How come does a system become more stable when a zero is added to a system.. I mean i doesn't not change the location of the pole, it is still the same?

An example: Looking a closed loop system consisting of an controller $G_c(s)$, and a system $G(s) = \frac{1}{s(s+1)}$

If the controller is a PD it the close cloop transfer function $T(s)$ will look like this.

$G_c(s) = Kp + Kds$ $T(s) = \frac{\frac{Kp}{s(s+1)} + \frac{Kds}{s(s+1)}}{1 + \frac{Kp}{s(s+1)} + \frac{Kds}{s(s+1)}}$

With adding a PD controller the close loop transfer function receives an zero, but how come does make a system more stable.

  • $\begingroup$ Not sure what you mean exactly, but it could be that the zero cancels the pole. $\endgroup$ – Matt L. May 14 '14 at 20:11
  • $\begingroup$ In PID control, when you add the D-part it should slow down the system, since the a zero is added. but adding a zero wouldn't affect the placement of the pole. it only affects the root locus, but that is only a problem when i add a gain. $\endgroup$ – newbiemath May 14 '14 at 20:28
  • $\begingroup$ Check out this document: ctms.engin.umich.edu/CTMS/… In Eq. (10) you see that the D-part ($K_d$) does change the pole location. $\endgroup$ – Matt L. May 14 '14 at 21:06
  • $\begingroup$ Further explanation on how it does would be nice. The close loop transfer function doesn't resemble the normal form.. $\endgroup$ – newbiemath May 14 '14 at 21:53
  • $\begingroup$ Maybe it would be a good idea if you added some formulas to your question, e.g. what you call the "normal form" of the closed loop TF etc. Then it should be possible to see what each part of the controller adds to it. $\endgroup$ – Matt L. May 15 '14 at 7:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.