convolution of two image in frequency domain?

Question

TULIKA 2014-8-23

0
链接

此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/152083-convolution-of-two-image-in-frequency-domain

评论： Image Analyst 2021-12-13

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 个评论
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Linas Svilainis 2021-12-13

Are you sure this code has no erros? kernel1=fft2(kernel);

BUT

Imagef=kernel*grayImage1;

NOT kernel1?

Why * not .*?

Image Analyst 2021-12-13

@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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Answer 1

Image Analyst 2014-8-23

0
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/152083-convolution-of-two-image-in-frequency-domain#answer_149632

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 个评论
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TULIKA 2014-8-24

在 MATLAB Online 中打开

Sir I did....i hope now i have done it correctly...
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, []);
imagefinal=ifft2(fftshift(imagef));
image3=abs(imagefinal);subplot(2,3,2);
imshow (image3, []);
image4=imag(imagefinal);subplot(2,3,3);
imshow (image3, []);
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

TULIKA 2014-8-25

在 MATLAB Online 中打开

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 2014-8-25

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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