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Laplace Transform of Product of Two Functions

  1. Dec 12, 2011 #1
    1. The problem statement, all variables and given/known data

    Laplace Transform of u(t-∏/2)et

    (u is unit step function)

    2. Relevant equations

    Laplace Transform Table (any)

    3. The attempt at a solution

    I tried using the Laplace transform for the unit step function and the exponential function.

    L{u(t-∏/2)} = e-(∏s)/2

    L{et} = 1/(s-1)

    L{u(t-∏/2)et} = (e-(∏s)/2)(1/(s-1))

    When I check my answer this is all correct except I'm missing an e∏/2 term. (The correct answer is (e-(∏s)/2)(1/(s-1))(e∏/2). Does anyone know where this extra term comes from?
     
  2. jcsd
  3. Dec 12, 2011 #2

    LCKurtz

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    I assume you have the formula[tex]
    \mathcal L(f(t-a)u(t-a) = e^{-as}\mathcal Lf(t)[/tex]Write your function as[tex]
    e^{(t-\frac \pi 2)}e^{\frac \pi 2}[/tex]and use that.
     
  4. Dec 12, 2011 #3

    Ray Vickson

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    The Laplace transform of a product is NOT the product of the transforms. Why don't you just do the problem directly: u(t - π/2) is 0 for t < π/2 and 1 if t > π/2, so you just have a simple integral of an exponential.

    RGV
     
  5. Dec 12, 2011 #4
    Thanks, I got it now. I was thinking that the transform of the product was the product of the transforms.
     
    Last edited: Dec 12, 2011
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