Partial Fraction Expansion (Inverse Z-Transform)

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SUMMARY

The discussion focuses on the application of partial fraction expansion in the context of the Inverse Z-Transform for the function H(z) = (6 - z^{-1}) / (1 + 0.5z^{-1}) + 2 / (1 - 0.4z^{-1}). The key point is the evaluation of A at z^{-1} = -2, which raises questions about the methodology of setting z^{-1} to specific values. Participants seek clarification on how to properly utilize the terms in the denominators for the evaluation process.

PREREQUISITES
  • Understanding of Inverse Z-Transform
  • Familiarity with partial fraction expansion techniques
  • Knowledge of complex variable theory
  • Basic proficiency in digital signal processing (DSP)
NEXT STEPS
  • Study the properties of Inverse Z-Transform in detail
  • Learn advanced techniques for partial fraction decomposition
  • Explore applications of partial fraction expansion in digital signal processing
  • Review examples of evaluating functions at specific points in Z-domain analysis
USEFUL FOR

Students and professionals in electrical engineering, particularly those specializing in digital signal processing, who are looking to deepen their understanding of the Inverse Z-Transform and its applications in system analysis.

ElfenKiller
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Homework Statement



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}}

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

Homework Equations



Partial fraction expansion.

The Attempt at a Solution



Why is z^{-1} set equal to -2? I thought normally you use the terms in the denominators for z^{-1} ...
 
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What do you mean by "use the terms in the denominator for z^{-1}"? Use them how?
 
vela said:
What do you mean by "use the terms in the denominator for z^{-1}"? Use them how?

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