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Complex Partial Fraction Expansion

  1. Sep 26, 2011 #1
    1. The problem statement, all variables and given/known data
    I'm trying to expand the following with partial fractions:

    [tex]H(z) = \frac{z^2 + 1.5932z + 1}{z^2 + 0.9214z + 0.5857}[/tex]


    2. Relevant equations



    3. The attempt at a solution

    Just going through my normal procedure, I end up with the following:

    [tex]H(z) = \frac{0.3359-j0.0857518}{z-(-0.4607+j0.6111)}+\frac{0.3359+j0.0857518}{z-(-0.4607-j0.6111)}+1[/tex]

    but it turns out, to facilitate performing an inverse-z transform (which is a different thread), we need to perform the partial fraction a little different:

    [tex]\frac{H(z)}{z} = \frac{z+1.5932+z^{-1}}{z^2+0.9214z+0.5857}[/tex]

    I follow my normal procedures again:

    [tex]\frac{H(z)}{z} = \frac{-0.354-j0.283}{z-(-0.4607+j0.6111)}+\frac{-0.354+j0.283}{z-(-0.4607-j0.6111)}[/tex]

    Now this is apparently wrong, and according to my professor, I forgot to add on a 1.7074 to this answer.

    My question is: Why do I need to add on the 1.7074? I thought if the degree of the denominator is higher than the numerator, there is no extra term added? And for that matter, how would I actually go about calculating this 1.7074 term?
     
  2. jcsd
  3. Sep 26, 2011 #2
    Looks like I posted too soon...because I just figured it out.

    Turns out, having a negative power in the partial fraction was a bad idea, so I multiply the top and bottom by z, giving me one extra expansion term. Sure enough, it ends up being

    [tex]\frac{1.707}{z}[/tex]

    And when I finally multiply H(z) by z to get the final output, the z on the bottom goes away.

    Guess I just needed to look at it in fantastic LaTex
     
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