Hello! I have some questions related to this post from 2015: https://public.kitware.com/pipermail/rtk-users/2015-February/009750.html
First, the referenced website "http://www.openrtk.org/Doxygen/classrtk_1_1NormalizedJosephBackProjectionImageFilter.html"; seems to not exist anymore. Second, I have some issues concerning backprojection and adjointness: For the following I am using these parameters: sid = 500 sdd = 1000 n_projections = 4 angular_increment = 90.0 I am using the python version of ITK and tried forward projection (CUDAForwardProjectionImageFilter) followed by backprojection (JosephBackProjectionImageFilter/CudaBackProjectionImageFilter/CudaRayCastBackProjectionImageFilter) of an image with a single pixel value set to one: img = np.zeros(image_size) img[image_size[0] // 2, image_size[1] // 2, image_size[2] // 2] = 1.0 1. I do not get expected results for the 2D case: image_size = [3, 1, 3] pixel_size_mm = 1.0 slice_thickness_mm = 1.0 projection_size = [5, 1] projection_spacing = [1.0, 1.0] After forward projection, the backprojected image by JosephBackProjectionImageFilter is all zeros. Is this expected behavior? 1. According to the definition of adjoint operator, the following should hold: <Ax, y> = <x, adjoint(A)y> (with <.,.> indicating the scalar product) This works for 3D CPU case (JosephBackProjectionImageFilter): image_size = [3, 3, 3] pixel_size_mm = 1.0 slice_thickness_mm = 1.0 projection_size = [5, 5] projection_spacing = [1.0, 1.0] <Ax, y> = <fp, fp> = 9.0 <x, adjoint(A)y> = <img, bp> = 9.000001907348633 This discrepancy is within tolerance. 1. However, in the 3D GPU case: <x, adjoint(A)y> = <img, bp> = 4.0 for CudaBackProjectionImageFilter and <x, adjoint(A)y> = <img, bp> = 0.33841705322265625 for CudaRayCastBackProjectionImageFilter What kind of filtering or normalization has been used on the projections/images? I could not verify adjointment with the aforementioned method of backprojecting a sinogram of ones and applying division. Is there a GPU backprojection filter for which adjointness holds? Thank you in advance for your reply and suggestions. Best regards, Laura
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