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<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)} \ </math>

Revision as of 04:44, 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)} \ $

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