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Laplace transform
MrOletta
Registered User regular
Registered User regular
Hey I'm working on a problem and I just can't figure it out. I'm to prove..
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!
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
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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?