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Convolution Integral and Differential Equation

  1. Apr 23, 2013 #1

    I am really confused where to start with this problem. I know about convolutions somewhat. We have done them a little. Where is a good place to begin with this problem?
  2. jcsd
  3. Apr 23, 2013 #2
    What does the convolution theorem state about two functions? What are those two functions in this problem?

  4. Apr 23, 2013 #3
    if f and g are piecewise continuous then the convolution of the two is
    [itex]\int f(\tau)g(t-\tau)d\tau[/itex] from 0 to t
    [itex]\int f(t-\tau)g(\tau)d\tau[/itex] from 0 to t


    F(s)G(s)=Laplace Transform of the convolution of f and g

    in this case y(t-w) is f(t-[itex]\tau[/itex]) and [itex]e^{-10w}[/itex] is g([itex]\tau[/itex])

    So now how can I use this too get y(t)?
  5. Apr 23, 2013 #4
    Do you see that the integral in the problem you gave is the convolution of y(t) and [itex]e^{-10t [/itex] ?

    How can you apply the convolution theorem? You will obtain an algebraic equation in s. Solve it for Y(s).

  6. Apr 23, 2013 #5
    So what I've figured so far is s*Y(s)-25* inverse laplace of {Y(s)G(s)} = 1

    How can I solve for Y(s) when it is in the inverse laplace transform?
  7. Apr 24, 2013 #6
    Taking the Laplace transform of every term in the differential equation should give you an algebraic equation in Y(s). Solve for this in terms of s, then take the inverse transform.

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