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Sp11_midterm2_sol

# Sp11_midterm2_sol - ECE 483 Midterm 2 Solution Problem 1(25...

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ECE 483 Midterm 2 Solution Problem 1 (25 points) In the above system, T = 0 . 5 s, and D ( z ) = 1 z 1 , G ( z ) = Z bracketleftbigg 1 e sT s G p ( s ) bracketrightbigg = 0 . 125 ( z 1)( z 0 . 5) . (a) (5 pts) Is the closed-loop system stable? Solution: The characteristic equation is 1 + D ( z ) G ( z ) = 1 + (1 z 1 ) 0 . 125 ( z 1)( z 0 . 5) = 1 + 0 . 125 z ( z 0 . 5) = 0 , which has two solutions 0 . 25 ± j 0 . 25. These are the closed-loop poles and are both inside the unit circle: | 0 . 25 ± j 0 . 25 | = 0 . 3536 < 1. Therefore, the system is stable. (b) (10 pts) What are the steady-state errors of the system for tracking the unit step and the unit ramp input, respectively? Solution: Since the system is stable, from the open-loop transfer function D ( z ) G ( z ) = 0 . 125 z ( z 0 . 5) , we conclude that the system is of type 0, with K dc = lim z 1 D ( z ) G ( z ) = 0 . 25 . Thus, the steady state tracking error for unit step input is e ss = 1 1 + K dc = 0 . 8 , and the steady state tracking error for unit ramp input is infinity. (c) (10 pts) What are the approximate settling time (2% criterion) and maximum overshoot of the system’s step response? Solution: Pick any of the two closed-loop pole, say, p = 0 . 25 + j 0 . 25. We have | p | = 0 . 3536 and negationslash p = 0 . 7854 rad. Thus, the corresponding pole on the s -plane is q = ln | p | T + j negationslash p T = 2 . 0794 + j 1 . 5708. This pole q has the damping ratio ζ = 2 . 0794 / 2 . 0794 2 + 1 . 5708 2 = 0 . 7979. Therefore, the approximate settling time and maximum overshot are

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Sp11_midterm2_sol - ECE 483 Midterm 2 Solution Problem 1(25...

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