Vibrations Rao 4thsi Ch14

Chapter 14\nRandom Vibration\n\n14.1\nSolution:\n p (x) = { k (1 - x/30) ; 20 ≤ x ≤ 30\n 0 ; elsewhere\nNormalization:\n ∫_20^30 p (x) dx = k ∫_20^30 (1 - x/30) dx = k (x - x^2/60) |^30_20 = 1\n k = 0.6\n P (x ≥ 28) = ∫_28^30 p (x) dx = k ∫_28^30 (1 - x/30) dx = k (x - x^2/60) |^30_28\n = k (4/60) = 0.6 (4/60) = 0.04\n\n14.2\np (x,y) = { x^2 / 9 ; 0 ≤ x ≤ 2, 0 ≤ y ≤ 3\n 0 ; elsewhere\nNormalization:\n ∫_0^2 ∫_0^3 p (x,y) dx dy = ∫_0^2 (x^2 / 9) dx ∫_0^3 dy = ∫_0^2 (x^2 / 9) dx (3) = 1\n = 1/9 ∫_0^2 x^2 dx = 1/9 [ (x^3)/3 ] |^2_0 = 1 (satisfied) (a) Marginal density functions\n pX (x) = ∫_0^3 p (x,y) dy = ∫_0^3 (x^2 / 9) dy = x^2 / 2 ; 0 ≤ x ≤ 2\n pY (y) = ∫_0^2 p (x,y) dx = ∫_0^2 (x^2 / 9) dx = 2/9 y ; 0 ≤ y ≤ 3\n(b)\n μX = x = E [f (x) . x dx] = ∫_0^2 p (x,y) . x dx = 1/2 ∫_0^2 x^2 dx = 1/2 (1/3) = 4/9\n μY = Ȳ = E [f (x,y) . y dy] = ∫_0^3 p (x,y) dy dx = 1/9 ∫_0^3 (y^3 / 3) dy = 2\n(c)\n σX^2 = E [ (x - μX)^2 ] = ∫_0^2 (x - μX)^2 p (x) dx = ∫_0^2 (x - (4/9))^2 (x^2 / 9) dx = 2/9 14.3\n E [x^2] = x̄ = ∫_0^∞ x . p (x) dx = ∫_0^.5 x dx = 1.0\n E [x] = ∫_0^∞ x^2 p (x) dx = ∫_0^.5 x^2 dx = 1.333\n σX^2 = ∫_0^∞ (x - x̄)^2 p (x) dx = ∫_0^0.5 (x - 0.5)^2 (0.5) dx = 0.5 ∫_0^x^2 dx = 0 - 0.3333\n σX = 0.5773\n\n14.4\n z = x + y , \n E [z] = E [x + y] = E [x^2 + y^2 + 2xy] = E [x^2] + E [y^2] + 2 E [x] E [y]\n E [x^*] = E [x^'] + E [y^*] + 2 E [x] E [y] 147\na) x(t) = x0 sin \\omega t, \\ x(t + \\tau) = x0 sin (\\omega (t + \\tau))\n x(t) x(t + \\tau) = x0^2 (sin^2 \\omega t + sin \\omega t \\cdot sin \\omega \\tau)\nR(\\tau) = \\lim_{T \\rightarrow \\infty} \\frac{x0^2}{T} [\\int_{-T/2}^{T/2} (1 - \\omega^2 \\tau) \\omega \\cdot \\sin \\omega t \\cdot \\sin \\omega(t + \\tau)] dt\n= \\lim_{T \\rightarrow \\infty} \\frac{x0^2}{T} [\\frac{1}{2} \\tau \\cdot \\cos \\omega \\tau + (0) \\sin \\omega \\tau]\n = \\frac{x0^2}{2} \\cdot R(\\tau)\n \\frac{x0}{2}\n\t\\tau\n\n(b)\\n\n\\tau/2\n\\rho(t) = x(t + \\tau)\\nR(\\tau)\n0\n\t\\tau 148\n\nx(t) = {\\begin{cases} (2x0 \\cdot t / c); & 0 \\leq t \\leq \\frac{T}{2} \\\\ (2x0 / \\gamma) - 2x0; & \\frac{T}{2} < t \\leq T \\end{cases}\\\\\nc_n = \\frac{1}{T} \\int_0^{T} x(t) e^{-in\\omega t} dt = \\frac{1}{T} \\left[ \\frac{2x0 t}{2} e^{-in\\omega t} \\right]_{0}^{T/2}\n+ \\int_{T/2}^{T} \\left( \\frac{2x0 \\cdot t}{2} - 2x0 \\right) e^{-in\\omega t} dt \\\\= 2\\cdot \\frac{x0}{T^2} \\left[ e^{-in\\omega t} \\cdot ( in\\omega t + e^{-in\\omega t}) \\right]_{\\frac{T}{2}}\\n\n\\frac{2x0}{\\pi} \\sum_{n=-\\infty}^{\\infty} \\frac{(-1)^{n}}{n^2} e^{in\\omega t} + \\frac{2x0}{\\pi} \\sum_{n=-\\infty}^{\\infty} e^{in\\omega t}\\n 149\nFor x(t) = x \\sin \\cos(t), R_x(\\tau) = \\frac{x^2}{1} \\cos \\tau [from Problem 14.7]\\\\\nFor square wave:\n\\ x(t) \\ x(t + \\tau)\\n\\t \\\\ + \\ \\cos \\omega t + \\sin \\omega t\\\\\n\\ x(t) = \\ \\frac{\\sin(zt)}{z}\\\\\nFor z < 0:\n\\ x = \\left {\\begin{array}{}\n\\lambda e^{-\\lambda t} ; t \\geq 0\\\\\n0 ; t < 0 \\\\ \\end{array}\\right . \\\\\ \nP(t) = \\int_{-\\infty}^{\\infty} \\phi(t') dt' = \\int_{0}^{t} \\tau=1 - e^{-\\lambda t}\\\\\n\\sigma_T^2 = \\int_0^{\\infty} (t - \\frac{1}{\\lambda})^2 \phi(t) dt = \\lambda \\int_0^{\\infty} e^{-\\lambda t} dt = \\lambda [ -\\frac{t^2}{\\lambda} e^{-\\lambda t} -\\frac{2 men]{\\cdots} ]_{0}^{\\infty} + 1 = \\frac{1}{\\lambda} \\\\n\\sigma_T = \\frac{1}{\\lambda} 14.11\n\nx(t) = x0 sin (πt/T )\n⟨x(t)⟩ = \n\nT/2\n \n∫T/0 x0 sin (πt/T ) dt = lim \nT→∞ \n = lim \nT→∞ x0 \n\nπ T\n0 (cos (πT/2) + cos (-πT/2)) = 0 \n\n⟨x2(t)⟩ = lim \nT→∞ \n \nT/2 \n \n∫T/0 x0 sin (πt/T )2 dt\n\n = lim \nT→∞ \nT/2\n \n\n= lim \nT→∞ \n\nx20 T/2 1 − _ \nπ sin (πT/2)\n 2\nT\n \n\n = lim \nT→∞ (x20 T/2)(sin (πT/2))/(πT) = \n\nx20 \nπ 2 \n\n 14.12\n\nc9 = 2π(100)/60 = 188.496 rad/sec.\n k = (x̄, σx2) = (2.5×10^6, 0.225×10^6) N/m\n (normally distributed)\n cω3 = √(k/m) = √(k/100) = √(k)/10 rad/sec\n P(ωn ≥ ω) = P[ωω2 ≥ ω2]\n P [ k ≥ 100(188.496)2] = P [k ≥ 3553074.202] (E1)\n Defining standard normal variate z as z = ( k − μk)/σk\n E̅k (E̅k) can be rewritten as\n P(ωn > ω) = P{ k - k̄ / σk ≥ 3.55371 × 10^6 − 2.25 × 10^6 \n 0.225 × 10^6 }\n = P{z ≥ 5.7916} = 0.3316 × 10^−8 \n from standard normal distribution tables\n [see, for example, Ref. 14.5]\n\n14.13\n\nx(t) = {A; −a ≤ t ≤ a \n0 ; elsewhere\nx(ω) = ∫−∞x(t)e^−iωtdt = A ∫−a^a e^−iωtdt = A(e−iaω − eiaω)\n = A/iω [e−iaω − eiaω] = A/iω \n = A( −(iω − −cos ωa + i sin ωa))\n = 2Aω sin ωa (E1)\n E̅0(E̅1) shows that X(ω) = 2Aω as ω → 0.\n 2Aω x(ω)\n x(ω)\n sin(ωa)\n 14.14\n\nx(t) = {A e−at; t > 0\n0; i else where\nX(ω) = ∫−∞x(t)e^−iωtdt = A ∫e^(−at)e^−iωtdt \n = A ∫−∞ e^(−(a + iω)t)dt \n = A/a + iω \n(−(iω − a)e^(−(iω − a)t) |−∞^0\n = A/a + iω + A/a + iω\n = A a + iω + A a − iω + A \n = 2A/(a2 + ω2 ) \n = x(ω)\n 2A/a \n \n14.15\nx(t) = A ∫−a^a e−a |t|\n = A ∫−a^a e−btn - i + A ∫−e^−bt = a(1 - e^−bt)\n + A \n + \n = A/a + iω \n + A − i a e^−|x| \n + A/(a + iω)\n + A(a + iω)\n + A/(d + ω^2)\n 14.16\n\nx(t) = δ(t − a)\nX(ω) = ∫−∞δ(t − a)e−iωtdt = e^(−iaω)\n since, by definition, the Dirac delta function is zero \n everywhere except at t = a. At t = a, e^−iaω = e^−iaω\n\n14.17\n\nF(t), N\n 500\n 0 |____________________| \n t1 t2 t3 t4 t5 t6 t7 t8 t9 t10\n -500\n-\n j = 0 1 2 3 4 5 6 7 8 9 10\n ts (sec) = 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0\n F ∫(t), N = 0 200 300 400 500 −400 −300 −200 0\n(i) Spectrum of F(t):\nFrom Eq. (14.46), cn = c1/ τ = ∫F(t) e^−iω0tdt\n where τ = time period = 2π/w0 and w0 = fundamental frequency, τ = 1.0 sec.\nw0 = 2π rad/sec.\n Since the integration in Eq. (E1) is over one-time period, cn can be expressed as\n cn = 1/N ∑_{n=1}^{N} F(tj)e^−j(2π/N)tj/\tau Eq. (E3) 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 \n 0 50 0 0.25 x 10^1\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-1} |c_n|^2 = 85000.00\\]\n\n14.18\nR_X(t) = ae^{-b|t|}\nS_X(\\omega) = \\frac{1}{2\\pi} \\int_{-\\infty}^{\\infty} R_X(t) e^{-j\\omega t} dt\n= \\frac{1}{2\\pi} \\int_{0}^{\\infty} ae^{-b t} e^{-j\\omega t} dt\n\\quad + \\frac{1}{2\\pi} \\int_{-\\infty}^{0} ae^{b t} e^{-j\\omega t} dt\n= \\frac{a}{2\\pi} \\cdot \\frac{1}{(i\\omega+b)} \\left[ e^{-(i\\omega+b)t} \\right]_{0}^{\\infty} + \\frac{a}{(2\\pi)} \\cdot \\frac{1}{(-i\\omega+b)} \\left[ e^{-(\\infty) t} \\right]_{-\\infty}^{0}\n= -\\frac{a}{2\\pi(i\\omega-b)} + \\frac{a}{2\\pi(q(i\\omega+b))}\n= \\frac{ab}{\\pi(b^2+\\omega^2)}