# Partial Fraction Expansion (Inverse Z-Transform)

## 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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vela
Staff Emeritus
What do you mean by "use the terms in the denominator for $z^{-1}$"? Use them how?
What do you mean by "use the terms in the denominator for $z^{-1}$"? Use them how?