Solution manual for system dynamics 3rd edition by palm

Solutions Manual c to accompany System Dynamics, Third Edition by William J Palm III University of Rhode Island Solutions to Problems in Chapter Two c Solutions Manual Copyright 2014 The McGraw-Hill Companies All rights reserved No part of this manual may be displayed, reproduced, or distributed in any form or by any means without the written permission of the publisher or used beyond the limited distribution to teachers or educators permitted by McGraw-Hill for their individual course preparation Any other reproduction or translation of this work is unlawful rom https://testbankgo.eu/p/Solution-Manual-for-System-Dynamics-3rd-Edition-by-Palm 2.1 a) Nonlinear because of the y yă term b) Nonlinear because of the sin y term c) √ Nonlinear because of the y term d) Variable coefficient, but Linear e) Nonlinear because of the sin y term f) Variable coefficient, but linear c 2014 McGraw-Hill This work is only for non-profit use by instructors in courses for which the textbook has been adopted Any other use without publisher’s consent is unlawful rom https://testbankgo.eu/p/Solution-Manual-for-System-Dynamics-3rd-Edition-by-Palm 2.2 a) x t dx = t dt x(t) = + t2 b) x t e−4t dt dx = x(t) = 3.1 − 0.1e−4t c) Let v = x ˙ v t dv = t dt dx = + t2 dt x t dx = + t2 dt x(t) = + 7t + t3 18 v(t) = d) Let v = x ˙ v t dv = e−2t dt 23 −2t − e v(t) = 8 x t 23 dx = − e−2t dt 8 57 23 + t + e−2t x(t) = 16 16 e) x˙ = C1 , but x ă(0) = 5, so C1 = x = 5t + C2 , but x(0) = 2, so C2 = Thus x = 5t + c 2014 McGraw-Hill This work is only for non-profit use by instructors in courses for which the textbook has been adopted Any other use without publisher’s consent is unlawful rom https://testbankgo.eu/p/Solution-Manual-for-System-Dynamics-3rd-Edition-by-Palm 2.3 a) t dx = dt = t 25 − 5x √ √ √ 5x dx = arctanh − arctanh 25 − 5x2 25 5 x x Let =t √ C = arctanh Solve for x to obtain x= √ √ tanh(5 5t + C) b) x 10 t dx = 36 + 4x2 x tan−1 12 dt = t x =t 10 x(t) = tan(12t + C) C = tan−1 10 c) x x − ln(x + 5) x x dx = 5x + 25 = t dt x − ln(x + 5) − + ln = t 5 x − ln(x + 5) = 5t + − ln So a closed form solution does not exist (continued on the next page) c 2014 McGraw-Hill This work is only for non-profit use by instructors in courses for which the textbook has been adopted Any other use without publisher’s consent is unlawful rom https://testbankgo.eu/p/Solution-Manual-for-System-Dynamics-3rd-Edition-by-Palm Problem 2.3 continued: d) t dx = −2 e−4t dt x −4t ln x|x5 = e −1 x −4t ln = e −1 5 −4t x(t) = √ e e e x c 2014 McGraw-Hill This work is only for non-profit use by instructors in courses for which the textbook has been adopted Any other use without publisher’s consent is unlawful rom https://testbankgo.eu/p/Solution-Manual-for-System-Dynamics-3rd-Edition-by-Palm 2.4 From the transform definition, we have T L[mt] = lim T →∞ T mte−st dt = m lim T →∞ te−st dt The method of integration by parts states that T u dv = uv|T0 − T v du Choosing u = t and dv = e−st dt, we have du = dt, v = −e−st /s, and T L[mt] = m lim T →∞  = m lim  t T →∞ T e−st −s −  e−st te−st dt = m lim  t T →∞ −s e−st (−s)2 T   = m lim T →∞ T T − 0  e−st  dt −s T e−sT e−sT e0 + −0− −s (−s)2 (−s)2 m s2 because, if we choose the real part of s to be positive, then = lim T e−sT = T →∞ c 2014 McGraw-Hill This work is only for non-profit use by instructors in courses for which the textbook has been adopted Any other use without publisher’s consent is unlawful rom https://testbankgo.eu/p/Solution-Manual-for-System-Dynamics-3rd-Edition-by-Palm 2.5 From the transform definition, we have T L[t2 ] = lim T →∞ t2 e−st dt The method of integration by parts states that T T u dv = uv|T0 − v du Choosing u = t2 and dv = e−st dt, we have du = 2t dt, v = −e−st /s, and T L[t2 ] = lim T →∞ = lim T →∞ −T  e−st t2 e−st dt = lim  t2 T →∞ −s e−st + s s T te−st dt = lim T →∞ T T − −T  e−st 2t dt −s e−st + s s s2 s3 because, if we choose the real part of s to be positive, then, = lim T e−sT = T →∞ c 2014 McGraw-Hill This work is only for non-profit use by instructors in courses for which the textbook has been adopted Any other use without publisher’s consent is unlawful rom https://testbankgo.eu/p/Solution-Manual-for-System-Dynamics-3rd-Edition-by-Palm 2.6 a) X(s) = 10 + s s b) X(s) = + (s + 5) s+3 c) From Property 8, X(s) = − dY (s) ds where y(t) = e−3t sin 5t Thus Y (s) = 5 = (s + 3)2 + 52 s + 6s + 34 10s + 30 dY (s) =− ds (s + 6s + 34)2 Thus X(s) = (s2 10s + 30 + 6s + 34)2 d) X(s) = e−5s G(s), where g(t) = t Thus G(s) = 1/s2 and X(s) = e−5s s2 c 2014 McGraw-Hill This work is only for non-profit use by instructors in courses for which the textbook has been adopted Any other use without publisher’s consent is unlawful rom https://testbankgo.eu/p/Solution-Manual-for-System-Dynamics-3rd-Edition-by-Palm 2.7 f (t) = 5us (t) − 7us (t − 6) + 2us (t − 14) Thus F (s) = e−6s e−14s −7 +2 s s s c 2014 McGraw-Hill This work is only for non-profit use by instructors in courses for which the textbook has been adopted Any other use without publisher’s consent is unlawful rom https://testbankgo.eu/p/Solution-Manual-for-System-Dynamics-3rd-Edition-by-Palm 2.8 a) sin 3t b) cos 2t + sin 2t c) 2e−2t sin 3t d) 5 e−3 t − 3 e) e−3 t e−7 t − 2 f) e−3 t e−7 t + 2 c 2014 McGraw-Hill This work is only for non-profit use by instructors in courses for which the textbook has been adopted Any other use without publisher’s consent is unlawful rom https://testbankgo.eu/p/Solution-Manual-for-System-Dynamics-3rd-Edition-by-Palm 2.53 P (t) = 3us (t) − 3us (t − 5) From Property 6, P (s) = X(s) = 3 −5s − e s s P (s) − e−5s − e−5s = = 4s + s(4s + 1) s(s + 1/4) Let F (s) = 1 =3 − s(s + 1/4) s s + 1/4 Then f (t) = − e−t/4 Since X(s) = F (s) − e−5s we have x(t) = f (t) − f (t − 5)us (t − 5) = − e−t/4 − − e−(t−5)/4 us (t − 5) c 2014 McGraw-Hill This work is only for non-profit use by instructors in courses for which the textbook has been adopted Any other use without publisher’s consent is unlawful rom https://testbankgo.eu/p/Solution-Manual-for-System-Dynamics-3rd-Edition-by-Palm 2.54 Let f (t) = t + Then F (s) = t3 2t5 + 15 16 s4 + 2s2 + 16 + + = s2 s4 s6 s6 From the differential equation, X(s) = = F (s) s4 + 2s2 + 16 = s+1 s6 (s + 1) 19 16 16 18 18 19 19 − + − + − + s s s s s s s+1 Thus t − t + 3t3 − 9t2 + 19t − 19 + 19e−t 15 On a plot of this and the solution obtained from the lower-order approximation, the two solutions are practically indistinguishable x(t) = c 2014 McGraw-Hill This work is only for non-profit use by instructors in courses for which the textbook has been adopted Any other use without publisher’s consent is unlawful rom https://testbankgo.eu/p/Solution-Manual-for-System-Dynamics-3rd-Edition-by-Palm 2.55 From the derivative property of the Laplace transform, we know that ∞ L[x(t)] ˙ = −st x(t)e ˙ dt = sX(s) − x(0) Therefore ∞ lim [sX(s)] = lim x(0) + s→∞ = lim x(0) + lim s→∞ s→∞ s→∞ −st x(t)e ˙ dt lim →0+ −st x(t)e ˙ dt 0 + lim →0+ −st lim x(t)e ˙ dt s→∞ The limits on and s can be interchanged because s is independent of t Within the interval [0, 0+], e−st = 1, and so lim [sX(s)] = x(0) + lim s→∞ s→∞ lim →0+ x(t) ˙ dt + lim →0+ −st lim x(t)e ˙ dt s→∞ = x(0) + x(t)|t=0+ t=0 + = x(0+) This proves the theorem c 2014 McGraw-Hill This work is only for non-profit use by instructors in courses for which the textbook has been adopted Any other use without publisher’s consent is unlawful rom https://testbankgo.eu/p/Solution-Manual-for-System-Dynamics-3rd-Edition-by-Palm 2.56 From the derivative property of the Laplace transform, we know that ∞ L[x(t)] ˙ = −st x(t)e ˙ dt = sX(s) − x(0) Therefore, ∞ lim [sX(s)] = lim x(0) + lim s→0 s→0 ∞ = x(0) + s→0 −st x(t)e ˙ dt ∞ −st lim x(t)e ˙ dt = x(0) + s→0 x(t) ˙ dt because s is independent of t and lims→0 e−st = Thus T lim [sX(s)] = x(0) + lim T →∞ s→0 x(t) ˙ dt = x(0) + lim T →∞ x(t)|t=T t=0 = x(0) + lim x(T ) − x(0) = lim x(T ) = lim x(t) T →∞ T →∞ t→∞ This proves the theorem c 2014 McGraw-Hill This work is only for non-profit use by instructors in courses for which the textbook has been adopted Any other use without publisher’s consent is unlawful rom https://testbankgo.eu/p/Solution-Manual-for-System-Dynamics-3rd-Edition-by-Palm 2.57 Let t g(t) = x(t) dt Then t L t x(t) dt = L[g(t)] = g(t)e−st dt To use integration by parts we define u = g and dv = e−st dt, which give du = dg = x(t) dt and v = −e−st /s Thus t −st g(t)e =0+ g(t)e−st dt = −s g(0) + s s = s ∞ t=∞ ∞ − t=0 x(t)e−st dt = x(t) dt + t=0 e−st x(t) dt −s g(0) X(s) + s s X(s) s This proves the property If there is an impulse in x(t) at t = 0, then g(0) equals the strength of the impulse If there is no impulse at t = 0, then g(0) = c 2014 McGraw-Hill This work is only for non-profit use by instructors in courses for which the textbook has been adopted Any other use without publisher’s consent is unlawful rom https://testbankgo.eu/p/Solution-Manual-for-System-Dynamics-3rd-Edition-by-Palm 2.58 a) [r,p,k] = residue([8,5],[2,20,48]) The result is r = [10.7500, -6.7500], p = [-6.0000, -4.0000], and k = [ ] The solution is x(t) = 10.75e−6t − 6.75e−4t b) [r,p,k] = residue([4,13],[1,8,116]) The result is r = [2.0000 - 0.1500i, 2.0000 + 0.1500i], p = [-4.0000 + 10.0000i, -4.0000 - 10.0000i], and k = [ ] The solution is x(t) = (2 − 0.15j)e(−4+10j)t + (2 + 0.15j)e(−4−10j)t The solution is x(t) = 2e−4t (2 cos 10t + 0.15 sin 10t) c) [r,p,k] = residue([3,2],[1,10,0,0]) The result is r = [ -0.2800, 0.2800, 0.2000], p = [-10, 0, 0], and k = [ ] The solution is x(t) = −0.28e−10t + 0.28 + 0.2t (continued on the next page) c 2014 McGraw-Hill This work is only for non-profit use by instructors in courses for which the textbook has been adopted Any other use without publisher’s consent is unlawful rom https://testbankgo.eu/p/Solution-Manual-for-System-Dynamics-3rd-Edition-by-Palm Problem 2.58 continued: d) [r,p,k] = residue([1,0,1,6],[1,2,0,0,0,0]) The result is r = [-0.2500, 0.2500, 0.5000, -1.0000, 3.0000], p =[ -2, 0, 0, 0, 0], and k = [ ] The solution is 1 x(t) = −0.25e−2t + 0.25 + 0.5t − t2 + t3 2 e) [r,p,k] = residue([4,3],[1,6,34,0]) The result is r = [-0.0441 - 0.3735i, -0.0441 + 0.3735i, 0.0882], p = [-3.0000 + 5.0000i, -3.0000 - 5.0000i, 0], and k = [ ].The solution is x(t) = (−0.0441 − 0.3735j)e(−3+5j)t + (−0.0441 + 0.3735j)e(−3−5j)t + 0.0882 The solution is x(t) = 2e−3t (−0.0441 cos 5t + 0.3735 sin 5t) + 0.0882 (continued on the next page) c 2014 McGraw-Hill This work is only for non-profit use by instructors in courses for which the textbook has been adopted Any other use without publisher’s consent is unlawful rom https://testbankgo.eu/p/Solution-Manual-for-System-Dynamics-3rd-Edition-by-Palm Problem 2.58 continued: f) [r,p,k] = residue([5,3,7],[1,12,44,48]) The result is r = [21.1250 -18.7500 2.6250], p = [ -6, -4, -2], and k = [ ] The solution is x(t) = 21.125e−6t − 18.75e−4t + 2.625e−2 c 2014 McGraw-Hill This work is only for non-profit use by instructors in courses for which the textbook has been adopted Any other use without publisher’s consent is unlawful rom https://testbankgo.eu/p/Solution-Manual-for-System-Dynamics-3rd-Edition-by-Palm 2.59 a) [r,p,k] = residue(5,conv([1,8,16],[1,1])) The result is r = [-0.5556, -1.6667, 0.5556], p = [-4.0000, -4.0000, -1.0000], k = [ ] The solution is x(t) = −0.5556e−4t − 1.6667te−4t + 0.5556e−t b) [r,p,k] = residue([4,9],conv([1,6,34],[1,4,20])) The result is r = [-0.1159 + 0.1073i, -0.1159 - 0.1073i, 0.1159 - 0.1052i, 0.1159 + 0.1052i], p = -3.0000 + 5.0000i, -3.0000 - 5.0000i, -2.0000 + 4.0000i, -2.0000 - 4.0000i], and k = [ ] The solution is x(t) = (−0.1159 + 0.1073j)e(−3+5j)t + (−0.1159 − 0.1073j)e(−3−5j)t + (0.1159 − 0.1052j)e(−2+4j)t + (0.1159 + 0.1052j)e(−2−4j)t The solution is x(t) = 2e−3t (−0.1159 cos 5t − 0.1073 sin 5t) + 2e−2t (0.1159 cos 4t + 0.1052 sin 4t) c 2014 McGraw-Hill This work is only for non-profit use by instructors in courses for which the textbook has been adopted Any other use without publisher’s consent is unlawful rom https://testbankgo.eu/p/Solution-Manual-for-System-Dynamics-3rd-Edition-by-Palm 2.60 a) sys = tf(1,[3,21,30]); step(sys) b) sys = tf(1,[5,20, 65]); step(sys) c) sys = tf([3,2],[4,32,60]); step(sys) c 2014 McGraw-Hill This work is only for non-profit use by instructors in courses for which the textbook has been adopted Any other use without publisher’s consent is unlawful rom https://testbankgo.eu/p/Solution-Manual-for-System-Dynamics-3rd-Edition-by-Palm 2.61 a) sys = tf(1,[3,21,30]); impulse(sys) b) sys = tf(1,[5,20, 65]); impulse(sys) c 2014 McGraw-Hill This work is only for non-profit use by instructors in courses for which the textbook has been adopted Any other use without publisher’s consent is unlawful rom https://testbankgo.eu/p/Solution-Manual-for-System-Dynamics-3rd-Edition-by-Palm 2.62 sys = tf(5,[3,21,30]); impulse(sys) c 2014 McGraw-Hill This work is only for non-profit use by instructors in courses for which the textbook has been adopted Any other use without publisher’s consent is unlawful rom https://testbankgo.eu/p/Solution-Manual-for-System-Dynamics-3rd-Edition-by-Palm 2.63 sys = tf(5,[3,21,30]); step(sys) c 2014 McGraw-Hill This work is only for non-profit use by instructors in courses for which the textbook has been adopted Any other use without publisher’s consent is unlawful rom https://testbankgo.eu/p/Solution-Manual-for-System-Dynamics-3rd-Edition-by-Palm 2.64 a) sys = tf(1,[3,21,30]); t = [0:0.001:1.5]; f = 5*t; [x,t] = lsim(sys,f,t); plot(t,x) b) sys = tf(1,[5,20,65]); t = [0:0.001:1.5]; f = 5*t; [x,t] = lsim(sys,f,t); plot(t,x) c) sys = tf([3,2],[4,32,60]); t = [0:0.001:1.5]; f = 5*t; [x,t] = lsim(sys,f,t); plot(t,x) c 2014 McGraw-Hill This work is only for non-profit use by instructors in courses for which the textbook has been adopted Any other use without publisher’s consent is unlawful rom https://testbankgo.eu/p/Solution-Manual-for-System-Dynamics-3rd-Edition-by-Palm 2.65 a) sys = tf(1,[3,21,30]); t = [0:0.001:6]; f = 6*cos(3*t); [x,t] = lsim(sys,f,t); plot(t,x) b) sys = tf(1,[5,20,65]); t = [0:0.001:6]; f = 6*cos(3*t); [x,t] = lsim(sys,f,t); plot(t,x) c) sys = tf([3,2],[4,32,60]); t = [0:0.001:6]; f = 6*cos(3*t); [x,t] = lsim(sys,f,t); plot(t,x) c 2014 McGraw-Hill This work is only for non-profit use by instructors in courses for which the textbook has been adopted Any other use without publisher’s consent is unlawful rom https://testbankgo.eu/p/Solution-Manual-for-System-Dynamics-3rd-Edition-by-Palm ... been adopted Any other use without publisher’s consent is unlawful rom https://testbankgo.eu/p /Solution-Manual-for-System-Dynamics-3rd-Edition-by-Palm 2.2 a) x t dx = t dt x(t) = + t2 b) x t... been adopted Any other use without publisher’s consent is unlawful rom https://testbankgo.eu/p /Solution-Manual-for-System-Dynamics-3rd-Edition-by-Palm 2.3 a) t dx = dt = t 25 − 5x √ √ √ 5x dx... been adopted Any other use without publisher’s consent is unlawful rom https://testbankgo.eu/p /Solution-Manual-for-System-Dynamics-3rd-Edition-by-Palm Problem 2.3 continued: d) t dx = −2 e−4t