Skip to main content

Problem with InverseFourier of a 2d FFT of a non-squared image


I have the following 8bit grey scale image:


enter image description here



The 2d FFT of this image, showing the color coded Abs[fft] values, is:


enter image description here


The code to obtain the FFT image is:


img = Import["https://i.stack.imgur.com/CA0nv.png"];

dimimg = ImageDimensions[img];
rdimimg = Reverse[dimimg];

fft = Fourier[ImageData[img]];
fftRotated = RotateLeft[fft, Floor[Dimensions[fft]/2]];


fftAbsData = Abs[fftRotated];

minc = 140;
myColorTable = 
  Flatten@{Table[{Blend[{Blue, Green, Yellow, Orange}, x]}, {x, 
      1/minc, 1, 1/minc}], 
    Table[{Blend[{Orange, Red, Darker@Red}, x]}, {x, 1/(256 - minc), 
      1, 1/(256 - minc)}]};


g = Colorize[
   ImageResize[Image[fftAbsData], {rdimimg[[1]], rdimimg[[2]]}], 
   ColorFunction -> (Blend[myColorTable, #] &)];

xfrequencies = (Range[rdimimg[[1]]] - Round[rdimimg[[1]]/2])/
   rdimimg[[1]];
yfrequencies = (Range[rdimimg[[2]]] - Round[rdimimg[[2]]/2])/
   rdimimg[[2]];

minmaxxf = MinMax[xfrequencies];

minmaxyf = MinMax[yfrequencies];

dy = minmaxyf[[2]] - minmaxyf[[1]];
dx = minmaxxf[[2]] - minmaxxf[[1]];

scaleFactor = 600;

imagefft = ImageResize[g, scaleFactor*{dx, dy}]

Questions:



Now I would like to cut out the red ring and make a backward FFT to see which objects of the original image belong to the high amplitude fft data, seen in red.


How can I cout out a circular region in the fft image?


Even without cutting out a part I am not able to reproduce the original image from fft:


inverse=Image[InverseFourier[fft]]

gives me:


enter image description here


Why does InverseFourier not reproduce the original image?



Answer



Using your image, you can first check that InverseFourier does indeed reproduce the original image:



inverse = Image[Chop@InverseFourier[fft]];
ImageDistance[inverse, img]

returns 3.90437*10^-6



Now I would like to cut out the red ring and make a backward FFT to see which objects of the original image belong to the high amplitude fft data, seen in red.



Well, don't expect "objects" - the Fourier transform is a global operation, so you'll basically see the result of a linear (bandpass) filter by masking in the FFT domain.


But you can try easily enough. Simply create a binary mask:


mask = Array[Boole[.15 < Norm[{##}] < .25] &, 

   Dimensions[fftRotated], {{-1., 1.}, {-1., 1.}}];

HighlightImage[Image[Rescale@Log[Abs@fftRotated]], Image[mask]]

enter image description here


Multiply it with the FFT and apply inverse transform:


fftMasked = RotateRight[mask*fftRotated, Floor[Dimensions[fft]/2]];

And display the real part of the result:


Image[Rescale[Chop[Re@InverseFourier[fftMasked]]]]


enter image description here


Comments

Post a Comment

Popular posts from this blog

functions - Get leading series expansion term?

Given a function f[x] , I would like to have a function leadingSeries that returns just the leading term in the series around x=0 . For example: leadingSeries[(1/x + 2)/(4 + 1/x^2 + x)] x and leadingSeries[(1/x + 2 + (1 - 1/x^3)/4)/(4 + x)] -(1/(16 x^3)) Is there such a function in Mathematica? Or maybe one can implement it efficiently? EDIT I finally went with the following implementation, based on Carl Woll 's answer: lds[ex_,x_]:=( (ex/.x->(x+O[x]^2))/.SeriesData[U_,Z_,L_List,Mi_,Ma_,De_]:>SeriesData[U,Z,{L[[1]]},Mi,Mi+1,De]//Quiet//Normal) The advantage is, that this one also properly works with functions whose leading term is a constant: lds[Exp[x],x] 1 Answer Update 1 Updated to eliminate SeriesData and to not return additional terms Perhaps you could use: leadingSeries[expr_, x_] := Normal[expr /. x->(x+O[x]^2) /. a_List :> Take[a, 1]] Then for your examples: leadingSeries[(1/x + 2)/(4 + 1/x^2 + x), x] leadingSeries[Exp[x], x] leadingSeries[(1/x + 2 + (1 - 1/x...
Post a Comment
Read more

mathematical optimization - Minimizing using indices, error: Part::pkspec1: The expression cannot be used as a part specification

I want to use Minimize where the variables to minimize are indices pointing into an array. Here a MWE that hopefully shows what my problem is. vars = u@# & /@ Range[3]; cons = Flatten@ { Table[(u[j] != #) & /@ vars[[j + 1 ;; -1]], {j, 1, 3 - 1}], 1 vec1 = {1, 2, 3}; vec2 = {1, 2, 3}; Minimize[{Total@((vec1[[#]] - vec2[[u[#]]])^2 & /@ Range[1, 3]), cons}, vars, Integers] The error I get: Part::pkspec1: The expression u[1] cannot be used as a part specification. >> Answer Ok, it seems that one can get around Mathematica trying to evaluate vec2[[u[1]]] too early by using the function Indexed[vec2,u[1]] . The working MWE would then look like the following: vars = u@# & /@ Range[3]; cons = Flatten@{ Table[(u[j] != #) & /@ vars[[j + 1 ;; -1]], {j, 1, 3 - 1}], 1 vec1 = {1, 2, 3}; vec2 = {1, 2, 3}; NMinimize[ {Total@((vec1[[#]] - Indexed[vec2, u[#]])^2 & /@ R...
Post a Comment
Read more

plotting - Plot 4D data with color as 4th dimension

I have a list of 4D data (x position, y position, amplitude, wavelength). I want to plot x, y, and amplitude on a 3D plot and have the color of the points correspond to the wavelength. I have seen many examples using functions to define color but my wavelength cannot be expressed by an analytic function. Is there a simple way to do this? Answer Here a another possible way to visualize 4D data: data = Flatten[Table[{x, y, x^2 + y^2, Sin[x - y]}, {x, -Pi, Pi,Pi/10}, {y,-Pi,Pi, Pi/10}], 1]; You can use the function Point along with VertexColors . Now the points are places using the first three elements and the color is determined by the fourth. In this case I used Hue, but you can use whatever you prefer. Graphics3D[ Point[data[[All, 1 ;; 3]], VertexColors -> Hue /@ data[[All, 4]]], Axes -> True, BoxRatios -> {1, 1, 1/GoldenRatio}]
Post a Comment
Read more