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I am studying control systems and I have seen that a non-minimum phase zero imposes limitation in bandwidth , which is that the bandwidth has to be lower than a frequency which is half the frequency of the RHP pole.

I have also seen that it imposes in the step response an undershoot, but as far as I know there are no other influences of RHP zero in the time domain.

Does a RHP zero imposes some limitation in the time domain?

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If a RHP zero imposes limitation on the bandwidth in the frequency domain, it also imposes limitation in the time domain. This is due to time-frequency duality.

For example, if you want to have a really fast step response, a slow RHP zero will prevent you from reaching that goal.

I really recommand Murray's book for more information http://www.cds.caltech.edu/~murray/amwiki/index.php/Second_Edition

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  • $\begingroup$ Thanks for answering. So the limitations regard only the speed of the response and the undershoot? Thanks again. $\endgroup$ – J.D. Feb 11 at 22:27
  • $\begingroup$ It is a pretty important limitation, don't you think? It could prevent you from reaching your specifications. $\endgroup$ – Ben Feb 11 at 22:28
  • $\begingroup$ I have found this paper that talks about a time domain integral when it is present a RHP zero: sciencedirect.com/science/article/pii/S1474667015385694. It makes a differentiation about when the zero is stable or when it is unstable. So, it is possible to compute a time until whihìch the step response has better behaviour? Sorry, maybe is not a good question, but I want to understand well. Thanks. $\endgroup$ – J.D. Feb 12 at 14:59
  • $\begingroup$ For example, if I have a RHP zero at $s=1$, does it change something if I plot the step response starting at $t=0$ or at $t=5$? Thanks in advance. $\endgroup$ – J.D. Feb 12 at 15:33
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    $\begingroup$ If it is time invariant, then if I shift the input , I obtain the same output shifted by the same quantity, so it means that I have the same step response but shifted. So, even if I have a RHP zero, it impses limitation on the badwidth, and so it imposes limitation on the speed and adds an undershoot in the step response. If it is time invariant it should have the same characteristics but shifted in time. I thisnk that this solves my doubts. Thanks. $\endgroup$ – J.D. Feb 12 at 16:02

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