# Inverse Multiscale Discrete Radon Transform by Filtered Backprojection
## 2 Proposed Method
### Deconvolving Different Impulse Responses  
By using Fourier, we can only deconvolve quickly if all the positions generate the same impulse response, but that is not the case, as we have shown that we can have \(N/4\) different impulse responses affecting the horizontal half and \(N/4\) affecting to the vertical half of the vicinities of a point. Our task now is to create a method that takes into consideration this diversity of impulse responses. It works in our favor that the responses and the positions where they happen are known. We could say that this system is partially LTI: it is not completely LTI, but its variability is limited enough that it can be deconvolved with a couple of iterations of simple deconvolutions.  
Let \(f(x)\) be an unidimensional unaltered signal; and let \(m(x)\) be the measured signal that results by altering \(f(x)\) by some known finite impulse responses \(q(x)\) or \(r(x)\), each impulse response affecting respectively to samples at positions \(P_{q}(x)\) or \(P_{r}(x)\)  
Figure 7: **Left**, unique impulse responses for size \(N=8\). **Right**, the equivalent for \(N=16\).  
those functions will be \(1\) where the system responds with the response \(q\) and \(r\) respectively, or \(0\) otherwise; and let \(P_{q}(x)+P_{r}(x)=1,\forall x\). Then \(m(x)=[f(x)\cdot P_{q}(x)]*q(x)+[f(x)\cdot P_{r}(x)]*r(x)\), the \((*)\) symbolizes the convolution operand.  
Let us denote the inverse of the impulse response \(q(x)\), with \(\tilde{q}(x)=\tilde{q}^{-1}\big{(}\frac{1}{\tilde{q}(x)}\big{)}\), and let us use it to deconvolve the whole \(m(x)\) signal. The values at positions \(P_{q}(x)\) will be back, mostly, to their original values, \(f(x)|_{P_{q}(x)=1}\), except for the tails of adjacent \(r(x)*\tilde{q}(x)\) elements at positions \(P_{r}(x)\) that are perturbing them: \(m_{q}(x)=m(x)*\tilde{q}(x)=[f(x)\cdot P_{q}(x)]+[f(x)\cdot P_{r}(x)]*r(x)* \tilde{q}(x)\). The next step is to eliminate these perturbations with a second deconvolution only performed at position \(P_{r}(x)\) with the compound of \(r(x)*\tilde{q}(x)\). Let us denote by \(r\tilde{q}(x)=(r(x)*\tilde{q}(x))\), and with \(r\tilde{q}(0)\) the central value of that kernel. Then, it be can written:  
\[f_{0}(x)=m_{q}(x)\]  
\[f_{n}(x)=m_{q}(x)-\big{[}f_{n-1}(x)\cdot P_{r}(x)/r\tilde{q}(0)\big{|}_{n=1} \big{]}*r\tilde{q}(x)+[f_{n-1}(x)\cdot P_{r}(x)/r\tilde{q}(0)\big{|}_{n=1}]\]  
\[f(x)\approx f_{n}(x)\qquad n\to\infty\]  
The subtracting term eliminates the wrongly deconvolved elements at positions \(P_{r}(x)\), and their surroundings. Then we reincorporate our best guess, up to that iteration, just at the main spike of the kernel at those positions. Notice that the central value of the kernel may differ from one, so we are compensating this effect with the divisions by \(r\tilde{q}(0)\), but only in the first iteration. For the sort of impulse responses that backprojection DRT generates, the just proposed method recovers \(f(x)\) from \(m(x)\) in \(2\) or \(3\) iterations, and can be easily extended to more dimensions and for more than two impulse responses.