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Laplace transform dirac delta

  1. Nov 8, 2014 #1
    1. The problem statement, all variables and given/known data
    [itex]x(t) = cos(3πt)[/itex]
    [itex]h(t) = e-2tu(t)[/itex]

    Find [itex]y(t) = x(t) * h(t) [/itex](ie convolution)

    2. Relevant equations
    Y(s) = X(s)H(s) and then take inverse laplace tranform of Y(s)

    3. The attempt at a solution
    [itex] L(x(t)) = [πδ(ω - 3π) + πδ(ω + 3π)] [/itex]
    [itex] L(h(t)) = \frac{1}{s+2} [/itex]

    Laplace Transform inverse :
    [itex] y(t) = \frac{1}{2} \int \frac{e^{st}}{s+2} * δ(w-3π) [/itex] + [itex] \frac{1}{2} \int \frac{e^{st}}{s+2} * δ(w+3π)dt = \frac{1}{2} \frac {e^{3sπ} + e^{-3sπ}}{s+2} [/itex]

    Im not sure about direc delta integration..
     
    Last edited: Nov 8, 2014
  2. jcsd
  3. Nov 8, 2014 #2

    vela

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    The Laplace transform of ##\cos 2\pi t## isn't what you said it is. (It should be a function of ##s##, not ##\omega##.) You're thinking of the Fourier transform.
     
  4. Nov 8, 2014 #3
    Is it [itex] \frac {s}{s^2 + (3π)^2}[/itex] ?
     
  5. Nov 8, 2014 #4

    vela

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    Yes.
     
  6. Nov 8, 2014 #5
    But isn't that the LT of cos(3pi*t)u(t)?
     
  7. Nov 8, 2014 #6

    vela

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    Yes. Usually, it's assumed you're working with the unilateral Laplace transform, and you essentially ignore what's going on before t=0. Are you supposed to be using the bilateral Laplace transform?
     
  8. Nov 8, 2014 #7

    Ray Vickson

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    The "bilateral" transform of ##\cos(3 \pi t)## does not exist, because the integral
    [tex] \int_{-\infty}^{\infty} e^{-st} \cos(3 \pi t) \, dt [/tex]
    is not convergent. However, the "unilateral" LT of ##u(t) \cos(3 \pi t)## certainly does exist.
     
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