# Tidying up my final answer on z-transform

Discussion in 'Homework Help' started by u-will-neva-no, Jan 14, 2012.

1. ### u-will-neva-no Thread Starter Member

Mar 22, 2011
230
2
Hey!

The sequence of: $x[n] = 1, \frac{1}{2}^1,\frac{1}{2}^2,\frac{1}{2}^3,...$

was asked to be expressed using the z-transform:

$X(z) = 1 +\frac{1}{2}^1z^-^1 +\frac{1}{2}^2z^-^2+\frac{1}{2}^3z^-^3+...$

What I can't do is express the final result into the one on the sheet which is:
$\frac{z}{z-0.5}$

The notes say to use the linear difference equation:

$\frac{x[n]+x[n-1]}{2}$

I found x[n-1] just by subtracting one from each value for x[n] above so when I added the numerator I got 1 and the denominator as 2...clearly it is wrong.

Thank you in advance for helping!

2. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
790
This is the sum of a geometric series.

$X(z)=1+\frac{1}{2z}+\frac{1}{4z^2}+\frac{1}{8z^3} + ....$

Since the series is infinite we may write ...

$\frac{1}{2z}X(z)=X(z)-1$

$\frac{1}{2z}X(z)-X(z)=-1$

$(\frac{1}{2z}-1)X(z)=-1$

$(\frac{1-2z}{2z})X(z)=-1$

$X(z)=\frac{-2z}{1-2z}$

$X(z)=\frac{2z}{(2z-1)}$

$X(z)=\frac{z}{(z-0.5)}$

u-will-neva-no likes this.