# Poles and zeros of a transfer function

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

$$H(z)=z+2+z^{-1}$$

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 :-)

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.
• 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? – Likely May 28 '17 at 20:54
• @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. – Matt L. May 28 '17 at 21:19
• @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 $\infty$? – Likely May 28 '17 at 21:59
• @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. – Likely May 29 '17 at 0:46