Step in fourier transform derivation

In summary, the Fourier transform can be derived from the Fourier series by taking the limit of the period going to infinity. This leads to the substitution of a double integral that can be rewritten as a single integral, resulting in a factor of 1/4 in the final equation. The development of the Fourier transform from the Fourier series is described in Morse and Feshbach's "Methods of Theoretical Physics", which takes some time to fully understand.
  • #1
Nick R
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Looking at how the Fourier transform comes about from the Fourier series when the period goes to infinity, they make the following step

[tex] h \left( x \right) = \frac{2}{\pi}\int^{\infty}_{0} \left( dk \right) \left[ \int^{\infty}_{0} h \left( \varsigma \right) sin \left( k \varsigma \right) sin \left( x \varsigma \right) d\varsigma \right] [/tex]

and the limits of integration can be changed to

[tex] h \left( x \right) = \frac{1}{2 \pi}\int^{\infty}_{-\infty} \left( dk \right) \left[ \int^{\infty}_{-\infty} h \left( \varsigma \right) sin \left( k \varsigma \right) sin \left( x \varsigma \right) d\varsigma \right] [/tex]

I'm trying to understand why this factor of 1/4 arises...

My understand is this
- in the first case, one integral gives us a Fourier coefficent [tex] A_{k} [/tex] the other integration is effectively doing this; [tex]\sum^{\infty}_{k=0} A_{k} sin \left( kx \right)[/tex]

- in the second case, one integral gives [tex] 2 A_{k} [/tex], because it integrates across 2 periods, but it can't matter which period you integrate across to find the Fourier coefficent. The other integral is effectively [tex]2 \sum^{\infty}_{k=0} A_{k} sin \left( kx \right)[/tex] since it is summing over 2 periods instead of 1.

Is this a correct way of looking at it? I'm more confident I know what's going on after writing this post and having to articulate it...
 
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  • #2
Nick R said:
Looking at how the Fourier transform comes about from the Fourier series when the period goes to infinity, they make the following step

[tex] h \left( x \right) = \frac{2}{\pi}\int^{\infty}_{0} \left( dk \right) \left[ \int^{\infty}_{0} h \left( \varsigma \right) sin \left( k \varsigma \right) sin \left( x \varsigma \right) d\varsigma \right] [/tex]

and the limits of integration can be changed to

[tex] h \left( x \right) = \frac{1}{2 \pi}\int^{\infty}_{-\infty} \left( dk \right) \left[ \int^{\infty}_{-\infty} h \left( \varsigma \right) sin \left( k \varsigma \right) sin \left( x \varsigma \right) d\varsigma \right] [/tex]

I'm trying to understand why this factor of 1/4 arises...

My understand is this
- in the first case, one integral gives us a Fourier coefficent [tex] A_{k} [/tex] the other integration is effectively doing this; [tex]\sum^{\infty}_{k=0} A_{k} sin \left( kx \right)[/tex]

- in the second case, one integral gives [tex] 2 A_{k} [/tex], because it integrates across 2 periods, but it can't matter which period you integrate across to find the Fourier coefficent. The other integral is effectively [tex]2 \sum^{\infty}_{k=0} A_{k} sin \left( kx \right)[/tex] since it is summing over 2 periods instead of 1.

Is this a correct way of looking at it? I'm more confident I know what's going on after writing this post and having to articulate it...
The simplest way to see it is to note that in the first equation you cite, the integrals are both going form 0 to [itex]\infty[/itex] while in the second equation, they are going from [itex]-\infty[/itex] to [itex]\infty[/itex].
[tex]\int_{-\infty}^\infty} f(x)dx= \int_{-\infty}^0 f(x)dx= \int_0^\infty f(x)dx[/tex]

and, since both integrands are symmetric,
[tex]\int_{-\infty}^0 f(x)dx= \int_0^\infty f(x)dx[/tex]
which gives
[tex]\int_{-\infty}^\infty f(x)dx= 2\int_0^\infty f(x)dx[/tex]
or
[tex]\int_0^\infty f(x)dx= \frac{1}{2}\int_{-\infty}^\infty f(x) dx[/tex]
for each integral.

It has nothing to do with "periods" because, in the Fourier transform, the functions are not necessarily periodic.
 
  • #3
The factor of 4 is easy enough to explain. I'm a little more concerned with where the :

[tex] \frac{2}{\pi} \int^{\infty}_{0} \left( dk \right) [/tex]

bit comes from. That's bit is clearly infinite.

BTW. I'm reasonably familiar with the development of the Fourier transform from the Fourier series but I haven't seen that method before. Do you have a either a link or the equations/development that precedes the lines shown above?
 
  • #4
uart said:
The factor of 4 is easy enough to explain. I'm a little more concerned with where the :

[tex] \frac{2}{\pi} \int^{\infty}_{0} \left( dk \right) [/tex]

bit comes from. That's bit is clearly infinite.

BTW. I'm reasonably familiar with the development of the Fourier transform from the Fourier series but I haven't seen that method before. Do you have a either a link or the equations/development that precedes the lines shown above?

This all comes from Morse and Feshbach's "Methods of Theoretical Physics". They make a substitution, take the limit of the period going to infnity and the summation becomes an integral. Its the best description I've found but its taking me a while to assimulate it. Most explanations online are "heres the formula - magnificent!".

Also that whole thing could be rewritten [tex] \frac{2}{\pi} \int^{infty}_{0} \int^{infty}_{0} \left( \right) d\varsigma dk [/tex]

So its not infinity. Its just a double integral.

edit: also I made a mistake in those formulas in the first post but it shouldn't matter now.
 

1. What is the purpose of the Fourier transform?

The Fourier transform is a mathematical operation that decomposes a function into its constituent frequencies. It is commonly used in signal processing and image processing to analyze and manipulate data in the frequency domain.

2. How is the Fourier transform derived?

The Fourier transform can be derived by using the Fourier series, which is a representation of a periodic function as a sum of sinusoidal functions. The Fourier transform extends this concept to non-periodic functions by allowing the period to approach infinity.

3. Can you explain the steps involved in deriving the Fourier transform?

The derivation of the Fourier transform involves the use of complex numbers, Euler's formula, and integration techniques. The basic steps include representing the function as a sum of complex exponentials, applying the Fourier series formula, and taking the limit as the period approaches infinity.

4. What is the difference between Fourier transform and inverse Fourier transform?

The Fourier transform decomposes a function into its constituent frequencies, while the inverse Fourier transform takes the transformed function and reconstructs the original function. In other words, the Fourier transform is used for analysis, and the inverse Fourier transform is used for synthesis.

5. How is the Fourier transform used in real-world applications?

The Fourier transform has a wide range of applications, including signal processing, image processing, data compression, and solving differential equations. It is also used in various fields such as physics, engineering, and finance for analyzing and manipulating data in the frequency domain.

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