How can poles and zeros exist at infinity?Can anybody explain using a system function?


2 Answers 2


consider a general rational transfer function of order $N$, first with an equal number of zeros and poles:

$$ \begin{align} H(z) & = A \prod_{n=1}^N \frac{z - q_n}{z - p_n} \\ & = A \frac{\prod_{n=1}^N z - q_n}{\prod_{n=1}^N z - p_n} \\ & = A \frac{\prod_{n=1}^N q_n - z }{\prod_{n=1}^N p_n - z} \\ & = B \frac{\prod_{n=1}^N 1 - \frac{z}{q_n} }{\prod_{n=1}^N 1 - \frac{z}{p_n}} \\ \end{align} $$

where $ B = A \prod_{n=1}^N \frac{q_n}{p_n}$ .

now suppose that the number of zeros is actually less than the number of poles. we could express the transfer function as

$$ H(z) = C \frac{\prod_{n=1}^M 1 - \frac{z}{q_n} }{\prod_{n=1}^N 1 - \frac{z} {p_n}} $$

where $M<N$. or we can express it as

$$ H(z) = B \frac{\prod_{n=1}^N 1 - \frac{z}{q_n} }{\prod_{n=1}^N 1 - \frac{z} {p_n}} $$

where $(N-M)$ zeros have values of $\infty$ which make $\frac{z}{q_n}$ disappear (for those zeros), leaving only $1$ as a factor in the transfer function.

at the moment, i am not sure what to do with the $A$, $B$, or $C$ factors which might have an $\infty$ in them. i'll worry about that later.


I will show you two simple examples to build up some intuition. Imagine a causal impulse response with $h[0]=0$:


In this case you get

$$\lim_{z\rightarrow \infty}H(z)=0$$

This is of course due to the negative powers of $z$ in (1).

Now imagine a non-causal impulse response that starts at $n=-1$, such as:


Now you have


This pole is of course caused by the positive power of $z$ in (2).

Note that the transfer function of a causal impulse response can never have poles at infinity because you don't get any positive powers of $z$.

  • $\begingroup$ so Matt, are you saying that, for a causal discrete-time system, the number of zeros can exceed the number of poles? $\endgroup$ Commented Oct 9, 2014 at 16:17
  • $\begingroup$ well, it's a salient point that you almost made but did not. normally, i would consider that a tad "incomplete". it's your answer. it speaks to an issue my answer did not (i didn't say why there might be fewer zeros than poles or fewer poles than zeros) but didn't quite finish the point. $\endgroup$ Commented Oct 9, 2014 at 17:01
  • $\begingroup$ BTW, if i were you and worried about misinterpretation of an answer, i would worry a lot more about the other two questions. despite the case you make, you would almost certainly get the exam question wrong if it were "true or false, $y[n] = p \cdot y[n-1] + x[n]$ is an LTI system." you would protest to the professor, but you would still get that question wrong because, at least in the EE discipline, it would not be interpreted as you do. $\endgroup$ Commented Oct 9, 2014 at 17:06

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