# Is it impossible to determine the inverse Z-transform without any other information?

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:

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.

• Very sorry, I forgot to accept the answer! – Novicegrammer Nov 11 at 21:34
• No problem, thanks! – Matt L. Nov 12 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

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

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 $$n<0$$.