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Proof of the convolution theorem for laplace transform

  1. Dec 28, 2007 #1
    My textbook provides a proof but there's one thing about the proof i do not understand

    it starts assuming L{f(t)} = the laplace integral with the f(t) changed to f(a)

    same goes with L{g(t)} as it changes it to g(b)

    i understand the big picture>>starting from a product of 2 L transforms and working ur way back to an expression in terms of the two functions transformed, but how can u change the variable for f and g when we started off stating f of T and g of T.???
  2. jcsd
  3. Dec 28, 2007 #2
    Am I not presenting my questions clearly?
  4. Dec 28, 2007 #3

    [tex]F(s)=\mathcal{L}\{f(t)\}=\int_0^{+\infty}e^{-s\,t}\,d\,t\,f(t) \quad \text{and} \quad
    G(s)=\mathcal{L}\{g(t)\}=\int_0^{+\infty}e^{-s\,t}\,g(t)\,d\,t [/tex]

    be the two Laplace transormations and let's denote

    [tex](f\star g)(t)=\int_0^{+\infty}f(\tau)\,g(t-\tau)\,d\,\tau [/tex]

    the convolution of [tex]f,\,g[/tex].

    Thus the Laplace transform for the convolution would be

    [tex]\mathcal{L}\{(f\star g)(t)\}=\int_0^{+\infty}e^{-s\,t}\left( \int_0^{+\infty}f(\tau)\,g(t-\tau)\,d\,\tau \right)\,d\,t =\int_0^{+\infty}\int_0^{+\infty}e^{-s\,t}f(\tau)\,g(t-\tau)\,d\,\tau \,d\,t \quad (1)[/tex]

    The above double integral is to be evaluated in the domain [tex] \mathcal{D}=\{(t,\tau): t \in (0,+\infty),\tau \in (0,+\infty)\} [/tex].

    Performing the change of variables
    [tex](t,\tau)\rightarrow (u,v): t=u+v, \, \tau=v [/tex] with [tex] u \in (0,+\infty),v \in (0,+\infty) [/tex] we have [tex]d\,\tau\,d\,t=d\,u\,d\,v[/tex] since the Jacobian equals to 1. Thus (1) becomes

    [tex]\mathcal{L}\{(f\star g)(t)\}=\int_0^{+\infty}\int_0^{+\infty}e^{-s\,(u+v)}f(u)\,g(v)\,d\,u\,d\,v= \int_0^{+\infty}e^{-s\,u}f(u)\,d\,u \cdot \int_0^{+\infty}e^{-s\,v}\,g(v)\,d\,v[/tex]

    yielding to

    [tex]\mathcal{L}\{(f\star g)(t)\}=\mathcal{L}\{f(t)\} \cdot \mathcal{L}\{g(t)\} [/tex]

    Nothing but double integrals! :smile:
  5. May 10, 2010 #4
    Isn't the laplace convolution between 0 and t rather than 0 and infinity?
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