Suppose I give you this as my transfer function $H(z)$:

$$ H(z) = \frac{1} { 1 - az^{-1}}$$

With no other information given, is it even possible to determine the inverse Z-transform?

The reason I'm asking is because on a particular Z-transform table I have, there are two possible resolutions for this, depending on the desired region of convergence:

h[n] =

As one can see we have two choices,

  1. $a^nu[n]$
  2. $-a^nu[-n-1]$

So I think I need more information about the system to make the right decision, that is, whether the response is supposed to be causal, stable, or both.

  • $\begingroup$ Very sorry, I forgot to accept the answer! $\endgroup$ – Novicegrammer Nov 11 '19 at 21:34
  • $\begingroup$ No problem, thanks! $\endgroup$ – Matt L. Nov 12 '19 at 8:01

It's true that the algebraic expression for the $\mathcal{Z}$-transform is generally not sufficient for computing the corresponding time-domain sequence. The additional information we need is the region of convergence (ROC). The ROC together with the expression for the $\mathcal{Z}$-transform uniquely determines the corresponding time-domain sequence.

For the given example


it is clear that the expression $(1)$ can be obtained in two ways. First,

$$\begin{align}-\sum_{n=-\infty}^{-1}a^nz^{-n}&=-\sum_{n=1}^{\infty}a^{-n}z^n\\&=-\frac{\frac{z}{a}}{1-\frac{z}{a}}\\&=\frac{1}{1-az^{-1}},\quad |z|<|a|\end{align}$$

corresponding to the sequence $h[n]=-a^nu[-n-1]$, and, secondly,

$$\sum_{n=0}^{\infty}a^nz^{-n}=\frac{1}{1-az^{-1}},\quad |z|>|a|$$

corresponding to $h[n]=a^nu[n]$. In both cases I've used the formula for the geometric series.

The only difference is the ROC, so it's crucial to know the ROC in addition to the expression for the $\mathcal{Z}$-transform. The ROC $|z|>|a|$ indicates a right-sided sequence, whereas the ROC $|z|<|a|$ corresponds to a left-sided sequence. For higher order systems, you can also have an annulus $|a|<|z|<|b|$ as ROC, corresponding to a two-sided sequence.

If the sequence is interpreted as the impulse response of a system, then the system is stable if the ROC contains the unit circle $|z|=1$. It is causal if the sequence is zero for $n<0$.

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