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?

  • $\begingroup$ Maths formula look better in $\LaTeX$. Here (meta.math.stackexchange.com/questions/5020/…) is a quick tutorial. $\endgroup$ – Τίμων Apr 6 '16 at 19:27
  • $\begingroup$ Thanks, will keep that in mind! $\endgroup$ – Tusike Apr 6 '16 at 19:28
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    $\begingroup$ If $A(z)=B(z) = z-1$ then ${B \over A} = 1$ is minimum phase but $z-1$ is a factor of $A+KB$. $\endgroup$ – copper.hat Apr 6 '16 at 19:40
  • $\begingroup$ Indeed, that would be a great counterexample, but I forgot to mention that the system is irreducible (no pole-zero cancellations allowed, it's already a minimal realization). Thanks for pointing it out, I've edited the question to include this assumption. $\endgroup$ – Tusike Apr 6 '16 at 19:47

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 $A$.

  • $\begingroup$ Thank you, that's a perfect example! I was trying graphically and I did not think of inverting the sign of the highest order coefficient; I checked and this inverts the normal rules I learned about root-locus plots. Out of curiosity, do you think there would be a counterexample with the assumptions imposed on the coefficients of $z^n$? So not $A(0) = 1$ and $B(0) > 0$, but $a_n = 1$ and $b_n > 0$? $\endgroup$ – Tusike Apr 6 '16 at 20:55
  • $\begingroup$ I added another example. $\endgroup$ – copper.hat Apr 7 '16 at 1:00
  • $\begingroup$ But for it to work you had to drop the assumption of the orders being equal? $\endgroup$ – Tusike Apr 7 '16 at 5:22
  • $\begingroup$ My solution is that it won't, because due to $z^n$ having the same sign in both $A$ and $B$ and no solutions being outside the unit circle, $A(z)$ and $B(z)$ will also have the same sign outside the unit circle. Thus the expression $A(z)+K \cdot B(z)$ must remain of that same sign regardless of the value of $K$ and can never reach zero. Is this a correct train of thought? $\endgroup$ – Tusike Apr 7 '16 at 10:29
  • $\begingroup$ @Tusike: $A,B$ are complex valued. what do you mean by having the same sign? $\endgroup$ – copper.hat Apr 7 '16 at 11:42

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