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 the product of two functions

  1. Dec 8, 2012 #1

    I am trying to figure out in my notes how my professor did

    L[(e^-3t)(sin2t)] = 2/(s+3)^2 +4

    I think she just did it in her head and wrote it, so I don't actually know how to solve it. I am looking at my table of laplace transforms and there is none for a product of an exponential and sin/cos.

    I tried solving this with the definition, and I came across a integration by parts that looks particularly nasty, not even sure if solvable.

    Anyways, can anyone point me in the right direction to this one? I know the product of a laplace transforms is not the laplace of the products, so thats out.
  2. jcsd
  3. Dec 8, 2012 #2


    User Avatar
    Science Advisor

    You're over-thinking it. If ##\tilde{f}(s)## is the Laplace transform of ##f(t)## then
    ##\mathcal{L}({e^{-at}f(t)}) = \int_0^\infty e^{-st} \, e^{-at} \, f(t) \, dt = \int_0^\infty e^{-(s+a)t}\,f(t)\,dt = \tilde{f}(s+a)##
  4. Dec 8, 2012 #3
    I know that one, but then I get a cos and exponential in the same integral, and I don't think there is a good way to go about solving that, right?

    Excuse me for not knowing latex..but what are you saying is f(t)? Just the sin2t, or the entire thing? Then once you get there..is that thing solved by parts? If so, I got it and it looks really nasty to solve..not eloquent
  5. Dec 8, 2012 #4


    User Avatar
    Science Advisor

    Of course there is, every first year calculus course covers Euler's formula.

    ##f(t) = \sin 2t## so ##\tilde{f}(s) = \frac{2}{s^2+2^2}##.
    Also ##a = 3##.

    This is why I'm saying you are over-thinking it. Write out the definition of the Laplace transform, and replace "s" with "s+a".

    Wait a sec. Do understand either of the following:
    1. The Laplace transform of sin ωt is equal to ω/(s^2 + ω^2).
    2. The Laplace transform of e^{-at}f(t) is equal to ##\tilde{f}(s+a)##.
    Last edited: Dec 8, 2012
  6. Dec 8, 2012 #5
    Ok allow me to go about this

    L[f(t)] = ∫e^-st f(t)dt

    f(t) = e^(-3t)sin(2t)

    ∫e^-(st) e^(-3t)sin(2t) dt

    At this point I am not seeing how to solve that integral
  7. Dec 8, 2012 #6


    User Avatar
    Science Advisor

    The problem has two parts that are proved separately: the sine and the phase shift. Please read my edit in post 4. Do you understand how to do one or the other?
  8. Dec 8, 2012 #7
    I understand 1, but not sure about 2. In fact I definitely don't understand 2
  9. Dec 8, 2012 #8


    User Avatar
    Science Advisor

    Okay, you know the definition of the Laplace:
    ##\tilde{f}(s) = \int_0^\infty e^{-st} \, f(t) \, dt##
    Then replacing "s" with "s+a" we have
    ##\tilde{f}(s+a) = \int_0^\infty e^{-(s+a)t} \, f(t) \, dt##
    Does that make sense?
  10. Dec 8, 2012 #9
    yes, got it
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook