Homework2solutions_mae143b

# Homework2solutions_mae143b - MAE143B Homework Set 2...

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MAE143B Homework Set 2 - Solutions Question 1 FPE-N Question 3.33: Sketch the step response of a system with transfer function G ( s ) = s/ 2 + 1 ( s/ 40 + 1)[( s/ 4) 2 + s/ 4 + 1] . Justify your answer on the basis of locations of poles and zeros. (Do not ﬁnd the inverse Laplace transform. [Although feel free to compute approximate residue values in your answer.]) To assist you, I have plotted the response using matlab. Describe as much as you can about the curve and the evidence for this from the poles and zero positions. Figure 1: Step response for Question 1. The system has three poles; one at s = - 40 and the others as a complex conjugate pair at s = - 2 . 0000 ± 3 . 4641 j. But I choose to leave these poles described by s 2 + 2 ζω n s + ω 2 n , with undamped natural frequency ω n = 4 and damping ratio ζ = 1 / 2 . The system has a zero at s = - 2 . The point of this question is that the pole at s = 40 can effectively be neglected in the step response, since with a step input the residue at s = 40 will be roughly 20 / (100 × 40) , which is pretty small. It is clear from the way the transfer function is factored that the steady state response to a step will be one. Thus, the system is dominated by the complex conjugate pair of poles. We can appeal to the approximate expressions for rise time ( τ r 1 . 8 ω n = 0 . 45 s), overshoot M p = e - πζ/ 1 - ζ 2 = 0 . 16 or 16 percent, peak time t p = π/ω d = π/ [ ω n p 1 - ζ 2 ] = 0 . 9069 s, settling time t s = 4 . 6 / ( ζω n ) = 2 . 3 s. This is, of course, all approximate and fails to take account of the role of the zero in the calculation. Hence, the overshoot is poorly estimated. The peak time is a little better and the setting time is very

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## This note was uploaded on 02/02/2010 for the course MAE MAE 143b taught by Professor Callafon during the Fall '09 term at UCSD.

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Homework2solutions_mae143b - MAE143B Homework Set 2...

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