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                                                 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

Alumni Liaison

Ph.D. 2007, working on developing cool imaging technologies for digital cameras, camera phones, and video surveillance cameras.

Buyue Zhang