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Lec_10

# Lec_10 - Lecture 10 Stability Tests Motivating Examples...

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1 Lecture 10: Stability Tests Motivating Examples Stability of Close-Loop Systems Routh- Hurwitz’s Stability Criterion Jury’s Stability Test Reading: Chapter 7.1-7.5 A Motivating Example C ( s ) K s ( s +1) R ( s ) + ¡ T 1 ¡ e ¡ T s s G ( z ) = K ( e ¡ T + T ¡ 1) z +(1 ¡ e ¡ T ¡ Te ¡ T ) ( z ¡ 1)( z ¡ e ¡ T ) System is of type one. Hence the steady state error for tracking unit step input is 0, and the steady state error for tracking unit ramp input is e ss = T K dc = T KT = 1 K Let T =1; K =1. Then e ss = 1 0 2 4 6 8 10 12 14 16 0 2 4 6 8 10 12 14 16 Closed-loop poles: p 1 ; 2 = 0 : 5 § j 0 : 618

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2 Decrease e ss by Increasing gain K ? If we set K =5, supposedly e ss = 1 K = 0 : 2 But in reality: 0 1 2 3 4 5 6 7 8 9 10 0 2 4 6 8 10 12 14 16 18 20 Closed-loop poles: p 1 ; 2 = ¡ 0 : 2358 § j 1 : 2781 With Less Frequent Samplings Even if K =1 remains the same, but with less frequent samplings: T =10 0 5 10 15 20 25 30 35 0 10 20 30 40 50 60 Closed-loop poles: p 1 = ¡ 7 : 873, p 2 = ¡ 0 : 127.