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<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 | ||
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| + | Shortcut to computing equivalent to complex integration formula's | ||
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| + | 1) Write x(z) as a power series. | ||
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| + | <math>x(z)= \sum_{n =-\infty}^{\infty} \C_n z^n \ </math> ,series must converge for all z's in the ROC of x(z) | ||
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| + | 2) Observe that | ||
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| + | <math>x(z) = \sum_{n =-\infty}^{\infty} x[n] z^{-n} \ </math> | ||
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| + | i.e., | ||
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| + | <math>x(z) = \sum_{n =-\infty}^{\infty} x[-n] z^n \ </math> | ||
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| + | 3) By comparoson | ||
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| + | <math>x[-n]= \C_n \ </math> | ||
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| + | or | ||
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| + | <math>x[n]= \C_{-n} \ </math> | ||
Revision as of 08:51, 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
Shortcut to 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 in 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 comparoson
$ x[-n]= \C_n \ $
or
$ x[n]= \C_{-n} \ $
