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Unformatted text preview: 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 = . 125 ( z 1)( z . 5) . (a) (5 pts) Is the closedloop system stable? Solution: The characteristic equation is 1 + D ( z ) G ( z ) = 1 + (1 z 1 ) . 125 ( z 1)( z . 5) = 1 + . 125 z ( z . 5) = 0 , which has two solutions 0 . 25 j . 25. These are the closedloop poles and are both inside the unit circle:  . 25 j . 25  = 0 . 3536 < 1. Therefore, the system is stable. (b) (10 pts) What are the steadystate errors of the system for tracking the unit step and the unit ramp input, respectively? Solution: Since the system is stable, from the openloop transfer function D ( z ) G ( z ) = . 125 z ( z . 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 systems step response? Solution: Pick any of the two closedloop pole, say, p = 0 . 25 + j . 25. We have  p  = 0 . 3536 and negationslash p = 0 . 7854 rad. Thus, the corresponding pole on the splane is q = ln  p  T + j negationslash...
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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 UniversityWest Lafayette.
 Spring '08
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