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1)  Write x(z) as a power series.
 
1)  Write x(z) as a power series.
  
<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)
+
<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)
  
 
2) Observe that  
 
2) Observe that  
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3) By comparoson
 
3) By comparoson
  
<math>x[-n]= \C_n \ </math>
+
<math>x[-n]= C_n \ </math>
  
 
or
 
or
  
<math>x[n]= \C_{-n} \ </math>
+
<math>x[n]= C_{-n} \ </math>

Revision as of 07:52, 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} \ $

Alumni Liaison

Recent Math PhD now doing a post-doctorate at UC Riverside.

Kuei-Nuan Lin