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.
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)
(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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- Winter '09
- 1 g, GD, Unstable zero/pole cancellation