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Invserse Z-Transform

  1. Apr 12, 2014 #1
    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. jcsd
  3. Apr 12, 2014 #2

    vela

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    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?
     
  4. Apr 12, 2014 #3
    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
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