# Partial Fraction Expansion of Transfer Function

How do I go from:

$$\frac{3(1+0.2z^{-1})(1+z^{-1})}{(1+0.5z^{-1})(1-0.4z^{-1})}$$

to

$$-3 + \frac{7}{1-0.4z^{-1}} - \frac{1}{1+0.5z^{-1}}$$

I understand that the first form can be expanded as

$$\frac{A_1}{1-0.4z^{-1}} + \frac{A_2}{1+0.5z^{-1}}$$

so the $7$ and $-1$ don't bother me, but I don't understand where the first term $(-3)$ in the second equation comes from.

Thank you

The degree of the polynomial (in the variable $x=z^{-1}$) in the numerator and denominator are the same. You have to do the division first, and then apply partial fractions on the remainder term.
$${3(1+.2x)(1+x)\over (1+.5 x)(1-.4x)}=-3+{3.9x+6\over (1+.5x)(1-.4x)}.$$
Then write $${3.9x+6\over (1+.5x)(1-.4x)} ={A\over 1+.5x}+{B\over 1-.4x}.$$
• @Aeon Write the left hand side as $.6x^2+3.6x+3\over -.2x^2+.1x+1$; then divide: $$-.2x^2+.1x+1 |\overline{.6x^2+3.6x+3}$$ The first step in the division is to say $-.2x^2$ goes into $.6x^2$ a total of $-3$ times... See here for an example of polynomial long division. – David Mitra Apr 18 '12 at 17:42