| Line 11: | Line 11: | ||
<math>f(X)= \sum_{n=0}^\infty \frac{f^n X_0 (X-X_0)^n}{n!} \ </math> , near <math>X_0</math> | <math>f(X)= \sum_{n=0}^\infty \frac{f^n X_0 (X-X_0)^n}{n!} \ </math> , near <math>X_0</math> | ||
| − | <math>\frac{1}{(1-x)} \ </math> | + | <math>\frac{1}{(1-x)} = \sum_{n=0}^\infty X^n \ </math> , geometric series where |X|=1 |
Revision as of 04:47, 23 September 2009
Inverse Z-transform
$ x[n] = \oint_C {X(Z)}{Z ^ (n-1)} , dZ \ $
where C is a closed counterwise countour inside the ROC of the Z- transform and around the origin.
$ = \sum_{poles a_i ( X(Z) Z ^ (n-1))} Residue ( X(Z) Z ^ (n-1)) \ $ $ = \sum_{poles a_i ( X(Z) Z ^ (n-1))} Coefficient of degree (-1) term on the power series expansion of ( X(Z) Z ^ (n-1)) about a_i \ $
So inverting X(Z) involves power series.
$ f(X)= \sum_{n=0}^\infty \frac{f^n X_0 (X-X_0)^n}{n!} \ $ , near $ X_0 $
$ \frac{1}{(1-x)} = \sum_{n=0}^\infty X^n \ $ , geometric series where |X|=1
