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main.tex (6255B)

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 \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 dirac comb can be represented in a series of discrete dirac delta's
 \begin{align}
     \Sha_m[n] = \sum_{l=-N}^N \delta[n - lm],
 \end{align}
 where $\delta[s] = 1$ if $s = 0$ else $0$, for $s \in \mathbb{Z}$.
 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(
             \sum_{l=-N}^N \delta(n-lm)
             \right)
             e^{-2\pi i \frac{k}{N}n},
 \end{align}
 where the summation happens exactly $\frac{N}{m}$ times, then
 \begin{align}
     &\frac{1}{m}\sum_{l=-N}^N e^{-2\pi i \frac{k}{N}lm}=\\
     &= \frac{1}{m} \sum_{l=-N}^N \delta[k - l\cdot \frac{N}{m}]\qquad
     \text{(Poisson's summation formula)} \\
     &= \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}^2}\xi \varphi(x) e^{2\pi i\xi x}\ dxd\xi\\
     &= \int_{\mathbb{R}^2}\varphi(x) \xi e^{2\pi i \xi x}\ dxd\xi\\
     &=\int_{\mathbb{R}^2}\varphi(x)\frac{i}{2\pi} \frac{\partial}{\partial x}
      e^{2\pi i \xi x}\ dxd\xi =\\
     &=\frac{i}{2\pi}\int_{\mathbb{R}^2}\varphi(x)\delta'(x)\ dx=\\
     &=\frac{i}{2\pi} \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}
 
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