Vibração - Rao - Solução - Cap 14 - 4 ed

Chapter 14\nRandom Vibration\n\n14.1 Solution:\np(x) = {k(1 - x/30); 20 ≤ x ≤ 30\n0; elsewhere\n\nNormalization:\n\\int_{20}^{30} p(x) dx = k \\int_{20}^{30} (1 - \\frac{x^{2}}{30^{2}}) dx = k \\left( \\frac{x - \\frac{x^{2}}{60}}{30} \\right)^{30}_{20} = 1\n\nk = 0.6\nP(x ≥ 28) = \\int_{28}^{30} p(x) dx = k \\int_{28}^{30} (1 - \\frac{x^{2}}{30^{2}}) dx = k \\left( \\frac{x - \\frac{x^{2}}{60}}{28} \\right)^{30}_{28}\n\n= k \\left( \\frac{4}{60} \\right) = 0.6 \\cdot \\frac{4}{60} = 0.04\n\n14.2 p(x,y) = {\\frac{x^{2}}{9}; 0 ≤ x ≤ 2, 0 ≤ y ≤ 3\n0; elsewhere\n\nNormalization:\n\\int_{0}^{2} \\int_{0}^{3} p(x,y) dx dy = \\int_{0}^{2} \\frac{x^{2}}{9} dx \\int_{0}^{3} dy = \\int_{0}^{2} \\frac{x^{2}}{9} \\cdot 3 dx = \\frac{1}{3} \\int_{0}^{2} x^{2} dx\n\n= \\frac{1}{3} \\cdot \\left( \\frac{x^{3}}{3} \\right)_{0}^{2} = 1 (satisfied) (a) Marginal density functions \\phi_{x}(x) = \\int_{0}^{3} p(x,y) dy = \\frac{1}{9} \\int_{0}^{2} x dx = \\frac{1}{9} \\cdot \\frac{x^{2}}{2} \\Big|_{0}^{2} = \\frac{4}{9}\n\\phi_{y}(y) = \\int_{0}^{2} p(x,y) dx = \\frac{1}{9} \\int_{0}^{3} y^{3} dy = \\frac{1}{9} \\cdot \\frac{y^{4}}{4} \\Big|_{0}^{3} = \\frac{27}{36} = 2\n\n(b) \\mu_{x} = \\int_{0}^{2} x \\phi_{x}(x) dx = \\int_{0}^{2} \\frac{x^{2}}{9} \\cdot x dx = \\frac{1}{9} \\cdot \\int_{0}^{2} x^{2} dx = \\frac{8}{27}\n\\mu_{y} = \\int_{0}^{3} y \\phi_{y}(y) dy = \\int_{0}^{3} \\frac{y^{4}}{9} dy = \\frac{1}{9} \\cdot \\frac{81}{5} = \\frac{9}{5}\n\n(c) \\sigma_{x}^{2} = E[(x - \\mu_{x})^{2}] = \\int_{0}^{2} (x - \\mu_{x})^{2} p(x) dx = \\int_{0}^{2} (x - \\frac{4}{9})^{2} \\frac{x^{2}}{9} dx = \\frac{1}{9} \\cdot \\int_{0}^{2} (x^{2} - 8/9x + \\frac{16}{81}) x^{2} dx = 0.6\n\nE[(y-\\mu_{y})^{2}] = \\int_{0}^{3} (y - \\mu_{y})^{2} p(y,x) dy = \\int_{0}^{3} \\left(y - \\frac{3/2}{1/9} \\right)^{2} \\cdot \\frac{y^{3}}{9} dy\n\nS_{x,y} = 0\n\n14.3 E[x^{2}] = \\mu_{x}^{2} = \\int_{0}^{\\infty} x p(x) dx = \\int_{0}^{\\infty} 0.5x dx = 1.0\nE[x^{3}] = \\int_{0}^{\\infty} x^{3} p(x) dx = \\int_{0}^{\\infty} 0.5x^{2} dx = 1.333\n\n\\sigma_{x}^{2} = \\int_{0}^{\\infty} (x - \\mu_{x})^{2} p(x) dx = .0 = 0.5777 14.7\na) x(t) = x0 sin ωt, \n x(t + τ) = x0 sin ω(t + τ)\n R(τ) = lim T→∞ x0/T*[ (1/T) ∫(sin ωt sin ω(t + τ) dt] \n = Lim T→∞ x0/T[\[ ( T I 0 sin ωt sdt + 1/T sin ωt × sin ω(t + τ) dt\n\t\t=> \frac{x0}{2} [R(τ) = \frac{x0}{2} ] \n\t\t where R(τ) = \frac{x0}{2}\n\ttb)\n} 14.8\nx(t) = { (2 x0 t/T) ; 0 ≤ t ≤ T/2 \n (2 x0/T) - 2x0 ; T/2 < t ≤ T \ncn = 1/T ∫0T x(t) e^(-inωt) dt = 1/T [ (2 x0 t/2) e^(-inωt) |0T/2 \n + ∫T/2T (2 x0 - 2x0) e^(-inωt) dt ]\n= 2 x0/π \[-(1/n^2) (e^(-inωt) - e^{inωt})/nT/2 \n= 2 x0/π \[ i/n e^{-inωt} - e^{inωt} \]\n= 2 x0 π/n^2 (−1)^n\n= 2 x0/n^2 ∑_{n=-∞}^{∞} e^{inωt}\n( n=even)\n + i (2 x0/π) ∑_{n=-∞}^{∞} (-1)^{(n/n=even)} e^{inωt} \n+ (2 x0/π) ∑_{n=-∞}^{-1} (-1)^{(n=odd)} e^{inωt} \n 14.9\nFor x(t) = x0 sin ωt, R_z(x) = x1 cos ωt [from Problem 14.7]\nFor square wave:\n x(t) x(t + τ)\nx(t) = x(t) = { x0 sin ωt ; for x (t) > 0\n x(t + τ)\n\n14.10\nh(t) = { λ e^(-λt) ; t ≥ 0\n 0 ; t < 0\n(i) P(t) = ∫0∞ h(t') dt' = λ ∫0t e^(-λt') dt' = 1 - e^(-λt)\n(ii) τ = ∫0∞ t h(t) dt = λ ∫0∞ t e^(-λt) dt = -λ/(−λ)²[e^(-λt)]_0∞\n(iii) σ_t² = ∫0∞ (t - 1)² h(t) dt\ngive us: σ_t = 1/λ 1411\nx(t) = x0 sin ( π t / T )\n⟨ x(t) ⟩ = T / 2\n = lim T -> ∞ x0 / T ∫ T / 0 sin ( π t / T ) dt = lim T -> ∞ ( - x0 cos ( π T / T ) + x0 cos ( 0 ) )\n T / 2\n = 0\n⟨ x2(t) ⟩ = lim T -> ∞ T / 2\n = lim T -> ∞ ( x0 2 / T ) ∫ T / 0 sin 2 ( π t / T ) dt\n = lim T -> ∞ ( x0 2 / T ) ( T / 2 )\n = lim T -> ∞ x0 2 / 2\n = x0 2 / 2\n = ( x0 2 / 2 )\n\n1412\nc9 = 2π (1100) / 60 = 188.496 rad/sec\nk = ( x̅, σx 2 ) = ( 1.0 2.5 x 10 6 , 0.225 x 10 6 ) N/m\n (normally distributed)\nωn = √( k / m ) = √( k / 100 ) rad/sec\nP ( ωn ≥ ω ) = P ( ω2 n ≥ ω2 )\n= P [ k ≥ 100(188.496)2 ] = P [ k ≥ 3553074.202 ]\n (E1)\nDefining standard normal variate z as z = ( k - μ ) / σk\nEμ ( Eσ ) can be rewritten as\nP [ ( ωn > ω ) ] = P [ { k - μ / σk } ≥ 3.5537 x 10 6 - 2.25 x 10 6 ]\n 0.225 x 10 6\n = P [ z ≥ 5.7916 ] = 0.3316 x 10 -8\n [ from standard normal distribution tables ]\n [ see, for example, Ref. 14.5 ] 1413\nx(t) = {\n A ; - a ≤ t ≤ a\n 0 ; elsewhere\n}\n\nX(ω) = ∫ -∞∞ x(t) e^{-i ω t} dt = A ∫ -a a e^{-i ω t} dt\n = A / - i ω [ e^{-i ω t} ] _ -a ^ a \n = A / ( a + i ω ) - A / ( a - i ω )\n\n= A / ( a + i ω ) + A / a + i ω = A ( a + i ω ) / a2 + ω2\n= 2A ω / a2 + ω2 1416\nx(t) = δ(t - a)\nX(ω) = ∫ -∞∞ δ(t - a) e^{-i ω t} dt = e^{-i ω a}\n\nsince, by definition, the Dirac delta function is zero everywhere except at t = a. At t = a, δ(t - a) = e^{-i ω a}\n\n1417\nF(ω), N\n\n 0 1 2 3 4 5 6 7 8 9 10\n t1 (sec) 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0\nF(t) = F(t) N / 100 200 300 400 500 -400 -300 -200 0\n(i) Spectrum of F(t):\nFrom Eq. (14.46), c_n = 1 / τ [\u2026]\nwhere τ = time period = 2π / ω0 and ω0 = fundamental frequency, τ = 1.0 sec, ω0 = 2π rad/sec.\nSince the integration in Eq. (E1) is over one-time period, c_n can be expressed as\n c_n = 1 / N ∑_i=1^N F(i) e^{-i (2π / N) t_j}\nwhere t_j = j / N\n= c_n = 1 / N ∑_i=1^N F(i) e^{-i (2π / N) j} cos (2π nj / N) - i sin (2π nj / N) (E2) Eq. (E9) gives the following result for N = 10:\n\n n Real (c_n) Imag (c_n) Spectrum of F (t) = (c_n)^2 = Real (c_n)^2 + Imag (c_n)^2\n 0 50 0 0.25 x 10^4\n 1 50 -153.8842 2.6180 x 10^4\n 2 50 68.8191 0.7236 x 10^4\n 3 50 -36.3271 0.3820 x 10^4\n 4 50 16.2460 0.2764 x 10^4\n 5 50 0.0000 0.2500 x 10^4\n 6 50 -16.2460 0.2764 x 10^4\n 7 50 -36.3271 0.3820 x 10^4\n 8 50 68.8191 0.7236 x 10^4\n 9 50 -153.8842 2.6180 x 10^4\n\n(ii) Mean square value of F (t):\nUsing Eq. 14.49,\nF^2(t) = \\sum_{n=0}^{N-1} |c_n|^2 = \\sum_{n=0}^{N} |c_n|^2 = 85000.00\n\nR_X(t) = \\alpha e^{-b|t|}\nS_X(\\omega) = \\frac{1}{2\\pi} \\int_{-\\infty}^{\\infty} R_X(t) e^{-i\\omega t} dt\n= \\frac{1}{2\\pi} \\int_{0}^{\\infty} \\alpha e^{-b t} e^{-i\\omega t} dt\n+ \\frac{1}{2\\pi} \\int_{-\\infty}^{0} \\alpha e^{bt} e^{-i\\omega t} dt\n= \\frac{(\\alpha}{2\\pi} \\cdot \\frac{1}{(i\\omega + b)}[e^{-i(t)} - e^{-(i + b)}_{-\\infty}^{\\infty}}\n= -\\frac{\\alpha}{2\\pi(i\\omega - b)} + \\frac{\\alpha}{\\pi(i\\omega + b)} = \\frac{ab}{\\pi (b^2 + \\omega^2)}\n