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Laplacian Translation

  1. Apr 15, 2006 #1
    Given:
    F(s) = (s-1)/(s+1)^3

    Find:
    f(t)

    Solution:

    Using the equation that when F(s) = n!/(s-a)^(n=1), L^(-1){F(s)} = t^n*e^(at)

    So far I find that f(t) = e^(-t)*(-t^2+__)

    The book says that f(t) = e^(-t)*(t-t^2)
    How did they get the t?
     
  2. jcsd
  3. Apr 16, 2006 #2
    You should rewrite F(s)
    [tex]
    F(s) \equiv \frac{s-1}{(s+1)^3} \equiv \frac{1}{(s+1)^2} - \frac{2}{(s+1)^3}
    [/tex]
    and apply the inverse laplace transform
    when F(s) = n!/(s-a)^(n=1), L^(-1){F(s)} = t^n*e^(at)
    for each term.
     
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