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Fourier Transform of cos(2*pi*t)

  1. May 31, 2009 #1
    I have a practice question, which is to find the Fourier Transform of cos(2^pi^t)

    By substitution into the FT formula, and use of eulers formular,I have managed to reduced to:

    INTEGRALOF ( (cos(2*pi*t) * ( cos(2*pi*F*t) - j*sin(2*pi*F*t) ) )

    By plotting the frequency graph of the original function, I know that the answer I am looking for is: delta(1) + delta(-1)

    I have also been told that the integral of two trig functions multiplied together equals 0 if the functions have different frequencies. This indicates that the above formula is only non-zero where F = 1.

    My problem is that I don't know how to get from the above formula to delta(1) and delta (-1). Can anybody help?

    Also, I'm relatively new to Fourier Transforms, so as much detail as possible in answers will be appreciated!

    thanks in advance for any help
     
  2. jcsd
  3. May 31, 2009 #2

    tiny-tim

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    Welcome to PF!

    Hi BriWel! Welcome to PF! :smile:

    (have a pi: π and a delta: δ and an integral: ∫ and try using the X2 tag just above the Reply box :wink:)
    You need to learn your https://www.physicsforums.com/library.php?do=view_item&itemid=18"

    in this case, 2cosxcosy = cos(x+y) + cos(x-y) and 2sinxcosy = sin(x+y) + sin(x-y) …

    and then you should be able to prove that the ∫0 is 0 unless x = ±y :wink:
     
    Last edited by a moderator: Apr 24, 2017
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