω 0 where K 100 p 12 and Ω 0 π4 This response can be written in the partialfraction form Yz Kz pp cos Ω 0 p e j Ω 0 p e j Ω 0 z p transient response Az B z 2 2z cos Ω 0 1 forced response rob28124ch13650679indd 658 041216 201 pm 135 System Responses to Standard Signals 659 Using 132 yn Z 1 100z 1212 cosπ4 12 e jπ4 12 e jπ4 z 12 100 e jπ4 e jπ4 12 cos Ω 0 n 100 e jπ4 e jπ4 12 un yn 1907 12 n 13572 cosπn4 05 un 133 Figure 1310 Figure 1310 Causal cosine and system response 5 10 15 20 5 10 15 20 1 1 150 150 n xn n yn For comparison lets find the system response to a true cosine applied at time n using the discretetime Fourier transform DTFT The transfer function expressed as a func tion of radian frequency Ω using the relationship z e jΩ is H e jΩ 100 e jΩ e jΩ 12 The DTFT of x n is X e jΩ π δ 2π Ω Ω 0 δ 2π Ω Ω 0 Therefore the response is Y e jΩ π δ 2π Ω Ω 0 δ 2π Ω Ω 0 100 e jΩ e jΩ 12 or Y e jΩ 100π k e jΩ e jΩ 12 δΩ Ω 0 2πk k e jΩ e jΩ 12 δΩ Ω 0 2πk rob28124ch13650679indd 659 041216 201 pm C h a p t e r 13 zTransform System Analysis 660 Using the equivalence property of the impulse Y e jΩ 100π k e j Ω 0 2πk e j Ω 0 2πk 12 δΩ Ω 0 2πk e j Ω 0 2πk e j Ω 0 2πk 12 δΩ Ω 0 2πk Since e j Ω 0 2πk e j Ω 0 and e j Ω 0 2πk e j Ω 0 for integer values of k Y e jΩ 100π k e j Ω 0 δΩ Ω 0 2πk e j Ω 0 12 e j Ω 0 δΩ Ω 0 2πk e j Ω 0 12 or Y e jΩ 100π e j Ω 0 δ 2π Ω Ω 0 e j Ω 0 12 e j Ω 0 δ 2π Ω Ω 0 e j Ω 0 12 Finding a common denominator applying Eulers identity and simplifying Y e jΩ 100π 54 cos Ω 0 1 12 cos Ω 0 δ 2π Ω Ω 0 δ 2π Ω Ω 0 j2 sin Ω 0 δ 2π Ω Ω 0 δ 2π Ω Ω 0 Finding the inverse DTFT yn 50 54 cos Ω 0 1 12 cos Ω 0 2 cos Ω 0 n sin Ω 0 sin Ω 0 n or since Ω 0 π4 yn 11906 cosπn4 65113 sinπn4 13572 cosπn4 05 This is exactly the same except for the unit sequence un as the forced response in 133 136 SIMULATING CONTINUOUSTIME SYSTEMS WITH DISCRETETIME SYSTEMS zTRANSFORMLAPLACETRANSFORM RELATIONSHIPS We explored in earlier chapters important relationships between Fourier transform methods In particular we showed that there is an information equivalence between a discretetime signal xn xn T s formed by sampling a continuoustime signal and a continuoustime impulse signal x δ t xt δ T s t formed by impulse sampling the same continuoustime signal with f s 1 T s We also derived the relationships between the DTFT of xn and the continuoustime Fourier transform CTFT of x δ t in Chap ter 10 Since the z transform applies to a discretetime signal and is a generalization of the DTFT and a Laplace transform applies to a continuoustime signal and is a generalization of the CTFT we should expect a close relationship between them also Consider two systems a discretetime system with impulse response hn and a continuoustime system with impulse response h δ t and let them be related by h δ t n hnδt n T s 134 This equivalence indicates that everything that happens to xn in the discretetime system happens in a corresponding way to x δ t in the continuoustime system rob28124ch13650679indd 660 041216 201 pm 136 Simulating ContinuousTime Systems with DiscreteTime Systems 661 Figure 1311 Therefore it is possible to analyze discretetime systems using the Laplace transform with the strengths of continuoustime impulses representing the values of the discretetime signals at equally spaced points in time But it is notation ally more convenient to use the z transform instead The transfer function of the discretetime system is Hz n0 hn z n and the transfer function of the continuoustime system is H δ s n0 hn e n T s s If the impulse responses are equivalent in the sense of 134 then the transfer func tions must also be equivalent The equivalence is seen in the relationship H δ s Hz z e s T s It is important at this point to consider some of the implications of the transfor mation z e s T s One good way of seeing the relationship between the s and z complex planes is to map a contour or region in the splane into a corresponding contour or region in the z plane Consider first a very simple contour in the splane the contour s jω j2πf with ω and f representing real radian and cyclic frequency respectively This contour is the ω axis of the splane The corresponding contour in the z plane is e jω T s or e j2πf T s and for any real value of ω and f must lie on the unit circle However the mapping is not as simple as the last statement makes it sound To illustrate the complication map the segment of the imaginary axis in the splane π T s ω π T s that corresponds to f s 2 f f s 2 into the corresponding contour in the z plane As ω traverses the contour π T s ω π T s z traverses the unit circle from e jπ to e jπ in the counterclockwise direction making one complete traversal of the unit circle Now if we let ω traverse the contour π T s ω 3π T s z traverses the unit circle from e jπ to e j3π which is exactly the same contour again because e jπ e jπ e j3π e j2n 1π n any integer Therefore it is apparent that the transformation z e s T s maps the ω axis of the splane into the unit circle of the z plane infinitely many times Figure 1312 This is another way of looking at the phenomenon of aliasing All those segments of the imaginary axis of the splane of length 2π T s look exactly the same when trans lated into the z plane because of the effects of sampling So for every point on the Figure 1311 Equivalence of a discretetime and a continuoustime system n xn Ts t xδt n yn hn hδt t yδt rob28124ch13650679indd 661 041216 201 pm C h a p t e r 13 zTransform System Analysis 662 imaginary axis of the splane there is a corresponding unique point on the unit circle in the z plane But this unique correspondence does not work the other way For every point on the unit circle in the z plane there are infinitely many corresponding points on the imaginary axis of the splane Carrying the mapping idea one step farther the left half of the splane maps into the interior of the unit circle in the z plane and the right half of the splane maps into the exterior of the unit circle in the z plane infinitely many times in both cases The corresponding ideas about stability and pole locations translate in the same way A stable continuoustime system has a transfer function with all its poles in the open left half of the splane and a stable discretetime system has a transfer function with all its poles in the open interior of the unit circle in the z plane Figure 1313 Figure 1312 Mapping the ω axis of the splane into the unit circle of the z plane z s Imz ω ω Rez b a b a z s Imz σ σ b a Rez b a 3π Ts π Ts π Ts 3π Ts π Ts π Ts Figure 1313 Mapping of the regions of the splane into regions in the z plane s z Rez Imz s z Rez Imz s z Rez Imz s z Rez Imz 3π Ts π Ts π Ts 3π Ts 3π Ts π Ts π Ts 3π Ts 3π Ts π Ts π Ts 3π Ts 3π Ts π Ts π Ts 3π Ts σ σ σ σ ω ω ω ω IMPULSE INVARIANCE In Chapter 10 we examined how continuoustime signals are converted to discretetime signals by sampling We found that under certain conditions the discretetime signal was a good representation of the continuoustime signal in the sense that it preserved rob28124ch13650679indd 662 041216 201 pm 136 Simulating ContinuousTime Systems with DiscreteTime Systems 663 all or practically all of its information A discretetime signal formed by properly sam pling a continuoustime signal in a sense simulates the continuoustime signal In this chapter we examined the equivalence between a discretetime system with impulse response h n and a continuoustime system with impulse response h δ t n hnδt n T s The system whose impulse response is h δ t is a very special type of system because its impulse response consists only of impulses As a practical matter this is impossible to achieve because the transfer function of such a system being periodic has a nonzero response at frequencies approaching infinity No real continuoustime system can have an impulse response that contains actual impulses although in some cases that might be a good approximation for analysis purposes To simulate a continuoustime system with a discretetime system we must first address the problem of forming a useful equivalence between a discretetime sys tem whose impulse response must be discrete and a continuoustime system whose impulse response must be continuous The most obvious and direct equivalence between a discretetime signal and a continuoustime signal is to have the values of the continuoustime signal at the sampling instants be the same as the values of the discretetime signal at the corresponding discrete times xn xn T s So if the ex citation of a discretetime system is a sampled version of a excitation of a continu oustime system we want the response of the discretetime system to be a sampled version of the response of the continuoustime system Figure 1314 Figure 1314 Signal sampling and system discretization xt Sampling Sampling Discretization yt ht xn yn hn The most natural choice for hn would be hn hn T s Since hn is not actually a signal occurring in this system but rather a function that characterizes the system we cannot accurately say that Figure 1314 indicates a sampling process We are not sam pling a signal Instead we are discretizing a system The choice of impulse response for the discretetime system hn hn T s establishes an equivalence between the impulse responses of the two systems With this choice of impulse response if a unit continuoustime impulse excites the continuoustime system and a unit discretetime impulse of the same strength excites the discretetime system the response yn is a sampled version of the response yt and yn yn T s But even though the two sys tems have equivalent impulse responses in the sense of hn hn T s and yn yn T s that does not mean that the system responses to other excitations will be equivalent in the same sense A system design for which hn hn T s is called an impulse invariant design because of the equivalence of the system responses to unit impulses It is important to point out here that if we choose to make hn hn T s and we excite both systems with unit impulses the responses are related by yn yn T s but we can not say that xn xn T s as in Figure 1314 Figure 1314 indicates that the discretetime rob28124ch13650679indd 663 041216 201 pm C h a p t e r 13 zTransform System Analysis 664 excitation is formed by sampling the continuoustime excitation But if the continuoustime excitation is an impulse we cannot sample it Try to imagine sampling a continuoustime im pulse First if we are sampling at points in time at some finite rate to try to catch the impulse when it occurs the probability of actually seeing the impulse in the samples is zero because it has zero width Even if we could sample exactly when the impulse occurs we would have to say that δn δn T s but this makes no sense because the amplitude of a continuoustime impulse at its time of occurrence is not defined because it is not an ordinary function so we cannot establish the corresponding strength of the discretetime impulse δn SAMPLEDDATA SYSTEMS Because of the great increases in microprocessor speed and memory and large reductions in the cost of microprocessors modern system design often uses discretetime subsys tems to replace subsystems that were traditionally continuoustime subsystems to save money or space or power consumption and to increase the flexibility or reliability of the system Aircraft autopilots industrial chemical process control manufacturing pro cesses automobile ignition and fuel systems are examples Systems that contain both discretetime subsystems and continuoustime subsystems and mechanisms for convert ing between discretetime and continuoustime signals are called sampleddata systems The first type of sampleddata system used to replace a continuoustime sys tem and still the most prevalent type comes from a natural idea We convert a continuoustime signal to a discretetime signal with an analogtodigital converter ADC We process the samples from the ADC in a discretetime system Then we con vert the discretetime response back to continuoustime form using a digitaltoanalog converter DAC Figure 1315 Figure 1315 A common type of sampleddata simulation of a continuoustime system t xt ht t yt hn ADC DAC t xt t xn t yn t ydt The desired design would have the response of the sampleddata system be very close to the response that would have come from the continuoustime system To do that we must choose hn properly and in order to do that we must also understand the actions of the ADC and DAC It is straightforward to model the action of the ADC It simply acquires the value of its input signal at the sampling time and responds with a number proportional to that signal value It also quantizes the signal but we will ignore that effect as negligible in this analysis The subsystem with impulse response hn is then designed to make the sampleddata system emulate the action of the continuoustime system whose impulse response is ht The action of the DAC is a little more complicated to model mathematically than the ADC It is excited by a number from the discretetime subsystem the strength of an impulse and responds with a continuoustime signal proportional to that number rob28124ch13650679indd 664 041216 201 pm 136 Simulating ContinuousTime Systems with DiscreteTime Systems 665 which stays constant until the number changes to a new value This can be modeled by thinking of the process as two steps First let the discretetime impulse be converted to a continuoustime impulse of the same strength Then let the continuoustime impulse excite a zeroorder hold first introduced in Chapter 10 with an impulse response that is rectangular with height one and width T s beginning at time t 0 h zoh t 0 t 0 1 0 t T s 0 t T s rect t T s 2 T s Figure 1316 The transfer function of the zeroorder hold is the Laplace transform of its impulse response h zoh t which is H zoh s 0 h zoh t e st dt 0 T s e st dt e st s 0 T s 1 e s T s s The next design task is to make hn emulate the action of ht in the sense that the overall system responses will be as close as possible The continuoustime system is excited by a signal xt and produces a response y c t We would like to design the cor responding sampleddata system such that if we convert xt to a discretetime signal xn xn T s with an ADC process that with a system to produce the response yn then convert that to a response y d t with a DAC then y d t y c t Figure 1317 Figure 1317 Desired equivalence of continuoustime and sampleddata systems xt ydt hn xn ht yn DAC ADC yct Figure 1316 Equivalence of a DAC and a discretetimetocontinuoustime impulse conversion followed by a zeroorder hold xn n t xt DA xn n t t xδt xt ZeroOrder Hold δn δt This cannot be accomplished exactly except in the theoretical limit in which the sampling rate approaches infinity But we can establish conditions under which a good approximation can be made one that gets better as the sampling rate is increased rob28124ch13650679indd 665 041216 201 pm C h a p t e r 13 zTransform System Analysis 666 As a step toward determining the impulse response hn of the subsystem first consider the response of the continuoustime system not to xt but rather to x δ t defined by x δ t n xn T s δt n T s xt δ T s t The response to x δ t is yt ht x δ t ht m xn T s δt m T s m xmht m T s where xn is the sampled version of xt xn T s The response at the nth multiple of T s is yn T s m xmhn m T s 135 Compare this to the response of a discretetime system with impulse response hn hn T s to xn xn T s which is yn xn hn m xm hn m 136 By comparing 135 and 136 it is apparent that the response yt of a continuoustime system with impulse response ht at the sampling instants n T s to a continuoustime impulsesampled signal x δ t n xn T s δt n T s can be found by finding the response of a system with impulse response hn hn T s to xn xn T s and making the equivalence yn T s yn Figure 1318 Figure 1318 Equivalence at continuous times n T s and corresponding discrete times n of the responses of continuoustime and discrete time systems excited by continuoustime and discretetime signals derived from the same continuoustime signal ht hn hnTs Impulse Modulation Ts t xt Analogto Digital Conversion t xδt n xn t n yt yn Now returning to our original continuoustime and sampleddata systems modify the continuoustime system as illustrated in Figure 1319 Using the equivalence in Figure 1318 yn yn T s rob28124ch13650679indd 666 041216 201 pm 136 Simulating ContinuousTime Systems with DiscreteTime Systems 667 Now change both the continuoustime system and discretetime system impulse responses by multiplying them by the time between samples T s Figure 1320 In this modified system we can still say that yn yn T s where now yt x δ t T s ht n xn T s δt n T s ht T s n xn T s ht n T s T s 137 yn m xm hn m m xm T s hn m T s Figure 1320 Continuoustime and sampleddata systems when their impulse responses are multiplied by the time between samples T s xt ydt xn Tsht yn xδt DAC ADC xn ADC yt hn TshnTs δn δt Figure 1319 Continuoustime and sampleddata systems when the continuoustime system is excited by x δ t instead of xt xt ydt xn ht yn xδt DAC ADC xn ADC yt hn hnTs δn δt The new subsystem impulse response is hn T s hn T s and ht still represents the im pulse response of the original continuoustime system Now in 137 let T s approach zero In that limit the summation on the righthand side becomes the convolution integral first developed in the derivation of convolution in Chapter 5 lim T s 0 yt lim T s 0 n xn T s ht n T s T s xτht τ dτ which is the signal y c t the response of the original continuoustime system in Figure 1317 to the signal xt Also in that limit yn y c n T s So in the limit the spacing between points T s approaches zero the sampling instants n T s merge into a continuum t and there is a onetoone correspondence between the signal values yn and the signal values y c t The response of the sampleddata system y d t will be indistinguishable from the response y c t of the original system to the signal xt Of course in practice we can never sample at an infinite rate so the correspondence yn y c n T s can never be exact but it does establish an approximate equivalence between a continuoustime and a sampleddata system There is another conceptual route to arriving at the same conclusion for the discretetimesystem impulse response hn T s hn T s In the development above we formed a continuoustime impulse signal x δ t n xn T s δt n T s rob28124ch13650679indd 667 041216 201 pm C h a p t e r 13 zTransform System Analysis 668 whose impulse strengths were equal to samples of the signal xt Now instead form a modified version of this impulse signal Let the new correspondence between xt and x δ t be that the strength of an impulse at n T s is approximately the area under xt in the sampling interval n T s t n 1 T s not the value at n T s The equivalence between xt and x δ t is now based on approximately equal areas Figure 1321 The ap proximation gets better as the sampling rate is increased Figure 1321 A comparison of value sampling and area sampling Value Sampling Area Sampling t xt t xt t xδt t xδt The area under xt is approximately T s xn T s in each sampling interval Therefore the new continuoustime impulse signal would be x δ t T s n xn T s δt n T s If we now apply this impulse signal to a system with impulse response ht we get ex actly the same response as in 137 yt n xn T s ht n T s T s and of course the same result that yn y c n T s in the limit as the sampling rate ap proaches infinity All we have done in this development is associate the factor T s with the excitation instead of with the impulse response When the two are convolved the result is the same If we sampled signals setting impulse strengths equal to signal areas over a sampling interval instead of setting them equal to signal values at sampling instants then the correspondence hn hn T s would be the design correspondence between a continuoustime system and a sampleddata system that simulates it But since we dont sample that way because most ADCs do not work that way we instead associate the factor T s with the impulse response and form the correspondence hn T s hn T s ExamplE 135 Design of a sampleddata system to simulate a continuoustime system A continuoustime system is characterized by a transfer function H c s 1 s 2 40s 300 rob28124ch13650679indd 668 041216 201 pm 136 Simulating ContinuousTime Systems with DiscreteTime Systems 669 Design a sampleddata system of the form of Figure 1315 to simulate this system Do the de sign for two sampling rates f s 10 and f s 100 and compare step responses The impulse response of the continuoustime system is h c t 120 e 10t e 30t ut The discretetimesubsystem impulse response is then h d n T s 20 e 10n T s e 30n T s un and the corresponding zdomain transfer function is H d z T s 20 z z e 10 T s z z e 30 T s The step response of the continuoustime system is h 1c t 2 3 e 10t e 30t 600 ut The response of the subsystem to a unit sequence is h 1d n T s 20 e 10 T s e 30 T s 1 e 10 T s 1 e 30 T s e 10 T s e 10 T s 1 e 10n T s e 30 T s e 30 T s 1 e 30n T s un and the response of the DA converter is h 1d t n0 yn rect t T s n 12 T s Figure 1322 Figure 1322 Comparison of the step responses of a continuoustime system and two sampleddata systems that simulate it with different sampling rates fs 10 fs 100 h1dt 1 300 05 t h1ct 1 300 h1ct 1 300 05 t 50 n h1dn 1 300 h1dn 1 300 5 n 05 t h1dt 1 300 05 t For the lower sampling rate the sampleddata system simulation is very poor It approaches a forced response value that is about 78 percent of the forced response of the continuoustime rob28124ch13650679indd 669 041216 201 pm C h a p t e r 13 zTransform System Analysis 670 system At the higher sampling rate the simulation is much better with a forced response ap proaching a value that is about 99 percent of the forced response of the continuoustime system Also at the higher sampling rate the difference between the continuoustime response and the sampleddatasystem response is much smaller than at the lower sampling rate We can see why the disparity between forced values exists by examining the expression yn T s 20 e 10 T s e 30 T s 1 e 10 T s 1 e 30 T s e 10 T s e 10 T s 1 e 10n T s e 30 T s e 30 T s 1 e 30n T s un The forced response is y forced T s 20 e 10 T s e 30 T s 1 e 10 T s 1 e 30 T s If we approximate the exponential functions by the first two terms in their series expansions as e 10 T s 1 10 T s and e 30 T s 1 30 T s we get y forced 1300 which is the correct forced response However if T s is not small enough the approximation of the exponential function by the first two terms of its series expansion is not very good and actual and ideal forced values are significantly different When f s 10 we get e 10 T s 0368 and 1 10 T s 0 and e 30 T s 00498 and 1 30 T s 2 which are terrible approximations But when f s 100 we get e 10 T s 0905 and 1 10 T s 09 and e 30 T s 0741 and 1 30 T s 07 which are much better approximations 137 STANDARD REALIZATIONS OF SYSTEMS The realization of discretetime systems very closely parallels the realization of continu oustime systems The same general techniques apply and the same types of realizations result CASCADE REALIZATION We can realize a system in cascade form from the factored form of the transfer function Hz A z z 1 z p 1 z z 2 z p 2 z z M z p M 1 z p M1 1 z p M2 1 z p N where the numerator order is M N Figure 1323 Figure 1323 Overall cascade system realization z1 z2 p1 Xz p2 pN1 pN A Yz z1 z1 z1 z1 PARALLEL REALIZATION We can express the transfer function as the sum of partial fractions Hz K 1 z p 1 K 2 z p 2 K N z p N and realize the system in parallel form Figure 1324 rob28124ch13650679indd 670 041216 202 pm 138 Summary of Important Points 671 Discretetime systems are actually built using digital hardware In these systems the signals are all in the form of binary numbers with a finite number of bits The operations are usually performed in fixedpoint arithmetic That means all the signals are quantized to a finite number of possible values and therefore are not exact repre sentations of the ideal signals This type of design usually leads to the fastest and most efficient system but the roundoff error between the ideal signals and the actual signals is an error that must be managed to avoid noisy or in some cases even unstable system operation The analysis of such errors is beyond the scope of this text but generally speaking the cascade and parallel realizations are more tolerant and forgiving of such errors than the Direct Form II canonical realization 138 SUMMARY OF IMPORTANT POINTS 1 It is possible to do analysis of discretetime systems with the Laplace transform through the use of continuoustime impulses to simulate discrete time But the z transform is notationally more convenient 2 Discretetime systems can be modeled by difference equations or block diagrams in the time or frequency domain 3 A discretetime LTI system is stable if all the poles of its transfer function lie in the open interior of the unit circle 4 The three most important types of system interconnections are the cascade connection the parallel connection and the feedback connection 5 The unit sequence and sinusoid are important practical signals for testing system characteristics 6 Discretetime systems can closely approximate the actions of continuoustime systems and the approximation improves as the sampling rate is increased 7 The Direct Form II cascade and parallel realizations are important standard ways of realizing systems Figure 1324 Overall parallel system realization Xz Yz p1 z1 z1 z1 K1 p2 K2 pN KN rob28124ch13650679indd 671 041216 202 pm C h a p t e r 13 zTransform System Analysis 672 EXERCISES WITH ANSWERS Answers to each exercise are in random order Stability 1 Evaluate the stability of the systems with each of these transfer functions a H z z z 2 b H z z z2 78 c H z z z2 3z 2 98 d Hz z2 1 z3 2z2 375z 05625 e Hz z 1 z2 2 z 1 2 f H z z 2 z 1 2 z e j π 3 z ej π 3 g H z z z 08 j08z 08 j08 Answers 1 Stable and 6 Unstable 2 Determine whether these systems are stable or unstable and explain how you know a A system described by 2yn 3yn 1 xn b A feedback system with H1z 07z z2 06z 05 and H2z z1 Answers Both Unstable 3 A discretetime feedback system has a forward path transfer function H 1 z Kz z 05 and a feedback path transfer function H 2 z 4 z 1 For what range of values of K is the system stable Answer 18 K 38 Parallel Cascade and Feedback Connections 4 A feedback discretetime system has a transfer function H z K 1 K z z 09 For what range of Ks is this system stable Answer K 01 or K 19 5 Find the overall transfer functions of the systems in Figure E5 in the form of a single ratio of polynomials in z rob28124ch13650679indd 672 041216 202 pm Exercises with Answers 673 a z1 Xz Yz 03 b Xz 03 Yz 09 z1 z1 Figure E5 Answers z z 03 z 2 z 2 12z 027 Response to Standard Signals 6 Find the timedomain responses yn of the systems with these transfer functions to the unitsequence excitation xn un a H z z z 1 b H z z 1 z 12 Answers 12 n u n y n ramp n 1 7 A discretetime system has a transfer function Hz zz 05 z 02z 08 a What is the final value of its response to a unitsequence excitation b If a discretetime impulse of strength 8 is applied at time n 7 and that is the only excitation of the system for all time find the value of the response at time n 12 Answers 34086 03472 8 A discretetime system has a transfer function Hz 05z2 z2 12z 027 If a unit sequence un is applied as an excitation to this system what are the numerical values of the responses y0 y1 and y2 Answers 0485 01 05 9 Find the responses yn of the systems with these transfer functions to the excitation xn cos2πn8 un Then show that the steadystate response is the same as would have been obtained by using DTFT analysis with an excitation xn cos2πn8 a H z z z 09 b Hz z2 z2 16z 063 Answers 13644 cos 2πn8 10518 19293 cos 2πn8 13145 rob28124ch13650679indd 673 041216 202 pm C h a p t e r 13 zTransform System Analysis 674 Root Locus 10 Draw a root locus for each system with the given forward and feedback path transfer functions a H1 z K z 1 z 1 2 H2 z 4z z 08 b H1 z K z 1 z 1 2 H2 z 4 z 08 c H1 z K z z 1 4 H2 z z 1 5 z 3 4 d H1 z K z z 1 4 H2 z z 2 z 3 4 e H1 z K 1 z2 1 3 z 2 9 H2 z 1 Answers Rez Imz Rez Imz Rez Imz Rez Imz Rez Imz 11 Draw a root locus for each of the polezero diagrams of loop transfer functions of feedback systems in Figure E11 1 0 1 z 05 05 05 05 1 0 1 z 05 05 05 05 1 0 1 1 0 1 1 0 1 1 0 1 z 05 05 05 05 Figure E11 rob28124ch13650679indd 674 041216 202 pm Exercises with Answers 675 Answers 1 0 1 1 0 1 z 05 05 05 05 1 0 1 1 0 1 z 05 05 05 05 1 0 1 1 0 1 z 05 05 05 05 12 Sketch a root locus for each polezero map of a loop transfer function in Figure E12 Then for each one indicate whether the system is unstable at a finite positive value of the gain constant K z 1 1 1 1 z 1 1 1 1 a b Figure E12 Answers One Unstable and One Stable LaplaceTransformzTransform Relationship 13 Sketch regions in the z plane corresponding to these regions in the splane a 0 σ 1 Ts 0 ω 2π T s b 1 T s σ 0 π T s ω 0 c σ 0 ω 2π T s Answers The entire z plane z Rez Imz 1 2718 z Rez Imz 1 0368 SampledData Systems 14 Using the impulseinvariant design method design a discretetime system to approximate the continuoustime systems with these transfer functions at the rob28124ch13650679indd 675 041216 202 pm C h a p t e r 13 zTransform System Analysis 676 sampling rates specified Compare the impulse and unitstep or sequence responses of the continuoustime and discretetime systems a H s 6 s 6 f s 4 Hz b H s 6 s 6 f s 20 Hz Answers t 025 1 ht 6 Impulse Response n 5 20 hn 8 Impulse Response t 025 1 h1t 1 Unit Step Response n 5 20 h1n 25 Unit Sequence Response t 05 1 ht 6 Impulse Response n 2 4 hn 8 Impulse Response t 05 1 h1t 1 Unit Step Response n 2 4 h1n 8 Unit Sequence Response System Realization 15 Draw a cascadeform block diagram for each of these system transfer functions a Hz z z 13z 34 b Hz z 1 4z3 2z2 2z 3 Answers Xz Yz 0888 1 0388 08446 025 z 1 z 1 z 1 Xz Yz 13 34 z 1 z 1 rob28124ch13650679indd 676 041216 202 pm Exercises without Answers 677 16 Draw a parallelform block diagram for each of these system transfer functions a H z z z 13 z 34 b Hz 8z3 4z2 5z 9 7z3 4z2 z 2 Answers z1 z1 Xz Yz 13 413 34 913 Xz Yz 08212 02599 z1 z1 z1 02497 03479 09646 1278 1143 EXERCISES WITHOUT ANSWERS Stability 17 A discretetime feedback system has a forwardpath transfer function H 1 z K and a feedbackpath transfer function H 2 z 3 1 2 z 1 For what range of real values of K is this system stable 18 If 11 n cos 2πn16 u n 𝒵 H 1 z and H 2 z H 1 az and H 1 z and H 2 z are transfer functions of systems 1 and 2 respectively what range of values of a will make system 2 stable and physically realizable Root Locus 19 The loop transfer function of a discretetime feedback system is Tz zz 02 z 01z 08 z2 06 What regions of the real axis in the z plane are part of the root locus Parallel Cascade and Feedback Connections 20 A feedback system has a forward path transfer function H1 z Kz z 05 and a feedback path transfer function H2 z 4z1 For what range of values of K is the system stable Response to Standard Signals 21 If the system below is excited by a unit impulse what are the values of y 0 y 1 y 2 and y 7 xn yn D 05 2 08 12 22 A system has a transfer function Hz z z2 z 024 rob28124ch13650679indd 677 041216 202 pm C h a p t e r 13 zTransform System Analysis 678 If a unit sequence un is applied to this system what are the values of the responses y 0 y 1 and y 2 23 Find the responses yn of the systems with these transfer functions to the unit sequence excitation xn un a Hz z z2 18z 082 b Hz z2 1932z 1 zz 095 24 In Figure E24 are six polezero diagrams for six discretetime system transfer functions Answer the following questions about them 1 2 3 Re Im z Re Im z 2 Re Im z 2 4 5 6 Re Im z 2 Re Im z 2 Re Im z Figure E24 a Which of these systems have an impulse response that is monotonic Monotonic means always moving in the same direction always rising or always falling not oscillating or ringing b Of those systems which have a monotonic impulse response which one has the fastest response to a unit sequence Fastest means approaching its final value more quickly c Of those systems which have an oscillatory or ringing impulse response which one system rings at the fastest rate and has the largest overshoot in its response 25 Answer the following questions a A digital filter has an impulse response hn 06n un If it is excited by a unit sequence what is the final value of the response b A digital filter has a transfer function Hz 10z z 05 At what discretetime radian frequency Ω is its magnitude response a minimum z e jΩ c A digital filter has a transfer function Hz 10z 1 z 03 At what radian frequency Ω is its magnitude response a minimum z e jΩ d A digital filter has a transfer function Hz 2z z 07 What is the magnitude of its response at a discretetime radian frequency of Ω π2 z e jΩ rob28124ch13650679indd 678 041216 202 pm Exercises without Answers 679 LaplaceTransformzTransform Relationship 26 For any given sampling rate fs the relationship between the s and z planes is given by z esTs where Ts 1fs Let fs 100 a Describe the contour in the z plane that corresponds to the entire negative σ axis in the splane b What is the minimum length of a line segment along the ω axis in the splane that corresponds to the entire unit circle in the z plane c Find the numerical values of two different points in the splane s1 and s2 that correspond to the point z 1 in the z plane SampledData Systems 27 Using the impulseinvariant design method design a discretetime system to approximate the continuoustime systems with these transfer functions at the sampling rates specified Compare the impulse and unitstep or sequence responses of the continuoustime and discretetime systems a Hs 712s s2 46s 240 fs 20 Hz b Hs 712s s2 46s 240 fs 200 Hz System Realization 28 Draw a parallelform block diagram for each of these system transfer functions a Hz 1 z1 18 z 01z 07 b Hz z z 1 1 z z 1 z2 z2 12 General 29 In Figure E29 are some descriptions of discretetime systems in different forms Answer the following questions about them z1 07 xn yn xn yn D 11 xn yn D A B C H z z 1 z 1 y n x n x n 1 2y n y n 1 x n D E F H z z 2 z 1 z 2 Y z X z 08 z 1 Y z 11 z 2 Y z G H Figure E29 a Which of these systems are unstable including marginally stable b Which of these systems have one or more zeros on the unit circle rob28124ch13650679indd 679 041216 202 pm 680 141 INTRODUCTION AND GOALS One of the most important practical systems is the filter Every system is in one sense a filter because every system has a frequency response that attenuates some frequencies more than others Filters are used to tailor the sound of music according to personal tastes to smooth and eliminate trends from signals to stabilize otherwise unstable systems to remove undesirable noise from a received signal and so on The study of the analysis and design of filters is a very good example of the use of transform methods CHAPTER GOA L S 1 To become familiar with the most common types of optimized continuoustime filters to understand in what sense they are optimal and to be able to design them to meet specifications 2 To become familiar with the filter design and analysis tools in MATLAB 3 To understand how to convert one type of filter to another through a change of variable 4 To learn methods of simulating optimized continuoustime filters with discretetime filters and to understand the relative advantages and disadvantages of each method 5 To explore both infinitedurationimpulseresponse and finitedurationimpulse response discretetime filter designs and to understand the relative advantages and disadvantages of each method 142 ANALOG FILTERS In this chapter continuoustime filters will be referred to as analog filters and dis cretetime filters will be referred to as digital filters Also when discussing both ana log and digital filters the subscript a will be used to indicate functions or parameters applying to analog filters and the subscript d will be used similarly for functions or parameters applying to digital filters 14 C H A P T E R Filter Analysis and Design rob28124ch14680734indd 680 041216 205 pm 142 Analog Filters 681 BUTTERWORTH FILTERS Normalized Butterworth Filters A very popular type of analog filter is the Butterworth filter named after British applied physicist Stephen Butterworth who invented it An nth order lowpass Butterworth filter has a frequency response whose squared magnitude is H a jω 2 1 1 ω ω c 2n The lowpass Butterworth filter is designed to be maximally flat for frequencies in its passband ω ωc meaning its variation with frequency in the passband is monotonic and approaches a zero derivative as the frequency approaches zero Figure 141 illustrates the frequency response of a Butterworth filter with a corner frequency of ωc 1 for four different orders n As the order is increased the filters magnitude frequency response approaches that of an ideal lowpass filter Figure 141 Butterworth filter magnitude frequency responses for a corner frequency ωc 1 and four different orders 5 4 3 2 1 1 2 3 4 5 Ha jω n 1 n 2 n 4 n 8 2 1 1 ω Figure 142 Butterworth filter pole locations n 2 σ ω ωc 90 60 60 n 1 σ ω ωc n 3 σ ω ωc The poles of a lowpass Butterworth filter lie on a semicircle of radius ωc in the open left halfplane Figure 142 The number of poles is n and the angular spacing between poles for n 1 is always πn If n is odd there is a pole on the negative real axis and all the other poles occur in complex conjugate pairs If n is even all the poles occur in complex conjugate pairs Using these properties the transfer function of a unitygain lowpass Butterworth filter can always be found and is of the form H a s 1 1 sp11 sp21 spn k1 n 1 1 spk k1 n pk s pk where the pks are the pole locations The MATLAB signal toolbox has functions for designing analog Butterworth filters The MATLAB function call zapaka buttapN returns the finite zeros in the vector za the finite poles in the vector pa and the gain coefficient in the scalar ka for an Nth order unitygain Butterworth lowpass filter with rob28124ch14680734indd 681 041216 205 pm C h a p t e r 14 Filter Analysis and Design 682 a corner frequency ωc 1 There are no finite zeros in a lowpass Butterworth filter transfer function so za is always an empty vector and since the filter is unitygain ka is always one The zeros and gain are included in the returned data because this form of returned data is used for other types of filters for which there may be finite zeros and the gain may not be one zapaka buttap4 za za pa pa 03827 09239i 03827 09239i 09239 03827i 09239 03827i ka ka 1 Filter Transformations Once a design has been done for a lowpass Butterworth filter of a given order with a corner frequency ωc 1 the conversion of that filter to a different corner frequency andor to a highpass bandpass or bandstop filter can be done with a change of the frequency variable MATLAB allows the designer to quickly and easily design an nth order lowpass Butterworth filter with unity gain and a corner frequency ωc 1 Denormalizing the gain to a nonunity gain is trivial since it simply involves changing the gain coefficient Changing the corner frequency or the filter type is a little more involved To change the frequency response from a corner frequency ωc 1 to a general corner frequency ωc 1 make the independentvariable change s sωc in the trans fer function For example a firstorder unitygain normalized Butterworth filter has a transfer function H norm s 1 s 1 If we want to move the corner frequency to ωc 10 the new transfer function is H 10 s H norm s10 1 s10 1 10 s 10 This is the transfer function of a unitygain lowpass filter with a corner frequency ωc 10 The real power of the filter transformation process is seen in converting a lowpass filter to a highpass filter If we make the change of variable s 1s then H HP s H norm 1s 1 1s 1 s s 1 and HHPs is the transfer function of a firstorder unitygain highpass Butterworth filter with a corner frequency ωc 1 We can also simultaneously change the corner frequency by making the change of variable s ωcs We now have a transfer function rob28124ch14680734indd 682 041216 205 pm 142 Analog Filters 683 with one finite pole and one finite zero at s 0 In the general form of the transfer function of a normalized lowpass Butterworth filter H norm s k1 n pk s pk when we make the change of variable s 1s we get H HP s k1 n pk s pk s1s k1 n pk 1s pk k1 n pks pks 1 k1 n s s 1pk The poles are at s 1pk They are the reciprocals of the normalized lowpass filter poles all of which have a magnitude of one The reciprocal of any complex number is at an angle that is the negative of the angle of the complex number In this case since the magnitudes of the poles have not changed the poles move to their complex conjugates and the overall constellation of poles is unchanged Also there are now n zeros at s 0 If we make the change of variable s ωcs the poles have the same angles but their magnitudes are now all ωc instead of one Transforming a lowpass filter into a bandpass filter is a little more complicated We can do it by using the change of variable s s 2 ω L ω H s ω H ω L where ωL is the lower positive corner frequency of the bandpass filter and ωH is the higher positive corner frequency For example lets design a firstorder unitygain bandpass filter with a passband from ω 100 to ω 200 Figure 143 H BP s H norm s 2 ω L ω H s ω H ω L 1 s 2 ω L ω H s ω H ω L 1 s ω H ω L s 2 s ω H ω L ω L ω H H BP jω jω ω H ω L ω 2 jω ω H ω L ω L ω H Figure 143 Magnitude frequency response of a unitygain firstorder bandpass Butterworth filter 1000 1 ω 1000 200 100 100 200 HBP jω 1 2 Simplifying and inserting numerical values H BP jω j100ω ω 2 j100ω 20000 j100ω jω 50 j1322 jω 50 j1322 rob28124ch14680734indd 683 041216 205 pm C h a p t e r 14 Filter Analysis and Design 684 The peak of the bandpass response occurs where the derivative of the frequency response with respect to ω is zero d dω H BP jω ω 2 jω ω H ω L ω L ω H j ω H ω L jω ω H ω L 2ω j ω H ω L ω 2 jω ω H ω L ω L ω H 2 0 ω 2 jω ω H ω L ω L ω H 2 ω 2 jω ω H ω L 0 ω 2 ω L ω H 0 ω ω L ω H So the natural radian frequency is ω n ω L ω H Also to conform to the standard secondorder system transfer function form j2ζ ω n ω jω ω H ω L ζ ω H ω L 2 ω L ω H So the damping ratio is ζ ω H ω L 2 ω H ω L Finally we can transform a lowpass filter into a bandstop filter with the transformation s s ω H ω L s 2 ω L ω H Notice that for a lowpass filter of order n the degree of the denominator of the transfer function is n but for a bandpass of order n the degree of the denominator of the trans fer function is 2n Similarly for a highpass filter the denominator degree is n and for a bandstop filter the degree of the denominator is 2n MATLAB Design Tools MATLAB has commands for the transformation of normalized filters They are lp2bp Lowpass to bandpass analog filter transformation lp2bs Lowpass to bandstop analog filter transformation lp2hp Lowpass to highpass analog filter transformation lp2lp Lowpass to lowpass analog filter transformation The syntax for lp2bp is numtdent lp2bpnumdenw0bw where num and den are vectors of coefficients of s in the numerator and denomina tor of the normalized lowpass filter transfer function respectively w0 is the center frequency of the bandpass filter and bw is the bandwidth of the bandpass filter both in rads and numt and dent are vectors of coefficients of s in the numerator and denominator of the bandpass filter transfer function The syntax of each of the other commands is similar As an example we can design a normalized lowpass Butterworth filter with buttap zpk buttap3 z z rob28124ch14680734indd 684 041216 205 pm 142 Analog Filters 685 p p 05000 08660i 10000 05000 08660i k k 1 This result indicates that a thirdorder normalized lowpass Butterworth filter has the frequency response H L P s 1 s 1s 05 j0866s 05 j0866 We can convert this to a ratio of polynomials using MATLAB systemobject commands numden tfdatazpkzpkv num num 0 0 0 1 den den 10000 20000 00000i 20000 00000i 10000 00000i This result indicates that the normalized lowpass frequency response can be written more compactly as H L P s 1 s 3 2 s 2 2s 1 Using this result we can transform the normalized lowpass filter to a denormalized bandpass filter with center frequency ω 8 and bandwidth Δω 2 numtdent lp2bpnumden82 numt numt Columns 1 through 4 0 00000 00000i 00000 00000i 80000 00000i Columns 5 through 7 00000 00000i 00000 00000i 00000 00000i dent dent 10e05 Columns 1 through 4 00000 00000 00000i 00020 00000i 00052 00000i Columns 5 through 7 01280 00000i 01638 00000i 26214 00000i bpf tfnumtdent bpf rob28124ch14680734indd 685 041216 205 pm C h a p t e r 14 Filter Analysis and Design 686 Figure 144 Typical magnitude frequency responses of Butterworth Chebyshev and Elliptic filters 60 40 20 0 Lowpass Butterworth Analog Filter Order 6 Corner at 400 Hz 102 60 40 20 0 f Hz 102 f Hz Lowpass Chebyshev Type 1 Analog Filter Order 6 Corner at 400 Hz 102 60 40 20 0 f Hz Lowpass Chebyshev Type 2 Analog Filter Order 6 Corner at 400 Hz 102 60 40 20 0 f Hz Lowpass Elliptic Analog Filter Order 6 Corner at 400 Hz Ha f dB Ha f dB Ha f dB Ha f dB Transfer function 1542e 14 s5 232e 13 s4 8 s3 3644e 11 s2 9789e 11 s 9952e 10 s6 4 s5 200 s4 520 s3 128e04 s2 1638e04 s 2621e05 This result indicates that the bandpassfilter transfer function can be written as H BP s 8 s 3 s 6 4 s 5 200 s 4 520 s 3 12800 s 2 16380s 262100 The extremely small nonzero coefficients in the numerator of the transfer function reported by MATLAB are the result of roundoff errors in the MATLAB calculations and have been neglected Notice they were zero in numt CHEBYSHEV ELLIPTIC AND BESSEL FILTERS We have just seen how the MATLAB command buttap can be used to design a normalized Butterworth filter and how to denormalize it to other Butterworth filters There are several other MATLAB commands that are useful in analog filter design There are four other ap commands cheb1ap cheb2ap ellipap and besselap that design normalized analog filters of optimal types other than the Butterworth filter The other optimal analog filter types are the Chebyshev sometimes spelled Tchebysheff or Tchebischeff filter the Elliptic filter sometimes called the Cauer filter and the Bessel filter Each of these filter types optimizes the performance of the filter according to a different criterion The Chebyshev filter is similar to the Butterworth filter but has an extra degree of design freedom Figure 144 rob28124ch14680734indd 686 041216 205 pm 142 Analog Filters 687 The Butterworth filter is called maximally flat because it is monotonic in the pass and stopbands and approaches a flat response in the passband as the order is increased There are two types of Chebyshev filter types one and two The typeone Chebyshev has a frequency response that is not monotonic in the passband but is monotonic in the stopband Its frequency response ripples varies up and down with frequency in the passband The presence of ripple in the passband is usually not in itself desirable but it allows the transition from the passband to the stopband to be faster than a Butterworth filter of the same order In other words we trade passband monotonicity for a narrower transition band The more ripple we allow in the passband the narrower the transition band can be The typetwo Chebyshev filter is just the opposite It has a monotonic passband and ripple in the stopband and for the same filter order also allows for a narrower transition band than a Butterworth filter The Elliptic filter has ripple in both the passband and stopband and for the same filter order it has an even narrower transition band than either of the two types of Chebyshev filter The Bessel filter is optimized on a different basis The Bessel filter is optimized for linearity of the phase in the passband rather than flat magnitude response in the passband andor stopband or narrow transition band The syntax for each of these normalized analog filter designs is given below zpk cheb1apNRp zpk cheb2apNRs zpk ellipapNRpRs zpk besselapN where N is the order of the filter Rp is allowable ripple in the passband in dB and Rs is the minimum attenuation in the stopband in dB Once a filter has been designed its frequency response can be found using either bode which was introduced earlier or freqs The function freqs has the syntax H freqsnumdenw where H is a vector of responses at the real radianfrequency points in the vector w and num and den are vectors containing the coefficients of s in the numerator and denomi nator of the filter transfer function ExamplE 141 Comparison of fourthorder bandstop Butterworth and Chebyshev filters using MATLAB Using MATLAB design a normalized fourthorder lowpass Butterworth filter transform it into a denormalized bandstop filter with a center frequency of 60 Hz and a bandwidth of 10 Hz then compare its frequency response with a typeone Chebyshev bandstop filter of the same order and corner frequencies and an allowable ripple in the pass band of 03 dB Butterworth design Design a normalized fourthorder Butterworth lowpass filter and put the zeros poles and gain in zb pb and kb zbpbkb buttap4 rob28124ch14680734indd 687 041216 205 pm C h a p t e r 14 Filter Analysis and Design 688 Use MATLAB system tools to obtain the numerator and denominator coefficient vectors numb and denb numbdenb tfdatazpk zbpbkbv Set the cyclic center frequency and bandwidth and then set the corresponding radian center frequency and bandwidth f0 60 fbw 10 w0 2pif0 wbw 2pifbw Denormalize the lowpass Butterworth to a bandstop Butterworth numbsbdenbsb lp2bsnumbdenbw0wbw Create a vector of cyclic frequencies to use in plotting the frequency response of the filter Then create a corresponding radianfrequency vector and compute the frequency response wbsb 2pi400280 Hbsb freqsnumbsbdenbsbwbsb Chebyshev design Design a normalized fourthorder typeone Chebyshev lowpass filter and put the zeros poles and gain in zc pc and kc zcpckc cheb1ap403 wc wb Use MATLAB system tools to obtain the numerator and denominator coefficient vectors numc and denc numcdenc tfdatazpkzcpckcv Denormalize the lowpass Chebyshev to a bandstop Chebyshev numbscdenbsc lp2bsnumcdencw0wbw Use the same radianfrequency vector used in the Butterworth design and compute the frequency response of the Chebyshev bandstop filter wbsc wbsb Hbsc freqsnumbscdenbscwbsc The magnitude frequency responses are compared in Figure 145 Notice that the Butterworth filter is monotonic in the passbands while the Chebyshev filter is not but that the Chebyshev filter has a steeper slope in the transition between pass and stopbands and slightly better stopband attenuation rob28124ch14680734indd 688 041216 205 pm 143 Digital Filters 689 143 DIGITAL FILTERS The analysis and design of analog filters is a large and important topic An equally large and important topic maybe even more important is the design of digital fil ters that simulate some of the popular kinds of standard analog filters Nearly all discretetime systems are filters in a sense because they have frequency responses that are not constant with frequency SIMULATION OF ANALOG FILTERS There are many optimized standard filter design techniques for analog filters One very popular way of designing digital filters is to simulate a proven analog filter design All the commonly used standard analog filters have sdomain transfer functions that are ratios of polynomials in s and therefore have impulse responses that endure for an infinite time This type of impulse response is called an infiniteduration impulse response IIR Many of the techniques that simulate the analog filter with a digital filter create a digital filter that also has an IIR infiniteduration impulse response and these types of digital filters are called IIR filters Another popular design method for digital filters produces filters with a finiteduration impulse response and these filters are called FIR filters In the following discussion of simulation of analog filters with digital filters the analog filters impulse response will be hat its transfer function will be Has the digital filters impulse response will be hdn and its transfer function will be Hdz FILTER DESIGN TECHNIQUES IIR Filter Design TimeDomain Methods ImpulseInvariant Design One approach to digitalfilter design is to try to make the digital filter response to a standard digital excitation a sampled version of the analog filter response to the corresponding standard continuoustime excitation This idea leads to the impulseinvariant and stepinvariant design procedures Impulse invariant design makes the response of the digital filter to a discretetime unit impulse a Figure 145 Comparison of the Butterworth and Chebyshev magnitude frequency responses f 80 60 40 Ha f 1 Chebyshev Butterworth Butterworth rob28124ch14680734indd 689 041216 205 pm C h a p t e r 14 Filter Analysis and Design 690 sampled version of the response of the analog filter to a continuoustime unit impulse Stepinvariant design makes the response of the digital filter to a unit sequence a sampled version of the response of the analog filter to a unit step Each of these design processes produces an IIR filter Figure 146 Figure 146 The impulseinvariant and stepinvariant digitalfilter design techniques ImpulseInvariant Design StepInvariant Design hat hdn hdn t yt n yn n yn hat t yt t ut t δt 1 n δn 1 1 n un 1 We know from sampling theory that we can impulse sample the analog filter impulse response hat to form hδt whose Laplace transform is Hδs and whose continuoustime Fourier transform CTFT is H δ jω f s k H a jω k ω s where Has is the analog filters transfer function and ωs 2π fs We also know that we can sample hat to form hdn whose z transform is Hdz and whose discretetime Fourier transform DTFT is H d e jΩ f s k H a j f s Ω 2πk 141 So it is apparent that the digitalfilters frequency response is the sum of scaled aliases of the analog filters frequency response and to the extent that the aliases overlap the two frequency responses must differ As an example of impulseinvariant design let Has be the transfer function of a secondorder Butterworth lowpass filter with lowfrequency gain of A and cutoff frequency of ωc radians per second H a s A ω c 2 s 2 2 ω c s ω c 2 Then inverse Laplace transforming h a t 2 A ω c e ω c t 2 sin ω c t 2 ut Now sample at the rate fs to form h d n 2 A ω c e ω c n T s 2 sin ω c n T s 2 un Figure 147 and H d z 2 A ω c z e ω c T s 2 sin ω c T s 2 z 2 2 e ω c T s 2 cos ω c T s 2 z e 2 ω c T s 2 rob28124ch14680734indd 690 041216 205 pm 143 Digital Filters 691 or H d e jΩ 2 A ω c e jΩ e ω c T s 2 sin ω c T s 2 e j2Ω 2 e ω c T s 2 cos ω c T s 2 e jΩ e 2 ω c T s 2 142 Equating the two forms 141 and 142 H d e jΩ f s k A ω c 2 j f s Ω 2πk 2 j 2 ω c f s Ω 2πk ω c 2 2 A ω c e jΩ e ω c T s 2 sin ω c T s 2 e j2Ω 2 e ω c T s 2 cos ω c T s 2 e jΩ e 2 ω c T s 2 If we let A 10 and ωc 100 and sample at a rate of 200 samplessecond then H d e jΩ 2000 k 1 j2Ω 2πk 2 j2 2 Ω 2πk 1 or H d e jΩ 1000 2 e jΩ e 12 2 sin 12 2 e j2Ω 2 e 12 2 cos 12 2 e jΩ e 1 2 343825 e jΩ e j2Ω 131751 e jΩ 049306 As a check compare the two forms at Ω 0 The complete digitalfilter frequency response is shown in Figure 148 The heavy line is the actual frequency response and the light lines are the individual scaled aliases of the analog filters frequency response The difference between the analog filters response at zero frequency and the digital filters response at zero frequency is about 2 due to the effects of aliasing This filter can be realized directly from its transfer function in Direct Form II H d z Y d z X d z 343825z z 2 131751z 049306 or z 2 Yd z 131751zYd z 049306Yd z 343825z X d z Figure 147 Analog and digital filter impulse responses 0 002 004 006 008 0 100 200 300 400 t s hat 0 5 10 15 0 100 200 300 400 n hdn rob28124ch14680734indd 691 041216 205 pm C h a p t e r 14 Filter Analysis and Design 692 Rearranging and solving for Ydz Yd z 343825 z 1 X d z 131751 z 1 Yd z 049306 z 2 Yd z Then inverse z transforming yd n 343825 x d n 1 131751 yd n 1 049306 yd n 2 Figure 149 Figure 148 Digitalfilter frequency response showing the effects of aliasing 15 10 5 0 5 10 15 0 500 1000 1500 2000 2500 Ω radianssample Scaled Aliases of the Analog Filter Frequency Response DigitalFilter Frequency Response Hde jΩ Figure 149 Block diagram of a lowpass filter designed using the impulseinvariant method z1 z1 Xz Yz 343825 049306 13175 To illustrate a subtlety in this design method consider a firstorder lowpass analog filter whose transfer function is H a s A ω c s ω c H a jω A ω c jω ω c with impulse response h a t A ω c e ω c t ut Sample at the rate fs to form h d n A ω c e ω c n T s un and H d z A ω c z z e ω c T s H d e jΩ A ω c e jΩ e jΩ e ω c T s 143 rob28124ch14680734indd 692 041216 205 pm 143 Digital Filters 693 and the frequency response can be written in the two equivalent forms H d e jΩ f s k A ω c j f s Ω 2πk ω c A ω c e jΩ e jΩ e ω c T s Let a 10 ωc 50 and fs 100 and again check the equality at Ω 0 f s k A ω c j f s Ω 2πk ω c k 50000 j200πk 50 10207 A ω c e jΩ e jΩ e ω c T s 500 1 1 e 12 12707 These two results which should ideally be equal differ by almost 25 at Ω 0 The two frequency responses are illustrated in Figure 1410 15 10 5 0 5 10 15 0 200 400 600 800 1000 1200 1400 DigitalFilter Frequency Response Sum of Aliases Hde jΩ Ω radianssample Figure 1410 Digitalfilter frequency response showing an apparent error between two frequency responses that should be equal The question of course is why are they different The error arises from the statement above that the digital filter impulse response found by sampling the analog filter impulse response is h d n A ω c e ω c n T s un The analog impulse response has a discontinuity at t 0 So what should the sample value be at that point The impulse response h d n A ω c e ω c n T s un implies that the sample value at t 0 is Aωc But why isnt a sample value of zero just as valid since the discontinuity extends from zero to Aωc If we replace the first sample value of Aωc with Aωc 2 the average of the two limits from above and below of the analog filters impulse response at t 0 then the two formulas for the digitalfilter frequency response agree exactly So it would seem that when sampling at a discontinuity the best value to take is the average value of the two limits from above and below This is in accord with Fourier transform theory for which the Fourier transform representation of a discontinuous signal always goes through the midpoint of a discontinuity This problem did not arise in the previous analysis of the secondorder Butterworth lowpass filter because its impulse response is continuous Given the error in the firstorder lowpass digitalfilter design due to sampling at a discontinuity one might suggest that to avoid the problem we could simply delay the analog filters impulse response by some small amount less than the time between rob28124ch14680734indd 693 041216 207 pm C h a p t e r 14 Filter Analysis and Design 694 samples and avoid sampling at a discontinuity That can be done and the two forms of the digitalfilters frequency response again agree exactly The MATLAB signal toolbox has a command impinvar that does impulseinvariant digitalfilter design The syntax is bdad impinvarbaaafs where ba is a vector of coefficients of s in the numerator of the analog filter transfer function aa is a vector of coefficients of s in the denominator of the analog filter transfer function fs is the sampling rate in samplessecond bd is a vector of coefficients of z in the numerator of the digitalfilter transfer function and ad is a vector of coefficients of z in the denominator of the digitalfilter transfer function Its transfer function is not iden tical to the impulseinvariant design result given here It has a different gain constant and is shifted in time but the impulse response shape is the same see Example 142 ExamplE 142 Digital bandpass filter design using the impulseinvariant method Using the impulseinvariant design method design a digital filter to simulate a unitygain secondorder bandpass Butterworth analog filter with corner frequencies 150 Hz and 200 Hz and a sampling rate of 1000 samplessecond The transfer function is H a s 987 10 4 s 2 s 4 4443 s 3 2467 10 6 s 2 5262 10 8 s 1403 10 12 and the impulse response is h a t 24607 e 12241t cos 11994t 148 2005 e 9974t cos 97727t 1683 ut Compare the frequency responses of the analog and digital filters This impulse response is the sum of two exponentially damped sinusoids with time constants of about 82 ms and 10 ms and sinusoidal frequencies of 119942π 1909 and 977272π 15554 Hz For a reasonably accurate simulation we should choose a sampling rate such that the sinusoid is oversampled and there are several samples of the exponential decay per time constant Let the sampling rate fs be 1000 samplessecond Then the discretetime impulse response would be h d n 24607 e 012241n cos 11994n 148 2005 e 009974n cos 097727n 1683 un The z transform of this discretetime impulse response is the transfer function H d z 484 z 3 1077 z 2 5146z z 4 1655 z 3 2252 z 2 1319z 06413 The analog and digital filters impulse responses are illustrated in Figure 1411 Figure 1411 Analog and digital filter impulse responses 0 002 004 100 0 100 t s hat Analog Filter Impulse Response 0 20 40 100 0 100 n hd n Digital Filter Impulse ResponseImpulse Invariant rob28124ch14680734indd 694 041216 207 pm 143 Digital Filters 695 The magnitude frequency responses of the analog and digital filters are illustrated in Figure 1412 and their polezero diagrams are in Figure 1413 Two things immediately stand out about this design First the analog filter has a response of zero at f 0 and the digital filter does not The digitalfilters frequency response at Ω 0 is about 085 of its peak frequency response Since this filter is intended as a bandpass filter this is an undesirable design result The gain of the digital filter is much greater than the gain of the analog filter The gain could be made the same as the analog filter by a simple adjustment of the multiplication factor in the expression for Hdz Also although the frequency response does peak at the right frequency the attenuation of the digital filter in the stopband is not as good as the analog filters attenuation If we had used a higher sampling rate the attenuation would have been better Doing this design with MATLABs impinvar command bdad impinvar987e4 0 01 4443 2467e6 5262e8 1403e121000 bd 2 σ ω s 1224077 997364 11994107 9772666 9772666 11994107 Rez Imz z 03211 05062 069427 15316 082447 075028 075028 082447 Figure 1413 Polezero diagrams of the analog filter and its digital simulation by the impulseinvariant method Figure 1412 Magnitude frequency responses of the analog filter and its digital simulation by the impulseinvariant method 2000 1000 0 1000 0 05 1 f Hz Bandpass Butterworth Analog Filter Order 2 Corners at 150 200 Hz 2 0 2 0 500 1000 Bandpass Butterworth Digital Filter Impulse Invariant Sampling Rate 1000 samplessecond Ha f HdejΩ Ω radianssample rob28124ch14680734indd 695 041216 207 pm C h a p t e r 14 Filter Analysis and Design 696 00000 00484 01077 00515 ad 10000 16547 22527 13188 06413 The resulting transfer function is H M z Yz Xz 00484 z 2 01077z 00515 z 4 16547 z 3 22527 z 2 13188z 06413 Compare this to the result above H d z 484 z 3 1077 z 2 5146z z 4 1655 z 3 2252 z 2 1319z 06413 The relation between them is H M z z 1 f s H d z So MATLABs version of impulseinvariant design divides the transfer function by the sam pling rate changing the filters gain constant and multiplies the transfer function by z1 delay ing the impulse response by one unit in discrete time Multiplication by a constant and a time shift are the two things we can do to a signal without distorting it Therefore the two impulse responses although not identical have the same shape StepInvariant Design A closely related design method for digital filters is the stepinvariant method In this method the unitsequence response of the digital filter is designed to match the unitstep response of the analog filter at the sampling instants If an analog filter has a transfer function Has the Laplace transform of its unitstep response is Hass The unitstep response is the inverse Laplace transform h 1a t 1 H a s s The corresponding discretetime unitsequence response is then h 1d n h 1a n T s Its z transform is the product of the zdomain transfer function and the z transform of a unit sequence Z h 1d n z z 1 H d z We can summarize by saying that given an sdomain transfer function Has we can find the corresponding zdomain transfer function Hdz as H d z z 1 z Z 1 H a s s tn T s n In this method we sample the analog unitstep response to get the digital unitsequence response If we impulse sample the analog filter unitstep response h1at we form h1δt whose Laplace transform is H1δs and whose CTFT is H 1δ jω f s k H 1a jω k ω s rob28124ch14680734indd 696 041216 207 pm 143 Digital Filters 697 where H1as is the Laplace transform of the analog filters unitstep response and ωs 2πfs We also know that we can sample h1at to form h1dn whose z transform is H1dz and whose DTFT is H 1d e jΩ f s k H 1a j f s Ω 2πk 144 Relating this result to the analog and digital transfer functions H 1d e jΩ e jΩ e jΩ 1 H d e jΩ and H 1a jω H a jωjω H d e jΩ e jΩ 1 e jΩ H 1d e jΩ e jΩ 1 e jΩ f s k H a j f s Ω 2πk j f s Ω 2πk ExamplE 143 Digital bandpass filter design using the stepinvariant method Using the stepinvariant method design a digital filter to approximate the analog filter whose transfer function is the same as in Example 142 H a s 987 10 4 s 2 s 4 4443 s 3 2467 10 6 s 2 5262 10 8 s 1403 10 12 with the same sampling rate fs 1000 samplessecond The unitstep response is h 1a t 02041 e 122408t cos 11994t 31312 02041 e 9974t cos 97727t 001042 ut The unitsequence response is h 1d n 02041 08847 n cos 11994n 31312 02041 09051 n cos 097727n 00102 un The digitalfilter transfer function is H d z 003443 z 3 003905 z 2 002527z 002988 z 4 1655 z 3 2252 z 2 1319z 06413 The step responses magnitude frequency responses and polezero diagrams of the analog and digital filters are compared in Figure 1414 Figure 1415 and Figure 1416 Figure 1414 Step responses of the analog filter and its digital simulation by the stepinvariant method 0 002 004 01 0 01 t s h1at Analog Filter Step Response 0 20 40 01 0 01 n h1d n Digital Filter Step ResponseStep Invariant rob28124ch14680734indd 697 041216 207 pm C h a p t e r 14 Filter Analysis and Design 698 In contrast with the impulse invariant design this digital filter has a response of zero at Ω 0 Also the digital filter peak passband frequency response and the analog filter peak pass band frequency response differ by less than 01 FiniteDifference Design Another method for designing digital filters to simulate an alog filters is to approximate the differential equation describing the linear system with a difference equation The basic idea in this method is to start with a desired transfer function of the analog filter Has and find the differential equation corresponding to it in the time domain Then continuoustime derivatives are approximated by finite dif ferences in discrete time and the resulting expression is a digitalfilter transfer function approximating the original analog filter transfer function For example suppose that H a s 1 s a Figure 1415 Magnitude frequency responses of the analog filter and its digital simulation by the stepinvariant method 2000 1000 0 1000 Ω radianssample 0 05 1 f Hz Bandpass Butterworth Analog Filter Order 2 Corners at 150 200 Hz 0 2 0 05 1 Bandpass Butterworth Digital Filter Step Invariant Sampling Rate 1000 samplessecond Hde jΩ Ha f Figure 1416 Polezero diagrams of the analog filter and its digital simulation by the stepinvariant method 2 σ ω s 1224077 997364 11994107 9772666 9772666 11994107 Rez Imz z 086698 03211 05062 099991 10012 082447 075028 075028 082447 rob28124ch14680734indd 698 041216 207 pm 143 Digital Filters 699 Since this is a transfer function it is the ratio of the response Yas to the excitation Xas Y a s X a s 1 s a Then Y a ss a X a s Taking the inverse Laplace transform of both sides d dt y a t a y a t x a t A derivative can be approximated by various finitedifference expressions and each choice has a slightly different effect on the approximation of the digital filter to the analog filter Let the derivative in this case be approximated by the forward difference d dt y a t y d n 1 y d n T s Then the differenceequation approximation to the differential equation is y d n 1 y d n T s a y d n x d n and the corresponding recursion relation is y d n 1 x d n T s 1 a T s y d n The digitalfilter transfer function can be found by z transforming the equation into z Y d z y d 0 T s X d z 1 a T s Y d z Transfer functions are computed based on the assumption that the system is initially in its zero state Therefore yd 0 0 and H d z Y d z X d z T s z 1 a T s 145 A block diagram realization of this filter is illustrated in Figure 1417 Figure 1417 Block diagram of a digitalfilter designed by approximating a differential equation with a difference equation using forward differences Ts 1 aTs D D xdn ydn The digital filter could also have been based on a backwarddifference approxima tion to the derivative d dt y a t y d n y d n 1 T s rob28124ch14680734indd 699 041216 207 pm C h a p t e r 14 Filter Analysis and Design 700 or a central difference approximation to the derivative d dt y a t y d n 1 y d n 1 2 T s We can systematize this method by realizing that every s in an sdomain expression represents a corresponding differentiation in the time domain d dt x a t s X a s again with the filter initially in its zero state We can approximate derivatives with forward backward or central differences d dt x a t x a t T s x a t T s x d n 1 x d n T s d dt x a t x a t x a t T s T s x d n x d n 1 T s or d dt x a t x a t T s x a t T s 2 T s x d n 1 x d n 1 2 T s The z transforms of these differences are x d n 1 x d n T s Z z 1 T s X d z x d n x d n 1 T s Z 1 z 1 T s X d z z 1 z T s X d z or x d n 1 x d n 1 2 T s Z z z 1 2 T s X d z z 2 1 2z T s X d z Now we can replace every s in an sdomain expression with the corresponding zdomain expression Then we can approximate the sdomain transfer function H a s 1 s a with a forwarddifference approximation to a derivative H d z 1 s a s z1 T s 1 z 1 T s a T s z 1 a T s 146 which is exactly the same as 145 This avoids the process of actually writing the differential equation and substituting a finite difference for each derivative There is one aspect of finitedifference digitalfilter design that must always be kept in mind It is possible to approximate a stable analog filter and create an unstable digital filter using this method Take the transfer function in 145 as an example It has a pole at z 1 aTs The analog filters pole is at s a If the analog filter is rob28124ch14680734indd 700 041216 207 pm 143 Digital Filters 701 stable a 0 and 1 aTs is at a location z Rez 1 on the real axis of the z plane If aTs is greater than or equal to two the zplane pole is outside the unit circle and the digital filter is unstable A digital filters transfer function can be expressed in partial fractions one for each pole and some poles may be complex A pole at location s s0 in the s plane maps into a pole at z 1 s0Ts in the z plane So the mapping s0 1 s0Ts maps the ω axis of the s plane into the line z 1 and the left half of the s plane into the region of the z plane to the left of z 1 For stability the poles in the z plane should be inside the unit circle Therefore this mapping does not guarantee a stable digitalfilter design The s0s are determined by the analog filter so we cannot change them Therefore to solve the insta bility problem we could reduce Ts which means we would increase the sampling rate If instead of using a forward difference we had used a backward difference in 146 we would have gotten the digitalfilter transfer function H d z 1 s a s z1 z T s 1 z 1 z T s a z T s z 1 az T s 1 1 a T s z T s z 11 a T s Now the pole is at z 11 aTs The mapping a 11 aTs maps positive values of a for stable analog filters into the real axis of the z plane between z 0 and z 1 The pole is inside the unit circle and the system is stable regardless of the values of a and Ts More generally if the analog filter has a pole at s s0 the digital filter has a pole at z 11 s0Ts This maps the ω axis in the s plane into a circle in the z plane of radius 12 centered at z 12 and maps the entire lefthalf of the s plane into the interior of that circle Figure 1418 Although this mapping of poles guarantees a stable digital filter from a sta ble analog filter it also restricts the type of digital filter that can be effectively de signed using this method Lowpass analog filters with poles on the negative real axis of the s plane become lowpass digital filters with poles on the real axis of the z plane in the interval 0 z 1 If the analog filter has poles at σ0 jω0 with ω0 σ0 meaning the analog filter is tuned to strongly respond at frequencies near ω0 and if ω0Ts 1 Figure 1418 Mapping z 11 s0Ts s z z 1 1 jω0Ts 1 ω0Ts 1 ω ω σ ω0Ts 1 s z 1 σ rob28124ch14680734indd 701 041216 207 pm C h a p t e r 14 Filter Analysis and Design 702 the zplane poles will not lie close to the unit circle and its response will not be nearly as strong near the equivalent discretetime frequency ExamplE 144 Bandpass filter design using the finitedifference method Using the differenceequation design method with a backward difference design a digital filter to simulate the analog filter of Example 142 whose transfer function is H a s 987 10 4 s 2 s 4 4443 s 3 2467 10 6 s 2 5262 10 8 s 1403 10 12 using the same sampling rate fs 1000 samplessecond Compare the frequency responses of the two filters If we choose the same sampling rate as in Example 142 fs 1000 the zdomain transfer function is H d z 0169 z 2 z 1 2 z 4 1848 z 3 1678 z 2 07609z 01712 The impulse responses magnitude frequency responses and polezero diagrams of the analog and digital filters are compared in Figure 1419 Figure 1420 and Figure 1421 Figure 1419 Impulse responses of the analog filter and its digital simulation using the finitedifference method 0 002 004 100 0 100 t s hat Analog Filter Impulse Response 0 20 40 001 0 001 n hd n Digital Filter Impulse ResponseFinite Difference Figure 1420 Magnitude frequency responses of the analog filter and its digital simulation using the finitedifference method 2000 1000 0 1000 0 05 1 f Hz Bandpass Butterworth Analog Filter Order 2 Corners at 150 200 Hz 2 0 2 0 002 004 006 Bandpass Butterworth Digital Filter Finite Difference Sampling Rate 1000 samplesseconds Ha f Hde jΩ Ω radianssample The digital filter impulse response does not look much like a sampling of the analog filter impulse response and the width of the digital filter passband is much too large Also the attenu ation at higher frequencies is very poor This result is much worse than the previous two designs rob28124ch14680734indd 702 041216 207 pm 143 Digital Filters 703 ExamplE 145 Lowpass filter design using the finitedifference method Using the differenceequation design method with a forward difference design a digital filter to simulate the analog filter whose transfer function is H a s 1 s 2 600s 4 10 5 using a sampling rate fs 500 samplessecond The zdomain transfer function is H d z 1 z 1 T s 2 600 z 1 T s 4 10 5 or H d z T s 2 z 2 600 T s 2z 1 600 T s 4 10 5 T s 2 or H d z 4 10 6 z 2 08z 14 This result looks quite simple and straightforward but the poles of this zdomain transfer function are outside the unit circle and the filter is unstable even though the sdomain transfer function is stable Stability can be restored by increasing the sampling rate or by using a backward difference Figure 1421 Polezero diagrams of the analog filter and its digital simulation using the finitedifference method 2 s 1224077 997364 11994107 9772666 9772666 11994107 2 2 Rez Imz z 041596 050809 04515 044449 044449 04515 ω σ rob28124ch14680734indd 703 041216 207 pm C h a p t e r 14 Filter Analysis and Design 704 FrequencyDomain Methods Direct Substitution and the Matched zTransform A different approach to the design of digital filters is to find a direct change of variable from s to z that maps the s plane into the z plane converts the poles and zeros of the sdomain transfer function into appropriate corresponding locations in the z plane and converts stable analog filters into stable digital filters The most common techniques that use this idea are the matchedz transform direct substitution and the bilinear transformation This type of design process produces an IIR filter Figure 1422 The direct substitution and matched ztransform methods are very similar These methods are based on the idea of simply mapping the poles and zeros of the sdomain transfer function into the z domain through the relationship z e s T s For example to transform the analog filter frequency response H d s 1 s a which has a pole at s a we simply map the pole at a to the corresponding loca tion in the z plane Then the digital filter pole location is e a T s The direct substitution method implements the transformation s a z e a T s while the matched ztransform method implements the transformation s a 1 e a T s z 1 The zdomain transfer functions that result in this case are Direct Substitution H d z 1 z e a T s z 1 1 e a T s z 1 with a pole at z e a T s and no finite zeros Matched zTransform H d z 1 1 e a T s z 1 z z e a T s with a pole at z e a T s and a zero at z 0 Notice that the matched ztransform result is exactly the same result as was obtained using the impulseinvariant method and the direct substitution result is the same except for a single sample delay due to the z 1 factor For more complicated sdomain transfer functions the results of these methods are not so similar These methods do not directly involve any timedomain analysis The design is done entirely in the s and z domains The transformations s a z eaT and s a 1 eaT z1 both map a pole in the open lefthalf of the s plane into a pole in the open interior of the unit circle in the z plane Therefore stable analog filters are transformed into stable digital filters Figure 1422 Mapping of poles and zeros from the s plane to the z plane s z Rez Imz σ ω rob28124ch14680734indd 704 041216 207 pm 143 Digital Filters 705 ExamplE 146 Digital bandpass filter design using the matchedz transform Using the matchedz transform design method design a digital filter to simulate the analog filter of Example 142 whose transfer function is H a s 987 10 4 s 2 s 4 4443 s 3 2467 10 6 s 2 5262 10 8 s 1403 10 12 using the same sampling rate fs 1000 samplessecond Compare the frequency responses of the two filters This transfer function has a double zero at s 0 and poles at s 997 j978 and at s 1224 j11986 Using the mapping s a 1 e aT z 1 we get a zdomain double zero at z 1 a double zero at z 0 and poles at z 05056 j07506 and 03217 j08242 and a zdomain transfer function H d z z 2 98700 z 2 197400z 98700 z 4 1655 z 3 2252 z 2 1319z 06413 or H d z 98700 z 2 z 1 2 z 4 1655 z 3 2252 z 2 1319z 06413 The impulse responses magnitude frequency responses and polezero diagrams of the analog and digital filters are compared in Figure 1423 Figure 1424 and Figure 1425 If this design had been done using the direct substitution method the only differences would be that the zeros at z 0 would be removed the impulse response would be the same except delayed two units in discrete time the magnitude frequency response would be exactly the same and the phase of the frequency response would have a negative slope with a greater magnitude Figure 1423 Impulse responses of the analog filter and its digital simulation by the matchedz transform method 0 002 004 100 0 100 t s hat Analog Filter Impulse Response 0 20 40 2 0 2 105 n hd n Digital Filter Impulse Response Matched zTransform rob28124ch14680734indd 705 041216 207 pm C h a p t e r 14 Filter Analysis and Design 706 Figure 1424 Frequency responses of the analog filter and its digital simulation by the matchedz transform method 2000 1000 0 1000 0 05 1 f Hz Bandpass Butterworth Analog Filter Order 2 Corners at 150 200 Hz 2 0 2 0 5 10 15 105 Ω radianssample Bandpass Butterworth Digital Filter Matched zTransform Sampling Rate 1000 samplessecond Ha f Hde jΩ Figure 1425 Polezero diagrams of the analog filter and its digital simulation by the matchedz transform method 2 s 1224077 997364 11994107 9772666 9772666 11994107 2 2 Rez Imz z 03211 05062 082447 075028 075028 082447 σ ω The Bilinear Method The impulseinvariant and stepinvariant design techniques try to make the digital filters discretetimedomain response match the corresponding analog filters continuoustimedomain response to a corresponding standard excitation Another way to approach digitalfilter design is to try to make the frequency response of the digital filter match the frequency response of the analog filter But just as a discretetimedomain response can never exactly match a continuoustimedomain response the frequency response of a digital filter cannot exactly match the frequency response of an analog filter One reason mentioned earlier that the frequency responses cannot exactly match is that the frequency response of a digital filter is inherently periodic When a sinusoidal continuoustime signal is sampled to create a sinusoidal discretetime excitation if the frequency of the continuoustime signal is changed by an integer multiple of the sampling rate the discretetime signal does not change at all The digital filter cannot tell the difference and responds the same way as it would respond to the original signal Figure 1426 rob28124ch14680734indd 706 041216 207 pm 143 Digital Filters 707 According to the sampling theorem if a continuoustime signal can be guaranteed never to have any frequency components outside the range f fs 2 then when it is sampled at the rate fs the discretetime signal created contains all the information in the continuoustime signal Then when the discretetime signal excites a digital filter the response contains all the information in a corresponding continuoustime signal So the design process becomes a matter of making the digitalfilter frequency response match the analog filter frequency response only in the frequency range f fs 2 not outside In general this still cannot be done exactly but it is often possible to make a good approximation Of course no signal is truly bandlimited Therefore in practice we must arrange to have very little signal power beyond half the sampling rate instead of no signal power Figure 1427 If a continuoustime excitation does not have any frequency components outside the range f fs 2 any nonzero response of an analog filter outside that range would have no effect because it has nothing to filter Therefore in the design of a digital filter to simulate an analog filter the sampling rate should be chosen such that the response of the analog filter at frequencies f fs 2 is approximately zero Then all the filtering action will occur at frequencies in the range f fs 2 So the starting point for a frequencydomain design process is to specify the sampling rate such that X f 0 and H a f 0 f f s 2 or X jω 0 and H a jω 0 ω π f s ω s 2 Now the problem is to find a digitalfilter transfer function that has approxi mately the same shape as the analog filter transfer function we are trying to simulate in the frequency range f fs 2 As discussed earlier the straightforward method to accomplish this goal would be to use the transformation e s T s z to convert a desired transfer function Has into the corresponding Hdz The transformation Figure 1427 Magnitude spectrum of a continuoustime signal and a discretetime signal formed by impulse sampling it f X f Xδ f fs fs fs 2 fs 2 fs 2 fs fs fs 2 f A Afs Figure 1426 Two identical discretetime signals formed by sampling two different sinusoids 1 2 3 4 5 6 7 8 9 10 n 1 x1n cos 2πn 5 1 1 2 3 4 5 6 7 8 9 10 n 1 1 x2n cos 12πn 5 rob28124ch14680734indd 707 041216 207 pm C h a p t e r 14 Filter Analysis and Design 708 e s T s z can be turned around into the form s lnzTs Then the design process would be H d z H a s s 1 T s lnz Although this development of the transformation method is satisfying from a theoretical point of view the functional transformation s lnzTs transforms an analog filter transfer function in the common form of the ratio of two polynomials into a digitalfilter transfer function which involves a ratio of polynomials not in z but rather in lnz making the function transcendental with infinitely many poles and zeros So although this idea is appealing it does not lead to a practical digitalfilter design At this point it is common to make an approximation in an attempt to simplify the form of the digitalfilter transfer function One such transformation arises from the series expression for the exponential function e x 1 x x 2 2 x 3 3 k0 x k k We can apply that to the transformation e s T s z yielding 1 s T s s T s 2 2 s T s 3 3 z If we approximate this series by the first two terms we get 1 sTs z or s z 1 T s The approximation e s T s 1 s T s is a good approximation if Ts is small and gets better as Ts gets smaller and of course fs gets larger That is this approximation becomes very good at high sampling rates Examine the transformation s z 1Ts A multiplication by s in the s domain corresponds to a differentiation with respect to t of the corresponding function in the continuoustime domain A multiplication by z 1Ts in the z domain corresponds to a forward difference divided by the sampling time Ts of the corresponding function in the discretetime domain This is a forwarddifference approximation to a derivative As mentioned in the finitedifference method the two operations multiplication by s and by z 1Ts are analogous So this method has the same problem as the finitedifference method using forward differences A stable analog filter can become an unstable digital filter A very clever modification of this transformation solves the problem of creating an unstable digital filter from a stable analog filter and at the same time has other advan tages We can write the transformation from the s domain to the z domain as e s T s e s T s 2 e s T s 2 z approximate both exponentials with an infinite series 1 s T s 2 s T s 2 2 2 s T s 2 3 3 1 s T s 2 s T s 2 2 2 s T s 2 3 3 z rob28124ch14680734indd 708 041216 207 pm 143 Digital Filters 709 and then truncate both series to two terms 1 s T s 2 1 s T s 2 z yielding s 2 T s z 1 z 1 or z 2 s T s 2 s T s This mapping from s to z is called the bilinear z transform because the numerator and denominator are both linear functions of s or z Dont get the terms bilinear and bilateral z transform confused The bilinear z transform transforms any stable analog filter into a stable digital filter because it maps the entire open left half of the s plane into the open interior of the unit circle in the z plane This was also true of matchedz transform and direct substitution but the correspondences are different The mapping z e s T s maps any strip ω0Ts ω ω0 2πTs of the s plane into the entire z plane The mapping from s to z is unique but the mapping from z to s is not unique The bilinear mapping s 2Tsz 1z 1 maps each point in the s plane into a unique point in the z plane and the inverse mapping z 2 sTs 2 sTs maps each point in the z plane into a unique point in the s plane To see how the mapping works consider the contour s jω in the s plane Setting z 2 sTs 2 sTs we get z 2 jω T s 2 jω T s 12 tan 1 ω T s 2 e j2 tan 1 ω T s 2 which lies entirely on the unit circle in the z plane Also the contour in the z plane is traversed exactly once for ω For the more general contour s σ0 jω σ0 a constant the corresponding contour is also a circle but with a different radius and centered on the Rez axis such that as ω approaches z approaches 1 Figure 1428 Figure 1428 Mapping of an splane region into a corre sponding zplane region through the bilinear z transform 20 10 10 20 s z Ts 1 2 2 a a b b c c d d f f e e 1 1 1 1 ω σ rob28124ch14680734indd 709 041216 207 pm C h a p t e r 14 Filter Analysis and Design 710 As the contours in the s plane move to the left the contours in the z plane become smaller circles whose centers move closer to the z 1 point The mapping from s to z is a onetoone mapping but the distortion of regions becomes more and more severe as s moves away from the origin A higher sampling rate brings all poles and zeros in the s plane nearer to the z 1 point in the z plane where the distortion is minimal That can be seen by taking the limit as Ts approaches zero In that limit z approaches 1 The important difference between the bilinear z transform method and the impulse invariant or matched ztransform methods is that there is no aliasing using the bilinear z transform because of the unique mapping between the s and z planes However there is a warping that occurs because of the way the s jω axis is mapped into the unit circle z 1 and vice versa Letting z e jΩ Ω real determines the unit circle in the z plane The corresponding contour in the s plane is s 2 T s e jΩ 1 e jΩ 1 j 2 T s tan Ω 2 and since s σ jω σ 0 and ω 2Ts tanΩ2 or inverting the function Ω 2 tan1ωTs 2 Figure 1429 Figure 1429 Frequency warping caused by the bilinear transformation ωTs 20 20 Ω π π For low frequencies the mapping is almost linear but the distortion gets progres sively worse as we increase frequency because we are forcing high frequencies ω in the s domain to fit inside the range π Ω π in the z domain This means that the as ymptotic behavior of an analog filter as f or ω approaches positive infinity occurs in the z domain at Ω π which through Ω ωTs 2π f Ts is at f fs 2 the Nyquist frequency Therefore the warping forces the full infinite range of continuoustime frequencies into the discretetime frequency range π Ω π with a nonlinear invertible function thereby avoiding aliasing The MATLAB signal toolbox has a command bilinear for designing a digital filter using the bilinear transformation The syntax is bdad bilinearbaaafs or zdpdkd bilinearzapakafs where ba is a vector of numerator coefficients in the analog filter transfer function aa is a vector of denominator coefficients in the analog filter transfer function bd is a vector of numerator coefficients in the digitalfilter transfer function ad is a vector of denominator coefficients in the digitalfilter transfer function za is a vector of analog filter zero locations pa is a vector of analog filter pole locations ka is the gain factor rob28124ch14680734indd 710 041216 207 pm 143 Digital Filters 711 of the analog filter fs is the sampling rate in samplessecond zd is a vector of digital filter zero locations pd is a vector of digital filter pole locations and kd is the gain factor of the digital filter For example za pa 10 ka 1 fs 4 zdpdkd bilinearzapakafs zd zd 1 pd pd 01111 kd kd 00556 ExamplE 147 Comparison of digital lowpass filter designs using the bilinear transformation with different sampling rates Using the bilinear transformation design a digital filter to approximate the analog filter whose transfer function is H a s 1 s 10 and compare the frequency responses of the analog and digital filters for sampling rates of 4 Hz 20 Hz and 100 Hz Using the transformation s 2 T s z 1 z 1 H d z 1 2 T s z 1 z 1 10 T s 2 10 T s z 1 z 2 10 T s 2 10 T s For a 4 samplessecond sampling rate H d z 1 18 z 1 z 1 9 For a 20 samplessecond sampling rate H d z 1 50 z 1 z 3 5 For a 100 samplessecond sampling rate H d z 1 210 z 1 z 19 21 Figure 1430 rob28124ch14680734indd 711 041216 207 pm C h a p t e r 14 Filter Analysis and Design 712 ExamplE 148 Digital bandpass filter design using the bilinear transformation Using the bilinearz transform design method design a digital filter to simulate the analog filter of Example 142 whose transfer function is H a s 987 10 4 s 2 s 4 4443 s 3 2467 10 6 s 2 5262 10 8 s 1403 10 12 using the same sampling rate fs 1000 samplessecond Compare the frequency responses of the two filters Using the transformation s 2Tsz 1z 1 and simplifying H d z 1238 z 4 2477 z 2 1238 z 4 1989 z 3 2656 z 2 1675z 0711 or H d z 1238 z 1 2 z 1 2 z 4 1989 z 3 2656 z 2 1675z 0711 The impulse responses magnitude frequency responses and polezero diagrams of the analog and digital filters are compared in Figure 1431 Figure 1432 and Figure 1433 Hde j2π fTs f 4 4 f 4 4 f 20 20 f 20 20 Hdej2π fTs 01 01 Ha f Ha f 01 01 fs 20 fs 4 f 100 100 f 100 100 Hdej2π fTs 01 01 Ha f fs 100 Figure 1430 Magnitude frequency responses of the analog filter and three digitalfilters designed using the bilinear transform and three different sampling rates Figure 1431 Impulse responses of the analog filter and its digital simulation by the bilinearz transform method 0 002 004 100 0 100 t s hat Analog Filter Impulse Response 0 20 40 100 0 100 n hd n Digital Filter Impulse ResponseBilinear rob28124ch14680734indd 712 041216 207 pm 143 Digital Filters 713 FIR Filter Design Truncated Ideal Impulse Response Even though the commonly used analog filters have infiniteduration impulse responses because they are stable systems their impulse responses approach zero as time t approaches positive infinity Therefore another way of simulating an analog filter is to sample the impulse response as in the impulse invariant design method but then truncate the impulse response beginning at discrete time n N where it has fallen to some low level creating a finiteduration impulse response Figure 1434 Digital filters that have finiteduration impulse responses are called FIR filters The technique of truncating an impulse response can also be extended to approxi mating noncausal filters If the part of an ideal filters impulse response that occurs be fore time t 0 is insignificant in comparison with the part that occurs after time t 0 then it can be truncated forming a causal impulse response It can also be truncated after some later time when the impulse response has fallen to a low value as previously described Figure 1435 2000 1000 0 1000 0 05 1 f Hz Bandpass Butterworth Analog Filter Order 2 Corners at 150 200 Hz 2 0 2 0 500 1000 Ω radianssample Bandpass Butterworth Digital Filter Bilinear Sampling Rate 1000 samplessecond Ha f Hde jΩ Figure 1432 Magnitude frequency responses of the analog filter and its digital simulation by the bilinear method Figure 1433 Polezero diagrams of the analog filter and its digital simulation by the bilinearz transform method 2 2 2 σ ω s 1224077 997364 11994107 9772666 9772666 11994107 Rez Imz z 042846 056582 080725 072877 072877 080725 rob28124ch14680734indd 713 041216 207 pm C h a p t e r 14 Filter Analysis and Design 714 Of course the truncation of an IIR response to an FIR response causes some difference between the impulse and frequency responses of the ideal analog and actual digital filters but that is inherent in digitalfilter design So the problem of digitalfilter design is still an approximation problem The approximation is just done in a different way in this design method Once the impulse response has been truncated and sampled the design of an FIR filter is quite straightforward The discretetime impulse response is in the form of a finite summation of discretetime impulses h N n m0 N1 a m δn m and can be realized by a digital filter of the form illustrated in Figure 1436 Figure 1435 Truncation of a noncausal impulse response to a causal FIR impulse response N hat hdn hNn t n n Figure 1434 Truncation of an IIR impulse response to an FIR impulse response hat hdn hNn t n n N Figure 1436 Prototypical FIR filter D xn yn a0 a1 D a2 D aN1 rob28124ch14680734indd 714 041216 207 pm 143 Digital Filters 715 One essential difference between this type of filter design and all the others presented so far is that there is no feedback of the response to combine with the exci tation to produce the next response This type of filter has only feedforward paths Its transfer function is H d z m0 N1 a m z m This transfer function has N 1 poles all located at z 0 and is absolutely stable regardless of the choice of the coefficients a This type of digital filter is an approximation to an analog filter It is obvious what the difference between the two impulse responses is but what are the differences in the frequency domain The truncated impulse response is h N n h d n 0 n N 0 otherwise h d nwn and the DTFT is H N e jΩ H d e jΩ W e jΩ Figure 1437 As the nonzero length of the truncated impulse increases the frequency response approaches the ideal rectangular shape The similarity in appearance to the conver gence of a CTFS is not accidental A truncated CTFS exhibits the Gibbs phenomenon in the reconstructed signal In this case the truncation occurs in the continuoustime domain and the ripple which is the equivalent of the Gibbs phenomenon occurs in the frequency domain This phenomenon causes the effects marked as passband rip ple and as side lobes in Figure 1437 The peak amplitude of the passband ripple does not diminish as the truncation time increases but it is more and more confined to the region near the cutoff frequency We can reduce the ripple effect in the frequency domain without using a longer truncation time by using a softer truncation in the time domain Instead of window ing the original impulse response with a rectangular function we could use a differently shaped window function that does not cause such large discontinuities in the truncated impulse response There are many window shapes whose Fourier transforms have less ripple than a rectangular windows Fourier transform Some of the most popular are the following 1 von Hann or Hanning wn 1 2 1 cos 2πn N 1 0 n N 2 Bartlett w n 2n N 1 0 n N 1 2 2 2n N 1 N 1 2 n N rob28124ch14680734indd 715 041216 207 pm C h a p t e r 14 Filter Analysis and Design 716 3 Hamming wn 054 046 cos 2πn N 1 0 n N 4 Blackman wn 042 05 cos 2πn N 1 008 cos 4πn N 1 0 n N Figure 1437 Three truncated ideallowpassfilter discretetime impulse responses and their associated magnitude frequency responses N n hNn 02 04 1 Passband Ripple Ω Ω 2π HNejΩ 2π 2π 2π Passband Transition Band 100 Side Lobes Side Lobes HNe jΩdB N n hNn 02 04 1 Passband Ripple Ω 2π HNejΩ 2π 2π 2π Ω 100 Side Lobes Side Lobes HNe jΩdB n hNn 02 04 1 Passband Ripple Ω 2π HNe jΩ 2π 2π 2π Passband Transition Band Ω 100 Side Lobes Side Lobes N HNe jΩdB Passband Transition Band rob28124ch14680734indd 716 041216 207 pm 143 Digital Filters 717 5 Kaiser wn I 0 ω a N 1 2 2 n N 1 2 2 I 0 ω a N 1 2 where I0 is the modified zeroth order Bessel function of the first kind and ωa is a pa rameter that can be adjusted to trade off between transitionband width and sidelobe amplitude Figure 1438 Figure 1438 Window functions N 32 wn n 1 n wn n 1 Rectangular n 1 wn n 1 wn n 1 wn von Hann Bartlett Hamming Blackman 1 wn 1 wn 1 1 wn wn n Kaiser ωa Kaiser ωa n n Kaiser ωa Kaiser ωa 1 1 8 1 4 1 2 The transforms of these window functions determine how the frequency response will be affected The magnitudes of the transforms of these common window functions are illustrated in Figure 1439 Looking at the magnitudes of the transforms of the window functions it is ap parent that for a fixed N two design goals are in conflict When approximating ideal filters with FIR filters we want a very narrow transition band and very high attenua tion in the stopband The transfer function of the FIR filter is the convolution of the ideal filters transfer function with the transform of the window function So the ideal window function would have a transform that is an impulse and the corresponding window function would be an infinitewidth rectangle That is impossible so we must compromise If we use a finitewidth rectangle the transform is the Dirichlet function and we get the transform illustrated in Figure 1439 for a rectangle which makes a relatively fast transition from the peak of its central lobe to its first null but then the sinc function rises again to a peak that is only about 13 dB below the maximum When we convolve it with an ideal lowpass filter frequency response the transition band rob28124ch14680734indd 717 041216 207 pm C h a p t e r 14 Filter Analysis and Design 718 is narrow compared to the other windows but the stopband attenuation is not very good Contrast that with the Blackman window The central lobe width of its transform magnitude is more than twice that of the rectangle so the transition band will not be as narrow But once the magnitude goes down it stays more than 60 dB down So its stopband attenuation is much better Another feature of an FIR filter that makes it attractive is that it can be designed to have a linear phase response The general form of the FIR impulse response is h d n h d 0 δn h d 1 δn 1 h d N 1 δn N 1 its z transform is H d z h d 0 h d 1 z 1 h d N 1 z N1 and the corresponding frequency response is H d e jΩ h d 0 h d 1 e jΩ h d N 1 e jN1Ω The length N can be even or odd First let N be even and let the coefficients be chosen such that h d 0 h d N 1 h d 1 h d N 2 h d N2 1 h d N2 Figure 1440 We jΩdB 120 120 120 Rectangular We jΩdB Bartlett We jΩdB We jΩdB 200 We jΩdB We jΩdB 200 200 π π 200 We jΩdB Kaiser ωa 1 π π π π π π π π π π π π We jΩdB 120 120 von Hann We jΩdB Hamming Blackman π π π π Kaiser ωa 1 8 Kaiser ωa 1 4 Kaiser ωa 1 2 Ω Ω Ω Ω Ω Ω Ω Ω Ω Figure 1439 Magnitudes of z transforms of window functions N 32 Figure 1440 Example of a symmetric discretetime impulse response for N 8 N 8 n N hNn Center Point rob28124ch14680734indd 718 041216 207 pm 143 Digital Filters 719 This type of impulse response is symmetric about its center point Then we can write the frequency response as H d e jΩ h d 0 h d 0 e j N1 Ω h d 1 e jΩ h d 1 e jN2Ω h d N2 1 e jN21Ω h d N2 1 e jNΩ2 or H d e jΩ e j N1 2 Ω h d 0 e j N1 2 Ω e j N1 2 Ω h d 1 e j N3 2 Ω e j N3 2 Ω h d N2 1 e jΩ e jΩ or H d e jΩ 2 e j N1 2 Ω h d 0 cos N 1 2 Ω h d 1 cos N 3 2 Ω h d N2 1 cos Ω This frequency response consists of the product of a factor ejN 12Ω that has a linear phase shift with frequency and some other factors which have real values for all Ω Therefore the overall frequency response is linear with frequency except for jumps of π radians at frequencies at which the sign of the real part changes In a similar manner it can be shown that if the filter coefficients are antisymmetric meaning h d 0 h d N 1 h d 1 h d N 2 h d N2 1 h d N2 then the phase shift is also linear with frequency For N odd the results are similar If the coefficients are symmetric h d 0 h d N 1 h d 1 h d N 2 h d N 3 2 h d N 1 2 or antisymmetric h d 0 h d N 1 h d 1 h d N 2 h d N 3 2 h d N 1 2 h d N 1 2 0 the phase frequency response is linear Notice that in the case of N odd there is a cen ter point and if the coefficients are antisymmetric the center coefficient hdN 12 must be zero Figure 1441 rob28124ch14680734indd 719 041216 207 pm C h a p t e r 14 Filter Analysis and Design 720 ExamplE 149 Digital lowpass FIR filter design by truncating the ideal impulse response Using the FIR method design a digital filter to approximate a singlepole lowpass analog filter whose transfer function is H a s a s a Truncate the analog filter impulse response at three time constants and sample the truncated impulse response with a time between samples that is onefourth of the time constant forming a discretetime function Then divide that discretetime function by a to form the discretetime impulse response of the digital filter a Find and graph the magnitude frequency response of the digital filter versus discretetime radian frequency Ω b Repeat part a for a truncation time of five time constants and a sampling rate of 10 samples per time constant The impulse response is hat aeat ut The time constant is 1a Therefore the truncation time is 3a the time between samples is 14a and the samples are taken at discrete times 0 n 12 The FIR impulse response is then h d n a e n4 un un 12 a m0 11 e m4 δn m The zdomain transfer function is H d z a m0 11 e m4 z m and the frequency response is H d e jΩ a m0 11 e m4 e jΩ m a m0 11 e m14jΩ Figure 1441 Examples of symmetric and antisymmetric discretetime impulse responses for N even and N odd N 8 N 8 N 7 N 7 Symmetric Antisymmetric N Even N n N Odd hNn Center Point N n hNn Center Point N n hNn Center Point N n hNn Center Point rob28124ch14680734indd 720 041216 207 pm 143 Digital Filters 721 For the second sampling rate in part b the truncation time is 5a the time between sam ples is 110a and the samples are taken at discrete times 0 n 50 The FIR impulse response is then h d n a e n10 un un 50 a m0 49 e m4 δn m The zdomain transfer function is H d z a m0 49 e m10 z m and the frequency response is H d e jΩ a m0 49 e m10 e jΩ m a m0 49 e m110jΩ Figure 1442 The effects of truncation of the impulse response are visible as the ripple in the frequency response of the first FIR design with the lower sampling rate and shorter truncation time ExamplE 1410 Communicationchannel digitalfilter design A range of frequencies between 900 MHz and 905 MHz is divided into 20 equalwidth channels in which wireless signals may be transmitted To transmit in any of the channels a transmitter Figure 1442 Impulse responses and frequency responses for the two FIR designs hdn hdn a a HdejΩ HdejΩ 4a Hde jΩ HdejΩ π π 3 Samples per Time Constant Truncation Time 3 Time Constants 10a 10 Samples per Time Constant Truncation Time 5 Time Constants n 5 20 n 5 60 2π π π 2π 2π 2π 2π 2π 2π 2π Ω Ω Ω Ω rob28124ch14680734indd 721 041216 207 pm C h a p t e r 14 Filter Analysis and Design 722 must send a signal whose amplitude spectrum fits within the constraints of Figure 1443 The transmitter operates by modulating a sinusoidal carrier whose frequency is the center frequency of one of the channels with the baseband signal Before modulating the carrier the baseband signal which has an approximately flat spectrum is prefiltered by an FIR filter which ensures that the transmitted signal meets the constraints of Figure 1443 Assuming a sampling rate of 2 million samplessecond design the filter We know the shape of the ideal baseband analog lowpass filters impulse response h a t 2A f m sinc2 f m t t 0 where fm is the corner frequency The sampled impulse response is h d n 2A f m sinc2 f m n T s t 0 We can set the corner frequency of the ideal lowpass filter to about halfway between 100 kHz and 125 kHz say 115 kHz or 575 of the sampling rate Let the gain constant A be one The time between samples is 05 μs The filter will approach the ideal as its length approaches infinity As a first try set the meansquared difference between the filters impulse response and the ideal filters impulse response to be less than 1 and use a rectangular window We can iteratively determine how long the filter must be by computing the meansquared difference between the filter and a very long filter Enforcing a meansquared error of less than 1 sets a filter length of 108 or more This design yields the frequency responses of Figure 1444 Figure 1443 Specification for spectrum of the transmitted signal fc 05 dB 0 dB 40 dB 250 kHz 200 kHz Figure 1444 Frequency response of an FIR filter with a rectangular window and less than 1 error in impulse response Lowpass Rectangular Window f 200000 f 200000 Hdej2πf TsdB 140 Hdej2πfTsdB 1 1 Pass Stop Pass Stop rob28124ch14680734indd 722 041216 207 pm 143 Digital Filters 723 This design is not good enough The passband ripple is too large and the stopband atten uation is not great enough We can reduce the ripple by using a different window Lets try a Blackman window with every other parameter remaining the same Figure 1445 Figure 1445 Frequency response of an FIR filter with a Blackman window and less than 1 error in impulse response Lowpass Blackman Window f 200000 f 200000 Hdej2πf TsdB 140 Hdej2πf TsdB 1 1 Pass Stop Pass Stop Figure 1446 Frequency response of an FIR filter with a Blackman window and less than 025 error in impulse response Lowpass Blackman Window f 200000 f 200000 Hde j2πf TsdB 140 Hde j2πf TsdB 1 1 Pass Stop Pass Stop This design is also inadequate We need to make the meansquared error smaller Making the meansquared error less than 025 sets a filter length of 210 and yields the magnitude fre quency response in Figure 1446 This filter meets specifications The stopband attenuation just barely meets the specifi cation and the passband ripple easily meets specification This design is by no means unique Many other designs with slightly different corner frequencies meansquared errors or windows could also meet the specification Optimal FIR Filter Design There is a technique for designing filters without win dowing impulse responses or approximating standard analog filter designs It is called ParksMcClellan optimal equiripple design and was developed by Thomas W Parks and James H McClellan in the early 1970s It uses an algorithm developed in 1934 by Evgeny Yakovlevich Remez called the Remez exchange algorithm An explanation of the method is beyond the scope of this text but it is important enough that students should be aware of it and able to use it to design digital filters rob28124ch14680734indd 723 041216 207 pm C h a p t e r 14 Filter Analysis and Design 724 The ParksMcClellan digitalfilter design is implemented in MATLAB through the command firpm with the syntax B firpmNFA where B is a vector of N1 real symmetric coefficients in the impulse response of the FIR filter which has the best approximation to the desired frequency response described by F and A F is a vector of frequency band edges in pairs in ascending order between 0 and 1 with 1 corresponding to the Nyquist frequency or half the sampling frequency At least one frequency band must have a nonzero width A is a real vector the same size as F which specifies the desired amplitude of the frequency response of the resultant filter B The desired response is the line connecting the points FkAk and Fk1Ak1 for odd k firpm treats the bands between Fk1 and Fk2 for odd k as transition bands Thus the desired amplitude is piecewise linear with transition bands This description serves only as an introduction to the method More detail can be found in MATLABs help description ExamplE 1411 ParksMcClellan design of a digital bandpass filter Design an optimal equiripple FIR filter to meet the magnitude frequency response specification in Figure 1447 Figure 1447 Bandpass filter specification Ω 06 07 22 23 25 dB 05 dB 05 dB Hde jΩdB Figure 1448 Frequency response of an optimal equiripple FIR bandpass filter with N 70 0 05 1 15 2 25 3 40 20 0 1 15 2 2 1 0 1 2 0 1 2 3 30 25 20 Hde jΩdB Hde jΩdB Hde jΩdB Ω Ω Ω The band edges are at Ω 006072223π and the desired amplitude responses at those band edges are A 001100 The vector F should therefore be F Ωπ 001910222807003073211 After a few choices of N it was found that a filter with N 70 met the specification Figure 1448 rob28124ch14680734indd 724 041216 207 pm 143 Digital Filters 725 MATLAB Design Tools In addition to the MATLAB features already mentioned in earlier chapters and in earlier sections of this chapter there are many other commands and functions in MATLAB that can help in the design of digital filters Probably the most generally useful function is the function filter This is a function that actually digitally filters a vector of data representing a finitetime piece of a discretetime signal The syntax is y filterbdadx where x is the vector of data to be filtered and bd and ad are vectors of coefficients in the recursion relation for the filter The recursion relation is of the form ad1yn bd1xn bd2xn 1 bdnb1xn nb ad2yn 1 adna1yn na written in MATLAB syntax which uses for arguments of all functions without making a distinction between continuoustime and discretetime functions A related function is filtfilt It operates exactly like filter except that it filters the data vector in the normal sense and then filters the resulting data vector backward This makes the phase shift of the overall filtering operation identically zero at all frequencies and doubles the magnitude effect in dB of the filtering operation There are four related functions each of which designs a digital filter The function butter designs an Nth order lowpass Butterworth digital filter through the syntax bdad butterNwn where N is the filter order and wn is the corner frequency expressed as a fraction of half the sampling rate not the sampling rate itself The function returns filter coefficients bd and ad which can be used directly with filter or filtfilt to filter a vector of data This function can also design a bandpass Butterworth filter simply by making wn a row vector of two corner frequencies of the form w1w2 The passband of the filter is then w1 w w2 in the same sense of being fractions of half the sampling rate By adding a string high or stop this function can also design highpass and bandstop digital filters Examples bdad butter301 lowpass thirdorder Butterworth filter corner frequency 05fs bdad butter401 02 bandpass fourthorder Butterworth filter corner frequencies of 005fs and 01fs bdad butter4002high highpass fourthorder Butterworth filter corner frequency 01fs bdad butter2032 034stop bandstop secondorder Butterworth filter corner frequencies 016fs and 017fs There are also alternate syntaxes for butter Type help butter for details It can also be used to do analog filter design The other three related digitalfilter design functions are cheby1 cheby2 and ellip They design Chebyshev and Elliptical filters Chebyshev and Elliptical filters have a narrower transition region for the same filter order than Butterworth filters but do so at the expense of passband andor stopband ripple Their syntax is similar except that maximum allowable ripple in dB must also be specified in the passband and the minimum attenuation in dB must be specified in the stop band Several standard window functions are available for use with FIR filters They are bartlett blackman boxcar rectangular chebwin Chebyshev ham ming hanning von Hann kaiser and triang similar to but not identical to bartlett rob28124ch14680734indd 725 041216 207 pm C h a p t e r 14 Filter Analysis and Design 726 The function freqz finds the frequency response of a digital filter in a manner similar to the operation of the function freqs for analog filters The syntax of freqz is HW freqzbdadN where H is the complex frequency response of the filter W is a vector of discretetime frequencies in radians not radians per second because it is a discretetime frequency at which H is computed bd and ad are vectors of coefficients of the numerator and de nominator of the digitalfilter transfer function and N is the number of points The function upfirdn changes the sampling rate of a signal by upsampling FIR filtering and down sampling Its syntax is y upfirdnxhpq where y is the signal resulting from the change of sampling rate x is the signal whose sampling rate is to be changed h is the impulse response of the FIR filter p is the factor by which the signal is upsampled by zero insertion before filtering and q is the factor by which the signal is downsampled decimated after filtering These are by no means all of the digital signal processing capabilities of MATLAB Type help signal for other functions ExamplE 1412 Filtering a discretetime pulse with a highpass Butterworth filter using MATLAB Digitally filter the discretetime signal xn un un 10 with a thirdorder highpass digital Butterworth filter whose discretetime radian corner fre quency is π6 radians Use 30 points to represent the excitation x and the response y N 30 Generate the excitation signal n 0N 1 x uDTn uDTn 10 Design a thirdorder highpass digital Butterworth filter bdad butter316high Filter the signal y filterbdadx The excitation and response are illustrated in Figure 1449 rob28124ch14680734indd 726 041216 207 pm Exercises with Answers 727 144 SUMMARY OF IMPORTANT POINTS 1 The Butterworth filter is maximally flat in both the pass and stopbands and all its poles lie on a semicircle in the left half of the s plane 2 A lowpass Butterworth filter can be transformed into a highpass bandpass or bandstop filter by appropriate variable changes 3 Chebyshev Elliptic and Bessel filters are filters optimized on a different basis than Butterworth filters They can also be designed as lowpass filters and then transformed into highpass bandpass or bandstop filters 4 One popular design technique for digital filters is to simulate a proven analog filter design 5 Two broad classifications of digital filters are inifiniteduration impulse response IIR and finiteduration impulse response FIR 6 The most popular types of IIR digitalfilter design are the impulse invariant step invariant finite difference direct substitution matched z and bilinear methods 7 FIR filters can be designed by windowing ideal impulse responses or by the ParksMcClellan equiripple algorithm EXERCISES WITH ANSWERS Answers to each exercise are in random order ContinuousTime Filters 1 Using only a calculator find the transfer function of a thirdorder n 3 lowpass Butterworth filter with cutoff frequency ω c 1 and unity gain at zero frequency Answer 1 s 3 2 s 2 2s 1 Figure 1449 Excitation and response of a thirdorder highpass digital Butterworth filter xn 1 1 yn 1 1 30 n 30 n rob28124ch14680734indd 727 041216 207 pm C h a p t e r 14 Filter Analysis and Design 728 2 Using MATLAB find the transfer function of an eighthorder lowpass Butterworth filter with cutoff frequency ω c 1 and unity gain at zero frequency Answer 1 s 8 5126 s 7 131371 s 6 218462 s 5 256884 s 4 218462 s 3 131371 s 2 5126s 1 3 What are the numerical splane finite pole and finite zero locations for Butterworth filters of order N with corner frequencies f c in Hz or ω c in radians second For repeated poles or zeros list them multiple times a Lowpass N 2 ω c 25 b Highpass N 2 f c 5 Answers Poles at 10π e j3π4 222 1 j and Double Zero at zero Poles at 25 e j3π4 1768 1 j No finite zeros 4 Using MATLAB design Chebyshev Type 1 and Elliptic fourthorder analog highpass filters with a cutoff frequency of 1 kHz Let the allowed ripple in the passband be 2 dB and let the minimum stopband attenuation be 60 dB Graph the magnitude Bode diagram of their frequency responses on the same scale for comparison How wide is the transition band for each filter Answers The transition band for the Elliptic filter goes from 445 Hz to 1000 Hz for a width of 555 Hz The transition band for the Chebyshev Type 1 filter goes from 274 Hz to 1000 Hz for a width of 726 Hz 101 102 103 104 105 160 140 120 100 80 60 40 20 0 Frequency f Hz Chebyshev Elliptic Ha f dB FiniteDifference Filter Design 5 What is the transition from s to z that is used in the differenceequation design technique to approximate a backward difference Answer s 1 z 1 T s rob28124ch14680734indd 728 041216 207 pm Exercises with Answers 729 6 Using the finitedifference technique with backward differences design a digital filter that approximates the lowpass filter whose transfer function is H s 1 s 1 Is there a finite sampling rate for which the digital filter is unstable If so provide one Answer z T s z 1 T s 1 Filter is absolutely stable 7 Using the finitedifference method and all backward differences design digital filters to approximate analog filters with these transfer functions In each case if a sampling frequency is not specified choose a sampling frequency that is 10 times the magnitude of the distance of the farthest pole or zero from the origin of the s plane Graphically compare the step responses of the digital and analog filters a H a s s f s 1 MHz b H a s 1s f s 1 kHz c H a s 2 s 2 3s 2 Answers t n 1 1 103 h1dn h1dn h1dn h1at h1at h1at t n 1 106 7 t 02 12 n 25 12 Matched zTransform and Direct Substitution Filter Design 8 A continuoustime filter with transfer function H s 10 s 8 s 2 7s 12 is approximated by a digital filter using the matched ztransform technique It is desired that all the zeros and poles of the digitalfilters transfer function H z be located within a distance of 02 of the z 1 point in the z plane What is the minimum numerical sampling rate required Answer 4388 9 A digitalfilter designed by the matched ztransform technique using a sampling rate of 10 samplessecond has a pole at z 05 The sampling rate is changed to 50 samplessecond What is the new numerical location of the pole rob28124ch14680734indd 729 041216 207 pm C h a p t e r 14 Filter Analysis and Design 730 Answer 08706 t 14 h1at 005 02 n 12 h1dn 05 25 Bilinear zTransform Filter Design 10 The transfer function H s s 2 s s 4 is approximated by a digitalfilter designed using the bilinear transformation with a sampling rate of 10 samples second The digital filter can be realized by a block diagram of the form shown in Figure E10 Enter into the block diagram the numbers in the empty rectangles Xz Yz z1 z1 Figure E10 Answers 1 01667 0667 00375 04125 008333 FIR Filter Design 11 Using a rectangular window of width 50 and a sampling rate of 10000 samples second design an FIR digital filter to approximate the analog filter whose transfer function is H a s 2000s s 2 2000s 2 10 6 Compare the frequency responses of the analog and digital filters Answers h d n 2000 2 09048 n cos 01n 07854 u n u n 50 5000 Frequency f Hz Analog 0 0 02 1000 2000 3000 4000 04 06 08 1 Ha f Hde jΩ 35 Frequency Ω Digital 00 2000 05 1 15 2 25 3 4000 6000 8000 10000 12000 rob28124ch14680734indd 730 041216 207 pm Exercises without Answers 731 5000 Frequency f Hz 0 0 1 2 1 05 2 1 1000 2000 3000 4000 35 Frequency Ω 0 05 1 15 2 25 3 Ha f Hde jΩ 0 05 1 12 Design a digital filter approximation to each of these ideal analog filters by sampling a truncated version of the impulse response and using the specified window In each case choose a sampling frequency which is 10 times the highest frequency passed by the analog filter Choose the delays and truncation times such that no more than 1 of the signal energy of the impulse response is truncated Graphically compare the magnitude frequency responses of the digital and ideal analog filters using a dB magnitude scale versus linear frequency Answers a LowpassRectangular Window Hde j2πf TsdB Hde j2πf TsdB f 5 120 f 1 5 f 5 Hde j2πfTsdB Hde j2πfTsdB 20 120 5 f 1 b LowpassVon Hann Window DigitalFilter Design Method Comparison 13 A continuoustime filter has a transfer function Hs 4 s 2 s s 2 It is approximated by three digitalfilter design methods matched ztransform direct substitution and bilinear z transform using a sampling rate f s 2 What are the numerical pole and zero locations of these digital filters Answers 2718 1 0368 3 1 1 0333 0 2718 1 0368 EXERCISES WITHOUT ANSWERS Analog Filter Design 14 Among Butterworth Chebyshev Type 1 Chebyshev Type 2 and Elliptic filters identify which type or types of lowpass filter have the characteristic described a Magnitude frequency response in the passband monotonically approaches a slope of zero at zero frequency Monotonic means always moving in the rob28124ch14680734indd 731 041216 207 pm C h a p t e r 14 Filter Analysis and Design 732 same direction That is always moving up or always moving down or not rippling b Magnitude frequency response in the stopband monotonically approaches zero as frequency approaches infinity c Fastest transition from passband to stopband for a given filter order d Slowest transition from passband to stopband for a given filter order 15 Thermocouples are used to measure temperature in many industrial processes A thermocouple is usually mechanically mounted inside a thermowell a metal sheath which protects it from damage by vibration bending stress or other forces One effect of the thermowell is that its thermal mass slows the effective time response of the thermocouplethermowell combination compared with the inherent time response of the thermocouple alone Let the actual temperature on the outer surface of the thermowell in Kelvins be Τst and let the voltage developed by the thermocouple in response to temperature be vt t The response of the thermocouple to a oneKelvin step change in the thermowell outersurface temperature from T 1 to T 1 1 is vt t K T1 1 e t 02 ut where K is the thermocouple temperaturetovoltage conversion constant a Let the conversion constant be K 40 µVK Design an active filter that processes the thermocouple voltage and compensates for its time lag making the overall system have a response to a oneKelvin step thermowellsurface temperature change that is itself a step of voltage of 1 mV b Suppose that the thermocouple also is subject to electromagnetic interference EMI from nearby highpower electrical equipment Let the EMI be modeled as a sinusoid with an amplitude of 20 µV at the thermocouple terminals Calculate the response of the thermocouple filter combination to EMI frequencies of 1 Hz 10 Hz and 60 Hz How big is the apparent temperature fluctuation caused by the EMI in each case 16 Design a Chebyshev Type 2 bandpass filter of minimum order to meet these specifications Passband 4 kHz to 6 kHz Gain between 0 dB and 2 dB Stopband 3 kHz and 8 kHz Attenuation 60 dB What is the minimum order Make a Bode diagram of its magnitude and phase frequency response and check it to be sure the pass and stopband specifications are met Make a polezero diagram What is the time of occurrence of the peak of its impulse response ImpulseInvariant and StepInvariant Filter Design 17 Using the impulseinvariant design method design a discretetime system to approximate the continuoustime systems with these transfer functions at the rob28124ch14680734indd 732 041216 207 pm Exercises without Answers 733 sampling rates specified Compare the impulse and unitstep or sequence responses of the continuoustime and discretetime systems a Has 712s s 2 46s 240 f s 20 Hz b Has 712s s 2 46s 240 f s 200 Hz FiniteDifference Filter Design 18 Using the differenceequation method and all backward differences design digital filters to approximate analog filters with these transfer functions In each case choose a sampling frequency which is 10 times the magnitude of the distance of the farthest pole or zero from the origin of the s plane Graphically compare the step responses of the digital and analog filters a Has s 2 s 2 3s 2 b Has s 60 s 2 120s 2000 c Has 16s s 2 10s 250 Matched zTransform and Direct Substitution Filter Design 19 A continuoustime filter with a transfer function H s K s 10 is approximated by a digitalfilter designed using both the direct substitution and matched ztransform methods Then the digital filters are used as the forwardpath transfer functions H 1 z in two discretetime feedback systems both with feedback transfer functions H 2 z 1 The gain K is gradually increased from 0 upward Which system will go unstable at a finite value of K Explain your answer Bilinear zTransform Filter Design FIR Filter Design 20 Graph the frequency response of an FIR filter designed using the ParksMcClellan algorithm that meets the specification in Figure E20 with the shortest possible impulse response 0 05 1 15 2 25 3 60 50 40 30 20 10 0 10 20 Radian Frequency Ω Filter Gain Magnitude dB 40 4 50 4 40 3 3 04 065 115 14 185 20 26 275 Figure E20 rob28124ch14680734indd 733 041216 207 pm C h a p t e r 14 Filter Analysis and Design 734 DigitalFilter Design Method Comparison 21 A lowpass digital filter is designed first using the impulseinvariant technique and then using the stepinvariant technique Which one is guaranteed to have the correct response magnitude at zero frequency and why 22 In designing a digital filter to approximate an analog lowpass filter which design technique guarantees that the digital filters magnitude response is zero at Ω π 23 A digital filter has a transfer function H 1 z z 2 z 2 az b If it is excited by a unit sequence the first three values of its response y 1 n are y10 3 y1 1 4 y1 2 2 If a digital filter with a transfer function H 2 z z 1 H 1 z is excited by a unit sequence what are the first three numerical values of its response y 2 n 24 Of the three window types Rectangular von Hann and Blackman if they are all used to design the same type of digital FIR filter with the same number of samples in the impulse response a Which one yields the narrowest transition from passband to stopband b Which one yields the greatest stopband attenuation 25 What is a disadvantage of finitedifference digitalfilter design using forward differences 26 Generally in order to design a digital filter with a certain passband gain transition bandwidth and stopband attenuation which type of digital filter accomplishes the design goal with fewer multiplications and additions IIR or FIR 27 We have studied six methods for approximating an analog filter by an IIR digital filter Impulse Invariant Step Invariant Finite Difference Matchedz Direct Substitution and Bilinear a Which methods can be done without using any timedomain functions in the process b Which methods squeezes the entire continuoustime frequency from zero to infinity into the discretetime radianfrequency range zero to π c Which two methods are almost the same differing only by a timedomain delay d Which method can design an unstable digital filter while trying to approximate a stable analog filter and if constrained to only stable designs restricts the flexibility of the design to only certain types of filters rob28124ch14680734indd 734 041216 207 pm A1 e x 1 x x 2 2 x 3 3 x 4 4 sinx x x 3 3 x 5 5 x 7 7 cosx 1 x 2 2 x 4 4 x 6 6 cosx cosx and sinx sinx e jx cosx j sinx sin 2 x cos 2 x 1 cosx cos y 1 2 cosx y cosx y sinx sin y 1 2 cosx y cosx y sinx cos y 1 2 sinx y sinx y cosx y cosx cos y sinx sin y sinx y sinx cos y cosx sin y A cosx B sinx A 2 B 2 cosx tan 1 BA d dx tan 1 x 1 1 x 2 u dv uv v du A P P E N D I X I Useful Mathematical Relations rob28124appA1A25indd 1 051216 217 pm A p p e n d ix I Useful Mathematical Relations A2 x n sinx dx x n cosx n x n1 cosx dx x n cosx dx x n sinx n x n1 sinx dx x n e ax dx e ax a n 1 ax n n ax n1 nn1 ax n2 1 n1 nax 1 n n n 0 e ax sinbx dx e ax a 2 b 2 a sinbx b cosbx e ax cosbx dx e ax a 2 b 2 a cosbx b sinbx dx a 2 bx 2 1 ab tan 1 bx a dx x 2 a 2 1 2 ln x x 2 a 2 1 2 0 sinmx x dx π2 m 0 0 m 0 π2 m 0 π 2 sgnm Z 2 Z Z n 0 N1 r n 1 r N 1r r 1 N r 1 n 0 r n 1 1r r 1 n k r n r k 1r r 1 n 0 n r n r 1r 2 r 1 rob28124appA1A25indd 2 051216 217 pm Appe n di x I Useful Mathematical Relations A3 e jπn e jπn N 0 drcl n N 0 N 0 δ N 0 n n and N 0 integers drcl n 2m 1 2m 1 δ 2m1 n n and m integers rob28124appA1A25indd 3 051216 217 pm A4 II A P P E N D I X ContinuousTime Fourier Series Pairs Continuoustime Fourier series CTFS for a periodic function with fundamental period T 0 1 f 0 2π ω 0 represented over the period T xt k c x k e j2πktT ℱ𝒮 T c x k 1 T T xt e j2πktT dt In these pairs k n and m are integers t Rext 1 Imxt 1 1 1 T0 k cxk 1 k m π π cxk t xt 1 1 T0 k cxk 1 k m m π π cxk t xt 1 1 T0 k cxk 1 k π π m m cxk e j2πktT0 ℱ𝒮 mT0 δk m cos2πtT0 ℱ𝒮 mT0 12δk m δk m sin2πtT0 ℱ𝒮 mT0 j2δk m δk m rob28124appA1A25indd 4 051216 217 pm A5 Appe n d i x II ContinuousTime Fourier Series Pairs t xt 1 k 2m m m 2m 2m m m 2m cxk 1 k π π T0 cxk t xt 1 T0 k cxk 1 k π π cxk t xt T0 w w 1 k cxk 1 k π π w T0 T0 cxk t xt T0 w w 1 k cxk k π π w T0 T0 1 cxk 1 ℱ𝒮 T δk T is arbitrary δT0 t ℱ𝒮 mT0 f0δmk 1w recttw δT0 t ℱ𝒮 T0 f0 sincwk f0 1w tritw δT0 t ℱ𝒮 T0 f0 sinc2 wk f0 rob28124appA1A25indd 5 051216 217 pm A6 A p p e n d ix II ContinuousTime Fourier Series Pairs t xt T0 1 k cxk k π π 1 2M1 M cxk t xt 1 w T0 k cxk k π π 2T0 w cxk t xt T0 w w1 k cxk 1 k π π w T0 T0 cxk 1w sinctw δT0 t ℱ𝒮 T0 f0 rectwk f0 drcl f0 t 2M 1 ℱ𝒮 T0 un M u n M 1 2M 1 M an integer t w ut ut w δT0 t ℱ𝒮 T0 1 wT0 j2πkwT0 1ej2πkwT0 1 2πkwT02 rob28124appA1A25indd 6 051216 217 pm A7 A P P E N D I X III Discrete Fourier Transform Pairs Discrete Fourier Transform DFT for a periodic discretetime function with funda mental period N 0 represented over the period N xn 1 N kN Xk e j2πknN 𝒟ℱ𝒯 N Xk nN xn e j2πknN In all these pairs k n m q Nw N0 N n0 and n1 are integers n Rexn 1 Imxn 1 1 1 N0 k Xk mN0 k π π mN0 1 mN0 1 m Xk n xn 1 1 N0 k Xk mN0 mN0 k π π m m Xk n xn 1 1 N0 m m mN0 k Xk mN0 k π π Xk e j2πnN0 𝒟ℱ𝒯 mN0 mN0 δmN0 k m cos2πnN0 𝒟ℱ𝒯 mN0 mN0 2 δmN0 k m δmN0 k m sin2πnN0 𝒟ℱ𝒯 mN0 jmN0 2 δmN0 k m δmN0 k m rob28124appA1A25indd 7 051216 217 pm A8 A p p e n d ix III Discrete Fourier Transform Pairs n xn 1 1 N0 k Xk mN0 k π π mq mN0 Xk n xn 1 1 N0 k Xk mN0 k π π mq mN0 Xk n xn 1 N k Xk N k π π N N Xk n xn 1 N0 N0 k Xk k π π m m mN0 Xk n xn 1 N0 Nw k Xk k π π N0 N0 2Nw1 Xk cos2πqnN0 𝒟ℱ𝒯 mN0 mN0 2 δmN0 k mq δmN0 k mq sin2πqnN0 𝒟ℱ𝒯 mN0 jmN0 2 δmN0 k mq δmN0 k mq 1 𝒟ℱ𝒯 N NδN k N is arbitrary δN0 n 𝒟ℱ𝒯 mN0 mδmN0 k un Nw un Nw 1 δN0 n 𝒟ℱ𝒯 N0 2Nw 1 drclkN0 2Nw 1 Nw an integer rob28124appA1A25indd 8 051216 217 pm A9 Appe n d i x III Discrete Fourier Transform Pairs n xn 1 n0 n1 n0Nw k k Xk π π N0 n1n0 Xk n xn 1 N0 w k w Xk k π π N0 Xk n xn 1 N0 k w Xk k π π N0 Xk n xn 1 N0 n Xk N0 M n N0 M N0 2M1 Xk un n0 un n1 δN0 n 𝒟ℱ𝒯 N0 ejπkn1n0N0 ejπkN0 n1 n0 drclkN0 n1 n0 trinw δN0 n 𝒟ℱ𝒯 N0 w sinc2 wkN0 δN0 k trinNw δN0 n 𝒟ℱ𝒯 N0 Nw drcl2 kN0 Nw Nw an integer sincnw δN0 n 𝒟ℱ𝒯 N0 wrect wkN0 δN0 k drclnN0 2M 1 𝒟ℱ𝒯 N0 un M un M 1 2M 1 N0 δN0 k M an integer rob28124appA1A25indd 9 051216 217 pm A10 IV A P P E N D I X ContinuousTime Fourier Transform Pairs xt X f e j2πft df ℱ X f xt e j2πft dt xt 1 2π X jω e jωt dω ℱ X jω xt e jωt dt For all the periodic time functions the fundamental period is T0 1f0 2π ω0 t xt 1 π π f 2 2 X f f 2 2 1 2 X f t xt 1 4π 4π 4π 4π π π ω X jω π ω X jω ut ℱ 12δ f 1j2π f ut ℱ πδω 1jω rob28124appA1A25indd 10 051216 217 pm A11 Appe n d i x IV ContinuousTime Fourier Transform Pairs t 1 1 xt 1 4 4 f 1 1 ω 2π 2π X f and X jω 1 f 1 1 π π ω 2π 2π X f and X jω t 1 2 1 2 xt 1 f ω 4 4 8π 8π X f and X jω 1 f 4 4 π π ω 8π 8π X f and X jω t 1 4 1 4 xt 1 f ω 2π 2π 1 1 2π 2π 1 1 X f and X jω 1 f π π ω X f and X jω t xt 1 1 2 1 2 f 4 4 ω 8π 8π X f and X jω 1 f 4 4 π π ω 8π 8π X f and X jω rectt ℱ sinc f rectt ℱ sincω2π sinct ℱ rect f sinct ℱ rectω2π trit ℱ sinc2 f trit ℱ sinc2ω2π sinc2t ℱ tri f sinc2t ℱ triω2π rob28124appA1A25indd 11 051216 217 pm A12 A p p e n d ix IV ContinuousTime Fourier Transform Pairs t xt 1 1 X f or X jω f or ω f or ω π π X f or X jω 1 xt t f X f 1 f π π X f 1 xt t ω X jω 2π ω π π X jω t Rext 1 Imxt 1 1 1 T0 f X f 1 f π π f0 X f t xt ab ab b f f π π X f and X jω ω ω ab a1 a1 a1 a1 a2π a2π a2π a2π X f and X jω a b 2 tri 2t a b a b 2 tri 2t a b ℱ ab sincaf sincbf a b 2 tri 2t a b a b 2 tri 2t a b ℱ ab sinc aω 2π sinc bω 2π a b 0 δt ℱ 1 1 ℱ δ f 1 ℱ 2πδω e j2πf0t ℱ δ f f0 rob28124appA1A25indd 12 051216 217 pm A13 Appe n d i x IV ContinuousTime Fourier Transform Pairs t xt 1 1 f 2 2 X f 1 f 2 2 π π X f t xt 1 1 ω 4π 4π X jω 2π ω 4π 4π π π X jω t xt 1 T0 f X f f π π f0 f0 f0 X f t xt 1 T0 ω X jω ω π π ω0 ω0 ω0 X jω t Rext 1 Imxt 1 1 1 T0 ω X jω 2π ω π π ω0 X jω e jω0t ℱ 2πδω ω0 sgnt ℱ 1jπf sgnt ℱ 2jω δT0t ℱ f0δf0 f f0 1T0 δT0t ℱ ω0δω0ω ω0 2πT0 rob28124appA1A25indd 13 051216 217 pm A14 A p p e n d ix IV ContinuousTime Fourier Transform Pairs t xt 1 1 T0 ω X jω 2π π ω π π ω0 ω0 ω0 ω0 X jω t xt 1 1 T0 f X f 1 f π π f0 f0 f0 f0 1 2 X f t xt 1 1 T0 ω X jω 2π π ω π π ω0 ω0 ω0 ω0 X jω t xt 1 1 T0 f X f 1 f π π f0 f0 f0 f0 1 2 X f cos2πf0t ℱ 1 2 δ f f0 δ f f0 cosω0t ℱ πδω ω0 δω ω0 sin2πf0t ℱ j 2 δ f f0 δ f f0 sinω0t ℱ jπδω ω0 δω ω0 rob28124appA1A25indd 14 051216 217 pm A15 Appe n d i x IV ContinuousTime Fourier Transform Pairs t xt 1a e1 a 1 1 12a2 1a2 1 1 f ω ω π π a a 2π f a a 2π X jω and X f X jω and X f t xt lnab ab ω ω π π f f X jω and X f 1 ab X jω and X f t xt ωc 2π ω ω π π π f f X jω and X f fn fn ωn ωn 2 X jω and X f t xt 1 1a f ω 1a ω π π a a a 2π 1 2 π 4 f a a 2π X jω and X f X jω and X f eat ut ℱ 1 jω a Rea 0 eat ut ℱ 1 j2πf a Rea 0 teat ut ℱ 1 jω a2 Rea 0 teat ut ℱ 1 j2πf a2 Rea 0 eat ebt b a ut ℱ 1 jω a jω b Rea 0 Reb 0 a b eat ebt b a ut ℱ 1 j2πf a j2πf b Rea 0 Reb 0 a b eat sinωct ut ℱ ωc jω α2 ω c 2 eζωn t sin ωn 1 ζ2t ut ℱ ωc jω2 jω2ζωn ω n 2 ωc ωn 1 ζ2 α ζωn rob28124appA1A25indd 15 051216 217 pm A16 A p p e n d ix IV ContinuousTime Fourier Transform Pairs t xt 1 ωc 2π ω ω π π f f X jω and X f fn fn ωn ωn X jω and X f t xt 1 1a e1 2 f f ω ω π π a a X jω and X f 1a X jω and X f t 2 2 xt 1 1 2π e12 2 2 4π 4π 1 ω π π f f X f and X jω 4π 4π ω 2 2 1 2π 2π X f and X jω e12 t xt 1 lnab ab ω f ω π π f X jω and X f X jω and X f aeat bebt a b ut ℱ jω jω a jω b Rea 0 Reb 0 a b aeat bebt a b ut ℱ j2πf j2πf a j2πf b Rea 0 Reb 0 a b eat cosωct ut ℱ jω α jω α2 ω c 2 eζωnt cos ωn 1 ζ2t ut ℱ jω ζωn jω2 jω2ζωn ω n 2 ωc ωn 1 ζ2 α ζωn ea t ℱ 2a ω2 a2 Rea 0 ea t ℱ 2a 2πf 2 a2 Rea 0 eπt2 ℱ eπf 2 eπt2 ℱ eω24π rob28124appA1A25indd 16 051216 217 pm A17 A P P E N D I X V DiscreteTime Fourier Transform Pairs x n 1 XF e j2πFn dF ℱ XF n x n e j2πFn xn 1 2π 2π X e jΩ e jΩn dΩ ℱ X e jΩ n x n e jΩn For all the periodic time functions the fundamental period is N0 1F0 2πΩ0 In all these pairs n NW N0 n0 and n1 are integers n xn 1 F XF 1 F π π 1 2 2 1 1 2 2 1 XF Ω 2π Ω π π π 2π 2π π π 2π 2π π X X e jΩ e jΩ n xn 1 1 ℱ δ1F 1 ℱ 2πδ2πΩ rob28124appA1A25indd 17 051216 217 pm A18 A p p e n d ix V DiscreteTime Fourier Transform Pairs n xn 1 n0 n1 F 2 2 F 2 2 π π XF and XejΩ Ω 4π 4π Ω 4π 4π n1n0 n1n0 1 XF and Xe jΩ n 16 16 xn 1 w F F 2 2 2 2 π π Ω Ω 4π 4π 4π 4π w XF and XejΩ XF and XejΩ n xn w 1 w F 2 2 F 2 2 Ω Ω w w 1 π π 4π 4π 4π 4π XF and XejΩ XF and Xe jΩ n 1 xn 1 F Ω Ω F π XF and XejΩ XF and XejΩ un n0 un n1 ℱ ejπFn1n0 ejπF n1 n0 drclF n1 n0 un n0 un n1 ℱ ejΩn1n02 ejΩ2 n1 n0 drcl Ω 2π n1 n0 trinw ℱ w drcl2 Fw trinw ℱ w drcl2 Ω2πw sincnw ℱ w rect wF δ1F sincnw ℱ w rect w Ω2π δ2πΩ δn ℱ 1 rob28124appA1A25indd 18 051216 217 pm A19 Appe n d i x V DiscreteTime Fourier Transform Pairs n xn 1 F 2 2 1 F 2 2 π π XF XF Ω Ω π π 2π 2π 4π 4π 2π 2π 4π 4π 2π N0 2π N0 XejΩ XejΩ n xn 1 N0 4π 4π 4π 4π Ω π Ω π π XejΩ Xe jΩ n xn 1 n xn 1 N0 F F π π 1 2 2 1 1 2 2 1 N0 1 N0 1XF XF un ℱ 1 1 ej2πF 1 2 δ1F un ℱ 1 1 ejΩ πδ2πΩ δN0 n ℱ 1N0δ1N0F F0δF0 F δN0 n ℱ 2πN0δ2πN0Ω Ω0δΩ0 Ω rob28124appA1A25indd 19 051216 217 pm A20 A p p e n d ix V DiscreteTime Fourier Transform Pairs n xn 1 1 N0 Ω 2π Ω π π 2π 4π 4π 2π 2π 4π 4π 2π N0 2π N0 2π XejΩ Xe jΩ Ω 2π Ω π π 2π 4π 4π 2π 2π 4π 4π 2π N0 2π N0 2π XejΩ XejΩ n xn 1 1 N0 n xn 1 1 N0 F 1 F π π 1 2 2 1 1 2 2 1 N0 1 N0 1 XF XF n xn 1 1 N0 F 1 F π π 1 2 2 1 1 2 2 1 1 N0 N0 1 XF XF cos2πF0n ℱ 1 2 δ1F F0 δ1 F F0 cosΩ0n ℱ πδ2πΩ Ω0 δ2π Ω Ω0 sin2πF0n ℱ j 2 δ1F F0 δ1 F F0 sinΩ0n ℱ jπδ2πΩ Ω0 δ2π Ω Ω0 rob28124appA1A25indd 20 051216 217 pm A21 Appe n d i x V DiscreteTime Fourier Transform Pairs n xn 1 F 2 2 F 2 2 π π 4π 4π Ω 4π 4π Ω 1 1α XejΩ and XF Xe jΩ and XF n 12 12 xn 1 n 4 20 xn 1 1 n 4 20 xn 1 1 F 2 2 F 2 2 F 2 2 π π Ω 4π 4π Ω 4π 4π XejΩ and XF XejΩ and XF F 2 2 F 2 2 π π Ω 4π 4π Ω 4π 4π XejΩ and XF XejΩ and XF F 2 2 F 2 2 π π 4π 4π Ω 4π 4π Ω 1α 1α XF and XejΩ XF and XejΩ αn un ℱ 1 1 αejΩ αn un ℱ 1 1 αej2πF αn sinΩnn un ℱ α sinΩnejΩ 1 2α cosΩnejΩ α2ej2Ω αn sin2πFnn un ℱ α sin2πFnej2πF 1 2α cos2πFnej2πF α2ej4πF α 1 αn cosΩnn un ℱ 1 α cosΩnejΩ 1 2α cosΩnejΩ α2ej2Ω αn cos2πFn n un ℱ 1 α cos2πFnej2πF 1 2α cos2πFnej2πF α2ej4πF α 1 α n ℱ 1 α2 1 2α cos2πF α2 α n ℱ 1 α2 1 2α cosΩ α2 α 1 α 1 rob28124appA1A25indd 21 051216 217 pm A22 VI A P P E N D I X Tables of Laplace Transform Pairs CAUSAL FUNCTIONS δt 1 All s ut 1 s Res 0 u n t ut ut n1 convolutions 1 s n Res 0 t ut 1 s 2 Res 0 e αt ut 1 s α Res α t n ut n s n1 Res 0 t e αt ut 1 s α 2 Res α t n e αt ut n s α n1 Res α sin ω 0 t ut ω 0 s 2 ω 0 2 Res 0 cos ω 0 t ut s s 2 ω 0 2 Res 0 e αt sin ω c t ut ω c s α 2 ω c 2 Res α rob28124appA1A25indd 22 051216 217 pm A23 Appe n d i x VI Tables of Laplace Transform Pairs e αt cos ω c t ut s α s α 2 ω c 2 Res α e αt A cos ω c t B Aα β sin ω c t ut As B s α 2 ω c 2 e αt A 2 B Aα ω c 2 cos ω c t tan 1 B Aα A ω c ut As B s α 2 ω c 2 e C 2 t A cos D C 2 2 t 2B AC 4D C 2 sin D C 2 2 t ut As B s 2 Cs D e C 2 t A 2 2B AC 4D C 2 2 cos D C 2 2 t tan 1 2B AC A 4D C 2 ut As B s 2 Cs D ANTICAUSAL FUNCTIONS ut 1 s Res 0 e αt ut 1 s α Res α t n ut n s n1 Res 0 NONCAUSAL FUNCTIONS e α t 1 s α 1 s α α Res α rectt e s2 e s2 s All s trit e s2 e s2 s 2 All s rob28124appA1A25indd 23 051216 217 pm A24 VII A P P E N D I X zTransform Pairs CAUSAL FUNCTIONS δ n 𝒵 1 All z un 𝒵 z z 1 1 1 z 1 z 1 α n un 𝒵 z z α 1 1 α z 1 z α n un 𝒵 z z 1 2 z 1 1 z 1 2 z 1 n 2 un 𝒵 zz 1 z 1 3 1 z 1 z1 z 1 z 1 n α n un 𝒵 zα z α 2 α z 1 1 α z 1 2 z α n m α n un 𝒵 z m d m d z m z z α z α nn 1n 2 n m 1 m α nm u n 𝒵 z z α m 1 z α sin Ω 0 n un 𝒵 z sin Ω 0 z 2 2z cos Ω 0 1 sin Ω 0 z 1 1 2 cos Ω 0 z 1 z 2 z 1 cos Ω 0 n un 𝒵 z z cos Ω 0 z 2 2z cos Ω 0 1 1 cos Ω 0 z 1 1 2 cos Ω 0 z 1 z 2 z 1 rob28124appA1A25indd 24 051216 217 pm A25 Appe n di x VII zTransform Pairs α n sin Ω 0 n un 𝒵 zα sin Ω 0 z 2 2αz cos Ω 0 α 2 α sin Ω 0 z 1 1 2α cos Ω 0 z 1 α 2 z 2 z α α n cos Ω 0 n un 𝒵 zz α cos Ω 0 z 2 2αz cos Ω 0 α 2 1 α cos Ω 0 z 1 1 2α cos Ω 0 z 1 α 2 z 2 z α ANTICAUSAL FUNCTIONS un 1 𝒵 z z 1 z 1 α n un 1 𝒵 z z α z α n α n un 1 𝒵 αz z α 2 z α NONCAUSAL FUNCTIONS α n 𝒵 z zα z z1α α z 1α rob28124appA1A25indd 25 051216 217 pm Analog Filters Huelsman L and Allen P Introduction to the Theory and Design of Active Filters New York NY McGrawHill 1980 Van Valkenburg M Analog Filter Design New York NY Holt Rinehart and Winston 1982 Basic Linear Signals and Systems Brown R and Nilsson J Introduction to Linear Systems Analysis New York NY John Wiley and Sons 1966 Chen C Linear System Theory and Design New York NY Holt Rinehart and Winston 1984 Cheng D Analysis of Linear Systems Reading MA AddisonWesley 1961 ElAli T and Karim M Continuous Signals and Systems with MATLAB Boca Raton FL CRC Press 2001 Gajic Z Linear Dynamic Systems and Signals Upper Saddle River NJ Prentice Hall 2003 Gardner M and Barnes J Transients in Linear Systems New York NY John Wiley and Sons 1947 Gaskill J Linear Systems Fourier Transforms and Optics New York NY John Wiley and Sons 1978 Haykin S and VanVeen B Signals and Systems New York NY John Wiley Sons 2003 Jackson L Signals Systems and Transforms Reading MA AddisonWesley 1991 Kamen E and Heck B Fundamentals of Signals and Systems Upper Saddle River NJ Prentice Hall 2007 Lathi B Signal Processing and Linear Systems Carmichael CA BerkeleyCambridge 1998 Lathi B Linear Systems and Signals New York NY Oxford University Press 2005 Lindner D Introduction to Signals and Systems New York NY McGrawHill 1999 Neff H Continuous and Discrete Linear Systems New York NY Harper Row 1984 Oppenheim A and Willsky A Signals and Systems Upper Saddle River NJ Prentice Hall 1997 Phillips C and Parr J Signals Systems and Transforms Upper Saddle River NJ Prentice Hall 2003 Schwartz R and Friedland B Linear Systems New York McGrawHill 1965 Sherrick J Concepts in System and Signals Upper Saddle River NJ Prentice Hall 2001 Soliman S and Srinath M Continuous and Discrete Signals and Systems Englewood Cliffs NJ Prentice Hall 1990 Varaiya L Structure and Implementation of Signals and Systems Boston MA AddisonWesley 2003 Ziemer R Tranter W and Fannin D Signals and Systems Continuous and Discrete Upper Saddle River NJ Prentice Hall 1998 Circuit Analysis Dorf R and Svoboda J Introduction to Electric Circuits New York NY John Wiley and Sons 2001 Hayt W Kemmerly J and Durbin S Engineering Circuit Analysis New York NY McGrawHill 2002 Irwin D Basic Engineering Circuit Analysis New York NY John Wiley and Sons 2002 Nilsson J and Riedel S Electric Circuits Upper Saddle River NJ Prentice Hall 2000 Paul C Fundamentals of Electric Circuit Analysis New York NY John Wiley and Sons 2001 Thomas R and Rosa A The Analysis and Design of Linear Circuits New York John Wiley and Sons 2001 Communication Systems Couch L Digital and Analog Communication Systems Upper Saddle River NJ Prentice Hall 2007 Lathi B Modern Digital and Analog Communication Sys tems New York NY Holt Rinehart and Winston 1998 Roden M Analog and Digital Communication Systems Upper Saddle River NJ Prentice Hall 1996 Shenoi K Digital Signal Processing in Telecommunications Upper Saddle River NJ Prentice Hall 1995 Stremler F Introduction to Communication Systems Reading MA AddisonWesley 1982 Thomas J Statistical Communication Theory New York NY John Wiley and Sons 1969 Ziemer R and Tranter W Principles of Communications New York NY John Wiley and Sons 1988 DiscreteTime Signals and Systems and Digital Filters Bose N Digital Filters Theory and Applications New York NY NorthHolland 1985 Cadzow J DiscreteTime Systems Englewood Cliffs NJ Prentice Hall 1973 Childers D and Durling A Digital Filtering and Signal Processing St Paul MN West 1975 BIBLIOGRAPHY B1 rob28124bibB1B2indd 1 051216 226 pm B2 DeFatta D Lucas J and Hodgkiss W Digital Signal Processing A System Design Approach New York NY John Wiley and Sons 1988 Gold B and Rader C Digital Processing of Signals New York NY McGrawHill 1969 Hamming R Digital Filters Englewood Cliffs NJ Prentice Hall 1989 Ifeachor E and Jervis B Digital Signal Processing Harlow England Prentice Hall 2002 Ingle V and Proakis J Digital Signal Processing Using MATLAB ThomsonEngineering 2007 Kuc R Introduction to Digital Signal Processing New York NY McGrawHill 1988 Kuo B Analysis and Synthesis of SampledData Control Systems Englewood Cliffs NJ Prentice Hall 1963 Ludeman L Fundamentals of Digital Signal Processing New York NY John Wiley and Sons 1987 Oppenheim A Applications of Digital Signal Processing Englewood Cliffs NJ Prentice Hall 1978 Oppenheim A and Shafer R Digital Signal Processing Englewood Cliffs NJ Prentice Hall 1975 Peled A and Liu B Digital Signal Processing Theory Design and Implementation New York NY John Wiley and Sons 1976 Proakis J and Manolakis D Digital Signal Processing Principles Algorithms and Applications Upper Saddle River NJ Prentice Hall 1995 Rabiner L and Gold B Theory and Application of Digital Sig nal Processing Englewood Cliffs NJ Prentice Hall 1975 Roberts R and Mullis C Digital Signal Processing Reading MA AddisonWesley 1987 Shenoi K Digital Signal Processing in Telecommunications Upper Saddle River NJ Prentice Hall 1995 Stanley W Digital Signal Processing Reston VA Reston Publishing 1975 Strum R and Kirk D Discrete Systems and Digital Signal Processing Reading MA AddisonWesley 1988 Young T Linear Systems and Digital Signal Processing Englewood Cliffs NJ Prentice Hall 1985 The Fast Fourier Transform Brigham E The Fast Fourier Transform Englewood Cliffs NJ Prentice Hall 1974 Cooley J and Tukey J An Algorithm for the Machine Computation of the Complex Fourier Series Mathematics of Computation Vol 19 pp 297301 April 1965 Fourier Optics Gaskill J Linear Systems Fourier Transforms and Optics New York NY John Wiley and Sons 1978 Goodman J Introduction to Fourier Optics New York NY McGrawHill 1968 Related Mathematics Abramowitz M and Stegun I Handbook of Mathematical Functions New York NY Dover 1970 Churchill R Operational Mathematics New York NY McGrawHill 1958 Churchill R Brown J and Pearson C Complex Variables and Applications New York NY McGrawHill 1990 Craig E Laplace and Fourier Transforms for Electrical Engineers New York NY Holt Rinehart and Winston 1964 Goldman S Laplace Transform Theory and Electrical Transients New York NY Dover 1966 Jury E Theory and Application of the zTransform Method Malabar FL R E Krieger 1982 Kreyszig E Advanced Engineering Mathematics New York NY John Wiley and Sons 1998 Matthews J and Walker R Mathematical Methods of Physics New York NY W A Benjamin 1970 Noble B Applied Linear Algebra Englewood Cliffs NJ Prentice Hall 1969 Scheid F Numerical Analysis New York NY McGrawHill 1968 Sokolnikoff I and Redheffer R Mathematics of Physics and Modern Engineering New York NY McGrawHill 1966 Spiegel M Complex Variables New York NY McGrawHill 1968 Strang G Introduction to Linear Algebra Wellesley MA WellesleyCambridge Press 1993 Random Signals and Statistics Bendat J and Piersol A Random Data Analysis and Measurement Procedures New York NY John Wiley and Sons 1986 Cooper G and McGillem C Probabilistic Methods of Signal and System Analysis New York NY Oxford University Press 1999 Davenport W and Root W Introduction to the Theory of Random Signals and Noise New York NY John Wiley and Sons 1987 Fante R Signal Analysis and Estimation New York John Wiley and Sons 1988 LeonGarcia A Probability and Random Processes for Elec trical Engineering Reading MA AddisonWesley 1994 Mix D Random Signal Processing Englewood Cliffs NJ Prentice Hall 1995 Papoulis A and Pillai S Probability Random Variables and Stochastic Processes New York NY McGrawHill 2002 Thomas J Statistical Communication Theory New York NY WileyIEEE Press 1996 Specialized Related Topics DeRusso P Roy R and Close C State Variables for Engi neers New York NY John Wiley and Sons 1998 Bibliography rob28124bibB1B2indd 2 051216 226 pm A A matrix diagonalizing 780781 absolute bandwidth 517 accumulation or summation 9496 accumulation property 416 422423 424 acoustic energy 736 737 acquisition of signals 446 active filters 536545 active highpass filter design of 540542 active integrator 538 active RLC realization of a biquadratic filter 545 ADC response 447 additive system 133134 air pressure variations 14 aliases 452 aliasing 454457 462463 495 661 almostideal discretetime lowpass filter 565 alternate statevariable choices 777 ambiguity problem 94 American Standard Code for Information Interchange ASCII 5 amplifier 124 145 amplifier transfer function 602 amplitude modulation 41 738753 755757 amplitude scaling 36 37 4344 89 analog and digital filter impulse responses 694 analog filters 680689 analog modulation and demodulation 738753 analog multiplier 141 142 analog recording device 446 analog signals 3 analog voltage converting to a binary bit pattern 448 analogtodigital converter ADC 4 447 664 angle modulation 744752 exercises 757758 antialiasing filter RC filter as 456457 anticausal signal 139 423 antiderivative of a function of time 48 antisymmetric filter coefficients 719720 aperiodic convolution 244 479480 aperiodic function 54 56 aperiodic signals 255 323324 aperture time 447 area property of the convolution integral 179 area sampling compared to value sampling 668 arguments of functions 20 34 in MATLAB 25 artificial systems 118 associativity property of convolution 179 181 196 200 asymptotes 525 asynchronous demodulation 743 asynchronous transmission 5 attenuated signal 547 attenuation 540 audio amplifier 253 511 audio compact disk CD 461462 audio range 509 audioamplifier controls 510 automobile suspension system model of 119 axial mode spacing 612 B backward difference approximation 699 of a discretetime function 94 96 bandlimited periodic signals 467470 exercise 495 bandlimited signals 242 453 457458 517 exercises 492 bandpass Butterworth analog filter 694 bandpass Butterworth digital filter 694 bandpass discretetime filter 556 bandpass filter design 702703 bandpass filters 128 390 392 511 512 519 535 683 See also causal band pass filter bandpass signals sampling 461463 bandpassfilter transfer function 686 bandstop discretetime filter 556 bandstop filters 128 511 512513 519 684 See also causal bandstop filter bandwidth 517 Bartlett window function 715 717 bartlett window function in MATLAB 725 baseband signal relation with modulated carrier 743 transmission 739 basis vectors 313 beat frequency 741 belB 521 Bell Alexander Graham 521 Bessel filter 686689 Bessel function of the first kind 751 besselap command 687 best possible approximation 242 BIBO stable system 138 200 BIBO unstable system 152 154 bilateral Laplace transform 380 bilinear command in MATLAB 710711 bilinear method 706711 bilinear transformation 704 711713 bilinear z transform 709 exercise 730 binary numbers 671 biological cell as a system 120 biquadratic RLC active filter 544546 biquadratic transfer function 513 Blackman window 718 Blackman window function 716 717 blackman window function in MATLAB 725 block diagrams 145146 of convolution 173 of discretetime systems 651 representing systems 124126 Bode Hendrik 523 bode command in MATLAB 389 Bode diagrams 521531 623 exercises 568569 Bode plot 523 bounded excitation producing an unbounded response 138139 154 boundedinputboundedoutput BIBO sta ble system See BIBO stable system boxcar rectangular window function in MATLAB 725 bridgedT network response of 384385 brightness of top row of pixels 551 buttap command in MATLAB 684685 butter function in MATLAB 725 Butterworth filters 681686 687 688 Butterworth lowpass filter 457 C capacitor values 541 capacitor voltage 763 capacitors 128 533 540 carrier modulating 738 cascade and parallel connections 599 cascade connection exercises 672 of system 330331 358 409 of two systems 181 200 cascade realization 630631 670 causal bandpass filter 519 550 causal bandstop filter 519 550 INDEX I1 rob28124idxI1I12indd 1 061216 955 pm I2 Index causal discretetime system as BIBO stable 651 causal energy signal sampling 478 causal exponential z transform of 420421 causal exponentially damped sinusoid z transform of 420421 causal highpass filter 519 550 causal lowpass filter 519 550 causal signal 139 causal sinusoid 431 657658 causal system 139 causality 139 140 548 causallyfiltered brightness 551 central difference approximation 700 centroid of the root locus 615 change of period property 245 316 changeofscaleinz property 416 419 changeofscale property 421 channel 1 cheb1ap command 687 cheb2ap command 687 chebwin Chebyshev window function in MATLAB 725 cheby1 function in MATLAB 725 cheby2 function in MATLAB 725 Chebyshev Tchebysheff or Tchebischeff filter 686689 725 checkerboard pattern filtered 553 554 chopperstablilized amplifier 761762 circuits 127 clipped signal 515 clock driving a computer 81 closedloop system 125 code 447 combinations of even and odd signals 100101 comment lines in MATLAB 25 communication between people 16 time delay 39 communication system analysis 735755 communication systems 1 735737 communicationchannel digital filter design 721722 commutativity property 179 196 compact trigonometric Fourier series 237239 complementary root locus 617 complex conjugate pair of poles 530532 complex CTFS 234 complex exponential excitation 182183 203204 complex exponential excitation and response 357 408 complex exponentials 22 144 complex sinusoids 22 55 145 184 230 components system as an assembly of 124 compound interest accruing 154 computers as discretetime systems 81 conjugation property 245 316 331 416 constant as special case of sinusoid 231232 309 constantK bandpass filter 542543 continuous independent variables signals as functions of 17 continuous signals 239 continuous time 164186 207208 329 contiguouspulse approximation 169 continuousspace function of spatial coordinates 551 continuoustime Butterworth filters exer cises 727 728 continuoustime causality exercise 567 continuoustime communication systems 735752 continuoustime convolution 169187 479480 continuoustime derivatives approximating 698 continuoustime exponential 85 continuoustime feedback systems 1213 126127 continuoustime filters 510542 554 566 continuoustime Fourier methods 229274 continuoustime Fourier series See CTFS continuoustime Fourier transform See CTFT continuoustime frequency response exer cise 567568 continuoustime functions 2021 continuoustime ideal filters exercises 567 continuoustime impulse function 471 continuoustime LTI system as BIBO stable 138 continuoustime numerical convolution 198 continuoustime practical active filters exercises 574575 continuoustime practical passive filters exercises 570574 continuoustime pressure signal 1415 continuoustime problem solving 147 continuoustime sampling 447478 continuoustime signal functions summary of 34 continuoustime signals 34 5 7 compared to discretetime 80 estimating CTFT of 480 graphing convolution of 198 mathematical description of 1956 sampling 7980 451 continuoustime sinusoids 8283 continuoustime state equations exercises 791793 continuoustime system response exercise 793 continuoustime systems 119142 763781 approximate modeling of 146 as BIBO stable 182 feedback in 12 frequency response of 185186 interpretation of the root locus 653 response to periodic excitation 246248 simulating with discretetime systems 660669 continuousvalue signal 4 continuums 4 control toolbox in MATLAB 393395 conv command in MATLAB 196 conv function in MATLAB 198 198199 convD function in MATLAB 562564 convergence 239241 327 convergence factor 260 356 convolution 6 164 229 in discrete time 196 exercises 209213 215219 finding response of a system using 201203 graphical and analytical examples of 173177 192194 in time property 377 as two general procedures 174 of two unit rectangles 180 convolution integral 32 173 179 667 convolution method 186188 convolution operator 173 convolution properties 178180 196197 333 416 convolution result graphing 193 convolution sum computing with MATLAB 197198 for system response 191 Cooley James 322 coordinated notation for singularity functions 33 corner frequency 525 cosine accumulation graphing 98 cosinewave frequency modulation 752 cosines 52 carriers modulated by 749 sampled 464 Cramers rule 385 critical damping 627 critical radian frequency 144 CTFS continuoustime Fourier series 230255 DFT approximating 250252 properties 244 245 relation to CTFT exercises 293 CTFS harmonic function 340 750 computing with DFT 478 from a DFT harmonic function 468 estimating 248 exercises 281284 rob28124idxI1I12indd 2 061216 955 pm I3 Index CTFS harmonic function Continued of a periodic signal using CTFT 266 of a rectangular wave 238 CTFS pairs 236 244 CTFS representation of a continuous periodic signal 239 CTFT continuoustime Fourier transform 6 255280 approximating with DFT 478 of convolution of signals 272 DFT approximating 274276 of an impulsesampled signal 454 limitations of 354 of a modulated sinusoid 266 of scaled and shifted rectangle 272 of the signum and unitstep functions 262263 of a single continuoustime rectangle 340 system analysis using 277281 of timescaled and timeshifted sines 271 total area under a function using 271 of the unitrectangle function 264 using differentiation property 270271 CTFT pairs 259 CTFTCTFSDFT relationships exercises 495496 CTFTDFT relationship 470471 CTFTDTFT relationship 471474 cumsum function in MATLAB 49 96 cumulative integral 48 cup anemometer 120 D damped sinusoid 145 damping factor 144 damping ratio 144 decaying exponential shape signal with 44 deci prefix 522 decibel dB 521523 decimation 9091 481 482 definite integral 48 delay 145 demodulation 739740 744 derivation 169174 189192 derivative generalized exercises 67 derivative of the phase controlling 746 derivatives of even and odd functions 53 derivatives of functions exercises 66 deterministic signal 4 DFT discrete Fourier transform 249 310311 approximating CTFS 250252 approximating CTFS harmonic function 273 approximating CTFT 274276 defined 334 exercises 342344 497500 of a periodically repeated rectangular pulse 316317 properties 315321 signal processing using 470480 using to find a system response 338340 DFT harmonic function 310 based on one fundamental period 323 of a discretetime function 468 period of 323 DFT pairs 320 DFT transform pair 315 diagonalization 779782 exercise 794 diff function in MATLAB 96 difference 94 difference equations describing discretetime systems 650 for a discretetime system 203 exercises solving 438439 with initial conditions 424 modeling discretetime systems 146151 solution of 424425 differenceequation description exercise 794 differencing and accumulation 94 exercises 109110 differencing property of the convolution sum 196 differential equations approximating difference equation 699 exercises solving 399 with initial conditions 383385 modeling systems using 120127 solution of 121 differentialequation description exercise 794 differentiation 4750 differentiation property of the convolution integral 179 of the CTFT 269270 z transform using 422 differentiators 527 digital bandpass filter design bilinear transformation 712713 impulseinvariant method 694696 matchedz transform 705706 ParksMcClellan 724 stepinvariant method 697699 digital filters 446 680 689722 creating unstable 700701 frequency response as periodic 706 frequency response matching analog filter 706707 functions designing 725 digital hardware 671 digital image processing on computers 8 digital lowpass filter designs 711712 720721 digital signal processing DSP 446 digital signals 4 56 digital simulation by impulseinvariant method 695 digitalfilter frequency response 691692 digitaltoanalog converter DAC 448 664665 diode as statically nonlinear component 140 141 Direct Form II 359 realization 359360 419 system 511 system realization 409410 410 system realization exercise 396 435 direct substitution 704 direct substitution method 704705 direct terms vector of 374 diric function in MATLAB 320321 Dirichlet conditions 237 Dirichlet function 319 717 discontinuities functions with 2332 discontinuous function 21 discontinuous signals 240241 discrete Fourier transform See DFT discrete independent variable signals as functions of 18 discrete time 186203 329 exercises 214 discretespace function 551 discretetime causality exercise 576 discretetime convolution 189204 479480 discretetime delay 11 discretetime DSBSC modulation 754 discretetime exponentials 8586 discretetime feedback system 151 discretetime filters 546564 discretetime Fourier methods 307338 discretetime Fourier series See DTFS discretetime Fourier transform See DTFT discretetime frequency response 555 exercises 575576 discretetime functions 80 continuoustime singularity functions and 8689 domain of 90 examples 81 graphing 81 9294 summations of 101 discretetime ideal filters exercises 576 discretetime impulses MATLAB function for 86 discretetime numerical convolution 196 discretetime practical filters exercises 577578 rob28124idxI1I12indd 3 061216 955 pm I4 Index discretetime pulse filtering 726 discretetime radian frequency representing 425 discretetime sampling 481485 discretetime signal functions summary of 89 discretetime signals 3 6 from continuoustime signals 454 examples 80 sampling 481 simulating continuoustime signals 663 discretetime sinusoidalcarrier amplitude modulation 753754 discretetime sinusoids 8284 discretetime state equations exercises 794795 discretetime system objects 428429 discretetime system response exercise 795796 discretetime system stability analysis 653 discretetime systems 1112 145154 330 331 equivalence with continuoustime sys tems 660 feedback in 12 frequency response of 425 modeled by block diagrams 651 periodic frequency response 426 548 properties of 152 realization of 670671 simulating continuoustime systems 660669 statespace analysis of 782790 discretetime time scaling 482 discretetime unit ramp 96 discretevalue signals 4 discretizing a system 663 distortion 515516 547 distortionless system 516 547 distributivity property of convolution 181 201 divideandconquer approach to solving linearsystem problems 134 domain of a function 20 doublesideband suppressedcarrier DSBSC modulation 738741 signal sampling 485 doublesideband suppressed carrier modulation 378 doublesideband transmitted carrier DSBTC 741744 downsampling 483 DTFS discretetime Fourier series 307310 341 DTFS harmonic function 310 DTFT discretetime Fourier transform 323338 of any discretetime signal 454 approximating with DFT 478 compared to other Fourier methods 340 convergence 327 of a decimated signal 483 defined 334 derivation and definition 324325 derived from the z transform 546 of a discretetime function 547 of a discretetime signal 481 482 exercises 344 generalized 326327 generalizing 407408 of modulation carrier and modulated carrier 754 numerical computation of 334338 of a periodic impulse 330 properties 327333 of a system response 430431 of a window function 473 DTFT pairs 325 326327 dynamic system 140 E earbrain system 229 eig command in MATLAB 781782 eigenfunctions 22 121 electromagnetic energy propagation 736 electromechanical feedback system 772 ellip function in MATLAB 725 ellipap command 687 Elliptic filter Cauer filter 686689 725 encoded response 448 encoding 448 encoding signals 45 energy signals 59 60 energy spectral density 270 envelope detector 742 equalization filter 601 equalization system 762 equation of motion 9 equivalence of continuoustime and discretetime systems 666 equivalence property of the impulse 30 279 331 error signal 599 Eulers identity 22 184 230 even and odd functions combinations of 5253 exercises 110111 even and odd parts of a function 5051 even and odd signals 98101 exercises 6769 even function 49 excitation harmonic function 248 excitations 1 176 exercises 157158 existence of z transform exercise 435 exponentials 82 8586 exponentials exp 21 F F117 stealth fighter 12 604 fast Fourier transform FFT 250 321322 feedback 12 feedback connection 599 beneficial effects of feedback 601604 exercises 672 instability caused by feedback 604608 rootlocus method 612615 stability feedback effects on 600 stable oscillation using feedback 608612 of systems 652 terminology and basic relationships 599600 tracking errors in unitygain feedback systems 618621 feedback systems 1214 149152 feedbackpath transfer function 599 618 feedbacksystem transfer function 602 604 feedforward paths 715 fft algorithm implementing DFT on computers 338340 fft function in MATLAB 250 318 322 fftshift function in MATLAB 237 275 filter classifications 516 548554 filter function in MATLAB 725 filter transformations 682684 filtering images 549552 filters 509 510 continuoustime 510542 design techniques 689721 effects on signals 558560 processing signals 6 uses of 680 filtfilt function in MATLAB 725 final value theorem 416 finite difference design 698700 729 finitedifference method 702703 703 finiteduration impulse response 689 713 FIR filter design 713722 exercises 730731 FIR filters 689 firpm command in MATLAB 724725 first backward difference property 416 first time derivative property 381 firstorder hold 460461 firstorder systems 143 fixedpoint arithmetic 671 fluid system 910 fluidmechanical system modeling 122123 FM frequency modulation 746 747 forced response 139 145 432 forced response of the system 624 rob28124idxI1I12indd 4 061216 955 pm I5 Index forced response values 669670 forcing function 121122 165 forward and inverse discretetime Fourier transforms exercises 345348 forward and inverse Laplace transforms exercises 286293 396397 forward and inverse z transforms examples of 418422 exercises 435438 forward CTFT 478 forward DFT 311 311314 321 forward difference of a discretetime function 94 95 forward Fourier transform 355 forward Laplace transform 355 forward transfer function 621 forward z transform defined 407 forwardpath output signal 600 forwardpath transfer function 599 600 618 Fourier Jean Baptiste 230 Fourier method comparisons 340 Fourier methods matrix 340 Fourier series 230 of even and odd periodic functions 243 exercises 281282 extending to aperiodic signals 255260 numerical computation of 248255 Fourier transform 255 alternate definitions of 355356 generalized 260264 generalizing 355357 as not a function of time 258 numerical computation of 273280 Fourier transform pairs 258 264 Fourier transform properties 265270 Fourier transform representation of a discontinuous signal 693 Fourierseries tables and properties 244248 freqresp function in MATLAB 623 freqs function 687 frequency 15 453 frequency compression 268 frequency differentiation property 265 frequency domain 6 229 frequency modulation FM 746 747 frequency multiplexing 737738 frequency responsees 184185 204205 of a bandpass filter 562 of discretetime and continuoustime lowpass filters 555 of discretetime systems 425 in everyday life 509 of a filter 512 of ideal filters 517 518 548 of a lowpass filter 554555 phase of 386 from polezero diagram 387389 shaping 510 of a system 270 331332 from a transfer function 427428 frequency scaling property 265 268 330 382 frequency shifting 38 frequency shifting property 245 265 266 268 316 329 frequency warping 710 frequencydomain methods 704711 frequencydomain resolution 334 frequencyindependent gain 389 527528 frequencyscaling property 419 freqz function in MATLAB 726 fullwave rectifier as not invertible 143144 functions combinations of 3436 with discontinuities 2332 even and odd parts 98 exercises 105107 fundamental period of 101 graphing accumulation of in MATLAB 97 graphing combinations 3536 with integrals 48 sums products and quotients of 35 types of 20 fundamental cyclic frequency 53 fundamental period 53 247 of CTFS representation 236 of a function 101 of a signal 5556 fundamental radian frequency 53 G gain as opposite of attenuation 540 gate function unit rectangle function as 33 gcd function in MATLAB 56 generalized CTFT 326 generalized derivative 29 generalized DTFT 326327 generalized Fourier transform 260264 356 generalized Fouriertransform pair 261 Gibbs Josiah Willard 240 Gibbs phenomenon 240 Gibbs phenomenon 715 graphic equalizer 513514 546 graphing function scaling and shifting with MATLAB 4546 greatest common divisor GCD 55 H halfpower bandwidth 517 Hamming window function 716 717 hamming window function in MATLAB 725 hanning von Hann window function in MATLAB 725 harmonic function 233 harmonic number 233 310 harmonic response 246248 Heaviside Oliver 25 highpass active filters cascade of two inverting 541 highpass discretetime filter 556 highpass filters 128 387 392 511 512 519 535 See also causal highpass filter design of active 540542 frequency response of 390 response to sinusoids 557558 highspatialfrequency information in an image 553 highway bridge as a system 120 homeentertainment audio system 509 homogeneity 131132 homogeneous solution 121 164 homogeneous system 131 human body as a system 120 human ear response to sounds 509510 I ideal bandpass filter 516 548 ideal discretetime filters 548 ideal filters 509 515520 discretetime 547553 frequency responses 516517 impulse and frequency responses of 548 as noncausal 518 ideal highpass filter 516 548 ideal interpolation 458459 ideal lowpass filter 515 516 548 ideal operational amplifier 537 ideallowpassfilter impulse response 564 IIR filter design 689710 IIR filters 689 imageprocessing techniques application of 8 images 78 549552 impedance 533534 impedance concept of circuit analysis 593 impinvar command in MATLAB 694695 695 impulse invariance 662664 impulse invariant design 663 impulse modulation 452 impulse responses 164168 173 181 182 186188 of any discretetime system 554 of continuoustime systems 165168 of discretetime and RC lowpass filters 555 of a distortionless system 516 547 exercises 208 215 of a filter 547 of ideal filters 517 518 549550 rob28124idxI1I12indd 5 061216 955 pm I6 Index of an LTI system 173 for the movingaverage filter 562 of an RC lowpass filter 176 526 of the RLC bandpass filter 536537 of a system 188189 201 at three outputs 558 time delay in 566 truncating ideal 720721 of a zeroorder hold 460 impulse sample 690 impulse sampling 452 exercises 489491 interpolation and 458 impulse train 32 impulseinvariant design 690694 exercise 489491 MATLABs version of 696 impulseinvariant method digital bandpass filter design 694696 impulses graphical representations of 30 indefinite integral 48 independent variable 34 inductor current 764 inductors equations for 533 infinite energy 58 59 infiniteduration impulse response IIR 689 See also IIR filters infinitely many samples availability of 459 information 15 inhomogeneous system 131 initial value theorem 416 inner product of complex sinusoids 235 inphase part 465 input signals 1 119 inputs 1 instantaneous frequency 745 instrumentation system in an industrial process 514 integer multiple of the fundamental frequency 467 integrals of even and odd functions 53 exercises 66 of functions 48 integration 4750 integration property 277 748 integrators 124 125 527 538 interference 16 interpolation 90 458461 481 483485 exercises 493494 intrinsic functions in MATLAB 21 invariant functions 49 inverse CTFT 263264 479 inverse DFT 310311 approximating the inverse DTFT 336 defined 334 335 of a periodic function 474 inverse DTFT exact and approximate 336 MATLAB program finding 337338 of a periodically repeated rectangle 333334 of two periodic shifted rectangles 328329 using the DFT 334335 inverse Fourier transform 357 inverse Fourier transform integral 263264 inverse Laplace transform 360 365366 using partialfraction expansion 367368 368 371372 inverse unilateral Laplace transform 380 inverse z transform 410 415416 432 433 654 658 inverse ztransform methods 417422 invertibility 142143 invertible system 142 inverting amplifier 537 K Kaiser window function 717 718 kaiser window function in MATLAB 725 Kirchhoffs voltage law 128 Kronecker delta function 86 L Laplace Pierre Simon 355 Laplace system analysis 592 goals 592 standard realizations of systems 630632 system analysis using MATLAB 621623 system connections 599621 system representations 592596 system responses to standard signals 623629 system stability 596598 Laplace transform 184 analysis of dynamic behavior of continuoustime systems 650 counterpart to 406 development of 355358 exercises 395 existence of 360362 generalizing CTFT 407 making Fourier transform more directly compatible with 258 of a noncausal exponential signal 365366 properties 377379 of the system response 624 of timescaled rectangular pulses 378 Laplace transform pairs 357 362366 379 Laplacetransformztransform relationship exercise 675 lcm function in MATLAB 55 leakage minimizing 473 reducing 474 least common multiple LCM 54 leftsided signal 362 411 412 Leibnizs formula 138 LHôpitals rule 167 239 320 light waves Doppler shift with 41 linear timeinvariant system 134135 linear algebra theory 778 linear system 134 linear system dynamics 383 linearity 134 171 linearity property 245 265 271 316 332 377 416 418 423 linearizing a system 137 local oscillator 741 logamplified signal 547 logarithmic graphs 523 exercises 568 logarithmic scale uniform spacing on 514 logmagnitude graph 523 loop transfer function 600 612 loop transmission 600 lowpass Butterworth filter converting to a highpass 682 maximally flat 681 transforming into a bandpass filter 683 transforming into a bandstop filter 684 lowpass discretetime filter 754 lowpass filter 128 390 511 519 520 532534 538539 See also causal lowpass filter lowpass filter design 703 LTI discretetime system 153 LTI systems 134 excited by sinusoids 230 frequency response of a cascade of 270 impulse responses of 354 response of 375377 response to a complexexponential excitation 357 system and output equations of 766 testing for causality 139 M magnitude Bode diagrams 523 526 530 magnitude spectrum of a general bandpass signal 462 magnitudefrequencyresponse Bode diagram 523 marginally stable system 597 matchedz transform 704 705706 matchedz transform and direct substitution filter design exercise 729 mathematical functions describing signals 19 35 mathematical model 9 rob28124idxI1I12indd 6 061216 955 pm I7 Index mathematical voltagecurrent relations 128 MATLAB arguments 26 bartlett window function 725 bilinear command 710711 blackman window function 725 bode command 389 boxcar rectangular window function 725 buttap command 684685 butter function 725 chebwin Chebyshev window function 725 cheby1 function 725 cheby2 function 725 comment lines 25 computing convolution sum 197198 control toolbox 393395 conv command 197 conv function 198199 convD function 562564 creating functions in 25 cumsum function 49 96 design tools 684685 725726 designing analog Butterworth filters 681 diff function 4142 87 dirac function 31 diric function 320 eig command 781782 ellip function 725 exponentials and sinusoids in 21 fft function 250 322 fftshift command 275 fftshift function 252 filter function 725 filtfilt function 725 finding inverse DTFT 337338 firpm command 724725 freqresp function 623 freqz function 726 function for discretetime impulses 86 gcd function 56 graphic function scaling and shifting 4546 graphing function combinations 3536 hamming window function 725 hanning von Hann window function 725 heaviside intrinsic function 25 impinvar command 694695 695 int function 48 intrinsic functions 21 invoking a function 34 kaiser window function 725 lcm function 56 m file for the ramp function 27 minreal command 623 name 25 NaN constant 25 numerical integration functions in 49 pzmap command 389 residue function 374375 rlocus command 623 sign function 24 simulating a discretetime system 11 stem command 81 systemobject commands 685686 system objects 393395 428429 tf transfer function command 393394 tfdata command 394 tools for statespace analysis 790 transformation of normalized filters 684 triang window function 725 upfirdn function 726 use of 18 zpk command 393 zpkdata command 394 matrix transfer function 775 maximally flat Butterworth filter 681 McClellan James H 723 McLaurin series 28 measurement instruments 120 mechanical systems 9 modeling 120122 statespace analysis of 772775 memory 139140 minimum error of Fourierseries partial sums 242244 minimum sampling rate reducing 461 minreal command in MATLAB 623 modified CTFS harmonic functions 256 for rectangularwave signals 257 modulated carrier 739 modulation 738 753 modulation index 741 movingaverage digital filter 194195 movingaverage filter 196 560563 multipath distortion 762 multiple bandstop filter 561 multiplicationconvolution duality 330 multiplicationconvolution duality property 244 245 265 266 316 475 N name in MATLAB 25 narrowbandpasssignal spectrum 461 narrowband FM 747 narrowband PM 747 natural radian frequency 144 natural response 624 natural systems 118 negative amplitudescaling factor 37 negative feedback 599 negative sine function signal shape of 44 noise 1 4 16 17 520521 noise removal 520521 nonadditive system 133134 noncausal filter 155 noncausal lowpass filter 553 noncausal signalprocessing systems 155 noninverting amplifier 537 noninverting amplifier transfer function 537 nonlinear systems 137 140 normalized analog filter designs 687 normalized Butterworth filters 681683 normalized filters MATLAB commands for transformation of 684 null bandwidth 517 numerical computation of discretetime Fourier transform 334338 of Fourier series 248255 of Fourier transform 273280 numerical convolution 196 numerical CTFT exercise 294 numerical integration cumsum function 49 numerical integration functions in MATLAB 49 Nyquist Harry 453 Nyquist frequency 454 Nyquist rates 453 exercise 491 of signals 456457 sinusoids sampled above below and at 464466 O octave intervals filters spaced at 514 odd functions 49 53 Ohms law 140 onefinitepole onefinitezero highpass filter 390 onefinitepole lowpass filter 389 onepole system unitsequence response 655 onerealpole system 525 onerealzero system 526 onesided Laplace transform 380 open left halfplane 597 open loop system 125 operational amplifiers 537538 saturation in real 141 optimal FIR filter design 723724 order of a system 764 orthogonal basis vectors 312313 orthogonal complex sinusoids 234 orthogonality exercises 282 342 harmonic function and 234236 oscillator feedback system 609 output equations 764 765 766 output signals 1 outputs 1 rob28124idxI1I12indd 7 061216 955 pm I8 Index overdamped case 627 overmodulation 743 oversampled signal 453 P parallel cascade and feedback connections exercises 635637 672 parallel connections of systems 652 of two systems 181 200201 parallel realization 632 670 parallel response ADC 447 parallel RLC circuit 764 parentheses indicating a continuoustime function 81 Parks Thomas W 723 ParksMcClellan design of a digital bandpass filter 724 ParksMcClellan optimal equiripple design 723 Parseval des Chênes MarcAntoine 270 Parsevals theorem 245 265 270 288 316 332 partialfraction expansion 367377 418 passband 510 filter distortionless within 516 ripple 687 715 716 signal transmission 739 passive filters 532535 pendulum analyzing 137138 period of a function 53 in a periodic signal 255 periodic convolution 244 479 periodic even signal 243 periodic excitation response of a continu oustime system and 246248 periodic functions 54 101102 exercises 111112 periodic impulse 32 periodic odd function 244 periodic signals 5355 101102 139 average signal power calculation 58 with discontinuities 240241 exercises 6970 as power signals 58 periodically repeated sinc function 318 periodicimpulse sampling 481483 periodicity of the DTFT 331 periodicrepetition relationship sampling and 474478 phase 267 269 phase Bode diagram 524 526 530 phase modulation PM 745 phaselocked loop 741 photographs 551 physical systems as filters 536 picket fencing 476 pitch 15 pixels 551 PM phase modulation 747 point spread function 554 pole of a Laplace transform 364 of an analog filter 695 exercises 439440 frequency response and 385393 425427 polezero diagrams 364 of system transfer functions 426 using the z transform 656657 polezero plots 421 427428 power of signals finding 59 power signals 59 60 power spectral density 1516 power spectrum 514 520 practical filters 532544 554565 practical interpolation 459 propagation delay in ordinary conversation 38 prototype feedback system 609 public address system 606 607 pulse amplitude modulation exercises 487 pure sinusoids 431 pzmap command in MATLAB 389 623 Q quadrature part 465 qualitative concepts 449450 quantization 448 quantized response 448 quantizing signals 45 R radian frequency 547 ramp function 26 random signals 4 6 558560 range of a function 20 rate 453 rational function 183 RC circuit frequency response of 528529 RC filter as an antialiasing filter 456457 RC lowpass filter 135 170 526 532 real exponential functions 21 real systems eigenfunctions of 135 realization 630 670671 realtime filtering of time signals 552553 realvalued sines and cosines replacing 234235 realvalued sinusoids 21 receiver 1 739 rectangular pulses convolution of 177 rectangularrule integration 171 rectangular wave CTFS harmonic function of 238 recursion 419 784785 red shift 41 regenerative travellingwave light amplifier 610 region of convergence ROC 361 362 364365 380 412 Remez Evgeny Yakovlevich 723 Remez exchange algorithm 723 residue function of MATLAB 374375 residues vector of 374 resistive voltage divider 140 resistors 128 533 540 resonant frequency 535 response harmonic function 247 responses 1 result in MATLAB 25 reverberation 605 RF signal transmission 739 rightsided signal 361362 411 ripple effect reducing in the frequency domain 715 RLC circuit 143 rlocus command in MATLAB 623 ROC region of convergence 361 362 364365 380 412 root locus for discretetime feedback system 653 exercises 637639 674 rootlocus method 612615 rootlocus plot 613 running integral 4849 S Sa function 239 SallenKey bandpass filter 542543 sampleandhold SH 447 sampleddata systems 664670 designing 668669 exercise 675 sampled sinc function 264 sampling 7980 446 at a discontinuity 693 exercises 487488 a signal 3 sampling methods 447449 sampling period or interval 80 sampling property of the impulse 3132 sampling rate 449450 461 463464 706707 sampling signals 45 sampling theorem 449453 satellite communication system propagation delay 38 scaled aliases of an analog filters frequency response 690 691 692 scaling 3645 89 exercises 6265 107109 scaling property 31 179 272 script file 51 sdomain differentiation 377378 378379 rob28124idxI1I12indd 8 061216 955 pm I9 Index sdomain shifting property 378 secondorder complex pole pair 531 secondorder complex zero pair 532 secondorder system 143144 655 secondorder system transfer function 625 sequentialstate machines 81 serial response ADC 447 Shannon Claude 450 shifting 3746 89 exercises 6265 107109 shifting property 31 side lobes 715 717 sidebands 738 signal energy 5657 exercises 7071 finding signal power using MATLAB 103105 finding using MATLAB 5960 per unit cyclic frequency 271 of a signal 102103 of a sinc signal 332333 signal energy and power exercises 112113 signal functions exercises 61 signal power 58 103 signal processing using the DFT 470480 signal reconstruction 460 461 signal transmission types of 739 signals 1 approximated by constants 231 approximated by periodic functions 54 examples of 19 finding Nyquist rates of 456457 spatially separating 737 switching on or off 23 system responses to standard 654660 types of 38 signaltonoise ratio SNR 16 17 521 signum function 2425 87 simultaneous shifting and scaling 4344 sinc function carriers modulated by 749 definition of 239 similarity to Dirichlet function 320 sinc signal signal energy of 332333 sines 52 465 sinewave phase of a carrier 746 singleinput singleoutput system 1 124 singlenegativerealzero subsystem 526 singlesideband suppressedcarrier SSBSC modulation 743744 singularity functions 23 33 8689 sinusoid response 627629 sinusoidal signal signal power of 58 sinusoids 22 8284 adding to constants 231 in discretetime signal and system analysis 82 multiplied by unit sequences 432 real and complex 230 responses to 229 sampling 464466 signal as burst of 44 system responses to 431432 smoothing filter 561 sound 14 229 space functions of 7 space shifting 38 spatial dimension independent variable as 38 spatial variables 8 spectra of PM and FM signals 747748 spectrum analyzer 520 splane region mapping 708 709 spontaneous emission 610 square brackets indicating a discretetime function 81 in MATLAB 87 square wave representing 135 squarewave phase of a carrier 746 ss function 782 790 SSBSC singlesideband suppressedcarrier modulation 743744 ssdata function 790 ss2ss function 782 790 stability 138 181 exercises 213214 219220 635 672 impulse response and 200 types of 598 stable analog filter becoming unstable digital filter 708 standard realizations of systems 630 cascade realization 630631 parallel realization 632 standard signals response to exercises 673 standard signals system responses to 623 exercises 640641 sinusoid response 627629 unitstep response 624627 start bit 5 state equations diagonalizing using MATLAB 781 state space 764 state transition matrix 767 785 state variables 763 777 778 state vector 764 statespace analysis characteristics of 764 MATLAB tools for 782 of a mechanical system 772775 of a twoinput twooutput system 769772 using state variables 763 static nonlinear components 140 static nonlinearity 140141 static system 140 statically nonlinear system 154155 stem command in MATLAB 81 step response 181 182 stepinvariant design 689 696 stepinvariant method 696700 stop bits 5 stopbands 510 straightline signal reconstruction 460 strength of an impulse 30 strictly bandlimited signals 453 517 subfunctions 51 sum property of the convolution sum 196 summing junction 11 124125 145 superposition applying to find approximate system re sponse 171 applying to linear systems 134 finding response of a linear system 136 finding response to a square wave 135 for LTI systems 145 suppressed carrier 739 symbolic integration int function 48 symmetric impulse response 719 synchronous demodulation 741 synthetic division 417 system analysis using CTFT 277281 system and output equations 764775 783787 system connections 181182 200201 652653 system discretization signal sampling and 663 system equations 765766 system modeling 119121 122 145154 system models exercises 156157 system objects in MATLAB 393395 428429 system properties 127140 152155 exercises 158160 system realization 359 exercises 641 676 system response exercises 294 to standard signals 654660 to system excitation 191 using DTFT and DFT 336338 system stability 651652 systemobject commands in MATLAB 685686 systems 1 3 defining 118 examples of 814 T tf transfer function command in MATLAB 393394 tfdata command in MATLAB 394 rob28124idxI1I12indd 9 061216 955 pm I10 Index thermocouples 732 thermostat 12 119 thermowell 732 time compression for discretetime functions 9091 time constant 625 time derivative properties 381 time differentiation property 245 265 377 time expansion 9091 time expansion property 416 time expression 268 time index 83 time integration property 245 265 377 379 381 time invariance 132133 time invariant system 132 time limited signals 58 457458 time multiplexing 737 time reversal property 245 316 416 time reversed function 39 time scaling 3943 8993 329 time scaling property 245 265 272 316 377 381 382 time shifting property 245 316 416 time shifting 3739 4243 89 time signals 7 time translation 37 time variant system 132 153 timedomain block diagram of a system 651 timedomain methods 689693 timedomain response of a onepole system 655 timedomain system analysis 164203 timelimited signals 361 410411 exercises 492 timescaling property 268 330 timeshifted signal 515 timeshifted unitstep function 38 timeshifting property 265 267268 272 273 277 317 378 381 418 419 422 tonal sound 15 tone 15 Toricellis equation 123 146 147 148 total area property 265 total harmonic distortion THD 253255 total system response 430431 tracking errors in unitygain feedback systems exercises 639640 trajectory 764 transfer function 182183 358 775777 787 common kind of 385 for discretetime systems 204 408 exercises 633634 frequency response and 184 205207 using timeshifting property 418419 transform method comparisons 430434 transformation 6 transformations 778 787 transient response 624 transmitted carrier 741 transmitter 1 travellingwave light amplifier 610 triang window function in MATLAB 725 triangular pulses convolution of 178 trigonometric form of the CTFS 233 trigonometric Fourier series 237 truncated ideal impulse response 713718 Tukey John 322 tuning a radio receiver 741 twodimensional signal images as 549 twofinitepole lowpass filter 391 twofinite pole system 390 twoinput twooutput system statespace analysis of 769772 twoinput OR gate in a digital logic system 154155 twopole highpass filter 540 twopole system See secondorder system twosided Laplace transform 380 twostage active filter frequency response of 538539 typeone Chebyshev bandstop filter 687688 typeone Chebyshev filter 687 typetwo Chebyshev filter 687 U unbounded response 138139 unbounded zerostate response 152 uncertainty principle of Fourier analysis 269 undamped resonance 393 underdamped case 627 underdamped highpass filter 391 392 underdamped low pass filter 391 underdamped system 144 undersampled signal 453 undersampling ambiguity caused by 465 uniform sampling 80 unilateral Laplace transform 379385 unilateral Laplace transform integral exercise 399 unilateral Laplacetransform pairs 382 unilateral z transform 423424 unilateral ztransform properties exercises 438 unit discretetime periodic impulse or impulse train 88 unit doublet 33 unit function 239 unit impulse 2930 unit pulse response of an RC lowpass filter 170 unit ramp function 2627 unit rectangle function 33 264 272 unit rectangles convolution of 180 unit sequence defined 96 unit step integral relationship with unit ramp 27 unit triangle function 180 unit triplet 33 unitarea rectangular pulse 28 unitarea triangular pulse 29 unitimpulse function 8687 unitpulse response 170 unitramp function 8889 unitsample function See unitimpulse function unitsequence function 8788 unitsequence response as accumulation of unitimpulse response 201 impulse response and 201 at three outputs 559 using the z transform 654655 656657 in the z domain 654 unitsinc function 238 unitstep function 2425 29 unitstep response 624627 of an RC lowpass filter 176 of a onepole continuoustime system 655 unitygain feedback systems tracking errors in 618621 exercises 639640 unitygain system 618 unstable digital filter avoiding 708 upfirdn function in MATLAB 726 upsampling 483 V value returned by a function 20 value sampling compared to area sampling 668 vector of state variables 766 voiced sound 15 voltage divider RC lowpass filter as 533 voltage gain 604 voltage response determinants of 129 voltage signal ASCIIencoded 5 voltagecurrent relationships for resistors capacitors and inductors 533 von Hann or Hanning window function 715 717 W water level differential equations for 123 versus time for volumetric inflows 10 weight of an impulse 30 wideband FM spectrum with cosine modulation 752753 window function 472473 window shapes 715 windowing 473 windows exercise 497 rob28124idxI1I12indd 10 061216 955 pm I11 Index Z z transform 406432 analysis of dynamic behavior of dis cretetime systems 650 existence of 410413 as a generalization of DTFT 407 of a noncausal signal 413414 of the state transition matrix 786 of a unitsequence response 655656 z transform pairs 413416 zdomain block diagram of a system 651 zdomain differentiation property 416 zdomain response to a unitsequence 655 ztransform pair 407 ztransform properties 416 ztransformLaplacetransform relationships 660662 zero of a Laplace transform 364 zero padding 334 zeroinput response 122 149150 zeroorder hold 460 zerostate response 122 130 562 of a discretetime system 787790 of a system 629 to a unitsequence excitation 153 zpk command in MATLAB 393 zpkdata command in MATLAB 394 rob28124idxI1I12indd 11 061216 955 pm