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Unformatted text preview: ECE 483 Quiz # 6 Solution R ( s ) Y ( s ) T D ( z ) G p ( s ) + H ( s ) 1 − e sT s Problem 1. In the above system, it is known that T = 0 . 1 s, H ( s ) = 1, D ( z ) = K for K ≥ 0, and G ( z ) = Z bracketleftbigg 1 e − sT s G p ( s ) bracketrightbigg = 2 z 2 ( z . 5) . (4 pts) Write the characteristic equations first on the zplane and then on the wplane. You do not have to simplify the equations. Solution: The characteristic equation on the zplane is 1 + D ( z ) G ( z ) = 1 + 2 K z 2 ( z . 5) = 0 . By the bilinear transformation z = 1+( T/ 2) w 1 − ( T/ 2) w = 20+ w 20 − w , the characteristic equation on the wplane is 1 + 2 K ( 20+ w 20 − w ) 2 ( 20+ w 20 − w . 5) = 0 ⇔ 2 K (20 w ) 3 + (20 + w ) 2 (10 + 1 . 5 w ) = 0 . (8 pts) Determine the range of K ≥ 0 so that the closedloop system is stable. Solution: The characteristic equation on the zplane is equivalent to Q ( z ) = z 3 . 5 z 2 + 2 K = 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.
 Spring '08
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