(New page: This page would give an example of how to perform the z-transform. Suppose <math>x[n] = \frac{-u[-n-1]}{2^n}</math> Using the definition of z-transform: :<math>X(Z) = \sum_{n=-\infty}...)
 
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Using the definition of z-transform:
 
Using the definition of z-transform:
  
:<math>X(Z) = \sum_{n=-\infty}^{\infty}x[n]z^-n<math>
+
:<math>X(Z) = \sum_{n=-\infty}^{\infty}x[n]z^{-n}</math>
 +
 
 +
:<math>X(Z) = \sum_{n=-\infty}^{\infty}\frac{-u[-n-1]}{2^n}z^{-n}</math>
 +
 
 +
:<math>X(Z) = \sum_{n=-\infty}^{-1}-\frac{z^{-n}}{2^n}</math>
 +
 
 +
by letting m = -n
 +
 
 +
:<math>X(Z) = \sum_{m=1}^{\infty}-\frac{z^m}{2^{-m}}</math>

Revision as of 16:24, 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}} $

Alumni Liaison

Correspondence Chess Grandmaster and Purdue Alumni

Prof. Dan Fleetwood