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Unformatted text preview: 4/08/10 ECE 483 Midterm 2 Sampler (A/D) Data hold (D/A) e ( t ) c ( t ) G p ( s ) H ( s ) r ( ) t Digital Filter D ( z ) + Problem 1 (30 points) Assume in the system above, T = 2 s, H ( s ) = 1, and, G ( z ) = Z bracketleftbigg 1 − e − sT s G p ( s ) bracketrightbigg = z 2 ( z − 1)( z − 2) . (a) (10 pts) Suppose D ( z ) = 0 . 5. What are the steady state errors of the system output when tracking unit step and unit ramp inputs, respectively? Justify your answers. Solution: The characteristic equation 1 + D ( z ) G ( z ) H = 0 is equivalent to the quadratic equation ( z − 1)( z − 2) + 0 . 5 z 2 = 0, which has two solutions 1 ± j . 5774. Thus, the closed loop system is unstable, and steadystate error does not exist. (b) (10 pts) Suppose D ( z ) = 30. What are the steady state errors of the system output when tracking unit step and unit ramp inputs, respectively? Justify your answers. Solution: The characteristic equation ( z − 1)( z − 2)+30 z 2 = 0 has two roots 0 . 0484 ± j . 2493. So the closedloop system is stable. The openloop transfer function D ( z ) G ( z ) = 30 z 2 ( z − 1)( z − 2) is of type one, with K dc = lim z → 1 ( z − 1) D ( z ) G ( z ) = − 30. For tracking unit step input, e ss = 0; For tracking unit ramp input, e ss = T K d c = 2 / ( − 30) = − . 0667. (c) (10 pts) Suppose D ( z ) = 8. Find the (approximate) settling time and maximum overshoot of the system response to unit step input. Solution: The characteristic equation ( z − 1)( z − 2) + 8 z 2 = 0 has two stable roots 0 . 1667 ± j . 4410. So the closedloop system is stable, with dominant poles 0 . 1667 ± j . 4410. Take q = 0 . 1667 + j . 4410 = 0...
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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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