- #1
Drao92
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- 0
Homework Statement
H(z)=1/(1-2/3*z-1)
h(n)=?
Homework Equations
I tried to use the residue method to solve this problem but it doesn't give me the good result and i am not sure if i don't know how to use the method or this problem can't 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
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?