Difference between revisions of "Why does Z≠0 correspond to instability?"
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First recall some definitions: | First recall some definitions: | ||
+ | |||
* the "loop transfer function" is <math>L(s)=P(s)C(s)</math> | * the "loop transfer function" is <math>L(s)=P(s)C(s)</math> | ||
* the closed loop system is described by <math>\frac{L(s)}{1+L(s)}</math> | * the closed loop system is described by <math>\frac{L(s)}{1+L(s)}</math> | ||
+ | * <math>Z=#\text{RHP zeros of}1+L(s)</math> | ||
+ | |||
+ | So we see that if a point is a zero of <math>1+L(s)</math>, then it is a pole of the closed-loop system. Now, if that pole lies in the right half-plane, then the closed-loop system will be unstable. Thus if the function <math>1+L(s)</math> has any RHP zeros, the closed loop system around the loop transfer function is unstable. |
Revision as of 01:10, 13 November 2007
First recall some definitions:
- the "loop transfer function" is
- the closed loop system is described by
- Failed to parse (lexing error): Z=#\text{RHP zeros of}1+L(s)
So we see that if a point is a zero of , then it is a pole of the closed-loop system. Now, if that pole lies in the right half-plane, then the closed-loop system will be unstable. Thus if the function has any RHP zeros, the closed loop system around the loop transfer function is unstable.