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Fourier Transform of a Triangular Voltage Pulse

  1. Mar 15, 2012 #1
    1. The problem statement, all variables and given/known data
    JR9TZ.jpg
    So this is a physics problem, but this question doesn't really have to do with the "physics" part of it as much as simply calculating the Fourier transform. (This is a second year physics course and our prof is trying to briefly teach us math tools like this in learning quantum mechanics).


    2. Relevant equations

    [tex] \tilde{g}(\omega) = \frac{1}{\sqrt2\pi} \int g(t) e^{-i \omega t} dt [/tex]

    3. The attempt at a solution
    I have done the calculation of g(ω) several times and got an answer

    [tex] \frac{2}{(\tau \omega ^2 \sqrt2 \pi)} (1 - cos(\omega \tau)) [/tex]

    I believe it is right, but since the work to get it is extensive I don't want to type it up unless someone thinks I made an error. My actual concern is that I have a problem sketching the transform. I graphed it on Wolfram so I have a general idea, but I really have no idea how to find the amplitude, width, and whether it should be centred at ω=0 or at a k0 value. Any insight would be greatly appreciated.
     
    Last edited: Mar 15, 2012
  2. jcsd
  3. Mar 15, 2012 #2
    Try factoring [itex](1 - cos(\omega \tau))[/itex] in terms of sin(ωτ/2) and identify the celebrated 'sinc' function
     
  4. Mar 15, 2012 #3
    I had to google 'sinc function' to find out what it was... We have not been taught this so surely there is a way to find the attributes of the transform by simply looking at it. Our prof mentioned something about finding the general root of the transform (here it would be whenever the cos term is 1), but I don't know how to relate that to the centre or anything
     
  5. Mar 16, 2012 #4
    If you haven't been taught sinc function, then learn it, it's simple enough. Your expression can be written as the square of the sinc function, where behavior of sinc is well understood. I certainly don't know what your prof had in mind, but the sinc function is centered at zero.
     
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