prb6.tex (5077B)
1 \include{preamble.tex} 2 3 \begin{document} 4 \maketitle 5 \tableofcontents 6 7 \section{Sheet 6} 8 \subsection{Fourier Transform of the convolution} 9 Consider the function $f(x)$, which has a Fourier Transform $\hat{f}(\xi)$, 10 now let us compute the Fourier transform of 11 \begin{align} 12 h(x) = f(3x-1) \sin(x) . 13 \end{align} 14 We know that the Fourier transform of the convolution is (we use somewhat of 15 the inverse convolution theorem). 16 \begin{align} 17 \widehat{(f(3x-1)*g(x))} = \widehat{f(3x-1)} \cdot \hat{g}(\xi). 18 \end{align} 19 The Fourier transform of $f(3x-1)$ is simply done by substituting a new 20 variable 21 \begin{align} 22 \widehat{f(3x-1)} = \frac{1}{3}e^{2\pi i\frac{\xi}{3}}\ 23 f\left(\frac{\xi}{3}\right). 24 \end{align} 25 The Fourier transform of $\sin(x)$ can be calculated when looking at the 26 Fourier transform of the Dirac-delta function 27 \begin{align} 28 \widehat{\delta(ax-b)} 29 &=\int_\mathbb{R} \delta(ax-b) e^{-2\pi i x \xi}\ dx 30 \;\;\;\;\;\;\; (y = ax-b)\\ 31 &=\int_\mathbb{R} \delta(y) e^{-2\pi i (y+b)\frac{\xi}{a}}\frac{dy}{a}\\ 32 &=\frac{1}{a} e^{-2\pi i \xi \frac{b}{a}}. 33 \end{align} 34 We may plug in $\sin(x)$ in the definition of the Fourier transformation and 35 observe where we can use the Dirac-delta to to the inverse Fourier transform 36 \begin{align} 37 \widehat{\sin(x)} 38 &=\int_\mathbb{R} \sin(x)e^{-2\pi i x\xi}\ dx=\\ 39 &=\frac{1}{2i}\int_\mathbb{R} (e^{ix} - e^{-ix})e^{-2\pi i \xi x}\ dx\\ 40 &=\frac{1}{2i}\left( 41 \int_\mathbb{R} e^{ix} e^{-2\pi i \xi x}\ dx+ 42 \int_\mathbb{R} e^{-ix} e^{-2\pi i \xi x}\ dx 43 \right). 44 \end{align} 45 Here we may use the above formula for the Fourier transform of the Dirac 46 delta. We choose $a=1$, $b= \pm \frac{1}{2\pi}$ and do some $y=-x$ 47 substitutions and thereby get the following result 48 \begin{align} 49 \widehat{\sin(x)} = \frac{1}{2i} \left( 50 \delta(\xi - \frac{1}{2\pi}) 51 -\delta(\xi + \frac{1}{2\pi}) 52 \right) 53 \end{align} 54 The whole result is thereby 55 \begin{align} 56 \widehat{f(3x-1)} * \widehat{sin(x)} 57 =& \frac{1}{6i} \bigg( 58 e^{2\pi 59 i(\frac{\xi}{3}-\frac{1}{6\pi})}\hat{f}\big(\frac{\xi}{3}-\frac{1}{6\pi}\big)- 60 e^{2\pi 61 i(\frac{\xi}{3}+\frac{1}{6\pi})}\hat{f}\big(\frac{\xi}{3}+\frac{1}{6\pi}\big) 62 \bigg) 63 \end{align} 64 \subsection{More Fourier Transforms} 65 Consider the function 66 \begin{align} 67 f(x) = e^{-|x|} 68 \end{align} 69 The Fourier transform of this function is 70 \begin{align} 71 \hat{f}(\xi) 72 &=\int_\mathbb{R} e^{-|x| e^{-2\pi i x \xi}}\ dx\\ 73 &= \int_{-\infty}^0 e^x e^{-2\pi i x \xi}\ dx 74 + \int_0^\infty e^{-x} e^{-2\pi i x \xi}\ dx=\\ 75 &= \frac{1}{1-2\pi i \xi} e^{(1-2\pi i \xi) x}\bigg|_{-\infty}^0+ 76 \frac{-1}{1+2\pi i \xi} e^{-(1+2\pi i \xi) x}\bigg|_{-\infty}^0 = \\ 77 &= \frac{1}{1-2\pi i \xi} + \frac{1}{1 + 2\pi i \xi} =\\ 78 &= \frac{2}{1+(2\pi \xi)^2}. 79 \end{align} 80 Let us use this result to solve the following integral 81 \begin{align} 82 \int_\mathbb{R} \frac{\cos(a\xi)}{(2\pi \xi)^2 + 1}\ d\xi = 83 \frac{1}{2}\int_\mathbb{R} \hat{f}(\xi) \text{Re}(e^{ia\xi})\ dx,\\ 84 \end{align} 85 where we used the fact that $\text{Re}(e^{ia\xi}) = \cos(a\xi)$ and 86 $\hat{f}(\xi) = \frac{2}{1+(2\pi \xi)^2}$, thereby 87 \begin{align} 88 \frac{1}{2}\int_\mathbb{R} \hat{f}(\xi) \text{Re}(e^{ia\xi})\ dx 89 &= \frac{1}{2}\text{Re}\left( 90 \int_\mathbb{R}\hat{f}(\xi)e^{ia\xi}\ d\xi 91 \right)=\\ 92 &= \frac{1}{2}\text{Re}\left( 93 \int_\mathbb{R} \hat{f}(\xi) e^{2\pi i \frac{a}{2\pi}\xi}\ d\xi 94 \right)=\\ 95 &= \frac{1}{2}\text{Re}\left(f(\frac{a}{2\pi})\right)=\\ 96 &= \frac{1}{2} e^{-\frac{|a|}{2\pi}}. 97 \end{align} 98 \subsection{Finite discrete Fourier transform} 99 Consider $s\in \mathbb{C}^N$ with entries 100 \begin{align} 101 s[n] = \sin\left(2\pi\xi_0\frac{n}{N}\right), 102 \end{align} 103 for same $0 < \xi_0 < N$. The finite discrete Fourier transform of $s$ is 104 \begin{align} 105 \hat{s}[k] &= \frac{1}{N} \sum_{n=0}^{N-1} \sin\left(2\pi\xi_0\frac{n}{N}\right) 106 e^{-2\pi i \frac{k}{N} n} =\\ 107 &=\frac{1}{2iN}\left( 108 \sum_{n=0}^{N-1}e^{2\pi i \frac{n}{N}(\xi_0 -k)} - e^{-2\pi i 109 \frac{n}{N}(\xi_0 +k)} 110 \right). 111 \end{align} 112 If we consider $\xi_0 \in \mathbb{Z}$, we have 113 \begin{align} 114 \hat{s}[k] = 115 \begin{cases} 116 \frac{1}{2i}\;\;\;\;\;\; \xi_0 = k\\ 117 -\frac{1}{2i}\;\;\;\;\;\; \xi_0 = -k\\ 118 0 \;\;\;\;\;\; \text{else} 119 \end{cases} 120 \end{align} 121 \subsection{Discrete Matrix Notation} 122 The convolution of two vectors $f, g \in \mathbb{C}^N$, can be expressed by a 123 circulate matrix applied to f 124 \begin{align} 125 (f * g) [n] = \sum_{k=0}^{N-1} f[k] g[n-k]. 126 \end{align} 127 Consider $g=s$, then the matrix takes the following values 128 \begin{align} 129 s[n-k] = s_{nk} = \sin\left(2\pi \xi_0 \frac{n-k}{N}\right). 130 \end{align} 131 The convolution with an impulse input $f=\delta_{0k}$, a vector that is $1$ 132 for $k=0$ and else 0 reads 133 \begin{align} 134 \sum_k s_{nk}f_k &= \sum_k s_{nk} \delta_{0k} =\\ 135 &= \sin\left(2\pi \xi_0 \frac{n}{N}\right). 136 \end{align} 137 138 %\printbibliography 139 \end{document}