Maybe the title of my question is a quite confusing but I am going to explain better. I have an expression like this: $$\frac{Z_1}{Z_2+Z_3+Z_4+Z_5+Z_6},$$ where every parameter $Z_x$ is an expression whit poles and zeros, like $\dfrac{sa+b}{sc+d}$ with $s=j\omega$.

Can I find all the poles and zeros of the first expression without resolve it if I know the poles/zeros of the singles $Z_x$?


The zeros? yes. The poles? no.

Zeros can only occur where the numerator is $0$ or the denominator is $\infty$. And the denominator can only be $\infty$ at a pole of one or more terms in the denominator. You do have to check these points to make sure cancellation does not occur that will result in something non-zero. But that is easier than fully "resolving" the denominator.

Poles can only occur when the numerator is $\infty$ or the denominator is $0$. Unfortunately, identifying where the denominator is $0$ will require "resolving" it. Therefore these potential poles (they too will need checked) are much harder to find.

  • $\begingroup$ Thanks Paul for your answer. Yes, the zeros are more simple cause substantialy the are the the poles of the espressions at denominator (correct me if I'm wrong). But can you explain me your statement "And the denominator can only be ∞ at a pole of one or more terms in the denominator" ? $\endgroup$ – thoraz Jun 23 '18 at 9:53
  • $\begingroup$ If each of the $Z_i$ in the denominator is finite, then so is their sum. So the only way for the entire denominator to have a pole is at a pole of one of the individual $Z_i$. So if you know the poles of the each of the $Z_i$, you also know where the poles of the denominator as a whole have to be. But you do have to be careful of the possibility of poles of two terms canceling out. For example $$\frac {x-1}x - \frac{x^2 - 1}x$$ has no pole, though each term does. $\endgroup$ – Paul Sinclair Jun 23 '18 at 17:17

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