I have a discrete system for which the root locus equation is given as:
$$A(z) + K\cdot B(z) = 0$$
They are such that $A(0) = 1, B(0) > 0$, and $K>0$. $\frac B A$ is minimum phase and a minimal realization, so all the poles and zeros lie strictly inside the unit circle and there can be no pole-zero cancellations.. Furthermore, the degree of $\mathbf A$ and $\mathbf B$ is the same, so the number of zeros and poles is the same and we cannot have any asymptotes going off into infinity. Is it still possible to give some combination of zeros and poles inside the unit circle for which the root locus can become unstable? (i.e. there exists a $K$ such that the equation is satisfied for some $z$ outside the unit circle).
My guess is that this is not possible, but only because I haven't been able to find a counterexample. Can someone verify this? How could I prove it?
Let $A(z) = 2z+1, B(z) = 1-2z, K=2$. Then $A({3 \over 2}) + K B({3 \over 2}) = 0$.
Here is an example with different degrees:
$A$ as above, $B(z) = 2z^2-2z+1$, $K={1 \over 2}$. Then the roots of $B$ have magnitude ${1 \over \sqrt{2}}$ and $A({1 \over 2} (-1+i\sqrt{5})) + K B({1 \over 2} (-1+i\sqrt{5})) = 0$. Divide $A,B$ by $2$ to normalise the leading coefficient of $A$.