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

Fourier transform and singularities

  1. Jan 19, 2012 #1
    Consider the Fourier transform of a complex function f(t):
    [tex]f(t)=\int_{-\infty}^\infty F(\omega)e^{-i\omega t}[/tex]
    Here t and [itex]\omega[/itex] are on real axis. Let's suppose f(t) is square integrable. Here are my questions:

    1) Since f(t) is square integrable, so we have [itex]\text{limit}_{|\omega|\rightarrow\infty}F(\omega)=0[/itex]. Now, if [itex]F(\omega)[/itex] is analytically extended onto the complex plane, i.e., now [itex]\omega\in\mathbb C[/itex], then does the property [itex]\text{limit}_{|\omega|\rightarrow\infty}F(\omega)=0[/itex] still hold at infinity on the complex plane?

    2) The width of f(t) and [itex]F(\omega)[/itex] in real t and [itex]\omega[/itex] domain respectively satisfy the uncertainty relation. When calculating [itex]f(t)=\int_{-\infty}^\infty F(\omega)e^{-i\omega t}[/itex], if the answer to the first question is "yes," then for different sign of t, the contour can be closed by half circles on the complex plane for [itex]\omega[/itex]. To evaluate f(t), we only need to know the information about the singularity points of [itex]F(\omega)[/itex] on complex plane. So, the question is: is there any simple rule of thumb relation between the width of f(t) and the position of the singularity of [itex]F(\omega)[/itex] and the corresponding residues?

    Thanks!
     
  2. jcsd
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Can you offer guidance or do you also need help?
Draft saved Draft deleted



Similar Discussions: Fourier transform and singularities
  1. Fourier Transform (Replies: 2)

  2. Fourier transform? (Replies: 3)

  3. Fourier Transformations (Replies: 10)

Loading...