commit dcabfda0a4c4fb548221e8061b759a7d7f685a6c parent d27e2526a41b97ed9625081e935e1d99b776bf6c Author: miksa234 <milutin@popovic.xyz> Date: Sun, 16 Jan 2022 14:01:13 +0100 done some exercises Diffstat:
18 files changed, 714 insertions(+), 0 deletions(-)
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The Fourier coefficients are +\begin{align} + a_n &= 2\int_{-\frac{1}{2}}^{\frac{1}{2}} f(t) \cos(2\pi n t)\ dt = 0 + \;\;\;\;\; \text{(odd: g(-t) = -g(t))},\\ + \nonumber\\ + b_n &= 2\int_{-\frac{1}{2}}^{\frac{1}{2}} f(t) \sin(2\pi n t)\ dt = \\ + &= 2 \left(-\frac{1}{2\pi n} \cos(2\pi n + t)\bigg|_{-\frac{1}{2}}^{\frac{1}{2}} + \int_{\frac{1}{2}}^{\frac{1}{2}} \frac{1}{2 \pi n}\cos(2\pi n t)\ dt + \right) =\\ + &= -\frac{1}{\pi n}\left( -\cos(\pi n) + \frac{1}{\pi n }\sin(\pi + n)\right) = + \frac{\sin(\pi n) - \pi n \cos(\pi n)}{(\pi n)^2}. +\end{align} +Thereby the Fourier series of $f(t) = t$ is +\begin{align} + f(t) = \sum_{n=1}^\infty \left(\frac{\sin(\pi n) - \pi n \cos(\pi n)}{(\pi + n)^2}\right) \sin(2\pi n t) = t +\end{align} +\subsection{Truncation Error} +The truncation error of the trigonometric polynomial $(Sf_N)$ of degree $N$ is +\begin{align} + \sum_{|k| > N} |\hat{f}(k)|^2 = \lVert f - S_N\rVert_2^2 = + \int_{-\frac{1}{2}}^{\frac{1}{2}} |E_N(t)|^2 dt. +\end{align} +Computations for $N = 3, 9$ were done in python with a integration error of +around $10^{-15}$, resulting in the overall truncation errors of +\begin{align} + \sum_{|k| > 3} |\hat{f}(k)|^2 = 0.0053,\\ + \sum_{|k| > 9} |\hat{f}(k)|^2 = 0.0143. +\end{align} +To achieve $\lVert E_N\rVert^2_2 < 0.1 \lVert f \rVert^2_2$, the number of +coefficients needed are about $61$. This was done using a while loop and +evaluating $\lVert E_N\rVert^2_2$ for $N$ until the above condition is met. + +\subsection{Orthonormal Bases} +Here we will go through the most important properties of orthonormal bases. +So let $\{b_n\}_{n\in \mathbb{N}}$ be an ONB of a vector space $\mathcal{H}$, +then for every $x\in \mathcal{H}$ we may write +\begin{align} + x = \sum_{b_n} \langle b_n, x\rangle b_n, +\end{align} +and +\begin{align} + \lVert x \rVert^2 = \sum_{b_n} |\langle b_n, x\rangle|^2. +\end{align} +For any $x, y \in \mathcal{H}$ we can write the scalar product as +\begin{align} + \langle x, y\rangle = \sum_{b_n} \langle b_n, x\rangle \langle b_n, + y\rangle, +\end{align} +Furthermore there exists a linear projection $\Phi\ : \mathcal{H} +\rightarrow l^2(\{b_n\}_n)$ such that +\begin{align} + \langle \Phi(x), \Phi(y)\rangle = \langle x, y \rangle\;\;\; \forall x, y + \in \mathcal{H}. +\end{align} + +An example of an orthonormal basis, which spans $L^2([-\frac{p}{2}, +\frac{p}{2}])$ is $\mathcal{T}_p = \{e_n := \frac{e^{2\pi i +\frac{n}{p}x}}{\sqrt{p}}\}_{n\in\mathbb{Z}}$. The $e_n$ are orthonormal in $L^2$ which +can be easily seen by using the scalar product of $L^2$, so for $n, m \in +\mathbb{Z}$ +\begin{align} + \langle e_n, e_m\rangle_{L^2([-\frac{p}{2}, \frac{p}{2})} &= + \frac{1}{p}\int_{[-\frac{p}{2}, \frac{p}{2}]}e_n \cdot \bar{e}_m \ dx=\\ + &=\frac{1}{p}\int_{[-\frac{p}{2}, \frac{p}{2}]} e^{2\pi i \frac{(n-m)}{p} x} \ dx=\\ + &=\frac{\sin(\pi (n-m))}{\pi(n-m)} = + \begin{cases} + 0 \;\;\;\; n\neq m\\ + 1 \;\;\;\; n=m + \end{cases} +\end{align} +\subsection{Dirichlet Kernel} +The function +\begin{align} + D_t(x) := \sum_{\lVert k \rVert_\infty \leq t} e_k(x), \;\;\;\;\; x\in + \mathbb{R}^d +\end{align} +is called the Dirichlet Kernel. For $0 < t \in \mathbb{N}$ we have +\begin{align} + (S_tf)(x) = \int_{I^d} f(y) D_t(x-y) dy, +\end{align} +where $S_t$ represents the orthogonal projection onto the trigonometric +polynomials $\Pi_t$ of degree $t$, by +\begin{align} + &S_t:\ L^1(\mathbb{T}^d) \rightarrow \Pi_t \\ + &f \mapsto \sum_{\lVert k \rVert \leq t} \langle f, + e_k\rangle_{L^2(\mathbb{T}^d)} e_k \;\;\;\;\; k \in \mathbb{Z}^d +\end{align} +And furthermore the Dirichlet Kernel satisfies +\begin{align} + D_t(x) = \prod_{i=1}^d \frac{e_{t+1}(x_i) - e_{-t}(x_i)}{e_1(x_i) - 1} +\end{align} +We start off by applying the orthogonal projection into the trigonometric +polynomials $S_t$ onto a function $f \in L(\mathbb{T}^d)$ +\begin{align} + (S_tf) &= \sum_{\lvert k\rVert_\infty \leq t} \int_{I^d} f(y) e^{-2\pi i + \langle k, y\rangle}\ dy\ e^{2\pi i\langle k, x\rangle} =\\ + &= \int_{I^d}f(y) \sum_{\lvert k\rVert_\infty \leq t} e^{2\pi i \langle + k, (x- y)\rangle}\ dy =\\ + &= (f * D_t) (x) = \int_{I^d} f(y) D_t(x - y)\ dy. +\end{align} +The Dirichlet kernel can be simply rewritten +\begin{align} + \sum_{\lVert k \rVert_\infty \leq t} e^{2\pi i \langle k , x\rangle} &= + \prod_{i=1}^d \sum_{k_i = -t}^t e^{2\pi i k_i x_i} =\\ + &= \prod_{i=1}^d e^{-2\pi i t x_i} \sum_{k_i = 0}^{2t} e^{2\pi i k_i + x_i}=;\;\;\;\; \text{(trigonometric series)}\\ + &= \prod_{i=1}^d e^{-2\pi i t x_i} \frac{e^{2\pi i (2t +1}x_i - + 1}{e^{2\pi i x_i} - 1} =\\ + &= \prod_{i = 1} \frac{e_{t+1}(x_i) - e_{-t}(x_i)}{e_1(x_i) - 1}. +\end{align} +%\printbibliography +\end{document} diff --git a/appl_ana/sesh5/main.pdf b/appl_ana/sesh5/main.pdf Binary files differ. diff --git a/appl_ana/sesh5/main.tex b/appl_ana/sesh5/main.tex @@ -0,0 +1,145 @@ +\documentclass[a4paper]{article} + + +\usepackage[T1]{fontenc} +\usepackage[utf8]{inputenc} +\usepackage{mlmodern} + +%\usepackage{ngerman} % Sprachanpassung Deutsch + +\usepackage{graphicx} +\usepackage{geometry} +\geometry{a4paper, top=15mm} + +\usepackage{subcaption} +\usepackage[shortlabels]{enumitem} +\usepackage{amssymb} +\usepackage{amsthm} +\usepackage{mathtools} +\usepackage{braket} +\usepackage{bbm} +\usepackage{graphicx} +\usepackage{float} +\usepackage{yhmath} +\usepackage{tikz} +\usetikzlibrary{patterns,decorations.pathmorphing,positioning} +\usetikzlibrary{calc,decorations.markings} + +%\usepackage[backend=biber, sorting=none]{biblatex} +%\addbibresource{uni.bib} + +\usepackage[framemethod=TikZ]{mdframed} + +\tikzstyle{titlered} = + [draw=black, thick, fill=white,% + text=black, rectangle, + right, minimum height=.7cm] + + +\usepackage[colorlinks=true,naturalnames=true,plainpages=false,pdfpagelabels=true]{hyperref} +\usepackage[parfill]{parskip} +\usepackage{lipsum} + + +\usepackage{tcolorbox} +\tcbuselibrary{skins,breakable} + +\pagestyle{myheadings} + +\markright{Popović\hfill Applied Analysis\hfill} + + +\title{University of Vienna\\ Faculty of Mathematics\\ +\vspace{1cm}Applied Analysis Problems +} +\author{Milutin Popovic} + +\begin{document} +\maketitle +\tableofcontents + +\section{Sheet 5} +\subsection{Fourier Transform} +In this section we prove the linearity of the Fourier Transform $\mathcal{F}$ on +$L^1(\mathbb{R}^d)$. For $f, g \in L^1(\mathbb{R}^d)$ and $\lambda, \mu \in +\mathbb{R}$ the linearity condition for $\mathcal{F}$ is the following +\begin{align} + \mathcal{F}(\lambda f + \mu g) = \lambda \mathcal{F}(f) + \mu + \mathcal{F}(g). +\end{align} +We start by using the Fourier transform definition for $x, \xi \in \mathbb{R}^d$ +\begin{align} + \mathcal{F}(\lambda f + \mu g)(\xi) &= \int_{\mathbb{R}^d} (\lambda f(x)+ + \mu g(x)) e^{-2\pi i \langle x, \xi\rangle}\ dx =\\ + &= \int_{\mathbb{R}^d} \lambda f(x) e^{-2\pi i\langle x,\xi\rangle} + \mu + g(x) e^{-2\pi i\langle x,\xi\rangle}\ dx =\\ + &= \int_{\mathbb{R}^d} \lambda f(x) e^{-2\pi i\langle x,\xi\rangle}\ dx+ + \int_{\mathbb{R}^d} \mu + g(x) e^{-2\pi i\langle x,\xi\rangle}\ dx =\\ + &= \lambda \int_{\mathbb{R}^d} f(x) e^{-2\pi i\langle x,\xi\rangle}\ dx+ + \mu \int_{\mathbb{R}^d} + g(x) e^{-2\pi i\langle x,\xi\rangle}\ dx =\\ + &= \lambda \mathcal{F}(f)(\xi) + \mu \mathcal{F}(g)(\xi) +\end{align} +\subsection{Identities of the Fourier transform} +The following are three identities of the Fourier transform + +\begin{table}[h!] +\centering +\begin{tabular}{| l | c | c |} +\hline + & $g(x)$ & $\hat{g}(\xi)$ \\ \hline \hline +(1) & $f(x-x_0)$ & $e^{-2\pi ix_0 \xi} \hat{f}(\xi)$ \\ \hline +(2) & $e^{2\pi i \xi_0 x} f(x)$ & $f(\xi - \xi_0)$ \\ \hline +(3) & $f(ax)$ & $\frac{1}{a} \hat{f}(\frac{\xi}{a})$\\ \hline +\end{tabular} + \caption{Identities of the Fourier transform for $0 < a\in\mathbb{R}, + \xi_0, x \in \mathbb{R}$} +\end{table} +We start with (1) +\begin{align} + \widehat{f(x-x_0)} + &= \int_\mathbb{R} f(x-x_0) e^{-2\pi i x \xi}\ dx= + \;\;\;\;\;\; (y = x-x_0)\\ + &= \int_\mathbb{R} f(y) e^{-2\pi i (y+x_0) \xi}\ + dy=\\ + &= e^{-2\pi i x_0 \xi} \int_\mathbb{R}f(y)e^{-2\pi i y + \xi}\ dy=\\ + &= e^{-2\pi i x_0 \xi} \hat{f}(\xi). +\end{align} +For (2) we have +\begin{align} + \widehat{e^{2\pi i x \xi_0} f(x)} + &= \int_\mathbb{R} e^{2\pi i x \xi_0} f(x) e^{-2\pi i x \xi}\ dx =\\ + &= \int_\mathbb{R} f(x) e^{-2\pi i x (\xi -\xi_0)}\ dx=\\ + &= \hat{f}(\xi - \xi_0). +\end{align} +For (3) we have +\begin{align} + \widehat{f(ax)} + &= \int_\mathbb{R} f(ax) e^{-2\pi i \xi x}\ dx = \;\;\;\;\; (y=ax)\\ + &= \int_\mathbb{R} \frac{1}{a}f(y) e^{-2\pi i \frac{\xi}{a} y}\ dy=\\ + &= \frac{1}{a} \hat{f}(\frac{\xi}{a}). +\end{align} +\subsection{The Box-Function} +Consider the following Box-Function +\begin{align} + \Pi(x) := + \begin{cases} + 1\;\;\;\;\;\; -\frac{3}{2} < x < \frac{1}{2}\\ + 0\;\;\;\;\; \text{else} + \end{cases} +\end{align} +The Fourier transform of this function is +\begin{align} + \widehat{\Pi(x)} + &= \int_\mathbb{R} \Pi(x) e^{-2\pi i x\xi}\ dx=\\ + &= \int_{-\frac{3}{2}}^{\frac{1}{2}} e^{-2\pi i x \xi}\ dx + =\frac{-1}{2\pi i \xi} e^{-2\pi i x\xi} + \bigg|_{-\frac{3}{2}}^{\frac{1}{2}}=\\ + &= \frac{1}{2\pi i \xi} \left(e^{3\pi i \xi} - e^{-\pi i \xi}\right)=\\ + &= \frac{e^{\pi i \xi}\sin(2\pi\xi)}{\pi \xi}. +\end{align} + +%\printbibliography +\end{document} diff --git a/appl_ana/sesh6/main.pdf b/appl_ana/sesh6/main.pdf Binary files differ. diff --git a/appl_ana/sesh6/main.tex b/appl_ana/sesh6/main.tex @@ -0,0 +1,185 @@ +\documentclass[a4paper]{article} + + +\usepackage[T1]{fontenc} +\usepackage[utf8]{inputenc} +\usepackage{mlmodern} + +%\usepackage{ngerman} % Sprachanpassung Deutsch + +\usepackage{graphicx} +\usepackage{geometry} +\geometry{a4paper, top=15mm} + +\usepackage{subcaption} +\usepackage[shortlabels]{enumitem} +\usepackage{amssymb} +\usepackage{amsthm} +\usepackage{mathtools} +\usepackage{braket} +\usepackage{bbm} +\usepackage{graphicx} +\usepackage{float} +\usepackage{yhmath} +\usepackage{tikz} +\usetikzlibrary{patterns,decorations.pathmorphing,positioning} +\usetikzlibrary{calc,decorations.markings} + +%\usepackage[backend=biber, sorting=none]{biblatex} +%\addbibresource{uni.bib} + +\usepackage[framemethod=TikZ]{mdframed} + +\tikzstyle{titlered} = + [draw=black, thick, fill=white,% + text=black, rectangle, + right, minimum height=.7cm] + + +\usepackage[colorlinks=true,naturalnames=true,plainpages=false,pdfpagelabels=true]{hyperref} +\usepackage[parfill]{parskip} +\usepackage{lipsum} + + +\usepackage{tcolorbox} +\tcbuselibrary{skins,breakable} + +\pagestyle{myheadings} + +\markright{Popović\hfill Applied Analysis\hfill} + + +\title{University of Vienna\\ Faculty of Mathematics\\ +\vspace{1cm}Applied Analysis Problems +} +\author{Milutin Popovic} + +\begin{document} +\maketitle +\tableofcontents + +\section{Sheet 6} +\subsection{Fourier Transform of the convolution} +Consider the function $f(x)$, which has a Fourier Transform $\hat{f}(\xi)$, +now let us compute the Fourier transform of +\begin{align} + g(x) = f(3x-1) \sin(x). +\end{align} +We know that the Fourier transform of the convolution is +\begin{align} + (\widehat{f(3x-1)*g(x)}) = \widehat{f(3x-1)} \cdot \hat{g}(\xi). +\end{align} +The Fourier transform of $f(3x-1)$ is simply done by substituting a new +variable +\begin{align} + \widehat{f(3x-1)} = \frac{1}{3}e^{2\pi i\frac{\xi}{3}} f(\frac{\xi}{3}). +\end{align} +The Fourier transform of $\sin(x)$ can be calculated when looking at the +Fourier transform of the Dirac-delta function +\begin{align} + \widehat{\delta(ax-b)} + &=\int_\mathbb{R} \delta(ax-b) e^{-2\pi i x \xi}\ dx + \;\;\;\;\;\;\; (y = ax-b)\\ + &=\int_\mathbb{R} \delta(y) e^{-2\pi i (y+b)\frac{\xi}{a}}\frac{dy}{a}\\ + &=\frac{1}{a} e^{-2\pi i \xi \frac{b}{a}}. +\end{align} +We may plug in $\sin(x)$ in the definition of the Fourier transformation and +observe where we can use the Dirac-delta to to the inverse Fourier transform +\begin{align} + \widehat{\sin(x)} + &=\int_\mathbb{R} \sin(x)e^{-2\pi i x\xi}\ dx=\\ + &=\frac{1}{2i}\int_\mathbb{R} (e^{ix} - e^{-ix})e^{-2\pi i \xi x}\ dx\\ + &=\frac{1}{2i}\left( + \int_\mathbb{R} e^{ix} e^{-2\pi i \xi x}\ dx+ + \int_\mathbb{R} e^{-ix} e^{-2\pi i \xi x}\ dx + \right). +\end{align} +Here we may use the above formula for the Fourier transform of the Dirac +delta. We choose $a=1$, $b= \pm \frac{1}{2\pi}$ and do some $y=-x$ +substitutions and thereby get the following result +\begin{align} + \widehat{\sin(x)} = \frac{1}{2i} \left( + \delta(\xi - \frac{1}{2\pi}) + -\delta(\xi + \frac{1}{2\pi}) + \right) +\end{align} +The whole result is thereby +\begin{align} + \widehat{f(3x-1)} \cdot \widehat{sin(x)} + =& \frac{1}{6i} \left( + e^{2\pi + i(\frac{\xi}{3}-\frac{1}{2\pi})}\hat{f}(\frac{\xi}{3}-\frac{1}{2\pi})- + e^{2\pi i(\frac{\xi}{3}+\frac{1}{2\pi})}\hat{f}(\frac{\xi}{3}+\frac{1}{2\pi}) + \right) +\end{align} +\subsection{More Fourier Transforms} +Consider the function +\begin{align} + f(x) = e^{-|x|} +\end{align} +The Fourier transform of this function is +\begin{align} + \hat{f}(\xi) + &=\int_\mathbb{R} e^{-|x| e^{-2\pi i x \xi}}\ dx\\ + &= \int_{-\infty}^0 e^x e^{-2\pi i x \xi}\ dx + + \int_0^\infty e^{-x} e^{-2\pi i x \xi}\ dx=\\ + &= \frac{1}{1-2\pi i \xi} e^{(1-2\pi i \xi) x}\bigg|_{-\infty}^0+ + \frac{-1}{1+2\pi i \xi} e^{-(1+2\pi i \xi) x}\bigg|_{-\infty}^0 = \\ + &= \frac{1}{1-2\pi i \xi} + \frac{1}{1 + 2\pi i \xi} =\\ + &= \frac{2}{1+(2\pi \xi)^2}. +\end{align} +Let us use this result to solve the following integral +\begin{align} + \int_\mathbb{R} \frac{\cos(a\xi)}{(2\pi \xi)^2 + 1}\ d\xi &= + \frac{1}{2}\int_\mathbb{R} \hat{f}(\xi) \text{Re}(e^{ia\xi})\ dx=\\ + &= \frac{1}{2}\text{Re}\left( + \int_\mathbb{R}\hat{f}(\xi)e^{ia\xi}\ d\xi + \right)=\\ + &= \frac{1}{2}\text{Re}\left( + \int_\mathbb{R} \hat{f}(\xi) e^{2\pi i \frac{a}{2\pi}\xi}\ d\xi + \right)=\\ + &= \frac{1}{2}\text{Re}\left(f(\frac{a}{2\pi})\right)=\\ + &= \frac{1}{2} e^{-\frac{|a|}{2\pi}} +\end{align} +\subsection{Finite discrete Fourier transform} +Consider $s\in \mathbb{C}^N$ with entries +\begin{align} + s[n] = \sin\left(2\pi\xi_0\frac{n}{N}\right), +\end{align} +for same $0 < \xi_0 < N$. The finite discrete Fourier transform of $s$ is +\begin{align} + \hat{s}[k] &= \frac{1}{N} \sum_{n=0}^{N-1} \sin\left(2\pi\xi_0\frac{n}{N}\right) + e^{-2\pi i \frac{k}{N} n} =\\ + &=\frac{1}{2iN}\left( + \sum_{n=0}^{N-1}e^{2\pi i \frac{n}{N}(\xi_0 -k)} + e^{-2\pi i + \frac{n}{N}(\xi_0 +k)} + \right). +\end{align} +If we consider $\xi_0 \in \mathbb{Z}$, we have +\begin{align} + \hat{s}[k] = + \begin{cases} + \frac{1}{2i}\;\;\;\;\;\; \xi_0 = k\\ + -\frac{1}{2i}\;\;\;\;\;\; \xi_0 = -k\\ + 0 \;\;\;\;\;\; \text{else} + \end{cases} +\end{align} +\subsection{Discrete Matrix Notation} +The convolution of two vectors $f, g \in \mathbb{C}^N$, can be expressed by a +circulate matrix applied to f +\begin{align} + (f * g) [n] = \sum_{k=0}^{N-1} f[k] g[n-k]. +\end{align} +Consider $g=s$, then the matrix takes the following values +\begin{align} + s[n-k] = s_{nk} = \sin\left(2\pi \xi_0 \frac{n-k}{N}\right). +\end{align} +The convolution with an impulse input $f=\delta_{0k}$, a vector that is $1$ +for $k=0$ and else 0 reads +\begin{align} + \sum_k s_{nk}f_k &= \sum_k s_{nk} \delta_{0k} =\\ + &= \sin\left(2\pi \xi_0 \frac{n}{N}\right). +\end{align} + +%\printbibliography +\end{document} diff --git a/appl_ana/sesh7/main.pdf b/appl_ana/sesh7/main.pdf Binary files differ. diff --git a/appl_ana/sesh7/main.tex b/appl_ana/sesh7/main.tex @@ -0,0 +1,191 @@ +\documentclass[a4paper]{article} + + +\usepackage[T1]{fontenc} +\usepackage[utf8]{inputenc} +\usepackage{mlmodern} + +%\usepackage{ngerman} % Sprachanpassung Deutsch + +\usepackage{graphicx} +\usepackage{geometry} +\geometry{a4paper, top=15mm} + +\usepackage{subcaption} +\usepackage[shortlabels]{enumitem} +\usepackage{amssymb} +\usepackage{amsthm} +\usepackage{mathtools} +\usepackage{braket} +\usepackage{bbm} +\usepackage{graphicx} +\usepackage{float} +\usepackage{yhmath} +\usepackage{tikz} +\usetikzlibrary{patterns,decorations.pathmorphing,positioning} +\usetikzlibrary{calc,decorations.markings} + +%\usepackage[backend=biber, sorting=none]{biblatex} +%\addbibresource{uni.bib} + +\usepackage[framemethod=TikZ]{mdframed} + +\tikzstyle{titlered} = + [draw=black, thick, fill=white,% + text=black, rectangle, + right, minimum height=.7cm] + + +\usepackage[colorlinks=true,naturalnames=true,plainpages=false,pdfpagelabels=true]{hyperref} +\usepackage[parfill]{parskip} +\usepackage{lipsum} + +\usepackage[OT2,T1]{fontenc} +\DeclareSymbolFont{cyrletters}{OT2}{wncyr}{m}{n} +\DeclareMathSymbol{\Sha}{\mathalpha}{cyrletters}{"58} + +\usepackage{tcolorbox} +\tcbuselibrary{skins,breakable} + +\pagestyle{myheadings} + +\markright{Popović\hfill Applied Analysis\hfill} + + +\title{University of Vienna\\ Faculty of Mathematics\\ +\vspace{1cm}Applied Analysis Problems +} +\author{Milutin Popovic} + +\begin{document} +\maketitle +\tableofcontents + +\section{Sheet 7} +\subsection{Dirac Comb} +The Dirac train or Dirac comb on defined in the following way +\begin{align} + \Sha_m[n] = + \begin{cases} + 1\;\;\;\;\;\; n = 0, \pm m, \pm 2m,\dots\\ + 0\;\;\;\;\;\; \text{else} + \end{cases} +\end{align} +The discrete Fourier transform of the Dirac comb in $\mathbb{C}^N$ is +\begin{align} + \widehat{\Sha_m[n]} + &=\frac{1}{N}\sum_{n=0}^{N-1} \Sha_m[n] e^{-2\pi i \frac{k}{N}n}=\\ + &=\frac{1}{N}\sum_{n=0}^{N-1} + \left( + m\sum_{l\in\mathbb{Z}} \delta(n-lm) + \right) + e^{-2\pi i \frac{k}{N}n}=\\ + &=\frac{m}{N}\sum_{l\in\mathbb{Z}} e^{-2\pi i \frac{k}{N}lm}= + \;\;\;\;\;\;\;\;\;\;\;\;\; (m = \frac{N}{m'})\\ + &= \frac{1}{m'} \sum_{l\in\mathbb{Z}} e^{-2\pi i \frac{k}{m'}l} =\\ + &= \frac{1}{m}\Sha_{\frac{N}{m}}[k] +\end{align} +\subsection{Schwartz Space} +The Schwartz space $\mathcal{S}(\mathbb{R}^d)$, for $d \in \mathbb{N}$ is +defined as +\begin{align} + &\mathcal{S} := +\bigg\{ + f\in\mathcal{C}^\infty(\mathbb{R}^d): + \forall\alpha,\beta\in\mathbb{N}^d\;\; \lVert f \rVert_{\alpha,\beta} + < \infty +\bigg\},\\ +&\lVert f \rVert_{\alpha, \beta} := +\sup_{x\in\mathbb{R}^d}\left|x^\alpha (D^\beta f) (x) \right|. +\end{align} +Our aim is to show that if $f\in\mathcal{S}(\mathbb{R})$ then $\hat{f} \in +\mathcal{S}(\mathbb{R})$. The condition is obviously +\begin{align} + &\lVert \hat{f} \rVert_{\alpha, \beta} = + \sup_{\xi\in\mathbb{R}}\left|\xi^\alpha (D^\beta \hat{f}) (\xi) + \right|<\infty, +\end{align} +for all $\alpha, \beta \in \mathbb{N}$. +We can start with what we know about the Fourier transform +\begin{align} + \xi^\alpha \hat{f}(\xi) &= \mathcal{F}\left(\frac{1}{(2\pi + i)^\alpha}(D^{\alpha}f)(x)\right)\\ + D^{\beta}\hat{f}(\xi) &= \mathcal{F}\left( + (-2\pi i x)^\beta f(x) + \right). +\end{align} +Combining the two relations above we get +\begin{align} + \xi^\alpha (D^\beta \hat{f})(\xi) = + \mathcal{F}\left(\frac{(-2\pi i x)^\beta}{(2\pi + i)^\alpha}x^\beta(D^{\alpha}f)(x)\right)=: \mathcal{F}(g(x))\\ +\end{align} +If we call this function $g$, then $g\in\mathcal{S}(\mathbb{R})$ and +$g\in L^1(\mathbb{R})$. Applying the Riemann-Lebesgue Lemma we get +\begin{align} + \hat{g}(\xi) = \int_\mathbb{R} g(x) e^{-2\pi i x \xi}\ dx \longrightarrow 0 + \;\;\; \text{as $|\xi| \rightarrow \infty$ } +\end{align} +Thereby $\hat{g} \in \mathcal{S}(\mathbb{R})$ and thus $\hat{f} \in +\mathcal{S}(\mathbb{R})$. +\subsection{Tempered Distributions} +Tempered distributions are the elements of +\begin{align} + \mathcal{S}'(\mathbb{R}^d) := + \bigg\{ + L: \mathcal{S}(\mathbb{R}^d) \rightarrow \mathbb{C} | \text{$L$ is + linear and continuous} + \bigg\}. +\end{align} +Consider $\xi$ as a tempered distribution, buy acting on $\varphi \in +\mathcal{S}(\mathbb{R})$ we have +\begin{align} + \xi(\phi) = \int_\xi \xi \varphi(\xi)\ d\xi. +\end{align} +The Fourier transform of $\xi$ is +\begin{align} + \hat{\xi}(\varphi) + &=\xi(\hat{\varphi}) + = \int_\mathbb{R} \xi \hat{\varphi}(\xi)\ d\xi=\\ + &= \int_\mathbb{R}\xi \int_\mathbb{R} \varphi(x) e^{-2\pi i \xi x}\ dx\ + d\xi=\\ + &=\int_\mathbb{R}\int_\mathbb{R} \xi e^{-2\pi i \xi x}\ d\xi \varphi(x)\ + dx=\\ + &=\int_\mathbb{R} \frac{i}{2\pi}\frac{d}{dx}\int_\mathbb{R}e^{-2\pi i \xi +x} \ d\xi \varphi(x)\ dx=\\ + &= \int_\mathbb{R}i\frac{1}{2\pi} \frac{d}{dx}\left(2\pi \delta(x)\right) + \varphi(x)\ dx=\\ + &= \int_\mathbb{R} i\delta'(x)\varphi(x)\ dx\\ + &= i \delta'(\varphi). +\end{align} +\subsection{Fourier transform of the Dirac Comb} +The general case of the Dirac Comb as a distribution is +\begin{align} + \Sha_T = \sum_{n \in \mathbb{Z}} \delta_{nT}. +\end{align} +The Fourier transform of the $\Sha_T$ distribution for $\varphi \in +\mathcal{S}(\mathbb{R})$ is +\begin{align} + \widehat{\Sha_T}(\varphi) + &= \sum_{n\in\mathbb{Z}} \hat{\delta}_{nT}(\varphi)\\ + &= \sum_{n\in\mathbb{Z}} \delta_{n\omega_0}(\varphi)\\ + &=\Sha_{\omega_0}(\varphi). +\end{align} +The Fourier transform, transforms the period of the combs. +\subsection{Shannon Sampling} +The Fourier transform of $1_{[-\frac{a}{2}, \frac{a}{2}]}(x)$ is +\begin{align} + \mathcal{F}\left(1_{[-\frac{a}{2}, \frac{a}{2}]}\right)(\xi) + &= \int_\mathbb{R} 1_{[-\frac{a}{2}, \frac{a}{2}]} e^{-2\pi i x \xi}\ + dx\\ + &= \int_{-\frac{a}{2}}^{\frac{a}{2}} e^{-2\pi i x\xi}\ dx\\ + &= \frac{-1}{2\pi i \xi} e^{-2\pi i x + \xi}\bigg|_{-\frac{a}{2}}^{\frac{a}{2}}\\ + &= \frac{1}{\pi \xi} \frac{1}{2i}\left( + e^{pi i a \xi} - e^{-\pi i a \xi} + \right)\\ + &= \frac{\sin(\pi \xi a)}{\pi \xi} +\end{align} + +%\printbibliography +\end{document}