Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Competing definitions of the Fourier transform

  1. Jun 8, 2012 #1
    Just began a serious study of the Fourier transform with a couple of books. One of them defines the Fourier transform on [itex]\mathbb R[/itex] as

    [tex]
    \hat f(\xi) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^\infty f(x) e^{-i\xi x}dx.
    [/tex]

    Another defines it as

    [tex]
    \hat f(\xi) = \int_{-\infty}^\infty f(x) e^{-2\pi i \xi x} dx.
    [/tex]

    A few questions:

    (1) Are these definitions somehow equivalent? I cannot seem to obtain the second from the first by making a simple change of variables.

    (2) Why worry about the factors of [itex]2\pi[/itex] in the definitions? What does that do for us? Why not leave those out altogether?
     
  2. jcsd
  3. Jun 8, 2012 #2
    See M.L.Boas - "Mathematical Methods" for disambiguation. I had the same problem and it really helped me.
     
  4. Jun 8, 2012 #3
    The definitions are equivalent, and the factor of [itex]1/\sqrt{2\pi}[/itex] is added in to ensure that applying the transform and its inverse doesn't mulitply the result by a constant factor other than 1. You need to know whether the [itex]2\pi[/itex] is in the exponential or not to figure out this normalization constant.
     
  5. Jun 8, 2012 #4

    AlephZero

    User Avatar
    Science Advisor
    Homework Helper

    If you leave them out of the definition of the Fourier transform, they appear in the inverse transform. If you try to leave them all out, then IFT ( FT [ f(x) ] ) doesn't equal f(x).

    Unfortunately there isn't an "obvious" place to put the ##2\pi## that everybody agrees on, so you have to check what convention any particular book or paper is using.

    Similar issues apply to the discrete Fourier transform, and computer software routines that calculate it. There you also have to watch out for factors of n and 1/n, where n is the number of samples in the DFT.
     
  6. Jun 10, 2012 #5
    Oh. that is easily solved. Make change of variables in the familiar transform you understand clearly to include the 2[itex]\pi[/itex] in the exponent, THEN see if the product of coefficients of Fourier and inverse Fourier transforms gives the factor claimed. if they are the same, this is correct, if it is not, it gotta be wrong.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Competing definitions of the Fourier transform
  1. Fourier Transforms (Replies: 8)

Loading...