# How to calculate Inverse Z-Transform by long division

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, 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 $z$. –  user53153 Dec 30 '12 at 20:51
@PavelM, good pointing! Now I could understand what happened. Thank you. –  Lourenco Dec 31 '12 at 18:06