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Fourier Transform

by Radarithm
Tags: fourier, transform
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Radarithm
#1
Oct17-13, 10:02 AM
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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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jtbell
#2
Oct17-13, 10:22 AM
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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."
Radarithm
#3
Oct17-13, 10:45 AM
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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 + ...)

stevendaryl
#4
Oct17-13, 11:42 AM
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P: 2,109
Fourier Transform

Quote Quote by Radarithm View Post
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]
Radarithm
#5
Oct17-13, 12:40 PM
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P: 154
Quote Quote by stevendaryl View Post
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!


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