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:

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


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

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?


2 Answers 2


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.

  • 3
    $\begingroup$ 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. $\endgroup$ 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.

  • $\begingroup$ Just curious: how does this work at $z=0$ where $z/z$ is undefined? $\endgroup$
    – Hilmar
    May 30, 2022 at 11:21
  • $\begingroup$ 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? $\endgroup$
    – DaBler
    May 30, 2022 at 11:29
  • $\begingroup$ That's only true if $K=L$; generally $H_1(z)\neq H_2(z)$. $\endgroup$
    – Matt L.
    May 30, 2022 at 17:38

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