Complex Partial Fraction Expansion

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SUMMARY

The discussion centers on the complex partial fraction expansion of the transfer function H(z) = (z² + 1.5932z + 1) / (z² + 0.9214z + 0.5857). The user initially attempts to expand H(z) using standard procedures but encounters an issue when performing an inverse-z transform. The key conclusion is that an additional term of 1.7074 must be added to the expansion, which is derived from multiplying the numerator and denominator by z to eliminate the negative power in the partial fraction. This adjustment leads to the final expression of H(z) being simplified correctly.

PREREQUISITES
  • Understanding of complex numbers and their operations
  • Familiarity with partial fraction decomposition
  • Knowledge of inverse-z transforms
  • Proficiency in using LaTeX for mathematical expressions
NEXT STEPS
  • Study advanced techniques in partial fraction decomposition for complex functions
  • Learn about inverse-z transforms and their applications in signal processing
  • Explore the use of LaTeX for presenting mathematical solutions clearly
  • Investigate the implications of negative powers in transfer function expansions
USEFUL FOR

Students and professionals in electrical engineering, particularly those focusing on signal processing and control systems, will benefit from this discussion. It is also valuable for anyone looking to enhance their understanding of complex function expansions and inverse transforms.

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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?
 
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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

\frac{1.707}{z}

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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