notizen.tex

\section{Notizen}
\subsection{Gleichungen}
$arg(\frac{27}{(1+j)^8}) = arg(27) - arg((1+j)^8) = 2k\pi - 8\cdot arg(1+j) = 2k\pi -8(\frac{\pi}{4}+2l\pi) = 2k\pi - 2\pi - 16l\pi = 2\pi\overbrace{(k-8l-1)}^{m} = 2m\pi$\\ \vspace{\baselineskip}
$|e^{-j\frac{\pi}{6}}| = 1 (Betrag!)$\\ \vspace{\baselineskip}
$\frac{[cos(5) + j\cdot sin(5)]^6}{e^{j(30+40\pi)}} = \frac{[cjs(5)]^6}{e^{30j+40\pi j}} = \frac{(e^{5j})^6}{e^{30j}\cdot (e^{2\pi j})^{20}} = \frac{e^{30j}}{e^{30j}} = 1 $\\
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$z^5 + z^3 + 8z^2 + 8 = 0 \rightarrow 0 = z^3(z^2 + 1) + 8(z^2 + 1) = (z^3 + 8)(z^2 + 1) \rightarrow z^3 + 8 = 0 \text{ oder }  z^2 + 1 = 0$\\
$z^3 + 8 = 0: z = \sqrt[3]{-8} = \sqrt[3]{|-8|}\cdot cjs(\frac{arg(-8)}{3} + k\frac{2\pi}{3}) = 2[cos(\frac{\pi}{3} + k\frac{2\pi}{3}) + j\cdot sin(\frac{\pi}{3} + k\frac{2\pi}{3})] \rightarrow k = 0 ...; k = 1 ...; k = 2 ...$\\
$z^2+1 = 0 : z^2 = -1 \rightarrow z = \sqrt{-1} \rightarrow z_4 = j; z_5 = -j$\\
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$1 = \frac{e^z + e^{-z}}{2} \leftrightarrow 2 = e^z + e^{-z} \rightarrow{\cdot e^z}\rightarrow (e^z)^2 - 2e^z + 1 = 0 \rightarrow (e^z - 1)^2 = 0 \rightarrow e^z = 1 \rightarrow z = ln(1) + j\cdot (arg(1)+2k\pi) = j\cdot 2k\pi$\\
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$\frac{e^{j\frac{a+b}{2}}}{2j} = \frac{e^{ja} + e^{jb}}{4j}   $



\subsection{Fourierreihen}
$c_n = \frac{1}{2\pi}\int\limits_{0}^{2\pi} sin(1.5t)\cdot e^{-jnt} dt -> \text{ Euler } -> \frac{1}{4\pi j} \int\limits_{0}^{2\pi} e^{(1.5-n)jt} - e^{(-1.5-n)jt} dt = \frac{1}{4\pi j}[\frac{e^{(1.5-n)jt
}}{(1.5-n)j}]_{0}^{2\pi} - \frac{1}{4\pi j} [\frac{e^{(-1.5-n)jt}}{(-1.5-n)j}]_{0}^{2\pi} = \frac{e^{3\pi j
}\cdot e^{-2\pi nj} -1}{-2(3-2n)\pi} + \frac{e^{-3\pi j} \cdot e^{-2\pi nj} -1}{-2(3+2n)\pi} = \frac{1}{(3-2n)\pi} + \frac{1}{(3+2n)\pi} = \frac{(3+2n) + (3-2n)}{(3-2n)(3+2n)\pi} = \frac{6}{(9-4n^2)\pi}$\\
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