Quiz_5_sol

# Quiz_5_sol - for tracking unit step input we have e ss = 0...

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ECE 483 Quiz # 5 Solution Problem 1. (20 points) In the above sampled-data feedback system, assume T = 0 . 5 s, and Z b 1 - e sT s G p ( s ) B = z + 1 . 5 ( z - 1)( z + 1) . (a) (4 pts) Find the open-loop poles and the closed-loop poles of the system. Solution: The open-loop poles are ± 1 and the closed-loop poles are the solutions of the characteristic equation 1 + z + 1 . 5 ( z - 1)( z + 1) = 0 z 2 + z + 0 . 5 = 0 , namely, 0 . 5 ± j 0 . 5. (b) (4 pts) What is the dominant closed-loop pole (or poles)? Is the system stable? Solution: The dominant closed-loop poles are 0 . 5 ± j 0 . 5. Since both are inside the unit circle, the system is stable. (c) (6 pts) Determine the steady-state error e ss of the system when tracking the unit step input and the unit ramp input, respectively. Solution: The system is stable and of type one as there is one open-loop pole at 1. Therefore,
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Unformatted text preview: for tracking unit step input, we have e ss = 0; and for tracking unit ramp input, we have e ss = T K dc = . 5 2 . 5 / 2 = 0 . 4 . (d) (6 pts) Estimate approximately the settling time (2% criterion) and the maximum overshoot of the system’s step response. Solution: Based on either of the dominant closed-loop pole locations, say, p = 0 . 5 + j . 5 = | p | e j n p = √ 2 2 e j π 4 , we have the corresponding poles on the s-plane: q = ln | p | T + j n p T =-. 6931 + 1 . 5708 , which corresponds to a damping ratio ζ = . 6931 √ . 6931 2 +1 . 5708 2 = 0 . 4037. Thus, t s = 4 . 6931 = 5 . 7712 , M p = e − ζ √ 1-ζ 2 π = 25% . 1...
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## This note was uploaded on 04/15/2011 for the course ECE 483 taught by Professor Evens during the Spring '08 term at Purdue.

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