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So inverting X(Z) involves power series. | So inverting X(Z) involves power series. | ||
| − | <math>f(X)= \sum_{n=0}^\infty \frac{ | + | <math>f(X)= \sum_{n=0}^\infty \frac{e^3/x}{x^2}\!} \ </math> , near <math>X_0</math> |
<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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Computing equivalent to complex integration formula's | Computing equivalent to complex integration formula's | ||
| − | 1) Write X(Z) as a power series. | + | 1) Write X(Z) as a power <math>Insert formula here</math>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) | <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) | ||
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<math>X[n] = \ C_ -n</math> | <math>X[n] = \ C_ -n</math> | ||
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Example 1: | Example 1: | ||
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| + | <math>X(Z) = </math> | ||
Revision as of 05:08, 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{e^3/x}{x^2}\!} \ $ , 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 $ Insert formula here $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 $
Example 1:
$ X(Z) = $
