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<math> = \sum_{poles a_i ( X(Z) Z ^ (n-1))} Residue ( X(Z) Z ^ (n-1)) \ </math> | <math> = \sum_{poles a_i ( X(Z) Z ^ (n-1))} Residue ( X(Z) Z ^ (n-1)) \ </math> | ||
<math> = \sum_{poles a_i ( X(Z) Z ^ (n-1))} \ </math> Coefficient of degree (-1) term on the power series expansion of <math> ( X(Z) Z ^ (n-1)) \ </math> <math> about a_i \ </math> | <math> = \sum_{poles a_i ( X(Z) Z ^ (n-1))} \ </math> Coefficient of degree (-1) term on the power series expansion of <math> ( X(Z) Z ^ (n-1)) \ </math> <math> about a_i \ </math> | ||
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So inverting X(Z) involves power series. | So inverting X(Z) involves power series. | ||
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<math>\frac{1}{(1-x)} = \sum_{n=0}^\infty X^n \ </math> , geometric series where |X|=1 | <math>\frac{1}{(1-x)} = \sum_{n=0}^\infty X^n \ </math> , geometric series where |X|=1 | ||
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Revision as of 06:01, 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) \ $
i.e.,
$ X(Z) = \sum_{n=-\infty}^{\infty} \ x[-n] Z^n \ $
3) By comparison
$ X[-n] = \ C_n \ $
or
$ X[n] = \ C_ -n $
