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 $k<0$.