# Invserse Z-Transform

1. Apr 12, 2014

### Drao92

1. The problem statement, all variables and given/known data
H(z)=1/(1-2/3*z-1)
h(n)=?

2. Relevant equations
I tried to use the residue method to solve this problem but it doesnt give me the good result and i am not sure if i dont know how to use the method or this problem cant be solved using this method.
Also, its pretty obvious that the result is h(n)=(2/3)n because it looks like a geometric progression

3. The attempt at a solution
I used the formula of residue method.
h(n)=residue(H(z)*zn-1) when z=2/3
The result is 3*(2/3)n
Its this result wrong because when we calculate Z transform of (2/3)n which is (1-(2/3)n*z-n )/(1-(2/3)*z-1) we ignore (2/3)n because is 0 when n=infinity?

2. Apr 12, 2014

### vela

Staff Emeritus
You have
$$z^{n-1} H(z) = \frac{z^{n-1}}{1-\frac{2}{3z}} = \frac{z^n}{z-\frac{2}{3}}.$$ How did you get $h(n)=3(2/3)^n$ from that?

3. Apr 12, 2014

### Drao92

I think i found what what im doing wrong.
I made H(z)=3z/(3z-2) then i multiply by zn-1 and simplify by (3z-2) but i must simplify by (3z-2)/3 when calculating the residue... From now ill stick only to z-x form for denominator.
Eveyrthing is clear now, thanks!

Last edited: Apr 12, 2014