ex-6: rearranged the RL deconvolution description

notes/sections/6.md

@@ -107,14 +107,14 @@ The convolution was implemented as follow. Consider the definition of
 convolution of two functions $f(x)$ and $g(x)$:
 
 $$
-  f*g (x) = \int \limits_{- \infty}^{+ \infty} dy f(y) g(x - y)
+  f \otimes g (x) = \int \limits_{- \infty}^{+ \infty} dy f(y) g(x - y)
 $$
 
 Since a histogram is made of discrete values, a discrete convolution of the
 signal $s$ and the kernel $k$ must be computed. Hence, the procedure boils
-down to a dot product between $s$ and the reverse histogram of $k$ for each
+down to an element wise product between $s$ and the reverse histogram of $k$
-relative position of the two histograms. Namely, if $c_i$ is the $i^{\text{th}}$
+for each relative position of the two histograms. Namely, if $c_i$ is the
-bin of the convoluted histogram:
+$i^{\text{th}}$ bin of the convoluted histogram:
 
 $$
   c_i = \sum_j k_j s_{i - j}
@@ -186,7 +186,7 @@ one.
   \node [above] at (1.95,0)    {$s_{i-3}$};
   \node [below] at (1.75,-1)   {$k_3$};
 \end{tikzpicture}
-\caption{Dot product as a step of the convolution between the original signal
+\caption{Element wise product as a step of the convolution between the original signal
          (above) and the kernel (center). The final result is the lower
          fledging histogram.}\label{fig:dot_conv}
 }
@@ -200,7 +200,7 @@ Fast Fourier Transform. This method is based on the property of the Fourier
 transform according to which, given two functions $f(x)$ and $g(x)$:
 
 $$
-  \hat{F}[f*g] = \hat{F}[f] \cdot \hat{F}[g]
+  \hat{F}[f \otimes g] = \hat{F}[f] \cdot \hat{F}[g]
 $$
 
 where $\hat{F}[\quad]$ stands for the Fourier transform of its argument.  
@@ -209,8 +209,8 @@ trasform of the smeared signal and the kernel, the ratio between their
 transforms and the anti-transformation of the result:
 
 $$
-  \hat{F}[s*k] = \hat{F}[s] \cdot \hat{F}[k] \thus
+  \hat{F}[s \otimes k] = \hat{F}[s] \cdot \hat{F}[k] \thus
-  \hat{F} [s] = \frac{\hat{F}[s*k]}{\hat{F}[k]}
+  \hat{F} [s] = \frac{\hat{F}[s \otimes k]}{\hat{F}[k]}
 $$
 
 Being the histogram a discrete set of data, the Discrete Fourier Transform (DFT)
@@ -302,10 +302,10 @@ the negative backwards from the end of the array (see @fig:reorder).
 }
 \end{figure}
 
-When $\hat{F}[s*k]$ and $\hat{F}[k]$ are computed, their normal format must be
+When $\hat{F}[s \otimes k]$ and $\hat{F}[k]$ are computed, their normal format
-restored in order to use them as standard complex numbers and compute the ratio
+must be restored in order to use them as standard complex numbers and compute
-between them. Then, the result must return in the half-complex format for the
+the ratio between them. Then, the result must return in the half-complex format
-inverse DFT application.  
+for the inverse DFT application.  
 GSL provides the function `gsl_fft_halfcomplex_unpack` which passes the vectors
 from half-complex format to standard complex format. The inverse procedure,
 required to compute the inverse transformation of $\hat{F}[s]$, which is not
@@ -330,42 +330,43 @@ image can be represented in terms of a transition matrix
 $P$ operating on an underlying image:
 
 $$
- d_i = \sum_{j} P_{i, j} u_j
+ d_i = \sum_{j} u_j \, P_{i, j} 
 $$
 
 where $u_j$ is the intensity of the underlying image at pixel $j$ and $d_i$ is
-the detected intensity at pixel $i$. In general, a matrix whose elements are
+the detected intensity at pixel $i$. Hence, the matrix describes the portion of
-$P_{i,j}$ describes the portion of signal from the source pixel $j$ that is
+signal from the source pixel $j$ that is detected in pixel $i$.  
-detected in pixel $i$.  
+In one dimension, the transfer function can be expressed in terms of the
-In one dimension, the transfer function
+distance between the source pixel $j$ and the observed $i$:
-can be expressed in terms of the distance between the source pixel $j$ and the
-observed $i$:
 
 $$
-  P_{i, j} = \widetilde{P}(i-j)
+  P_{i, j} = \widetilde{P}(i-j) = P_{i - j}
 $$
 
-In order to estimate $u_j$ given the observed $d_i$ and a known $\widetilde{P}$,
+In order to estimate $u_j$ given {$d_i$} and $\widetilde{P}$, the following
-the following iterative procedure for the estimate  $\hat{u}^t_j$ of $u_j$ can
+iterative procedure can be applied for the estimate $\hat{u}^t_j$ of $u_j$,
-be applied. The $t^{\text{th}}$ iteration is updated as follows:
+where $t$ stands for the iteration number. The $t^{\text{th}}$ step is updated
+as follows:
 
 $$
-  \hat{u}^{t+1}_j = \hat{u}^t_j \sum_i \frac{d_i}{c_i} \, P_{ij}
+  \hat{u}^{t+1}_j = \hat{u}^t_j \sum_i \frac{d_i}{c_i} \, P_{i - j}
-  \with c_i = \sum_j P_{ij} {\hat{u^t}}_j.
+  \with c_i = \sum_j \hat{u}^t_j \, P_{i - j}
 $$
 
+where $c_i$ is thereby an estimation of the blurred signal obtained with the
+previous estimation of the clean signal.  
 It has been shown empirically that if this iteration converges, it converges to
-the maximum likelihood solution for $u_j$.  
+the maximum likelihood solution for $u_j$. Writing it in terms of convolution,
-Writing this more generally in terms of convolution with a point spread function
+it becomes:
-$\tilde{P}$ it becomes:
 
 $$
-   \hat{u}^{t+1} = \hat {u}^{t} \cdot \left( \frac{d}{{\hat{u}^{t}} \otimes
+   \hat{u}^{t+1} = \hat {u}^{t} \cdot \left( \frac{d}{{\hat{u}^{t}} \otimes P}
-   \widetilde{P}} \otimes \widetilde{P}^{\star} \right)
+   \otimes P^{\star} \right)
 $$
 
 where the division and multiplication are element wise, and
-$\widetilde{P}^{\star}$ is the flipped point spread function.
+$P^{\star}$ is the flipped point spread function.
+
 
 ---