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Partial Fraction Expansion (Inverse Z-Transform)

  1. Nov 20, 2011 #1
    1. The problem statement, all variables and given/known data

    [itex]H(z) = \frac{6-z^{-1}}{1+0.5z^{-1}} + \frac{2}{1-0.4z^{-1}} = k + \frac{A}{1+0.5z^{-1}} + \frac{2}{1-0.4z^{-1}}[/itex]

    where [itex]A = (6-z^{-1})[/itex] is evaluated at [itex]z^{-1}=-2[/itex].

    2. Relevant equations

    Partial fraction expansion.

    3. The attempt at a solution

    Why is [itex]z^{-1}[/itex] set equal to -2? I thought normally you use the terms in the denominators for [itex]z^{-1}[/itex] ...
     
  2. jcsd
  3. Nov 20, 2011 #2

    vela

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    What do you mean by "use the terms in the denominator for [itex]z^{-1}[/itex]"? Use them how?
     
  4. Nov 20, 2011 #3
    Like here: http://dspcan.homestead.com/files/Ztran/zinvpart.htm" [Broken]
     
    Last edited by a moderator: May 5, 2017
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