hwsln-421-s06-05

# hwsln-421-s06-05 - ECE 421 Spring 2006 Solutions to HW...

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ECE 421, Spring 2006 Solutions to HW Assignment #5 Problem #B-5-23 From the given G ( s ) , the closed-loop transfer function and closed-loop characteristic equation are T CL ( s )= C ( s ) R ( s ) = G ( s ) 1+ G ( s ) = K s ( s +1)( s +2)+ K = K s 3 +3 s 2 +2 s + K (1) ( s s 3 s 2 s + K (2) and the Routh array is s 3 : 1 2 s 2 : 3 K s 1 : 6 K 3 s 0 : K For closed-loop stability, there must be no sign changes in the f rst column of the array. Checking the s 0 and s 1 rows (the only ones involving K ) produces the following limits on K for stability. s 0 : K> 0 ,s 1 :6 0 K< 6 (3) 0 <K< 6 (4) Problem #B-5-24 G ( s ) , the closed-loop transfer function and closed-loop characteristic equation are T ( s C ( s ) R ( s ) = G ( s ) G ( s ) = 10 s ( s 1) (2 s +3)+10 = 10 2 s 3 + s 2 3 s +10 (5) ( s )=2 s 3 + s 2 3 s (6) It is obvious by inspection of ( s ) that the closed-loop system is unstable since not all the coe cients have the same sign. Forming the Routh array to check this conclusion gives the following result. s 3 : 2 3 s 2 : 1 10 s 1 : 23 s 0 : 10 Since there are 2 sign changes in the f rst column of the array, there are 2 closed-loop poles in the right-half plane. Therefore, the closed-loop system is unstable. Problem #B-5-25 From the given closed-loop characteristic ( s s 4 s 3 +(4+ K ) s 2 +9 s +25=0 (7) the Routh array is 1

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s 4 : 1 (4 + K ) 25 s 3 : 2 9 s 2 : 2 K 1 2 25 s 1 : 18 K 109 2 K 1 s 0 : 25 For closed-loop stability, there must be no sign changes in the f rst column, so the s 2 and s 1 rows must be checked. s 2 :2 K 1 > 0 K> 0 . 5 (8) s 1 :1 8 K 109 > 0 109 / 18 = 6 . 056 (9) Since the constraint from the s 1 row is more restrictive than the costraint from the s 2 row, the limit on K for closed-loop stability is 6 . 056 .
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hwsln-421-s06-05 - ECE 421 Spring 2006 Solutions to HW...

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