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I am studying Feedback Control of Computing Systems. (specifically using Hellerstein's book, section 3.1.4, page 74)

An inverse Z-Tranform also can be obtained by a long division. In the book there is an example I poorly understood. Let $$ U(z) = \frac{2}{(z-1)^2} = \frac{2}{z^2-2z+1} $$ and the long division is: (doubt equation) $$ \qquad\qquad {\atop{z^2-2z+1)}} \frac{\qquad\qquad\quad 2z^{-2} + 4z^{-3} + 6z^{-4} + \cdots} {2+0z^{-1}+0z^{-2}+0z^{-3}+0z^{-4}+\cdots} \\ \frac{2-4z^{-1}+2z^{-2}}{\;\;\;\quad4z^{-1}-2z^{-2}}\\ \qquad\qquad\quad\;\frac{4z^{-1}-8z^{-2}+4z^{-3}}{\qquad\quad\;6z^{-2}-4z^{-3}}\\ \qquad\qquad\qquad\qquad\qquad\qquad\;\frac{6z^{-2}-12z^{-3}+6z^{-4}}{\qquad\quad\;8z^{-3}-6z^{-4}}\\ \qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\vdots $$ That is, $u(2) = 2$, $u(3) = 4$ e $u(4) = 6$. And it is consistent with the previously known time-domain function defined as $$u(k) = 2(k-1)$$

Does anyone explain what exactly is happening (step-by-step) in the doubt equation?

Assumption: all signals have a value of $0$ for $k<0$.

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    It's just like [long division of polynomials](http://en.wikipedia.org/wiki/Polynomial_long_division), except that negative exponents are allowed. This way there is no remainder, and the process goes on forever, creating an infinite series with negative powers of $z$.2012-12-30
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    @PavelM, good pointing! Now I could understand what happened. Thank you.2012-12-31

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The doubt equation is about Long Division of Polynomials, as pointed by @PavelM.

In fact, the equation is the result of an algorithm.

This video is a good source to understand the algebraic step-by-step.

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    +1 Thanks for adding the video link. Technical point: links can be included in the convenient format [text](url); I edited your post accordingly. Also, you can accept your answer by clicking the checkmark on the left; this will make it clear to others that the issue is resolved.2012-12-31
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    @PavelM, thank you for your tips. I still should wait 23 hours to accept my own answers, after that I will do so.2012-12-31