Sinogram reconstruction with fourier transform
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I want to reconstruct this sinogram.
I do the 1D fourier on the sinogram and get this.
for y = 1:height
slice = i(y,:);
slice = fftshift(ifft2(ifftshift(slice)));
subTargetImage(y, 1:width) = slice(1,:);
end
Then I create the fourier space
for x = 1:height
for y = 1:width
[tX, tY] = pol2cart(x, y - width/2);
if(round(tX) + targetImageHeight/2 == 0)
tX = tX + 1;
end
if(round(tY) + targetImageWidth/2 == 0)
tY = tY + 1;
end
targetImage(round(tX) + targetImageHeight/2, round(tY) + targetImageWidth/2) = subTargetImage(x, y);
targetImage = fftshift(ifft2(ifftshift(targetImage)));
But when I do the reverse transformation I just get a black Image. How do I get a cleare image without using the filtered backprojection?
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Answers (3)
David Ding
on 29 Dec 2016
Hello Egor,
It appears that the statement "targetImage = fftshift(ifft2(ifftshift(targetImage)))", is not the inverse transformation of "slice = fftshift(ifft2(ifftshift(slice)))", as the second operation of the original transformation is "ifft2". Perhaps use "fft2" for the "targetImage" operation?
I conducted a similar workflow by applying fftshift(ifft2(ifftshift(.))) on a sample image and obtained similar output as your last screenshot. Then I applied fftshift(fft2(ifftshift(.))) and I do not have a black image anymore.
Thanks,
David
Sean de Wolski
on 1 Mar 2018
iradon or ifanbeam?
1 Comment
Sergio Serra Sánchez
on 1 Mar 2018
Thank you for your answer!
The main idea of this experiment it's to try avoid use these functions. It have to be implemented with 1D Fourier of each angle on the sinogram and then to build the 2D transformation for all the projection. Do you know how do it?
Thank again.
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