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 $$ {3.9x+6\over (1+.5x)(1-.4x)} ={A\over 1+.5x}+{B\over 1-.4x}. $$