Quiz_6_sol

# Quiz_6_sol - ECE 483 Quiz 6 Solution R(s T D(z 1esT s G p(s...

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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 - 0 . 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 - 0 . 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 - 0 . 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 - 0 . 5 z 2 + 2 K = 0 . We construct the Jury’s array as: z 0 z 1 z 2 z 3 2 K 0 - 0 . 5 1 1 - 0 . 5 0 2 K 4 K 2 - 1 0 . 5 - K For the system to be stable, we must have: Q (1) = 2 K + 0 . 5 > 0 K > - 0 . 25 ( - 1) 3 Q ( - 1) = 1 - ( - 0 . 5) + 0 - 2 K = 1 . 5 - 2 K > 0 K < 0 . 75 | a 0 | = | 2 K | < a 3 = 1 - 0 . 5 < K < 0 . 5 | b 0 | = | 4 K 2 - 1 | > | b 2 | = | K | .

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