1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Invserse Z-Transform

  1. Apr 12, 2014 #1
    1. The problem statement, all variables and given/known data

    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


    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    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
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted