Step in fourier transform derivation

  • Thread starter 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 whats going on after writing this post and having to articulate it...
 

HallsofIvy

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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 whats 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.
 

uart

Science Advisor
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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?
 
70
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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.
 

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