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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, 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

1 Answer 1

up vote 1 down vote accepted

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. –  user53153 Dec 31 '12 at 18:18
    
@PavelM, thank you for your tips. I still should wait 23 hours to accept my own answers, after that I will do so. –  Lourenco Dec 31 '12 at 21:42

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