Difference between revisions of "Why does Z≠0 correspond to instability?"

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* 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>
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* Z=#RHP zeros of <math>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.
 
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.
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[[User:Hines|Hines]] 17:12, 12 November 2007 (PST)
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[[Category:CDS101/110

Revision as of 01:12, 13 November 2007

First recall some definitions:

  • the "loop transfer function" is L(s)=P(s)C(s)
  • the closed loop system is described by {\frac  {L(s)}{1+L(s)}}
  • Z=#RHP zeros of 1+L(s)

So we see that if a point is a zero of 1+L(s), 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 1+L(s) has any RHP zeros, the closed loop system around the loop transfer function is unstable.

Hines 17:12, 12 November 2007 (PST)

[[Category:CDS101/110