Lecture7

S 3s 5 case 2 gs is unstable if w presents any

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Unformatted text preview: pression to the left. Stability: Open loop transfer function L(s) := G(s)D(s) is stable. Case 1: G(s) is stable. Any stable D(s) makes L(s) stable. For example, G(s) = 2 , s+3 D (s ) = s+1 s+5 ⇒ L(s ) = 2(s + 1) . (s + 3)(s + 5) Case 2: G(s) is unstable. If w presents, any small w makes y (t) diverge. If w is absent, can L(s) be stabilized? No. For example, G(s) = 1 , s−1 D (s ) = s−1 s+2 ⇒ L(s ) = s−1 , (s − 1)(s + 2) poles: 1, -2. Unstable. Unstable zero/pole cancellation does not work. Tracking: For example, tracking a step command, G(s) = 1 , s+1 D (s ) = 10(s + 1) , s + 10 ⇒ >problem is to cause the output to follow the reference input as closely as possible. L(s ) = 10(s + 1) 10 = . (s + 1)(s + 10) s + 10 The DC gain L(0) = 1. So the final value of unit step response is 1. >proble...
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