Inverse Z transform - Rhea

                                                  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 $