# Pole-zero cancellation Paradox

Suppose we have an open-loop transfer function $$G(s) = \frac{1}{s(s+a)(s+b)}$$ If we plot the root locus for the closed-loop system we will get roughly something like this : Now the question is when I add a new zero to the system which is at $-a$ then the book says that we should plot the root-locus without cancelling the pole-zero ( $-a$ in this case ) . I have a practical doubt , in real systems suppose we add a new zero somehow obviously it will not be exactly at $-a$ but at some $-a+\epsilon$ where $\epsilon$ is very small , even in that case the root-locus will become something like this : Because now the aysmptotes will change cause $$n-m = 2$$ . Therefore the new asymptotes are at $$\frac{\pi}{2}$$ rad and $$\frac{3\pi}{2}$$ rad . These two plots become completely different , if I go strictly by the book then my system (even after the addition of new zero ) becomes unstable for some value of gain $K$ but if I consider the situation practically then my system is always stable . Please help me out , which one is correct .

## 2 Answers

It's not a paradox. Adding that zero makes it a completely different system. I checked in MATLAB and both plots look right (for an arbitrary a and b, |b|>|a|). That new zero annihilates with a pole which is why you are getting a different root locus plot. Adding LHP zeroes, among other things, improves the stability of your system, though at the cost of transient performance.

This all theoretically makes sense. Your plots are correct and what not but the answer to your question is probably more to do with the practicality of adding that extra zero in. If you're thinking in terms of a mechanical system, then it is difficult to add a zero which is so close to a pole.