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Suppose a closed feedback system has the transfer function

[tex]\frac{C(s)}{R(s)} = \frac{G(s)}{1 + G(s)H(s)}[/tex]

In order to employ Nyquist's method to judge the stability of the system, I consider the loop gain

[tex]L(s) = G(s)H(s)[/tex]

and map the s-plane contour into the [itex]L(s)[/itex] plane. But what if there is a pole of G(s) common with a zero of H(s) (or a zero of G(s) common with a pole of H(s))?

Since the Nyquist plot depends only on the loop gain and not on G and H separately, the effect of this canceled pole-zero pair will not be evident in the L(s)-plane contour. What is the physical significance of this? Does it have a direct relation to the notion of observability in control theory?

In other words, what is the effect of a pole-zero cancellation in the feedforward and feedback transfer functions, on the system vis a vis its stability and other time/frequency domain characteristics?

Thanks in advance.

EDIT: Strictly, I can cancel the pole-zero pair only when I am not at that point. Therefore, the reduced expression (after cancellation) will hold only for those neighborhoods of the s-plane that do not enclose the pole/zero pair. Suppose the pole-zero pair is at s = 0, i.e. G(s) has a pole of order 1 at s = 0 and H(s) has a zero of order 1 at s = 0. Then, L(s) will have no pole or zero at s = 0. But am I supposed to consider a semicircular contour at s = 0 while constructing the Nyquist s-plane contour?

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# Effect of Pole Zero Cancellation on Nyquist Plot/Stability Criterion

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