Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

  1. Feb 21, 2006 #1
    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...?
  2. jcsd
  3. Feb 22, 2006 #2
    Do you mean to find the Laplace transform of [itex]\cos (at) * \cos (bt)[/itex] ? Where the [itex]*[/itex] denotes convolution? Or did you meant it as multiplication?
  4. Feb 22, 2006 #3
    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
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Laplace Transform of cos(at) * cos(bt)
  1. Laplace transforms (Replies: 1)

  2. Dft of cos function (Replies: 5)