analistica/notes/sections/6.md

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:slit;
• 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.}\label{fig:slit} } \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.

{#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. Some results in terms of various \sigma are shown in @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. As can be seen, 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}

Unfolding with FFT

Two different unfolding routines were implemented, one of which exploiting the 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]

where \hat{F}[\quad] stands for the Fourier transform of its argument.
Thus, the implementation of this tecnique lies in the computation of the Fourier trasform of the signal and the kernel, the product of their transforms and the anti-transformation of the result. Being the histogram a discrete set of data, the Discrete Fourier Transform (DFT) was emploied.

In order to accomplish this procedure, every histogram was transformed into a vector. The kernel vector was 0-padded in order to make its length the same as the one of the signal, making it feasable to implement the dot product between the vectors.

FFT are efficient algorithms for calculating the DFT. Namely, given a set of $n$ values {$z_i$}, each one is transformed into:

  x_j = \sum_{k=0}^{n-1} z_k \exp \left( - \frac{2 \pi i j k}{n} \right)

The evaluation of the DFT is a matrix-vector multiplication W \vec{z}. A general matrix-vector multiplication takes O(n^2) operations. FFT algorithms, instad, use a divide-and-conquer strategy to factorize the matrix into smaller sub-matrices. If n can be factorized into a product of integers n_1, $n_2 \ldots n_m$, then the DFT can be computed in $O(n \sum n_i) < O(n^2)$ operations, hence the name.
The inverse Fourier transform is thereby defined as:

  z_j = \frac{1}{n}
        \sum_{k=0}^{n-1} x_k \exp \left( \frac{2 \pi i j k}{n} \right)

In GSL, gsl_fft_complex_forward() and gsl_fft_complex_inverse() are functions which allow to compute the foreward and inverse transform, respectively.

In this special case, the sequence which must be transformed is made of real numbers, but the Fourier transform is not real: it is a complex sequence wich satisfies:

  z_k = z^*_{n-k}

Where z^* is the conjugate of z. A sequence with this symmetry is called 'half-complex'. This structure requires a particular storage layouts for the forward transform (from real to half-complex) and inverse transform (from half-complex to real). As a consequence, the routines are divided into two sets: gsl_fft_real and gsl_fft_halfcomplex. The symmetry of the half-complex sequence implies that only half of the complex numbers in the output need to be stored. This works for all lengths: when the length is even, the middle value where k = n/2 is real. Thus, only n real numbers are required to store the half-complex sequence (real and imaginary parts).
The ratio between the kernel and the sigmal transformations where therefore made between complex numbers. If the time-step of the DFT is \Delta, then the frequency-domain includes both positive and negative frequencies, ranging from -1 / (2 \Delta) to +1 / (2 \Delta). The positive frequencies are stored from the beginning of the array up to the middle, and the negative frequencies are stored backwards from the end of the array. \textcolor{red}{Useful?}

---

{width=7cm} {width=7cm}

{width=7cm} {width=7cm}

{width=7cm} {width=7cm}

{width=7cm} {width=7cm}

Signal convolved with kernel on the left and deconvolved signal on the right. Increasing values of \sigma from top to bottom. \Delta \theta is the bin width.

As can be seen from the plots on the left in @fig:convolved, increasig the value of \sigma implies a stronger smoothing of the curve, while the deconvolution process seems not to be affected by \sigma amplitude changes: it always gives the same outcome.

It was also implemented the possibility to add a Poisson noise to the distribution to check weather the deconvolution is affected or not by this kind of noise.