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      1 \include{preamble.tex}
      2 
      3 \begin{document}
      4 \maketitle
      5 \tableofcontents
      6 
      7 \section{Sheet 7}
      8 \subsection{Dirac Comb}
      9 The Dirac train or Dirac comb on defined in the following way
     10 \begin{align}
     11     \Sha_m[n] =
     12     \begin{cases}
     13         1\;\;\;\;\;\; n = 0, \pm m, \pm 2m,\dots\\
     14         0\;\;\;\;\;\; \text{else}
     15     \end{cases}
     16 \end{align}
     17 The dirac comb can be represented in a series of discrete dirac delta's
     18 \begin{align}
     19     \Sha_m[n] = \sum_{l=-N}^N \delta[n - lm],
     20 \end{align}
     21 where $\delta[s] = 1$ if $s = 0$ else $0$, for $s \in \mathbb{Z}$.
     22 The discrete Fourier transform of the Dirac comb in $\mathbb{C}^N$ is
     23 \begin{align}
     24     \widehat{\Sha_m[n]}
     25     &=\frac{1}{N}\sum_{n=0}^{N-1} \Sha_m[n] e^{-2\pi i \frac{k}{N}n}=\\
     26     &=\frac{1}{N}\sum_{n=0}^{N-1}
     27         \left(
     28             \sum_{l=-N}^N \delta(n-lm)
     29             \right)
     30             e^{-2\pi i \frac{k}{N}n},
     31 \end{align}
     32 where the summation happens exactly $\frac{N}{m}$ times, then
     33 \begin{align}
     34     &\frac{1}{m}\sum_{l=-N}^N e^{-2\pi i \frac{k}{N}lm}=\\
     35     &= \frac{1}{m} \sum_{l=-N}^N \delta[k - l\cdot \frac{N}{m}]\qquad
     36     \text{(Poisson's summation formula)} \\
     37     &= \frac{1}{m}\Sha_{\frac{N}{m}}[k]
     38 \end{align}
     39 \subsection{Schwartz Space}
     40 The Schwartz space $\mathcal{S}(\mathbb{R}^d)$, for $d \in \mathbb{N}$ is
     41 defined as
     42 \begin{align}
     43     &\mathcal{S} :=
     44 \bigg\{
     45     f\in\mathcal{C}^\infty(\mathbb{R}^d):
     46     \forall\alpha,\beta\in\mathbb{N}^d\;\; \lVert f \rVert_{\alpha,\beta}
     47     < \infty
     48 \bigg\},\\
     49 &\lVert f \rVert_{\alpha, \beta} :=
     50 \sup_{x\in\mathbb{R}^d}\left|x^\alpha (D^\beta f) (x) \right|.
     51 \end{align}
     52 Our aim is to show that if $f\in\mathcal{S}(\mathbb{R})$ then $\hat{f} \in
     53 \mathcal{S}(\mathbb{R})$. The condition is obviously
     54 \begin{align}
     55     &\lVert \hat{f} \rVert_{\alpha, \beta} =
     56     \sup_{\xi\in\mathbb{R}}\left|\xi^\alpha (D^\beta \hat{f}) (\xi)
     57     \right|<\infty,
     58 \end{align}
     59 for all $\alpha, \beta \in \mathbb{N}$.
     60 We can start with what we know about the Fourier transform
     61 \begin{align}
     62     \xi^\alpha \hat{f}(\xi) &= \mathcal{F}\left(\frac{1}{(2\pi
     63     i)^\alpha}(D^{\alpha}f)(x)\right)\\
     64             D^{\beta}\hat{f}(\xi) &= \mathcal{F}\left(
     65         (-2\pi i x)^\beta f(x)
     66     \right).
     67 \end{align}
     68 Combining the two relations above we get
     69 \begin{align}
     70     \xi^\alpha (D^\beta \hat{f})(\xi) =
     71     \mathcal{F}\left(\frac{(-2\pi i x)^\beta}{(2\pi
     72     i)^\alpha}x^\beta(D^{\alpha}f)(x)\right)=: \mathcal{F}(g(x))\\
     73 \end{align}
     74 If we call this function $g$, then $g\in\mathcal{S}(\mathbb{R})$ and
     75 $g\in L^1(\mathbb{R})$. Applying the Riemann-Lebesgue Lemma we get
     76 \begin{align}
     77     \hat{g}(\xi) = \int_\mathbb{R} g(x) e^{-2\pi i x \xi}\ dx \longrightarrow 0
     78     \;\;\;  \text{as $|\xi| \rightarrow \infty$ }
     79 \end{align}
     80 Thereby $\hat{g} \in \mathcal{S}(\mathbb{R})$ and thus $\hat{f} \in
     81 \mathcal{S}(\mathbb{R})$.
     82 \subsection{Tempered Distributions}
     83 Tempered distributions are the elements of
     84 \begin{align}
     85     \mathcal{S}'(\mathbb{R}^d) :=
     86     \bigg\{
     87         L: \mathcal{S}(\mathbb{R}^d) \rightarrow \mathbb{C} | \text{$L$ is
     88         linear and continuous}
     89     \bigg\}.
     90 \end{align}
     91 Consider $\xi$ as a tempered distribution, buy acting on $\varphi \in
     92 \mathcal{S}(\mathbb{R})$ we have
     93 \begin{align}
     94     \xi(\phi) = \int_\xi \xi \varphi(\xi)\ d\xi.
     95 \end{align}
     96 The Fourier transform of $\xi$ is
     97 \begin{align}
     98     \hat{\xi}(\varphi)
     99     &=\xi(\hat{\varphi})
    100     = \int_\mathbb{R} \xi \hat{\varphi}(\xi)\ d\xi\\
    101     &= \int_{\mathbb{R}^2}\xi \varphi(x) e^{2\pi i\xi x}\ dxd\xi\\
    102     &= \int_{\mathbb{R}^2}\varphi(x) \xi e^{2\pi i \xi x}\ dxd\xi\\
    103     &=\int_{\mathbb{R}^2}\varphi(x)\frac{i}{2\pi} \frac{\partial}{\partial x}
    104      e^{2\pi i \xi x}\ dxd\xi =\\
    105     &=\frac{i}{2\pi}\int_{\mathbb{R}^2}\varphi(x)\delta'(x)\ dx=\\
    106     &=\frac{i}{2\pi} \delta'(\varphi).
    107 \end{align}
    108 \subsection{Fourier transform of the Dirac Comb}
    109 The general case of the Dirac Comb as a distribution is
    110 \begin{align}
    111     \Sha_T = \sum_{n \in \mathbb{Z}} \delta_{nT}.
    112 \end{align}
    113 The Fourier transform of the $\Sha_T$ distribution for $\varphi \in
    114 \mathcal{S}(\mathbb{R})$ is
    115 \begin{align}
    116     \widehat{\Sha_T}(\varphi)
    117     &= \sum_{n\in\mathbb{Z}} \hat{\delta}_{nT}(\varphi)\\
    118     &= \sum_{n\in\mathbb{Z}} \delta_{n\omega_0}(\varphi)\\
    119     &=\Sha_{\omega_0}(\varphi).
    120 \end{align}
    121 The Fourier transform, transforms the period of the combs.
    122 \subsection{Shannon Sampling}
    123 The Fourier transform of $1_{[-\frac{a}{2}, \frac{a}{2}]}(x)$ is
    124 \begin{align}
    125     \mathcal{F}\left(1_{[-\frac{a}{2}, \frac{a}{2}]}\right)(\xi)
    126     &= \int_\mathbb{R} 1_{[-\frac{a}{2}, \frac{a}{2}]} e^{-2\pi i x \xi}\
    127     dx\\
    128     &= \int_{-\frac{a}{2}}^{\frac{a}{2}} e^{-2\pi i x\xi}\ dx\\
    129     &= \frac{-1}{2\pi i \xi} e^{-2\pi i x
    130         \xi}\bigg|_{-\frac{a}{2}}^{\frac{a}{2}}\\
    131     &= \frac{1}{\pi \xi} \frac{1}{2i}\left(
    132         e^{pi i a \xi} - e^{-\pi i a \xi}
    133     \right)\\
    134     &= \frac{\sin(\pi \xi a)}{\pi \xi}
    135 \end{align}
    136 
    137 %\printbibliography
    138 \end{document}