# Tidying up my final answer on z-transform

#### u-will-neva-no

Joined Mar 22, 2011
230
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!

#### t_n_k

Joined Mar 6, 2009
5,455
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