Pole/Zero existence at infinity

Question

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

Answer 1

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.

Answer 2

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

$$H(z)=h[1]z^{-1}+h[2]z^{-2}+\ldots\tag{1}$$

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:

$$H(z)=h[-1]z+h[0]+h[1]z^{-1}\tag{2}$$

Now you have

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

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 $z$.