ch04 - F O U R Time Response SOLUTIONS TO CASE STUDIES...

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F O U R Time Response SOLUTIONS TO CASE STUDIES CHALLENGES Antenna Control: Open-Loop Response The forward transfer function for angular velocity is, G(s) = ω 0 (s) V P (s) = 24 (s+150)(s+1.32) a. ω 0 (t) = A + Be -150t + Ce -1.32t b. G(s) = 24 s 2 +151.32s+198 . Therefore, 2 ζω n =151.32, ω n = 14.07, and ζ = 5.38. c. ω 0 (s) = 24 s(s 2 +151.32s+198) = Therefore, ω 0 (t) = 0.12121 + .0010761 e -150t - 0.12229e -1.32t . d. Using G(s), ω 0 •• + 151.32 0 + 198 0 = 24 v p ( t ) Defining, x 1 = 0 x 2 = 0 Thus, the state equations are, x 1 = x 2 x 2 =− 198 x 1 151.32 x 2 + 24 v p ( t ) y = x 1 In vector-matrix form, x = 01 198 151.32 x + 0 24 v p ( t ); y = 10 [] x
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Solutions to Case Studies Challenges 75 e. Program: 'Case Study 1 Challenge (e)' num=24; den=poly([-150 -1.32]); G=tf(num,den) step(G) Computer response: ans = Case Study 1 Challenge (e) Transfer function: 24 ------------------- s^2 + 151.3 s + 198 Ship at Sea: Open-Loop Response a. Assuming a second-order approximation: ω n 2 = 2.25, 2 ζω n = 0.5. Therefore ζ = 0.167, ω n = 1.5. T s = 4 ζω n = 16; T P = π ω n 1- ζ 2 = 2.12 ; %OS = e - ζ π / 1 - ζ 2 x 100 = 58.8%; ω n T r = 1.169 therefore, T r = 0.77. b. θ s 2.25 s s 2 0.5 s 2.25 ++ = = 1 s s 0.5 + s 2 0.5 s 2.25 = 1 s s 0.25 + 0.25 2.1875 2.1875 + s 0.25 + 2 2.1875 +
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76 Chapter 4: Time Response = 1 s s 0.25 + 0.16903 1.479 + s 0.25 + 2 2.1875 + Taking the inverse Laplace transform, θ (t) = 1 - e -0.25t ( cos1.479t +0.16903 sin1.479t) c. Program: 'Case Study 2 Challenge (C)' '(a)' numg=2.25; deng=[1 0.5 2.25]; G=tf(numg,deng) omegan=sqrt(deng(3)) zeta=deng(2)/(2*omegan) Ts=4/(zeta*omegan) Tp=pi/(omegan*sqrt(1-zeta^2)) pos=exp(-zeta*pi/sqrt(1-zeta^2))*100 t=0:.1:2; [y,t]=step(G,t); Tlow=interp1(y,t,.1); Thi=interp1(y,t,.9); Tr=Thi-Tlow '(b)' numc=2.25*[1 2]; denc=conv(poly([0 -3.57]),[1 2 2.25]); [K,p,k]=residue(numc,denc) '(c)' [y,t]=step(G); plot(t,y) title('Roll Angle Response') xlabel('Time(seconds)') ylabel('Roll Angle(radians)') Computer response: ans = Case Study 2 Challenge (C) ans = (a) Transfer function: 2.25 ------------------ s^2 + 0.5 s + 2.25 omegan = 1.5000 zeta = 0.1667 Ts = 16
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Solutions to Case Studies Challenges 77 Tp = 2.1241 pos = 58.8001 Tr = 0.7801 ans = (b) K = 0.1260 -0.3431 + 0.1058i -0.3431 - 0.1058i 0.5602 p = -3.5700 -1.0000 + 1.1180i -1.0000 - 1.1180i 0 k = [] ans = (c)
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78 Chapter 4: Time Response ANSWERS TO REVIEW QUESTIONS 1. Time constant 2. The time for the step response to reach 67% of its final value 3. The input pole 4. The system poles 5. The radian frequency of a sinusoidal response 6. The time constant of an exponential response 7. Natural frequency is the frequency of the system with all damping removed; the damped frequency of oscillation is the frequency of oscillation with damping in the system. 8. Their damped frequency of oscillation will be the same. 9. They will all exist under the same exponential decay envelop. 10. They will all have the same percent overshoot and the same shape although differently scaled in time. 11. ζ , ω n , T P , %OS, T s 12. Only two since a second-order system is completely defined by two component parameters 13. (1) Complex, (2) Real, (3) Multiple real 14. Pole's real part is large compared to the dominant poles, (2) Pole is near a zero 15. If the residue at that pole is much smaller than the residues at other poles 16. No; one must then use the output equation 17. The Laplace transform of the state transition matrix is (sI - A ) -1 18. Computer simulation 19. Pole-zero concepts give one an intuitive feel for the problem.
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This note was uploaded on 04/04/2008 for the course MECH COntrol Sy taught by Professor Khurshid during the Spring '08 term at Michigan State University.

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ch04 - F O U R Time Response SOLUTIONS TO CASE STUDIES...

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