This page would give an example of how to perform the z-transform.

Suppose

$x[n] = \frac{-u[-n-1]}{2^n}$

Using the definition of z-transform:

$X(Z) = \sum_{n=-\infty}^{\infty}x[n]z^{-n}$
$X(Z) = \sum_{n=-\infty}^{\infty}\frac{-u[-n-1]}{2^n}z^{-n}$
$X(Z) = \sum_{n=-\infty}^{-1}-\frac{z^{-n}}{2^n}$

by letting m = -n

$X(Z) = \sum_{m=1}^{\infty}-\frac{z^m}{2^{-m}}$
$X(Z) = -\sum_{m=1}^{\infty}(2z)^{m}$
$X(Z) = -\left(\sum_{m=0}^{\infty}(2z)^{m}-1\right)$

if |2z| is greater than or equal to 1 then x(z) diverges, else:

$x(z) = -\left(\frac{1}{1-2z}-1\right)$
$x(z) = \frac{2z}{2z-1}$

Therefore, the z-transform of $\frac{-u[-n-1]}{2^n}$ is

$\frac{2z}{2z-1}$

## Alumni Liaison

Correspondence Chess Grandmaster and Purdue Alumni

Prof. Dan Fleetwood