This may be a very simple question and I think I know the answer but I would like it to be confirmed by somebody else.

My questions regard the zeros of a systems (the roots of the numerator of a Transfer Function).

Question 1) Is it possible to move zeros around (kind of how one can do with poles)?

My current answer to this (correct me if I'm wrong) is no. The reason is because the zeros of the closed loop transfer function are at least the same with those of the plant (of course more zeros may be presented by a controller). As a results, regardless of the controller method, the zeros of the plant/system will not move. If this is correct, this leads me to my second questions.

Question 2) If the Plant/system has an unstable zeros, doesn't this automatically mean that, as far as loop gain is concerned, Robust Stability is never achieved?

The reason I think this is because if one plots the Root Locus, a pole will always move to that zero (assuming at least proper closed loop transfer function). So from a control engineering point of view, the only thing one can do, as far as stability is concerned, is to move the poles as far away from the imaginary axis as possible. The further away the poles, the more gain can be presented in the system that would not lead to instability.

Am I correct to think the above?

P.S. as far as Robust Stability, this is what I currently understand it to be. That being said, I am not very sure if it is correct.

  • $\begingroup$ With robust stability are you referring to the gain margin? And why would you think that in order to be robust you need an infinite gain margin? $\endgroup$ – fibonatic Feb 23 at 11:38
  • $\begingroup$ @fibonatic Yes I was referring to the gain margin. As for you second point, I din't think about it like that. This is an good point. So robustness can be a case by case thing? If one says "I want my system to be stable for gains up to 100" and we find that it is stable for up to 200, we can say that it has Robust Stability? $\endgroup$ – Dimitris Pantelis Feb 28 at 11:19

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.