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Rudolf Rabenstein edited untitled.tex
almost 9 years ago
Commit id: 61f13cb5685070b2c9e4f4280b4a6533e77410b7
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Aus~\eqref{eq:1} folgt durch Ausmultiplizieren direkt
\begin{equation}
Y(z) = \frac14\, y[0]\frac{1}{z^2-\frac14} + y[1] \frac{z}{z^2-\frac14}
\right) \; .
\label{eq:2}
\end{equation}
Die beiden von $z$ abhängigen Terme haben die folgenden Partialbruchzerlegungen
\begin{align}
\frac{1}{z^2-\frac14} &= \frac{1}{(z-\frac12)(z+\frac12)} = \frac{1}{(z-\frac12)} - \frac{1}{(z+\frac12)}
\\
\frac{z}{z^2-\frac14} &= \frac{z}{(z-\frac12)(z+\frac12)} = \frac12\frac{1}{(z-\frac12)} - \frac12\frac{1}{(z+\frac12)}
\end{align}