ex-6: section results completed
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@ -304,7 +304,7 @@ $$
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](images/1-landau-kde.pdf)
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On the other hand, obtaining a good estimate of the FWHM from a sample is much
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more difficult. In principle, it could be measured by binning the data and
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@ -181,7 +181,7 @@ where 'floor' is the function which gives the bigger integer smaller than its
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argument and the bins are counted starting from zero.
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Then, a vector in which the $j^{\text{th}}$ entry contains both the sum $S_j$
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of all the $|P_v|$s relative to each $P_h$ fallen into the $j^{\text{th}}$ bin
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itself and the number num$_j$ of the bin entries was reiteratively updated. At
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itself and the number num$_j$ of the bin entries was iteratively updated. At
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the end, the average value of $|P_v|_j$ was computed as $S_j / \text{num}_j$.
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For the sake of clarity, for each sampled couple the procedure is the
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following. At first $S_j = 0 \wedge \text{num}_j = 0 \, \forall \, j$, then:
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@ -68,7 +68,7 @@ In this case $f(x) = e^{x}$ and $\Omega = [0,1]$, hence $V = 1$.
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Since the proximity of $I_N$ to $I$ is related to $N$, the accuracy of the
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method lies in how many points are generated, namely how many function calls
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are executed when the reiterative method is implemented. In @fig:MC and
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are executed when the iterative method is implemented. In @fig:MC and
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@fig:MI, results obtained with the plain MC method are shown in red. In
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@tbl:MC, some of them are listed: the estimated integrals $I^{\text{oss}}$ are
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compared to the expected value $I$ and the differences diff between them are
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@ -333,7 +333,7 @@ the largest contribution to the integral.
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As stated before, in practice it is impossible to sample points from the best
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distribution $P^{(y)}$: only a good approximation can be achieved. In GSL, the
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VEGAS algorithm approximates the distribution by histogramming the function $f$
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in different subregions with a reiterative method [@lepage78], namely:
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in different subregions with a iterative method [@lepage78], namely:
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- a fixed number of points (function calls) is evenly generated in the whole
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region;
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@ -371,7 +371,7 @@ where $I_i$ and $\sigma_i$ are are the integral and standard deviation
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estimated in each interval $\Delta x_i$ of the last iteration using the plain
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MC technique.
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For the results to be reliable, the chi-squared per degree of freedom
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$\chi_r^2$ must be consistent with 1. While performing the reiterations, if
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$\chi_r^2$ must be consistent with 1. While performing the eiterations, if
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the $\chi_r^2$ value exceed 1.5, the cycle breaks.
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Clearly, once again a better estimation is achieved with a greater number of
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@ -85,21 +85,21 @@ omitted:
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The so obtained sample was binned and stored in a histogram with a customizable
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number $n$ of bins (default set $n = 150$) ranging from $\theta = 0$ to $\theta
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= \pi/2$ bacause of the system symmetry. In @fig:original an example is shown.
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= \pi/2$ because of the system symmetry. In @fig:original an example is shown.
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{#fig:original}
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## Gaussian convolution {#sec:convolution}
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In order to simulate the instrumentation response, the sample was then to be
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smeared with a Gaussian function with the aim to recover the original sample
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afterwards, implementing a deconvolution routine.
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In order to simulate the instrumentation response, the sample was then smeared
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with a Gaussian function with the aim to recover the original sample afterwards,
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implementing a deconvolution routine.
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For this purpose, a 'kernel' histogram with an even number $m$ of bins and the
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same bin width of the previous one, but a smaller number of them ($m \sim 6\%
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\, n$), was created according to a Gaussian distribution with mean $\mu$,
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\, n$), was generated according to a Gaussian distribution with mean $\mu = 0$
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and variance $\sigma$. The reason why the kernel was set this way will be
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descussed lately.
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discussed shortly.
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Then, the original histogram was convolved with the kernel in order to obtain
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the smeared signal. As an example, the result obtained for $\sigma = \Delta
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\theta$, where $\Delta \theta$ is the bin width, is shown in @fig:convolved.
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@ -121,11 +121,11 @@ down to an element wise product between $s$ and the flipped histogram of $k$
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histograms. Namely, if $c_i$ is the $i^{\text{th}}$ bin of the convolved
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histogram:
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$$
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c_i = \sum_j k_j s_{i - j} = \sum_{j'} k_{m - j'} s_{i - m + j'}
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\with j' = m - j
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c_i = \sum_{j = 0}^{m - 1} k_j s_{i - j}
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= \sum_{j' = m - 1}^{0} k_{m - 1 - j'} s_{i - m + 1 + j'}
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\with j' = m - 1 - j
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$$
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where $j$ runs over the bins of the kernel.
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For a better understanding, see @fig:dot_conv: the third histogram turns out
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with $n + m - 1$ bins, a number greater than the original one.
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@ -187,8 +187,8 @@ with $n + m - 1$ bins, a number greater than the original one.
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% nodes
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\node [above] at (2.25,-5.5) {$c_i$};
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\node [above] at (3.25,0) {$s_i$};
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\node [above] at (1.95,0) {$s_{i-3}$};
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\node [below] at (1.75,-1) {$k_3$};
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\node [above] at (1.95,0) {$s_{i-j}$};
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\node [below] at (1.75,-1) {$k_j$};
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\end{tikzpicture}
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\caption{Element wise product as a step of the convolution between the original signal
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(above) and the kernel (center). The final result is the lower
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@ -210,13 +210,13 @@ $$
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where $\hat{F}[\quad]$ stands for the Fourier transform of its argument.
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Being the histogram a discrete set of data, the Discrete Fourier Transform (DFT)
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was appied. When dealing with arrays of discrete values, the theorems still
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holds if the two arrays have the same lenght and a cyclical convolution is
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was applied. When dealing with arrays of discrete values, the theorem still
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holds if the two arrays have the same length and a cyclical convolution is
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applied. For this reason the kernel was 0-padded in order to make it the same
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lenght of the original signal. Besides, the 0-padding allows to avoid unpleasant
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length of the original signal. Besides, the 0-padding allows to avoid unpleasant
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side effects due to the cyclical convolution.
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In order to accomplish this procedure, every histogram was transformed into a
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vector. The implementation lies in the computation of the Fourier trasform of
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In order to accomplish this procedure, both histograms were transformed into
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vectors. The implementation lies in the computation of the Fourier transform of
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the smeared signal and the kernel, the ratio between their transforms and the
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anti-transformation of the result:
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$$
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@ -225,16 +225,17 @@ $$
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$$
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The FFT are efficient algorithms for calculating the DFT. Given a set of $n$
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values {$z_i$}, each one is transformed into:
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values {$z_j$}, each one is transformed into:
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$$
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x_j = \sum_{k=0}^{n-1} z_k \exp \left( - \frac{2 \pi i j k}{n} \right)
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$$
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where $i$ is the imaginary unit.
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The evaluation of the DFT is a matrix-vector multiplication $W \vec{z}$. A
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general matrix-vector multiplication takes $O(n^2)$ operations. FFT algorithms,
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instad, use a divide-and-conquer strategy to factorize the matrix into smaller
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sub-matrices. If $n$ can be factorized into a product of integers $n_1$, $n_2
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\ldots n_m$, then the DFT can be computed in $O(n \sum n_i) < O(n^2)$
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instead, use a *divide-and-conquer* strategy to factorize the matrix into
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smaller sub-matrices. If $n$ can be factorized into a product of integers $n_1$,
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$n_2 \ldots n_m$, then the DFT can be computed in $O(n \sum n_i) < O(n^2)$
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operations, hence the name.
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The inverse Fourier transform is thereby defined as:
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$$
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@ -243,13 +244,13 @@ $$
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$$
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In GSL, `gsl_fft_complex_forward()` and `gsl_fft_complex_inverse()` are
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functions which allow to compute the foreward and inverse transform,
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functions which allow to compute the forward and inverse transform,
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respectively.
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The inputs and outputs for the complex FFT routines are packed arrays of
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floating point numbers. In a packed array the real and imaginary parts of
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each complex number are placed in alternate neighboring elements. In this
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special case, the sequence of values to be transformed is made of real numbers,
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but the Fourier transform is not real: it is a complex sequence wich satisfies:
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floating point numbers. In a packed array, the real and imaginary parts of each
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complex number are placed in alternate neighboring elements. In this special
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case, the sequence of values to be transformed is made of real numbers, hence
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Fourier transform is a complex sequence which satisfies:
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$$
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z_k = z^*_{n-k}
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$$
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@ -260,7 +261,7 @@ forward transform (from real to half-complex) and inverse transform (from
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half-complex to real). As a consequence, the routines are divided into two sets:
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`gsl_fft_real` and `gsl_fft_halfcomplex`. The symmetry of the half-complex
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sequence requires only half of the complex numbers in the output to be stored.
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This works for all lengths: when the length is even, the middle value is real.
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This works for all lengths: when the length is odd, the middle value is real.
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Thus, only $n$ real numbers are required to store the half-complex sequence
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(half for the real part and half for the imaginary).
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@ -293,7 +294,7 @@ Thus, only $n$ real numbers are required to store the half-complex sequence
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\end{scope}
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\end{tikzpicture}
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\caption{The histogram on the right shows how the real numbers histogram on the
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left is handled by the dedicated GSL functions`.}\label{fig:reorder}
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left is handled by the dedicated GSL functions.}\label{fig:reorder}
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}
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\end{figure}
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@ -305,7 +306,7 @@ the array up to the middle and the negative backwards from the end of the array
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Whilst do not matters if the convolved histogram has positive or negative
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values, the kernel must be centered in zero in order to compute a correct
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convolution. This requires the kernel to be made of an ever number of bins
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in order to be possible to cut it into two same-lenght halves.
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in order to be possible to cut it into two same-length halves.
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When $\hat{F}[s * k]$ and $\hat{F}[k]$ are computed, they are given in the
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half-complex GSL format and their normal format must be restored in order to
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@ -318,24 +319,24 @@ code.
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At the end, the external bins which exceed the original signal size are cut
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away in order to restore the original number of bins $n$. Results will be
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discussed in @sec:conv_Results.
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discussed in @sec:conv_results.
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## Unfolding with Richardson-Lucy
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The Richardson–Lucy (RL) deconvolution is an iterative procedure tipically used
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The Richardson–Lucy (RL) deconvolution is an iterative procedure typically used
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for recovering an image that has been blurred by a known 'point spread
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function'.
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Consider the problem of estimating the frequeny distribution $f(\xi)$ of a
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Consider the problem of estimating the frequency distribution $f(\xi)$ of a
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variable $\xi$ when the available measure is a sample {$x_i$} of points
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dronwn not by $f(x)$ but by another function $\phi(x)$ such that:
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drown not by $f(x)$ but by another function $\phi(x)$ such that:
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$$
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\phi(x) = \int d\xi \, f(\xi) P(x | \xi)
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$$ {#eq:conv}
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where $P(x | \xi) \, d\xi$ is the probability (presumed known) that $x$ falls
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in the interval $(x, x + dx)$ when $\xi = \xi$. If the so-colled point spread
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in the interval $(x, x + dx)$ when $\xi = \xi$. If the so-called point spread
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function $P(x | \xi)$ follows a normal distribution with variance $\sigma$,
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namely:
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$$
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@ -371,12 +372,12 @@ $$
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f(\xi) = \int dx \, \phi(x) Q(\xi | x)
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$$ {#eq:second}
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Since $Q (\xi | x)$ depends on $f(\xi)$, @eq:second suggests a reiterative
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Since $Q (\xi | x)$ depends on $f(\xi)$, @eq:second suggests an iterative
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procedure for generating estimates of $f(\xi)$. With a guess for $f(\xi)$ and
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a known $P(x | \xi)$, @eq:first can be used to calculate and estimate for
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$Q (\xi | x)$. Then, taking the hint provided by @eq:second, an improved
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estimate for $f (\xi)$ is generated, using the observed sample {$x_i$} to give
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an approximation for $\phi$.
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estimate for $f (\xi)$ can be generated, using the observed sample {$x_i$} to
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give an approximation for $\phi$.
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Thus, if $f^t$ is the $t^{\text{th}}$ estimate, the $t^{\text{th + 1}}$ is:
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$$
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f^{t + 1}(\xi) = \int dx \, \phi(x) Q^t(\xi | x)
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@ -401,15 +402,15 @@ $$
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where $P^{\star}$ is the flipped point spread function [@lucy74].
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In this special case, the Gaussian kernel which was convolved with the original
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histogram stands for the point spread function and, dealing with discrete
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values, the division and multiplication are element wise and the convolution is
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to be carried out as described in @sec:convolution.
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histogram stands for the point spread function. Dealing with discrete values,
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the division and multiplication are element wise and the convolution is to be
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carried out as described in @sec:convolution.
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When implemented, this method results in an easy step-wise routine:
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- create a flipped copy of the kernel;
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- choose a zero-order estimate for {$f(\xi)$};
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- create a flipped copy of the kernel;
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- compute the convolutions, the product and the division at each step;
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- proceed until a given number of reiterations is achieved.
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- proceed until a given number $r$ of iterations is reached.
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In this case, the zero-order was set $f(\xi) = 0.5 \, \forall \, \xi$. Different
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number of iterations where tested. Results are discussed in
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@ -450,10 +451,9 @@ $$
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W (P, Q, F) = \sum_{i = 1}^m \sum_{j = 1}^n f_{ij} d_{ij}
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$$
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The fact is that the $Q$ region is to be considerd empty at the beginning: the
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'dirt' present in $P$ must be moved to $Q$ in order to reach the same
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distribution as close as possible. Namely, the following constraints must be
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satisfied:
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The $Q$ region is to be considered empty at the beginning: the 'dirt' present in
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$P$ must be moved to $Q$ in order to reach the same distribution as close as
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possible. Namely, the following constraints must be satisfied:
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\begin{align*}
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&\text{1.} \hspace{20pt} f_{ij} \ge 0 \hspace{15pt}
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@ -474,7 +474,7 @@ second limits the amount of dirt moved by each position in $P$ in order to not
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exceed the available quantity; the third sets a limit to the dirt moved to each
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position in $Q$ in order to not exceed the required quantity and the last one
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forces to move the maximum amount of supplies possible: either all the dirt
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present in $P$ has be moved, or the $Q$ distibution is obtained.
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present in $P$ has been moved or the $Q$ distribution has been obtained.
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The total moved amount is the total flow. If the two distributions have the
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same amount of dirt, hence all the dirt present in $P$ is necessarily moved to
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$Q$ and the flow equals the total amount of available dirt.
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@ -487,9 +487,9 @@ $$
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{\sum_{i = 1}^m \sum_{j=1}^n f_{ij}}
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$$
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In this case, where the EMD has to be applied to two same-lenght histograms, the
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procedure simplifies a lot. By representing both histograms with two vectors $u$
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and $v$, the equation above boils down to [@ramdas17]:
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In this case, where the EMD is to be measured between two same-length
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histograms, the procedure simplifies a lot. By representing both histograms
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with two vectors $u$ and $v$, the equation above boils down to [@ramdas17]:
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$$
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\text{EMD} (u, v) = \sum_i |U_i - V_i|
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@ -497,7 +497,7 @@ $$
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where the sum runs over the entries of the vectors $U$ and $V$, which are the
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cumulative vectors of the histograms. In the code, the following equivalent
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recursive routine was implemented.
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iterative routine was implemented.
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$$
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\text{EMD} (u, v) = \sum_i |\text{EMD}_i| \with
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@ -526,35 +526,202 @@ In fact:
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\end{align*}
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This simple formula enabled comparisons to be made between a great number of
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results.
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results.
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In order to make the code more flexible, the data were normalized before
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computing the EMD: in doing so, it is possible to compare even samples with a
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different number of points.
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## Results comparison {#sec:conv_results}
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As can be seen, increasing the value of $\sigma$ implies a stronger smoothing of
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the curve. The FFT deconvolution process seems not to be affected by $\sigma$
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amplitude changes: it always gives the same outcome, which is exactly the
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original signal. In fact, the FFT is the analitical result of the deconvolution.
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In the real world, it is unpratical, since signals are inevitably blurred by
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noise.
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The same can't be said about the RL deconvolution, which, on the other hand,
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looks heavily influenced by the variance magnitude: the greater $\sigma$, the
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worse the deconvolved result. In fact, given the same number of steps, the
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deconvolved signal is always the same 'distance' far form the convolved one:
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if it very smooth, the deconvolved signal is very smooth too and if the
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convolved is less smooth, it is less smooth too.
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The original signal is shown below for convenience.
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### Noiseless results {#sec:noiseless}
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Along with the analysis of the results obtained varying the convolved Gaussian
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width $\sigma$, the possibility to add a Gaussian noise to the convolved
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histogram was also implemented to check weather the deconvolution is affected
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or not by this kind of interference. This approach is described in the next
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subsection, while the noiseless results are given in this one.
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The two methods were compared for three different values of $\sigma$:
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$$
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\sigma = 0.1 \, \Delta \theta \et
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\sigma = 0.5 \, \Delta \theta \et
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\sigma = \Delta \theta
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$$
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Since the RL method depends on the number $r$ of performed rounds, in order to
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find out how many of them it was sufficient or necessary to compute, the earth
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mover's distance between the deconvolved signal and the original one was
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measured for different $r$s for each of the three tested values of $\sigma$.
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To achieve this goal, a number of 1000 experiments (default and customizable
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value) were simulated and, for each of them, the original signal was convolved
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with the kernel, the appropriate $\sigma$ value set, and then deconvolved with
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the RL algorithm with a given $r$ and the EMD was measured. Then, an average of
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the so-obtained EMDs was computed together with the standard deviation. This
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procedure was repeated for a few tens of different $r$s till a flattening or a
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minimum of the curve became evident. Results in @fig:rounds-noiseless.
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The plots in @fig:rless-0.1 show the average (red) and standard deviation (grey)
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of the measured EMD for $\sigma = 0.1 \, \Delta \theta$. The number of
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iterations does not affect the quality of the outcome (those fluctuations are
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merely a fact of floating-points precision) and the best result is obtained
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for $r = 2$, meaning that the convergence of the RL algorithm is really fast and
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this is due to the fact that the histogram was modified pretty poorly. In
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@fig:rless-0.5, the curve starts to flatten at about 10 rounds, whereas in
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@fig:rless-1 a minimum occurs around \SI{5e3}{} rounds, meaning that, whit such a
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large kernel, the convergence is very slow, even if the best results are close
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to the one found for $\sigma = 0.5$.
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The following $r$s were chosen as the most fitted:
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\begin{align*}
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\sigma = 0.1 \, \Delta \theta &\thus n^{\text{best}} = 2 \\
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\sigma = 0.5 \, \Delta \theta &\thus n^{\text{best}} = 10 \\
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\sigma = 1 \, \Delta \theta &\thus n^{\text{best}} = \SI{5e3}{}
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||||
\end{align*}
|
||||
|
||||
Note the difference between @fig:rless-0.1 and the plots resulting from $\sigma =
|
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0.5 \, \Delta \theta$ and $\sigma = \, \Delta \theta$ as regards the order of
|
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magnitude: the RL deconvolution is heavily influenced by the variance magnitude:
|
||||
the greater $\sigma$, the worse the deconvolved result.
|
||||
On the other hand, the FFT deconvolution procedure is not affected by $\sigma$
|
||||
amplitude changes: it always gives the same outcome, which would be exactly the
|
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original signal, if the floating point precision would not affect the result. In
|
||||
fact, the FFT is the analytical result of the deconvolution.
|
||||
|
||||
<div id="fig:rounds-noiseless">
|
||||
{#fig:rless-0.1}
|
||||
|
||||
{#fig:rless-0.5}
|
||||
|
||||
{#fig:rless-1}
|
||||
|
||||
EMD as a function of RL rounds for different kernel $\sigma$ values. The
|
||||
average is shown in red and the standard deviation in grey. Noiseless results.
|
||||
</div>
|
||||
|
||||
<div id="fig:emd-noiseless">
|
||||
{#fig:eless-0.1}
|
||||
|
||||
{#fig:eless-0.5}
|
||||
|
||||
{#fig:eless-1}
|
||||
|
||||
EMD distributions for different kernel $\sigma$ values. The plots on the left
|
||||
show the results for the FFT deconvolution, the central column the results for
|
||||
the RL deconvolution and the third one shows the EMD for the convolved signal.
|
||||
Noiseless results.
|
||||
</div>
|
||||
|
||||
For this reason, the EMD obtained with the FFT can be used as a reference point
|
||||
against which to compare the EMDs measured with RL.
|
||||
As described above, for a given $r$, a thousands of experiments were simulated:
|
||||
for each of this simulations, an EMD was computed. Besides computing their
|
||||
average and standard deviations, those values were used to build histograms
|
||||
showing the EMD distribution.
|
||||
Once the most fitted numbers of rounds $r^{\text{best}}$ were found, their
|
||||
histograms were compared to the histograms of the FFT results, started from the
|
||||
same convolved signals, and the EDM of the convolved signals themselves, in
|
||||
order to check if an improvement was truly achieved. Results are shown in
|
||||
@fig:emd-noiseless.
|
||||
|
||||
As expected, the FFT results are always of the same order of magnitude,
|
||||
\SI{1e-15}{}, independently from the kernel width, whereas the RL deconvolution
|
||||
results change a lot, ranging from \SI{1e-16}{} for $\sigma = 0.1 \, \Delta
|
||||
\theta$ to \SI{1e-4}{} for $\sigma = \Delta \theta$.
|
||||
The first result was quite unexpected: for very low values of $\sigma$, the RL
|
||||
routine gives better results with respect to the FFT. This is because of the
|
||||
math which lies beneath the two methods: apparently, the RL technique is less
|
||||
subject to round-off errors.
|
||||
Then, as regards the comparison with the convolved signal, which shows always
|
||||
an EMD of \SI{1e-2}{}, both RL and FFT always return results closer to the
|
||||
original signal, meaning that the deconvolution is working.
|
||||
|
||||
|
||||
### Noisy results
|
||||
|
||||
In order to observe the effect of the Gaussian noise upon these two
|
||||
deconvolution methods, a certain value of $\sigma$ ($\sigma = 0.8 \, \Delta
|
||||
\theta$) was arbitrary chosen. The noise was then applied to the convolved
|
||||
histogram as follow.
|
||||
|
||||
{#fig:noisy}
|
||||
|
||||
<div id="fig:rounds-noise">
|
||||
{#fig:rnoise-0.005}
|
||||
|
||||
{#fig:rnoise-0.01}
|
||||
|
||||
{#fig:rnoise-0.05}
|
||||
|
||||
EMD as a function of RL rounds for different noise $\sigma_N$ values with the
|
||||
kernel $\sigma = 0.8 \Delta \theta$. The average is shown in red and the
|
||||
standard deviation in grey. Noisy results.
|
||||
</div>
|
||||
|
||||
<div id="fig:emd-noisy">
|
||||
{#fig:enoise-0.005}
|
||||
|
||||
{#fig:enoise-0.01}
|
||||
|
||||
{#fig:enoise-0.05}
|
||||
|
||||
EMD distributions for different noise $\sigma_N$ values. The plots on the left
|
||||
show the results for the FFT deconvolution, the central column the results for
|
||||
the RL deconvolution and the third one shows the EMD for the convolved signal.
|
||||
Noisy results.
|
||||
</div>
|
||||
|
||||
For each bin, once the convolved histogram was computed, a value $v_N$ was
|
||||
randomly sampled from a Gaussian distribution with standard deviation
|
||||
$\sigma_N$, and the value $v_n \cdot b$ was added to the bin itself, where $b$
|
||||
is the content of the bin. An example with $\sigma_N = 0.05$ of the new
|
||||
histogram is shown in @fig:noisy.
|
||||
The following three values of $\sigma_N$ were tested:
|
||||
$$
|
||||
\sigma_N = 0.005 \et
|
||||
\sigma_N = 0.01 \et
|
||||
\sigma_N = 0.05
|
||||
$$
|
||||
|
||||
The same procedure followed in @sec:noiseless was then repeated for noisy
|
||||
signals. Hence, in @fig:rounds-noise the EMD as a function of the RL rounds is
|
||||
shown, this time varying $\sigma_N$ and keeping $\sigma = 0.8 \, \Delta \theta$
|
||||
constant.
|
||||
In @fig:rnoise-0.005, the flattening is achieved around $r = 20$ and in
|
||||
@fig:rnoise-0.01 it is sufficient $\sim r = 15$. When the noise becomes too
|
||||
high, on the other hand, the more $r$ increases, the more the algorithm becomes
|
||||
inefficient, requiring a very small number of iterations.
|
||||
Hence, the most fitting values were chosen as:
|
||||
\begin{align*}
|
||||
\sigma_N = 0.005 &\thus r^{\text{best}} = 20 \\
|
||||
\sigma_N = 0.01 &\thus r^{\text{best}} = 15 \\
|
||||
\sigma_N = 0.05 &\thus r^{\text{best}} = 1
|
||||
\end{align*}
|
||||
|
||||
As regards the order of magnitude, as expected, the more $\sigma_n$ increases,
|
||||
the more the EMD rises, ranging from $\sim$ \SI{2e-4}{} in @fig:rnoise-0.005
|
||||
to $\sim$ \SI{1.5e-3}{} in @fig:rnoise-0.05.
|
||||
Since the FFT is no more able to return the original signal as close as before,
|
||||
it is no more assumed to be a reference point. In fact, as shown in @fig:noisy,
|
||||
the FFT algorithm, when dealing with noisy signals whose noise shape is
|
||||
unknown, is as efficient as the RL, since they share the same order of magnitude
|
||||
regarding the EMD.
|
||||
An increasing noise entails the shift of the convolved histogram to slightly
|
||||
greater values, still remaining in the same order of magnitude of the noiseless
|
||||
signals. However, the deconvolution still works, being the EMD of the convolved
|
||||
histogram an order of magnitude greater than the worst obtained with both
|
||||
methods.
|
||||
|
||||
In conclusion, when dealing with smeared signals with a known point spread
|
||||
function, the FFT proved to be the best deconvolution method, restoring
|
||||
perfectly the original signal to the extent permitted by the floating point
|
||||
precision. When the point spread function is particularly small, this precision
|
||||
problem is partially solved by the RL deconvolution process, which employs a
|
||||
more stable math.
|
||||
However, in the real world signals are inevitably blurred by unknown-shpaed
|
||||
noise. When the kernel is not known a-priori, either of them turns out to be as
|
||||
good as the FFT in the aforementioned situation: only a poor approximation of
|
||||
the original signal can be derived.
|
||||
|
||||
It was also implemented the possibility to add a Poisson noise to the
|
||||
convolved histogram to check weather the deconvolution is affected or not by
|
||||
this kind of interference. It was took as an example the case with $\sigma =
|
||||
\Delta \theta$. In @fig:poisson the results are shown for both methods when a
|
||||
Poisson noise with mean $\mu = 50$ is employed.
|
||||
In both cases, the addition of the noise seems to partially affect the
|
||||
deconvolution. When the FFT method is applied, it adds little spikes nearly
|
||||
everywhere on the curve and it is particularly evident on the edges, where the
|
||||
expected data are very small. On the other hand, the Richardson-Lucy routine is
|
||||
less affected by this further complication.
|
||||
|
||||
|
||||
@ -239,7 +239,7 @@ $$ {#eq:perc}
|
||||
|
||||
The training was performed as follow. Initial values were set as $w = (0,0)$ and
|
||||
$b = 0$. From these, the perceptron starts to improve their estimations. The
|
||||
sample was passed point by point into a reiterative procedure a grand total of
|
||||
sample was passed point by point into a iterative procedure a grand total of
|
||||
$N_c$ calls: each time, the projection $w \cdot x$ of the point was computed
|
||||
and then the variable $\Delta$ was defined as:
|
||||
|
||||
|
||||
Loading…
Reference in New Issue
Block a user