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TOPIC: calibration of experimental and simulation results

calibration of experimental and simulation results 2 years 6 months ago #34

Hi,

I am using Pluto SLM having resolution (1920 X 1080) of pixel size 8um. I want to generate a beam at Fourier Transform (FT) plane of a lens (focal length f=10cm) by displaying an amplitude grating mask on to the SLM. At FT plane I used a CCD camera to record the beam. I want to generate the same equal size and shape of the beam experimentally recorded beam in simulation using Matlab. CCD camera has the resolution of 1600 x 1200 and pixel size of 4.4um.

For this I make a mask of size equal to the resoLultion (1920 X 1080) of SLM with pixel size 8um. Then I took FFT of the generated mask which provide me FT profile at FT plane. I used relation for spatial coordinate at FT PLANE x2=lambda*f*fx, where, lambda= wavelength of laser beam, fx = spatial frequency.

so what should be the size of image at FT plane which give same size and shape of the beam as in experiment? Is FT PLANE coordinates should be equal to the camera resolutions and pixels ? How should I display the images in simulation?
Please give your suggestion.
Thanks.

calibration of experimental and simulation results 2 years 5 months ago #37

  • sebastien.popoff
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Hi Brijesh,

First of all, the Pluto SLM is a phase only SLM, you cannot directly display amplitude grating masks.

Then, if I understand correctly, your question is how to link simulations to experiments?

This is purely a geometrical problem that you seem to have already solved.

You calculate the FFT using Matlab/Python, you end up with an image where each pixel corresponds to a spatial frequency (it goes from -f_s/2 to f_s/2, where f_s is the spatial sampling frequency, for more information look that the doc of the function you use).

Then, your formula is correct, x~lambda*f*fx, just apply this coordinate transformation to have the position in the spatial coordinates.

You have everything. If you have a 1920 x 1080, you calculate the FFT and using the above formula you have the image on the CCD.

If you want a given pattern on you CCD plane, use the inverse operation to find out what is the correct amplitude and phase mask to apply on the SLM plane. As you cannot modulate in amplitude and phase but just in phase, a first approximation is to just send the phase of the calculated field. This is not optimal at all, to find a better solution, you can use a Gerchberg–Saxton algorithm (See en.wikipedia.org/wiki/Gerchberg%E2%80%93Saxton_algorithm).

One tip, usually the pattern you want to see in the Fourier plane is quite close to the center of the origin of your FFT image. You may need to increase the number of output points of your FFT function to be able to zoom in and have the details of the image you want (look at the docs of the FFT function).

Best,

Sebastien

calibration of experimental and simulation results 2 years 5 months ago #38

Thanks a lot! for your comments and suggestions. Your reply clear my confusion.

calibration of experimental and simulation results 9 months 3 weeks ago #62

  • Piter
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I am also a bit confused with this topic.
In work of Jesacher (page 28) www.uibk.ac.at/exphys/photonik/downloads/thesis_jesacher.pdf
The following procedure of relating experimental and computer generated pictures is outlined.

Mholo - diameter of hologram pattern displayed on SLM in pixels
Marray - side length of array used in calculations.
So we need to place a correct aperture on the Marray matrix
---copied from Jesacher---
The choice of the right diameter of the aperture is crucial. Experimentally, it can easily be determined by subsequently displaying two gratings on the SLM, the grating vectors of which differ by Δk (in inverse pixels), and measuring the corresponding spatial shift of the laser spot on the CCD. The desired aperture diameter A (in pixel units) is then determined by

A = Mholo*Marray*Δk/(2*pi*Δl)

where Δl is the focus-shift on the CCD in pixel units, MArray the side length of the data array, and MHolo the diameter of the hologram on the SLM, both measured in pixels.

end of copy---

I display a Fresnel lens (F = 50 cm) with two different gratings k*x with k = 4 or 8 and x equal to pixel number.
It shifts focus by 100 pixels.
I am confused with ¨Δk in inverse pixels¨, shall I use it as 4 or 1/4?
If I use 4 the aperture is too big to be real, and if I use 1/4 resulting intensities are still far from experimental. What do I miss?

May be you can point me out where from this relation comes from?

Thanks in advance.
Petro.

calibration of experimental and simulation results 9 months 3 days ago #63

  • sebastien.popoff
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Hi,

I did not understand exactly what is all this for, and to be honest, I am not ready to read the thesis for this.

However, about the Δk issue, the text seems pretty clear for me. When you say you have a grating with k*x with k=4 or 8 with x in pixel units, then k has the dimension of the inverse of pixels, so I would use 4 not 1/4.

About the formula itself, I have some doubts. If Mholo, Marray and Δl are in pixel units and Δk in inverse of pixels, then A is not in pixel units, it has no dimension...
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calibration of experimental and simulation results 8 months 3 weeks ago #64

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Hi Sebastien,

Thank you for reply.

I have figured it out with help of one of colleagues.
If D is hologram/grating periodicity then
Δk = 2pi/D1 - 2pi/D2 = 2pi * (1/D1 - 1/D2) = 2pi (1/4 - 1/8) = 2pi/8 = pi/4

Now it looks much better.
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