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1 \documentclass[a4paper]{article} 2 3 4 \usepackage[T1]{fontenc} 5 \usepackage[utf8]{inputenc} 6 \usepackage{mlmodern} 7 8 %\usepackage{ngerman} % Sprachanpassung Deutsch 9 10 \usepackage{graphicx} 11 \usepackage{geometry} 12 \geometry{a4paper, top=15mm} 13 14 \usepackage{subcaption} 15 \usepackage[shortlabels]{enumitem} 16 \usepackage{amssymb} 17 \usepackage{amsthm} 18 \usepackage{mathtools} 19 \usepackage{braket} 20 \usepackage{bbm} 21 \usepackage{graphicx} 22 \usepackage{float} 23 \usepackage{yhmath} 24 \usepackage{tikz} 25 \usetikzlibrary{patterns,decorations.pathmorphing,positioning} 26 \usetikzlibrary{calc,decorations.markings} 27 28 %\usepackage[backend=biber, sorting=none]{biblatex} 29 %\addbibresource{uni.bib} 30 31 \usepackage[framemethod=TikZ]{mdframed} 32 33 \tikzstyle{titlered} = 34 [draw=black, thick, fill=white,% 35 text=black, rectangle, 36 right, minimum height=.7cm] 37 38 39 \usepackage[colorlinks=true,naturalnames=true,plainpages=false,pdfpagelabels=true]{hyperref} 40 \usepackage[parfill]{parskip} 41 \usepackage{lipsum} 42 43 44 \usepackage{tcolorbox} 45 \tcbuselibrary{skins,breakable} 46 47 \pagestyle{myheadings} 48 49 \markright{Popović\hfill Applied Analysis\hfill} 50 51 52 \title{University of Vienna\\ Faculty of Mathematics\\ 53 \vspace{1cm}Applied Analysis Problems 54 } 55 \author{Milutin Popovic} 56 57 \begin{document} 58 \maketitle 59 \tableofcontents 60 61 \section{Sheet 6} 62 \subsection{Fourier Transform of the convolution} 63 Consider the function $f(x)$, which has a Fourier Transform $\hat{f}(\xi)$, 64 now let us compute the Fourier transform of 65 \begin{align} 66 h(x) = f(3x-1) \sin(x) . 67 \end{align} 68 We know that the Fourier transform of the convolution is (we use somewhat of 69 the inverse convolution theorem). 70 \begin{align} 71 \widehat{(f(3x-1)*g(x))} = \widehat{f(3x-1)} \cdot \hat{g}(\xi). 72 \end{align} 73 The Fourier transform of $f(3x-1)$ is simply done by substituting a new 74 variable 75 \begin{align} 76 \widehat{f(3x-1)} = \frac{1}{3}e^{2\pi i\frac{\xi}{3}}\ 77 f\left(\frac{\xi}{3}\right). 78 \end{align} 79 The Fourier transform of $\sin(x)$ can be calculated when looking at the 80 Fourier transform of the Dirac-delta function 81 \begin{align} 82 \widehat{\delta(ax-b)} 83 &=\int_\mathbb{R} \delta(ax-b) e^{-2\pi i x \xi}\ dx 84 \;\;\;\;\;\;\; (y = ax-b)\\ 85 &=\int_\mathbb{R} \delta(y) e^{-2\pi i (y+b)\frac{\xi}{a}}\frac{dy}{a}\\ 86 &=\frac{1}{a} e^{-2\pi i \xi \frac{b}{a}}. 87 \end{align} 88 We may plug in $\sin(x)$ in the definition of the Fourier transformation and 89 observe where we can use the Dirac-delta to to the inverse Fourier transform 90 \begin{align} 91 \widehat{\sin(x)} 92 &=\int_\mathbb{R} \sin(x)e^{-2\pi i x\xi}\ dx=\\ 93 &=\frac{1}{2i}\int_\mathbb{R} (e^{ix} - e^{-ix})e^{-2\pi i \xi x}\ dx\\ 94 &=\frac{1}{2i}\left( 95 \int_\mathbb{R} e^{ix} e^{-2\pi i \xi x}\ dx+ 96 \int_\mathbb{R} e^{-ix} e^{-2\pi i \xi x}\ dx 97 \right). 98 \end{align} 99 Here we may use the above formula for the Fourier transform of the Dirac 100 delta. We choose $a=1$, $b= \pm \frac{1}{2\pi}$ and do some $y=-x$ 101 substitutions and thereby get the following result 102 \begin{align} 103 \widehat{\sin(x)} = \frac{1}{2i} \left( 104 \delta(\xi - \frac{1}{2\pi}) 105 -\delta(\xi + \frac{1}{2\pi}) 106 \right) 107 \end{align} 108 The whole result is thereby 109 \begin{align} 110 \widehat{f(3x-1)} * \widehat{sin(x)} 111 =& \frac{1}{6i} \bigg( 112 e^{2\pi 113 i(\frac{\xi}{3}-\frac{1}{6\pi})}\hat{f}\big(\frac{\xi}{3}-\frac{1}{6\pi}\big)- 114 e^{2\pi 115 i(\frac{\xi}{3}+\frac{1}{6\pi})}\hat{f}\big(\frac{\xi}{3}+\frac{1}{6\pi}\big) 116 \bigg) 117 \end{align} 118 \subsection{More Fourier Transforms} 119 Consider the function 120 \begin{align} 121 f(x) = e^{-|x|} 122 \end{align} 123 The Fourier transform of this function is 124 \begin{align} 125 \hat{f}(\xi) 126 &=\int_\mathbb{R} e^{-|x| e^{-2\pi i x \xi}}\ dx\\ 127 &= \int_{-\infty}^0 e^x e^{-2\pi i x \xi}\ dx 128 + \int_0^\infty e^{-x} e^{-2\pi i x \xi}\ dx=\\ 129 &= \frac{1}{1-2\pi i \xi} e^{(1-2\pi i \xi) x}\bigg|_{-\infty}^0+ 130 \frac{-1}{1+2\pi i \xi} e^{-(1+2\pi i \xi) x}\bigg|_{-\infty}^0 = \\ 131 &= \frac{1}{1-2\pi i \xi} + \frac{1}{1 + 2\pi i \xi} =\\ 132 &= \frac{2}{1+(2\pi \xi)^2}. 133 \end{align} 134 Let us use this result to solve the following integral 135 \begin{align} 136 \int_\mathbb{R} \frac{\cos(a\xi)}{(2\pi \xi)^2 + 1}\ d\xi = 137 \frac{1}{2}\int_\mathbb{R} \hat{f}(\xi) \text{Re}(e^{ia\xi})\ dx,\\ 138 \end{align} 139 where we used the fact that $\text{Re}(e^{ia\xi}) = \cos(a\xi)$ and 140 $\hat{f}(\xi) = \frac{2}{1+(2\pi \xi)^2}$, thereby 141 \begin{align} 142 \frac{1}{2}\int_\mathbb{R} \hat{f}(\xi) \text{Re}(e^{ia\xi})\ dx 143 &= \frac{1}{2}\text{Re}\left( 144 \int_\mathbb{R}\hat{f}(\xi)e^{ia\xi}\ d\xi 145 \right)=\\ 146 &= \frac{1}{2}\text{Re}\left( 147 \int_\mathbb{R} \hat{f}(\xi) e^{2\pi i \frac{a}{2\pi}\xi}\ d\xi 148 \right)=\\ 149 &= \frac{1}{2}\text{Re}\left(f(\frac{a}{2\pi})\right)=\\ 150 &= \frac{1}{2} e^{-\frac{|a|}{2\pi}}. 151 \end{align} 152 \subsection{Finite discrete Fourier transform} 153 Consider $s\in \mathbb{C}^N$ with entries 154 \begin{align} 155 s[n] = \sin\left(2\pi\xi_0\frac{n}{N}\right), 156 \end{align} 157 for same $0 < \xi_0 < N$. The finite discrete Fourier transform of $s$ is 158 \begin{align} 159 \hat{s}[k] &= \frac{1}{N} \sum_{n=0}^{N-1} \sin\left(2\pi\xi_0\frac{n}{N}\right) 160 e^{-2\pi i \frac{k}{N} n} =\\ 161 &=\frac{1}{2iN}\left( 162 \sum_{n=0}^{N-1}e^{2\pi i \frac{n}{N}(\xi_0 -k)} - e^{-2\pi i 163 \frac{n}{N}(\xi_0 +k)} 164 \right). 165 \end{align} 166 If we consider $\xi_0 \in \mathbb{Z}$, we have 167 \begin{align} 168 \hat{s}[k] = 169 \begin{cases} 170 \frac{1}{2i}\;\;\;\;\;\; \xi_0 = k\\ 171 -\frac{1}{2i}\;\;\;\;\;\; \xi_0 = -k\\ 172 0 \;\;\;\;\;\; \text{else} 173 \end{cases} 174 \end{align} 175 \subsection{Discrete Matrix Notation} 176 The convolution of two vectors $f, g \in \mathbb{C}^N$, can be expressed by a 177 circulate matrix applied to f 178 \begin{align} 179 (f * g) [n] = \sum_{k=0}^{N-1} f[k] g[n-k]. 180 \end{align} 181 Consider $g=s$, then the matrix takes the following values 182 \begin{align} 183 s[n-k] = s_{nk} = \sin\left(2\pi \xi_0 \frac{n-k}{N}\right). 184 \end{align} 185 The convolution with an impulse input $f=\delta_{0k}$, a vector that is $1$ 186 for $k=0$ and else 0 reads 187 \begin{align} 188 \sum_k s_{nk}f_k &= \sum_k s_{nk} \delta_{0k} =\\ 189 &= \sin\left(2\pi \xi_0 \frac{n}{N}\right). 190 \end{align} 191 192 %\printbibliography 193 \end{document}