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\usepackage[pdfstartview=FitH, bookmarks, colorlinks=false, pdftitle={Mitschrift Systhemtheorie Prof. Hoffmann}, pdfauthor={Fabian Kurz}, pdfsubject={Systemtheorie}, pdfkeywords={Mathematik Elektrotechnik Systemtheorie}, linkbordercolor={1 1 1}]{hyperref}
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\date{Zuletzt aktualisiert:\\\today}
\author{Fabian~Kurz\\\href{http://fkurz.net/}{http://fkurz.net/}}
\title{Systemtheorie II -- SS 05\\Prof. Dr.-Ing. habil. Hoffmann, TU Dresden\\Mitschrift}
\begin{document}
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\subsection{Systembeschreibung im Bildbereich}

\subsubsection{Lösung der Zustandsgleichungen}

Gleichungen aus 5.3.1:

\bigskip

\begin{minipage}{8cm}
\begin{empheq}[innerbox=\fbox]{align*}
\dot z(t) &= A \cdot z(t) + B \cdot x(t)\\
y(t) &= C \cdot z(t) + D \cdot x(t)
\end{empheq}
\end{minipage}%
\begin{minipage}{8cm}
Zustandsgleichungen eines linearen Systems
\end{minipage}

\bigskip

$\mathcal L$-Transformation der ersten Gleichung: $\mathcal L\{z(t)\} = Z(s)$,
$\mathcal L\{x(t)\} = X(s)$, 

Anfangszustand $z(0) = z(+0)$.
\begin{align*}
s \cdot Z(s) - z(0) &= A \cdot Z(s) + B \cdot X(s)\\
s \cdot Z(s) - A \cdot Z(s) &= B \cdot X(s) + z(0)\\
(s \cdot E - A) \cdot Z(s) &= z(0) + B \cdot X(s)
\end{align*}
\[\Rightarrow \quad Z(s) = \underbrace{(s \cdot E - A)^{-1}}_{\displaystyle\Phi(s)} \cdot z(0) + (s \cdot E - A)^{-1} \cdot B \cdot X(s)\]
$\Phi(s)$: \emph{Fundamentalmatrix} im Bildbereich

\bigskip


$\mathcal L$-Transformation der zweiten Gleichung mit $\mathcal L\{y(t)\} = Y(s)$:
\[Y(s) = C \cdot Z(s) + D \cdot X(s)\]
$Z(s)$ einsetzen:
\[Y(s) = C \cdot \Phi(s) \cdot z(0) + \underbrace{\left[C \cdot (S \cdot E - A)^{-1} \cdot B + D\right]}_{\displaystyle G(s)} \cdot  X(s)\]
$G(s)$: \emph{Übertragungsmatrix}

\begin{minipage}{8cm}
\[\boxed{\vphantom{\bigg|}\quad Y(s) = C \cdot \Phi(s) \cdot z(0) + G(s) \cdot X(s)\quad}\]
\end{minipage}%
\begin{minipage}{8cm}
\bigskip 
\center Input-Output-Gleichung im Bildbereich
\end{minipage}

\bigskip

Inverse Laplace-Transformation:

\begin{minipage}{8.2cm}
\[\boxed{\vphantom{\bigg|}\quad y(t) = C \cdot \varphi(t) \cdot z(0) + \int\limits_0^t g(t-\tau) \cdot x(\tau)\,d\tau \quad}\]
\end{minipage}%
\begin{minipage}{7.8cm}
\bigskip 
\center Input-Output-Gleichung im Zeitbereich 
\end{minipage}

$\hspace{1.7cm} \underbrace{\hphantom{ C \cdot \varphi(t) \cdot z(0)}}_{\text{freie Ausgabe}}\hphantom{+}\underbrace{\hphantom{\int\limits_0^t g(t-\tau) \cdot x(\tau)}}_{\text{erzwungene Ausgabe}}$

\bigskip

$\varphi(t)$: Fundamentalmatrix im Zeitbereich

$g(t)$: Gewichtsmatrix


\paragraph{Hierbei gilt:}
\begin{enumerate}
\item $\varphi(t) = \mathcal L^{-1} \{\Phi(s)\} = e^{At}$ \quad (Exponentialmatrix)

\textbf{Beweis:} Es ist zu zeigen, daß $\mathcal L\{e^{At}\} = (s \cdot E - A)^{-1}$.
\begin{align*}
\varphi(t) &= e^{At} = E + \frac{A \cdot t}{1!} + \frac{A^2\cdot t^2}{2!} + \frac{A^3\cdot t^3}{3!} + \cdots \qquad \text{insbes: } \varphi(0) = 0\\[1ex]
\dot{\varphi}(t) &= A + A^2 \cdot t + \frac{A^3 \cdot t^2}{2!} + \cdots = A \cdot \left(E + \frac{A}{1!} +  \frac{A^2\cdot t^2}{2!} + \cdots\right)\\
&= A \cdot \varphi(t)
\end{align*}
\[s \cdot \Phi(s) - \underbrace{\varphi(0)}_{E} = A \cdot \Phi(s) \quad \Rightarrow \quad (s \cdot E - A) \cdot \Phi(s) = E \quad \Rightarrow  \quad \underline{\underline{\Phi(s) = (s \cdot E - A)^{-1}}}\]

\item $g(t) = \mathcal L^{-1}\{G(s)\} = \mathcal L^{-1} \{C \cdot \Phi(s)\cdot B + D\} = C \cdot \varphi(t) \cdot B + D\cdot \delta(t)$
\end{enumerate}

\bigskip

\begin{minipage}{10cm}
\paragraph{Wichtiger Sonderfall:} System im Nullzustand, $z(0) = 0$, $l = m = 1 \to$ nur ein Ein- und Ausgang.

\bigskip

\end{minipage}%
\begin{minipage}{6cm}
\center


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\put(70,15){\makebox(0,0)[c]{$g(t)$}}
\put(70,35){\makebox(0,0)[c]{$z(0)=0$}}
\end{picture}
\end{minipage}

\bigskip

\begin{minipage}{8cm}
\begin{empheq}[innerbox=\fbox]{align*}
\vphantom{\Big|}Y(s) &= G(s) \cdot X(s)\\
y(t) &= \int\limits_0^t g(t - \tau) x(\tau)\,d\tau
\end{empheq}
\end{minipage}%
\begin{minipage}{8cm}
$G(s)$: Übertragungsfunktion

\bigskip

$g(t)$: Gewichtsfunktion
\end{minipage}

\paragraph{Veranschaulichung:}\hspace*{1cm}\\
\smallskip

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\[Y(s) = G(s) \cdot X(s) = G(s) \cdot \mathcal L\{\delta(t)\} = 1, \quad y(t) = g(t)\]

\paragraph{Folgerung:} Die Gewichtsfunktion $g$ ist die Reaktion des Systems auf das Impulssignal $\delta(t)$ und heißt deshalb auch \emph{Impulsantwort}.

\smallskip

Ist die Reaktion des Systems auf das Impulssignal bekannt, d.h. ist $g(t)$ gegeben, kann die Reaktion des Systems auf ein beliebiges Eingangssignal $x(t)$ berechnet werden:
\[y(t) = (g * x)(t)\]

\subsubsection{Übertragungsfunktion}

Es gelte weiterhin: $z(0) = 0$, $l = m = 1$.

\subsubsection*{Ermittlung von $G(s)$}

\begin{enumerate}
\item Aus Zustandsgleichungen: $G(s) = C \cdot (s \cdot E - A)^{-1} \cdot B + D$.
\item Aus Eingabe und Ausgabe: $G(s) = \dfrac{Y(s)}{X(s)} = \dfrac{\mathcal L\{\text{Wirkung}\}}{\mathcal L \{\text{Ursache}\}}$
\item Aus der Differentialgleichung (dem Blockschaltbild aus 5.3.2):
\begin{itemize}
\item $\mathcal L$ der Gleichungen (1) bis ($n+1$).
\item $z_1(0) = z_2(0) = \ldots = z_n(0) = 0$
\item $z_1(s)$, $z_2(s)$, \ldots \ schrittweise durch Einsetzen eliminieren
\item Abschließend umordnen zu $G(s) = \dfrac{Y(s)}{X(s)}$:
\[G(s) = \dfrac{a_n s^{-n} + a_{n-1} s^{-(n-1)} + \cdots + a_1 s^{-1} + a_0}{b_n s^{-n} + b_{n-1} s^{-(n-1)} + \cdots + b_1 s^{-1} + 1} \qquad (b_0 = 1)\]
\end{itemize}
\end{enumerate}

\subsubsection*{Ergebnis:}
\begin{center}
$\xymatrix{
\text{DGL} \ar[rr] \ar[rd] &       & \ar[dl] G(s)\\
            & \text{kanonische Realisierung (Blockschaltbild)}
}$
\end{center}

\begin{align*}
G(s) &= \dfrac{a_n s^{-n} + a_{n-1} s^{-(n-1)} + \cdots + a_1 s^{-1} + a_0}{b_n
s^{-n} + b_{n-1} s^{-(n-1)} + \cdots + b_1 s^{-1} + 1} = 
\frac{a_n + a_{n-1} d + \cdots + a_1 s^{n-1} + a_0 s^n}{b_n + b_{n-1}s
+ \cdots + b_1s^{n-1} + 1s^n} \\[1ex]
&= a_0 \cdot \frac{(s-{s_1}') (s-{s_2}') \cdot \cdots \cdot (s-{s_n})}{(s-{s_1}) (s-{s_2}) \cdot \cdots \cdot (s-{s_n})} \qquad
{s_i}':\text{  Nullstellen},\ \circ \qquad s_i: \text{ Polstellen},\ \times
\end{align*}

\subsubsection*{Darstellung zweckmäßig in Pol-Nullstellen-Plan  (PN-Plan)}

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\end{minipage}%
\begin{minipage}{10cm}
\begin{itemize}
\item Pole und Nullstellen reell oder paarweise konjugiert komplex
\item können auch mehrfach sein
\item aus PN-Plan können wichtige Systemeigenschaften abgelesen werden (z.B. Stabilität)
\end{itemize}
\end{minipage}

\subsubsection{Verallgemeinerte symbolische Methode}
\paragraph{Ziel:} Berechnungen von $G(s)$ für lineare RLC-Netzwerke

\subsubsection*{Allgemeiner Netzwerkzweig}

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\begin{minipage}{7cm}
\begin{description}
\item[Voraussetzung:] Für $t < 0$ sei:

 $u_e(t) = 0$, $u_C(0) = 0$, $i_L(0) = 0$.
\end{description}
\end{minipage}

\bigskip

\[u(t) = i(t) \cdot R + L \cdot \dfrac{di(t)}{dt} + \frac 1 C \cdot \int\limits_0^t i(\tau)\,d\tau + u_C(t)\]
$\mathcal L$-Transformation:
\begin{align*}
U(s) &= I(s) \cdot R + L\cdot (s \cdot I(s) - \underbrace{i_L(0)}_{0}) + \frac 1 C \cdot 1 s \cdot I(s) + U_e(s) \\
 &= \underbrace{\left(R + s\cdot L + \frac{1}{s\cdot C}\right)}_{\displaystyle\text{Zweigimpedanz }Z(s)} \cdot I(s) + U_e(s)
\end{align*}

\subsubsection*{Symbolischer Netzwerkzweig}
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\subsubsection*{Zuordnungen:}

\smallskip

\begin{minipage}{8cm}
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\end{picture}& $R$ & $i(t)$ & $\longrightarrow$ & $I(s)$\\[2ex]
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\end{picture}& $L$ & $u(t)$&$\longrightarrow$&$U(s)$\\[2ex]
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\end{picture}& $C$ & $u_e(t)$&$\longrightarrow$&$U_e(s)$
\end{tabular}
\end{minipage}%
\hfill
\begin{minipage}{7cm}
Damit gelten Regeln, wie im Gleichstromnetzwerk:
\begin{itemize}
\setlength{\itemsep}{-5pt}
\item Spannungsteilerregel
\item Stromteilerregel
\item Zweipoltheorie etc.
\end{itemize}
\end{minipage}


\paragraph{Beispiel:} Übertragung eines Rechteckimpulses über einen Tiefpass

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\bigskip

Nebenbedingung: $\left(\frac{R}{2L}\right)^2 < \frac{1}{LC}$ (s. weiter unten)

\begin{enumerate}
\renewcommand{\labelenumi}{\bf\arabic{enumi}.}
\item \textbf{Berechnung von $\mathbf{U_1(s)}$}

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\[u_1(t) = \underbrace{u_0 \cdot \mathbbm 1(t)}_{\diamond} - \underbrace{u_0 \cdot \mathbbm (t-t_0)}_{\times}\]
\[\Rightarrow U_1(s) = \frac{U_0}{s} - \frac{U_0}{s}\cdot e^{-st_0}\]
\end{minipage}

\smallskip

\item \textbf{Berechnung von $\mathbf{G(s)}$: Spannungsteilerregel}

\[G(s) = \frac{U_1(s)}{U_2(s)} = \frac{\frac{1}{sC}}{R + sL + \frac 1{sC}}  = \frac{1}{s^2LC + s RC + 1} \]

\item \textbf{Berechnung von $\mathbf{U_2(s)}$}

\[U_2(s) = G(s) \cdot U_1(s) = \underbrace{\frac{1}{s^2LC + s RC + 1} \cdot \frac{U_0}{s}}_{F(s)} - \underbrace{\frac{1}{s^2LC + s RC + 1} \cdot \frac{U_0}{s}}_{F(s)}\cdot e^{-st_0}\]

\item \textbf{Berechnung von $\mathcal L\mathbf{^{-1}\{F(s)\}}$ }

\[F(s) = \frac{U_0}{sLC\left(s^2 + s\frac R L + \frac 1 {LC}\right)} = \frac{U_0}{LC} \cdot \frac{1}{s\cdot(s-s_1)(s-s_2)}\]
mit $s_{1/2} = -\frac{R}{2L} \pm \sqrt{\left(\frac{R}{2L}\right) - \frac 1 {LC}}$

\smallskip

\textbf{Abkürzungen:} $\frac{R}{2L} = \sigma_0$, \quad $\sqrt{\left(\frac{R}{2L}\right) - \frac 1 {LC}} = \omega_0$, \quad $s_{1/2} = - \sigma_0 \pm j\omega_0$
\begin{align*}
\mathcal L^{-1}\{F(s)\} &= \sum\mathrm{Res}\,\left(F(s) \cdot e^{st}\right) = \sum   \mathop{\res_{s=0}}_{s_{1/2}}    \left[\frac{U_0}{LC} \cdot \frac{1}{s\cdot(s-s_1)(s-s_2)} \cdot  e^{st}\right]\\
&= \frac{U_0}{LC}\left[\frac{1}{s_1\cdot s_2} + \underline{\frac{e^{s_1t}}{s_1(s_1-s_2)}} +  \underline{\frac{e^{s_2t}}{s_2(s_2-s_1)}} \right] \qquad \underline{\qquad}: \text{ konj. kompl.}\\
&= \frac{U_0}{LC}\left[\frac{1}{s_1\cdot s_2} + 2\,\mathrm{Re}\,\left(\frac{e^{s_1t}}{s_1(s_1-s_2)}\right)\right] \qquad\qquad \text{da } z + z* = 2\,\mathrm{Re}\,(z)\\
 & = U_0 \left[1 + \frac{2}{LC} \cdot \mathrm{Re}\,\left(\frac{e^{-\sigma_0 t + j\omega_0 t}}{(-\sigma_0 + j\omega_0) \cdot 2j\omega_0}\right)\right] \qquad\qquad \frac{1}{s_1s_2} = LC\\
 & = U_0 \left[1 - \frac 1 {LC} \cdot \mathrm{Re}\,\left(\frac{e^{-\sigma_0t}\cdot e^{j\omega_0 t}}{(j\sigma_0 + \omega_0)\cdot \omega_0}  \right)\right]\\
 &= U_0\left[1 - \frac{e^{-\sigma_0 t}}{\omega_0 LC} \cdot \mathrm{Re}\,\left(\frac{e^{j\omega_0t}}{e^{j\arctan\frac{\sigma_0}{\omega_0}} \cdot \sqrt{{\sigma_0}^2 + {\omega_0}^2}}\right) \right] \qquad \sqrt{{\sigma_0}^2 + {\omega_0}^2} = \frac{1}{\sqrt{LC}}\\
 &= U_0\left[1 - \frac{e^{-\sigma_0 t}}{\omega_0\sqrt{LC}} \cdot \mathrm{Re}\,\left(e^{j\omega_0 t - \arctan\frac{\sigma_0}{\omega_0}}\right) \right] \\
 &= U_0\left[1 - \frac{e^{-\sigma_0 t}}{\omega_0\sqrt{LC}} \cdot \cos\left(\omega_0 t - \arctan\frac{\sigma_0}{\omega_0}\right) \right] \qquad \qquad t > 0\\
\end{align*}
\item \textbf{Berechnung von $\mathbf{u_2(t)}$ (Verschiebungssatz)}
\begin{align*}
u_2(t) &=   U_0\left[1 - \frac{e^{-\sigma_0 t}}{\omega_0\sqrt{LC}} \cdot \cos\left(\omega_0 t - \arctan\frac{\sigma_0}{\omega_0}\right) \right]  \cdot \mathbbm1 (t)\\
& -  U_0\left[1 - \frac{e^{-\sigma_0 t}}{\omega_0\sqrt{LC}} \cdot \cos\left(\omega_0 t - \arctan\frac{\sigma_0}{\omega_0}\right) \right] \cdot \mathbbm 1 (t-t_0)
\end{align*}


\qquad
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\end{enumerate}


% 21. April 2005

\subsubsection{Stationärer und flüchtiger Vorgang}
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\end{minipage}%
\begin{minipage}{8cm}
\center
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\end{picture}
\end{minipage}%

\begin{description}
\item[Komplexe Amplituden:] $\underline X = \widehat X \cdot e^{j\varphi_x},\
\underline X^* = \widehat X \cdot e^{-j\varphi_x}$
\end{description}

\begin{align*}
x(t) &= \widehat X \cdot \cos (\omega_0 t + \varphi_x) \cdot \mathbbm 1 (t) =
\frac{\widehat X}{2}\cdot \left(e^{j\omega_0 t + j \varphi_x} + e^{-j\omega_0 t
- j \varphi_x}\right) \\
 &= \frac 1 2 \left(\underline X \cdot e^{j\omega_0 t} + \underline{X}^* \cdot
e^{-j\omega_0 t}\right)
\end{align*}
$\mathcal L$-Transformation:
\[X(s) = \frac 1 2 \left(\frac{\underline X}{s-j\omega_0} + \frac{\underline
X^*}{s+j\omega_0}\right), \qquad Y(s) = G(s) \cdot X(s)\]
\begin{align*}
y(t) &= \sum\res\left[\frac 1 2 \left( \frac{\underline X}{s-j\omega_0} +
\frac{\underline X^*}{s+j\omega_0}\right) \cdot G(s) \cdot e^{st}\right]\\[1ex]
&= \underbrace{\sum\limits_{s_i=\text{Pole von } G(s)} \res[\ldots]}_{\text{Flüchtiger Vorgang } g_{fl}(t)} + \underbrace{\sum\limits_{s=\pm j\omega_0}\res [\ldots]}_{\text{Stationärer Vorgang }y_{st}(t)}
\end{align*}

\bigskip

\begin{minipage}{5cm}
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\end{minipage}%
\begin{minipage}{11cm}
\begin{description}
\item[Flüchtiger Vorgang:] $\lim\limits_{t\to\infty} y_{ft}(t) = 0$ wegen $\mathrm{Re}\,(s_i) < 0$

$\Rightarrow$ Stabiles System, vgl. 5.5.1
\item[Stationärer Vorgang:]\hspace*{\fill}\\[1ex]
$\displaystyle y_{st}(t) = \sum\limits_{\begin{subarray}{1}s=j\omega_0\\s=-j\omega_0\end{subarray}} \res\left[\frac 1 2 \left(\frac{\underline X}{s-j\omega_0} +
\frac{\underline X^*}{s+j\omega_0}\right) \cdot G(s) \cdot e^{st}\right]$ 
\end{description}
\end{minipage}
\begin{align*}
\phantom{y_{st}(t)} &= \frac 1 2 \left(\underline X \cdot G(j\omega_0) \cdot
e^{j\omega_0 t} + \underline X^* \cdot G(-j\omega_0) \cdot e^{-j\omega_0
t}\right) 
= \frac{1}{\cancel{2}} \cdot \cancel 2 \cdot \mathrm{Re}\,\left(\underline X
\cdot G(j\omega_0) \cdot e^{j\omega_0 t}\right)\\
&= \mathrm{Re}\,\left(\widehat X \cdot e^{j\varphi_x} \cdot
\left|G(j\omega_0)\right| \cdot e^{j\arg G(j\omega_0)} \cdot e^{j\omega_0 t}
\right)\\
&= \widehat X \cdot \left|G(j\omega_0)\right| \cdot \cos\left(\omega_0 t +
\varphi_x + \arg G(j\omega_0)\right)
\end{align*}

\smallskip

Im eingeschwungenen (stationären) Fall erscheint als Systemreaktion

\smallskip

\begin{minipage}{7cm}
\[\boxed{\quad \vphantom{\Big|} y_{st}(t) = \widehat Y \cdot \cos(\omega_0 t +
\varphi_y) \quad }\]
\end{minipage}%
\begin{minipage}{9cm}
mit $\widehat Y = \widehat X \cdot |G(j\omega_0)|$ and $\varphi_y = \varphi_x +
\arg G(j\omega_0)$
\end{minipage}%

\paragraph{Anwendung:} Wechselstromlehre

Beispiel:

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\end{picture}
\end{minipage}%
\begin{minipage}{8cm}
\begin{align*}
x(t)  &\Leftrightarrow u(t) = \widehat U \cdot \cos(\omega_0t + \varphi_u)\\
y(t) &\Leftrightarrow i(t) = \ ?
\end{align*}

Ansatz: $i(t) = \widehat I \cdot \cos (\omega_0 t + \varphi_i)$
\end{minipage}%

\bigskip

\[G(s) = \frac{I(s)}{U(s)} = \frac{1}{R + sL} \quad \Rightarrow \quad
G(j\omega) = \frac{1}{R+j\omega_0 L}\]
\[|G(j\omega_0)| = \frac{1}{\sqrt{R^2 + (j\omega_0 L)^2}}, \quad \arg G(j \omega_0) = - \arctan \frac{\omega_0 L}{R}\]
\[\widehat I = \widehat U \cdot \frac{1}{\sqrt{R^2 + (j\omega_0 L)^2}}, \quad \varphi_i = \varphi_u - \arctan \frac{\omega_0 L}{R}\]
\paragraph{Ergebnis:} $\displaystyle i(t) = \frac{\widehat U}{\sqrt{R^2 + (j\omega_0 L)^2}} \cdot \cos\left(\omega_0 t + \varphi_u - \arctan \frac{\omega_0 L}{R}\right)$

\newpage  % naja :-)

\subsubsection{Frequenzcharakteristiken}

Es sei $\omega_0 = \omega$ variabel.

\subsubsection*{Definitionen:}
\begin{enumerate}
\item $G(j\omega)$ heißt \emph{komplexer Frequenzgang} des linearen Systems. Schreibweise auch $G(\omega)$, Bezeichnung auch \emph{Übertragungsfunktion}. Die Darstellung von $G(j\omega)$ in der komplexen Ebene heißt \emph{Ortskurve}.
\item $|G(j\omega)| = \frac{\widehat Y}{\widehat X} = A(\omega)$ heißt \emph{Amplitudenfrequenzgang}.
\item $\arg G(j\omega) = \varphi_y - \varphi_x = \varphi(\omega)$ heißt \emph{Phasenfrequenzgang}.
\item $a(\omega) = - \log A(\omega)$ heißt \emph{Dämpfungsmaß} in Neper,

$a(\omega) = - 20\,\lg A(\omega)$ heißt Dämpfungsmaß in Dezibel,

$b(\omega) = - \varphi(\omega)$ heißt \emph{Phasenmaß}.
\end{enumerate}

\begin{minipage}{10cm}
\subsubsection*{Beispiel:} 
\begin{align*}
G(s) &= \frac{s+3}{s^2 + s + 1}\\
     &= \frac{s+3}{\left(s + \frac 1 2 \left(1+ j\sqrt 3 \right)\right) \left(s
+ \frac 1 2 \left(1-j\sqrt 3\right)\right)}
\end{align*}
\end{minipage}%
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\end{picture}
\end{minipage}%

\bigskip

\[G(j\omega) = \frac{j\omega + 3}{-\omega^2 + j \omega + 1}, \quad |G(j\omega)|
= A(\omega) = \sqrt{\frac{\omega^2 + 9}{(1-\omega^2)^2 + \omega^2}}, \quad
\varphi(\omega) = \arg G(j\omega)\]

\bigskip

\begin{minipage}{6cm}
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\hfill
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\qbezier(60,45)(70,20)(80,20)
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\qbezier(95,22)(105,24)(110,24)  % früher war ich auch mal fleißiger..
\end{picture}
\end{minipage}%

\subsubsection*{Einteilung nach $A(\omega)$}

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Tiefpass

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\end{minipage}%
\begin{minipage}{4cm}
\center
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\put(20,15){\vector(1,0){80}}
\put(25,10){\vector(0,1){70}}
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\put(90,5){$\omega$}
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Hochpass

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\begin{minipage}{4cm}
\center
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\put(20,15){\vector(1,0){80}}
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Bandpass

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\end{minipage}%
\begin{minipage}{4cm}
\center
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\put(20,15){\vector(1,0){80}}
\put(25,10){\vector(0,1){70}}
\put(0,68){$A(\omega)$}
\put(90,5){$\omega$}
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\qbezier(45,35)(52,15)(60,15)
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\end{picture}

Bandsperre

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\end{minipage}%

\subsection{Systemeigenschaften und Klassifizierung}
\subsubsection{Stabilit\"at}

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\hfill
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\put(182,40){\makebox(0,0)[l]{$y$}}
\put(201,42){\makebox(0,0)[l]{$t$}}
\end{picture}
\end{minipage}

\paragraph{Definition:} Ein zeitkontinuierliches System heißt \emph{stabil},
falls
\[|x(t)| \leq k_1 \quad \Rightarrow \quad |y(t)| \leq k_2, \qquad (t \geq 0)\]
genauer: BIBO-Stabilit\"{a}t (bounded input - bounded output).

\subsubsection*{Voraussetzung}
\begin{enumerate}
\renewcommand{\labelenumi}{\arabic{enumi})}
\item $|x(t)|$ beschr\"{a}nkt $\Rightarrow$ $X(s)$ hat keine singul\"{a}ren
Stellen mit $\mathrm{Re}\,(s) > 0$. 
\item $Y(s) = G(s)\cdot X(s)$ sei rational in $s$ und verschwinde im
Unendlichen
\end{enumerate}
Dann gilt: $y(t) = \mathcal L^{-1}\{Y(s)\} = \sum\res[G(s)\cdot X(s) \cdot
e^{st}]$

\subsubsection*{Folgerung}
\begin{itemize}
\item[--] Das System ist stabil, wenn f\"{u}r alle Pole $s_i$ von $G(s)$ gilt:
$\mathrm{Re}\,(s_i) < 0$ 
\item[--] Das System ist instabil, wenn f\"{u}r mindestens einen Pol $s_i$ von
$G(s)$ gilt: $\mathrm{Re}\,(s_i) > 0$
\end{itemize}

Aus 5.4.2: $G(s) = \dfrac{a_0s^n + a_1 s^{n-1} + \cdots + a_{n-1} s +
a_n}{s^n + b_1 s^{n-1} + \cdots + b_{n-1} s + b_n}$
\raisebox{-7pt}{$\leftarrow$ \text{charakteristisches Polynom des Systems}}

\begin{description}
\item[Kriterium:] Hat das charakteristische Polynom nur Nullstellen mit
negativem Realteil, ist das System stabil.
Dann heißt das Polynom \textsc{Hurwitz}-Polynom.
\end{description}

\subsubsection{Stabilit\"{a}tskriterien}

\subsubsection*{a) Hurwitz-Kriterium}

Ein reellwertiges Polynom (mit $b_i > 0$, $i = 0,\,\ldots,\,
n$)\footnote{Folgerung: Ein Polynom mit welchselnden Vorzeichen kann
kein \textsc{Hurwitz}-Polynom sein}
\[f(s) = b_0s^n + b_1s^{n-1} + \ldots + b_{n-1}s + b_n\]
hat genau dann nur Nullstellen mit negativem Realteil, wenn die
Abschnittsdeterminanten $D_j$ ($j=1,\,\ldots,\, n$) der folgenden Determinate
$D$ positiv sind.

\smallskip

\begin{minipage}{8cm}
\[D = \begin{vmatrix}b_1 & b_3 & b_5 & b_7 & \cdots \\ b_0 & b_2 & b_4 & b_6 & \cdots
\\ 0 & b_1 & b_3 & b_5 & \cdots \\ 0 &  b_0 & b_2 & b_4 & \cdots \\ 0 & 0 & b_1
& b_3 & \cdots \\ 0 & 0 & b_2 & b_4 & \cdots \\ \vdots & \vdots &\vdots &\vdots & \ddots
\end{vmatrix}\]
\end{minipage}%
\begin{minipage}{8cm}
\[D_1 = b_1 > 0\]
\[D_2 = \begin{vmatrix}b_1&b_3\\b_0&b_2\end{vmatrix} > 0\]
\[D_3 = \begin{vmatrix}b_1&b_3 & b_5\\b_0&b_2&b_4\\0&b_1&b_3\end{vmatrix} > 0\]
\end{minipage}

\bigskip

\paragraph{Beispiel:} 
Welchen Wert darf $v$ annehmen, so daß das System stabil bleibt?

\begin{center}
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}
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\put(346,0){\circle{2}}
\end{picture}
\end{center}
Ablesen der \"Ubertragungsfunktion nach 5.4.2:
\[G(s) = \frac{1\cdot s^{-3} + 1 \cdot s^{-2} + 1 \cdot s^{-1} + 1}{10\cdot
s^{-3} + 2 \cdot s^{-2} + v \cdot s^{-1} + 1} = \frac{s^3 + s^2 + s^1 + 1}{s^3
+ v\cdot s^2 + 2 \cdot s^1 + 10}\]
$\Rightarrow b_0 = 1$, $b_1 = v$, $b_2 = 2$, $b_3 = 10$

\bigskip

Abschnittsdeterminaten:

\bigskip

$D_1 = b_1 = v > 0$

\bigskip

$D_2 = \begin{vmatrix}b_1 & b_3 \\b_0 & b_2\end{vmatrix} = \begin{vmatrix}v &
10 \\1 & 2\end{vmatrix} = 2v - 10 > 0 \quad \Rightarrow \quad \boxed{\ v >
5\ \vphantom{\big|}}$

\bigskip


$D_3 = \begin{vmatrix}b_1&b_3 & b_5\\b_0&b_2&b_4\\0&b_1&b_3\end{vmatrix} = 
\begin{vmatrix}v&10&0\\1&2&3\\0&v&10\end{vmatrix} = 10 \cdot D_2 > 0$
(erf\"ullt)


\subsubsection*{b) Routh-Kriterium}
Vermeidet die Auswertung großer Determinanten durch Berechnung eines
Koeffizientenschemas.

\bigskip

\begin{minipage}{8cm}
\begin{tabular}{cccccc}
$b_n$ & $b_{n-2}$ & $b_{n-4}$ & $\cdots$ &$b_0$ oder $b_1$ & $0$\\
$b_{n-1}$ & $b_{n-3}$ & $b_{n-5}$ & $\cdots$ & $b_0$ oder $0$ & $0$\\
\hline
$B_1$ & $B_2$ & $B_3$ & $\cdots$ & $0$ & $0$\\
$C_1$ & $C_2$ & $C_3$ & $\cdots$ & $0$ & $0$\\
$\vdots$ & $\vdots$ & $\vdots$ &  & $\vdots$ & $\vdots$\\
$K_1$ & $0$ & $0$ & $\cdots$ & $0$ & $0$\\
$L_1$ & $0$ & $0$ & $\cdots$ & $0$ & $0$
\end{tabular}
\end{minipage}%
\begin{minipage}{8cm}
\begin{description}
\item[Erste Zeile:] $T_1 = \dfrac{b_n}{b_{n-1}}$, $B_1 = b_{n-2} - T_1\cdot
b_{n-3}$, $B_2 = b_{n-4} - T_1 \cdot b_{n-5}$ usw.
\item[Zweite Zeile:] $T_2 = \dfrac{b_{n-1}}{B_1}$, $C_1 = b_{n-3} -  T_2\cdot
B_2$, $C_2 = b_{n-5} - T_2\cdot B_3$ usw.
\end{description}
\end{minipage}

\paragraph{Satz:} S\"amtliche Nullstellen des charakteristischen Polynoms haben
genau dann negative Realteile, wenn gilt:
\begin{enumerate}
\renewcommand{\labelenumi}{\alph{enumi})}
\item Alle Koeffizienten $b_i$ sind positiv
\item S\"amtliche Koeffizienten $B_1$, $C_1$, \ldots, $K_1$, $L_1$ des
Routh-Schemas sind positiv
\end{enumerate}

\paragraph{Beispiel:} (Fortsetzung)

\qquad $\left.\begin{array}{ccccc}
10&v&0\\ 2 & 1 & 0 \\ \cline{1-3} 
v-5 & 0 & 0 & & T_1 = 5\\ 1 & 0 & 0 & & T_2 = \frac 2 {v-5} 
\\ \cline{1-3}\cline{1-3}\end{array}\right\} \quad
\text{Stabilit\"atsbedingung: } v - 5 > 0 \Leftrightarrow \boxed{\ v > 5\ \vphantom{\big|}}$

\subsubsection*{c) Ortskurvenkriterium}

\begin{description}
\item[Gegeben:] charakteristisches Polynom des Systems: $f(s)$
\item[Kriterium:] Durchl\"auft die Ortskurve von $f(j\omega)$ f\"ur $0\leq
\omega \leq \infty$ genau $n$ Quadranten im mathematisch positiven Sinn, ist
das System stabil ($n$: Grad von $f(s)$).
\item[Beispiel:] $n=4$
\begin{center}
\begin{picture}(150,120)
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\put(0,60){\vector(1,0){140}}
\put(68,110){\makebox(0,0)[r]{$\mathrm{Im}(f(j\omega))$}}
\put(138,50){\makebox(0,0)[r]{$\mathrm{Re}(f(j\omega))$}}
\put(120,60){\circle*{2}}
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\put(100.2,6){\vector(2,-3){0}}
\put(80,102){\small stabil}
\put(75,75){\small instabil}
\end{picture}
\end{center}
\end{description}

\subsubsection{Allpass und Mindestphasensystem}

\begin{minipage}{10cm}
\paragraph{Definition:} Ein zeitkontinuierliches System  heißt \emph{Allpass}
(AP), wenn $|G(j\omega)| = 1$ ($\omega \in \mathbb R$).

\bigskip

Allgemeine Form: $G(s) = \dfrac{f(-s)}{f(s)}$

\bigskip

$\Rightarrow$ Pole und Nullstellen spiegelbildlich zur imagin\"aren Achse.
\end{minipage}%
\begin{minipage}{6cm}
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\put(0,45){\vector(1,0){140}}
\put(70,0){\vector(0,1){90}}
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\put(45,45){\makebox(0,0)[c]{$\times$}}
\put(95,45){\makebox(0,0)[c]{$\circ$}}
\end{picture}
\end{minipage}

\paragraph{Beispiel:} (\"ahnlich wie \"Ubungsaufgabe 5.19)

\begin{minipage}{5cm}
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\put(28,40){\makebox(0,0)[r]{$u_1(t)$}}
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\put(110,65){\vector(0,-1){50}}
\put(113,40){\makebox(0,0)[l]{$u_2(t)$}}
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\multiput(60,10)(0,60){2}{\multiput(0,0)(5,0){4}{\qbezier(0,0)(0,5)(2.5,5)\qbezier(2.5,5)(5,5)(5,0)}}
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\put(90,20){\line(1,-1){10}}
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\put(53,17){\line(1,1){15}}
\multiput(0,0)(15,15){2}{\qbezier(53,17)(53,17)(47,23)}
\multiput(70,2)(0,80){2}{\makebox(0,0)[c]{$L$}}
\multiput(45,35)(50,0){2}{\makebox(0,0)[c]{$R$}}
\end{picture}
\end{minipage}%
\begin{minipage}{7cm}
\[G(s) = \frac{U_2(s)}{U_1(s)} = \frac{R-sL}{R+sL}\]
%\[G(j\omega) = \frac{R-j\omega L}{R+j\omega L}\]

\smallskip

\[|G(j\omega)| = \frac{\sqrt{R^2 + (\omega L)^2}}{\sqrt{R^2 + (\omega L)^2}} =
1\]
\end{minipage}%
\begin{minipage}{4cm}
\begin{picture}(110,90)
\put(0,45){\vector(1,0){110}}
\put(55,0){\vector(0,1){90}}
\put(59,80){$\omega$}
\put(98,36){$\sigma$}
\put(25,45){\makebox(0,0)[c]{$\times$}}
\put(85,45){\makebox(0,0)[c]{$\circ$}}
\put(21,32){\makebox(0,0)[c]{$-\frac R L$}}
\put(85,32){\makebox(0,0)[c]{$\frac R L$}}
\end{picture}
\end{minipage}%

\paragraph{Definition:} Ein stabiles System, dessen \"Ubertragungsfunktion $G$
keine Nullstellen $s_i$ mit $\mathrm{Re}\,(s_i) > 0$ hat, heißt
Mindestphasensystem (MPS).

% den beispiel-PN-Plan habe ich mir mal gespart, das ist wirklich zu trivial

\subsubsection*{Schaltungsbeispiel} 
\begin{minipage}{5cm}
\begin{picture}(100,80)
\put(0,30){\makebox(0,0)[l]{$U_1(t)$}}
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\put(30,0){\line(1,0){80}}
\put(90,60){\line(1,0){20}}
\put(115,30){\makebox(0,0)[l]{$U_2(t)$}}
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\put(111,55){\vector(0,-1){50}}
\put(82,30){\makebox(0,0)[r]{$R$}}
\put(60,71){\makebox(0,0)[bc]{$C$}}
\end{picture}
\end{minipage}%
\begin{minipage}{7cm}
\center $G(s) = \dfrac{U_2(s)}{U_1(s)} = \dfrac{R}{R+\frac 1{sC}} = \dfrac{sRC}{1+sRC}$
\end{minipage}%
\begin{minipage}{4cm}
\hfill
\begin{picture}(100,75)
\put(0,20){\vector(1,0){100}}
\put(50,00){\vector(0,1){60}}
\put(53,47){$\mathrm{Im}\,s$}
\put(75,10){$\mathrm{Re}\,s$}
\put(25,20){\makebox(0,0)[c]{$\times$}}
\put(50,20){\makebox(0,0)[c]{$\circ$}}
\put(21,8){\makebox(0,0)[c]{$-\frac{1}{CR}$}}
\end{picture}
\end{minipage}

\paragraph{Satz:} Die \"Ubertragungsfunktion $G$ eines beliebigen linearen,
zeitkontinuierlichen Systems läßt sich in der Form $G(s) = G_A(s)\cdot  G_M(s)$ darstellen, wobei gilt:


\bigskip

\begin{minipage}{8cm}
\begin{description}
\setlength{\itemsep}{0ex}
\item[$G_A(s)$] \"Ubertragungsfunktion eines Allpasses
\item[$G_M(s)$] \"Ubertragungsfunktion eines MPS
\end{description}
\end{minipage}%
\begin{minipage}{8cm}
\hfill
\begin{picture}(150,30)
\multiput(4,20)(122,0){2}{\circle{2}}
\multiput(5,20)(50,0){3}{\line(1,0){20}}
\multiput(25,13)(50,0){2}{\line(1,0){30}}
\multiput(25,27)(50,0){2}{\line(1,0){30}}
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\multiput(75,13)(30,0){2}{\line(0,1){14}}
\put(40,20){\makebox(0,0)[c]{$G_A(s)$}} 
\put(90,20){\makebox(0,0)[c]{$G_M(\kern-0.2ex s\kern-0.2ex)$}}  % faulheit :)
\put(65,10){\vector(0,1){10}}
\put(65,0){\makebox(0,0)[bc]{\small R\"uckkopplungsfreie}}
\put(65,-10){\makebox(0,0)[bc]{\small Kettenschaltung}}  % ich wars nicht!
\put(65,28){\makebox(0,0)[bc]{$\overbrace{\hspace{80pt}}^{\displaystyle G(s)}$}}
\end{picture}
\end{minipage}

\subsubsection*{Beispiel}

\begin{minipage}{5cm}
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\put(55,0){\vector(0,1){90}}
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\put(98,36){$\sigma$}
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\put(128,45){\makebox(0,0){$\Rightarrow$}}
\end{picture}
\end{minipage}
\begin{minipage}{5cm}
\begin{picture}(130,90)
\put(0,45){\vector(1,0){110}}
\put(55,0){\vector(0,1){90}}
\put(59,80){$\omega$}
\put(98,36){$\sigma$}
\multiput(40,20)(0,50){2}{\makebox(0,0)[c]{$\times$}}
\multiput(70,20)(0,50){2}{\makebox(0,0)[c]{$\circ$}}
\put(128,45){\makebox(0,0){$\mathbf{\cdot}$}}
\end{picture}
\end{minipage}
\begin{minipage}{5cm}
\begin{picture}(110,90)
\put(0,45){\vector(1,0){110}}
\put(55,0){\vector(0,1){90}}
\put(59,80){$\omega$}
\put(98,36){$\sigma$}
\multiput(15,30)(0,30){2}{\makebox(0,0)[c]{$\times$}}
\put(20,45){\makebox(0,0)[c]{$\circ$}}
\put(30,45){\makebox(0,0)[c]{$\times$}}
\multiput(40,20)(0,50){2}{\makebox(0,0)[c]{$\circ$}}
\end{picture}
\end{minipage}

\paragraph{Aufgabe:} Berechnung von $G(s)$ aus $A(\omega) = |G(s)|$ bzw.
$a(\omega)$ $\to$ Filterentwurf.

Nichteindeutig l\"osbar, da $G_1(s)$ und $G_2(s)$ gleiches $A(\omega)$ haben,
wenn 
\[G_2(s) = G_1(s) \cdot G_A(s) \Rightarrow |G_2(j\omega)| = |G_1(j\omega)|
\cdot \underbrace{|G_A(j\omega)|}_{1} = A(\omega)\]

Aber: F\"ur MPS ist Berechnung von $G(s)$ aus $A(\omega)$ eindeutig. Dann ist
auch $b(\omega)$ eindeutig mit bestimmt. Der Zusammenhang zwischen $a(\omega)$
und $b(\omega)$ lautet:
\[b(\omega) = \frac 1 {\pi} \int\limits_{-\infty}^{\infty}
\frac{a(\omega_0)}{\omega_0 - \omega}\,d\omega_0 \qquad \qquad \text{(Hilbert
Transformation)}\]

\setcounter{secnumdepth}{-1}    % so erscheint es ohne nr. im inhaltsverz.
\chapter{Teil 3: Zeitdiskrete Systeme}
\setcounter{secnumdepth}{3}

\section{Zeitdiskrete Signale und Systeme}
\subsection{Signalbeschreibung im Zeitbereich}
\subsubsection{Zeitdiskrete Signale}

\begin{description}
\item[Zeitskala:] $T \subseteq \mathbb Z = \{\ldots,\,-2,\,-1,\,0,\,1, 
\,2,\,\ldots\}$

Sonderf\"alle: $T = \mathbb Z$, $T = \mathbb N_0$

\item[Alphabet:] $X \subseteq \mathbb C$, wichtiger Sonderfall: $X = \mathbb R$
\end{description}
\paragraph{Definition:} Ein zeitdiskretes Signal ist eine Abbildung
$x:T\longrightarrow X$, bei der jedem \emph{Zeitpunkt} $k \in T$ ein
\emph{Signalwert} $x(k)$ zugeordnet ist.

\subsubsection*{Darstellung}
\begin{enumerate}
\renewcommand{\labelenumi}{\alph{enumi})}
\item als Folge $x = (\ldots,\,x(-2),\,x(-1),\,x(0),\,x(1),\,x(2),\,\ldots)$
\item analytisch, z.B. $x(k) = \left\{\begin{array}{ll}a^k & l \geq 0\\ 0 &
\text{sonst}\end{array}\right.$
\item grafisch:

\begin{center}
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\put(70,10){\vector(0,1){90}}
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\put(120,22){\circle*{2}}
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\put(130,20){\circle*{2}}
\put(130,15){\line(0,1){5}}
\put(140,18){\circle*{2}}
\put(140,15){\line(0,1){3}}
\put(150,17){\circle*{2}}
\put(150,15){\line(0,1){2}}
\put(160,16.5){\circle*{2}}
\put(160,15){\line(0,1){1.5}}
\put(170,16){\circle*{2}}
\put(170,15){\line(0,1){1}}
\put(180,15.5){\circle*{2}}
\put(180,15){\line(0,1){0.5}}
\put(68,90){\makebox(0,0)[r]{$x(k)$}}
\put(187,7){\makebox(0,0)[l]{$k$}}
\end{picture}
\end{center}
\end{enumerate}

\subsubsection{Spezielle Signale}

\begin{enumerate}
\renewcommand{\labelenumi}{\alph{enumi})}
\begin{minipage}{3.5cm}
\item Impulssignal 
\end{minipage}%
\begin{minipage}{5.5cm}
\[x(k) = \delta(k) = \left\{\begin{array}{ll}1 & k=0 \\ 0 & k\ne
0\end{array}\right.\]

\smallskip
\end{minipage}%
\begin{minipage}{6cm}
\center
\begin{picture}(80,50)
\put(0,10){\vector(1,0){80}}
\put(30,5){\vector(0,1){45}}
\put(33,38){$\delta(k)$}
\put(70,0){$k$}
\put(28.5,30){\line(1,0){3}}
\put(30,30){\circle*{2}}
\put(28,30){\makebox(0,0)[r]{$1$}}
\multiput(10,8.5)(10,0){7}{\line(0,1){3}}
\multiput(10,10)(10,0){2}{\circle*{2}}
\multiput(40,10)(10,0){4}{\circle*{2}}
\end{picture}
\end{minipage}

\begin{minipage}{3.5cm}
\item Sprungsignal
\end{minipage}%
\begin{minipage}{5.5cm}
\[x(k) = \mathbbm 1(k) = \left\{\begin{array}{ll}1 & k\geq 0 \\ 0 & k<0\end{array}\right.\]

\smallskip
\end{minipage}%
\begin{minipage}{6cm}
\center
\begin{picture}(80,50)
\put(0,10){\vector(1,0){80}}
\put(30,5){\vector(0,1){45}}
\put(33,38){$\mathbbm 1(k)$}
\put(70,0){$k$}
\put(28.5,30){\line(1,0){3}}
\put(30,30){\circle*{2}}
\put(28,30){\makebox(0,0)[r]{$1$}}
\multiput(10,8.5)(10,0){7}{\line(0,1){3}}
\multiput(10,10)(10,0){2}{\circle*{2}}
\multiput(40,30)(10,0){4}{\circle*{2}}
\end{picture}
\end{minipage}
\end{enumerate}

\subsubsection*{Signaloperationen}
Einstellige Signaloperationen

\begin{enumerate}
\renewcommand{\labelenumi}{\alph{enumi})}
\begin{minipage}{3.5cm}
\item Skalarmultiplikation
\end{minipage}%
\begin{minipage}{4.5cm}
\center $y(k) = \alpha \cdot x(k)$
\smallskip
\end{minipage}%
\begin{minipage}{7cm}
\center
\begin{picture}(85,50)
\put(0,10){\vector(1,0){80}}
\put(20,5){\vector(0,1){45}}
\put(3,38){$\scriptstyle x(k)$}
\put(70,0){$k$}
\multiput(10,8.5)(10,0){7}{\line(0,1){3}}
\multiput(20,10)(40,0){2}{\circle*{2}}
\multiput(30,10)(20,0){2}{\line(0,1){10}}
\multiput(30,20)(20,0){2}{\circle*{2}}
\put(40,30){\circle*{2}}
\put(40,10){\line(0,1){20}}
\end{picture}
\begin{picture}(85,50)
\put(0,10){\vector(1,0){80}}
\put(20,5){\vector(0,1){45}}
\put(3,38){$\scriptstyle y(k)$}
\put(70,0){$k$}
\multiput(10,8.5)(10,0){7}{\line(0,1){3}}
\multiput(20,10)(40,0){2}{\circle*{2}}
\multiput(30,25)(20,0){2}{\circle*{2}}
\multiput(30,10)(20,0){2}{\line(0,1){15}}
\put(40,40){\circle*{2}}
\put(40,10){\line(0,1){30}}
\end{picture}
\end{minipage}

\begin{minipage}{3.5cm}
\item Translation
\end{minipage}%
\begin{minipage}{4.5cm}
\center
$y = s^n \cdot x$, d.h.\\[0.7ex]
$y(k) = x(k-n)$
\smallskip
\end{minipage}%
\begin{minipage}{7cm}
\center
\begin{picture}(85,50)
\put(0,10){\vector(1,0){80}}
\put(20,5){\vector(0,1){45}}
\put(3,38){$\scriptstyle x(k)$}
\put(70,0){$k$}
\multiput(10,8.5)(10,0){7}{\line(0,1){3}}
\multiput(20,10)(40,0){2}{\circle*{2}}
\multiput(30,10)(20,0){2}{\line(0,1){10}}
\multiput(30,20)(20,0){2}{\circle*{2}}
\put(40,30){\circle*{2}}
\put(40,10){\line(0,1){20}}
\end{picture}
\begin{picture}(85,50)
\put(0,10){\line(1,0){21}}
\put(28,10){\vector(1,0){52}}
\put(20,5){\vector(0,1){45}}
\put(3,38){$\scriptstyle y(k)$}
\put(70,0){$k$}
\multiput(10,8.5)(10,0){7}{\line(0,1){3}}
\multiput(30,10)(40,0){2}{\circle*{2}}
\multiput(40,20)(20,0){2}{\circle*{2}}
\multiput(40,10)(20,0){2}{\line(0,1){10}}
\put(50,30){\circle*{2}}
\put(50,10){\line(0,1){20}}
\put(21,10){\makebox(0,0)[l]{$\scriptscriptstyle \ldots$}}
\put(30,0){\makebox(0,0)[cb]{$n$}}
\end{picture}
\end{minipage}

\bigskip

\begin{minipage}{4.5cm}
\item Vorw\"artsdifferenz
\end{minipage}%
\begin{minipage}{7cm}
$y = \Delta x: \ y(k) = x(k+1) - x(k)$
\end{minipage}%

\begin{minipage}{4.5cm}
\item R\"uckw\"artsdifferenz
\end{minipage}%
\begin{minipage}{7cm}
$y = \nabla x: \ y(k) = x(k) - x(k-1)$
\end{minipage}%

\begin{minipage}{4.5cm}
\item Summation
\end{minipage}%
\begin{minipage}{7cm}
$y = \sum x:\ y(k) = \sum\limits_{i=-\infty}^k x(i)$
\end{minipage}%

\end{enumerate}


Zweistellige Operationen


\begin{enumerate}
\renewcommand{\labelenumi}{\alph{enumi})}
\begin{minipage}{4.5cm}
\item Signaladdition
\end{minipage}%
\begin{minipage}{7cm}
$y = x_1 + x_2:\ \,y(k) = x_1(k) + x_2(k)$ 
\end{minipage}%

\begin{minipage}{4.5cm}
\item Signalmultiplikation
\end{minipage}%
\begin{minipage}{7cm}
$y = x_1 \cdot x_2:\quad y(k) = x_1(k) \cdot x_2(k)$ 
\end{minipage}%

\begin{minipage}{4.5cm}
\item (diskrete) Faltung
\end{minipage}%
\begin{minipage}{10cm}
$y = x_1 * x_2:\quad y(k) = \sum\limits_{i=-\infty}^{\infty} x_1(i) x_2(k-i) =
\sum\limits_{-\infty}^{\infty} x_1(k-i) x_2(i)$ 
\end{minipage}%

Sonderfall: "`kausale Signale"' ($x_1 = x_2 = 0$ f\"ur $k < 0$): 

$y(k) = (x_1 *
x_2)(k) = \sum\limits_{i=0}^{\infty} x_1(k-i) x_2(i)$
\end{enumerate}

\subsubsection*{Veranschaulichung der Faltung}
\[y(k) = x_1(k)\cdot x_2(0) + x_1(k-1)\cdot x_2(1) + x_1(k-2)\cdot x_2(2) +
\cdots + x_1(1)\cdot x_2(k-1) + x_1(0) \cdot x_2(k)\]

\begin{center}
\hspace*{3.25cm}$\longleftarrow$ \begin{tabular}{|l}\hline $x_1(0)\ x_1(1)\ x_1(2)\ \cdots$ \\ \hline \end{tabular}

\begin{tabular}{l|}\hline $\cdots \ x_2(2)\ x_2(1)\ x_2(0)$ \\ \hline
\end{tabular} $\longrightarrow$ \hspace*{3.25cm}
\end{center}

Für $k=2$: $y(2) = x_1(0)\cdot x_2(2) + x_1(1)\cdot x_2(1) + x_1(2)\cdot x_2(0)$
\begin{center}
\hspace*{1.7cm}\begin{tabular}{|l}\hline $x_1(0)\ x_1(1)\ x_1(2)\ x_1(3)\ \cdots$ \\ \hline \end{tabular}

$\cdot \hspace{25pt} \cdot \hspace{25pt} \cdot\ $

\begin{tabular}{l|}\hline $\cdots \ x_2(3) \ x_2(2)\ x_2(1)\ x_2(0)$ \\ \hline
\end{tabular}\hspace*{1.7cm}
\end{center}

\newpage

\subsection{Statische zeitdiskrete Systeme}
\subsubsection{Elementarsysteme}
\paragraph{Grundbausteine:} (analog zu 4.2.1)

\bigskip

\begin{tabular}{l|l|l|l}
Signalabbildung & Gleichung & Schaltsymbol & Bezeichnung \\
\hline
\raisebox{10pt}{Skalarmultiplikation} & \raisebox{10pt}{$y(k) = \alpha \cdot x(k),\ \alpha \in \mathbb R$} & \begin{picture}(100,30)
\put(-2,9.5){$x(k)$}
\put(20,12.5){\circle{2}}
\put(21,12.5){\line(1,0){19}}
\put(40,0){\line(0,1){25}}
\put(40,0){\line(4,3){16.5}}
\put(43,9.5){$\alpha$}
\put(40,25){\line(4,-3){16.5}}
\put(56.5,12.5){\line(1,0){19}}
\put(76.5,12.5){\circle{2}}
\put(79,9.5){$y(k)$}
\end{picture}
& \raisebox{10pt}{Verstärker}\\
\hline
\raisebox{22pt}{Signaladdition}&\raisebox{22pt}{$y(k) = x_1(k) + x_2(k)$}&
\begin{picture}(80,52)
\put(47.5,25){\circle{15}}
\put(-2,7){$x_2(k)$}
\put(-2,37){$x_1(k)$}
\multiput(25,10)(0,30){2}{\circle{2}}
\multiput(26,10)(0,30){2}{\line(1,0){21.5}}
\put(47.5,10){\vector(0,1){7.5}}
\put(47.5,40){\vector(0,-1){7.5}}
\put(43,22){$+$}
\put(55,25){\line(1,0){20.5}}
\put(76.5,25){\circle{2}}
\put(79,22){$y(k)$}
\end{picture}
& \raisebox{22pt}{Verstärker}\\
\hline
\raisebox{10pt}{Signalmultiplikation}
&
\raisebox{10pt}{$y(k) = x_1(k) + x_2(k)$}
&
\begin{picture}(100,30)
\multiput(25,5)(0,15){2}{\circle{2}}
\multiput(26,5)(0,15){2}{\line(1,0){12}}
\multiput(38,0)(25,0){2}{\line(0,1){25}}
\multiput(38,0)(0,25){2}{\line(1,0){25}}
\put(-2,2){$x_2(k)$}
\put(-2,17){$x_1(k)$}
\put(63,12.5){\line(1,0){12.5}}
\put(76.5,12.5){\circle{2}}
\put(79,9.5){$y(k)$}
\put(46,9.5){$\times$}
\end{picture}
&
\raisebox{10pt}{Multiplizierglied}
\\
\hline
\raisebox{10pt}{Sonderfall}
&
\raisebox{10pt}{$y(k) = [x(k)]^n$}
&
\begin{picture}(100,30)
\put(25,12.5){\circle{2}}
\put(26,12.5){\line(1,0){12}}
\multiput(38,0)(25,0){2}{\line(0,1){25}}
\multiput(38,0)(0,25){2}{\line(1,0){25}}
\put(0,9.5){$x(k)$}
\put(63,12.5){\line(1,0){12.5}}
\put(76.5,12.5){\circle{2}}
\put(79,9.5){$y(k)$}
\put(39,9.5){$(...)^n$}
\end{picture}
&
\raisebox{10pt}{Potenzierglied}
\\
\hline
\end{tabular}

\subsubsection*{Zusammenschaltung(Beispiel)}
\begin{center}
\begin{picture}(300,125)
\put(29,20){\circle{2}}
\put(27,20){\makebox(0,0)[r]{$x_2(k)$}}
\put(30,20){\vector(1,0){20}}
\put(29,112.5){\circle{2}}
\put(27,112.5){\makebox(0,0)[r]{$x_1(k)$}}
\put(30,112.5){\vector(1,0){140}}
\multiput(50,5)(30,0){2}{\line(0,1){30}}
\multiput(50,5)(0,30){2}{\line(1,0){30}}
\put(65,20){\makebox(0,0){$(\ldots)^2$}}
\put(80,20){\vector(1,0){40}}
\multiput(0,0)(0,42.5){2}{
\multiput(120,12.5)(30,0){2}{\line(0,1){30}}
\multiput(120,12.5)(0,30){2}{\line(1,0){30}}
\put(150,27.5){\vector(1,0){20}}}
\multiput(0,0)(0,42.5){3}{
\put(170,12.5){\line(0,1){30}}
\put(170,12.5){\line(3,2){22.5}}
\put(170,42.5){\line(3,-2){22.5}}
}
\put(100,35){\line(0,1){77.5}}
\multiput(100,70)(0,42.5){2}{\circle*{2}}
\put(135,27.5){\makebox(0,0){$\times$}}
\put(135,70){\makebox(0,0){$(\ldots)^3$}}
\multiput(100,35)(0,35){2}{\vector(1,0){20}}
\put(177.5,27.5){\makebox(0,0){$c$}}
\put(177.5,70){\makebox(0,0){$b$}}
\put(177.5,112.5){\makebox(0,0){$a$}}
\multiput(192.5,27.5)(0,85){2}{\line(1,0){75}}
\put(192.5,70){\vector(1,0){30}}
\put(237.5,70){\line(1,0){30}}
\put(230,70){\circle{15}}
\put(230,27.5){\vector(0,1){35}}
\put(230,27.5){\circle*{2}}
\multiput(268.5,27.5)(0,42.5){3}{\circle{2}}
\put(272,27.5){\makebox(0,0)[l]{$y_3(k)$}}
\put(272,70){\makebox(0,0)[l]{$y_2(k)$}}
\put(272,112.5){\makebox(0,0)[l]{$y_1(k)$}}
\put(230,70){\makebox(0,0)[c]{$+$}}
\end{picture}
\end{center}

\begin{align*}
y_1(k) &= a \cdot x_1(k) &= f_1(x_1(k),x_2(k))\\
y_2(k) &= b \cdot {x_1}^3(k) + c \cdot x_1(k) \cdot {x_2}^2(k)&= f_2(x_1(k),x_2(k))\\
y_3(k) &= c \cdot x_1(k) \cdot {x_2}^2(k)&= f_1(x_1(k),x_2(k))
\end{align*}


\begin{minipage}{8cm}
\center
\begin{picture}(150,55)
\put(0,7){$x_2(k)$}
\put(0,37){$x_1(k)$}
\multiput(25,10)(0,30){2}{\circle{2}}
\multiput(26,10)(0,30){2}{\line(1,0){24}}
\multiput(50,0)(60,0){2}{\line(0,1){50}}
\multiput(50,0)(0,50){2}{\line(1,0){60}}
\multiput(110,10)(0,15){3}{\line(1,0){24}}
\multiput(135,10)(0,15){3}{\circle{2}}
\put(138,7){$y_3(k)$}
\put(138,22){$y_2(k)$}
\put(138,37){$y_1(k)$}
\put(71,17){\Huge$\Phi$}
\end{picture}
\end{minipage}
\begin{minipage}{7cm}
\[\Phi:\mathbb R^2 \to \mathbb R^3\]
\end{minipage}

\subsubsection{Alphabetabbildung}

\subsubsection*{Verallgemeinerung}

\begin{picture}(260,80)
\put(3,40){$x(k)\left\{\!\!\begin{array}{c}x_1(k)\\x_2(k)\\\vdots\\x_l(k)\end{array}\right.$}
\put(198,40){$\left.\begin{array}{c}y_1(t)\\y_2(k)\\\vdots\\y_m(k)\end{array}\!\!\right\}
y(k)$}
\put(26,0){
\multiput(0,0)(130,0){2}{\put(40,19){\circle{2}}
\multiput(40,52)(0,13){2}{\circle{2}}}
\multiput(0,0)(109,0){2}{
\multiput(41,52)(0,13){2}{\line(1,0){19}}
\put(41,19){\line(1,0){19}}}
\multiput(60,10)(90,0){2}{\line(0,1){64}}
\multiput(60,10)(0,64){2}{\line(1,0){90}}
\multiput(45,30.3)(116,0){2}{$\vdots$}
\put(95,35){\Huge$\Phi$}}
\end{picture}

\subsubsection*{Definitionen:}
\begin{enumerate}
\renewcommand{\labelenumi}{\alph{enumi})}
\item Eingangsalphabet $X = \mathbb R^l$, Buchstaben (Signalwerte) $x(k) =
\begin{pmatrix}x_1(k)\\\vdots\\x_l(k)\end{pmatrix} \in X$
\item Ausgabealphabet $Y = \mathbb R^m$, Buchstaben (Signalwerte) $y(k) =
\begin{pmatrix}y_1(k)\\\vdots\\y_m(k)\end{pmatrix} \in Y$
\item Alphabetabbildung $\Phi:\mathbb R^l \longrightarrow \mathbb R^m$,
$\Phi(X(k)) = g(k)$

\item Abstraktes statisches System $(X,Y,\Phi)$

\begin{center}
\begin{picture}(160,45)
\put(0,42){$X$}
\put(150,42){$Y$}
\multiput(30,20)(100,0){2}{\oval(60,40)}
\put(35,22){\circle*{3}}
\put(20,12){$x(k)$}
\put(135,15){\circle*{3}}
\put(135,8){$y(k)$}
\qbezier(35,22)(85,40)(134,15)
\put(134,15){\vector(2,-1){0}}
\put(75,35){$\Phi$}
\end{picture}
\end{center}
Beachte Analogie zu zeitkontinuierlichen Systemen!

\item $l$-dimensionales Eingabesignal $x:T\to X$ mit $X = \mathbb R^l$.
\[x = \begin{pmatrix}(\ldots ,\, x_1(-1),\, x_1(0),\, x_1(1),\,
x_1(2),\,\ldots) \\ \vdots  \\(\ldots ,\, x_2(-1),\, x_2(0),\, x_2(1),\,
x_2(2),\,\ldots) \end{pmatrix} = \begin{pmatrix}x_1 \\ \vdots \\
x_l\end{pmatrix}\]

\item $m$-dimensionmales Ausgabesignal: $y:T \to Y$ mit $Y = \mathbb R^m$. ($y
= \ldots$ analog) 

\item Signalräume

$X^* = X^T$ \quad\,Eingabesignalraum (Menge aller Eingabesignale)

$Y^* = Y^T$ \quad\ Ausgabesignalraum (Menge aller Ausgabesignale)

\item Signalabbildung: $\Phi:X^* \longrightarrow Y^*$. 

Für ein statisches System gilt: \quad $\boxed{\quad \vphantom{\Big|} y(k) =
\Phi(x(k))\quad}$
\end{enumerate}

\subsection{Dynamische zeitdiskrete Systeme}
\subsubsection{Zustandsbeschreibung}
Zu den in 6.2.1 betrachteten Elementarsystemen kommt der \textbf{Speicher} $S$:
\begin{center}
\begin{picture}(120,30)
\put(-13,4){$y(k+1)$}
\put(5,20){$x(k)$}
\multiput(29,15)(72,0){2}{\circle{2}}
\multiput(30,15)(50,0){2}{\line(1,0){20}}
\multiput(50,0)(30,0){2}{\line(0,1){30}}
\multiput(50,0)(0,30){2}{\line(1,0){30}}
\put(65,15){\makebox(0,0){$S$}}
\put(105,12){$y(k)$}
\end{picture}
\end{center}

\subsubsection*{Zusammenschaltung (Beispiel):}
\begin{center}
\begin{picture}(400,130)
\put(27,5){\makebox(0,0)[r]{$x_3(k)$}}
\put(29,5){\circle{2}}
\put(30,5){\line(1,0){147.5}}
\put(27,75){\makebox(0,0)[r]{$x_2(k)$}}
\put(29,75){\circle{2}}
\put(30,75){\vector(1,0){20}}
\put(57.5,75){\circle{15}}
\put(57.5,75){\makebox(0,0)[c]{$+$}}
\put(65,75){\line(1,0){35}}
\multiput(0,0)(0,-40){2}{
\multiput(100,60)(30,0){2}{\line(0,1){30}}
\multiput(100,60)(0,30){2}{\line(1,0){30}}}
\put(57.5,35){\vector(0,1){32.5}}
\put(57.5,35){\line(1,0){42.5}}
\put(82.5,78){\makebox(0,0)[bc]{$\scriptstyle z_1(k+1)$}}
\put(115,75){\makebox(0,0){$S$}}
\put(115,35){\makebox(0,0){$\times$}}
\put(130,75){\vector(1,0){40}}
\put(150,78){\makebox(0,0)[bc]{$\scriptstyle z_1(k)$}}
\put(177.5,75){\circle{15}}
\put(177.5,75){\makebox(0,0)[c]{$+$}}
\put(150,75){\circle*{2}}
\put(150,75){\line(0,-1){32.5}}
\put(150,42.5){\vector(-1,0){20}}
\put(170,27.5){\vector(-1,0){40}}
\put(177.5,27.5){\circle{15}}
\put(177.5,27.5){\makebox(0,0)[c]{$+$}}
\put(177.5,5){\vector(0,1){15}}
\multiput(185,75)(70,0){2}{\vector(1,0){40}}
\put(205,78){\makebox(0,0)[bc]{$\scriptstyle z_2(k+1)$}}
\put(275,78){\makebox(0,0)[bc]{$\scriptstyle z_2(k)$}}
\multiput(0,0)(70,0){2}{
\multiput(225,60)(30,0){2}{\line(0,1){30}}
\multiput(225,60)(0,30){2}{\line(1,0){30}}}
\put(240,75){\makebox(0,0){$S$}}
\put(310,75){\makebox(0,0){$(\ldots)^2$}}
\put(325,75){\line(1,0){39}}
\put(365,75){\circle{2}}
\put(345,75){\circle*{2}}
\put(345,75){\line(0,-1){47.5}}
\put(345,27.5){\vector(-1,0){90}}
\put(255,12.5){\line(-3,2){22.5}}
\put(255,12.5){\line(0,1){30}}
\put(255,42.5){\line(-3,-2){22.5}}
\put(245,27.5){\makebox(0,0){$-\frac{1}{10}$}}
\put(232.5,27.5){\vector(-1,0){47.5}}
\put(27,115){\makebox(0,0)[r]{$x_1(k)$}}
\put(29,115){\circle{2}}
\put(30,115){\vector(1,0){20}}
\put(50,100){\line(0,1){30}}
\put(50,100){\line(3,2){22.5}}
\put(50,130){\line(3,-2){22.5}}
\put(58,115){\makebox(0,0){$4$}}
\put(72.5,115){\line(1,0){291.5}}
\put(365,115){\circle{2}}
\put(177.5,115){\circle*{2}}
\put(177.5,115){\vector(0,-1){32.5}}
\put(368,115){\makebox(0,0)[l]{$y_1(k)$}}
\put(368,75){\makebox(0,0)[l]{$y_2(k)$}}
\end{picture}
\end{center}
\begin{align*}
z_1(k+1) &= x_2(k) + z_1(k) \cdot \left[x_3(k) - \frac{1}{10}
{z_2}^2(k)\right]  & y_1(k) &= 4 \cdot x_1(k)\\
z_2(k+1) &= z_1(k) + 4 \cdot x_1(k) & y_2(k) &= {z_2}^2(k)
\end{align*}

Es sei $z(0) = \begin{pmatrix}3\\5\end{pmatrix}$ und $x(k) =
\begin{pmatrix}x_1(k)\\x_2(k)\\x_3(k)\end{pmatrix} = \begin{pmatrix} k+1 \\
-3k^2 \\ 2^k\end{pmatrix} \cdot \mathbbm 1(k)$.

Gesucht: $y(k)$ f\"ur $k \geq 0$

\bigskip

\begin{minipage}{5cm}
$\begin{array}{l|rrr}
k & 0 & 1 & 2 \\
\hline
x_1(k) & 1 & 2 & \cdots \\
x_2(k) & 0 &-3 & \cdots \\
x_3(k) & 1 & 2 & \cdots \\
\hline
z_1(k) & 3 & -4{,}5 & \cdots \\
z_2(k) & 0 & 7 & \cdots \\
\hline
y_1(k) & 4 & 8 & \cdots \\
y_2(k) & 25& 49& \cdots 
\end{array}$
\end{minipage}%
\begin{minipage}{11cm}
\subsubsection*{Verallgemeinerung}
\begin{enumerate}
\renewcommand{\labelenumi}{\alph{enumi})}
\item Zustandsalphabet $Z = \mathbb R^n$

Zustand $z(k) = \begin{pmatrix}z_1(k)\\\vdots\\z_n(k)\end{pmatrix} \in Z$
\item $n$-dimensionales Zustandssignal $z:T \to Z$ mit $Z = \mathbb R^n$
\item Abstraktes dynamisches System $(X,\,Y,\,Z,\,f,\,g)$ mit:
\end{enumerate}
\end{minipage}%

\begin{minipage}{10cm}
\begin{empheq}[innerbox=\fbox]{align*}
\quad f:&Z \times X \longrightarrow Z,  & z(k+1) = f(z(k),x(k))
\vphantom{\Big|}\quad \\
\quad g:&Z \times X \longrightarrow Y,  & y(k) = f(z(k),x(k)) \vphantom{\Big|}
\quad
\end{empheq}
\end{minipage}%
\begin{minipage}{6cm}

\bigskip

Überführungsfunktion

\bigskip

Ergebnisfunktion
\end{minipage}%

\subsubsection{Lineares zeitdiskretes System 1. Ordnung}

\begin{center}
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\end{picture}
\end{center}
\begin{minipage}{10cm}
\begin{align*}
z(k+1) &= f(z(k),\,x(k)) = a \cdot z(k) + b \cdot x(k)\\
y(k) &= g(z(k),\,x(k)) = c \cdot z(k) + d \cdot x(k)
\end{align*}
\end{minipage}%
\hfill
\begin{minipage}{6cm}
$f$ und $g$ sind lineare Funktionen
\end{minipage}


\bigskip

Gegeben: $z(0)$ und $x(k)$ f\"ur $k \leq 0$, gesucht: $y(k)$ f\"ur $k > 0$.

\begin{align*}
k&=0: &  z(1) &= a \cdot z(0) + b \cdot x(0)\\
k&=1: &  z(2) &= a \cdot z(1) + b \cdot x(1) = a^2 \cdot z(0) + ab\cdot x(0) +
b \cdot x(1)\\
k&=2: &  z(3) &= a \cdot z(2) + b \cdot x(2) = a^3 \cdot z(0) + a^2b\cdot x(0)
+ ab\cdot x(1) + b\cdot x(2)
\end{align*}
\[\boxed{\quad z(k) = a^k \cdot z(0) + \sum\limits_{i=0}^{k-1} b \cdot a^{k-i-1}
x(i)\quad}\]

\[y(k) = c \cdot z(k) = d\cdot x(k) = \underbrace{c \cdot a^k\cdot
z(0) \vphantom{\sum\limits_{i=0}^{k-1}}  }_{\text{freie Ausgabe}} + \underbrace{ \sum\limits_{i=0}^{k-1} cb \cdot
a^{k-i-1} x(i) + d\cdot x(k)}_{\text{erzwungene Ausgabe}}\]


\paragraph{Sonderfall:} Autonomes System:  $x(k) = 0$ f\"ur alle $k$, es sei $c
= 1$.
\[\boxed{\quad y(k) = z(k) = a^k \cdot z(0)\quad}\]
$0 < a < 1$: fallende Ausgabe, \quad $a = 1$: konstante Ausgabe, \quad $1 < a < \infty$ : steigende Ausgabe 

\subsubsection{Nichtlineares zeitdiskretes System 1. Ordnung}

\subsubsection*{Autonomes Beispiel-System:}
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\end{center}
\paragraph{Zustandsgleichungen:} $z(k+1) = \lambda \cdot \left[z(k) -
z^2(k)\right]$, \quad $y(k) = z(k)$

\paragraph{Typische Frage:} Existenz von Fixpunkten, d.h. $z(k+1) = z(k)$.
\[z(k) = \lambda \cdot \left[z(k) - z^2(k)\right] \quad \Rightarrow \quad
\begin{array}{ll}z(k) = 0 & \text{(1. Fixpunkt)} \\ z(k) = 1 -
\frac{1}{\lambda} & \text{(2. Fixpunkt)}\end{array}\]

\subsubsection*{Verhaltensformeln f\"ur $\lambda > 0$ und $z(0) =0{,}2$:}

\begin{description}
\item[Fall I:] $0 < \lambda \leq 1$

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\end{picture}
\end{minipage}%
\hfill
\begin{minipage}{6cm}
\center 
$z(k)$ strebt gegen $z=0$ f\"ur $k \to \infty$

\bigskip

(1. Fixpunkt)
\end{minipage}

\item[Fall II:] $1 < \lambda \leq 3$

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\end{picture}
\end{minipage}%
\hfill
\begin{minipage}{6cm}
\center 
$z(k)$ strebt gegen $1-\dfrac{1}{\lambda}$ f\"ur $k \to \infty$

\bigskip

(2. Fixpunkt)
\end{minipage}

\item[Fall III:] $3 < \lambda \leq 1+\sqrt 6$

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\end{picture}
\end{minipage}%
\hfill
\begin{minipage}{6.5cm}
\center 
$z(k)$ pendelt f\"ur $k \to \infty$ zwischen zwei Werten
"`Periodenverdopplung"'.
\end{minipage}

\bigskip

F\"ur $k \to \infty$ gilt also: $z(k+2) = z(k)$.
\[z(k) = \lambda \cdot \left[\lambda(z(k) - z^2(k)) - \lambda^2(z(k) - z^2(k))^2\right]\]
$\Rightarrow 4$ L\"osungen: $z(k) = 0$, \ $z(k) = 1-\frac{1}{\lambda}$, \  $z(k) =
\frac1 2 \left(1+\frac 1 {\lambda}\right)\pm \frac{1}{2\lambda} \sqrt{\lambda^2
- 2\lambda - 3}$

\item[Fall IV:] $1+\sqrt 6 < \lambda \leq \lambda_{\infty} = 3{,}56995\ldots$

F\"ur $k \to \infty$ pendelt $z(k)$ zwischen 4, 8, 16, \ldots (mit steigendem
$\lambda$) Werten.

\item[Fall V:] $\lambda > \lambda_{\infty}$

$z(k)$ zeigt f\"ur $k \to \infty$ "`chaotisches"' Verhalten.
\end{description}
\subsubsection*{Zusammenfassung}
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\newpage

\section{Lineare zeitdiskrete Systeme}
\subsection{Signalbeschreibung im Bildbereich}
\subsubsection{Z-Transformation}

Sei $X^*$ die Menge aller zeitdiskreten Signale. Wir betrachten ${X_c}^* \leq
X^*$. Es gelte $x \in {X_c}^*$, wenn
\begin{enumerate}
\renewcommand{\labelenumi}{\alph{enumi})}
\item f\"ur $k < 0$ gilt: $x(k) = 0$
\item f\"ur $k \geq 0$ gilt: $|x(k)| < M \cdot e^{ck}$
\end{enumerate}

\paragraph{Satz:} F\"ur jedes zeitdiskrete Signal $x \in {X_c}^*$ l\"asst sich
eine Laurent-Reihe wie folgt angeben:
\[X(z) = \sum\limits_{k=0}^{\infty} x(k) \cdot z^{-k} = x(0) + \frac{x(1)}{z} +
\frac{x(2)}{z^2} + \frac{x(3)}{z^3} + \cdots, \qquad z \in \mathbb C\]
Kovergenzgebiet (siehe Funktionentheorie): $\mathbb C_R = \left\{z \ \Big| \
|z| > R = e^c\right\}$  % nichtsagende grafik des konvergenzgebietes gespart

$X(z)$ stellt im Konvergenzgebiet $\mathbb C_R$ eine regul\"are (analytische,
holomorphe) Funktion der komplexen Variablen $z$ dar.


\subsubsection*{Beispiele}

\begin{minipage}{10cm}
1. Impulssignal
\[x(k) = \left\{\begin{array}{ll}1 & k=0\\ 0 & k \ne 0\end{array}\right.\]
\[X(z) = \sum\limits_{k=0}^{\infty} x(k) \cdot z^{-k} = x(0) \cdot z^0 = 1\]
\end{minipage}%
\begin{minipage}{6cm}
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\multiput(40,10)(10,0){4}{\circle*{2}}
\end{picture}
\end{minipage}

\bigskip

\begin{minipage}{10cm}
2. Sprungsignal

\[x(k) = \mathbbm 1(k) = \left\{\begin{array}{ll}1 & k\geq 0 \\ 0 & k<0\end{array}\right.\]
\[X(z) = 1 + \frac 1 z + \frac 1 {z^2} + \cdots = \frac{1}{1-\frac 1
z}=\frac{z}{z-1} \text{ f\"ur }|z|>1\] % geometrische reihe. iss klar.
\end{minipage}%
\begin{minipage}{6cm}
\center
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\put(0,10){\vector(1,0){80}}
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\put(33,38){$\mathbbm 1(k)$}
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\end{picture}
\end{minipage}

\bigskip

\begin{minipage}{10cm}
3. Beispiel

\[x(k) = a^k \cdot \mathbbm 1(k)\]
\[X(z) = 1 + \frac a z + \frac{a^2}{z^2} + \frac{a^3}{z^3}+ \cdots = \frac{1}{1-\frac a z}=\frac{z}{z-a}\] % geometrische reihe. iss klar.
\end{minipage}%
\begin{minipage}{6cm}
\center
\begin{picture}(80,50)
\put(0,10){\vector(1,0){80}}
\put(30,5){\vector(0,1){45}}
\put(70,0){$k$}
\put(30,10){\circle*{2}}
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\put(5,38){$x(k)$}
\end{picture}
\end{minipage}

\bigskip

Weitere Korrespondenzen: $\to$ \"Ubungsheft. 

\paragraph{Frage:} Wie kann $x(k)$ aus $X(z)$ bestimmt werden?
\paragraph{Satz:} Jedes zeitdiskrete Signal $x \in {X_c}^*$ l\"asst sich
als folgendes komplexe Integral darstellen:
\[x(k) = \frac{1}{2\pi j} \oint X(z) z^{k-1}\,dz \qquad (k =
0,\,1,\,2,\,\ldots) \]
\paragraph{Beweis:} $X(z) z^{k-1} = x(0) z^{k-1} + x(1) z^{k-2} + x(2) z^{k-3}
+ \cdots$
\begin{multline*}
x(k) = \frac{1}{2\pi j}\oint\limits_C[X(0) z^{k-1} + x(1) z^{k-2} + x(2)
z^{k-3} + \cdots \\
\cdots + x(k-1)z^0 + x(k)z^{-1} + x(k+1)z^{-2} + \cdots]\,dz
\end{multline*}
Aus Funktionentheorie (Cauchy-Integral-Theorem):
\[\oint \frac{dz}{(z-z_0)^n} = \left\{\begin{array}{ll}2\pi j & n=1\\ 0 & n\ne
1 \end{array}\right.\]
\[x(k) = \frac{1}{2\pi j}\oint\limits_C x(k) z^{-1}\,dz = \frac{1}{2\pi j} x(k)
\cdot \underbrace{\oint\limits_C \frac 1 2\,dz}_{2\pi j} = x(k) \]

\subsubsection*{Zusammenfassung}

\begin{minipage}{8cm}
\center
\begin{picture}(180,45)
\put(-15,0){${X_c}^*$}
\put(160,0){${X_c}^{**}$}
\multiput(30,20)(100,0){2}{\oval(60,40)}
\put(35,22){\circle*{3}}
\put(23,15){$x$}
\put(125,22){\circle*{3}}
\put(130,15){$X$}
\qbezier(35,22)(85,40)(125,22)
\qbezier(35,22)(85,4)(125,22)
\put(125,22){\vector(2,-1){0}}
\put(35,22){\vector(-2,1){0}}
\put(75,35){$\Phi$}
\end{picture}
\end{minipage}%
\begin{minipage}{8cm}
Zu jedem Signal $x \in {X_c}^*$ existiert eine bijektive Abbildung mit den
Zuordnungen $x \to X$ bzw. $X \to x$. 

Diese Abbildung heißt \emph{$z$-Transformation} bzw. \emph{inverse
$z$-Transformation}.
\end{minipage}%

\begin{description}
\item[Symbolik:] $X(z) = \mathcal Z(x(k))$ oder $x(k) \laplace
X(z)$, \quad $x(k) = \mathcal Z^{-1}(X(z))$ oder $X(z) \Laplace x(k)$

\item[Transformationsgleichungen:]%\hspace*{0pt}\\
\begin{align*}
X(z) &= \mathcal Z(x(k)) = \sum\limits_{k=0}^{\infty} x(k) z^{-1}\\
x(k) &= \mathcal Z^{-1}(X(z)) = \frac{1}{2\pi j}\oint\limits_C X(z)z^{k-1}
\end{align*}
\item[Terminologie:]\hspace*{0pt}\\
\begin{tabular}{lp{13cm}}
$x$ & Originalsignal\\
$X$ & Bildsignal oder $z$-Transformierte von $x$\\
${X_c}^*$ & Originalbereich\\
${X_c}^{**}$ & \raggedright Bildbereich, ${X_c}^{**}=\{X \, |\, x(z) = \mathcal
Z(x(k)), x \in {X_c}^*\}$ (Menge aller $z$-Transformationen)
\end{tabular}
\end{description}

\subsubsection{Rechenregeln der z-Transformation}

\textbf{a) Linearit\"at}
\begin{align*}
\mathcal Z(\alpha x_1(k) + \beta x_2(k)) &= \sum\limits_{k=0}^{\infty}(\alpha
x_1(k) + \beta x_2(k)) z^{-k} = \sum\limits_{k=0}^{\infty}\alpha x_1(k) z^{-1}
+ \sum\limits_{k=0}^{\infty}\beta x_2(k) z^{-1} \\
&= \alpha
\sum\limits_{k=0}^{\infty} x_1(k) z^{-1} + \beta
\sum\limits_{k=0}^{\infty}z^{-1} = \alpha X_1(z) + \beta X_2(z)
\end{align*}
$\to z$-Transformation ist eine lineare Transformation

\bigskip

\newpage

\textbf{b) Verschiebungssatz 1 (Rechtsverschiebung)}

\begin{minipage}{8cm}
\begin{align*}
\mathcal Z(y(k)) &= \mathcal Z(x(k-m)) = \sum\limits_{k=0}^{\infty} x(k-m)
z^{-k} \\
& = \sum\limits_{k=m}^{\infty} x(k-m) z^{-k}
\end{align*}
\end{minipage}%
\begin{minipage}{8cm}
\begin{picture}(200,55)
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\multiput(170,20)(10,10){2}{\circle*{2}}
\put(135,0){$m$}
\end{picture}
\end{minipage}

\bigskip

Substitution: $k-m = k'$, $k = k' + m$ 
\[
= \sum\limits_{k'=0}^{\infty} x(k') z^{-(k'+m)} =
z^{-m}\sum\limits_{k'=0}^{\infty} x(k') z^{-k'} = z^{-m}\mathcal Z(x(k)) =
z^{-m} X(z) \qquad
\]
\textbf{Vgl.:} Zeitkontinuierlich: um $\tau$ nach rechts verschoben $\to$
Multiplikation mit $e^{-j\tau}$

\bigskip

\textbf{c) Verschiebungssatz 2 (Linksverschiebung)}

\begin{minipage}{8cm}
\begin{align*}
\mathcal Z(y(k)) &= \mathcal Z(x(k+m)) = \sum\limits_{k=0}^{\infty} x(k+m)
z^{-k} \\
& = \sum\limits_{k'=m}^{\infty} x(k') z^{-(k'-m)}
\end{align*}
\end{minipage}%
\begin{minipage}{8cm}
\begin{picture}(200,55)
\put(0,10){\vector(1,0){80}}
\put(30,5){\vector(0,1){45}}
\put(8,38){$x(k)$}
\put(74,0){$k$}
\put(184,0){$k$}
\put(30,30){\circle*{2}}
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\multiput(60,20)(10,10){2}{\circle*{2}}
% 2. diag.
\put(173,38){$y(k)$}
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\end{picture}
\end{minipage}
\[= z^{m} \left(\vphantom{\sum\limits_{k'}^1}\smash{\underbrace{\sum\limits_{k'=m}^{m-1} x(k') z^{-k'} +
\sum\limits_{k'=0}^{m-1} x(k') z^{k'}}_{X(z)}} - x\right) = z^m \left(X(z) -
\sum\limits_{i=0}^{m-1} x(i) z^{-i}\right)\quad\]

\bigskip

\textbf{d) Vorw\"artsdifferenz}

$\mathcal Z (x(k+1) - x(k)) = \sum\limits_{k=0}^{\infty} (x(k+1) -
x(k)) = \sum\limits_{k=0}^{\infty} x(k+1) z^{-k} -
\underbrace{\sum\limits_{k=0}^{\infty} x(k) z^{-k}}_{X(z)}$

\bigskip

Substitution: $k' = k+1$, $k = k'-1$

\[= \sum\limits_{k'=1}^{\infty} x(k') z^{-k'+1} = z(X(z) - x(0))- X(z) = (z-1)
(X(z) - zx(0))\]

\bigskip

\textbf{e) R\"uckw\"artsdifferenz}
\[\mathcal Z(x(k) - x(k-1)), \quad X(z) = (1-z^{-1})X(z)\]

Weitere Rechenregeln $\to$ Heft S. 58

\subsubsection*{Grenzwerts\"atze}
\begin{enumerate}
\item $x(0) = \lim\limits_{z\to \infty} X(z)$ \quad (vgl. Definition der
$z$-Transformation)
\item Falls $\lim\limits_{k\to \infty} x(k) = A$ existiert, gilt auch
$\lim\limits_{z\to 1} (z-1) X(z) = A$
\end{enumerate}

\newpage

\subsubsection{Inverse z-Transformation}
3 Methoden:
\paragraph{a) Polynomdivision:} Gegeben sei $X(z)$ als rationale Funktion
\[X(z) = \frac{a_0 + a_1 z^{-1} + a_2 z^{-2} + \cdots + a_m z^{-m}}{b_0 + b_1
z^{-1} + b_2 z^{-2} + \cdots + b_n z^{-n}}\]

\[ % Das sieht mal echt scheisse aus.
\begin{array}{llll}
\phantom{-}(a_0 + a_1 z^{-1} + \cdots)  : (b_0 + b_1
z^{-1} + \cdots ) & =
\underbrace{\frac{a_0}{b_0}}_{c_0} + \underbrace{\left(\frac{a_1}{b_0} -
\frac{a_0}{{b_0}^2}b_1\right)}_{c_1} z^{-1} + \cdots \\[-3.5ex]
-(a_0 + \frac{a_0}{b_0}b_1 z^{-1} + \frac{a_0}{b_0}b_2 z^{-2} + \cdots )\\[1ex]
\cline{1-1}
\\[-2ex]
\hspace{1cm}\phantom{-}\left(a_1 - \frac{a_0}{b_0}b_1\right) z^{-1} +\left(a_2 -
\frac{a_0}{b_0}b_1\right) z^{-2}  + \cdots \\[1ex]
\hspace{1.07cm}-\Big(\qquad \cdots \qquad \Big)
\end{array}
\]
\paragraph{Mit z-Transformation:} $X(z) = \sum\limits_{k=0}^{\infty} x(k) \cdot
z^{-k} = x(0) + x(1) z^{-1} + x(2) z^{-2} + \cdots$

Durch Koeffizientenvergleich: $\boxed{\quad \vphantom{\Big|} x(k) = c_k \quad}$

\paragraph{Beispiel:} $\displaystyle X(z) = \frac{z+3}{z^2-4} = \frac{z^{-1} +
3 z^{-2}}{1 - 4 z^{-2}} \quad \Rightarrow \quad 
(z^{-1} + 3 z^{-2}) : (1 - 4 z^{-2}) = z^{-1} + 3z^{-2} + 4 z^{-3}$

\bigskip

$\qquad \Rightarrow \quad x(0) = 0,\ x(1) = 1,\ x(2) = 3,\ x(3) = 4,\ \ldots $

\paragraph{b) Rekursionsformel:} Es sei $b_0 = 1$. Dann gilt die
Rekursionsformel 
\[\boxed{\quad x(k) = c_k = a_k - \sum\limits_{i=1}^{k} b_i \cdot c_{k-i}
\quad}\] %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% XXX
\paragraph{Beispiel:} $X(z) =  \dfrac{z^{-1} + 3 z^{-2}}{1 - 4 z^{-2}}$

\bigskip

$a_0 = 0, a_1 = 1, a_2 = 3, a_i = 0 \quad (i > 2)$, \qquad  $b_0 = 1, b_1 = 0, b_2 = -4, b_j = 0 \quad (j > 2)$

\[\begin{array}{ll}
x(0) = c_0 = a_0 & = 0\\
x(1) = c_1 = a_1 - b_1 \cdot c_0 & = 1\\
x(2) = c_2 = a_2 - b_1 \cdot c_1 - b_2 \cdot c_0 & = 3\\
\qquad \cdots
\end{array}\]

\subsubsection{c) Residuenmethode}
\[\boxed{\quad x(k) = \frac 1 {2\pi j} \oint X(z) \cdot z^{-k} \, dz =
\sum\limits_{z=z_i} \res\left[X(z) \cdot z^{k-1}\right]
\vphantom{\sum\limits^z} \quad}\]
$z_i$ \ldots singul\"are Stellen von $X(z) \cdot z^{k-1}$

\bigskip

Falls $z_i$ ein $m$-facher Pol ist:
\[\res\limits_{z=z_i} \left[X(z) \cdot z^{k-1}\right] = \frac{1}{(m-1)!} \cdot
\lim\limits_{z\to z_i} \frac{d^{(m-1)}}{dz^{m-1}} \left[X(z) \cdot z^{k-1}
\cdot (z-z_i)^m\right]\]

Falls $z_i$ ein einfacher Pol ist ($m = 1$):
\[\res\limits_{z=z_i} \left[X(z) \cdot z^{k-1}\right] = 
\lim\limits_{z\to z_i} \left[X(z) \cdot z^{k-1} \cdot (z-z_i)\right]\]

\paragraph{Beispiel:} $X(z) = \dfrac{z+3}{z^2-4} = \dfrac{z+3}{(z+2)(z-2)}$
\[x(k) = \sum\res\left[X(z) \cdot z^{k-1}\right] = \sum\res
\left[\dfrac{(z+3) z^{k-1}}{(z+2)(z-2)}\right]\] 
\textbf{NB:} $z^{k-1}$ liefert zus\"atzliche Polstelle f\"ur den Fall $k=0$.
\begin{description}
\item[Fall $k=0$:] $x(0) = \sum\limits_{\substack{z=\pm2\\z=0}} \res
\dfrac{z+3}{(z+2)(z-2)z} = \dfrac 5 8 + \dfrac 1 8 - \dfrac 3 4 = 0$

Eventuell einfacher mit Grenzwertsatz: $x(0) = \lim\limits_{z\to \infty} X(z) =
\lim\limits_{z\to \infty} \frac{z+3}{z^2 - 4}= 0$

\item[Fall $k=1,\,2,\,\ldots$:]
\[x(k) = \sum\limits_{\substack{z=\pm2\\z=0}} \res\dfrac{(z+3)z^{k-1}}{(z+2) 
(z-2)} = \frac 5 4 2^{k-1} - \frac 1 4 (-2)^{k-1}\]

\item[Zusammenfassung:]
\[x(k) = \left\{\begin{array}{ll}0 & k=0\\
\frac 5 4 2^{k-1} - \frac 1 4 (-2)^{k-1} & k=1,\,2,\,\ldots\end{array}\right.\]
\end{description}
Vorteil: Geschlossene L\"osung.

\subsection{Systembeschreibung im Zeitbereich}
\subsubsection{Zustandsgleichungen}
\begin{center}
\begin{picture}(260,75)
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\begin{align*}
z_1(k+1) &= f_1[z_1(k),\,\ldots,\,z_n(k),\,x_1(k),\,\ldots,\,x_l(k)] \\
         & = a_{11} z_1(k) + \cdots + a_{1n} z_n(k) + b_{11} x_1(k) + \cdots 
           + b_{1l} x_l(k)\\
& \ \ \vdots \\
z_n(k+1) &= f_n[z_1(k),\,\ldots,\,z_n(k),\,x_1(k),\,\ldots,\,x_l(k)] \\
         & = a_{n1} z_1(k) + \cdots + a_{nn} z_n(k) + b_{n1} x_1(k) + \cdots 
           + b_{nl} x_l(k)\\
\\
y_1(k) &= g_1[z_1(k),\,\ldots,\,z_n(k),\,x_1(k),\,\ldots,\,x_l(k)] \\
         & = c_{11} z_1(k) + \cdots + c_{1n} z_n(k) + d_{11} x_1(k) + \cdots 
           + d_{1l} x_l(k)\\
& \ \ \vdots \\
y_m(k) &= g_m[z_1(k),\,\ldots,\,z_n(k),\,x_1(k),\,\ldots,\,x_l(k)] \\
         & = c_{n1} z_1(k) + \cdots + c_{nn} z_n(k) + d_{n1} x_1(k) + \cdots 
           + d_{nl} x_l(k)
\end{align*}

\subsubsection*{In Matrizenform}
\[z(k) = \begin{pmatrix}z_1(k)\\\vdots\\z_n(k)\end{pmatrix}, \quad x(k) =
\begin{pmatrix}x_1(k)\\\vdots\\x_l(k)\end{pmatrix}, \quad y(k) =
\begin{pmatrix}y_1(k)\\\vdots\\y_m(k)\end{pmatrix}, \quad A = \begin{pmatrix} a_{11} & \cdots & a_{1n} \\ \vdots & \ddots & \vdots \\
a_{n1} & \cdots & a_{nn}\end{pmatrix}\]
\[B = \begin{pmatrix} b_{11} &
\cdots & b_{1n} \\ \vdots & \ddots & \vdots \\
b_{n1} & \cdots & b_{nn}\end{pmatrix}, \quad C = \begin{pmatrix} c_{11} &
\cdots & c_{1n} \\ \vdots & \ddots & \vdots \\
c_{n1} & \cdots & c_{nn}\end{pmatrix}, \quad D = \begin{pmatrix} d_{11} &
\cdots & d_{1n} \\ \vdots & \ddots & \vdots \\
d_{n1} & \cdots & d_{nn}\end{pmatrix}\]


\bigskip

\begin{minipage}{8cm}
\begin{empheq}[innerbox=\fbox]{align*}
z(k+1) &= A \cdot z(k) + B \cdot x(k)\\
y(k) &= C \cdot z(k) + D \cdot x(k)
\end{empheq}
\end{minipage}%
\begin{minipage}{8cm}
\center
Zustandsgleichungen des linearen
 zeitdiskreten Systems
\end{minipage}

\subsubsection{Differenzengleichung und Realisierung}
Im Weiteren gelte $l = m = 1$ (aber nach wie vor $n$ Speicher).

\begin{description}
\item[Satz:] Ein lineares zeitdiskretes dynamisches System (mit $l = m = 1$)
l\"asst sich durch eine \emph{Differenzengleichung} folgenden Typs beschreiben
($a_i, b_i \in \mathbb R, b_0 = 1$):
\begin{multline*}
y(k+n) + b_1 y(k+n -1) + b_2 y(k+n -2) + \cdots + b_{n-1} y(k+1) + b_n y(k) \\
= a_0 x(k+n) + a_1 x(k+n-1) + \cdots + a_{n-1} x(k+1) + a_n x(k)
\end{multline*}
\item[Beweisskizze:]
\[\text{Ansatz (Systemrealisierung)}\  \raisebox{-2pt}{$\stackrel{\scriptstyle
\nearrow}{\scriptstyle \searrow}$}
\begin{array}{l}\text{Differenzengleichung}\\ \text{Zustandsgleichungen}\end{array}\]
\end{description}
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\end{center}
\begin{align}
z_1(k+1) &= z_2(k) + a_1 \cdot x(k) - b_1 \cdot y(k)\\
z_2(k+1) &= z_3(k) + a_2 \cdot x(k) - b_2 \cdot y(k)\\
 &\ \ \vdots \notag\\
z_{n-1}(k+1) &= z_n(k) + a_{n-1} \cdot x(k) - b_{n-1}\cdot y(k) \tag{$n-1$}\\
z_{n}(k+1) &= a_{n} \cdot x(k) - b_{n}\cdot y(k) \tag{$n$}\\
y(k) &= z_1(k) + a_0 \cdot x(k) \tag{$n+1$}
\end{align}

\newpage

\begin{enumerate}
\renewcommand{\labelenumi}{{\Roman{enumi}.}}
\item \"Uberf\"uhrung des Ansatzes in die Differenzengleichung. Elimination von
$z_1,\,\ldots,\, z_n$ wie folgt:
\begin{itemize}
\item[-] in Gleichung $(n+1)$ wird $k$ durch $k+1$ substituiert und Gleichung
$(1)$ eingesetzt
\begin{equation}
y(k+1)= z_2(k) + a_1 \cdot x(k) - b_1 \cdot y(k) + a_0 \cdot x(k+1) \tag{$n+2$}
\end{equation}
\item[-] in Gleichung $(n+2)$ wird $k$ durch $k+1$ ersetzt und Gleichung $(2)$
eingesetzt
\item[-] nach $n$ Schritten sind alle $z_i$ substituiert $\to$
Differenzengleichung
\end{itemize}
\item \"Uberf\"uhrung des Ansatzes in die Zustandsgleichungen: Einsetzen von
Gleichung $(n+1)$ in die Gleichungen $(1)$, $(2)$, $\ldots$, $(n)$:
\"Uberf\"uhrungsgleichung:
\[
\begin{pmatrix}z_1(k+1)\\ z_2(k+1) \\ \vdots \\ z_{n-1}(k+1)\\ z_n(k+1)\end{pmatrix}
=
\begin{pmatrix}-b_1 & 1 & 0 & \cdots & 0\\ -b_2 & 0 & 1 &  \cdots  & 0 \\
\vdots & & \ddots &  &\vdots \\ -b_{n-1} & 0 & 0 & \cdots & 1 \\ -b_n & 0 & 0 & \cdots &
0\end{pmatrix}
\begin{pmatrix}z_1(k)\\ z_2(k) \\ \vdots \\ z_{n-1}(k)\\ z_n(k)\end{pmatrix}
+
\begin{pmatrix}a_1-b_1 \cdot  a_0\\ a_2 - b_2 \cdot a_0 \\ \vdots \\ a{n-1} -
b_{n-1}\cdot a_0 \\ a_n - b_n \cdot a_0\end{pmatrix}
x(k)
\]
\[y(k) = \begin{pmatrix}1 & 0 & 0 & \cdots & 0 & 0 \end{pmatrix}
\begin{pmatrix}z_1(k) \\ z_2(k) \\ \vdots \\ z_n(k)\end{pmatrix} +
\begin{pmatrix}a_0\end{pmatrix} x(k) \qquad\quad \text{Ergebnisgleichung}\]
\end{enumerate}

\subsection{Systembeschreibung im Bildbereich}
\subsubsection{L\"osung der Zustandsgleichungen}
Aus 7.2.1. (mit $X(z) = \mathcal Z\{x(k)\}$, $Y(z) = \mathcal Z\{y(k)\}$,
$Z(z) = \mathcal Z\{z(k)\}$):
\begin{align*}
z(k+1) &= A \cdot z(k) + B \cdot x(k)\\
y(k) &= C \cdot z(k) + D \cdot x(k)
\end{align*}
\paragraph{z-Transformation der 1. Gleichung:} (mit Regel 3, Verschiebungssatz)
\begin{align*}
z \cdot  Z(z) - z \cdot z(0) &= A \cdot  Z(z) + B \cdot X(z)\\
(zE - A) \cdot Z(z) &= z \cdot z(0) + B \cdot X(z) \\
Z(z) &= \underbrace{(zE-A)^{-1}}_{\Phi(z)} \cdot z(0) + (zE - A)^{-1} \cdot B
\cdot X(z)
\end{align*}
$\Phi(z)$ \ldots Fundamentalmatrix im Bildbereich

\paragraph{z-Transformation der 2. Gleichung:}
\begin{align*}
Y(z) &= C \cdot Z(z) + D \cdot X(z)\\
     &= C \cdot \Phi(z) \cdot z(0) + \underbrace{[C \cdot (zE-A)^{-1}\cdot B +
D]}_{G(z)} \cdot  X(z)
\end{align*}
$G(z)$\ldots \"Ubertragungsfunktion im Bildbereich

\begin{minipage}{8cm}
\[\boxed{\quad \vphantom{\Big|} Y(z) = C \cdot \Phi(z) \cdot z(0) + G(z) \cdot X(z) \quad}\]
\end{minipage}%
\begin{minipage}{8cm}
\center
\bigskip
Input-Output Gleichung 
im Bildbereich
\end{minipage}

\paragraph{Inverse z-Transformation:} (Regel 6, Faltungssatz)

\begin{minipage}{8cm}
\[\boxed{\quad \vphantom{\Big|} y(k) = C \cdot \varphi(k) \cdot z(0) +
\sum\limits_{i=0}^{k} g(k-i) \cdot x(i)\quad}\]
\vspace{-24pt}  % Kids, don't try this at home!
\[\quad \hphantom{y(k) =} \underbrace{\hphantom{C \cdot \varphi(k) \cdot
z(0)}}_{\text{freie Ausgabe}} \hphantom{+}
\underbrace{\hphantom{\sum\limits_{i=0}^{k} g(k-i) \cdot
x(i)}}_{\text{erzwungene Ausgabe}}\quad\]

\end{minipage}%
\begin{minipage}{8cm}
\center
Input-Output Gleichung im Zeitbereich
\end{minipage}

\subsubsection*{Anmerkungen}
\begin{enumerate}
\item $\varphi(k)$ heißt \emph{Fundamentalmatrix} im Zeitbereich. Es gilt:
$\varphi(k) = \mathcal Z^{-1}\{\Phi(z)\} = A^k$

\textbf{Beweis:} Es ist zu zeigen, daß $\mathcal Z\{A^k\} = (zE-A)^{-1}z$: 

$\varphi (k) = A^{k}$

$\varphi(k+1) = A^{k+1} = A \cdot A^k = A \cdot \varphi(k)$ 

$\mathcal Z: \quad z \cdot \Phi(z)- z \cdot \varphi(0) =  A \cdot
\Phi(z)$ \quad mit $\varphi(0) = A^0 = E$

$(zE - A) \Phi(z) = z \cdot E \quad \Leftrightarrow \quad \Phi(z) = (zE -
A)^{-1} \cdot  z \cdot \cancel{E}$

\item $g(k)$ heißt \emph{\"Ubertragungsmatrix} im Zeitbereich oder
\emph{Gewichtsmatrix}.
\[g(k) = \mathcal Z^{-1}\{G(z)\} = \left\{\begin{array}{ll}D & k=0\\ CA^{k-1}B &
k=1,\,2,\,3,\,\ldots \end{array}\right.\]
\end{enumerate}

\paragraph{Sonderfall f\"ur $l=m=1$, $z(0) = 0$:} (System mit einem Ein- und
Ausgang im Nullzustand)

\begin{minipage}{6cm}
\center
\begin{picture}(150,40)
\put(27,15){\makebox(0,0)[r]{$X(z)$}}
\multiput(29,15)(72,0){2}{\circle{2}}
\multiput(30,15)(50,0){2}{\line(1,0){20}}
\multiput(50,0)(30,0){2}{\line(0,1){30}}
\multiput(50,0)(0,30){2}{\line(1,0){30}}
\put(104,15){\makebox(0,0)[l]{$Y(z)$}}
\put(65,21){\makebox(0,0)[c]{$G(z)$}}
\put(65,9){\makebox(0,0)[c]{$\scriptstyle z(0)=0$}}
\end{picture}

\begin{picture}(150,40)
\put(27,15){\makebox(0,0)[r]{$x(k)$}}
\multiput(29,15)(72,0){2}{\circle{2}}
\multiput(30,15)(50,0){2}{\line(1,0){20}}
\multiput(50,0)(30,0){2}{\line(0,1){30}}
\multiput(50,0)(0,30){2}{\line(1,0){30}}
\put(104,15){\makebox(0,0)[l]{$y(k)$}}
\put(65,15){\makebox(0,0)[c]{$g(k)$}}
\end{picture}
\end{minipage}%
\begin{minipage}{10cm}
\[\boxed{\quad\vphantom{\Big|} Y(z) = G(z) \cdot X(z)  \quad}\qquad
G(z): \text{\"Ubertragungsfunktion}\] 
\[y(k) = \sum\limits_{i=0}^{k} g(k-i) \cdot x(i) = (g * x)(k)\]
\end{minipage}

\bigskip

\[y(k) = g(k) \cdot x(0) + g(k-1) \cdot x(1) + \cdots + g(0) \cdot x(k)\]
$g(n)$\ldots \ "`Gewichte"' f\"ur Eingabesignal
$\Rightarrow g(k)$ heißt (diskrete) Gewichtsfunktion des Systems

\subsubsection*{Veranschaulichung}
\begin{minipage}{5cm}
\center
\begin{picture}(80,50)
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\put(30,5){\vector(0,1){45}}
\put(8,38){$\delta(k)$}
\put(70,0){$k$}
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\put(30,30){\circle*{2}}
\put(28,30){\makebox(0,0)[r]{$1$}}
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\multiput(10,10)(10,0){2}{\circle*{2}}
\multiput(40,10)(10,0){4}{\circle*{2}}
\end{picture}
\end{minipage}%
\begin{minipage}{6cm}
\center
\begin{picture}(150,40)
\put(57,15){\makebox(0,0)[r]{$x(k)=\delta(k)$}}
\multiput(59,15)(72,0){2}{\circle{2}}
\multiput(60,15)(50,0){2}{\line(1,0){20}}
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\multiput(80,0)(0,30){2}{\line(1,0){30}}
\put(134,15){\makebox(0,0)[l]{$y(k)$}}
\put(95,15){\makebox(0,0)[c]{$g(k)$}}
\end{picture}
\end{minipage}%
\begin{minipage}{5cm}
\center
\begin{picture}(80,50)
\put(0,10){\vector(1,0){80}}
\put(30,5){\vector(0,1){45}}
\put(8,38){$y(k)$}
\put(70,0){$k$}
\put(30,25){\circle*{2}}
\put(40,32){\circle*{2}}
\put(50,22){\circle*{2}}
\put(60,15){\circle*{2}}
\put(70,13){\circle*{2}}
\multiput(10,8.5)(10,0){7}{\line(0,1){3}}
\end{picture}
\end{minipage}
\[y(k) = g(k) \cdot \underbrace{x(0)}_{1}+ g(k-1) \cdot \underbrace{x(1)}_0 +
\cdots + g(0) \cdot \underbrace{x(k)}_0\]

$\Rightarrow y(k) = g(k)$, deshalb heißt die Gewichtsfunktion auch
\emph{Impulsantwort} (Reaktion des Systems auf das Impulssignal $\delta$).

\subsubsection[\"Ubertragungsfunktion (l=m=1)]{\"Ubertragungsfunktion ($l=m=1$)}
Wie kann $G(z)$ berechnet werden?

\begin{enumerate}
\renewcommand{\labelenumi}{\alph{enumi})}
\item aus den Zustandsgleichungen:\quad $G(z) = C(zE-A)^{-1}\cdot B \cdot D$
\item aus Eingabe $X(z)$ und Ausgabe $Y(z)$: $G(z) = \dfrac{Y(z)}{X(z)}$ 
\item aus der Differenzengleichung (nach 7.2.2)
\begin{multline*}
y(k+m) + b_1 \cdot y(k+m-1) + \cdots + b_{m-1} \cdot y(k+1) + b_m \cdot y(k) = \\
a_0 \cdot x(k+n) + a_1 \cdot x(k+n-1) + \cdots + a_{n-1} \cdot x(k+1) + a_n
\cdot x(k)
\end{multline*}Im Weiteren sei $n=m$. Nach $z$-Transformation folgt unter
Beachtung der Voraussetzung $z(0) = 0$ (System im Nullzustand) der
Zusammenhang:
\[G(z) = \dfrac{Y(z)}{X(z)} = \dfrac{a_n \cdot z^{-n} + a_{n-1} \cdot
z^{-(n-1)} + \cdots + a_1 \cdot z^{-1} + a_0}{b_n \cdot z^{-n} + b_{n-1} \cdot
z^{-(n-1)} + \cdots + b_1 \cdot z^{-1} + 1}\]
\end{enumerate}

\paragraph{Beispiel:} Impulskompression

\bigskip

\begin{minipage}{5cm}
gegeben:
\center

\begin{picture}(80,55)
\put(0,10){\vector(1,0){115}}
\put(30,5){\vector(0,1){45}}
\put(8,38){$x(k)$}
\put(105,0){$k$}
\put(28.5,30){\line(1,0){3}}
\put(77,0){$5$}
\put(28,30){\makebox(0,0)[r]{$1$}}
\multiput(10,8.5)(10,0){10}{\line(0,1){3}}
\multiput(10,10)(10,0){2}{\circle*{2}}
\multiput(30,30)(10,0){6}{\circle*{2}}
\multiput(30,10)(10,0){6}{\line(0,1){20}}
\multiput(90,10)(10,0){2}{\circle*{2}}
\end{picture}
\end{minipage}
\begin{minipage}{5cm}
gew\"unscht:
\center

\begin{picture}(80,55)
\put(0,10){\vector(1,0){115}}
\put(30,5){\vector(0,1){45}}
\put(8,38){$x(k)$}
\put(105,0){$k$}
\put(28.5,30){\line(1,0){3}}
\put(28,30){\makebox(0,0)[r]{$3$}}
\multiput(10,8.5)(10,0){10}{\line(0,1){3}}
\put(77,0){$5$}
\multiput(10,10)(10,0){2}{\circle*{2}}
\multiput(30,10)(10,0){4}{\circle*{2}}
\multiput(70,30)(10,0){2}{\circle*{2}}
\multiput(70,10)(10,0){2}{\line(0,1){20}}
\multiput(90,10)(10,0){2}{\circle*{2}}
\end{picture}
\end{minipage}
\begin{minipage}{5cm}
gesucht:

\smallskip

Schaltung, 

\[G(z) = \dfrac{Y(z)}{X(z)}\]
\smallskip
\end{minipage}

\bigskip

\[X(z) = \sum\limits_{k=0}^{\infty} x(k) \cdot z^{-k} = 1 + \frac 1 z +
\frac{1}{z^2} + \frac{1}{z^3}+\frac{1}{z^4} + \frac{1}{z^5}, \quad Y(z) =
\frac{3}{z^4} + \frac{3}{z^5}\]
\[G(z) = \frac{Y(z)}{X(z)} = \frac{3z^{-4} + 3 z^{-5}}{1+z^{-1} + z^{-2} +
z^{-3} + z^{-4} + z^{-5}} = \frac{3z^{-4}}{z^{-4} + z^{-2} + 1}\]
$\Rightarrow$ $a_4 = 3,\ b_4 = 1,\ b_2 = 1,\ b_0 = 1$, die restlichen Glieder
fallen heraus:


\begin{center}
\begin{picture}(450,140)
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\put(50,24){\makebox(0,0)[b]{$\scriptstyle -b_{4}$}}
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\end{center}

Vereinfacht:
\begin{center}
\begin{picture}(400,100)

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\put(70,67.5){\vector(1,0){20}}

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\put(97,26){\makebox(0,0){$-1$}}
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\put(97.5,0){\line(1,0){242.5}}
\put(340,0){\line(0,1){67.5}}

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}
}
\end{picture}
\end{center}
\subsubsection*{Beachte:}
\begin{enumerate}
\item Die Realisierung eines linearen zeitdiskreten Systems wird vollst\"andig
durch die Koeffizienten $a_i$, $b_i$ beschrieben
\item Entwurf eines zeitdiskreten Systems bedeutet Bestimmung dieser
Koeffizienten (Digitalfilterentwurf)
\item Realisierung als Schaltung oder Algorithmus (Regelfall)
\end{enumerate}

\paragraph{Darstellung von $G(z)$:} (Pol-Nullstellen-Plan)
\begin{align*}
G(z) &= \frac{a_nz^{-n} + a_{n-1}z^{-n+1} + \cdots + a_nz^{-1} + a_0}
{b_nz^{-n} + b_{n-1}z^{-n+1} + \cdots + b_nz^{-1} + 1} = \frac{a_0 z^n + a_1
z^{n-1} + \cdots + a_{n-1} z + a_n}{z^n + b_1
z^{n-1} + \cdots + b_{n-1} z + b_n}\\[1ex]
&= a_0 \frac{(z-z_1')(z-z_2') \cdots (z-z_n')}{(z-z_1)(z-z_2) \cdots (z-z_n)}
\end{align*}

\begin{itemize}
\item Pole und Nullstellen entweder auf reeller Achse oder in konjugiert
komplexen Paaren 
\item Pole und Nullstellen k\"onnen auch mehrfach sein
\end{itemize}

\subsubsection*{Beispiel}
\begin{minipage}{7cm}
\quad \begin{picture}(150,100)
\put(0,50){\vector(1,0){150}}
\put(75,0){\vector(0,1){100}}
\qbezier(105,50)(105,62.426408)(96.213204,71.213204)
\qbezier(75,80)(87.426408,80)(96.213204,71.213204)
\qbezier(45,50)(45,37.573592)(53.786796,28.786796)
\qbezier(75,20)(62.573592,20)(53.786796,28.786796)
\qbezier(45,50)(45,62.426408)(53.786796,71.213204)
\qbezier(75,80)(62.573592,80)(53.786796,71.213204)
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\qbezier(75,20)(87.426408,20)(96.213204,28.786796)
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\put(106,40){$1$}
\put(76,82){$1$}
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\put(52,50){\makebox(0,0){$\circ$}}
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\multiput(30,25)(0,50){2}{\makebox(0,0){$\circ$}}
\put(73,90){\makebox(0,0)[r]{$\mathrm{Im}(z)$}}
\put(148,40){\makebox(0,0)[r]{$\mathrm{Re}(z)$}}
\end{picture}
\end{minipage}%
\begin{minipage}{9cm}
Wichtige Systemeigenschaften sind aus dem PN-Plan ablesbar (z.B. Stabilit\"at).
\end{minipage}

\subsubsection{Fl\"uchtiger und station\"arer Vorgang}

\begin{minipage}{8cm}
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\put(95,66){\makebox(0,0){$\Delta t$}}
\put(140,38){$k$}
\put(6,88){$x(k)$}
\end{picture}
\end{minipage}%
\begin{minipage}{8cm}
\center
\begin{picture}(170,40)
\put(22,20){\makebox(0,0)[r]{$x(k)$}}
\put(24,20){\circle{2}}
\put(25,20){\line(1,0){20}}
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\put(126,20){\circle{2}}
\put(129,20){\makebox(0,0)[l]{$y(k)=\ ?$}}
\put(75,20){\makebox(0,0)[c]{$G(z),\ g(z)$}}
\end{picture}
\end{minipage}%
\subsubsection*{Vorbetrachtung:} 
\begin{tabbing} % mmh, warum habe ich das frueher nie benutzt??
zeitkontinuierliches Signal: \=  $x(t) = \widehat X \cdot \cos(\omega t + \varphi)$
\\[1ex]
Abtastfrequenz: \> $\displaystyle f_A = \frac{1}{\Delta t}; \ 
\omega_A = \frac{2\pi}{\Delta t}$\\[1ex]
Abtastzeitpunkte: \> $\displaystyle t_k = k\Delta t = \frac k {f_A}, \quad
(k=0,\,1,\,2,\,\ldots)$\\[1ex]
Abtastwerte: \> $\displaystyle x(t_k) = \widehat X \cdot \cos (\omega t_k + \varphi_k) =
\widehat X \cdot \cos(\omega \frac{k}{f_A} + \varphi_k)$ \\[1ex]
\> $\displaystyle \phantom{x(t_k) }= \widehat X \cdot \cos (\Omega k + \varphi_k) = x(k)$
\end{tabbing}
$\Omega = \dfrac{\omega}{f_A}= \dfrac{2\pi f}{f_A}$ \ldots Normierte
Kreisfrequenz

\newpage

Das Abtasttheorem verlangt $f_A > 2f$, d.h. $\boxed{0 \leq \Omega \leq \pi}$

\subsubsection*{L"osung}
\begin{align*}
x(k) &= \widehat X \cdot \cos(\Omega k + \varphi_x) \cdot \mathbbm 1(k)\\
& = \frac 1 2 \widehat X \left( e^{j\Omega k}\cdot e^{j\varphi_x} + \cdot
e^{-j\Omega k}\cdot e^{-j\varphi_x}\right)\cdot \mathbbm 1(k) && \left(\text{komplexe
Amplituden: }\underline X = \widehat X \cdot e^{\varphi_x}\right)\\
& = \frac 1 2 \left(\underline X \cdot e^{j\Omega k} + \underline X^* \cdot
e^{-j\Omega k}\right) \cdot \mathbbm 1(k)
\end{align*}
$z$-Transformation lt. Korrespondenzentabelle (Zeile 10):
\[X(z) = \frac 1 2 \left(\frac{\underline X z}{z-e^{j\Omega}} +
\frac{\underline X^* z}{z - e^{-j\Omega}}\right)\]
$Y(z) = G(z) \cdot X(z)$ \quad inverse $z$-Transformation:
\begin{align*}
y(k) &= \sum\res \left[Y(z) z^{k-1}\right] = \sum\res \left[\frac 1 2
\left(\frac{\underline X}{z-e^{j\Omega}}+\frac{\underline X^*}{z -
e^{-j\Omega}}\right) \cdot G(z) \cdot z^k\right]\\[1ex]
&= \underbrace{\sum\res\limits_{\begin{subarray}{c}\text{Pole }z_i\\\text{von }
G(z)\end{subarray}} [\cdots]}_{\substack{\text{Fl\"uchtiger Vorgang
}\\y_{fl}(k)}} + \underbrace{\sum\res\limits_{z=e^{\pm j\Omega}}
[\ldots] \vphantom{\res\limits_{\begin{subarray}{c}\text{Pole }z_i\\\text{von }
G(z)\end{subarray}}}    }_{\substack{\text{Station\"arer Vorgang}\\y_{st}(k)}}
\end{align*}
F\"ur den \textbf{fl\"uchtigen Vorgang} $y_{fl}(k)$ gilt:
\[\lim\limits_{k\to\infty} y_{fl}(k) = 0,\quad \text{wenn alle } |z_i| < 1
\qquad \text{(stabiles System, vgl. 7.4.1)}\]
F\"ur den \textbf{station\"aren Vorgang} $y_{st}(k)$ gilt:
\begin{align*}
y_{st}(k) &= \sum\res\limits_{z=e^{\pm j\Omega}}\frac 1 2
\left(\frac{\underline X}{z-e^{j\Omega}}+\frac{\underline X^*}{z -
e^{-j\Omega}}\right) \cdot G(z) \cdot z^k\\
 &= \frac 1 2 \left(\underline X G(e^{j\Omega}) e^{j\Omega k} + \underline X^*
G(e^{-j\Omega}) e^{-\Omega k}\right)\\
&= \mathrm{Re}\left\{\widehat X e^{j\varphi_x} \left|G(e^{j\Omega})\right|
\cdot e^{j\arg(G(e^{j\Omega})} \cdot e^{j\Omega k} \right\}\\
&=\underbrace{\widehat X \left|G(e^{j\Omega})\right|}_{\widehat Y}\cdot
\cos\Big(\Omega k + \underbrace{\varphi_x + \arg
G(e^{j\Omega})}_{\varphi_y}\Big)
\end{align*}
Damit gilt nach hinreichend langer Zeit:

\smallskip

\begin{minipage}{8cm}
\[\boxed{\quad y(k) = \widehat Y \cdot \cos(\Omega k + \varphi_y)
\quad\vphantom{\Big|}}\]
\end{minipage}%
\begin{minipage}{8cm}
mit $\widehat Y = \widehat X \cdot \left|G(e^{j\Omega})\right|$ 
und $\varphi_y = \varphi_x + \arg G(e^{j\Omega})$
\end{minipage}%

\subsubsection{Frequenzcharakteristiken}

$\Omega$ sei nun variabel ($0 \leq \Omega \leq \pi$).
\subsubsection*{Definitionen}
\begin{enumerate}
\item $G(e^{j\Omega})$: Frequenzgang des linearen zeitdiskreten Systems
(Darstellung in der komplexen Ebene hei"st Ortskurve).
\item Amplitudenfrequenzgang: $|G(e^{j\Omega})| = A(\Omega) = \dfrac{\widehat
Y}{\widehat X} = \left.\sqrt{G(z) \cdot G(z^{-1})}\right|_{z=e^{j\Omega}}$
\item Phasenfrequenzgang: $\varphi(\Omega) = \varphi_y - \varphi_x = \arg
G(e^{j\Omega})$
\item 
\begin{tabbing}
$a(\Omega) = - \ln A(\Omega)$ \qquad\qquad \= D\"ampfungsma"s in $\mathrm{Np}$\\
$a(\Omega) = -20\log A(\Omega)$ \> D\"ampfungsma"s in $\mathrm{dB}$\\
$b(\Omega) = - \varphi(\Omega)$ \> Phasenma"s
\end{tabbing}
\end{enumerate}
\paragraph{Beispiel:} PN-Plan, Ortskurve

\[G(z) = \dfrac{z+1{,}5}{z^2  - z + 0{,}5} =
\dfrac{z+1{,}5}{(z-0{,}5(1+j)) (z-0{,}5(1-j))}\]
\[G(e^{j\Omega}) =\frac{e^{j\Omega} + 1{,}5}{e^{2j\Omega} - e^{j\Omega} + 0{,}5}
= \frac{(\cos \Omega + 1{,}5) + j \sin \Omega}{(\cos 2\Omega - \cos \Omega +
0{,}5) + j (\sin 2\Omega - \sin \Omega)}\]

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\end{minipage}

\bigskip 
\bigskip 


Amplitudenfrequenzgang: 
\begin{align*}
A(\Omega) &= |G(e^{j\Omega})| = \left.\sqrt{G(z)\cdot G(z^{-1})}\right|_{z= 
e^{j\Omega}}  = \left.\sqrt{\frac{z+1{,}5}{z^2 - z + 0{,}5} \cdot
\frac{z^{-1}+1{,}5}{z^{-2} - z^{-1} + 0{,}5}}\right|_{z=e^{j\Omega}}\\
 &=
\left.\sqrt{\frac{3{,}25 + 1{,}5 \cdot (z+z^{-1})}{2{,}25 - 1{,}5 (z+z^{-1}) + 
0{,}5 \cdot (z^{2}+z^{-2})}}\right|_{z=e^{j\Omega}} = \sqrt{\frac{3{,}25 + 3
\cdot \cos \Omega}{2{,}25 - 3 \cdot \cos \Omega + \cos 2\Omega}}
\end{align*}
Phasenfrequenzgang: $\varphi(\Omega) = \arg G(e^{j\Omega})$

\bigskip

\begin{minipage}{8cm}
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\end{picture}
\end{minipage}%

\bigskip

Absch\"atzung des Amplitudenfrequenzgangs mit der "`Zeltstangenmethode, entlang
des Einheitskreises auf dem PN-Plan.


\subsubsection*{Einteilung nach $A(\Omega)$:}


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Tiefpass

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\end{minipage}%
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\center
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\put(90,5){$\Omega$}
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\end{picture}

Hochpass

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\end{picture}
\end{minipage}%
\begin{minipage}{4cm}
\center
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\qbezier(95,15)(85,15)(75,40)
\end{picture}

Bandpass

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\center
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Bandsperre

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\end{minipage}%

\subsection{Systemeigenschaften und Klassifizierung}
\subsubsection{Stabilit\"at}
\begin{minipage}{6cm}
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\end{picture}
\end{minipage}%
\begin{minipage}{10cm}
Dieses zeitdiskrete System hei"st stabil (genau: BIBO-Stabil) wenn gilt:
\[|x(k)| \leq K_1 \Rightarrow |y(k)| \leq K2\]
$K_1$, $K_2$ endliche positive Konstanten ($k = 0,\,1,\,\ldots$)
\end{minipage}%
\paragraph{aus 7.3.1:} $\displaystyle y(k) = \sum\limits_{i=0}^{k} g(k-i) x(i) =
\sum\limits_{i=0}^{k} g(i) x(k-i)$

Absch"atzung: 
\[|y(k)| = \left|\sum\limits_{i=0}^{k} g(i) x(k-i)\right| \leq
\sum\limits_{i=0}^{k} \left|g(i)\right| \underbrace{\left|x(k-i)\right|}_{\leq
K_1} \leq K_1 \sum\limits_{i=0}^{K} \left|g(i)\right| \leq
\sum\limits_{i=0}^{\infty} |g(i)| \leq k < \infty\]
\begin{align*}
g(k) &= \mathcal Z^{-1} (G(z)) = \sum\res[G(z) \cdot z^{k-1}] =
\sum\res\limits_{\substack{z=0\\z=z_{\nu}}}  \left[
\frac{a_0z^n + a_1 z^{n-1} + \cdots + a_n}{z^n + b_1 z^{n-1} + \cdots + b_n}
z^{k-1} \right]\\
&= \underbrace{\vphantom{\sum\limits_{\nu}}C_0 \cdot z^0}_{< \infty \vphantom{|}} + \underbrace{\sum\limits_{\nu}
C_{\nu} {z_{\nu}}^k}_{< \infty \text{ f"ur } |z_{\nu}| < 1}
\end{align*}
Daraus folgt: Das System ist stabil, falls f"ur \textbf{alle} Pole $z_{\nu}$
von $G(z)$ gilt: \quad $\boxed{\ |z_{\nu}| < 1\ \vphantom{\Big|}}$ 

Das System ist instabil, falls f"ur wenigstens einen Pol von $G(z)$ gilt:
$|z_{\nu}| > 1$.

\subsubsection*{Beispiel:}
\begin{center}
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\end{center}
\begin{description}
\item["Ubertragungsfunktion:] $G(z) = \dfrac{5z-9}{z^2 + 8z + 12}$
\item[Pole:] $z_1 = -2$, $z_2 = -6$ $\Rightarrow$ System ist instabil, da
$|z_1| = 2 > 1$, $|z_2| = 2 > 1$
\end{description}
\[g(k) = \mathcal Z^{-1}\{G(z)\} = \sum\limits_{\substack{z=-2\\z=-6\\z=0}}
\left[\frac{5z-9}{(z+2)(z+6)} z^{k-1} \right]\]
\begin{description}
\item[$\mathbf{k=0}$:] $g(0) = 0$
\item[$\mathbf{k > 0}$:] $g(k) = \frac{39}{4}(-6)^{k-1} - \frac {19} 4
(-2)^{k-1}$
\end{description}
Mit $v_1 = -0{,}12$, $v_2 = -0{,}8 \quad \Rightarrow \quad G(z) =
\dfrac{5z-9}{z^2 + 0{,}8z + 0{,}12} = \dfrac{5z-9}{(z+0{,}2)(z+0{,}6)}$

\bigskip

$\leadsto$ Pole: $z_1 = -0{,}2$, $z_2 = - 0{,}6 \quad\Rightarrow$  Das System
ist stabil, da $|z_1 = 0{,}2 < 1$, $|z_2| =  0{,}6 < 1$.

\bigskip

$g(k) = \mathcal Z^{-1}\{G(z)\} = \left\{\begin{array}{ll} 0 & k=0\\ 30
(-0{,}6)^{k-1} - 25 (-0{,}2)^{k-1} & k = 1,\,2,\,\ldots\end{array}\right.$ 
\qquad
$\sum\limits_{k=0}^{\infty} |g(k)| < K < \infty$

\subsection{Allpass und Mindestphasensystem}
\paragraph{Definition 1:} Ein zeitdiskretes System, f"ur welches gilt
$|G(e^{j\Omega})| =  \mathrm{konst.},\ (\Omega \in \mathbb R)$
hei"st \emph{Allpass} (AP).

\bigskip

\begin{minipage}{9.5cm}
Allgemeine Form von $G(z)$:
\[G(z) = a_0 \cdot \frac{\cdots (z-{z_i}^{-1}) \cdots
(z-{z_j}^{-1})(z-{\overline z_j}^{-1}) \cdots}{ \cdots (z-{z_i}) \cdots
(z-{z_j})(z-{\overline z_j}) \cdots }\]

\paragraph{Eigenschaft des PN-Planes:} Pole und Nullstellen liegen
spiegelbildlich zu $|z| = 1$, und zwar Pole innerhalb des Eineitskreises $|z| =
1$, Nullstellen au"serhalb.
\end{minipage}%
\begin{minipage}{6.5cm}
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\newpage

\paragraph{Beispiel:} $\displaystyle G(z) = \frac{-9z^{-2} +1}{-\frac 1 9
z^{-2}+1} = \frac{(z+3)(z-3)}{(z+\frac 1 3)(z - \frac 1 3)}$ 

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\textbf{PN-Plan} $\rightarrow$ Allpass

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\begin{align*}
A(\Omega) &= \left|G(e^{j\Omega})\right| = \left.\sqrt{G(z) \cdot
G(z^{-1})}\right|_{z=e^{j\Omega}} = \left.\sqrt{\dfrac{(z^2-9)(z^{-2} -
9)}{(z^2-\frac 1 9)(z^{-2} - \frac 1 9)}}\right|_{z=e^{j\Omega}}\\
&=
\left.\sqrt{\dfrac{82 - 9(z^2 + z^{-2})}{\frac{82}{81} - \frac 1 9 (z^2 +
z^{-2})}}\right|_{z=e^{j\Omega}} = 9 = \text{konstant}
\end{align*}

\paragraph{Definition 2:} Ein stabiles System, dessen "Ubertragungsfunktion $G$
keine Nullstellen $z_i$ mit $|z_i| > 1$ hat, hei"st \emph{Mindestphasensystem}.

\paragraph{Beispiel:} $\displaystyle G(z) = \frac{z^2 - \frac 1 4}{z^2-\frac 1
9} = \frac{(z+\frac 1 2)(z-\frac 1 2)}{(z+\frac 1 3)(z - \frac 1 3)}$ 

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\textbf{PN-Plan} $\rightarrow$ Mindestphasensystem

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\paragraph{Satz:} Die "Ubertragungsfunktion eines beliebigen zeitdiskreten
Systems l"a"st sich wie folgt darstellen

\begin{minipage}{8cm}
\[\boxed{\vphantom{\Big|}\quad G(z) = G_A(z) \cdot G_M(z) \quad}\]
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\bigskip

$G_A(z)$: "Ubertragungsfunktion Allpass, \quad $G_M(z)$: "Ubertragungsfunktion
Mindestphasensystem

\newpage

\paragraph{Beispiel:} $G(z)\! =\! \dfrac{(z-0{,}2)(z-1{,}5)(z^2 - 2z +
2)}{z^3(z+0{,}5)} \!=\! \dfrac{(z-0{,}2)(z-1{,}5)(z-(1+j))(z-(1-j))}{z^3(z+0{,}5)}$

\bigskip

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Bedingungen f"ur Allpass und Mindestphasensystem durch Hinzuf"ugen von
kombinierten Pol- und Nullstellen ($\times\kern-1.5ex\circ$) schaffen:
\begin{itemize}
\item $|1+j| > 1 \ \to \ \frac{1}{1+j}$ \quad Pol-Nullstelle hinzuf"ugen 
\item $|1-j| > 1 \ \to \ \frac{1}{1-j}$ \quad Pol-Nullstelle hinzuf"ugen
\item $|1{,}5| > 1 \ \to \ \frac 2 3$ \hspace*{5.85ex} Pol-Nullstelle hinzuf"ugen
\end{itemize}
\end{minipage}

\bigskip
\bigskip

\begin{minipage}{8cm}%
\center
\textbf{Allpass}

\bigskip

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\center

\textbf{Mindestphasensystem}

\bigskip

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\begin{minipage}{8cm}%
\begin{align*}G_A(z) &= \frac{(z-1{,}5)(z-(1+j))(z-(1+j))}{(z-\frac 2
3)(z-\frac{1}{1+j})(z-\frac{1}{1-j})}\\
 &= \frac{(z-1{,}5)(z^2 - 2z + 2)}{(z-\frac 2 3)(z^2 - z + \frac 1 2)}
\end{align*}
\end{minipage}%
\begin{minipage}{8cm}%
\begin{align*}G_M(z) &= \frac{(z-0{,}2)(z- \frac 2
3)(z-\frac{1}{1+j})(z-\frac{1}{1-j})}{z^3(z+0{,}5)\vphantom{\frac 1 j}}\\
&= \frac{(z-0{,}2)(z-\frac 2 3)(z^2 - z + \frac 1 2)}{z^3
(z+0{,}5)\vphantom{\frac 1 j}}
\end{align*}
\end{minipage}%

\subsubsection{Rekursion und nicht-rekursive Systeme}
aus 7.2.2.: Kanonische Realisierung zeitdiskreter Systeme. Rekursives System
(allgemeiner Fall)
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\newpage

Sonderfall: $b_1 = b_2 = \cdots = b_n = 0 \quad \Rightarrow$ Nichtrekursives
System

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Hier gilt:
\[y(k) = a_0 \cdot x(k) + a_1 x(k-1) + \cdots + a_n x(k-n)\]
\[\Rightarrow y(z) = a_0 X(z) + a_1 z^{-1} X(z) + \cdots + a_n z^{-n}X(z) =
(a_0 + a_1z^{-1} + \cdots + a_n z^{-n}) \cdot X(z)\]
\[G(z) = \frac{Y(z)}{X(z)} = a_0 + a_1 z^{-1} + \cdots + a_n z^{-n} = \frac{a_0
z^n + a_1 z^{n-1} + \cdots a_n}{z^n}\]
Der PN-Plan hat eine $n$-fache Nullstelle bei $z = 0$.

\subsubsection*{Eigenschaften:}
\begin{enumerate}
\renewcommand{\labelenumi}{\alph{enumi})}
\item Nichtrekursive Systeme sind stets stabil
\item Die Impulsantowrt lautet $g = (a_0,\,a_1,\, \ldots ,\, a_n,\, 0,\,0,\,\ldots)$
\begin{itemize}
\item[$\rightarrow$] Die Impulsantwort ist endlich (FIR-Filter, \textbf Finite
\textbf Impulse \textbf Response)
\item[$\rightarrow$] Die Werte der diskreten Impulsantwort entsprechen den
Filterkoeffizienten
\end{itemize}
\end{enumerate}

\subsubsection{Linearphasige Systeme}
Ein nichtrekursives System, bei dem die Nullstellen von $G(z)$ spiegelbildlich
zum Einheitskreis liegen, ist ein \emph{linearphasiges System}.
% Grafik gespart

\subsubsection*{Eigenschaften}
\begin{itemize}
\item Phasenma"s $b(\Omega)$ h"angt linear von $\Omega$ ab
\item Gruppenlaufzeit $T_{gr} = \dfrac{d\,b(\Omega)}{d\Omega}$ ist konstant
\end{itemize}

\subsubsection*{Beispiel:} 
\begin{minipage}{10cm}
\begin{align*}
G(z) &= \dfrac{(z-1)^2(z-\frac 1 2)(z-2)}{z^4}\\
&= \frac{z^4 - 4{,}5z^3 + 7z^2 -
4{,}5z + 1}{z^4}\\
&= \frac 1 {z^2} \cdot \left(z^2 - 4{,}5z + 7 - 4{,}5 z^{-1} + z^{-2}\right)
\end{align*}
\begin{align*}
G(e^{j\Omega}) &= e^{-2j\Omega}\left(e^{2j\Omega} - 4{,}5 e^{j\Omega} + 7 -
4{,}5e^{-j\Omega} + e^{-2j\Omega}\right)\\
&= e^{-2j\Omega} (7-9\cos \Omega + 2 \cdot \cos 2\Omega)
\end{align*}
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\bigskip

Amplitudenfrequenzgang:
\[A(\Omega) = |G(e^{j\Omega})|  = \underbrace{|e^{-2j\Omega}|}_{1} \cdot
\underbrace{|7-9\cos \Omega + 2\cdot \cos 2\Omega|}_{\geq 1} = 7 - 9\cos \Omega
+ 2 \cos 2\Omega\]
Phasenma"s:
\[b(\Omega) = -\arg G(e^{j\Omega}) = -\arg (e^{-2j\Omega}) - \underbrace{\arg(7-
9\cos \Omega + 2\cdot \cos 2\Omega)}_{0} = 2\Omega \]

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\subsection[Zusammenh"ange zwischen zeitkontinuierlichen und zeitdiskreten
Systemen]{Zusammenh"ange zwischen zeitkontinuierlichen und\\zeitdiskreten Systemen}
\subsubsection{Zeitbereich}
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\begin{center}
$\boxed{\quad\displaystyle x(t) = \sum\limits_{k=-\infty}^{\infty} x(k) \cdot \mathrm{si}
\frac{\pi}{\Delta t} (t-k\Delta t)\quad}$
\qquad Sampling-Reihe
\end{center}

Das interpolierte Signal hat eine Grenzfrequenz von $\omega_g =
\frac{\pi}{\Delta t}$. Wegen $\omega_A = \frac{2\pi}{\Delta t}$ gilt: $\omega_g
= \frac{\omega_A}{2}$ oder $\omega_g \cdot \Delta t = \Omega_g = \pi$.

\textbf{Umkehrung:} Bei gegebenem $\omega_g$ ist die Abtastfrequenz
$\boxed{\omega_A \geq 2\omega_g}$ zu w"ahlen (\emph{Abtastbedingung}).
[siehe auch 7.3.3]

\subsubsection{Bildbereich (lineare Systeme)}
\ldots ist wohl nicht pr"ufungsrelevant und kommt daher auch nicht vor den
Semesterferien. 


\end{document}