Inverse Z-transform
$ x[n] = \frac{1}{2 \prod j} \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 $
Example 1:
$ X(z) = \frac{1}{(1-z)} \ $
Two possible ROC
Case 1: |z|<1
$ X(z) = \sum_{n=0}^\infty z^n \ $
$ = \sum_{k=-\infty}^{0} \ z^{-k} \ $ $ = \sum_{n=-\infty}^{\infty} \ u(-k) z^{-k} \ $
so, x[n]=u[-n]
Consistent as having inside a circle as ROC.
Case 2: |z|>1
$ X(z) = \frac{1}{(1-z)} \ $
$ = \frac{1}{z(\frac{1}{z}-1)} \ $ $ = \frac {-1}{z} {\frac{1}{1-\frac{1}{z}}} \ $, observe $ \ |\frac {1}{z}|< 1 \ $
Now by using the geometric series formula, the series can be formed as
$ = \frac {-1}{z} \sum_{n=0}^\infty (\frac {1}{z})^{n} \ $
$ = - \sum_{n=0}^\infty z^{-n-1} \ $
Let k=(n+1)
$ = - \sum_{k=1}^\infty z^{-k} \ $
$ = \sum_{k=1}^\infty -u(k-1) z^{-k} \ $
By coparison with the Z- transform formula
$ x[n]= -u[n-1] \ $
Consistent as having outside of circle as the ROC.
