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Consider a dynamic system $$\dot{x}=Ax+Bu \text{ and } y=Cx$$ The transfer function is $$sX(s) = AX(s)+BU(s),$$ so $$(sI-A)X(s)=BU(s)$$ and $$Y(s)=CX(s)$$ combining the two transfer equations, we have $$Y(s)=C(sI-A)^{-1}BU(s),$$ so $$G(s)=\frac{Y(s)}{U(s)}=C(sI-A)^{-1}B.$$

Now if, $G(s)$ has unstable zeros, which means zeros on the right half plane, section 1.3.4 "The Infinite Horizon LQ Problem" in book "Model Predictive Control: Theory, Computation, and Design 2nd Edition, Rawlings, et.al." said a LQR controller could invert this unstable zero to cause the system to become unstable. How is this possible? I know zeros of $G(s)$ is poles of $1/G(S)$, which is obvious. But how is this related to the system described above?

From the point of view of root-locus, infinite gain lead a pole to approach zero, but LQR always generate a finite value gain, what does the inversion of zeros mean in this context, could someone give some detailed explanation or maybe give some good reference?

sunxd
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  • "a book said". Please edit your question to properly cite which book, with page number, even if it's not in English. If it's available on the web, include a link. If the book is only available in print, you can quote exactly what it said (or a translation), with enough context that people can answer your question intelligently. – TimWescott Jun 05 '21 at 03:30
  • i wouldn't call zeros in the right half-plane "unstable". they're not minimum-phase. when a system is inverted, the poles and zeros swap roles and all of the right half-plane zeros become poles and those poles are unstable. – robert bristow-johnson Jun 05 '21 at 05:02
  • @TimWescott, I added the reference. – sunxd Jun 05 '21 at 11:14
  • In the standard form the LQR only considers the $A$ and $B$ matrices of the system (which are not enough information to define the zeros). Furthermore, the control signal would be defined as $u=-Kx$ with $x$ assumed to be known, so I don't see how it could cancel any zeros, since it would be just a static gain. I don't have access to that book, but is the relevant section maybe talking about zero- dynamics? – fibonatic Jun 05 '21 at 13:49

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When you invert a transfer function $H(s) = n(s)/d(s)$, all zeros become poles and all poles become zeros since inverting means $H(s)^{-1} = d(s)/n(s)$ where $n(s)$ and $d(s)$ refer to the original numerator and denominator polynomials.

Therefore, if you have a transfer function with zeros in the right half plane, if you invert that transfer function as a means of compensating for those zeros, then the zeros become unstable poles.

Dan Boschen
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