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Angular Spectrum Method and Fourier Transform

  1. Jun 1, 2015 #1
    Are the results of the Angular Spectrum Method and the Fourier Transform of a Fresnel Diffraction be different, or the same? Given the same distance between the input and output plane, and the same aperture.
     
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  3. Jun 2, 2015 #2

    blue_leaf77

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    Under the region where paraxial approximation is valid, they are equal. In fact the angular spectrum method is more exact compared to the Fresnel diffraction formula as the latter is obtained after a couple of approximations, while the former only relies on the validity of the scalar wave equation (which is of course not so exact either since by placing an aperture, the boundary conditions for free space are perturbed and the correct wave equation should involve coupling between different field components).
     
  4. Jun 2, 2015 #3
    I got these images from a numerical simulation of the Angular Spectrum Method and the Fourier Transform.
    Sample_Slit_100.png
    This is my aperture. A slit with a width of 100 pixels.
    ASpec_Slit_700_01.png
    This is the Fresnel Diffraction simulation using the Angular Spectrum Method; and
    Fourier_Slit_700_01.png
    This is the Fresnel Diffraction simulation using the Fourier Transform. Although they seem to be the same at the middle of the diffraction pattern, the pattern from the Fourier Transform seems to be from a rectangular aperture. There seems to be some sort of a "scaling" process done. Is this expected?
     
  5. Jun 3, 2015 #4

    blue_leaf77

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    Are they taken at the same distance? As for the lower picture, are you sure the requirements which allows the approximations leading to FT relation are satisified, e.g. the ratio of the distance to the observation plane to the largest size in the object plane must be sufficiently large. In fact your object is an aperture with one dimension very big (vertical), so I suggest that you use the form of Fresnel diffraction where the no approximations to the distance have been made.
     
  6. Jun 3, 2015 #5
    Both of them are taken at the same distance (400 m). The whole picture is 500 by 500 pixels, or 0.1 by 0.1 meters. The width of the aperture is 0.02 meters with a length of 0.1 meters. With these parameters, the ratio is 4000.
     
  7. Jun 3, 2015 #6

    blue_leaf77

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    Which ratio is this?

    Anyway, I'm not quite sure if what I think is the same as you, when you say using FT, did you actually transform the field aperture profile only? Such FT relation only holds within the far-field region in which the inequality ##z>>2D^2/\lambda## must be true.
    If that's not sufficiently fulfilled, then your experimental conditions might still be in the paraxial/Fresnel region, in which case the FT relation should be performed w.r.t. the aperture profile multiplied by a quadratic phase in the aperture plane ##\exp{\frac{jk}{2z}(\eta^2+\zeta^2)}##. That is, what must be transformed is not only the aperture profile, but there is an additional phase factor. In either case, knowledge of wavelength is necessary.

    Furthermore, it seems like you don't perform the FT correctly, it's obvious that the lower picture is not an FT of a rectangular object. Just for an advice, I don't think that numerical program can give accurate result of a FT computation if the transformed object is not bounded, for example the vertical size of your object extends to the end of the pixel array. What about simulating a more realistic rectangular aperture?
     
  8. Jun 3, 2015 #7
    It is the ratio between the distance to the observation plane, and the largest size in the object plane.

    Here is the equation that I transformed.

    ##E\left(x, y, z\right) = \frac{e^{ikz}}{i \lambda z} e^{i \frac{\pi}{\lambda z}\left(x^2 + y^2\right)}\mathcal{F}\left[E\left(x', y', 0\right) e^{i\frac{\pi}{\lambda z}\left(x'^2 + y'^2\right)}\right] ##

    where ##x## and ##y## are the coordinates for the output plane, and ##x'## and ##y'## for the input plane. And ##\lambda = 700 \times 10^{-9}## meters.

    This is the square aperture with 200 pixels on each side, or 0.04 meters on each side.
    Sample_Square_200.png
    This is the Angular Spectrum result:
    ASpec_Square_700_01.png
    And this is the Fourier Transform:
    Fourier_Square_700_01.png
    It's somehow the same with the slit. The diffraction pattern is smaller for the Fourier Transform.
     
  9. Jun 4, 2015 #8

    blue_leaf77

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    Ok assuming the other dimensions used are those given in your previous comment, I guess your experimental condition already satisfies the requirement for Fresnel approximation. Now what program do you use to calculate those images, MATLAB? If yes did you calculate the FT using the build-in function such as fft? As far as I know, when you use such build-in function, the coordinate scaling in the target plane is already determined by the function, therefore the scaling on the coordinate on both pictures above may be different, can you also check this?
    In any case, the Fresnel integral can be shown to coincide with the angular spectrum method provided paraxial approximation is valid within the problem considered, I don't see any reason why the two ways should differ even in scaling.
     
  10. Jun 4, 2015 #9
    I did use MatLab, and used the built-in function fft2. Does that mean that I need to code my own function for the Fourier Transform to remove the difference? How can I check the scaling on the coordinates?
     
  11. Jun 5, 2015 #10

    blue_leaf77

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    If you want you can setup your own 2D FFT matrix which does the FT to a given matrix but I think in most cases this manual way will be slower than when using fft2, for details see : http://fourier.eng.hmc.edu/e161/lectures/fourier/node11.html
    If you want to stay using fft2 then you need to know and show the coordinates in the target plane. For this purpose I found this link http://www.gaussianwaves.com/2014/0...b-fft-of-basic-signals-sine-and-cosine-waves/ can be hepful especially example no.4.
     
  12. Jun 15, 2015 #11
    I tried re-scaling the transformed equation, but it still remains the same (or maybe I just don't know on how to implement it).

    My sampling rate is the same size as the image, and according to MatLab, the fft2 of x can also be computed by:

    fft2 = fft(fft(x).').'

    I tried this one by putting the sampling rate on both Fourier Transform syntax, but still no changes. The size of the image (500 x 500 pixels) is still the same after the transform. Is it possible that the raw equation is the problem, not the Fourier Transform?
     
  13. Jun 15, 2015 #12

    blue_leaf77

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    What I basically suggested you to do is to display the correct coordinates (or x and y values) in the same unit in both the 2nd and 3rd pictures in comment #7. Can you do that? From that we can tell whether the results are actually in concord with each other or otherwise, there is something to investigate in the code you implemented.
     
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