# Inverse Multiscale Discrete Radon Transform by Filtered Backprojection
## 2 Proposed Method
### Characterization of the Impulse Responses
#### 2.2.1 Time Variant Impulse Responses  
As stated previously, the following are the recursive and non-recursive definitions of discrete line:  
\[l_{s}^{n}(u_{0},\,\ldots,\,u_{n-1})=l_{\left\lfloor s/2\right\rfloor}^{n-1}(u_ {0},\,\ldots,\,u_{n-2})+u_{n-1}\left\lfloor\frac{s+1}{2}\right\rfloor=\sum_{i= 0}^{n-1}u_{n-1-i}\cdot\left\lfloor\frac{\left\lfloor\frac{s}{2^{i}}\right\rfloor +1}{2}\right\rfloor.\]  
For example, when \(N=8\), \(n=\log_{2}(N)=3\), a 'discrete' line of slope \(s=5\), will traverse through points \(\{(u,\,l_{5}^{3}(u))\}\) with \(u=0\,...\,N-1\), _i.e._: \(\{(0,0),(1,1),(2,1),(3,2),\)\((4,3),(5,4),(6,4),(7,5)\}\). The lines with an odd ascent-slope will rise in the boundary between the values \(N/2-1\) and \(N/2\). This is the 'bit boundary' with the most significant bit weight: \(2^{n-1}=N/2\). In other words, whether a line ascends or not preceding \(N/2\) will be determined by the least significant bit of the slope, \(s_{0}\) (which is one for odd numbers). The next least significant 'bit boundary' occurs at \(N/4\) and \(N/2+N/4\), and the ascent in there will be determined by bit \(s_{1}\) of the slope. In general, the bit \(s_{i}\) of the slope will determine whether to ascend or not at positions whose greatest power of two divisor is \(2^{(n-1-i)}\). In the previous example, for \(n=3\) and \(s=5\), the discrete line has ascended in bit boundaries \(s_{0}\)\((u=3\leftrightarrow 4)\) and \(s_{2}\)\((u=0\leftrightarrow 1,2\leftrightarrow 3,4\leftrightarrow 5,6\leftrightarrow 7)\), and remained without ascension  
Figure 4: Extended DRT and original DRT, shown inserted on lower left corner. for \(s_{1}\)\((u=2\leftrightarrow 3,5\leftrightarrow 6)\); of course, in accordance with \(s=5\) expression in binary as \((1,0,1)\). The bits boundaries of the slopes for the extended DRT that we propose are shown in Figure 5. The \(N\) original data are located between positions \(N\) and \(2\cdot N\) when the domain is extended to \(4\cdot N\). The figure shows those positions, as well as \(N/2\) additional values to each side. An impulse that is located originally at the position 0 in one dimension, will be located at the position \(N\) in the extension and through it there will be \(N\) lines crossing, with slopes approximately equal to the multiples of 4 of the values from 0 to \(N-1\). Actually, the original slopes were \(s=0...\,N-1\), but now they will be \(\tilde{s}=4\cdot s\,+\,3\cdot(s\%2)\); this means that we add 3 to the odd slopes. This process at bit level is equivalent to shifting two places to the left the original bits of the slope and replicate in the two least significant additional bits the original \(s_{0}\) bit. In the Figure 5 this phenomenon is highlighted in the boundaries around \(N\) and \(2\cdot N\), limits where the original image is placed in, and that in terms of the extended slope it would be flanked by the bits \(\tilde{s}_{1}\) and \(\tilde{s}_{0}\), but to the effects of the lines that we are going to inject in the system these bits are equivalent to \(s_{0}\). With this preamble we can move on to describe how the combined response of the direct transform plus the adjoint transform is configured around an impulse in any position of a dimension of an image. The following description will be valid for the quadrant of positive slopes. The negative slopes are a horizontally inverted replica of the responses we will characterize now. The responses in the other dimension can be found with a transposition. The impulse response to a unitary input located originally in the position 0 (extended position N) in a problem of size \(N=16\) will find in the positive horizontal direction and by order of boundaries: \(s_{3},s_{2},s_{3},s_{1},s_{3},s_{2},s_{3},s_{0},s_{3},s_{2},s_{3},s_{1},s_{3}, s_{2},s_{3}\), see Figure 5. If the size of the problem was, instead, \(N=8\), the \(s_{3}\) boundaries would not exist. If the size of the problem was \(N=32\), there would be additional \(s_{4}\) boundaries interleaved between those of the \(N=16\) case, and so on.