discrete signals - Pole/Zero existence at infinity

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

Answer

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

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

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.