Fourier Transforms

[Stiganator]

The optics class I'm taking doesn't have a book, so I can't look it up and we never actually covered what a fourier transform is or its properties. I understand that it goes from time to spatial domain.

I'll try to walk you though the problem, its probably simple, but Wikipedia doesn't really clear this up for me.

U(x) = A(1+i*phi(x))

from wikipedia I gathered that linearity is maintained and constants become delta functions so

U(k) = A*delta + A*FT(i*phi(x))

That k is supposed to be spatial frequency I think, so I'm not really sure if that is the standard notation and if that changes it.


This is where I become lost. I don't have a book to reference, help me ATHT you're my only hope.

 

[QuixoticOne]

http://mathworld.wolfram.com/FourierTransform.html
http://mathworld.wolfram.com/FourierSeries.html
http://mathworld.wolfram.com/FourierMatrix.html
http://planetmath.org/encyclop.../FourierTransform.html
http://planetmath.org/encyclop...leOfFourierSeries.html
http://sharp.bu.edu/~slehar/fourier/fourier.html

FTs transform from spatial domain to frequency domain.
A simple thin double convex lens can perform an approximation of a fourier transform of its input image plane to its output image plane given coherent monochromatic input.

An point source radiating spherically expanding waves at the focal plane of a lens on the lens axis transforms to a collimated beam (constant function at every point in the output focal plane), and of course the reciprocal is true. The collimated beam can be envisioned as a plane wave, constant phase fronts across any plane normal to the lens at any distance from the lens.

A point source at the lens input focal plane but not on the lens axis transforms to a collimated beam on the output side of the lens, but the beam is pointed off-axis on the output side of the lens. So if you look at the output focal plane NORMAL to the lens you see a progressive phase front. If you look at a plane along the point source's propagation direction toward the lens you see a constant phase in a plane beyond the lens, i.e. the phase information is just encoding the propagation direction of the beam, the magnitude carries information on the spatial frequencies in the input image; it is the same collimated beam as if the point was axial to the lens, just with the phase fronts pointing in a different direction.
This is the shift invariance property; a shift in the position of the input image does not change the spatial frequencies of the output image, only its phase information.

A FT is a linear operation that is best understood in Fourier matrix form which is a unitary matrix. Since it is a linear operation the inverse transform of a FT is equal to the original spatial domain input; its conjugate transpose is equal to its inverse.

The transform of a convolution is equal to the product of the transforms of the terms, hence F[a*b] = F[a]F.

The transform of a product is equal to the convolution of the transforms:
F[ab] = F[a]*F.

 

[CycloWizard]

Hopefully QuixoticOne's post covers all the bases. I have a few cheap paperback optics books that I can recommend. One in particular would be very useful since it's literally just solutions to every sort of optics problem ever conceived. I don't have it here, but PM me and I can send you the info when I go into the lab tomororw.

 

[Stiganator]

Those * are just normal multiplication. I'm getting hung up on the fact that I don't know what the transform of i*phi(x), where phi(x) is the phase shift.

 

[Stiganator]

So then would this be correct?

FT( U(x) =A(1+iphi(x))) --> A(delta(k) + i Ph(k))

 

[toslat]

Originally posted by: Stiganator
Those * are just normal multiplication. I'm getting hung up on the fact that I don't know what the transform of i*phi(x), where phi(x) is the phase shift.

where \Phi(k) is the FT of \phi(x) and A & i are constants

 

[CP5670]

Originally posted by: Stiganator
So then would this be correct?

FT( U(x) =A(1+iphi(x))) --> A(delta(k) + i Ph(k))

Yeah, that's right. You can take the constants out by linearity. The only thing you may need to be careful with is the delta distribution in there, which isn't a function in the usual (pointwise) sense.