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Need Laplace transform help

  1. Apr 14, 2008 #1
    iv got a problem i cant seem to understand. if anyone could help me out it would be great

    f(t)= sin2tcos2t

    im just not sure what to do when i have the product of 2 trig functions.

    the correct answer is (2/(s^2 + 16))

    thanks for any help
     
    Last edited: Apr 14, 2008
  2. jcsd
  3. Apr 14, 2008 #2

    exk

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    Is there something you can do with f(t) to make it look different?
     
  4. Apr 14, 2008 #3
    yeah, like exk suggested, try to use this trig identity somehow:

    [tex] sin2x=2isnxcosx[/tex] can you figure it out how to transform your f(t) into a similar form?
     
  5. Apr 14, 2008 #4

    Vid

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    Use f(x) = 1/4sin^2(2t)
    f'(x) = sin(2t)cos(2t)
     
  6. Apr 14, 2008 #5
    I don't see how would this help!
    If i have gotten the op right, he just needst to take the laplace transform of

    [tex]f(t)=sin2tcos2t=\frac{1}{2}sin4t[/tex]
    so

    [tex]L{f(t)}=L{\frac{1}{2}sin4t}=\int_{0}^{\infty}\frac{1}{2}sin(4t)e^{-st}dt[/tex]
     
  7. Apr 14, 2008 #6

    Vid

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    I was thinking that sin^2(t) was a common laplace transform, but I was mistaken.
     
  8. Apr 14, 2008 #7

    exk

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    another strategy for this would be to look up in a transform table what your answer corresponds to and work backwards.

    sutupidmath's solution is correct.
     
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