Partial Fraction Expansion (Inverse Z-Transform)

  • #1

Homework Statement



[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].

Homework Equations



Partial fraction expansion.

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] ...
 

Answers and Replies

  • #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?
 
  • #3
What do you mean by "use the terms in the denominator for [itex]z^{-1}[/itex]"? Use them how?

Like here: http://dspcan.homestead.com/files/Ztran/zinvpart.htm" [Broken]
 
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