Difference between revisions of "Class notes on Inverse Z transform" - Rhea

Revision as of 08:34, 23 September 2009

                                                 Inverse Z-transform

$ x[n]= \frac{1}{2\pi 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 =0}^{\infty} \frac {f^{n} x_0 (x-x_0)^{n}} {n!} \ $

$ \frac{1}{1-x} = \sum_{n =0}^{\infty} x^{n} \ $ , geometric series when |x|< 1

Shortcut 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 comparison

$ x[-n]= C_n \ $

or

$ x[n]= C_{-n} \ $

Example 1: $ x(z) = \frac{1}{1-z} \ $

Two possible ROC's Case 1: |z|< 1

$ x(z)= \sum_{n =0}^{\infty} z^n \ $ , by the formula of goemetric series

$ = \sum_{k =-\infty}^{0} z^{-k} \ $

$ = \sum_{k =-\infty}^{\infty} u(-k) z^{-k} \ $

So x[n]=u[-n]

Consistent of having inside of 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

              $ = \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 =-\infty}^{\infty}  -u[k-1] z^{-k} \  $

By comparison with the Z- transform formula we have,

x[n]= -u[n-1]

Consistent with having outside of circle as ROC.