Number2Pencil
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Homework Statement
I'm trying to expand the following with partial fractions:
H(z) = \frac{z^2 + 1.5932z + 1}{z^2 + 0.9214z + 0.5857}
Homework Equations
The Attempt at a Solution
Just going through my normal procedure, I end up with the following:
H(z) = \frac{0.3359-j0.0857518}{z-(-0.4607+j0.6111)}+\frac{0.3359+j0.0857518}{z-(-0.4607-j0.6111)}+1
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:
\frac{H(z)}{z} = \frac{z+1.5932+z^{-1}}{z^2+0.9214z+0.5857}
I follow my normal procedures again:
\frac{H(z)}{z} = \frac{-0.354-j0.283}{z-(-0.4607+j0.6111)}+\frac{-0.354+j0.283}{z-(-0.4607-j0.6111)}
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?