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
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]
and the limits of integration can be changed to
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]
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 coefficient A_{k} the other integration is effectively doing this; \sum^{\infty}_{k=0} A_{k} sin \left( kx \right)
- in the second case, one integral gives 2 A_{k}, because it integrates across 2 periods, but it can't matter which period you integrate across to find the Fourier coefficient. The other integral is effectively 2 \sum^{\infty}_{k=0} A_{k} sin \left( kx \right) 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...
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]
and the limits of integration can be changed to
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]
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 coefficient A_{k} the other integration is effectively doing this; \sum^{\infty}_{k=0} A_{k} sin \left( kx \right)
- in the second case, one integral gives 2 A_{k}, because it integrates across 2 periods, but it can't matter which period you integrate across to find the Fourier coefficient. The other integral is effectively 2 \sum^{\infty}_{k=0} A_{k} sin \left( kx \right) 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...