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So inverting X(z) involves power series | So inverting X(z) involves power series | ||
| − | <math> f(x)= \sum_{n =-\infty}^{\infty} | + | <math> f(x)= \sum_{{n =-\infty}^{\infty}} \frac {f^{n} x_0 (x-x_0)^{n}} {n!} \ </math> |
<math>\frac{1}{1-x} = \sum_{n =-\infty}^{\infty} x^{n} \ </math> , geometric series when |x|< 1 | <math>\frac{1}{1-x} = \sum_{n =-\infty}^{\infty} x^{n} \ </math> , geometric series when |x|< 1 | ||
Revision as of 07:45, 23 September 2009
Inverse Z-transform
$ x[n]= \frac{1}{2\prod j} \oint_C X(z) z^{n-1} dz \ $
$ = \sum_ {poles a_i of X(z) z^{n-1}} \ $ Residue $ X(z) z^{n-1} \ $ $ = \sum_ {poles a_i of X(z) z^{n-1}} \ $ Coefficient of degree(-1) term in the power expansion of $ X(z) z^{n-1} \ $ about $ a_i $
So inverting X(z) involves power series
$ f(x)= \sum_{{n =-\infty}^{\infty}} \frac {f^{n} x_0 (x-x_0)^{n}} {n!} \ $
$ \frac{1}{1-x} = \sum_{n =-\infty}^{\infty} x^{n} \ $ , geometric series when |x|< 1
