What are the poles and zeros of this transfer function (in $z$):


and how would you approach the resolution of such problem?

Personally, I would write $$H(z)=\displaystyle\frac{z^2+2z+1}{z}=\frac{(z+1)^2}{z}$$ then I would say that there is a zero (double) at $z=-1$ (numerator $= 0$) and a pole at $z=0$ (denominator $= 0$).

  • But wouldn't we also have a pole at $z=\infty$ since it makes $H(z)\to \infty$ which corresponds to the definition of a pole?
  • And in that case wouldn't that be contradictory since we know that $H(z)$ can be seen as a transfer function of a FIR filter which we know is always stable, so how can the filter be stable if it has a pole at $z=\infty$ (outside unit circle)?

I'm sure I have a basic misunderstanding about poles and zeros otherwise there shouldn't be any contradiction and hopefully someone can help me clarify this :-)


1 Answer 1


The "poles-inside-unit-circle" stability criterion only applies to causal systems. Your system is not causal because it uses one sample from the future owing to the $z$ term.

The general technique to check for stability involves looking at the regions of convergence (ROC) of $H(z)$. If the ROC includes the unit circle, then the system is stable.

See also this answer to the question: What is the relationship between poles and system stability?

  • $\begingroup$ Ok! What about the pole at $z=\infty$? Do you agree that it is a pole for this function? And what about the statement that the number of poles is equal to the number of zeros for FIR filters? $\endgroup$
    – Likely
    May 28, 2017 at 20:54
  • 1
    $\begingroup$ @Likely: You're right: there are two zeros (at $z=-1$), and there are two poles, one at $z=0$ and one at $z=\infty$. Poles outside the unit circle say something about causality, and generally not about stability. It is only for causal systems that poles outside the unit circle imply instability. $\endgroup$
    – Matt L.
    May 28, 2017 at 21:19
  • $\begingroup$ @MattL. This means that when we check for poles we don't only check the $z$ values for which the denominator goes to $0$ but we must also check when the numerator goes to $\infty$. In that case, for the example I gave can we still conclude about the stability of $H(z)$ with this pole at $\infty$? $\endgroup$
    – Likely
    May 28, 2017 at 21:59
  • $\begingroup$ @MattL. I think I got it. The ROC here is all the z-plane except $0$ and $\infty$, hence, $H(z)$ is stable because the unit circle is inside this ROC. $\endgroup$
    – Likely
    May 29, 2017 at 0:46
  • $\begingroup$ @Likely: That's exactly it. $\endgroup$
    – Matt L.
    May 29, 2017 at 6:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.