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Can't find inverse Z transform

  1. May 21, 2017 #1
    • New poster has been reminded to use the Homework Help Template when starting threads in the schoolwork forums
    1. The problem statement, all variables and given/known data
    I got the laplace transfer function H(s) = 1/(s + 2) and I'm suppose to find the inverse Z transform by first converting to H(z) by s = Ts/2*(z-1)/(z+1)
    Then do inverse Z-transform using the "displacement rule" - Never heard of.

    2. Relevant equations
    H(s) = 1/(s + 2)
    s = Ts/2*(z-1)/(z+1)

    3. The attempt at a solution

    I can't get any serious answer, and the partial z-inverse that I manage to find it's incredibly complicated (see image).
    Namnl_s.png
    What is this 'displacement rule' and how do I use it here?
     
    Last edited: May 21, 2017
  2. jcsd
  3. May 21, 2017 #2

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    I don't have reference 17 (James G. Advanced modern engineering mathematics. Reading: Addison-Wesley; 1993.) that they refer to, but it looks like this paper ( https://www.ncbi.nlm.nih.gov/pmc/articles/PMC5240402/ ) uses it to associate the scaled/shift inside δ(γk-λtot(t)) of equation 13 with (1/γ)ztot(t)/γ in equation 15.
    Conversions like that are fairly common in transformations involving the delta function.
     
  4. May 22, 2017 #3
    The solution was to, after replacing 's' with the Z components, then put all under one fraction sign (no plus in nominator). Then multiply with Z^-1 in bottom and top until you got only Z^- terms. This = Y(x)/X(x) which can then be translated back to y[n] and y[n-1] = Z^-1*Y(x)
     
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