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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

1 Answers 1

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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}. $

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    I see, thanks. What a shame not to know that as an engineering student entering in Senior year.2012-04-18