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

Shape of fourier transform

  1. Aug 28, 2014 #1
    Hi! I am taking a second look on fourier transforms. While I am specifically asking about the shape of the fourier transform, I'd appreciate if you guys could also proof-read the question below as well, as I've written down allot of assumptions that I've gained, which might be wrong.

    OK.

    As far as I am aware, the nastier/"less smooth" (ie sharp, large and discontinious derivative and so on) a function ##f(x)## is (for ex. ##f(x) = \delta (x)##, the more "spread out" its fourier transform is. That is, the sines and cosines summed by the fourier-inverse integral get weighed increasingly equally irregardless of their frequencies, as ##f(x)## becomes less periodic.

    However if you got a very spread out and slowly changing ##f(x)## (for ex. ##f(x)=1##), the fourier transform will be narrow around 0, meaning only the low-frequency sines and cosines in the fourier-inverse integral will dominate.

    Why is this so? This applies to periodic functions as well, so let me rephrase the question in case you don't get me: Why can a "smooth and slow" function be described adequately with less terms in a fourier series, than a nasty one? Is it perhaps because the further you go out in a fourier series, the bigger the derivatives will be and thus these violent sines and cosines can adequately describe a swiftly changing function?
     
  2. jcsd
  3. Aug 28, 2014 #2
    And can somebody tell me how to evaluate the infinite integral of the complex exponential function so I can get something representing the dirac delta? (I didn't want to open a new thread for this question alone as it's related to the OP, so don't murder me moderators).

    I read that you can multiply it a converging term as a trick, like this, ##\lim_{\epsilon \to 0} \int_{-\infty}^{\infty} e^{-\epsilon x^2} \cdot e^{isx} dx##, but I don't remember how to evaluate a gaussian integral multiplied by another function.
     
  4. Aug 28, 2014 #3

    mathman

    User Avatar
    Science Advisor
    Gold Member

    ##\int_{-\infty}^{\infty} e^{-\epsilon x^2} \cdot cos(sx) dx=\sqrt{{\pi}/{\epsilon}}\cdot e^{{-s^2}/{4\epsilon}}##, from Gradshteyn and Ryzhik. Note that the imaginary part = 0.
     
  5. Sep 4, 2014 #4
    You confuse two things: number (amount) of non-zero terms, and their frequencies (i.e. this [itex]n[/itex] in your [itex]\cos nx[/itex] and [itex]\sin nx[/itex] or, in the standard presentation, [itex]\exp(inx)[/itex]).

    Also,
    • Do not use such terms as “nasty function”. Not clear.
    • Make distinction between Fourier transform of functions on ℝ and Fourier series on the circle (a.k.a. for periodic functions).
    • Learn to think in exponents, not ugly real trigonometry.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Shape of fourier transform
  1. Fourier transformation (Replies: 0)

  2. Fourier transform (Replies: 2)

  3. Fourier transforms (Replies: 1)

  4. Fourier transform (Replies: 1)

Loading...