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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??

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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

$$H(s)=s\tag{1}$$

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

$$H(s)=\frac{1}{s}\tag{2}$$

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

$$H(z)=z^{-1}\tag{3}$$

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

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

$$H(z)=z\tag{4}$$

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.

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  • $\begingroup$ 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?? $\endgroup$
    – DSP_CS
    Commented Sep 29, 2019 at 11:12
  • $\begingroup$ @abtj: Exactly that, they can't be causal and stable. An ideal differentiator is unstable. $\endgroup$
    – Matt L.
    Commented Sep 29, 2019 at 11:24
  • $\begingroup$ @MattL. can the systems with poles at infinity be marginally stable?? $\endgroup$
    – DSP_CS
    Commented Sep 13, 2020 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.
    Commented Sep 13, 2020 at 18:55
  • $\begingroup$ @MattL."Note that in continuous time as well as in discrete time, systems with poles at infinity cannot be causal and stable." But what about case of zero at infinity such as in case of integrator? $\endgroup$
    – DSP_CS
    Commented Sep 12, 2021 at 17:44

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