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Laplace transform

  1. May 26, 2009 #1
    to calculate the inverse laplace transform of a function F(s), s+2 was replaced with z for convenience. the inverse laplace transform of z was found--let's denote the function g(t).

    now, how do i prove that L-1{F(s)} = L-1{F(z-2)} = g(t) * e-2t

    i am attempting to prove by rewriting L-1{F(z-2)} in terms of L-1{F(z)*something} but i don't know what exactly to do.

    this is not a homework question, just attempt at a proof.
  2. jcsd
  3. May 27, 2009 #2
    I find it easiest to just show the shifting property using the integral definition and then using that information to show the inverse Laplace transform.

    [tex]F(s-a) = \int_0^\infty{e^{-(s-a)t}f(t)dt} = \int_0^\infty{e^{at}e^{-st}f(t)dt} = L[e^{at}f(t)][/tex]

    And therefore [tex]L^{-1}[F(s-a)] = e^{at}f(t) = e^{at}L^{-1}[F(s)][/tex]

    I am not sure if you were looking for this or something else in particular. It's 5am here and I'm having some trouble seeing straight.. :shy:
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