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Fourier transform of cos(100t)

  1. Aug 13, 2009 #1
    1. The problem statement, all variables and given/known data
    Find the fourier transform of cos(100t)

    3. The attempt at a solution
    now I know just from looking at a fourier transform table that if the equation is in the form cos(2Pi*k*t) then the answer is just 1/2(delta(f+k) + delta(f-k))

    So in this case is the answer 1/2(delta(f+100/2pi) + delta(f - 100/2pi)) ?

    I'm not that good at integrals so I haven't attempted to do this problem the traditional long way.

    Thanks :)
     
  2. jcsd
  3. Aug 13, 2009 #2

    dx

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    Write out the Fourier transform integral, and write cos(100t) as a sum of exponentials using Euler's formula. Then use the following fact:

    [tex] \frac{1}{2\pi} \int_{-\infty}^{\infty} e^{ixy} dy = \delta(x) [/tex]
     
  4. Aug 13, 2009 #3
    Ok starting with S e^(-2Pi*i*t) * (e^(i100/2pi*t) + e^(-i100/2pi*t))/2

    I got it down to 1/2PI S (e ^(-2Pi*i(t + 100/2Pi) + e ^ (-2Pi*i(t - 100/2pi))

    Which then equals delta(f + 100/2Pi) + delta (f - 100/2pi)

    I'm not sure if I did it correctly though or just worked my way backwards from the answer (I attempted this originally and then got stuck). Does that look like I am going in the right direction?
     
  5. Aug 13, 2009 #4

    dx

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    cos(100t) = (ei100t + e-i100t)/2

    I don't understand where you got the 2pi's in it.
     
  6. Aug 13, 2009 #5
    I did that originally so I could use the transform table, its ok I understand what to do now. I will have a shot at doing it from scratch and report back.

    Thanks guys.
     
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