Quiz_6_sol

Quiz_6_sol - ECE 483 Quiz # 6 Solution R ( s ) Y ( s ) T D...

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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 z-plane and then on the w-plane. You do not have to simplify the equations. Solution: The characteristic equation on the z-plane 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 w-plane 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 closed-loop system is stable. Solution: The characteristic equation on the z-plane 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.

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Quiz_6_sol - ECE 483 Quiz # 6 Solution R ( s ) Y ( s ) T D...

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