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Fourier transform of sin(t)/t

  1. Oct 4, 2009 #1
    1. The problem statement, all variables and given/known data

    Evaluate INT(|X(t)|^2) dt using parsevals theorem

    where x(t) = (sin(t)cos(10t))/(pi*t)


    2. Relevant equations

    parsevals theorem: int(|f(t)|^2 dt = (1/2*pi)INT(|F(W)|^2 dw



    3. The attempt at a solution

    So I've tried several attempts at this problem and this is my latest:

    first I use the fact that sin(x)*cos(y) = (sin(x+y)+sin(x-y)) /2
    to get sin(t)cos(10t)/pi*t = (sin(t+10t) + sin(t - 10t))/(2*pi*t)

    then I split it up into : sin(11t)/2t*pi + sin(-9t)/2t*pi

    then what I was going to do was take the fourier transform of each function here however, I can't figure out how in the world to take the fourier transform of sin(t)/t

    anyone have any ideas? thanks!
     
  2. jcsd
  3. Oct 4, 2009 #2

    jbunniii

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    Do you know a function whose Fourier transform is [itex]\sin(\omega)/\omega[/itex]?
     
  4. Oct 4, 2009 #3
    I checked on my transform table and looked around online a little and didn't see any transform that equals sin(w)/w. If there was i'd use the f(w) <-> F(t) rule and then it could work for me. Does that transform exist?
     
  5. Oct 4, 2009 #4
    Hint:

    Compute the (inverse) Fourier transform of f(x), defined as:

    f(x) = 1 for -L < x < L,

    otherwise f(x) = 0
     
  6. Oct 4, 2009 #5

    jbunniii

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    Yes, it exists and is one of the most fundamental transforms! If your table doesn't have it, I would get another table (seriously). Count Iblis' hint is right on the money.

    By the way, [itex]\sin(x)/x[/itex] is sometimes called [itex]sinc(x)[/itex], so look in your table for that. Caution: some authors define [itex]sinc(x) = \sin(\pi x)/(\pi x)[/itex].
     
    Last edited: Oct 4, 2009
  7. Oct 4, 2009 #6
    ooohhhh now it makes sense!!! thank you!! XD
     
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