Sinogram reconstruction with fourier transform

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?

Answer 1

[David Ding]

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

[Egor Fab]

Hey David I tried the fft2 and i'm not having a black image anymore.

But sadly this is not the reconstructed image.

Answer 2

[Sergio Serra Sánchez]

Any solution for this?

Answer 3

[Sean de Wolski]

iradon or ifanbeam?

[Sergio Serra Sánchez]

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.