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Fourier Transform

  1. Oct 17, 2013 #1

    Radarithm

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    Gold Member

    I understand the Fourier transform conceptually, but I am unable to reproduce it mathematically; I am very familiar with calculus and integration, but I am taking a QM course and I need to know how to apply it. No websites or videos are able to give me a good explanation as to how I can use it, so I decided to ask for help here.
    Thanks in advance.

    EDIT: Sorry if this is in the wrong section
     
    Last edited: Oct 17, 2013
  2. jcsd
  3. Oct 17, 2013 #2

    jtbell

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    Staff: Mentor

    It would greatly help people here if you could show an example that you don't understand, and say what you don't understand about it. Otherwise we're "shooting blindly in the dark."
     
  4. Oct 17, 2013 #3

    Radarithm

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  5. Oct 17, 2013 #4

    stevendaryl

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    The proof of that is a little complicated, but I can give you an intuitive argument.

    Start with the discrete version. Suppose we have a function [itex]f(x)[/itex] that is created by adding up a bunch of sines and cosines:

    • [itex]f(x) = \sum_n C_n e^{\frac{n \pi i x}{L}}[/itex]
    • [itex]C_n = \frac{1}{2 L} \int_{-L}^{L} f(x) e^{\frac{-n \pi i x}{L}} dx[/itex]

    Now, to make the forward and reverse transforms look more alike, let me introduce a new variable, [itex]k_n[/itex] defined by [itex]k_n = \frac{n \pi}{L}[/itex], then [itex]\delta k = k_{n+1} - k_n = \frac{\pi}{L}[/itex]

    In terms of [itex]k_n[/itex], you can write the forward and reverse transforms as:
    • [itex]f(x) = \frac{1}{2 \pi} \sum_n (2 L C_n) e^{i k_n x} \delta k[/itex]
    • [itex]2L C_n = \int_{-L}^{L} f(x) e^{-i k_n x} dx[/itex]

    Now, let's define [itex]F(k_n) = 2 L C_n[/itex], so we have:
    • [itex]f(x) = \frac{1}{2 \pi} \sum_n F(k_n) e^{i k_n x} \delta k[/itex]
    • [itex]F(k_n) = \int_{-L}^{L} f(x) e^{-i k_n x} dx[/itex]

    Now, as [itex]L \rightarrow \infty[/itex], [itex]\delta k \rightarrow 0[/itex]. Then the discrete sum for [itex]f(x)[/itex] approaches an integral:

    [itex]f(x) = \frac{1}{2 \pi} \int F(k) e^{i k x} dk[/itex]

    So in this limit, the forstward and reverse transformations look like:

    • [itex]f(x) = \frac{1}{2 \pi} \int F(k) e^{i k x} dk[/itex]
    • [itex]F(k) = \int f(x) e^{-i k x} dx[/itex]
     
  6. Oct 17, 2013 #5

    Radarithm

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    Made things a lot clearer, thanks! :biggrin:
     
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