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Inverse z transform

  1. Apr 24, 2015 #1

    etf

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    Hi!
    My task is to find discrete signal x(n), if z transform of that signal is $$X(z)=\frac{5}{(z-2)^{2}}$$. It is known that signal is causal. Here is what I have done. Since signal x(n) is causal, convergence of z transform of that signal will be outside of circle with radius r:

    Code_Cogs_Eqn_2.gif


    We have in bracket sum which represents z transform of signal:

    Code_Cogs_Eqn_3.gif

    But I don't know what to do next :(
     
    Last edited: Apr 24, 2015
  2. jcsd
  3. Apr 24, 2015 #2

    rude man

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    Homework Helper
    Gold Member

    Don't know about your approach.
    The standard approach is
    X(z) = 5/(z-2)2
    Take X(z)/z
    Form partial fraction expansion.
    One of the denominators will be (z-2)2 which has to be dealt with uniquely. My hint:
    multiply lhs and rhs of X(z)/z by z, then let z → ∞.
    Now you'd have all the numerator coefficients of X(z)/z.
    Now multiply by z to get X(z). Invert term-by-term. The term with (z-2)2 will probably throw you.
    Hint: Consider the theorem {n(n-1)(n-2) ... (n-m+1)/m!αmnu[n] ⇔ z/(z-α)(m+1).
     
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