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Laplace transform

  1. May 28, 2008 #1
    [SOLVED] Laplace transform

    1. The problem statement, all variables and given/known data

    Find the inverse laplace transform of [tex]\frac{e^{-2s} }{s^2 + s - 2}[/tex]

    2. Relevant equations

    3. The attempt at a solution

    I'm able to do about half of the problem
    using partial fractions, I've found

    I can find the inverse Laplace transform of the latter part of that expression [tex]1/3u_2(t) e^{t-2}[/tex] unfortunately, I don't know how I can modify the first part so that it's shifted by -2. Anyone know what I should do?
  2. jcsd
  3. May 28, 2008 #2
    Use calculus of residues (have you taken complex analysis yet?)
  4. May 28, 2008 #3

    Look in the wiki page about laplace transformations under 'time shifting'.
  5. May 28, 2008 #4
    Unfortunately I haven't studied complex analysis.
    I made an error in my original post (and I can't edit it for some reason)
    It's now worse than before
    [tex]s^2 + s - 2 = (s+2)(s-1)[/tex]
    [tex]-\frac{e^{-2s}}{3(s+2)} + \frac{e^{-2s}}{3(s-1)}[/tex]

    I see the time shift equation on wikipedia, but since the 's' is in the denominator I don't see how to shift either so that they match up with [tex]u_2(t)[/tex]
  6. May 28, 2008 #5
    Hum, you seem to be misreading it, do this for your two terms separately

    Step 1-- cover the exponential up with your hand. What's left is your 1/(s+2) (or 1/(s-1) for the other term) with your 3 and your signs of course. Call that F(s).

    Step 2-- look up on a table the f(t) such that L[f]=F(s) from Step 1. Write that f(t) down.

    Step 3-- now look on that exponential you were ignoring before (exp(-2s)) that's telling you that your answer will not be f(t), but instead be f(t-2)H(t-2) where H is the unit step function. So write that down.

    Now repeat for the other term and then add the two expressions you get from Step 3 together.
  7. May 28, 2008 #6
    got it now thanks, those steps are really handy
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