# Explanation for a Z-transform solution of a difference equation

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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Algebra-precalculus, really? –  mrf Dec 30 '12 at 23:09
@mrf, yes, I think it is about algebra inserted in the context of dynamical-systems as you can see in the accepted answer. –  Lourenco Dec 31 '12 at 18:11

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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Thank you! I just fixed the equation as you pointed. –  Lourenco Dec 30 '12 at 22:01
@Lourenco: Glad to help. Cheers! –  user26872 Dec 30 '12 at 22:01