How do I go from:

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


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

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

  • $\begingroup$ Okay, I think I got it. Give me a second to think about it. $\endgroup$ – Aeon Apr 18 '12 at 17:34
  • $\begingroup$ @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. $\endgroup$ – David Mitra Apr 18 '12 at 17:42
  • $\begingroup$ I see, thanks. What a shame not to know that as an engineering student entering in Senior year. $\endgroup$ – Aeon Apr 18 '12 at 17:44

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