Laplace transform

MrOlettaMrOletta Registered User regular
edited March 2007 in Help / Advice Forum
Hey I'm working on a problem and I just can't figure it out. I'm to prove..

equation.jpg

So this is basically convolution theorem.

We have f(t) * g(t), where f(t) is f(tau), and g(t) is our Bessel function which reduces to J[at].


L{f(t)*g(t)} = L{f(t)} L{g(t)}

The Laplace of the Bessel function is (s^2 + a^2) ^ (-1/2). The Laplace of Sin[a*t] is a/(s^2 + a^2)

So L{f(t)} * (s^2 + a^2) ^ (-1/2) = a/(s^2 + a^2)

I'm a little lost as to how to continue, or maybe my approach is totally wrong. Can someone point me in the right direction? L{f(t)} should be a/Sqrt[s^2 + a^2] for the problem to work out.

HELP!

MrOletta on

Posts

  • ecco the dolphinecco the dolphin Registered User regular
    edited March 2007
    If the problem that you're asking is correct, then I suspect that your approach might be all wrong.

    You're trying to say that L{f(t)} is a/Sqrt[s^2 + a^2]. This implies that f(t) = L^-1{a/Sqrt[s^2 + a^2]} which means that f(t) is a specific function.

    The way I read the problem is that it doesn't matter what f(t) is, the Laplace with the Bessel function always turns into Sin[a*t].

    It's a subtle difference - did I explain it clearly enough?

    ecco the dolphin on
    Penny Arcade Developers at PADev.net.
Sign In or Register to comment.