Fourier Transform

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Radarithm
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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
 
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  • #2
jtbell
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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."
 
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Radarithm
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  • #4
stevendaryl
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I am unable to understand how this: http://gyazo.com/5c78dc5774850d609ce200efa446cfbf
and this: http://gyazo.com/f182937cb662f5e2241bea977f2929ea
are equal, and how the Fourier transform is done. I do however understand what the Fourier transform is: A way to break down a certain wave into its sinusoidal wave components (ie. WAVE = sin wave1 + sin wave2 + ...)
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]
 
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  • #5
Radarithm
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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]
Made things a lot clearer, thanks! :biggrin:
 

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