convolution of two image in frequency domain?

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x= 1:150;y=1:150;
[p,q]=freqspace(150);
[X,Y]=meshgrid(p,q);
R=(X.^2 + Y.^2);
Lambda=633*10^-9;
dis=10*10^-3;
F = (exp(i.*pi.*R))./(Lambda.*dis);
kernel = imag(F); % Gray scale image = imaginary part.
mesh(kernel);
axis square
imshow(kernel, []);
axis on;
colormap(gray(256));
rgbImage = imread('image.jpg');
imshow(rgbImage);
axis on;
% R=rgbImage(:,:,1);
% G=rgbImage(:,:,2);
% B=rgbImage(:,:,3);
grayImage = rgb2gray(rgbImage);
imshow(grayImage, []);
axis on;
grayImage1=fft2(double(grayImage));
kernel1=fft2(kernel);
Imagef=kernel*grayImage1;
Imagefinal=ifft2(Imagef);
imshow(Imagefinal, []);
axis on;
This code has no errors. Its output should be like a hazy image of concentrating circle or may be overlapped, but it shows hazy straight line. Please help.
  2 Comments
Linas Svilainis
Linas Svilainis on 13 Dec 2021
Are you sure this code has no erros? kernel1=fft2(kernel);
BUT
Imagef=kernel*grayImage1;
NOT kernel1?
Why * not .*?
Image Analyst
Image Analyst on 13 Dec 2021
@Linas Svilainis I think they meant to say "This code has errors." The problem is to multiply the spectrum of a kernel times a spectrum of a filter, you'd need to use .* (dot star) instead of *, and you'd need to make sure they both have the same size (number of pixels).

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Accepted Answer

Image Analyst
Image Analyst on 23 Aug 2014
See my demos, attached. They're good examples that show you how it's done. Once you understand and run those I have no doubt you will be able to do the same thing with your image and kernel.
  13 Comments
TULIKA
TULIKA on 25 Aug 2014
this is the program what I've written last and got this output
x= 1:256;y=1:256;
[p,q]=freqspace(256);
[X,Y]=meshgrid(p,q);
R=(X.^2 + Y.^2);
Lambda=633*10^-9;
dis=10*10^-3;
F = (exp(i.*pi.*R))./(Lambda.*dis);
kernel = imag(F); % Gray scale image = imaginary part.
mesh(kernel);
axis square
imshow(kernel, []);
axis on;
colormap(gray(256));
rgbImage = imread('image.jpg');
imshow(rgbImage);
axis on;
% R=rgbImage(:,:,1);
% G=rgbImage(:,:,2);
% B=rgbImage(:,:,3);
grayImage = rgb2gray(rgbImage);
imshow(grayImage, []);
axis on;
fftOriginal = fft2(double(grayImage));
shiftedFFT = fftshift(fftOriginal);
frequencyimage=fftshift(fft2(kernel));
imagef=shiftedFFT.*frequencyimage;
subplot(2,3,1);
imshow(imagef, []);
title('fft of image');
imagefinal=ifft2(fftshift(imagef));
image3=real(imagefinal);subplot(2,3,2);
imshow (image3, []);
title('real part of image');
image3=imag(imagefinal);subplot(2,3,3);
imshow (image4, []);
title('imaginary part of image');
Warning: Displaying real part of complex input.
> In imuitools\private\imageDisplayValidateParams>validateCData at 149
In imuitools\private\imageDisplayValidateParams at 31
In imuitools\private\imageDisplayParseInputs at 79
In imshow at 199
Image Analyst
Image Analyst on 25 Aug 2014
This is one situation where clear all is good. See how you were able to display image4? But you never defined image4, so where did that image4 come from? Not from this running of the program. It must have existed already. Who knows how many other times this happened with other variables. Either call clear all or put it in the front of your script.

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