# Exercise 6

## Generating points according to Fraunhofer diffraction

The diffraction of a plane wave thorough a round slit must be simulated by
generating $N =$ 50'000 points according to the intensity distribution
$I(\theta)$ on a screen at a great distance $L$ from the slit itself:

$$
  I(\theta) = \frac{E^2}{2} \left( \frac{2 \pi a^2 \cos{\theta}}{L}
  \frac{J_1(x)}{x} \right)^2 \with x = k a \sin{\theta}
$$

where:

- $E$ is the electric field amplitude, default set $E = \SI{1e4}{V/m}$;
- $a$ is the radius of the slit aperture, default set $a = \SI{0.01}{m}$;
- $\theta$ is the angle specified in @fig:fenditure;
- $J_1$ is the Bessel function of first order;
- $k$ is the wavenumber, default set $k = \SI{1e-4}{m^{-1}}$;
- $L$ default set $L = \SI{1}{m}$.

\begin{figure}
\hypertarget{fig:slit}{%
\centering
\begin{tikzpicture}
  \definecolor{cyclamen}{RGB}{146, 24, 43}
  % Walls
  \draw [thick] (-1,3)  -- (1,3)  -- (1,0.3)  -- (1.2,0.3)  -- (1.2,3)
                       -- (9,3);
  \draw [thick] (-1,-3) -- (1,-3) -- (1,-0.3) -- (1.2,-0.3) -- (1.2,-3)
                       -- (9,-3);
  \draw [thick] (10,3) -- (9.8,3) -- (9.8,-3) -- (10,-3);
  % Lines
  \draw [thick, gray] (0.7,0.3)  -- (0.5,0.3);
  \draw [thick, gray] (0.7,-0.3) -- (0.5,-0.3);
  \draw [thick, gray] (0.6,0.3)  -- (0.6,-0.3);
  \draw [thick, gray] (1.2,0)    -- (9.8,0);
  \draw [thick, gray] (1.2,-0.1)  -- (1.2,0.1);
  \draw [thick, gray] (9.8,-0.1)  -- (9.8,0.1);
  \draw [thick, cyclamen] (1.2,0)  -- (9.8,-2);
  \draw [thick, cyclamen] (7,0) to [out=-90, in=50] (6.6,-1.23);
  % Nodes
  \node at (0,0) {$2a$};
  \node at (5.5,0.4) {$L$};
  \node [cyclamen] at (5.5,-0.4) {$\theta$};
  \node [rotate=-90] at (10.2,0) {screen};
\end{tikzpicture}
\caption{Fraunhofer diffraction.}
}
\end{figure}

Once again, the *try and catch* method described in @sec:3 was implemented and
the same procedure about the generation of $\theta$ was employed. This time,
though, $\theta$ must be evenly distributed on half sphere:

\begin{align*}
  \frac{d^2 P}{d\omega^2} = const = \frac{1}{2 \pi}
  &\thus d^2 P = \frac{1}{2 \pi} d\omega^2 =
  \frac{1}{2 \pi} d\phi \sin{\theta} d\theta \\
  &\thus \frac{dP}{d\theta} = \int_0^{2 \pi} d\phi \frac{1}{2 \pi} \sin{\theta}
  = \frac{1}{2 \pi} \sin{\theta} \, 2 \pi = \sin{\theta}
\end{align*}

\begin{align*}
  \theta = \theta (x) &\thus
  \frac{dP}{d\theta} = \frac{dP}{dx} \cdot \left| \frac{dx}{d\theta} \right|
  = \left. \frac{dP}{dx} \middle/ \, \left| \frac{d\theta}{dx} \right| \right. 
  \\
  &\thus \sin{\theta} = \left. 1 \middle/ \, \left|
  \frac{d\theta}{dx} \right| \right. 
\end{align*}

If $\theta$ is chosen to grew together with $x$, then the absolute value can be
omitted:

\begin{align*}
  \frac{d\theta}{dx} = \frac{1}{\sin{\theta}}
  &\thus d\theta \sin(\theta) = dx
  \\
  &\thus - \cos (\theta') |_{0}^{\theta} = x(\theta) - x(0) = x - 0 = x
  \\
  &\thus - \cos(\theta) + 1 =x
  \\
  &\thus \theta = \text{acos} (1 -x)
\end{align*}

The sample was binned and stored in a histogram with a customizable number $n$
of bins default set $n = 150$. In @fig:original an example is shown.

![Example of sorted points according to
  $I(\theta)$.](images/6_original.pdf){#fig:original}


## Gaussian noise convolution 

The sample must then be smeared with a Gaussian noise with the aim to recover
the original sample afterwards, implementing a deconvolution routine.  
For this purpose, a 'kernel' histogram with a odd number $m$ of bins and the
same bin width of the previous one, but a smaller number of them ($m < n$), was
filled with $m$ points according to a Gaussian distribution with mean $\mu$,
corresponding to the central bin, and variance $\sigma$.  
Then, the original histogram was convolved with the kernel in order to obtain
the smeared signal. The result is shown in @fig:convolved.

![Same sample of @fig:original convolved with the
  kernel.](images/6_convolved.pdf){#fig:convolved}

The convolution was implemented as follow. Consider the definition of
convolution of two functions $f(x)$ and $g(x)$:

$$
  f*g (x) = \int \limits_{- \infty}^{+ \infty} dy f(y) g(x - y)
$$

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
relative position of the two histograms. Namely, if $c_i$ is the $i^{\text{th}}$
bin of the convoluted histogram:

$$
  c_i = \sum_j k_j s_{i - j}
$$

where $j$ runs over the bins of the kernel.  
For a better understanding, see @fig:dot_conv. Thus, the third histogram was
obtained with $n + m - 1$ bins, a number greater than the initial one.

\begin{figure}
\hypertarget{fig:dot_conv}{%
\centering
\begin{tikzpicture}
  \definecolor{cyclamen}{RGB}{146, 24, 43}
  % original histogram
  \draw [thick, cyclamen, fill=cyclamen!05!white] (0.0,0) rectangle (0.5,2.5);
  \draw [thick, cyclamen, fill=cyclamen!05!white] (0.5,0) rectangle (1.0,2.8);
  \draw [thick, cyclamen, fill=cyclamen!25!white] (1.0,0) rectangle (1.5,2.3);
  \draw [thick, cyclamen, fill=cyclamen!25!white] (1.5,0) rectangle (2.0,1.8);
  \draw [thick, cyclamen, fill=cyclamen!25!white] (2.0,0) rectangle (2.5,1.4);
  \draw [thick, cyclamen, fill=cyclamen!25!white] (2.5,0) rectangle (3.0,1.0);
  \draw [thick, cyclamen, fill=cyclamen!25!white] (3.0,0) rectangle (3.5,1.0);
  \draw [thick, cyclamen, fill=cyclamen!05!white] (3.5,0) rectangle (4.0,0.6);
  \draw [thick, cyclamen, fill=cyclamen!05!white] (4.0,0) rectangle (4.5,0.4);
  \draw [thick, cyclamen, fill=cyclamen!05!white] (4.5,0) rectangle (5.0,0.2);
  \draw [thick, cyclamen, fill=cyclamen!05!white] (5.0,0) rectangle (5.5,0.2);
  \draw [thick, cyclamen] (6.0,0) -- (6.0,0.2);
  \draw [thick, cyclamen] (6.5,0) -- (6.5,0.2);
  \draw [thick, <->] (0,3.3) -- (0,0) -- (7,0);
  % kernel histogram
  \draw [thick, cyclamen, fill=cyclamen!25!white] (1.0,-1) rectangle (1.5,-1.2);
  \draw [thick, cyclamen, fill=cyclamen!25!white] (1.5,-1) rectangle (2.0,-1.6);
  \draw [thick, cyclamen, fill=cyclamen!25!white] (2.0,-1) rectangle (2.5,-1.8);
  \draw [thick, cyclamen, fill=cyclamen!25!white] (2.5,-1) rectangle (3.0,-1.6);
  \draw [thick, cyclamen, fill=cyclamen!25!white] (3.0,-1) rectangle (3.5,-1.2);
  \draw [thick, <->] (1,-2) -- (1,-1) -- (4,-1);
  % arrows
  \draw [thick, cyclamen, <->] (1.25,-0.2) -- (1.25,-0.8);
  \draw [thick, cyclamen, <->] (1.75,-0.2) -- (1.75,-0.8);
  \draw [thick, cyclamen, <->] (2.25,-0.2) -- (2.25,-0.8);
  \draw [thick, cyclamen, <->] (2.75,-0.2) -- (2.75,-0.8);
  \draw [thick, cyclamen, <->] (3.25,-0.2) -- (3.25,-0.8);
  \draw [thick, cyclamen,  ->] (2.25,-2.0) -- (2.25,-4.2);
  % smeared histogram
  \begin{scope}[shift={(0,-1)}]
  \draw [thick, cyclamen, fill=cyclamen!05!white] (-1.0,-4.5) rectangle (-0.5,-4.3);
  \draw [thick, cyclamen, fill=cyclamen!05!white] (-0.5,-4.5) rectangle ( 0.0,-4.2);
  \draw [thick, cyclamen, fill=cyclamen!05!white] ( 0.0,-4.5) rectangle ( 0.5,-2.0);
  \draw [thick, cyclamen, fill=cyclamen!05!white] ( 0.5,-4.5) rectangle ( 1.0,-1.6);
  \draw [thick, cyclamen, fill=cyclamen!05!white] ( 1.0,-4.5) rectangle ( 1.5,-2.3);
  \draw [thick, cyclamen, fill=cyclamen!05!white] ( 1.5,-4.5) rectangle ( 2.0,-2.9);
  \draw [thick, cyclamen, fill=cyclamen!25!white] ( 2.0,-4.5) rectangle ( 2.5,-3.4);
  \draw [thick, cyclamen] (3.0,-4.5) -- (3.0,-4.3);
  \draw [thick, cyclamen] (3.5,-4.5) -- (3.5,-4.3);
  \draw [thick, cyclamen] (4.0,-4.5) -- (4.0,-4.3);
  \draw [thick, cyclamen] (4.5,-4.5) -- (4.5,-4.3);
  \draw [thick, cyclamen] (5.0,-4.5) -- (5.0,-4.3);
  \draw [thick, cyclamen] (5.5,-4.5) -- (5.5,-4.3);
  \draw [thick, cyclamen] (6.0,-4.5) -- (6.0,-4.3);
  \draw [thick, cyclamen] (6.5,-4.5) -- (6.5,-4.3);
  \draw [thick, cyclamen] (7.0,-4.5) -- (7.0,-4.3);
  \draw [thick, cyclamen] (7.5,-4.5) -- (7.5,-4.3);
  \draw [thick, <->] (-1,-2.5) -- (-1,-4.5) -- (8,-4.5);
  \end{scope}
  % nodes
  \node [above] at (2.25,-5.5) {$c_i$};
  \node [above] at (3.25,0)    {$s_i$};
  \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
         (above) and the kernel (center). The final result is the lower
         fledging histogram.}\label{fig:dot_conv}
}
\end{figure}

\textcolor{red}{Missing various $\sigma$ comparison.}

## Unfolding