Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question

1 Answer 1

up vote 2 down vote accepted

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

share|improve this answer
    
Okay, I think I got it. Give me a second to think about it. –  Aeon Apr 18 '12 at 17:34
    
@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
    
I see, thanks. What a shame not to know that as an engineering student entering in Senior year. –  Aeon Apr 18 '12 at 17:44

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.