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:<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 <math>|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) = \frac{2z}{2z-1}</math> | ||
| + | |||
| + | Therefore, the z-transform of <math>\frac{-u[-n-1]}{2^n}</math> is | ||
| + | |||
| + | :<math>\frac{2z}{2z-1}</math> | ||
Revision as of 16:32, 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: :<math>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} $
