Lecture7

# The closed loop system is stable for k 1 s1 case 2 ds

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Unformatted text preview: (s)c(s) + b(s)d(s) = 0. (The closed-loop system is stable if all its poles are on LHP.) Zeros: a(s)c(s) = 0. Unstable systems can be stabilized by feedback control. Example 5.1 G(s) = Case 1: D(s) = k . 1 . s−1 GD k = 1 + GD s−1+k Poles: p = 1 − k . The closed-loop system is stable for k > 1. s−1 Case 2: D(s) = . (Try to cancel the RHP pole in the plant by the RHP zero in the controller ) s+1 Y (s ) GD s−1 = = R ( s) 1 + GD (s − 1) + (s − 1)(s + 1) Poles: (s − 1) + (s − 1)(s + 1) = 0 ⇒ p1 = 1, p2 = −2. The closed-loop system is unstable. Unstable zero/pole cancellation does not work. Y (s ) 1 = ⇒ y−y =u ˙ U ( s) s−1 U ( s) s−1 D (s ) = = ⇒ u+u=e−e ˙ ˙ E...
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## This note was uploaded on 01/27/2014 for the course MAE 171A taught by Professor Idan during the Winter '09 term at UCLA.

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