| Line 21: | Line 21: | ||
:<math>X(Z) = -\left(\sum_{m=0}^{\infty}(2z)^{m}-1\right)</math> | :<math>X(Z) = -\left(\sum_{m=0}^{\infty}(2z)^{m}-1\right)</math> | ||
| − | if | + | if |2z| is greater than or equal to 1 then x(z) diverges, else: |
:<math>x(z) = -\left(\frac{1}{1-2z}-1\right)</math> | :<math>x(z) = -\left(\frac{1}{1-2z}-1\right)</math> | ||
Latest revision as of 16:34, 28 November 2008
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} $
