hw2 solu 355

# hw2 solu 355 - Solutions to Problems 4-9 4 1 Cs 2 5 10 5 5...

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Solutions to Problems 4-9 4. Using voltage division, V C (s) V i (s) = 1 Cs (R+ 1 Cs ) = 2 (s+2) . Since V i (s) = 5 s , V C (s) = 10 s(s+2) = 5 s - 5 s+2 . Therefore v C (t) = 5 - 5e -2t . Also, T = 1 2 , T r = 2.2 a = 2.2 2 = 1.1, T s = 4 a = 4 2 = 2. 5. Program: clf num=2; den=[1 2]; G=tf(num,den) step(5*G) Computer response: Transfer function: 2 ----- s + 2 6. Writing the equation of motion, ( Ms 2 g14 8 s ) X ( s ) g32 F ( s ) Thus, the transfer function i s ,

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4-10 Chapter 4: Time Response X ( s ) F ( s ) g32 1 Ms 2 g14 8 s Differentiating to yield the transfer function in terms of velocity, sX ( s ) F ( s ) g32 1 Ms g14 8 g32 1/ M s g14 8 M Thus, the settling time, T s , and the rise time, T r , are given by T s g32 4 8/ M g32 1 2 M ; T r g32 2.2 8 / M g32 0.275 M 7. Program: Clf M=1 num=1/M; den=[1 8/M]; G=tf(num,den) step(G) pause M=2 num=1/M; den=[1 8/M]; G=tf(num,den) step(G) Computer response: Transfer function: M = 1 Transfer function: 1 ----- s + 8 M = 2 Transfer function: 0.5 ----- s + 4
Solutions to Problems 4-11 From plot, time constant = 0.125 s. From plot, time constant = 0.25 s. 8. a. Pole: -2; c(t) = A + Be -2t ; first-order response. b. Poles: -3, -6; c(t) = A + Be -3t + Ce -6t ; overdamped response. c. Poles: -10, -20; Zero: -7; c(t) = A + Be -10t + Ce -20t ; overdamped response.

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4-12 Chapter 4: Time Response d. Poles: (-3+j3 15 ), (-3-j3 15 ) ; c(t) = A + Be -3t cos (3 15 t + g73 ); underdamped. e. Poles: j3, -j3; Zero: -2; c(t) = A + B cos (3t + g73 ); undamped. f. Poles: -10, -10; Zero: -5; c(t) = A + Be -10t + Cte -10t ; critically damped. 9. Program: p=roots([1 6 4 7 2]) Computer response: p = -5.4917 -0.0955 + 1.0671i -0.0955 - 1.0671i -0.3173 10. G(s) = C (s I - A ) -1 B A g32 8 g16 4 1 g16 3 2 0 5 7 g16 9 g170g3 g172g3 g171g3 g171g3 g186g3 g188g3 g187g3 g187g3 ; B g32 1 3 7 g170g3 g172g3 g171g3 g171g3 g186g3 g188g3 g187g3 g187g3 ; C g32 2 8 g16 3 g62 g64 ( s I g16 A ) g16 1 g32 1 s 3 g16 s 2 g16 91 s g14 67 ( s 2 g14 7 s g16 18) g16 (4 s g14 29) ( s g16 2) g16 (3 s g14 27) ( s 2 g14 s g16 77) g16 3 5 s g16 31 7 s g16 76 ( s 2 g16 10 s g14 4) g170g3 g172g3 g171g3 g171g3 g186g3 g188g3 g187g3 g187g3 Therefore, G(s ) = 5 s 2 g14 136 s g16 1777 s 3 g16 s 2 g16 91 s g14 67 . Factoring the denominator, or using det(s I - A ), we find the poles to be 9.683, 0.7347, -9.4179. 11. Program: A=[8 -4 1;-3 2 0;5 7 -9] B=[1;3;7] C=[2 8 -3] D=0 [numg,deng]=ss2tf(A,B,C,D,1); G=tf(numg,deng) poles=roots(deng) Computer response: A = 8 -4 1 -3 2 0 5 7 -9 B = 1
4-14 Chapter 4: Time Response 14. The equation of motion is: (Ms 2 +f v s+K s )X(s) = F(s). Hence, X(s) F(s) = 1 Ms 2 +f v s+K s = 1 s 2 +s+5 . The step response is now evaluated: X(s) = 1 s(s 2 +s+5) = 1/5 s - 1 5 s + 1 5 (s+ 1 2 ) 2 + 19 4 = 1 5 (s+ 1 2 ) + 1 5 19 19 2 (s+ 1 2 ) 2 + 19 4 . Taking the inverse Laplace transform, x(t) = 1 5 - 1 5 e -0.5t ( cos 19 2 t + 1 19 sin 19 2 t ) = 1 5 g172 g170 g188 g186 1 - 2 5 19 e -0.5t cos ( 19 2 t - 12.92 o ) .

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