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Image Processing Mathematics Optics Physics Wolfram Language Modeling

fitting the Airy diffraction pattern with FindFit

Alex Mikh

Posted 12 years ago

Hello! I have 2D data (images) which are results of Gaussian type beam diffraction. According to the theory of diffraction the pattern can be discribed by function ~ J1(r)/r. I tried to fit my data with this function using the FIndFit routine, but had an error : Blockquote Power::infy: "Infinite expression 1/0. encountered" Blockquote data01 = Import["366.0000cm_660.0000nm.png"]; data02 = ImageData[data01]; min02 = Min[data02]; max02 = Max[data02]; (Dimensionalities of images) dim01a = Import["366.0000cm_660.0000nm.png", "ImageSize"] [[1]]; dim01b = Import["366.0000cm_660.0000nm.png", "ImageSize"] [[2]]; dim01 = Min[dim01a, dim01b]; ReducPix = 5; data03 = Table[ (data02[[i]][[j]][[1]] - min02)/max02, {i, 1, dim01b, ReducPix}, {j, 1, dim01a, ReducPix}]; (Dimensionalities) dim03a = Dimensions[data03][[1]]; dim03b = Dimensions[data03][[2]]; dim03 = Min[dim03a, dim03b] ListPlot3D[data03, PlotRange -> All] s[x_, y_] := Sqrt[a2 (x - a1)^2 + b2 (y - b1)^2] Bessel3D = Piecewise[ {{1/4, x == 0 && y == 0}, {(a0 BesselJ[1, s[x, 0]]/s[x, 0]), y == 0}, {(a0 BesselJ[1, s[0, y]]/s[0, y]), x == 0}}, {(a0 BesselJ[1, s[x, y]]/s[x, y])}] data3D = Flatten[MapIndexed[{#2[[1]], #2[[2]], #1} &, data03, {2}], 1]; fit = NonlinearModelFit[ data3D, (Bessel3D)^2, {{a0, 1}, {a1, 50}, {a2, 10}, {b1, 50}, {b2, 10}}, {x, y}]; I tried to use different method in the function (NMinimize, Gradient, so on ) but it did not help. It is really puzzling for me. Will appreciate any suggestions!!

Hello! I have 2D data (images) which are results of Gaussian type beam diffraction. enter image description here According to the theory of diffraction the pattern can be discribed by function ~ J1(r)/r. I tried to fit my data with this function using the FIndFit routine, but had an error :

Blockquote Power::infy: "Infinite expression 1/0. encountered" Blockquote

data01 = Import["366.0000cm_660.0000nm.png"];
data02 = ImageData[data01];
min02 = Min[data02];
max02 = Max[data02];

(*Dimensionalities of images*)
dim01a = Import["366.0000cm_660.0000nm.png", "ImageSize"] [[1]];
dim01b = Import["366.0000cm_660.0000nm.png", "ImageSize"] [[2]];
dim01 = Min[dim01a, dim01b];

ReducPix = 5;
data03 = Table[ (data02[[i]][[j]][[1]] - min02)/max02, {i, 1, dim01b, 
    ReducPix}, {j, 1, dim01a, ReducPix}];

(*Dimensionalities*)
dim03a = Dimensions[data03][[1]];
dim03b = Dimensions[data03][[2]];
dim03 = Min[dim03a, dim03b]

ListPlot3D[data03, PlotRange -> All]
s[x_, y_] := Sqrt[a2 (x - a1)^2 + b2 (y - b1)^2]
Bessel3D = Piecewise[
  {{1/4, x == 0 && y == 0},
   {(a0 BesselJ[1, s[x, 0]]/s[x, 0]), y == 0},
   {(a0 BesselJ[1, s[0, y]]/s[0, y]), x == 0}},
  {(a0 BesselJ[1, s[x, y]]/s[x, y])}]

data3D = Flatten[MapIndexed[{#2[[1]], #2[[2]], #1} &, data03, {2}], 1];
fit = NonlinearModelFit[
   data3D, (Bessel3D)^2, {{a0, 1}, {a1, 50}, {a2, 10}, {b1, 50}, {b2, 
     10}}, {x, y}];

I tried to use different method in the function (NMinimize, Gradient, so on ) but it did not help. It is really puzzling for me. Will appreciate any suggestions!!

POSTED BY: Alex Mikh

2 Replies

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Alex Mikh

Posted 12 years ago

POSTED BY: Alex Mikh

David Keith

Posted 12 years ago

Hi Alex, This is an interesting problem. The airy function has a singularity at 0 which can be removed by defining the value there as 1, which is the limit. I do that in the code below, and also make use of the radial symmetry to fit for only the central amplitude and the center. (Which is not all that is needed.) I have found that the code below works intermittently. It seems still dependent on choices of values made during the fit function, even though I think I have removed the singularity. I Hopefully someone with more knowledge will speak up. In[1]:= (2 BesselJ[1, r]/r)^2 /. r -> 0 During evaluation of In[1]:= Power::infy: Infinite expression 1/0^2 encountered. >> During evaluation of In[1]:= Infinity::indet: Indeterminate expression 0 ComplexInfinity encountered. >> Out[1]= Indeterminate In[2]:= Limit[(2 BesselJ[1, r]/r)^2, r -> 0] Out[2]= 1 In[3]:= airy[r_] := (2 BesselJ[1, r]/r)^2 /; r != 0. In[4]:= airy[0.] = 1 Out[4]= 1 In[5]:= airyData = Flatten[Table[{x, y, 5 airy[Sqrt[(x - 1)^2 + (y - 2)^2]]}, {x, -10, 10, .1}, {y, -10, 10, .1}], 1]; In[6]:= ListPlot3D[airyData]; In[7]:= NonlinearModelFit[airyData, a airy[Sqrt[(x - h)^2 + (y - k)^2]], {a, h, k}, {x, y}]

Hi Alex,

This is an interesting problem. The airy function has a singularity at 0 which can be removed by defining the value there as 1, which is the limit. I do that in the code below, and also make use of the radial symmetry to fit for only the central amplitude and the center. (Which is not all that is needed.)

I have found that the code below works intermittently. It seems still dependent on choices of values made during the fit function, even though I think I have removed the singularity. I Hopefully someone with more knowledge will speak up.

In[1]:= (2 BesselJ[1, r]/r)^2 /. r -> 0

During evaluation of In[1]:= Power::infy: Infinite expression 1/0^2 encountered. >>

During evaluation of In[1]:= Infinity::indet: Indeterminate expression 0 ComplexInfinity encountered. >>

Out[1]= Indeterminate

In[2]:= Limit[(2 BesselJ[1, r]/r)^2, r -> 0]

Out[2]= 1

In[3]:= airy[r_] := (2 BesselJ[1, r]/r)^2 /; r != 0.

In[4]:= airy[0.] = 1

Out[4]= 1

In[5]:= airyData = 
  Flatten[Table[{x, y, 5 airy[Sqrt[(x - 1)^2 + (y - 2)^2]]}, {x, -10, 
     10, .1}, {y, -10, 10, .1}], 1];

In[6]:= ListPlot3D[airyData];

In[7]:= NonlinearModelFit[airyData, 
 a airy[Sqrt[(x - h)^2 + (y - k)^2]], {a, h, k}, {x, y}]

POSTED BY: David Keith

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