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1) Write X(Z) as a power series. | 1) Write X(Z) as a power series. | ||
| − | + | <math>X(Z) = \sum_{n=-\infty}^{\infty} \ C_n Z^n \ </math> , series must converge for all Z's on the ROC of X(Z) | |
| + | |||
| + | 2) Observe that | ||
| + | |||
| + | <math>X(Z) = \sum_{n=-\infty}^{\infty} \ x[n] Z^(-n) \ </math> | ||
Revision as of 05:55, 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
Computing equivalent to complex integration formula's
1) Write X(Z) as a power series.
$ X(Z) = \sum_{n=-\infty}^{\infty} \ C_n Z^n \ $ , series must converge for all Z's on the ROC of X(Z)
2) Observe that
$ X(Z) = \sum_{n=-\infty}^{\infty} \ x[n] Z^(-n) \ $
