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Laplace Transform with Integral Convolution

  1. Oct 31, 2013 #1
    1. The problem statement, all variables and given/known data
    Determine the Laplace Transform of
    ∫(from 0 to t) (t-τ)cos(2(t-τ))e-4τ

    using Laplace Transform tables.


    2. Relevant equations
    I know the basic convolution theorem is
    (f*g)(t) = ∫f(τ)g(t-τ)dτ


    3. The attempt at a solution
    I'm not sure if this is double convolution or if I'm missing an easy substitution.
     
  2. jcsd
  3. Oct 31, 2013 #2

    LCKurtz

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    1. What are ##f(t)## and ##g(t)## in this problem?
    2. What is the formula for ##\mathcal L(f*g)##?
     
  4. Oct 31, 2013 #3
    Response:
    1. I'm not sure because I've never seen (t-tau) twice in one of these.
    2. Laplace(f*g) = F(s)G(s)
     
  5. Oct 31, 2013 #4

    LCKurtz

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    As an example, what would ##g(t-\tau)## be for ##g(t) = t^2\sin(t)##?
     
  6. Oct 31, 2013 #5
    In that problem, it would be setup so that f(t) = t^2 and g(t) = sin(t) making the convolution integral

    ∫ from 0 to t: t2*sin(t-τ) dτ

    and then integrated using trig identities
     
    Last edited: Oct 31, 2013
  7. Oct 31, 2013 #6

    LCKurtz

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    No. You didn't answer my question. I didn't ask you anything about convolution. Given ##g(t) = t^2\sin(t)## what would ##g(\pi/2)## be? ##g(s)##? ##g(t-\tau)##?
     
  8. Oct 31, 2013 #7
    [itex]\int_0^t (t-\tau)cos(2(t-\tau))e^{-4\tau} d\tau[/itex]

    it looks as if f(t) = t and g(t) = cos(2*f(t))

    Would it make sense to do a simple substitution and end up doing the integral

    [itex]\int_0^t u\cdot cos(2u) du[/itex]

    giving [itex]\frac{1}{2} u\cdot sin(2u) + cos(2u)[/itex]

    [itex]\frac{1}{2} t\cdot sin(2t) + cos(2t)[/itex]

    and then taking the laplace of those?
     
  9. Oct 31, 2013 #8
    I would say g(t-τ) would be (t-τ)2sin(t-τ)
     
  10. Oct 31, 2013 #9

    LCKurtz

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    No. You are missing the point of the problem. Just answer my question in my previous post for now. That will maybe give you an idea.
     
  11. Oct 31, 2013 #10

    LCKurtz

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    OK, so you see how you can have more than one ##t-\tau## for a function. Now can you identify ##f(t)## and ##g(t)## in your original problem?
     
  12. Oct 31, 2013 #11
    Yes, I do. Does that mean my attempt was incorrect?
     
  13. Oct 31, 2013 #12

    LCKurtz

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    If you mean your attempt in post #7, yes that is incorrect. Where did the exponential go in that post? But never mind that. What do you get for ##f(t)## and ##g(t)##? Do you see how to do the problem knowing those and using post #3?
     
  14. Oct 31, 2013 #13
    ahh, got it. thanks a lot.

    (sleep deprivation makes you miss things)
     
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