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I am studying Feedback Control of Computing Systems. (specifically using Hellerstein's book, section 3.1.4, page 76)

To solve difference equation Z-Tranform can be applied. In the book there is an example I did not understand how the solution was deduced. Given an integrator: (linear difference equation)

$ y(k+1) = y(k) + u(k) $

and the corresponding Z-transform:

$\sum_{k=0}^{\infty}{y(k+1)z^{-k}} = zY(z) - zy(0)$

So (begin)

$ zY(z)-zy(0)=Y(z)+U(z)$

Solving it: (end)

$ Y(z) = \frac{1}{z-1}[U(z)+zy(0)] $

What are the intermediate steps from the equation labeled as begin through the end one?

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

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    @mrf, yes, I think it is about algebra inserted in the context of dynamical-systems as you can see in the accepted answer.2012-12-31

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There is a sign error in your transformed equation. The correct equation is
$z Y(z) - z y(0) = Y(z) + U(z).$ The rest is algebra, $\begin{eqnarray*} zY(z)-zy(0) &=& Y(z)+U(z) \\ zY(z)-Y(z) &=& U(z) + z y(0) \\ (z-1)Y(z) &=& U(z) + z y(0) \\ Y(z) &=& \frac{1}{z-1}(U(z)+z y(0)). \end{eqnarray*}$

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    @Lourenco: Glad to help. Cheers!2012-12-30