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%% Title, authors and addresses

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\title{Fourier Transforms}

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\author{M.~Diehl}

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% main text
\begin{document}
\maketitle
% ----------------------------------------------------------------------------------------------------------------------------
\section{Discrete vs. continuous FT}
% ----------------------------------------------------------------------------------------------------------------------------
continuous Fourier transform
\begin{equation}
\hat{f}(k) = \int \limits_{-\pi}^{\pi} f(x) \cdot e^{-2\pi i k x} \inc x
\end{equation}
discrete Fourier transform
\begin{align}
\hat{f}_k &=  \frac{1}{d} \sum\limits_{n=0}^{N-1} f \left( x = \frac{n}{N}d \right) \cdot e^{\left(-2 \pi i \cdot \frac{k}{d} \cdot \frac{n}{N} \cdot d \right)} \cdot \frac{d}{N}\\
&= \frac{1}{N} \sum \limits_{n=0}^{N-1} f \left( x = \frac{n}{N} d \right) \cdot e^{-\frac{2 \pi i}{N} \cdot k \cdot n} 
\end{align}

% ----------------------------------------------------------------------------------------------------------------------------
\section{Differentation}
% ----------------------------------------------------------------------------------------------------------------------------
Expression in frequency and angular frequency
\begin{align}
\hat{f}(k) &= \frac{1}{d} \int \limits_0^d f(x) e^{\frac{-2 \pi i}{d} k x}\\
&= \frac{1}{d} \int \limits_0^d f(x) e^{-2 \pi i \xi x} \inc x
\end{align}
\begin{align}
\hat{f}'&= \frac{\partial}{\partial x} \left( \int \limits_{-\infty }^{\infty} \hat{f}(x) \cdot e^{i \xi x} \inc k \right)\\
&=  \int \limits_{-\infty}^{\infty} \frac{\partial}{\partial x} \left( \hat{f}(x) \cdot e^{i \xi x} \right) \inc k\\
&= \int \limits_{-\infty}^{\infty} i \xi \cdot \hat{f}(x) \cdot e^{i \xi x} \inc k
\end{align}

\section{Transform}
example with  $N=4$ and $x = \mathrm{sin}\left( \frac{n}{N} \cdot 2 \pi \right)$
\begin{align}
X_k &= \sum \limits_{n=0}^{N-1} x_n \cdot e ^{- \frac{2 \pi i }{N} \cdot  k \cdot n};~~~k = 0;1;..;N-1\\
&= \sum \limits_{n=0}^{N-1} x_n \cdot \left(\mathrm{cos}\left(- \frac{2 \pi}{N}\cdot k \cdot n \right) + i\cdot \mathrm{sin}\left(- \frac{2 \pi}{N} \cdot k \cdot n\right) \right)\\
X_0 &= \sum \limits_{n=0}^{N-1} x_n e ^\\
&= 0 + 1 + 0 +(-1) = 0 \\
X_1 &= \sum \limits_{n=0}^{N-1} x_n e^{-i \frac{2\pi}{N} \cdot 1 \cdot n} = 0 + e ^ {-i \frac{2\pi}{N} \cdot 1 \cdot 1} + 0 +e ^ {-i \frac{2\pi}{N} \cdot 1 \cdot 3}\\
&= 0 + (- 2i)\\
X_2 &= 0 + 0i\\
X_3 &= 0 + 2i
\end{align}
$X_2$ is Nyquist frequency and has only a real part, $X_3$ is conjugate complex of $X_1$ for real only input.
\section{Inverse Transform}
\begin{align}
x_n &= \frac{1}{N} \sum \limits_{k=0}^{N-1} X_k e^{\frac{2\pi i}{N} \cdot k \cdot n}\\
x_0 &= \frac{1}{4}\left(0 - 2ie^0 + 0 + 2ie^0 \right) = 0\\
x_1 &= \frac{1}{4}\left(0 - 2ie^{\frac{2\pi i}{4}\cdot 1 \cdot 1} + 0 + 2ie^{\frac{2\pi i}{4}\cdot 3 \cdot 1} \right) = 1\\
x_2 &= 0\\
x_1 &= -1
\end{align}
\end{document}
\endinput