Difference between revisions of "Inverse Z transform" - Rhea

Revision as of 07:20, 23 September 2009

                                                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.