Laplace Transform of cos(at) * cos(bt)

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SUMMARY

The Laplace transform of the product of cos(at) and cos(bt) can be expressed as a ratio of two polynomials. The discussion highlights the conversion of cosines into exponentials using Euler's formula, leading to terms like e^(jt(a+b)) and e^(-jt(a+b)). The confusion arises when attempting to find the Laplace transform of jsin(t(-a-b)), which should yield j*s / (s^2 + (-a-b)^2), but does not provide the correct result. The consensus is that the operation in question is multiplication, not convolution.

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I have to find the laplace transform of cos(at) * cos(bt) and express it as a ratio of two polynomials. I converted both of the cosines into exponentials, and took the laplace transform of those. I think I'm getting confused on the complex side of things.

I get a few things like e^(jt(a+b)), and e^(-jt(a+b)) so I use Euler's formula and get some cosines and jsines. So how do you find the laplace transform of say jsin(t(-a-b)). Shouldn't it just be j*s / (s^2 + (-a-b)^2)?

Except this does not yield the correct answer. Maybe I'm converting all of the transforms to one fraction incorrectly...?
 
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Do you mean to find the Laplace transform of \cos (at) * \cos (bt) ? Where the * denotes convolution? Or did you meant it as multiplication?
 
Corneo said:
Do you mean to find the Laplace transform of \cos (at) * \cos (bt) ? Where the * denotes convolution? Or did you meant it as multiplication?

I assume he meant it as multiplication, since he is doing all of that work. I thought the same thing myself. If it were convolution it sure would make the problem a lot easier. :p
 

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