Poles and zeros form of a transfer function

I know that a transfer function for a discrete-time LTI system can be written in the form

$$H(z) = \frac{Y(z)}{X(z)} = \frac { \displaystyle\sum_{m=0}^M {b_m z^{-m}}} {1 + \displaystyle\sum_{n=1}^N {a_n z^{-n}}}$$

Now I am interested in the poles and zeros of this transfer function. I have seen two different definitions:

$$H(z) = \frac{Y(z)}{X(z)} = A\frac {\displaystyle\prod_{k=1}^K (1 - \alpha_k z^{-1}) } {\displaystyle\prod_{l=1}^L (1 - \beta_l z^{-1}) } \tag{I}$$

and

$$H(z) = \frac{Y(z)}{X(z)} = B\frac {\displaystyle\prod_{k=1}^K (z - \alpha_k) } {\displaystyle\prod_{l=1}^L (z - \beta_l) } \tag{II}$$

where $$\alpha$$ are zeros and $$\beta$$ are poles.

These definitions are not the same. Thus, which of them is valid? If both, what is the difference between them?

The two expressions are generally not identical. In the special case $$K=L$$ they're equivalent, otherwise they differ by a (positive or negative) power of $$z$$:

$$\frac {\displaystyle\prod_{k=1}^K (1 - \alpha_k z^{-1}) } {\displaystyle\prod_{l=1}^L (1 - \beta_l z^{-1}) }=\frac {z^{-K}\displaystyle\prod_{k=1}^K (z - \alpha_k) } {z^{-L}\displaystyle\prod_{l=1}^L (z - \beta_l) }=z^{L-K}\cdot\frac {\displaystyle\prod_{k=1}^K (z - \alpha_k) } {\displaystyle\prod_{l=1}^L (z - \beta_l) }$$

If $$K\neq L$$, there are poles or zeros at the origin.

• Yes, yes. In my specific $K=L$ answer below I should have said "multiply $H_1(z)$ by $z^K/z^K$ (which is unity)." Thanks for the correct general answer Matt. May 31, 2022 at 9:39

@DaBler Ignoring the $$A$$ and $$B$$ factors, if you multiply $$H_1(z) = \frac {\displaystyle\prod_{k=1}^K (1 - \alpha_k z^{-1}) } {\displaystyle\prod_{l=1}^L (1 - \beta_l z^{-1}) }$$ by the ratio $$z/z$$ (which is unity) you obtain $$H_2(z) = \frac {\displaystyle\prod_{k=1}^K (z - \alpha_k) } {\displaystyle\prod_{l=1}^L (z - \beta_l) } .$$ So the above $$H_1(z) = H_2(z)$$. I find the $$H_2(z)$$ form to be the easiest to use in finding the values of poles and zeros.

• Just curious: how does this work at $z=0$ where $z/z$ is undefined? May 30, 2022 at 11:21
• I see. I assume I must multiply $z/z$ repeatedly according to $K$ or $L$ (whichever is greater). And in the end, I have multiple zeros or poles left at 0. Is that right? May 30, 2022 at 11:29
• That's only true if $K=L$; generally $H_1(z)\neq H_2(z)$. May 30, 2022 at 17:38