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Laplace transform sint * cost

  1. Jan 7, 2007 #1
    how do i go about solving the laplace transform of sint * cost ???

    i know the answer becomes 1/(s^2 + 4) but what is the method?

  2. jcsd
  3. Jan 7, 2007 #2
    do you know that
    [tex]\mathcal{L}\{f(t)*g(t)\} = \mathcal{L}\{f(t)\} \cdot \mathcal{L}\{g(t)\} = F(s) \cdot G(s)[/tex]

    and i think the answer comes out to be [tex]\frac{s}{{\left(s^2+1\right)}^2}[/tex]
    Last edited: Jan 7, 2007
  4. Jan 7, 2007 #3
    yes i did know that but i didnt think it was as simple as that. From reading the certain rules we can apply to problems e.g the shift rule to solve L(sin2t * e^3t) do we HAVE to use the shift rule or can we seperate each part and solve then multiply them together?
  5. Jan 7, 2007 #4
    i am sorry. do you mean convolution by the asterisk? or do you mean multiplication? what i said in my last post is only correct if the asterisk means convolution.
  6. Jan 7, 2007 #5
    i meant multiplication
  7. Jan 7, 2007 #6


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    Science Advisor

    Then use the definition
    [tex]\mathcal{L}(sin(x)cos(x)}= \int_0^{\infnty}e^{-st}sin(t)cos(t)dt[/tex]
    which can be done by integration by parts.
  8. Jan 7, 2007 #7
    Mybe it would help to notice that sin(t)cos(t) is equal to [tex]\frac{1}{2}sin(2t)[/tex]
  9. Mar 4, 2007 #8
    I wanna ask about a proof of this transform: L{(f(t)}= sF(s) - f(0) - f(a^+) - f(a^-) - exp^(-as) inwhich f(t) is continous except for an ordinary discontinuity (finite jump) at t=a, a>0
    Last edited: Mar 4, 2007
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