I have a question regarding the poles and zeros at infinity I often read here in DSP SE and also in some textbooks about poles and zeros at infinity

This question also answers somehow (but not in much easy/simple way) Pole/Zero existence at infinity

I am confused about how to realize the poles and zeros at infinity especially from observing/looking at system transfer function??


It's actually quite straightforward: positive powers of $s$ (or, in discrete-time, $z$), correspond to poles at infinity. Negative powers give you zeros at infinity.

Let's look at some examples. In continuous time, an ideal differentiator has the transfer function


Clearly, $\lim_{s\to\infty}H(s)=\infty$, hence you have a pole at infinity (and a zero at $s=0$).

The transfer function of an ideal integrator is


Here we have $\lim_{s\to\infty}H(s)=0$, i.e., a zero at infinity (and a pole at $s=0$).

In discrete-time, a delay of one sample corresponds to


and you get $\lim_{z\to\infty}H(z)=0$.

A (non-causal) advance by one sample is represented by the transfer function


In this case you have a pole at infinity: $\lim_{z\to\infty}H(z)=\infty$

Note that in continuous time as well as in discrete time, systems with poles at infinity cannot be causal and stable.

  • $\begingroup$ what do you mean by last line note??system with poles at infinity cannot be stable??can't be even marginally stable??as in the case of differentiator?? $\endgroup$ – abtj Sep 29 '19 at 11:12
  • $\begingroup$ @abtj: Exactly that, they can't be causal and stable. An ideal differentiator is unstable. $\endgroup$ – Matt L. Sep 29 '19 at 11:24
  • $\begingroup$ @MattL. can the systems with poles at infinity be marginally stable?? $\endgroup$ – engr Sep 13 '20 at 17:35
  • $\begingroup$ @engr: No, marginal stability means that there are single poles on the stability border, which in continuous time is the imaginary axis, while in discrete time it is the unit circle. $\endgroup$ – Matt L. Sep 13 '20 at 18:55

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