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\usepackage[bookmarks, colorlinks=false, pdftitle={Mitschrift Elektrotechnik III Prof. Dr.-Ing. Schwarz}, pdfauthor={Fabian Kurz}, pdfsubject={Elektrotechnik}, pdfkeywords={Mathematik Elektrotechnik TU Dresden}, linkbordercolor={1 1 1}]{hyperref}
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\date{Zuletzt aktualisiert:\\\today}
\author{Fabian~Kurz\\\href{http://fkurz.net/}{http://fkurz.net/}}
\title{Elektrotechnik III -- WS 04/05\\Prof. Dr.-Ing. Schwarz, TU Dresden\\Mitschrift}
\begin{document}
\def\thesubsection{\arabic{chapter}.\arabic{section}.\alph{subsection}}
\setcounter{secnumdepth}{3}
\def\thesubsubsection{\arabic{subsubsection}.}
\setlength{\parindent}{0pt}
\maketitle
\pagenumbering{Roman}
\tableofcontents
\newpage
\setcounter{chapter}{-1}
\chapter{Dynamische Netzwerke}
\pagenumbering{arabic}
\section{Problem}
\begin{minipage}{7.9cm}
gegeben:

\center

$RCLM$--Netzwerk

\end{minipage}
\begin{minipage}{7.9cm}
gesucht:

\vspace{0.8cm}
\end{minipage}

\begin{minipage}{7.9cm}
\center 

\begin{picture}(210,100)
\put(35,20){\line(0,1){60}}
\multiput(0,0)(120,0){2}{\multiput(35,20)(0,60){2}{\line(1,0){20}}}
\multiput(35,50)(140,0){2}{\circle{20}}
\put(0,47){$u(t)\Big\downarrow$}
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\put(165,50){\line(1,0){20}}
\put(187,47){$\Big\uparrow i(t)$}
\multiput(75,80)(40,0){2}{\line(1,0){20}}   %  R
\multiput(95,75)(20,0){2}{\line(0,1){10}}
\multiput(95,75)(0,10){2}{\line(1,0){20}}

\multiput(0,0)(0,-45){2}{
\multiput(75,63)(40,0){2}{\line(1,0){20}}                 % L
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\multiput(95,63)(5,0){4}{\qbezier(2.5,5)(5,5)(5,0)}
}

\multiput(75,50)(32,0){2}{\line(1,0){28}}                 % C
\thicklines\multiput(103,43)(4,0){2}{\line(0,1){14}}\thinlines

\multiput(75,33)(40,0){2}{\line(1,0){20}}                 % L
\multiput(95,33)(5,0){4}{\qbezier(0,0)(0,-5)(2.5,-5)}
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\multiput(74,18)(62,0){2}{\circle{2}}
\multiput(74,80)(62,0){2}{\circle{2}}

\end{picture}

\end{minipage}
\begin{minipage}{7.9cm}
Ströme und Spannungen im Netzwerk
\end{minipage}

\begin{description}
\item[Schreibweise:] $u = u(t)$, $i = i(t)$. Kleine Buchstaben bezeichnen zeitveränderliche Größen (Signale).
\end{description}

\begin{center}
\textbf{Netzwerkanalyse}
\end{center}
\begin{minipage}{7.9cm}
\center

Kirchhoffsche Gleichungen

\smallskip

Knotenspannungsanalyse

Maschenstromanalyse
\end{minipage}
\begin{minipage}{7.9cm}
\center

$u$--$i$--Relationen der Zweige
\end{minipage}
\begin{center}    % fies!
$\searrow$ \hspace{4cm} $\swarrow$

\bigskip

\fbox{\quad Netzwerkgleichungen \quad}

\smallskip

gewöhnliche Differentialgleichungen
\end{center}

\section{Klemmenverhalten der Grundschaltelemente}

\begin{minipage}{3.9cm}
\center

\begin{picture}(100,50)
\put(56,40){$R$}
\put(10,30){\vector(1,0){15}}
\put(15,33){$i$}
\multiput(30,30)(40,0){2}{\line(1,0){20}}   %  R
\multiput(50,25)(20,0){2}{\line(0,1){10}}
\multiput(50,25)(0,10){2}{\line(1,0){20}}
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\put(80,20){\vector(3,2){0}}
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\end{picture}
\end{minipage}
\begin{minipage}{3.9cm}
\center

\begin{picture}(100,50)
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\put(10,30){\vector(1,0){15}}
\put(15,33){$i$}
\multiput(30,30)(32,0){2}{\line(1,0){28}}                 % C
\thicklines\multiput(58,23)(4,0){2}{\line(0,1){14}}\thinlines
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\qbezier(40,20)(60,8)(80,20)
\put(80,20){\vector(3,2){0}}
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\end{picture}

\end{minipage}
\begin{minipage}{3.9cm}
\center

\begin{picture}(100,50)
\put(56,40){$L$}
\put(10,30){\vector(1,0){15}}
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\qbezier(40,20)(60,8)(80,20)
\put(80,20){\vector(3,2){0}}
\put(57,5){$u$}
\end{picture}

\end{minipage}
\begin{minipage}{3.9cm}
\center

\begin{picture}(100,50)
\put(75,45){$M$}
\multiput(25,23)(0,14){2}{\vector(-1,0){15}}
\put(14,13){$i_2$}
\put(14,40){$i_1$}
\multiput(30,23)(40,0){2}{\line(1,0){20}}   %  L
\multiput(29,23)(62,0){2}{\circle{2}}
\multiput(50,23)(5,0){4}{\qbezier(0,0)(0,5)(2.5,5)}
\multiput(50,23)(5,0){4}{\qbezier(2.5,5)(5,5)(5,0)}

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\end{picture}

\end{minipage}

\begin{minipage}{3.9cm}
\[u = R \cdot i\]
\end{minipage}
\begin{minipage}{3.9cm}
\[i = C \cdot\displaystyle \frac{du}{dt}\]
\end{minipage}
\begin{minipage}{3.9cm}
\[u = L \cdot \frac{di}{dt}\]
\end{minipage}
\begin{minipage}{3.9cm}
\[u_1 = L_1 \cdot \frac{di_1}{dt} + M \cdot \frac{di_2}{dt}\]
\[u_2 = M \cdot \frac{di_1}{dt} +L_2\cdot \frac{di_2}{dt}\]
\end{minipage}


\section{Netzwerk--Differentialgleichung}

\begin{minipage}{8cm}
\center 

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\end{picture}

\end{minipage}
\begin{minipage}{7.8cm}
Kirchhoffsche Gleichungen

\bigskip

MS: $u_R + u_C - u = 0$

\smallskip

KS: $i_R = i_C = i$
\end{minipage}

\begin{description}
\item[gesucht:] $u_C(t)$
\end{description}

$U$--$I$--Relationen: \quad $u_R = i \cdot R$, \qquad $i_C = C\cdot \displaystyle\frac{du_C}{dt}$

\[i \cdot R + u_C = u  \quad \Rightarrow \quad \underbrace{C\cdot R}_{\tau} \cdot \frac{du_C}{dt} + u_C = u\]
\begin{center}
$\Rightarrow$ \fbox{\quad $\displaystyle \vphantom{\int_a^b} \tau \cdot \frac{du_C}{dt} + u_C = u(t)$\quad } \qquad ($\tau = R\cdot C$, $u_C(0) = U_{C0}$)

\vspace{0.7cm}

\textbf{Spezialfälle für} $u(t)$

\smallskip

$\swarrow$ \hspace{4cm} $\downarrow$ \hspace{4cm} $\searrow$
\end{center}

\begin{minipage}{5.2cm}
\center

\emph{harmonische Funktion}
\end{minipage}
\begin{minipage}{5.2cm}
\center

\emph{periodische Funktion}
\end{minipage}
\begin{minipage}{5.2cm}
\center

\emph{Sprungform}
\end{minipage}


\begin{minipage}{5.2cm}
\[u = \hat{U} \cdot \cos(\omega t + \varphi_u)\]
\end{minipage}
\begin{minipage}{5.2cm}
\[u(t + k\cdot T) = u(t)\]
\[k = 0,\, \pm 1,\, \pm 2,\ldots\]

\smallskip 

\end{minipage}
\begin{minipage}{5.2cm}
\[u(t) = \left\{\begin{array}{ll}0 & t < 0\\U_0 & t \geq 0\end{array}\right.\]
\end{minipage}

\begin{minipage}{5.2cm}
\center
Netzwerke bei harmonischer Erregung
\end{minipage}
\begin{minipage}{5.2cm}
\center
Netzwerk bei periodischer Erregung
\end{minipage}
\begin{minipage}{5.2cm}
\center
Schaltvorgänge
\end{minipage}

\section[Lösung der Netzwerk-Differentialgleichung bei harmonischer Erregung]{Lösung der Netzwerk-Differentialgleichung bei\\harmonischer Erregung}

\[\tau \cdot \frac{du_C}{dt} + u_C = \left\{\begin{array}{ll}0&t < 0\\\hat{U} \cdot \cos(\omega t + \varphi_u) & t \geq 0 \qquad u_C(0) = U_{C0}\end{array}\right.\]

\[u_C(t) = \overbrace{U_{C0} \cdot e^{-\frac{t}{\tau}}}^{\textnormal{\tiny v. Anfangsw. abh.}} + \overbrace{\frac{\hat{U}}{\sqrt{1+(\omega\tau)^2}} \cdot \cos(\varphi_u + \varphi) \cdot e^{-\frac{t}{\tau}} +  \frac{\hat{U}}{\sqrt{1+(\omega\tau)^2}} \cdot \cos(\omega t + \varphi_u + \varphi)}^{\textnormal{\tiny von Erregung abhängig}}\]
$\hspace{2.3cm}\underbrace{\hspace{7.5cm}}_{k \cdot e^{\frac{t}{\tau}} \to 0\textrm{ \tiny für }t\to\infty}\hspace{0.5cm}\underbrace{\hspace{5.1cm}}_{A \cdot \cos(\omega t + \varphi)}$      % ganz ganz ganz ganz ganz ganz böse

$\varphi =\arctan \omega \tau$

\begin{minipage}{7.9cm}
\center

\begin{picture}(150,100)
\put(30,10){\vector(0,1){80}}
\put(25,15){\vector(1,0){125}}
\qbezier(30,75)(40,20)(140,17)
\put(20,82){$u$}
\put(140,5){$t$}
\end{picture}

transiente Lösung
\end{minipage}
\begin{minipage}{7.9cm}
\center

\begin{picture}(150,100)
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\put(25,50){\vector(1,0){125}}
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\qbezier(40,30)(47,30)(50,50)}
\put(20,82){$u$}
\put(140,40){$t$}
\end{picture}

stationäre Lösung
\end{minipage}

\section*{Verhalten stabiler linearer Netzwerke (SLN)}

\begin{itemize}
\item Wird auf ein SLN eine harmonische Erregung geschaltet, so sind nach einem Übergangsvorgang alle Ströme und Spannungen harmonische Zeitfunktionen
\item Ist in einem SLN ein Signal (Strom oder Spannung) harmonisch, so sind alle anderen Signale im stationären Zustand auch harmonisch
\item Interessiert nur der statische Fall, so braucht die Netzwerk-DGL nicht gelöst zu werden (symbolische Analyse)
\end{itemize}

\section{Bestimmung der stationären Lösung}

\begin{minipage}{5cm}
\center

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\put(37,59){$R$}
\put(52,23){$C$}
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\end{minipage}
\begin{minipage}{10.8cm}
\[\tau \frac{du_C}{dt} + u_C = \hat{U} \cdot \cos (\omega t + \varphi_u) \hspace{2cm} R\cdot C = \tau\]
\begin{description}
\item[gesucht:] stationäre Lösung, $u_C(t)$ für große $t$
\end{description}
\end{minipage}

\begin{description}
\item[Prinzipielle Form:] $u_C(t) = \hat{U} \cdot \cos(\omega t + \varphi_{u_C})$. Unbekannte: $\hat{U}$ und $\varphi_{u_C}$
\end{description}

\subsection*{Komplexe Differentialrechnung:}
\[\tau \frac{d\underline u_C}{dt} + \underline u_C = \hat{U} \cdot (\cos (\omega t + \varphi_u) + j \sin (\omega t + \varphi_u)) = \hat{U} \cdot e^{j(\omega t +\varphi_u)}\]
\[u_C(t) = \mathrm{Re}(\underline u_C(t))\]
\begin{center}
\fbox{\quad $\displaystyle \vphantom{\int_a^b} \tau \frac{d\underline u_C}{dt} 
+ \underline u_C = \underbrace{\hat{U}\cdot e^{j\varphi_u}}_{\underline{\hat{U}}} \cdot e^{j\omega t} = \underline{\hat{U}}e^{j\omega t}$ \quad}
\end{center}

\paragraph*{Ansatz:} $\underline{u}_C(t) = \underline{\hat{U}}_C \cdot e^{j\omega t}$ \qquad $ \underline{\hat{U}}_C =  \hat{U}_C \cdot e^{j\varphi_{u_C}}$
\[\frac{d\underline u_C}{dt} = \underbrace{\underline{\hat{U}}_C \cdot e^{j\omega t}}_{\underline u_C} \cdot j\omega = j\omega \cdot \underline u_C \]
einsetzen:
\begin{align*}
\tau \cdot j\omega \cdot \underline{\hat{U}}_C \cdot e^{j\omega t} + \underline{\hat{U}}_C \cdot e^{j\omega t} & = \underline{\hat{U}} \cdot e^{j\omega t} & \big| \cdot e^{-j\omega t}\\
\tau \cdot j\omega \cdot \underline{\hat{U}}_C + \underline{\hat{U}}_C & = \underline{\hat{U}}\\
(j\omega \tau + 1) \cdot \underline{\hat{U}}_C & = \underline{\hat{U}}
\end{align*}
\begin{equation}
\underline{\hat U}_C = \frac{\underline{\hat U}}{1 + j\omega \tau}
\end{equation}

(1) in Ansatz einsetzen:
\begin{align*}
\underline{u}_C(t) & = \underline{\hat U}_C \cdot e^{j\omega t} =  \frac{\underline{\hat U}}{1 + j\omega \tau} \cdot e^{j\omega t}\\
 & =  \frac{{\hat U} \cdot e^{j\varphi_u}}{1 + j\omega \tau} \cdot e^{j\omega t}
\end{align*}

Reeller Anteil:
\[u_C(t) = \mathrm{Re}\, (\underline u_C(t)) = \mathrm{Re}\left(\frac{\hat{U} \cdot e^{j\varphi_C}}{1 + j\omega \tau} \cdot e^{j\omega t}\right)\]

\begin{minipage}{9cm}
\begin{equation*}
\begin{split}
1 + j\omega \tau =& \sqrt{1 + (\omega \tau)^2} \cdot e^{j\varphi}\\
\varphi =& \arctan \frac{\omega \tau}{1}
\end{split}
\end{equation*}
\end{minipage}
\begin{minipage}{6cm}
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\put(3,80){Im}
\qbezier(40,15)(40,20)(36,25.5)
\put(41,21){$\varphi$}
\end{picture}
\end{minipage}

\begin{align*}
u_C(t) & = \mathrm{Re} \left(\frac{\hat{U} \cdot e^{j\varphi_u}}{\sqrt{1 + (\omega \tau)^2} \cdot e^{j\varphi}} \cdot e^{j\omega t}\right) = \mathrm{Re} \left(\frac{\hat{U}}{\sqrt{1 + (\omega \tau)^2}} \cdot e^{j(\omega t + \varphi_u - \varphi)}\right)\\
 & = \frac{\hat{U}}{\sqrt{1 + (\omega \tau)^2}} \cdot \cos(\omega t + \varphi_u - \arctan \omega \tau)
\end{align*}

\section{Deutung von Gleichung (1) aus dem Netzwerk}


\begin{minipage}{7cm}
\center 

\begin{picture}(150,80)
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\end{picture}
\end{minipage}
\begin{minipage}{6cm}
\[\underline{\hat U}_C = \frac{\underline{\hat U}}{1 + j\omega R\cdot C} = \frac{\frac{1}{j\omega C}}{R + \frac{1}{j\omega C}} \cdot \underline{\hat U}\]
\end{minipage}

\bigskip

$\underline{\hat U}_C$ kann aus einem symbolischen Gleichstromnetzwerk berechnet werden. Aufstellen und Lösen der Netzwerk--Differentialgleichung ist nicht erforderlich.

\chapter{Netzwerke bei harmonischer Erregung}
\section{Harmonische Zeitfunktionen}
\begin{center}
\fbox{\quad$\displaystyle \vphantom{\bigg|} y(t) = \hat{Y} \cdot \cos (\omega t + \varphi_y) = \hat{Y} \cdot \cos (\omega(t - t_y))$\quad}

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\multiput(175,49)(0,4){4}{\line(0,1){2}}

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\put(35,140){$y(t)$}
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\put(238,40){$\omega t$}
\put(30,65){$-t_y$}
\put(29,40){$-\varphi_y$}
\put(151,65){$\scriptstyle T\!-t_y$}
\put(146,40){$\scriptstyle 2\pi\!-\varphi_y$}
\put(171,65){$T$}
\put(171,40){$\varphi_y$}
\put(59,3){$\varphi_y = \omega t_y; \quad t_y = \frac{\varphi_y}{\omega} = \frac{\varphi_y}{2\pi}T$}
\end{picture}
\end{center}

\bigskip
\subsection*{Kenngrößen}
\begin{tabular}{ll|ll}
\multicolumn{2}{c}{\textbf{Wertkenngrößen}} \vline& \multicolumn{2}{c}{\textbf{Zeitkenngrößen}}\\
\hline
$\hat Y$ & Spitzenwert, Amplitude & $T$ & Periodendauer, Periode \\
$Y_{SS} = 2 \cdot \hat Y$ & Spitze--Spitze--Wert & $f = \frac 1 T$ &Frequenz\\
$Y = \frac{\hat Y}{\sqrt 2}$ & Effektivwert & $\omega = 2\pi f = \frac{2\pi}{T}$ & Kreisfrequenz\\
 & & $\varphi_y$ & Phasenwinkel, Phase
\end{tabular}

\section{Zeigerdarstellung von harmonischen Funktionen}

Bei gegebener Frequenz ($f$ bzw. $T$ bzw. $\omega$) kann einer harmonischen Zeitfunktion ein Zeiger (Phasor) zugeordnet werden.

\subsection{Komplexe Amplitude}

\begin{minipage}{7.9cm}
\begin{center}
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\setlength{\unitlength}{0.6pt}
\put(0,75){\vector(1,0){250}}
\put(55,0){\vector(0,1){150}}
\qbezier(40,125)(55,125)(70,75)
\qbezier(70,75)(85,25)(100,25)
\qbezier(100,25)(115,25)(130,75)
\qbezier(130,75)(145,125)(160,125)
\qbezier(160,125)(175,125)(190,75)
\multiput(40,74)(0,4){13}{\line(0,1){2}}
\multiput(40,20)(0,4){6}{\line(0,1){2}}
\multiput(160,20)(0,4){26}{\line(0,1){2}}
\put(40,20){\vector(1,0){120}}
\put(160,20){\vector(-1,0){120}}
\put(21,54){$-\frac{\varphi_y}{\omega}$}
\put(53,125){\line(1,0){4}}
\put(60,118){$\hat Y$}
\put(23,138){$y(t)$}
\put(235,60){$t$}
\put(97,4){$T$}
\end{picture}
\end{center}
\end{minipage}
\begin{minipage}{7.9cm}
\center

\begin{picture}(100,100)
\put(50,00){\vector(0,1){100}}
\put(00,50){\vector(1,0){100}}
\put(50,50){\vector(3,2){40}}
\qbezier(70,50)(70,58)(67,61.5)
\put(67,61.5){\vector(-2,3){0}}
\put(88,38){Re}
\put(53,88){Im}
\put(65,67){$\hat Y$}
\put(72,56){$\varphi_y$}
\end{picture}
\end{minipage}
\begin{minipage}{7.9cm}
\bigskip

\[y(t) = \hat Y \cdot \cos(\omega t + \varphi_y)\]
Umkehrung:
\[y(t) = \textrm{Re}\,(\underline{\hat Y} \cdot e^{j\omega t}) = \mathrm{Re}\,(\hat Y \cdot e^{j\varphi_y} \cdot e^{j\omega t})\]
\end{minipage}
\begin{minipage}{7.9cm}
\[\underline{\hat Y} = \hat Y \cdot e^{j\varphi_y} = \hat Y \cdot (\cos \varphi_y + j\sin \varphi_y)\]

\vspace{0.8cm}
\end{minipage}

\subsubsection*{geometrische Deutung von $\mathrm{Re}\,(\underline{\hat Y} \cdot e^{j\omega t})$:}
\begin{center}
\begin{picture}(100,150)
\put(50,0){\vector(0,1){150}}
\put(50,150){\vector(0,-1){150}}
\put(0,100){\vector(1,0){100}}
\put(50,100){\vector(2,1){31.5}}
\multiput(0,0)(-70,0){2}{\multiput(85,10)(0,4){23}{\line(0,1){2}}}

\multiput(81.5,100)(0,4){4}{\line(0,1){2}}
\qbezier(84,110)(83,115.097476)(74.748738,124.748738)
\multiput(49,20)(4,0){8}{\line(1,0){2}}
\put(80,121){$t$}
\put(42,03){$t$}
\put(40,17){$T$}
\put(87,90){\small Re}
\put(54,140){\small Im}
\put(61,115){$\underline{\hat Y}$}
\put(67,104){$\varphi_y$}
\put(74.75,124.75){\vector(-1,1){0}}
\put(0,-3){
\qbezier(85,110)(85,100)(50,90)
\qbezier(50,90)(15,80)(15,70)
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\qbezier(50,50)(85,40)(85,30)
\qbezier(85,30)(85,20)(50,10)
}
\end{picture}
\end{center}

\subsection{Komplexer Effektivwert}

\begin{center}
\fbox{\quad $\displaystyle \underline Y = \frac{\underline{\hat Y}}{\sqrt 2} = \frac{\hat Y}{\sqrt{2}} \cdot e^{j\varphi_y}$\quad} \qquad Definition komplexer Effektivert
\end{center}
Wird nachfolgend verwendet!

\newpage
\section{Lineare Operationen mit harmonischen Funktionen}

\begin{description}
\item[Motivation:] In linearen Differentialgleichungen treten drei lineare Operationen auf: 
\[\tau \underbrace{\cdot \vphantom{\frac a b}}_{\textnormal{\textcircled{\raisebox{-1pt}{1}}}} \underbrace{\frac{dy}{dt}}_{\textnormal{\textcircled{\raisebox{-1pt}{2}}}} \underbrace{+\vphantom{\frac a v}}_{\textnormal{\textcircled{\raisebox{-1pt}{3}}}}y = x\]
\end{description}

\subsubsection{Multiplikation mit konstantem Faktor}

\begin{tabular}{|c|c|}
\hline
\textbf{Zeitbereich} & \textbf{Bildbereich}\\
\hline
$K \cdot A \cdot \cos(\omega + \varphi)$ & $K \cdot \underline A = K \cdot A\cdot e^{j\varphi}$\\
\hline
\begin{picture}(215,100)
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\put(25,45){\vector(1,0){165}}
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\qbezier(145,65)(160,65)(175,45)
\thicklines
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\qbezier(145,80)(160,80)(175,45)
\thinlines
\put(47,82){$y(t)$}
\put(181,35){$t$}
\end{picture} &
\begin{picture}(215,100)
\put(105,0){\vector(0,1){90}}
\put(55,45){\vector(1,0){100}}
\put(130,75){$K \cdot \underline{A}$}
\put(114,62){$\underline{A}$}
\put(140,34){Re}
\put(89,80){Im}
\put(105,45){\vector(3,2){40}}
\thicklines
\put(105,45){\vector(3,2){20}}
\end{picture}\\
\hline
\end{tabular}

\subsubsection{Addition}

\begin{tabular}{|c|c|}
\hline
$A \cdot \cos(\omega t + \varphi_A) + B \cdot \cos(\omega t + \varphi_B)$ & $\underline A + \underline B = A \cdot e^{j\varphi_A} + B \cdot e^{j\varphi_B}$\\
\hline
\begin{picture}(215,100)
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\put(25,45){\vector(1,0){165}}
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}
\thicklines
\thicklines
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\qbezier(145,80)(160,80)(175,45)}
\thinlines
\put(47,82){$y(t)$}
\put(181,35){$t$}
\end{picture} &
\begin{picture}(215,100)
\put(105,0){\vector(0,1){90}}
\put(55,45){\vector(1,0){100}}
\put(120,78){$\underline{A} + \underline B$}
\put(108,69){$\underline{A}$}
\put(124,48){$\underline{B}$}
\put(140,34){Re}
\put(89,80){Im}
\put(105,45){\vector(2,1){20}}
\put(105,45){\vector(1,2){10}}
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\put(125,55){\line(1,2){10}}
\thicklines
\put(105,45){\vector(1,1){30}}
\end{picture}\\
\hline
\end{tabular}

\subsubsection{Differentiation}

\begin{tabular}{|c|c|}
\hline
$\vphantom{\bigg|}\displaystyle\frac{d}{dt}(A  \cos(\omega t + \varphi)) = \omega \cdot A \cos \left(\omega t + \varphi + \frac{\pi}{2}\right) $&$j\omega \cdot \underline A = \omega \cdot A \cdot e^{j(\omega + \pi/2)} $\\
\hline
\begin{picture}(215,100)
\put(45,0){\vector(0,1){90}}
\put(25,45){\vector(1,0){165}}
\put(30,0){
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\thicklines
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\qbezier(145,80)(160,80)(175,45)}
\thinlines
\put(47,82){$y(t)$}
\put(181,35){$t$}
\end{picture} &
\begin{picture}(215,100)
\put(105,0){\vector(0,1){90}}
\put(55,45){\vector(1,0){100}}
\put(125,75){$\underline{A}$}
\put(61,80){$j\omega \underline{A}$}
\put(93,65){$\frac{\pi}{2}$}
\put(140,34){Re}
\put(89,80){Im}
\put(105,45){\vector(1,1){25}}

\thicklines
\put(105,45){\vector(-1,1){30}}
\thinlines

\qbezier(105,60)(111.213204,60)(115.606602,55.606602)
\qbezier(105,60)(98.786796,60)(94.393398,55.606602)
\put(94.4,55.6){\vector(-3,-2){0}}

\thicklines
\end{picture}\\
\hline
\end{tabular}

\newpage
\section{Verhalten der linearen Grundschaltelemente}


\begin{tabular}{ccc}
\begin{picture}(100,50)
\put(56,40){$R$}
\put(10,30){\vector(1,0){15}}
\put(15,33){$i$}
\multiput(30,30)(40,0){2}{\line(1,0){20}}   %  R
\multiput(50,25)(20,0){2}{\line(0,1){10}}
\multiput(50,25)(0,10){2}{\line(1,0){20}}
\multiput(29,30)(62,0){2}{\circle{2}}
\qbezier(40,20)(60,8)(80,20)
\put(80,20){\vector(3,2){0}}
\put(57,5){$u$}
\end{picture}
&
\begin{picture}(100,50)
\put(56,40){$L$}
\put(10,30){\vector(1,0){15}}
\put(15,33){$i$}
\multiput(30,30)(40,0){2}{\line(1,0){20}}   %  L
\multiput(29,30)(62,0){2}{\circle{2}}
\multiput(50,30)(5,0){4}{\qbezier(0,0)(0,5)(2.5,5)}
\multiput(50,30)(5,0){4}{\qbezier(2.5,5)(5,5)(5,0)}
\qbezier(40,20)(60,8)(80,20)
\put(80,20){\vector(3,2){0}}
\put(57,5){$u$}
\end{picture}
&
\begin{picture}(100,50)
\put(56,40){$C$}
\put(10,30){\vector(1,0){15}}
\put(15,33){$i$}
\multiput(30,30)(32,0){2}{\line(1,0){28}}                 % C
\thicklines\multiput(58,23)(4,0){2}{\line(0,1){14}}\thinlines
\multiput(29,30)(62,0){2}{\circle{2}}
\qbezier(40,20)(60,8)(80,20)
\put(80,20){\vector(3,2){0}}
\put(57,5){$u$}
\end{picture}
\\
$u = R \cdot i$ & $u = L \cdot \frac{di}{dt}$ & $i = C \cdot \frac{du}{dt}$\\ \\
 & $i(t) = \hat I \cdot \cos(\omega t + \varphi_i)$\\ \\
$u(t) = R\cdot \hat I \cos (\omega t + \varphi_i)$ & $u(t) =\omega L \cdot\hat I \cos\left(\omega t + \varphi_i + \frac{\pi}{2}\right)$ & $u(t) = \frac{1}{\omega C} \cdot \hat I \cos \left(\omega t + \varphi_i - \frac{\pi}{2}\right)$\\ \\
$\hat U = R \cdot \hat I$ & $\hat U = \omega L \cdot \hat I$ & $\hat U = \frac{1}{\omega C}\cdot \hat I$\\ \\
$\varphi_u = \varphi_i$ & $\varphi_u = \varphi_i + \frac{\pi}{2}$ &  $\varphi_u = \varphi_i - \frac{\pi}{2}$ \\
\\
\begin{picture}(125,100)
\put(20,0){\vector(0,1){90}}
\put(10,45){\vector(1,0){110}}
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\qbezier(50,30)(60,30)(70,45)
\qbezier(70,45)(80,60)(90,60)
\put(93,57){$u$}
\put(93,67){$i$}
\end{picture} 
&
\begin{picture}(125,100)
\put(20,0){\vector(0,1){90}}
\put(10,45){\vector(1,0){110}}
\qbezier(10,70)(20,70)(30,45)
\qbezier(30,45)(40,20)(50,20)
\qbezier(50,20)(60,20)(70,45)
\qbezier(70,45)(80,70)(90,70)
\put(20,0){
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\qbezier(50,60)(60,60)(70,45)
}
\put(93,37){$u$}
\put(93,67){$i$}
\end{picture} 
&
\begin{picture}(125,100)
\put(20,0){\vector(0,1){90}}
\put(10,45){\vector(1,0){110}}
\qbezier(10,70)(20,70)(30,45)
\qbezier(30,45)(40,20)(50,20)
\qbezier(50,20)(60,20)(70,45)
\qbezier(70,45)(80,70)(90,70)
\put(20,0){
\qbezier(-10,45)(0,60)(10,60)
\qbezier(10,60)(20,60)(30,45)
\qbezier(30,45)(40,30)(50,30)
\qbezier(50,30)(60,30)(70,45)
}
\put(93,47){$u$}
\put(93,67){$i$}
\end{picture} \\
$u$ und $i$ \emph{in Phase}  & $u$ eilt $i$ um $\frac{\pi}{2}$ ($90^{\circ}$) \emph{vor} & $u$ eilt $i$ um $\frac{\pi}{2}$ ($90^{\circ}$) \emph{nach}\\
\\
\begin{picture}(100,50)
\put(56,40){$R$}
\put(10,30){\vector(1,0){15}}
\put(13,33){$\underline I$}
\multiput(30,30)(40,0){2}{\line(1,0){20}}   %  R
\multiput(50,25)(20,0){2}{\line(0,1){10}}
\multiput(50,25)(0,10){2}{\line(1,0){20}}
\multiput(29,30)(62,0){2}{\circle{2}}
\qbezier(40,22)(60,10)(80,22)
\put(80,22){\vector(3,2){0}}
\put(57,5){$\underline U$}
\end{picture}
&
\begin{picture}(100,50)
\put(51,40){$j\omega L$}
\put(10,30){\vector(1,0){15}}
\put(13,33){$\underline I$}
\multiput(30,30)(40,0){2}{\line(1,0){20}}   %  L
\multiput(29,30)(62,0){2}{\circle{2}}
\multiput(50,30)(5,0){4}{\qbezier(0,0)(0,5)(2.5,5)}
\multiput(50,30)(5,0){4}{\qbezier(2.5,5)(5,5)(5,0)}
\qbezier(40,22)(60,10)(80,22)
\put(80,22){\vector(3,2){0}}
\put(57,5){$\underline U$}
\end{picture}
&
\begin{picture}(100,50)
\put(50,44){$\frac{1}{j\omega C}$}
\put(10,30){\vector(1,0){15}}
\put(13,33){$\underline I$}
\multiput(30,30)(32,0){2}{\line(1,0){28}}                 % C
\thicklines\multiput(58,23)(4,0){2}{\line(0,1){14}}\thinlines
\multiput(29,30)(62,0){2}{\circle{2}}
\qbezier(40,22)(60,10)(80,22)
\put(80,22){\vector(3,2){0}}
\put(57,5){$\underline U$}
\end{picture}
\\
$\underline U = R\cdot \underline I$ & $\underline U = j\omega L \cdot \underline I$ & $\underline U = \frac{1}{j\omega C}\cdot \underline I$
\\ \\
& $\underline I = I \cdot e^{j\varphi_i}$ \\ \\
$U = R\cdot I$ & $U = \omega L \cdot I$ & $U = \frac{1}{\omega C}\cdot I$
\\ \\
$\varphi_u = \varphi_i$ & $\varphi_u = \varphi_i + \frac{\pi}{2}$ &  $\varphi_u = \varphi_i - \frac{\pi}{2}$ \\
\\
\begin{picture}(105,100)
\put(055,0){\vector(0,1){90}}
\put(0,45){\vector(1,0){110}}
\put(59,60){$\underline{I}$}
\put(71,75){$\underline{U}$}
\put(90,34){Re}
\put(39,80){Im}
\put(055,45){\vector(1,1){30}}
\qbezier(68,45)(68,50.3847768)(64.1923884,54.1923884)
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\put(64.3,54.2){\vector(-1,1){0}}
\put(79.8,69.7){\vector(-1,1){0}}
\put(69,50){$\varphi_i$}
\put(90,59){$\varphi_u$}
\thicklines
\put(055,45){\vector(1,1){15}}
\end{picture}
&
\begin{picture}(105,100)
\put(055,0){\vector(0,1){90}}
\put(0,45){\vector(1,0){110}}
\put(08,78){$\underline{U}$}
\put(71,75){$\underline{I}$}
\put(90,34){Re}
\put(39,80){Im}
\put(055,45){\vector(1,1){30}}
\qbezier(90,45)(90,59.497476)(79.748738,69.748738)

\qbezier(75,45)(75,53.284272)(69.142136,59.142136)
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\put(40.9,59.1){\vector(-1,-1){0}}
\put(79.8,69.7){\vector(-1,1){0}}
\put(39,69){$\varphi_u$}
\put(90,59){$\varphi_i$}
\thicklines
\put(055,45){\vector(-1,1){35}}
\end{picture}
&
\begin{picture}(105,100)
\put(055,0){\vector(0,1){90}}
\put(0,45){\vector(1,0){110}}
\put(92,03){$\underline{U}$}
\put(71,75){$\underline{I}$}
\put(90,34){Re}
\put(39,80){Im}
\put(055,45){\vector(1,1){30}}
\qbezier(90,45)(90,59.497476)(79.748738,69.748738)
\qbezier(75,45)(75,36.715728)(69.142136,30.857864)
\qbezier(75,45)(75,53.284272)(69.142136,59.142136)
\put(69.2,30.85){\vector(-1,-1){0}}
\put(90,45){\vector(0,-1){0}}
\put(74,32){$\varphi_u$}
\put(90,59){$\varphi_i$}
\thicklines
\put(055,45){\vector(1,-1){35}}
\end{picture}
\end{tabular}
\newpage

\section{Netzwerkanalyse bei harmonischer Erregung}
(Symbolische Methode, $j\omega$--Rechnung)

\begin{minipage}{7.8cm}
\begin{description}
\item[gegeben:] \quad \\ Netzwerk, harmonische Erregung
\item[gesucht:] \quad \\ Ströme Spannungen im stationären Fall
\end{description}
\end{minipage}
\begin{minipage}{7.8cm}
\center

Beispiel:

\begin{picture}(170,120)
\put(30,0){\line(1,0){110}}
\put(30,0){\line(0,1){100}}
\put(30,50){\circle{20}}
\put(5,47){$u\Big\downarrow$}
\multiput(30,100)(40,0){2}{\line(1,0){20}}
\multiput(50,95)(0,10){2}{\line(1,0){20}}
\multiput(50,95)(20,0){2}{\line(0,1){10}}
\multiput(90,0)(0,40){3}{\line(0,1){20}}
\multiput(85,20)(10,0){2}{\line(0,1){20}}
\multiput(85,20)(0,20){2}{\line(1,0){10}}
\multiput(90,60)(0,5){4}{
\qbezier(0,0)(5,0)(5,2.5)
\qbezier(5,2.5)(5,5)(0,5)
}
\multiput(140,0)(0,52){2}{\line(0,1){48}}
\put(90,100){\line(1,0){50}}
\put(54,108){$R_1$}
\put(97,27){$R_2$}
\put(97,67){$L$}
\put(150,47){$C$}
\put(146,27){$\downarrow i_2$}
\put(95,47){$\downarrow i_3$}
\put(10,72){$i_1 \uparrow$}
\multiput(90,0)(0,100){2}{\circle*{2}}
\thicklines
\multiput(133,48)(0,4){2}{\line(1,0){14}}
\end{picture}
\end{minipage}

\bigskip

Beispiel: gegeben: $u(t) = \hat U \cdot \cos(\omega t + \varphi_u)$, gesucht: $i_3 = \hat I_3 \cdot \cos(\omega t + \varphi_{i_3})$

\subsection{Lösungsprogramm}

\subsubsection{Transformation des Netzwerkes in den Bildbereich}

\begin{minipage}{4cm}
\center Zeitbereich
\end{minipage}
\begin{minipage}{0.5cm}
\end{minipage}
\begin{minipage}{5cm}
\center Bildbereich
\end{minipage}

\begin{minipage}{4cm}
\begin{picture}(100,60)
\put(56,38){$R$}
\put(10,30){\vector(1,0){15}}
\put(15,33){$i$}
\multiput(30,30)(40,0){2}{\line(1,0){20}}   %  R
\multiput(50,25)(20,0){2}{\line(0,1){10}}
\multiput(50,25)(0,10){2}{\line(1,0){20}}
\multiput(29,30)(62,0){2}{\circle{2}}
\qbezier(40,23)(60,11)(80,23)
\put(80,23){\vector(3,2){0}}
\put(57,5){$u$}
\end{picture}
\end{minipage}
\begin{minipage}{0.5cm}
$\Rightarrow$
\end{minipage}
\begin{minipage}{5cm}
\begin{picture}(100,60)
\put(56,38){$R$}
\put(10,30){\vector(1,0){15}}
\put(13,33){$\underline I$}
\multiput(30,30)(40,0){2}{\line(1,0){20}}   %  R
\multiput(50,25)(20,0){2}{\line(0,1){10}}
\multiput(50,25)(0,10){2}{\line(1,0){20}}
\multiput(29,30)(62,0){2}{\circle{2}}
\qbezier(40,23)(60,11)(80,23)
\put(80,23){\vector(3,2){0}}
\put(57,5){$\underline U$}
\end{picture}
\end{minipage}
\begin{minipage}{4cm}
$\underline U = R \cdot \underline I$
\end{minipage}

\begin{minipage}{4cm}
\begin{picture}(100,50)
\put(56,40){$L$}
\put(10,30){\vector(1,0){15}}
\put(15,33){$i$}
\multiput(30,30)(40,0){2}{\line(1,0){20}}   %  L
\multiput(29,30)(62,0){2}{\circle{2}}
\multiput(50,30)(5,0){4}{\qbezier(0,0)(0,5)(2.5,5)}
\multiput(50,30)(5,0){4}{\qbezier(2.5,5)(5,5)(5,0)}
\qbezier(40,23)(60,11)(80,23)
\put(80,23){\vector(3,2){0}}
\put(57,5){$u$}
\end{picture}
\end{minipage}
\begin{minipage}{0.5cm}
$\Rightarrow$
\end{minipage}
\begin{minipage}{5cm}
\begin{picture}(100,60)
\put(50,38){$j\omega  L$}
\put(10,30){\vector(1,0){15}}
\put(13,33){$\underline I$}
\multiput(30,30)(40,0){2}{\line(1,0){20}}   %  R
\multiput(50,25)(20,0){2}{\line(0,1){10}}
\multiput(50,25)(0,10){2}{\line(1,0){20}}
\multiput(29,30)(62,0){2}{\circle{2}}
\qbezier(40,23)(60,11)(80,23)
\put(80,23){\vector(3,2){0}}
\put(57,5){$\underline U$}
\end{picture}
\end{minipage}
\begin{minipage}{4cm}
$\underline U = j\omega L \cdot \underline I$
\end{minipage}


\begin{minipage}{4cm}
\begin{picture}(100,50)
\put(56,40){$C$}
\put(10,30){\vector(1,0){15}}
\put(15,33){$i$}
\multiput(30,30)(32,0){2}{\line(1,0){28}}                 % C
\thicklines\multiput(58,23)(4,0){2}{\line(0,1){14}}\thinlines
\multiput(29,30)(62,0){2}{\circle{2}}
\qbezier(40,23)(60,11)(80,23)
\put(80,23){\vector(3,2){0}}
\put(57,5){$u$}
\end{picture}
\end{minipage}
\begin{minipage}{0.5cm}
$\Rightarrow$
\end{minipage}
\begin{minipage}{5cm}
\begin{picture}(100,60)
\put(51,42){$\frac{1}{j\omega C}$}
\put(10,30){\vector(1,0){15}}
\put(13,33){$\underline I$}
\multiput(30,30)(40,0){2}{\line(1,0){20}}   %  R
\multiput(50,25)(20,0){2}{\line(0,1){10}}
\multiput(50,25)(0,10){2}{\line(1,0){20}}
\multiput(29,30)(62,0){2}{\circle{2}}
\qbezier(40,23)(60,11)(80,23)
\put(80,23){\vector(3,2){0}}
\put(57,5){$\underline U$}
\end{picture}
\end{minipage}
\begin{minipage}{4cm}
$\underline U = \displaystyle\frac{1}{j\omega C} \cdot \underline I$
\end{minipage}

\paragraph*{Ergebnis:} Gleichstromnetzwerk mit komplexen Größen
\begin{center}
\begin{picture}(170,120)
\put(30,0){\line(1,0){110}}
\put(30,0){\line(0,1){100}}
\put(30,50){\circle{20}}
\put(5,47){$\underline U\Big\downarrow$}
\multiput(30,100)(40,0){2}{\line(1,0){20}}
\multiput(50,95)(0,10){2}{\line(1,0){20}}
\multiput(50,95)(20,0){2}{\line(0,1){10}}
\multiput(90,0)(0,40){3}{\line(0,1){20}}
\multiput(0,0)(0,40){2}{
\multiput(85,20)(10,0){2}{\line(0,1){20}}
\multiput(85,20)(0,20){2}{\line(1,0){10}}
}
\multiput(140,0)(0,60){2}{\line(0,1){40}}
\multiput(135,40)(10,0){2}{\line(0,1){20}}
\multiput(135,40)(0,20){2}{\line(1,0){10}}
\put(90,100){\line(1,0){50}}
\put(54,108){$R_1$}
\put(97,27){$R_2$}
\put(97,67){$j\omega L$}
\put(150,47){$\displaystyle\frac{1}{j\omega C}$}
\put(146,80){$\downarrow \underline I_2$}
\put(95,47){$\downarrow \underline I_3$}
\put(10,72){$\underline I_1 \uparrow$}
\multiput(90,0)(0,100){2}{\circle*{2}}
\put(65,50){\circle{25}}
\put(63,38){\vector(-1,0){0}}
\put(58,47){$M_1$}
\put(58,20){$M_2$}
\put(65,10){\vector(-1,0){0}}
\put(85,50){\oval(75,80)}
\end{picture}
\end{center}

\subsubsection{Berechnen der gesuchten Größen im Bildbereich}

Analyse eines komplexen Gleichstromnetzwerkes (Maschensatz):

\[
\begin{array}{lll}
M_1: &\qquad (R_1 + R_2 + j\omega L) \cdot \underline I_3 + R_1 \cdot \underline I_2 &= \underline U\\
M_2: &\qquad R_1 \cdot \underline I_3 + \left(R_1 + \frac{1}{j\omega C}\right) \cdot \underline I_2 & = \underline U
\end{array}
\]

$\displaystyle \Rightarrow \underline I_3 = \frac{R_1 + \frac{1}{j\omega C} - R_1}{(R_1 + R_2 + j \omega L)\left(R_1 + \frac{1}{j\omega C}\right) -{R_1}^2} \cdot \underline U = \frac{\frac{1}{j\omega C}}{\frac{R_1 + R_2}{j\omega C} + R_1 R_2 + R_1 j\omega L + \frac L C} \cdot \underline U$

\bigskip

$\displaystyle \underline I_3 = I_3 \cdot e^{j\varphi_{i3}} = \frac{\underline A}{\underline B} \cdot \underline U \qquad \qquad \underline A = \frac{1}{j\omega C},\; \underline B = \frac{R_1 + R_2}{j\omega C} + R_1 R_2 + R_1 j\omega L + \frac L C$ 

\bigskip

\bigskip


$\displaystyle I_3 \cdot e^{j\varphi_{i3}} = \frac{A \cdot e^{j\varphi_A}}{B \cdot e^{j\varphi_B}} \cdot U \cdot e^{j\varphi_u} = \underbrace{\frac A B \cdot U}_{I_3} \cdot \underbrace{e^{j(\varphi_u + \varphi_A - \varphi_B)}}_{\varphi_{i_3}}$

\bigskip

$A = \displaystyle \frac{1}{\omega C}, \quad \varphi_A = \frac{1}{j} = -\frac{\pi}{2}$

\smallskip

$B = \sqrt{\left(R_1 R_2 + \frac L C\right)^2 + \left(\omega L R_1 - \frac{R_1 + R_2}{\omega C}\right)^2}, \quad \varphi_B = \arctan \frac{\omega L R_1 - \frac{R_1 + R_2}{\omega  C}}{R_1 R_2 + \frac L C}$

\begin{center}
\boxed{I_3 = \frac A B \cdot U, \qquad  \varphi_{i_3} = \varphi_u + \varphi_A - \varphi_B}
\end{center}

\subsubsection{Rücktransformation in den Zeitbereich}

\[i_3(t) = \text{Re}\,\left(\sqrt 2 \cdot \underline I \cdot e^{j\omega t}\right) = \underbrace{I_3 \cdot \sqrt{2}}_{\hat{I}_3} \cdot \cos (\omega t + \varphi_{i3})\]

\chapter{Komplexe Zweipole}
\section{Komplexer Widerstand und komplexer Leitwert}
\subsection{Lineare Zweipole bei harmonischer Erregung}

\begin{tabular}{|c|c|}
\hline
Zeitbereich $\vphantom{\Big|}$ & Bildbereich \\
\hline
lineares Netzwerk & komplexer Zweipol \\
(ZP aus $R$, $C$, $L$, $M$, gest. Quellen) & \\
\hline
\hspace{1.5cm}
\begin{picture}(100,60)
\put(10,30){\vector(1,0){15}}
\put(10,35){$i(t)$}
\multiput(30,30)(40,0){2}{\line(1,0){20}}   %  R
\multiput(50,25)(20,0){2}{\line(0,1){10}}
\multiput(50,25)(0,10){2}{\line(1,0){20}}
\multiput(29,30)(62,0){2}{\circle{2}}
\qbezier(40,23)(60,11)(80,23)
\put(80,23){\vector(3,2){0}}
\put(53,5){$u(t)$}
\end{picture}
\hspace{1.5cm}
&
\hspace{1.5cm}
\begin{picture}(100,60)
\put(10,30){\vector(1,0){15}}
\put(15,35){$\underline I$}
\multiput(30,30)(40,0){2}{\line(1,0){20}}   %  R
\multiput(50,25)(20,0){2}{\line(0,1){10}}
\multiput(50,25)(0,10){2}{\line(1,0){20}}
\multiput(29,30)(62,0){2}{\circle{2}}
\qbezier(40,23)(60,11)(80,23)
\put(80,23){\vector(3,2){0}}
\put(56,5){$\underline U$}
\end{picture}
\hspace{1.5cm}\ \\                 % fiese scheisse
\hline
$u(t) = \hat U \cdot \cos (\omega t + \varphi_u) \vphantom{\bigg|}$ &
$\underline U = U \cdot e^{j\varphi_u}, \qquad U = \frac{\hat U}{\sqrt 2}$ \\ 
\hline

$i(t) = \hat I \cdot \cos (\omega t + \varphi_i) \vphantom{\bigg|}$ &
$\underline I = I \cdot e^{j\varphi_i}, \qquad I = \frac{\hat I}{\sqrt 2}$\\
\hline
\end{tabular}

\subsection{Definitionen:}

\begin{center}
\boxed{\quad \displaystyle \underline Z = \frac{\phantom{.}\underline{U}\phantom{.}}{\underline I}\quad } \qquad Komplexer Widerstand (Impedanz)
\end{center}

\begin{minipage}{5cm}
\begin{picture}(130,85)
\put(10,15){\vector(1,0){80}}
\put(15,10){\vector(0,1){70}}
\put(15,15){\vector(3,2){60}}
\multiput(75,14)(0,4){10}{\line(0,1){2}}
\multiput(14,55)(4,0){15}{\line(1,0){2}}
\put(3,52){$X$}
\put(70,4){$R$}
\qbezier(43,15)(43,25)(40,32)
\put(40,32){\vector(-1,2){0}}
\put(45,23){$\varphi_Z$}
\put(73,59){$\underline Z$}
\end{picture}
\end{minipage}
\begin{minipage}{10.7cm}
\begin{tabular}{|c|c|}
\hline
Kartesische Koordinaten & Polarkoordinaten \\
\hline
$\underline Z = R + jX \vphantom{\Big|}$ & $\underline Z = Z \cdot e^{j\varphi_z}$\\
\hline
Wirkwiderstand, Resistanz & Scheinwiderstand \\
\hline
$X = \mathrm{Im}\,(\underline Z) = Z \sin \varphi_z  \vphantom{\Big|}$ & $\varphi_z = \arg (\underline Z) = \varphi_u - \varphi_i$\\
\hline
Blindwiderstand, Reaktanz & Phase der Impedanz\\
\hline
\end{tabular}
\end{minipage}


\begin{center}
\boxed{\quad \displaystyle \underline Y = \frac{\phantom{.}\underline{I}\phantom{.}}{\underline U}\quad } \qquad Komplexer Leitwert (Admittanz)
\end{center}

\begin{minipage}{5cm}
\begin{picture}(130,85)
\put(10,15){\vector(1,0){80}}
\put(15,10){\vector(0,1){70}}
\put(15,15){\vector(3,2){60}}
\multiput(75,14)(0,4){10}{\line(0,1){2}}
\multiput(14,55)(4,0){15}{\line(1,0){2}}
\put(3,52){$B$}
\put(70,4){$G$}
\qbezier(43,15)(43,25)(40,32)
\put(40,32){\vector(-1,2){0}}
\put(45,23){$\varphi_Y$}
\put(73,59){$\underline Y$}
\end{picture}
\end{minipage}
\begin{minipage}{10.7cm}
\begin{tabular}{|c|c|}
\hline
Kartesische Koordinaten & Polarkoordinaten \\
\hline
$\underline Y = G + jB \vphantom{\Big|}$ & $\underline Y = Y \cdot e^{j\varphi_y}$\\
\hline
Wirkleitwert, Konduktanz & Scheinleitwert \\
\hline
$Y = \mathrm{Im}\,(\underline Y) = Y \sin \varphi_y  \vphantom{\Big|}$ & $\varphi_y = \arg (\underline Y) = \varphi_i - \varphi_u$\\
\hline
Blindleitwert, Suszeptanz & Phase der Admittanz\\
\hline
\end{tabular}
\end{minipage}

\bigskip

\begin{description}
\item[Wichtig:] Die Bildung des Quotienten ist nur im Bildbereich sinnvoll. Im Zeitbereich sind die Quotienten $\displaystyle \frac{u(t)}{i(t)}$ und $\displaystyle \frac{i(t)}{u(t)}$ zeitabhängige Größen ohne Aussagekraft.
\end{description}

\subsection{Zusammenhänge}

\begin{align*}
\underline Z & = \frac{1}{\underline Y} & \underline Y = \frac{1}{\underline Z}
\end{align*}

\subsubsection*{Kartesische Koordinaten:}
\[R + jX = \frac{1}{G+jB} = \frac{G - jB}{G^2 + B^2} = \frac{G}{G^2 + B^2} - j\frac{B}{G^2 + B^2}\]
\begin{align*}
R & = \frac{G}{G^2 + B^2}  &  X &= -\frac{B}{G^2 + B^2}
\end{align*}

\smallskip

\[G + jB = \frac{1}{R+jX} = \frac{R - jX}{R^2 + X^2} = \frac{R}{R^2 + X^2} - j\frac{X}{R^2 + Y^2}\]
\begin{align*}
G & = \frac{R}{R^2 + X^2}  &  B &= -\frac{X}{R^2 + X^2}
\end{align*}

\subsubsection*{Polarkoordinaten}

\[\underline Z = Z \cdot e^{j\varphi_z} \qquad \underline Y = Y \cdot e^{j\varphi_y}\]

\[
\underline{Z}= \frac{1}{\underline Y} \longrightarrow Z \cdot e^{j\varphi_z} = \frac{1}{Y \cdot e^{j\varphi_y}} = \frac{1}{Y} \cdot e^{j\varphi_y} \quad \Rightarrow \quad Z = \frac{1}{Y}, \quad \varphi_z = -\varphi_y
\]


\newpage

\subsection*{Impedanz und Admittanz der Blindschaltelemente}
\addcontentsline{toc}{subsection}{Impedanz und Admittanz der Blindschaltelemente}

\begin{tabular}{|l|c|c|}
\hline
\begin{picture}(100,60)
\put(0,27){Schaltelement}
\end{picture}
 & 
\begin{picture}(100,60)
\put(56,40){$L$}
\put(10,30){\vector(1,0){15}}
\put(15,33){$i$}
\multiput(30,30)(40,0){2}{\line(1,0){20}}   %  L
\multiput(29,30)(62,0){2}{\circle{2}}
\multiput(50,30)(5,0){4}{\qbezier(0,0)(0,5)(2.5,5)}
\multiput(50,30)(5,0){4}{\qbezier(2.5,5)(5,5)(5,0)}
\qbezier(40,23)(60,11)(80,23)
\put(80,23){\vector(3,2){0}}
\put(57,5){$u$}
\end{picture}
&
\begin{picture}(100,60)
\put(56,40){$C$}
\put(10,30){\vector(1,0){15}}
\put(15,33){$i$}
\multiput(30,30)(32,0){2}{\line(1,0){28}}                 % C
\thicklines\multiput(58,23)(4,0){2}{\line(0,1){14}}\thinlines
\multiput(29,30)(62,0){2}{\circle{2}}
\qbezier(40,23)(60,11)(80,23)
\put(80,23){\vector(3,2){0}}
\put(57,5){$u$}
\end{picture}
\\
\hline
Impedanz & $\vphantom{\bigg|}\underline Z = j\omega L$ & $\underline Z = \dfrac{1}{j\omega C} = - j\dfrac{1}{\omega C}$\\
\hline
Betrag & $\vphantom{\bigg|}Z = \omega L$ & $Z = \dfrac{1}{\omega C}$\\
\hline
Phase & $\vphantom{\bigg|}\varphi_Z = \dfrac{\pi}{2}$ & $\varphi_Z = -\dfrac{\pi}{2}$ \\
\hline
Admittanz & $\vphantom{\bigg|}\underline Y = \dfrac{1}{j\omega L} = - j \dfrac{1}{\omega L}$ & $\underline Y = j \omega C$\\
\hline
Betrag & $\vphantom{\bigg|} Y = \dfrac{1}{\omega L}$ & $Y =  \omega C$\\
\hline
Phase & $\vphantom{\bigg|}\varphi_Y = -\dfrac{\pi}{2}$ & $\varphi_Y = \dfrac{\pi}{2}$ \\
\hline
\begin{picture}(100,105)
\put(0,49.5){Zeigerbilder}
\end{picture} &
\begin{picture}(150,105)
\put(75,10){\vector(0,1){80}}
\put(30,50){\vector(1,0){90}}
\put(58,81){Re}
\put(105,39){Im}
\put(83,70){$\underline Z$}
\put(83,22){$\underline Y$}
\thicklines
\put(75,50){\vector(0,1){30}}
\put(75,50){\vector(0,-1){30}}
\end{picture}
&
\begin{picture}(150,105)
\put(75,10){\vector(0,1){80}}
\put(30,50){\vector(1,0){90}}
\put(58,81){Re}
\put(105,39){Im}
\put(83,70){$\underline Y$}
\put(83,22){$\underline Z$}
\thicklines
\put(75,50){\vector(0,1){30}}
\put(75,50){\vector(0,-1){30}}
\end{picture}
\\
\hline
\begin{picture}(100,105)
\put(0,55.5){Frequenzgang des}
\put(0,43.5){Betrages}
\end{picture} &
\begin{picture}(150,105)
\put(35,10){\vector(0,1){80}}
\put(30,15){\vector(1,0){90}}
\put(108,7){$\omega$}
\put(92,25){$Y = \frac{1}{\omega L}$}
\put(80,80){$Z = {\omega L}$}
\thicklines
\put(35,15){\line(1,1){60}}
\qbezier(38,85)(40,20)(105,20)
\end{picture}
&
\begin{picture}(150,105)
\put(35,10){\vector(0,1){80}}
\put(30,15){\vector(1,0){90}}
\put(108,7){$\omega$}
\put(92,25){$Z = \frac{1}{\omega C}$}
\put(80,80){$Y = {\omega C}$}
\thicklines
\put(35,15){\line(1,1){60}}
\qbezier(38,85)(40,20)(105,20)
\end{picture}
\\
\hline
\begin{picture}(100,105)
\put(0,63.5){Frequenzgang in}
\put(0,49.5){doppeltlogarithm.}
\put(0,37.5){Darstellung}

\end{picture} &
\begin{picture}(150,105)
\put(35,10){\vector(0,1){80}}
\put(30,15){\vector(1,0){90}}
\put(76,4){$\log(\omega/\omega_0)$}
\put(92,25){$\log(Y/Y_0)$}
\put(80,80){$\log(Z/Z_0)$}
\thicklines
\put(35,15){\line(1,1){60}}
\put(35,75){\line(1,-1){60}}
\end{picture}
&
\begin{picture}(150,105)
\put(35,10){\vector(0,1){80}}
\put(30,15){\vector(1,0){90}}
\put(76,4){$\log(\omega/\omega_0)$}
\put(92,25){$\log(Z/Z_0)$}
\put(80,80){$\log(Y/Y_0)$}
\thicklines
\put(35,15){\line(1,1){60}}
\put(35,75){\line(1,-1){60}}
\end{picture}
\\
\hline
\end{tabular}

\subsection{Bestimmung von Impedanz und Admittanz}

\subsubsection{Aus den Zeitverläufen von $u$ und $i$}

\[\underline Z = Z \cdot e^{j\varphi} = \frac{\underline U}{\underline I} = \frac{U  \cdot e^{j\varphi_u}}{I \cdot e^{j\varphi_i}} = \frac{U}{I} \cdot e^{j(\varphi_u - \varphi_i)} \]
\[\Rightarrow  Z = \dfrac{U}{I}, \quad \varphi = \varphi_u - \varphi_i\]

\paragraph{Betrag (Scheinleitwert, Scheinwiderstand):} Durch Effektivwertmessung

\begin{minipage}{5cm}
\begin{picture}(120,80)
\put(0,27){$u(t)\!\Big\downarrow$}
\put(35,0){\line(0,1){60}}
\put(35,30){\circle{20}}
\put(35,0){\line(1,0){90}}
\put(35,60){\line(1,0){20}}
\put(65,60){\circle{20}}
\put(75,60){\line(1,0){50}}
\multiput(0,0)(30,0){2}{\multiput(95,0)(0,40){2}{\line(0,1){20}}}
\put(95,30){\circle{20}}
\multiput(120,20)(0,20){2}{\line(1,0){10}}
\multiput(120,20)(10,0){2}{\line(0,1){20}}
\put(61,56.5){A}
\put(91,26){V}
\multiput(95,0)(0,60){2}{\circle*{2}}
\put(133,32){dyn.}
\put(133,20){NW}
\end{picture}
\end{minipage}
\begin{minipage}{08cm}
\[Z = \frac{U}{I}, \quad Y = \frac I U \]
\end{minipage}

\paragraph{Phase:} Durch Zeitmessung (Oszilloskop)

\begin{minipage}{5cm}
\begin{picture}(140,120)
\put(15,10){\vector(0,1){100}}
\multiput(10,25)(0,35){2}{\vector(1,0){120}}
\qbezier(15,60)(25,90)(35,90)
\qbezier(35,90)(45,90)(55,60)
\qbezier(55,60)(65,30)(75,30)
\qbezier(75,30)(85,30)(95,60)
\qbezier(15,45)(18,48)(25,60)
\qbezier(25,60)(35,80)(45,80)
\qbezier(45,80)(55,80)(65,60)
\qbezier(65,60)(75,40)(85,40)
\qbezier(85,40)(90,40)(95,45)
\put(115,15){$\omega t$}
\put(122,50){$t$}
\put(28,95){$i(t)$}
\put(57,75){$u(t)$}
\multiput(55,24)(0,4){9}{\line(0,1){2}}
\multiput(65,24)(0,4){9}{\line(0,1){2}}
\put(56,28){$\scriptstyle T_v$}
\put(65,36){\vector(-1,0){10}}
\put(65,25){\vector(-1,0){10}}
\put(56,16){$\varphi$}
\put(92,64){$T$}
\put(95,58){\line(0,1){4}}
\end{picture}
\end{minipage}
\begin{minipage}{10cm}
\begin{align*}
T_v & = \frac{\varphi_u-\varphi_i}{\omega} = \frac{\varphi}{\omega} & \frac{T_v}{T} & = \frac{\varphi}{2\pi}
\end{align*}
\begin{align*}
\varphi & = \left\{\begin{array}{ll}2\pi \cdot \dfrac{T_v}{T} = 2 \pi \cdot f \cdot T_v & \text{in Radiant}\\ 360^{\circ} \cdot \dfrac{T_v}{T} = 360^{\circ} \cdot f \cdot T_v & \text{in Grad}\end{array}\right.
\end{align*}
\end{minipage}

\smallskip

$\varphi = \varphi_u - \varphi_i$, $\varphi$ ist positiv, wenn $u$ voreilt (\emph{induktiv}), negativ, wenn $i$ voreilt (\emph{kapazitiv}).

\paragraph{Phasenmessung mit Lissajous--Figur:} ($x$--$y$--Oszilloskop, $x = u(t)$, $y = i(t)$)

\begin{minipage}{5cm}
\begin{picture}(120,130)
\put(60,0){\vector(0,1){120}}
\put(0,60){\vector(1,0){120}}
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\qbezier(100.7193, 66.4757)(106.2898, 82.6202)(98.3022, 92.1394)
\qbezier(98.3022, 92.1394)(90.3147, 101.6586)(73.4482, 98.9762)
\qbezier(73.4482, 98.9762)(56.5817, 96.2939)(40.7164, 82.9813)
\qbezier(40.7164, 82.9813)(24.8511, 69.6688)(19.2807, 53.5243)
\qbezier(19.2807, 53.5243)(13.7102, 37.3798)(21.6978, 27.8606)
\qbezier(21.6978, 27.8606)(29.6853, 18.3414)(46.5518, 21.0238)
\qbezier(46.5518, 21.0238)(63.4183, 23.7061)(79.2836, 37.0187)
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\put(103.5,15){\vector(-1,0){86.5}}

\multiput(22,59)(0,4){4}{\line(0,1){2}}
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\put(98,75){\vector(-1,0){76}}
\put(65,78){$c'$}
\put(65,18){$c$}
\put(112,50){$x$}
\put(64,112){$y$}
\end{picture}
\end{minipage}
\begin{minipage}{10.1cm}
\[|\sin  \varphi | = \frac{c'}{c}\]

(komplette Herleitung:
\href{http://www.iee.et.tu-dresden.de/iee/ge/student/materialien/ET-III/folien/lissajous.pdf}{http://www.iee.et.tu-dresden.de/iee/}

\href{http://www.iee.et.tu-dresden.de/iee/ge/student/materialien/ET-III/folien/lissajous.pdf}{ge/student/materialien/ET-III/folien/lissajous.pdf})
\end{minipage}


\subsection{Beispiel}

\begin{minipage}{6cm}
\begin{picture}(100,60)
\put(56,40){$R$}
\put(10,30){\vector(1,0){15}}
\put(15,33){$i$}
\multiput(30,30)(40,0){2}{\line(1,0){20}}   %  R
\multiput(29,30)(102,0){2}{\circle{2}}
\multiput(50,25)(0,10){2}{\line(1,0){20}}
\multiput(50,25)(20,0){2}{\line(0,1){10}}
\qbezier(40,23)(80,5)(120,23)
\put(120,23){\vector(2,1){0}}
\put(77,5){$u$}
\multiput(70,30)(32,0){2}{\line(1,0){28}}                 % C
\thicklines\multiput(98,23)(4,0){2}{\line(0,1){14}}\thinlines
\put(96,40){$C$}
\end{picture}
\end{minipage}
\begin{minipage}{9.8cm}
\begin{description}
\item[gegeben:] $R = 200\,\Omega$, $C = 100\,\mathrm{nF}$, $f = 5\,\mathrm{kHz}$
\item[gesucht:] $\underline Z$, $\underline Y$ in allen Formen
\end{description}
\end{minipage}

\begin{minipage}{6cm}
\begin{picture}(100,60)
\put(56,40){$R$}
\put(10,30){\vector(1,0){15}}
\put(13,33){$\underline I$}
\multiput(29,30)(102,0){2}{\circle{2}}
\multiput(0,0)(40,0){2}{
\multiput(30,30)(40,0){2}{\line(1,0){20}}   %  R
\multiput(50,25)(0,10){2}{\line(1,0){20}}
\multiput(50,25)(20,0){2}{\line(0,1){10}}}   
\qbezier(40,23)(80,5)(120,23)
\put(120,23){\vector(2,1){0}}
\put(75,4){$\underline U$}
\put(91,42){$\frac{1}{j\omega C}$}
\end{picture}
\end{minipage}
\begin{minipage}{9.8cm}
\[\underline{Z} = R + \frac{1}{j\omega C} = R + jX \qquad R = 200\,\Omega\]
\[X = -\frac{1}{\omega C} = \frac{1 \cdot \mathrm s \cdot \mathrm V}{2\pi \cdot 5\cdot 10^3 \cdot 100 \cdot 10^{-9} \cdot \mathrm{As}} = -318\,\Omega\]
\end{minipage}

\begin{minipage}{6cm}
\begin{picture}(120,110)
\put(40,0){\vector(0,1){90}}
\put(35,70){\vector(1,0){100}}
\put(23,78){Im}
\put(120,58){Re}
\put(40,70){\vector(2,-3){40}}
\put(80,68){\line(0,1){4}}
\put(68,58){$200\,\Omega$}
\put(0,7){$-318\,\Omega$}
\put(38,10){\line(1,0){4}}
\multiput(40,10)(4,0){10}{\line(1,0){2}}
\multiput(80,10)(0,4){12}{\line(0,1){2}}
\end{picture}
\end{minipage}
\begin{minipage}{9.8cm}
\begin{align*}
\underline Z & = 200\,\Omega - j \,318\, \Omega = \sqrt{R^2  + \frac{1}{(\omega C)^2}} \cdot e^{j\arctan\frac{1}{\omega RC}} \\
 & = 376\, \Omega \cdot e^{-j58^{\circ}}
\end{align*}
\[\mathrm{Re}\,(\underline Z) = 200\,\Omega, \; \mathrm{Im}\,(\underline Z) = - 318\,\Omega, \; Z = 376 \, \Omega, \; \varphi = -58^{\circ}\]
\end{minipage}

Leitwert:
\[\underline Y = \frac{1}{\underline Z} = \frac{1}{376}\,\Omega \cdot e^{j58^{\circ}} = 2,\!66\,\mathrm{mS} \cdot e^{j58^{\circ}} = 2,\!66\,\mathrm{mS} \cdot (\cos 58^{\circ} + j \sin 58^{\circ}) = 1,\!41\,\mathrm{mS} + j\,2,26\,\mathrm{mS}\]
\[\mathrm{Re}\,(\underline Y) = 1,\!41\,\Omega, \; \mathrm{Im}\,(\underline Y) = 2,\!26\,\Omega, \; Y = 2,\!66 \, \Omega, \; \varphi_Y = 58^{\circ}\]

Strom:
\[\underline I = \frac{\underline U}{\underline Z} = \underline U \cdot \underline Y  = \frac{\hat U}{\sqrt{2}} \cdot 2,\!66\,\mathrm{mS} \cdot e^{j58^{\circ}} = 1,\!88\,\mathrm{mA} \cdot e^{j58^{\circ}}\]
\[I = 1,\!88\,\mathrm{mA}, \; \varphi_i = 58^{\circ}, \; \hat I = \sqrt 2 \cdot I = 2,\!66\,\mathrm{mA}, \; i(t) = 2,\!66\,\mathrm{mA} \cdot \cos(\omega t + 58^{\circ})\]

\section{Ersatzschaltungen für komplexe Zweipole}

In Gleichstromnetzwerken:

\begin{center}
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\qbezier(10,25)(40,05)(70,25)
\put(70,25){\vector(3,2){0}}
\put(36,2){$U$}
\end{picture}

$\swarrow$ \hspace{4cm} $\searrow$    % jaja, fies :-)
\end{center}

\begin{minipage}{7.9cm}
\center
passiver Zweipol

\begin{picture}(60,60)
\multiput(1,35)(58,0){2}{\circle{2}}
\multiput(2,35)(38,0){2}{\line(1,0){18}}
\multiput(20,30)(20,0){2}{\line(0,1){10}}
\multiput(20,30)(0,10){2}{\line(1,0){20}}
\put(25,44){$R_i$}
\put(0,5){$R = \dfrac{U}{I} = \dfrac{1}{G}$}
\end{picture}

\end{minipage}
\begin{minipage}{7.9cm}
\center 

aktiver Zweipol 

\begin{picture}(220,60)

\multiput(1,35)(99,0){2}{\circle{2}}
\put(2,35){\line(1,0){58}}
\put(30,35){\circle{20}}
\multiput(60,30)(20,0){2}{\line(0,1){10}}
\multiput(60,30)(0,10){2}{\line(1,0){20}}
\put(80,35){\line(1,0){19}}
\put(65,45){$R_i$}
\put(20,47){\vector(1,0){20}}
\put(24,51){$U_L$}

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\put(165,59){$R_i$}
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\put(170,20){\circle{20}}
\put(180,9){\vector(-1,0){20}}
\put(165,0){$I_K$}
\end{picture}
\end{minipage}

\smallskip

\subsection{Passiver Zweipol}

\begin{center}
\begin{picture}(60,45)
\multiput(1,25)(58,0){2}{\circle{2}}
\multiput(2,25)(38,0){2}{\line(1,0){18}}
\multiput(20,20)(20,0){2}{\line(0,1){10}}
\multiput(20,20)(0,10){2}{\line(1,0){20}}
\put(25.5,34){$\underline Z$}
\qbezier(10,20)(30,10)(50,20)
\put(50,20){\vector(2,1){0}}
\put(2,25){\vector(1,0){11}}
\put(8,30){$\underline I$}
\put(26,4){$\underline U$}
\end{picture}
\[\underline Z = \frac{\underline U}{\underline I} = \frac{1}{\underline Y}\]
$\swarrow$ \hspace{4cm} $\searrow$     % nicht schön, aber gut für den Umweltschutz (!?!)
\end{center}

\begin{minipage}{7.9cm}
\center
\[\underline Z = R_r + jX_r\]
Serienersatzschaltung

\begin{picture}(100,70)
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\multiput(60,30)(0,10){2}{\line(1,0){20}}
\put(25,44){$R_r$}
\put(61,44){$jX_r$}
\end{picture}

\hspace{0.3cm}

\begin{tabular}{c|c}
$X_r < 0$ & $X_r > 0$ \\
\hline
$\vphantom{\dfrac a b} jX_r = \frac{1}{j\omega C_r}$ & $jX_r = j\omega L_r$\\
$C_r = - \frac{1}{\omega X_r}$ & $L_r = \frac{X_r}{\omega}$\\
\begin{picture}(60,50)
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\put(13,27){$\scriptstyle R_r$}
\put(38,27){$\scriptstyle C_r$}
\end{picture}
 &
\begin{picture}(60,50)
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}
\put(13,27){$\scriptstyle R_r$}
\put(38,27){$\scriptstyle L_r$}
\end{picture}
\end{tabular}

\end{minipage}
\begin{minipage}{7.9cm}
\center
\[\underline Y = G_p + jB_p\]
Parallelersatzschaltung

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\put(45,29){$R_r$}
\put(41,59){$jX_r$}
\multiput(20,35)(60,0){2}{\circle*{2}}
\end{picture}

\hspace{0.3cm}

\begin{tabular}{c|c}
$B_p < 0$ & $B_p > 0$ \\
\hline
$\vphantom{\dfrac a b} jB_p = \frac{1}{j\omega L_p}$ & $jB_p = j\omega C_p$\\
$L_p = - \frac{1}{\omega B_p}$ & $C_p = \frac{B_p}{\omega}$\\

\begin{picture}(60,50) 
\multiput(1,20)(58,0){2}{\circle{2}}
\multiput(2,20)(50.5,0){2}{\line(1,0){5.5}}
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}
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\qbezier(2.5,3)(5,3)(5,0)
}
\put(25,37){$\scriptstyle G_p$}
\put(25,2){$\scriptstyle L_p$}
\end{picture}
&
\begin{picture}(60,50) 
\multiput(1,20)(58,0){2}{\circle{2}}
\multiput(2,20)(50.5,0){2}{\line(1,0){5.5}}
\multiput(7.5,10)(45,0){2}{\line(0,1){20}}
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\multiput(22.5,26.25)(15,0){2}{\line(0,1){7.5}}
\multiput(22.5,26.25)(0,7.5){2}{\line(1,0){15}}

\put(7.5,0){
\multiput(0,10)(24,0){2}{\line(1,0){21}}
\thicklines\multiput(21,5)(3,0){2}{\line(0,1){10}}\thinlines
}

\put(25,37){$\scriptstyle G_p$}
\put(25,-2){$\scriptstyle C_p$}   % igitt
\end{picture}
\end{tabular}
\end{minipage}

\subsubsection*{Beispiele}
\paragraph*{1.} gegeben: Zweipol mit $U = 230\,\mathrm{V}$, $I = 1\,\mathrm{A}$, $\varphi = \varphi_u - \varphi_i = 15^{\circ} \widehat{=} \dfrac{\pi}{12}$

\begin{description}
\item[Reihenersatzschaltung:]  $\underline Z = \dfrac{\underline U}{\underline I} = U \cdot e^{j\varphi} = Z \cdot \cos \varphi + j Z\cdot \sin \varphi = R_r + jX_r$

\begin{align*}
Z = \frac U I = \frac{230\,\mathrm V}{1\,\mathrm A} = 230\,\Omega & \to R_r = Z \cdot \cos \varphi = 230\,\Omega \cdot \cos\frac{\pi}{12} = 222\,\Omega\\
 & \to X_r = Z \cdot \sin \varphi = 230\,\Omega \cdot \sin \frac{\pi}{12} = 59,\!5\,\Omega
\end{align*}

\begin{minipage}{10cm}
$\displaystyle X_r = \omega L_r \quad \Rightarrow \quad L_r = \frac{X_r}{\omega} = \frac{Z \cdot \sin \varphi}{2\pi f} = 189,\!5\,\mathrm{mH} \quad \Rightarrow $
\end{minipage}
\begin{minipage}{4cm}
\begin{picture}(100,40)
\multiput(1,20)(103,0){2}{\circle{2}}
\multiput(2,20)(83,0){2}{\line(1,0){18}}
\put(40,20){\line(1,0){25}}
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}
\put(17,29){$222\,\Omega$}
\put(53,29){$189,\!5\,\mathrm{mH}$}
\end{picture}
\end{minipage}

\item[Parallelersatzschaltung:] $\underline Y = \dfrac{\underline I}{\underline U} = Y \cdot e^{-j\varphi} = Y \cdot \cos - j Y \cdot \sin \varphi = G_p + jB_p$
\begin{align*}
Y = \frac I U = 4,\!35\,\mathrm{mS} & \to G_p = Y\cdot \cos \varphi = 4,\!35\,\mathrm{mS} \cdot \cos{\frac{\pi}{12}} = 4,\!2\,\mathrm{mS} = \frac{1}{238\,\Omega} \\
 & \to  B_p = - Y\cdot \sin \varphi = -4,\!35\,\mathrm{mS} \cdot \sin{\frac{\pi}{12}} = -1,\!12\,\mathrm{mS}
\end{align*}

\begin{minipage}{10cm}
$\displaystyle B_p = - \frac{1}{\omega L_p} \quad \Rightarrow \quad L_p = - \frac{1}{\omega B_p} = \frac{1}{2\pi f Y \sin\varphi} = 2,\!83\,\mathrm{H}  \Rightarrow$
\end{minipage}
\begin{minipage}{4cm}
\begin{picture}(100,70)
\multiput(1,35)(98,0){2}{\circle{2}}
\multiput(2,35)(78,0){2}{\line(1,0){18}}
\multiput(20,20)(60,0){2}{\line(0,1){30}}
\multiput(40,45)(20,0){2}{\line(0,1){10}}
\multiput(40,45)(0,10){2}{\line(1,0){20}}

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\qbezier(2.5,5)(5,5)(5,0)
}
\end{picture}
\end{minipage}
\end{description}

\paragraph*{2.} gesucht: Parallelersatzschaltung folgender Schaltung:

\begin{minipage}{150pt}
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\put(50,37.5){\line(1,0){20}}
\put(70,34.5){$\Big\downarrow g_m \cdot \underline{U_1}$}
\put(2,17){$C$}
\put(2,47){$R$}
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\put(131,70){\vector(0,-1){65}}
\put(134,34.5){$\underline U$}
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\put(26.2,5){\vector(-2,-3){0}}
\put(35,17){$\underline{U_1}$}
\end{picture}
\end{minipage}
\begin{minipage}{10.62cm}
\[\underline I = \frac{\underline U}{R + \frac{1}{j\omega C}} + g_m \cdot \underline{C_1} \qquad \underline{U_1} = \frac{\frac{1}{j\omega C}}{R + \frac{1}{j\omega C}} \cdot \underline{U} = \frac{1}{1 + j\omega RC} \cdot \underline U\]
\[\underline I = \underline U \cdot \left(\frac{j\omega C}{1 + j\omega RC} + \frac{g_m}{1 + j\omega RC}\right) \]
\end{minipage}

\vspace{0.6cm}

\begin{align*}
\underline Y  &= \frac{\underline I}{\underline U} = \frac{j\omega C + g_m}{1 + j\omega RC} = \frac{(j\omega C + g_m) \cdot (1 - j\omega RC}{1 + (\omega RC)^2}  = \underbrace{\frac{(\omega C)^2 R + g_m}{1 + (\omega RC)^2}}_{\displaystyle =\frac{1}{R_p}} + \underbrace{j\omega C \cdot \frac{1 - g_m \cdot R}{1 + (\omega RC)^2}}_{\displaystyle = \frac{1}{j\omega L_p} \text{ für } g_mR > 1}
\end{align*}

\[\omega L_p = - \frac{1}{\omega C} \cdot \frac{1 + (\omega RC)^2}{1 - g_mR}  \quad \Rightarrow \quad L_p = \frac{1}{\omega^2 C}  \cdot \frac{1 + (\omega RC)^2}{1 - g_mR}, \qquad R_p = R \cdot \frac{1 + (\omega RC)^2}{g_mR + (\omega R C)^2}\]

\paragraph{Zahlenwerte:} $R = 1\,\mathrm k \Omega$, $g_m = 10\,\mathrm{mS}$, $C = 1\,\mu\mathrm{F}$, $f = 500\,\mathrm{Hz}$

\bigskip

$\omega RC = 2\pi \cdot 500\,\mathrm s^{-1} \cdot 10^3\,\frac{\mathrm V}{\mathrm A} \cdot 10^{-6}\,\frac{\mathrm As}{\mathrm V} = \pi$, \quad $g_m \cdot R = 10\,\mathrm{mS} \cdot 1\,\mathrm k\Omega = 10$

\bigskip

$\Rightarrow  R_p = 547\,\Omega, \quad L_p = 122\,\mathrm{mH}$

\subsection{Aktive Zweipole}
\begin{center}
\begin{picture}(80,50)
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\multiput(2,35)(53,0){2}{\line(1,0){23}}
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\qbezier(10,25)(40,05)(70,25)
\put(70,25){\vector(3,2){0}}
\put(36,2){$\underline U$}
\end{picture}

$\swarrow$ \hspace{4cm} $\searrow$    % jaja, fies :-)

\begin{picture}(260,75)

\multiput(1,35)(99,0){2}{\circle{2}}
\put(2,35){\line(1,0){58}}
\put(30,35){\circle{20}}
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\put(65,45){$\underline{Z_i}$}
\put(20,47){\vector(1,0){20}}
\put(24,53){$\underline{U_L}$}

\put(40,0){
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\put(165,61){$\underline{Z_i}$}
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\put(180,8){\vector(-1,0){20}}
\put(165,-2){$\underline{I_K}$}}
\end{picture}
\end{center}

\subsubsection*{Beispiel:}

\begin{minipage}{6cm}
\begin{picture}(150,75)
\put(1,27){$\underline{U_q}\Big\downarrow$}
\put(30,0){\line(0,1){60}}
\put(30,30){\circle{20}}

\put(30,0){\line(1,0){100}}
\put(30,60){\line(1,0){20}}
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\thicklines\multiput(83,28.5)(0,3){2}{\line(1,0){14}}\thinlines
\put(56,68){$R$}
\put(57,27){$\dfrac{1}{j\omega C}$}

\put(131,55){\vector(0,-1){50}}
\put(134,27){$\underline U$}

\put(115,60){\vector(1,0){5}}
\put(115,65){$\underline I$}
\end{picture}
\end{minipage}
\begin{minipage}{5cm}
\[\underline{U_q} = U_q \cdot e^{j\varphi_{u_q}}\]
\end{minipage}

\bigskip

\begin{minipage}{7.8cm}
\center

Spannungsquellenersatzschaltung

\bigskip

\begin{picture}(150,80)
\put(0,27){$\underline{U_L}\Big\downarrow$}
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\put(55,70){$\underline{Z_i}$}

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\put(114,27){$\underline U$}

\put(85,60){\vector(1,0){5}}
\put(85,65){$\underline I$}
\end{picture}
\end{minipage}
\begin{minipage}{7.8cm}
\center

Stromquellenersatzschaltung

\bigskip

\begin{picture}(150,80)
\put(0,27){$\underline{I_K}\Big\uparrow$}
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\put(78,27){$\underline{Z_i}$}
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\multiput(70,0)(0,60){2}{\circle*{2}}
 
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\put(114,27){$\underline U$}

\put(85,60){\vector(1,0){5}}
\put(85,65){$\underline I$}
\end{picture}

\end{minipage}



\begin{minipage}{7.8cm}
\begin{align*}
\underline{U_L} &= \underline U \Big|_{\underline I = 0} = \frac{\frac{1}{j\omega C}}{R + \frac{1}{j\omega C}}\cdot \underline{U_q}\\
 &= \frac{1}{1 + j\omega RC} \cdot U_q \cdot e^{j\varphi_{u_q}}\\
 &= \frac{U_q}{\sqrt{1 + (\omega RC)^2}} \cdot e^{j(\omega_{u_q} - \arctan \omega RC)}
\end{align*}
\end{minipage}
\begin{minipage}{7.8cm}
\[\underline{I_K} = \underline I \Big|_{\underline U = 0} = \frac{\underline{U_q}}{R} = \frac{U_q}{R} \cdot e^{j\varphi_{u_q}}\]
\end{minipage}

\bigskip

\[\underline{Z_i} = \frac{\underline{U_L}}{\underline{I_K}} \qquad \text{ bzw. }\qquad \underline{Z_i} = \frac{\underline U}{\underline I}\Big|_{\text{Quellen} = 0}\]

\bigskip

\[\underline{Z_i} = \frac{1}{j\omega R} \parallel R = \frac{1}{\frac 1 R + j\omega C} = \frac{R}{1 + j \omega RC} = \frac{R}{\sqrt{1 + (\omega RC)^2}} \cdot e^{-j\arctan \omega RC}\]

\chapter{Leistung bei Wechselstrom}

Bei Gleichstrom:

\begin{minipage}{6cm}
\center

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\put(8,23){$I$}
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\put(26,48){$U$}
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\put(50,38){\vector(3,-2){0}}
\end{picture}
\end{minipage}
\begin{minipage}{5cm}
\[P = U \cdot I = \dfrac{U^2}{R} = I^2 \cdot R\]

\vspace{0.4cm}    % hachja ..
\end{minipage}

\section{Leistungsbegriffe}
\subsection{Momentanleistung}
\begin{minipage}{6cm}
\center

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\put(10,35){$i(t)$}
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\end{picture}
\end{minipage}
\begin{minipage}{9.8cm}

\boxed{\quad\vphantom{\Big|} p(t) = u(t) \cdot i(t)\quad} \qquad Def.: Momentanleistung

\end{minipage}

\begin{minipage}{6cm}
\center

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\put(00,45){\vector(1,0){130}}

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\put(118,34){$\omega t$}
\end{picture}
\end{minipage}
\begin{minipage}{9.6cm}
\begin{align*}
u(t) & = \hat U \cdot \cos(\omega t + \varphi)\\
i(t) & = \hat I \cdot \cos(\omega t)
\end{align*}
\[p(t) = u(t) \cdot i(t) = \hat U \cdot \hat I \cdot \cos(\omega t + \varphi) \cdot \cos(\omega t)\]
\end{minipage}

\bigskip

Additionstheorem: $\cos \alpha \cdot \cos \beta = \frac 1 2 \left[\cos(\alpha + \beta) + \cos(\alpha - \beta)\right]$


\[\Rightarrow p(t) = \frac{\hat U \cdot \hat I}{2} \cdot\left[\cos \varphi + \cos(2\omega t + \varphi)\right]\]
mit $U = \dfrac{\hat U}{\sqrt 2}$ und $I = \dfrac{\hat I}{\sqrt 2}$:

\[p(t) = U \cdot I \cdot [\cos \varphi + \cos (2\omega t + \varphi)]\]

\begin{minipage}{6cm}
\begin{picture}(140,110)
\put(0,48){$\scriptstyle UI\cdot \cos \varphi$}
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\end{minipage}
\begin{minipage}{9.8cm}
\begin{itemize}
\item Sinusförmig, aber mit $2 \omega$
\item $p(t)$ kann Negativ werden (schraffiert)
\item weitere Zerlegung aufschlußreich
\end{itemize}
\end{minipage}



Additionstheorem: $\cos (2\omega t + \varphi) = \cos 2 \omega t \cdot \cos \varphi - \sin 2 \omega t \cdot  \sin \varphi$

\[\Rightarrow p(t) = U \cdot I \cdot \cos \varphi \cdot (1 + \cos 2 \omega t) - U \cdot I \cdot \sin \varphi \cdot \sin 2 \omega t\]

\begin{center}
$\swarrow$ \hspace{4cm} $\searrow$    %  und schon wieder
\end{center}

\begin{minipage}{7.8cm}
\center 
\begin{picture}(140,100)
\put(0,48){$\scriptstyle UI\cdot \cos \varphi$}
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\put(31,55){$\frac{T}{2}$}}
\end{picture}

Strom pulsierend in den Zweipol

\[\left.\begin{array}{l}\text{Mittelwert}\\\text{Amplitude}\end{array}\right\} U \cdot I \cdot \cos \varphi\]
$\Rightarrow$ \emph{Wirkleistung}
\end{minipage}
\begin{minipage}{7.8cm}
\center 
\begin{picture}(140,100)
\put(0,63){$\scriptstyle UI\cdot \sin \varphi$}
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}
\put(95,38){\line(0,1){4}}
\put(88,25){$\frac{T}{2}$}
\end{picture}

Strom pendelt zwischen Erzeuger und Verbraucher
\[\left.\begin{array}{ll}\text{Mittelwert:} & 0\\\text{Amplitude:} & U \cdot I \cdot \sin \varphi\end{array}\right.\]
$\Rightarrow$ \emph{Blindleistung}
\end{minipage}


\subsection[Wirkleistung]{Wirkleistung}

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\end{picture}
\end{minipage}
\begin{minipage}{9.8cm} Die Wirkleistung ist der zeitliche Mittelwert der Momentanleistung.
\[P = U \cdot I \cdot \cos \varphi, \qquad [P] = [U] \cdot [I] = 1\,\mathrm W\]
Sie hängt vom \emph{Leistungsfaktor} $\cos \varphi$ ab. Bei $\varphi = - \frac{\pi}{2}$ (reine Kapazität) und $\varphi = \frac{\pi}{2}$ (reine Induktivität) beträgt die Wirkleistung Null.
\end{minipage}

\smallskip

\subsection{Blindleistung}

\begin{minipage}{6cm}
\begin{picture}(150,130)

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\end{picture}
\end{minipage}
\begin{minipage}{9.8cm} 
Die Blindleistung ist die Amplitude der zwischen Erzeuger und Verbraucher periodisch ausgetauschten Leistung
\[Q = U \cdot I \cdot \sin \varphi\]
\[[Q] = 1 \,\mathrm{var} \qquad (\text{Voltampere reaktiv})\]

\end{minipage}

\subsection{Scheinleistung}

Die Scheinleistung ist das Produkt der Effektivwerte von Strom und Spannung

\begin{center}
\boxed{\quad \vphantom{\Big|} S = U  \cdot I \quad} \qquad  $[S] = 1 \,\mathrm{VA}$
\end{center}

\section{Komplexe Leistung}

\subsection{Definition}
$P = U \cdot I \cdot \cos \varphi, \quad Q = U \cdot I \cdot \sin \varphi$

\smallskip

\begin{center}
\boxed{\quad \vphantom{\Big|} \underline S  = P + jQ\quad} \qquad Definition komplexe Leistung
\end{center}

\smallskip

$\underline S = U \cdot I \cdot (\cos \varphi + j \sin \varphi) = U \cdot I \cdot e^{j\varphi} = S \cdot e^{j\varphi}$

\bigskip

$|\underline S| = S = U \cdot I = \sqrt{P^2 + Q^2}$ \qquad $P = \mathrm{Re}\,(\underline S), \quad Q = \mathrm{Im}\,(\underline S)$,\quad $\varphi = \arg(\underline S) = \varphi_u - \varphi_i$

\subsection[Zusammenhang mit U und I]{Zusammenhang mit $\underline U$ und $\underline I$}

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\end{minipage}
\begin{minipage}{11.8cm}
\begin{align*}
\underline S &= U \cdot I \cdot e^{j\varphi}, &\qquad \varphi = \varphi_u - \varphi_i\\
 & = U \cdot I \cdot e^{j(\varphi_u - \varphi_i)} = \underbrace{U\cdot e^{j\varphi_u}}_{\underline U} \cdot \underbrace{I\cdot e^{-j\varphi_i}}_{\underline I^*}
\end{align*}
\end{minipage}

\begin{center}
\boxed{\quad \vphantom{\Big|} \underline S = \underline U  \cdot \underline I^*\quad}
\end{center}

\subsubsection*{Beispiel:}

\begin{minipage}{5cm}
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\multiput(30,0)(0,40){2}{\line(0,1){20}}
\put(30,30){\circle{20}}
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\put(52,27){$R$}
\put(91,27){$C$}}
\end{picture}
\end{minipage}
\begin{minipage}{10.8cm}
\begin{description}
\item[gegeben:] $R = 1\,\Omega$, $C = 1\,\mathrm{mF}$, $f =50\,\mathrm{Hz}$, $I = 1\,\mathrm{A}$ 
\item[gesucht:] $\underline S$, $S$, $P$, $Q$
\end{description}
\[\underline S = \underline U \cdot \underline I^*, \quad \underline U = \underline I \cdot \underline Z = \dfrac{\underline I}{\underline Y} = \dfrac{\underline I}{G+jB}, \quad G = \dfrac{1}{R},\, B = \omega C\]

\end{minipage}

\bigskip

$\underline S = \dfrac{\underline I \cdot \underline I^*}{G + jB} = \dfrac{I^2}{G + jB} = \dfrac{I^2}{G + j\omega C} = \underbrace{\dfrac{G \cdot I^2}{(G^2 + (\omega C)^2}}_{P} + j \underbrace{\dfrac{-\omega C \cdot I^2}{G^2 + (\omega C)^2}}_{Q}$

\bigskip

$P = \dfrac{G}{G^2 + (\omega C)^2} \cdot I^2 = 0,\!717\,\mathrm{W}$ 

\bigskip

$Q = -\dfrac{\omega C}{G^2 + (\omega C)^2} \cdot I^2 = - 0,\!532\,\mathrm{var}$

\bigskip

$S = \dfrac{I^2}{\sqrt{G^2 + (\omega C)^2}} = \sqrt{P^2 + Q^2} = 0,\!893\,\mathrm{VA}$

\section{Leistungsübertragung im Grundstromkreis}

\subsection{(Wirk--)Leistungsübertragung}

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\end{picture}
\end{minipage}
\begin{minipage}{10.8cm}
\begin{description}
\item[Problem:] \quad \\ gegeben: Generator ($\underline{U_L}$, $\underline{Z_i}$)

gesucht: $\underline{Z_a}$ so, daß $P$ maximal wird.
\end{description}
\end{minipage}

$P = \mathrm{Re}\,(\underline S) = \mathrm{Re}\,(\underline U \cdot \underline I^*)$

\bigskip

$\underline U = \dfrac{\underline{Z_a}}{\underline{Z_i} + \underline{Z_a}} \cdot \underline{U_L}, \qquad \underline I = \dfrac{\underline{U_L}}{\underline{Z_i} + \underline{Z_a}}$ 

\bigskip

$\underline S = \underline{U} \cdot \underline{I}^* = \dfrac{\underline{Z_a} \cdot \underline{U_L} \cdot \underline{U_L}^*}{(\underline{Z_i} + \underline{Z_a}) \cdot {(\underline{Z_i} + \underline{Z_a})}^*} = \dfrac{\underline{Z_a}}{|Z_i + Z_a|} \cdot \underline{U_L}^2$

\bigskip

$\underline{Z_a} = R_a + j X_a$,  \qquad $\underline{Z_i} = R_i + j X_i$

\bigskip

$P = \mathrm{Re}\,(\underline S) = \dfrac{R_a \cdot {U_L}^2}{(R_i + R_a)^2 + (X_i + X_a)^2} \Rightarrow$ maximal


\bigskip

Maximum bei:

\smallskip

$
\left.\begin{array}{ll}
X_a = - X_i\\
P = \dfrac{R_a}{(R_i + R_a)^2} \cdot {U_L}^2  \longrightarrow  R_a = R_i
\end{array}\right\} \quad \boxed{\quad \vphantom{\Big|} \underline{Z_a}= R_i - jX_i = \underline{Z_i}^* \quad}$

\subsection{Blindleistungskompensation}

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\multiput(0,0)(0,40){2}{\line(0,1){20}}
\multiput(-5,20)(10,0){2}{\line(0,1){20}}
\multiput(-5,20)(0,20){2}{\line(1,0){10}}
}
\put(42,37){$\underline Y_K$}
\put(45,73){\vector(1,0){15}}
\put(50,77){$I$}
\put(70,73){\vector(1,0){15}}
\put(72,77){$I_V$}
\put(67,56){$\downarrow\! I_K$}
\put(122,56){$\downarrow\! I_B$}
\put(152,56){$\downarrow\! I_W$}
\put(127,37){$B_p$}
\put(157,37){$G_p$}

\multiput(0,0)(2,74){2}{\multiput(105,3)(4,0){18}{\line(1,0){2}}}
\multiput(0,0)(72,0){2}{\multiput(105,3)(0,4){19}{\line(0,1){2}}}
\put(112,83){Verbraucher}
\put(77,37){$\Longleftrightarrow$}
\put(82.5,27){$Q$}
\end{picture}
\end{minipage}
\begin{minipage}{8.5cm}
\bigskip

\subsubsection*{Problem}
\begin{description}
\item[gegeben:] Verbraucher am Netz, $\underline Y_K = G_p + jB_p$
\item[gesucht:] Kompensationszweipol, so daß $I$ maximal wird
\end{description}\
\end{minipage}

\bigskip

$\underline I = \underline U \cdot \underline Y = \underline U \cdot (\underline Y_K + \underline Y_V) = \underline U \cdot (\underbrace{G_K + jB_k}_{I_K} + \underbrace{G_p + jB_p}_{I_V})$

\smallskip

$I = |I| = U \cdot \sqrt{\smash[b]{(\underbrace{G_K}_{\geq 0} + G_p)^2 + (\underbrace{B_K}_{\lessgtr 0} + B_p)^2}} \quad \Rightarrow \mathrm{min}\,(G_K,B_K)$   % smash! danke herbert voss! :-)

\vspace{1cm}   % wegen smash

Mininum bei $G_K = 0 \quad \to I = U \cdot \sqrt{{G_p}^2+(B_K+B_p)^2}$ und $B_K = - B_p$. 

\bigskip

$\Rightarrow$ Minimaler Strom: $I = U \cdot G_p = I_W$

\subsubsection*{Ergebnis:}
\begin{itemize}
\item Verbraucher und Kompensationszweipol zusammen nehmen nur Wirkleistung auf

\item Blindleistung wird zwischen Verbraucher und Kompensationszweipol ausgetauscht

\item Verbraucher sind meistens induktiv $\longrightarrow$ Kompensationskapazität
\end{itemize}

\paragraph{Beispiel:} Leuchtstofflampe, gemessen: $U = 230\,\mathrm{V},\ P = 60\,\mathrm W,\ I = 0,\!52\,\mathrm A$

\begin{minipage}{11cm}
\textbf{Leistungen:}

\bigskip 

$S = U \cdot I = 230\,\mathrm V \cdot 0,\!52\,\mathrm A = 120\,\mathrm{VA}$

\smallskip

$P = U \cdot I \cdot \cos \varphi = S \cdot \cos \varphi \to \cos \varphi = \frac P S = 0,\!5\,\mathrm V,\ \varphi = 60^{\circ}$

\smallskip

$Q = U \cdot I \cdot \sin \varphi = \sqrt{S^2 - P^2} = 104\,\mathrm{var}$
\end{minipage}
\begin{minipage}{3cm}
\begin{picture}(75,110)
\put(9,13){$Q_K$}
\put(9,84){$Q_K$}
\put(51,38){$P_V$}
\put(58,89){$\underline S_V$}
\put(30,50){\vector(0,-1){39.5}}
\put(30,50){\vector(0,1){39.5}}
\put(30,50){\vector(1,0){26}}
\put(30,50){\vector(2,3){26}}
\multiput(30,89.5)(4,0){7}{\line(1,0){2}}
\multiput(56,50)(0,4){10}{\line(0,1){2}}
\qbezier(45,50)(47,62)(42,68)
\put(37,55){$\varphi$}
\end{picture}
\end{minipage}

\subsubsection*{Kompensationskapazität:}

$Q_K = -Q_V = - U^2 \cdot \omega C  \Rightarrow C_K = \dfrac{Q_V}{U^2 \cdot \omega} = 6,\!2\,\mu\mathrm F$

\paragraph*{Elemente der Ersatzschaltung:} (Modell des Verbrauchers)

\bigskip

$P = U^2 \cdot G_p \quad \Rightarrow \quad G_p = \dfrac{P}{U^2} = 1,\!1\,\mathrm{mS}\ \widehat{=}\ 909\,\Omega$

\bigskip

$Q = U^2 \cdot B_p \quad \Rightarrow \quad B_p = \dfrac{Q}{U^2} = 2\,\mathrm{mS} = \dfrac{1}{\omega L_p} \quad \Rightarrow \quad L_p = \dfrac{1}{\omega B_p} = 1,\!7\,\mathrm{H}$

\subsection*{Schaltung / Ersatzschaltung}

\quad 
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\put(48,93){$Dr$}
\put(0,63){$C_K$}
\put(145,42){$\Rightarrow$}      %  :-))
\end{picture}
\qquad 
\begin{picture}(150,100)
\put(4,42){$U\Big\downarrow$}
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\put(79,71){$\scriptstyle 0,43\,\mathrm{A}$}
\multiput(74,80)(32,0){2}{\vector(0,-1){15}}
\end{picture}

\subsection{Modelle technischer Bauelemente}

%  \begin{tabular}{p{0pt}@{\hspace{-6pt}}|c|c|}  (tja muß man erstmal drauf kommen, daß man im argument von multicolumn auch | angeben kann.. ich idiot.
\begin{tabular}{|c|c|} 
\hline
Spulen & Kondensatoren \\
\hline
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\qbezier(2.5,5)(5,5)(5,0)
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\put(34,20){\line(1,0){12}}
\put(22,30){$L$}
\put(52,30){$R$}
\end{picture}
&
\begin{picture}(80,60)
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\multiput(2,20)(58,0){2}{\line(1,0){18}}
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\thicklines \multiput(38,28)(4,0){2}{\line(0,1){14}}
\put(37,13){$R$}
\put(37,46){$C$}
\end{picture}
\\
\hline
 $\vphantom{\Big|} \underline Z = R_L + jX_L, \quad X_L = \omega L$ & $\underline Y = G_C + jB_C, \quad B_C = \omega C$\\
\hline
\begin{picture}(90,105)
\put(0,72){$X_L$}
\put(20,10){\vector(0,1){80}}
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\put(46,2){$R_L$}
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\put(52,79){$\underline Z$}
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\put(32,40){\vector(-1,1){0}}
\put(30,23){$\varphi$}
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\put(23,40){$\delta$}
\end{picture}
&
\begin{picture}(90,105)
\put(0,72){$B_C$}
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\put(36,48){\vector(2,-1){0}}
\put(23,40){$\delta$}
\end{picture}
\\
\hline
\multicolumn{2}{|c|}{Komplexe Leistung: \quad $\vphantom{\bigg|}\underline{S} = P + jQ = U \cdot I^* = I^2 \cdot \underline Z = U^2 \cdot \underline Y$} \\
\hline
$\vphantom{\bigg|} P = I^2\cdot R_L, \quad Q = I^2 \cdot X_L$ & $P = U^2\cdot G_C, \quad Q = U^2\cdot B_C$\\
\hline
\multicolumn{2}{|c|}{Verlustfaktor $\vphantom{\Bigg|} d = \dfrac{\text{Wirkleistung}}{|\text{Blindleistung}|} = \dfrac{P}{|Q|} = \tan \delta, \quad \delta = \text{Verlustwinkel}$ }\\
\hline
$\vphantom{\Bigg|} d_L = \dfrac{P}{Q} = \dfrac{R_L}{X_L} = \dfrac{R_L}{\omega L} = \tan \delta_L$ & $d_C = \dfrac{P}{Q} = \dfrac{G_C}{B_C} = \dfrac{G_C}{\omega C} = \tan \delta_C$ \\
\hline

\multicolumn{2}{|c|}{Güte $\vphantom{\Bigg|}Q = \dfrac{|\text{Blindleistung}|}{\text{Wirkleistung}} = \dfrac{|Q|}{P} = \dfrac{1}{d}$}\\
\hline
$\vphantom{\Bigg|} Q_L = \dfrac{1}{d_L} = \dfrac{X_L}{R_L} = \dfrac{\omega L}{R_L}$ & $Q_C = \dfrac{1}{d_C} = \dfrac{B_C}{G_C} = \dfrac{\omega C}{G_C}$\\
\hline
\multicolumn{1}{c}{\hspace{7cm}\quad} & \multicolumn{1}{c}{\hspace{7cm}\quad} % hachja..
\end{tabular}

\subsection{Kleinverbraucher am Netz}

Eine Lampe ($12\,\mathrm{V}$, $10\,\mathrm W$) ist am Netz ($230\,\mathrm V$) zu betreiben.

\bigskip

$I = \dfrac{P}{U} = \dfrac{10\,\mathrm W}{12\,\mathrm V} = 0,\!83\,\mathrm A$

\subsubsection*{Vorwiderstand}

\begin{minipage}{120pt}
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\put(64,68){$R$}
\put(45,27){$\Rightarrow P_N$}
\end{picture}
\end{minipage}
\begin{minipage}{11cm}
\[R = \frac{U - U_L}{I} = \frac{218\,\mathrm V}{0,\!83\,\mathrm A} = 262\,\Omega\]
\smallskip
\[P_N = U_N \cdot I = 230\,\mathrm V \cdot 0,\!83\,\mathrm A = 192\,\mathrm W\]
\end{minipage}

\subsubsection*{Serienkondensator}

\begin{minipage}{5.5cm}
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\end{picture}
\end{minipage}
\begin{minipage}{6cm}
\[U_C = \sqrt{{U_N}^2-{U_L}^2} = \frac{1}{\omega C}\]
\[C = \frac{1}{\omega \sqrt{{U_N}^2-{U_L}^2}} = 11,\!5\,\mu\mathrm F\]
\end{minipage}

\vspace{1cm}

$S = U_N \cdot I = 192\,\mathrm{VA}$, $P = 10\,\mathrm W$, $I_{KZ} = \dfrac P U = \dfrac{10\,\mathrm W}{230\,\mathrm{V}} = 0,\!043\,\mathrm{A}$

\chapter{Ortskurven}

\section{Grundbegriffe}

\begin{minipage}{6cm}
\begin{picture}(150,120)
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\put(15,15){\vector(1,0){135}}
\put(135,3){Re}
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\end{minipage}
\begin{minipage}{9.6cm}
Eine Ortskurve ist der geometrische Ort aller Endpunkte des Zeigers $\underline{F}(\lambda)$ bei Änderung von $\lambda$.

\smallskip

\begin{description}
\item[$\underline F$ kann sein:]  Spannung, Strom, komplexe Leistung, Impedanz, Admittanz, Spannungsverhältnis, Stromverhältnis. 
\item[$\lambda$ kann sein:] Frequenz, Widerstand, Kapazität, Induktivität
\end{description}
\end{minipage}

\subsubsection*{Beispiel 1}

\begin{minipage}{5.8cm}
\quad
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\put(50.5,15){\line(0,1){20}}
\put(60.5,11){\vector(-1,0){20}}
\put(47,1){$I$}
\multiput(0,0)(40,0){2}{
\qbezier(12,53)(30,45)(48,53)
\put(48,53.2){\vector(2,1){0}}}
\put(25,38){$\underline{U}_R$}
\put(65,38){$\underline{U}_C$}
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\put(88,70){\vector(4,-1){0}}
\put(46,78){$\underline U$}
\end{picture}
\end{minipage}
\begin{minipage}{10cm}
\begin{description}
\item[gegeben:] $R$, $C$, $\underline I = I$
\item[gesucht:] Ortskurven von $\underline U_R$, $\underline U_C$, $\underline U$, $\underline Z$ in Abhängigkeit von $\omega = 2 \pi f$
\end{description}
\end{minipage}

\bigskip

\begin{minipage}{5.3cm}
\quad
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\put(10,80){\vector(1,0){110}}
\put(105,70){Re}
\put(0,90){Im}
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\put(15,80){\vector(1,0){60}}
\put(15,80){\vector(1,-1){60}}
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\put(75,10){\line(0,1){70}}
\put(15,80){\vector(1,0){80}}
\put(-2,20){$\underline U_C$}
\put(0,40){$\omega\! \uparrow$}
\put(67,85){$\underline U_R$}
\put(89,85){$\underline I$}
\put(55,47){$\underline U$}
\put(78,40){$\uparrow \! \omega$}
\end{picture}
\end{minipage}
\begin{minipage}{5.5cm}
$\underline U_R = I \cdot R = U_R$ \quad (reell)

\smallskip

$\underline U_C = \dfrac{I}{j\omega C} = - j \dfrac{I}{\omega C}$

\smallskip

$\underline U = I \cdot R - j \dfrac{I}{\omega C} =I \cdot \underline{Z}$

\smallskip

$\underline{Z}(\omega) = R - j \dfrac{I}{\omega C}$

\end{minipage}
\begin{minipage}{5cm}
\quad
\begin{picture}(120,100)
\put(15,0){\vector(0,1){100}}
\put(10,80){\vector(1,0){110}}
\put(90,68){$\mathrm{Re}\,(\underline Z)$}
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\put(75,80){\vector(0,-1){60}}
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\put(15,80){\vector(1,-1){60}}
\put(75,10){\line(0,1){70}}
\put(78,20){$-\frac{j}{\omega C}$}
\put(68,85){$R$}
\put(48,53){$\underline Z(\omega)$}
\put(78,40){$\uparrow \! \omega$}
\end{picture}
\end{minipage}

\subsubsection*{Beispiel 2}

\begin{minipage}{5.8cm}
\quad
\begin{picture}(100,80)
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\put(1,60){\line(1,0){19}}
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\multiput(20,55)(0,10){2}{\line(1,0){20}}
\multiput(40,60)(32,0){2}{\line(1,0){28}}
\thicklines\multiput(68,53)(4,0){2}{\line(0,1){14}}\thinlines
\multiput(0,25)(101,0){2}{\line(0,1){34}}
\put(0,25){\line(1,0){101}}
\put(50.5,25){\circle{20}}
\put(46,1){$U$}
\put(65,15){\vector(3,1){0}}
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\put(48,53.2){\vector(2,1){0}}}
\put(25,38){$\underline{U}_R$}
\put(65,38){$\underline{U}_C$}
\put(0,65){\vector(1,0){15}}
\put(7,70){$\underline I$}
\end{picture}
\end{minipage}
\begin{minipage}{10cm}
\begin{description}
\item[gegeben:] $R$, $C$, $\underline U = U$
\item[gesucht:] Ortskurven von $\underline U_R$, $\underline U_C$, $\underline I$ in Abhängigkeit von $\omega = 2 \pi f$
\end{description}
\end{minipage}

\begin{minipage}{7cm}
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\put(20,75){\vector(1,1){50}}
%\put(70,125){\vector(1,-1){50}}
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\qbezier(40,95)(50,90)(60,95)
\put(50,97.5){\circle*{1}}
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\qbezier(40,55)(50,60)(60,55)
\put(50,52.5){\circle*{1}}
\put(33,82){$\scriptstyle \underline U_R$}
\put(34,63){$\scriptstyle \underline U_C$}
\put(57,82){$\scriptstyle \underline U_C$}
\put(54,64){$\scriptstyle \underline U_R$}
\qbezier(40,105)(50,110)(60,105)
\put(60,105){\vector(3,-1){0}}
\put(46,110){$\omega$}
\qbezier(40,45)(50,40)(60,45)
\put(40,45){\vector(-3,1){0}}
\put(46,35){$\omega$}
\put(83,78){$\scriptstyle U = \underline U$}
\put(20,75){\vector(1,0){100}}
\put(114,60){$\frac{U}{R}$}
\put(125,78){$\infty$}
\put(65,110){$\underline I$}
% Kreise: r = 30 um (50,45) und (50,105), und r = 50, um (70,75) bitte mit dem script in ~/qbezier/ erstellen!

\qbezier(80,75)(80,87.426408)(71.213204,96.213204)
\qbezier(50,105)(62.426408,105)(71.213204,96.213204)
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\qbezier(80,75)(80,62.573592)(71.213204,53.786796)
\qbezier(50,45)(62.426408,45)(71.213204,53.786796)

\qbezier(120,75)(120,95.71068)(105.35534,110.35534)
\qbezier(70,125)(90.71068,125)(105.35534,110.35534)
\qbezier(20,75)(20,95.71068)(34.64466,110.35534)
\qbezier(70,125)(49.28932,125)(34.64466,110.35534)

\end{picture}
\end{minipage}
\begin{minipage}{8.9cm}
$\underline U = \underline U_R + \underline U_C$, \quad $\underline U_R = I \cdot R,$ \quad $\underline U_C = - j \dfrac{I}{\omega C}$
\begin{description}
\item[Thaleskreis:] Die Ortskurven von $\underline U_C$, $\underline U_R$ und $I$ sind Kreise!
\item[Formelausdruck:]
\end{description}
\vspace{-0.5cm}  % tststs
\begin{align*}
\underline Z(\omega) &= R - j \dfrac{1}{\omega C} &\underline I(\omega) &= \dfrac{U}{\underline Z(\omega)} = \dfrac{U}{R - j\omega C}\\
\underline F(\lambda) &= \underline a + \underline b \cdot f(\lambda) &\underline F(\lambda) &= \dfrac{\underline b}{\underline a_1 + a_2 \cdot f(\lambda)}
\end{align*}
\end{minipage}

\section{Einfache Ortskurven}


\begin{minipage}{5cm}
\center

\begin{picture}(120,100)
\put(15,0){\vector(0,1){80}}
\put(10,10){\vector(1,0){100}}
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\thicklines
\put(15,10){\line(1,0){80}}
\put(0,70){Im}
\put(95,0){Re}
\put(95,74){\boxed{\underline x}}
\end{picture}
\end{minipage}
\begin{minipage}{5cm}
\[\underline y = F(\underline x)\]
\end{minipage}
\begin{minipage}{5cm}
\center

\begin{picture}(120,100)
\put(15,0){\vector(0,1){80}}
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\thicklines
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\thinlines
\put(0,70){Im}
\put(95,0){Re}
\put(95,74){\boxed{\underline y}}
\end{picture}
\end{minipage}

\subsubsection*{Gerade}

\begin{minipage}{5.9cm}
\qquad \begin{picture}(110,100)
\put(10,50){\vector(1,0){110}}
\put(20,0){
\put(0,90){Im}
\put(15,0){\vector(0,1){100}}
\put(15,50){\vector(2,1){20}}
\put(15,50){\vector(2,3){15}}
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\put(53,79){$\scriptstyle 1$}
\put(38,56.5){$\scriptstyle 0$}
\put(23,34){$\scriptstyle -1$}
\put(25,62.5){$\scriptstyle \underline a_0$}
\put(20,74){$\scriptstyle \underline a_1$}}
\put(108,40){Re}
\put(67,53){\vector(2,3){15}}
\put(77,57){$\lambda$}
\end{picture}
\end{minipage}
\begin{minipage}{10cm}
\center

\[f(\lambda) = \underline{a_0} + \underline{a_1} \cdot \lambda\]
Spezialfall: $\underline{a_0}= 0$
\[f(\lambda) = \underline{a_1} \cdot \lambda\]
$\to$ Gerade durch den Ursprung
\end{minipage}

\subsubsection*{Kreis um den Ursprung}
\begin{minipage}{5.9cm}
\qquad \begin{picture}(110,100)
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\put(55,10){\vector(0,1){80}}
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\end{minipage}
\begin{minipage}{10cm}
\center
\[\underline f (\lambda) = \underline{a_0} \cdot \frac{1 + \underline{a_1}^* \cdot  \lambda}{1 + \underline{{a_1}}  \cdot \lambda}\]
\[\underline f(\lambda) = \underline{a_0} \cdot e^{j\varphi(\lambda)}\]
\end{minipage}


\subsubsection*{Kreis durch den Ursprung}
\begin{minipage}{5.9cm}
\qquad \begin{picture}(110,100)
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\put(21,33){$\underline B$}
\put(38,47){$\underline B$}
\end{picture}
\end{minipage}
\begin{minipage}{10cm}
\center
\[\underline f (\lambda) = \frac{1}{\underline{a_0} + \underline{a_1} \cdot \lambda}\]
\[\underline f(\lambda) = \underline B \left(1 + e^{j\varphi(\lambda)}\right)\]
\end{minipage}



\subsubsection*{Kreis in allgemeiner Lage}
\begin{minipage}{5.9cm}
\qquad \begin{picture}(110,100)
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\put(32,43){$\underline A$}
\put(50,59){$\underline B$}
\end{picture}
\end{minipage}
\begin{minipage}{10cm}
\center
\[\underline f (\lambda) =  \frac{\underline{b_0} + \underline{b_1} \cdot \lambda}{\underline{a_0} + \underline{a_1} \cdot \lambda}\]
\[\underline f(\lambda) = \underline A + \underline B \cdot e^{j\varphi(\lambda)}\]

\end{minipage}

\section{Inversion}

Inversion einer komplexen Größe bedeutet die Bildung ihres Reziprokwertes.
\[\underline Y = \frac{1}{\underline Z}, \qquad Y \cdot e^{j\varphi_y} = \frac{1}{Z\cdot e^{j\varphi_z}} \quad \to \quad Y = \frac 1 Z, \ \varphi_y = - \varphi_z
\]
Der Zeiger $\underline Z$ aus der $\underline Z$-Ebene wird auf den Zeiger $\underline Y$ der $\underline Y$-Ebene abgebildet (konforme Abbildung) durch $\underline Y = \frac{1}{\underline Z}$.

\subsection{Inversion eines Zeigers}

\begin{minipage}{6.8cm}
\begin{picture}(160,150)
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\put(65,135){\line(1,-1){70}}

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\qbezier(95,75)(95,66.715728)(89.142136,60.857864)
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\put(83,69){$\scriptstyle \varphi_y$}
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\put(100,100){\circle*{2}}
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\put(98,40){$\underline Y$}
\put(77,127){$B$}
\put(125,131){$\underline Z$}
\put(127,117){$A$}
\put(76,94){$\frac{\underline Z}{|\underline Z|^2}$}
\end{picture}
\end{minipage}
\begin{minipage}{9cm}
\subsubsection{Betrag}
Die Dreiecke $0AB$ und $0BC$ sind ähnlich
\[\to \frac{Z}{1} = \frac 1 Y\]
\subsubsection{Phase}
\[\varphi_y = - \varphi_z\]
$\to$ Spiegelung an der reellen Achse
\end{minipage}


\subsubsection*{Maßstäbliche Konstruktion: Inversionsradius}

\begin{tabular}{cc}
\hspace{1.5cm}$Z$-Maßstab\hspace{1.5cm} &\hspace{1.5cm}$Y$-Maßstab\phantom{\hspace{1.5cm}}\\\\
$\underline z = m_z \cdot \underline Z$ & $\underline y = m_y \cdot \underline Y$\\\\
$[m_z] = \dfrac{\mathrm{cm}}{\Omega}$ & $[m_y] = \dfrac{\mathrm{cm}}{\mathrm{S}}$
\end{tabular}

\bigskip

\[\underline Z  \cdot \underline Y = 1 = \dfrac{\underline z}{m_z} \cdot \dfrac{\underline y}{m_y} = 1 \quad \to \quad \underline z \cdot \underline y = m_z \cdot m_y = r^2\]

$r$ -- Radius des Inversionskreises, Inversionsradius

\subsubsection*{Beispiel}

\begin{minipage}{9.3cm}
\begin{tabular}{c|c}
Darzustellender Bereich & gewählter Zeigermaßstab \\
\hline
$Z = 1 \ldots 10\,\Omega$ & $\vphantom{\Bigg|}m_z = \dfrac{1\,\mathrm{cm}}{\Omega}$\\
$Y = 0,\!1 \ldots 1\,\mathrm S$ & $\vphantom{\Bigg|}m_y = \dfrac{10\,\mathrm{cm}}{\Omega}$
\end{tabular}
\end{minipage}
\begin{minipage}{5.3cm}
\begin{align*}
\Rightarrow r &= \sqrt{m_z \cdot m_y} = \sqrt{10\,\mathrm{cm}^2}\\
 &= 3,\!2\,\mathrm{cm}
\end{align*}
\end{minipage}

\begin{picture}(120,100)
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\put(125,45){$\mathrm{Re}(\underline Y)$}
\put(125,28){$\mathrm{Re}(\underline Z)$}
\put(132,11){$\frac{R}{\Omega}$}
\put(132,59){$\frac{G}{\mathrm{S}}$}
\end{picture}


\subsection{Inversion von Ortskurven}

\begin{description}
\item[Problem:] gegeben: Ortskurve von $\underline F(\lambda)$
\qquad
gesucht: Ortskurve von $\dfrac{1}{\underline F(\lambda)}$
\item[Verfahren:] \

\begin{enumerate}
\item Punktweise Inversion: für verschiedene $\lambda$--Werte wird $\dfrac{1}{\underline F(\lambda)}$ aus $\underline F(\lambda)$ durch Zeigerinversion bestimmt.

\item Anwendung von \emph{Inversionsregeln} für einfache Ortskurven
\end{enumerate}
\item[Grundregel:]
 Bei der Inversion gehen Kreise in Kreise über $\longrightarrow$ Die Inversion ist \emph{Kreistreu}.

Beweis: $\underline F(\lambda) = \dfrac{\underline b_0 + \underline b_1\cdot \lambda}{\underline a_0 + \underline a_1 \cdot \lambda} \quad \to \quad  \dfrac{1}{\underline F(\lambda)} = \dfrac{\underline a_0 + \underline a_1\cdot \lambda}{\underline b_0 + \underline b_1 \cdot \lambda}$
\end{description}

\subsubsection*{Wichtige Fälle}
\begin{enumerate}
\item Gerade durch den Ursprung $\leftrightarrow$ Gerade durch den Urpsrung

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\put(117,5){$\frac{1}{\underline F(\lambda)}$}
\put(41,85){$-\frac 1 2$}
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\put(109,25){$\frac 1 2$}
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\put(95,5){$\lambda$}
\put(78,31){$\frac{1}{\underline a}$}
\end{picture}
\end{minipage}
\begin{minipage}{8.9cm}
\[\underline F(\lambda) = \underline a \cdot \lambda\]

\[\frac{1}{\underline F}(\lambda) = \frac{1}{a \cdot \lambda} = \frac{1}{\underline a} \cdot f(\lambda)\]
\end{minipage}

\item Gerade nicht durch den Ursprung $\leftrightarrow$ Kreis durch den Ursprung


\begin{minipage}{6cm}
\begin{picture}(150,105)
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\put(135,40){Re}
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\put(50,0){\vector(0,1){100}}
\put(50,50){\vector(1,1){15}}
\put(50,50){\line(1,1){16}}
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\put(62,0){$\lambda$}

\put(18,80){$\underline F(\lambda)$}
\put(20,33){$\frac{1}{\underline F(\lambda)}$}

\end{picture}
\end{minipage}
\begin{minipage}{8.9cm}

\[\underline F (\lambda) = \underline a + \underline b \cdot  \lambda\]

\[\frac{1}{\underline F} (\lambda) = \frac{1}{\underline a + \underline b \cdot \lambda}\]

\center

$\lambda_1$: minimaler Betrag von $\underline F(\lambda)$ 

$\to$ maximaler Betrag von $\frac{1}{\underline F} \to$ Radius
\end{minipage}

\item Kreis nicht durch den Ursprung $\leftrightarrow$ Kreis nicht durch den Ursprung

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\put(30,50){\vector(3,-2){50}}

% mp: 65,73.3
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\end{picture}
\end{minipage}
\begin{minipage}{8.9cm}
\[\underline F(\lambda)  = \frac{\underline a_0 + \underline a_1 \cdot \lambda}{\underline b_0 + \underline b_1 \cdot \lambda}\]

\[\frac{1}{\underline F(\lambda)} = \frac{\underline b_0 + \underline b_1 \cdot \lambda}{\underline a_0 + \underline a_1 \cdot \lambda}\]
\end{minipage}
\end{enumerate}

\subsection{Anwendungen}

\begin{enumerate}
\item $\underline Z$- und $\underline Y$-Ortskurven technischer Bauelemente in Abhängigkeit von $\omega$


\begin{minipage}{7.0cm}
\center

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\put(22,30){$L$}
\put(50,30){$R_L$}
\end{picture}
\[\underline Z(\omega) = R_L + j\omega L \vphantom{\frac{1}{\frac 1 1}}\]

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\put(68,65){\line(1,0){4}}
\put(74,62){$\omega_{45}$}

\put(75,30){\vector(0,1){20}}
\put(78,37){$\omega$}
\qbezier(40,15)(41,25)(35,30)
\put(30,19){$\varphi$}
\thicklines\put(70,14){\line(0,1){70}}
\end{picture}
\[R_L = \omega_{45} \cdot L, \ \omega_{45} = \frac{R_L}{L} \vphantom{\frac{1}{R_C}}\]
\[Z = \sqrt{{R_L}^2 + (\omega L)^2}, \ \varphi = \arctan\frac{\omega L}{R_L} \vphantom{\frac{1}{\sqrt{{(R_C)}^2}}}\]

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\put(-2,69){$Z$}
\put(-6,24){$R_L$}                         % *hust*
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\put(95,32){$\frac{\pi}{4}$}
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\multiput(115,14)(0,4){5}{\line(0,1){2}}
\put(110,6){$\omega_{45}$}
\end{picture}


\end{minipage}
\begin{minipage}{7.0cm}
\center

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\thicklines \multiput(38,28)(4,0){2}{\line(0,1){14}}
\put(34,13){$R_C$}
\put(37,46){$C$}
\end{picture}
\[\underline Z(\omega) = \frac{1}{\frac{1}{R_C} + j\omega C} = \frac{1}{G +j\omega C}\]

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\put(49,42){$\omega_{45}$}
\put(85,88){$R_L$}
\put(90,60){\vector(-1,-1){10}}
\put(90,48){$\omega$}
\end{picture}
\[G_C = \omega_{45} \cdot C, \ \omega_{45} = \frac{1}{C\cdot R_C}\]
\[Z = \frac{1}{\sqrt{{R_C}^2 + (\omega C)^2}}, \ \varphi = -\arctan\frac{\omega C}{G_C}\]

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\multiput(10,10)(95,0){2}{\vector(0,1){70}}
\multiput(5,15)(95,50){2}{\vector(1,0){80}}
\put(-2,69){$Z$}
\put(-6,50){$R_C$}                         % *hust*
\put(9,53){\line(1,0){71}}
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\qbezier(10,53)(25,50)(32,40)
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\put(95,70){$\varphi$}
\put(95,22){$\frac{\pi}{2}$}
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\put(145,25){\line(1,0){30}}
\end{picture}
\end{minipage}

\item Ortskurve in Abhängigkeit von einem Parameter

\begin{minipage}{6.5cm}
\begin{picture}(150,70)
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\put(154,27){\textsf B}
\put(58,34){\textsf C}
\put(90,0){\vector(1,1){20}}
\put(35,38){$R_1$}
\put(93,20){$R_2$}
\put(96,60){$C$}
\end{picture}
\end{minipage}
\begin{minipage}{8cm}
\begin{description}
\item[gesucht:] Ortskurven $\underline Z_{\textsf{AB}}(R_2),\ \underline{Y}_{\textsf{AB}}(R_2)$

($\underline Z_{\textsf{AB}}, \underline Y_{\textsf{AB}}$ formelmäßig berechnen)
\end{description}
\end{minipage}


\begin{minipage}{7.3cm}
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\put(15,75){\vector(1,0){180}}
\put(19,115){\line(1,0){70}}
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\put(2,112){$\omega C$}
\put(60,114){\line(0,1){2}}
\put(53,118){$R_{2_{45}}$}
\put(48,120){\vector(-1,0){20}}
\put(33,124){$R_2$}
\put(75,103){$\underline Y_{\textsf{CB}}$}
\put(19,25){\line(1,0){2}}
\put(4,22){$\frac{-1}{\omega C}$}
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\put(45,50){\line(1,0){2}}
\put(47,47){$R_{2_{45}}$}
\put(22,13){$\underline Z_{\textsf{CB}}$}
\put(100,74){\line(0,1){2}}
\put(95,79){$R_1$}
\put(95,13){$\underline Z_{\textsf{AB}}$}

\put(170,74){\line(0,1){2}}
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\put(167.5,65){$0$}
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\put(100,114){$Y_{\textsf{AB}}$}
\put(125,122){$\infty$}
\end{picture}
\end{minipage}
\begin{minipage}{7.5cm}
$\underline Y_{\textsf{CB}} = \frac{1}{R_2} + j\omega C, \quad \frac{1}{R_2} = \omega_{45} C,\ R_2 = \frac{1}{\omega_{45} C}$

\bigskip

$R_1$ in Serie zu $\underline Y_{\textsf{CB}} \to$ invertieren der OK
$\to Z_{\textsf{CB}}$, $R_1$ addieren $\to Z_{\textsf{AB}}$, invertieren $\to \underline Y_{\textsf{AB}}$.
\end{minipage}
\end{enumerate}

\section{Frequenzgänge}

Frequenzgänge sind die Darstellungen von Betrag und Phase einer komplexen Größe über die Frequenz.

\paragraph{Beispiel:} Spannungsverhältnis einer $RC$-Schaltung

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\put(126,27){$\underline U_2$}
\end{picture}
\end{minipage}
\begin{minipage}{10cm}

\bigskip

\begin{tabular}{ll}
$\underline G(j\omega) = \dfrac{\underline U_2}{\underline U_1}$ & (Spannungs-)Übertragungsfaktor\\
$A(\omega) = |\underline G(j\omega)| \vphantom{\dfrac A B}$ & Amplituden(frequenz)gang\\
$\varphi(\omega) = \arg(\underline G(j\omega)) \vphantom{\dfrac A B}$ & Phasen(frequenz)gang
\end{tabular}
\end{minipage}

\bigskip

\[\underline G(j\omega) = \frac{\frac{1}{j\omega C}}{R + \frac{1}{j\omega C}} = \frac{1}{1 + j\omega C R} = \frac{1}{j\omega \tau} = \frac{1}{1 + j \frac{\omega}{\omega_g}}\]

\[A(\omega) = |\underline G(j\omega C)| = \frac{1}{\sqrt{1 + \left(\frac{\omega}{\omega_g}\right)^2}} = \left\{\begin{array}{lll} \approx 1 & \text{für} & \omega \ll \omega_g\\ \phantom{\approx} \vphantom{\Bigg|}\frac{1}{\sqrt{2}} & \text{für} & \omega = \omega_g \\ \approx \frac{\omega_g}{\omega} & \text{für} & \omega \gg \omega_g\end{array}\right.\]

\smallskip

\[\varphi(\omega) = - \arctan \frac{\omega}{\omega_g}, \quad \omega_g = \frac 1 {\tau} = \frac 1 {RC}\]

\bigskip

\begin{minipage}{5.2cm}
\center

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\end{minipage}
\begin{minipage}{5.2cm}
\center 

\begin{picture}(100,80)
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\put(-4,68){$A(\omega)$}
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\put(35,6){$\omega_g$}
\put(95,6){$\omega$}
\end{picture}
\end{minipage}
\begin{minipage}{5.2cm}
\center 

\begin{picture}(100,80)
\put(15,55){\vector(1,0){90}}
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\put(-4,68){$\varphi(\omega)$}
\put(95,46){$\omega$}
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\multiput(19,37.5)(4,0){4}{\line(1,0){2}}
\put(10,17){$\frac{\pi}{2}$}
\put(10,34.5){$\frac{\pi}{4}$}
\multiput(34,56)(0,-4){5}{\line(0,-1){2}}
\put(30,59){$\omega_g$}
\end{picture}

\end{minipage}

\section{Bode Diagramm}

$\left.
\begin{array}{ll}
a(\omega) = 20 \log |\underline G(j\omega)| \, \mathrm{dB}\\
\varphi(\omega) 
\end{array}
\right\}$ dargestellt über $\log \frac{\omega}{\omega_g}$.

\smallskip

\begin{align*}
a(\omega) &= 20 \log\frac{1}{\sqrt{1+\left(\frac{\omega}{\omega_g}\right)^2}}\,\mathrm{dB} = - 20 \cdot \frac 1 2 \cdot \log\left(1+\left(\frac{\omega}{\omega_g}\right)^2\right)\,\mathrm{dB} \\
&= \left\{\begin{array}{rll}
0\,\mathrm{dB} & \text{für} & \frac{\omega}{\omega_g} \ll 1\\
-3\,\mathrm{dB} & \text{für} & \frac{\omega}{\omega_g} = 1\\
- 20 \cdot \log\frac{\omega}{\omega_g} \, \mathrm{dB} & \text{für} & \frac{\omega}{\omega_g} \gg 1\end{array}\right.
\end{align*}

{\centering%
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\put(72,84){$1$}
\put(97,84){$2$}
\put(122,84){$3$}
\put(144,84){$\cdots$}
\put(220,82){$\log\frac{\omega}{\omega_g}$}
\put(228,104){$\frac{\omega}{\omega_g}$}
\put(18,99){$0,\!1$}
\put(42,99){$1$}
\put(69,99){$10$}
\put(91,99){$100$}
\put(114,99){$1000$}
\put(144,99){$\cdots$}
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\put(12,47){$-20\,\mathrm{dB}$}
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\qbezier(75,70)(90,54)(115,0)
\put(105,47){Steigung: $20\,\mathrm{dB}/$Dekade}
\end{picture}
}

\subsubsection*{Ausgewählte Größenordnungen}

\quad \begin{tabular}{p{22pt}||p{22pt}|p{22pt}|p{22pt}|p{22pt}|p{22pt}|p{22pt}|p{22pt}}
{\centering $\dfrac{U_2}{U_1} \vphantom{\Bigg|}$} & {\centering $10^{-2}$} & {\centering $10^{-1}$} & {\centering  $\dfrac 1 {\sqrt{2}}$} & {\centering  1} &{\centering  $\sqrt 2$} & {\centering 10}&{\centering  100}\\
\hline
\centering $\dfrac{a}{\mathrm{dB}} \vphantom{\Bigg|}$ &\centering $-40$ &\centering $-20$ &\centering $-3$ & \centering $0$ & \centering $3$ & \centering $20$ & \centering $40$\\
\end{tabular}


\chapter{Resonanzkreise}

\section{Impedanz und Admittanz}

\begin{minipage}{7.8cm}
\center

Reihenresonanzkreis

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\end{picture}

\end{minipage}
\begin{minipage}{7.8cm}
\center

Parallelresonanzkreis

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\put(26,9){$L_p$}
\put(26,42){$C_p$}
\put(26,69){$R_p$}
\end{picture}
\end{minipage}

\smallskip

\begin{center}
Impedanz
\end{center}

\vspace{-0.5cm}

\begin{minipage}{7.8cm}
\begin{align*}
\underline Z_r &= R_r + j\omega L_r + \frac{1}{j\omega C_r}\\
             &= R_r + j\Big(\underbrace{\omega L_r - \frac{1}{\omega C_r}}_{X_r(\omega)}\Big)
\end{align*}
\end{minipage}
\begin{minipage}{7.8cm}
\begin{align*}
\underline Z_p &= \frac{1}{G_p + j\omega C_p + \frac{1}{j\omega L_p}}\\
             &= \frac{1}{G_p + j(\omega C_p - \frac 1 {\omega L_p})}
\end{align*}
\end{minipage}


\begin{minipage}{7.8cm}         % warum ich kaffee haag trinke? hmhmhm..
\center

\begin{picture}(155,115)
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\put(78,30){\vector(0,1){15}}
\put(81,34){$\omega$}
\put(120,20){$\omega < \omega_0$}
\put(120,74){$\omega > \omega_0$}
\end{picture}
\end{minipage}
\begin{minipage}{7.8cm}
\center

\begin{picture}(155,115)
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\put(103,53){$\omega_0$}
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\put(68,35){$\underline Z_p$}
\put(120,20){$\omega > \omega_0$}
\put(120,74){$\omega < \omega_0$}
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\put(99,80){$\omega$}
\end{picture}
\end{minipage}


\bigskip

\begin{center}
Admittanz
\end{center}

\vspace{-0.5cm}

\begin{minipage}{7.8cm}
\begin{align*}
\underline Y_r &= \frac{1}{R_r + j\omega L_r + \frac{1}{j\omega C_r}}\\
               &= \frac{1}{R_r + j(\omega L_r - \frac{1}{\omega C_r}}
\end{align*}
\end{minipage}
\begin{minipage}{7.8cm}
\begin{align*}
\underline Y_p &= G_p + j \omega C_p + \frac{1}{j\omega L_p}\\
             &= G_p + j(\omega C_p - \frac{1}{\omega L_p}
\end{align*}
\end{minipage}

\begin{minipage}{7.8cm}         % 2
\center

\begin{picture}(155,115)
\put(2,90){$\mathrm{Im}(\underline Y_r)$}
\put(130,38){$\mathrm{Re}(\underline Y_r)$}
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\put(99,80){$\omega$}
\end{picture}

\bigskip

Scheinwiderstand
\[Z = |\underline Z| = \sqrt{{R_r}^2 + \left(\omega L_r - \frac{1}{\omega C_r}\right)^2}\]


\end{minipage}
\begin{minipage}{7.8cm}
\center

\begin{picture}(155,115)
\put(2,90){$\mathrm{Im}(\underline Y_p)$}
\put(130,38){$\mathrm{Re}(\underline Y_p)$}
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\end{picture}

\bigskip

Scheinleitwert
\[Y = |Y| = \sqrt{{G_p}^2 + \left(\omega C_p - \frac{1}{\omega L_p}\right)^2}\]
\end{minipage}

\bigskip

\begin{center}
Phasenwinkel
\end{center}

\vspace{-0.5cm}

\begin{minipage}{7.8cm}
\center

\[\varphi_z = \arctan\frac{\omega L_r\vphantom{C_p} - \frac{1}{\omega C_r\vphantom{L_p}}}{R_r\vphantom{G_p}}\]

\begin{picture}(180,170)(0,-30)
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\put(15,-30){kapazitiv}
\put(94,-12){$\omega > \omega_0$}
\put(90,-30){induktiv}
\end{picture}


\end{minipage}
\begin{minipage}{7.8cm}
\center

\[\varphi_y = \arctan\frac{\omega C_p - \frac{1}{\omega L_p}}{G_p}\]
\begin{picture}(180,170)(0,-30)
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\put(90,-30){kapazitiv}
\end{picture}

\end{minipage}

\bigskip

\begin{minipage}{7.8cm}
\center

\begin{picture}(180,120)
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\end{picture}
\end{minipage}
\begin{minipage}{7.8cm}
\center

\begin{picture}(180,120)
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\put(54,108){$b_{\omega}$}
\end{picture}
\end{minipage}

\begin{center}
Resonanzfrequenz $\omega_0$
\end{center}


\begin{minipage}{7.8cm}
\center

Blindwiderstand
\[X_r(\omega_0) = 0\]
\[\omega_0 \cdot L_r = \frac 1 {\omega_0\cdot C_r},\ \omega_0 = \frac{1}{\sqrt{L_r \cdot C_r}}\]
\end{minipage}
\begin{minipage}{7.8cm}
\center

Blindleitwert
\[B_p(\omega_0) = 0\]
\[\omega_0\cdot C_p = \frac{1}{\omega_0\cdot L_p},\ \omega_0 = \frac{1}{\sqrt{C_p \cdot L_p}}\]
\end{minipage}

\newpage

\begin{center}   % hätte man das 
Imaginärteil von
\end{center}
\begin{minipage}{7.8cm}
\center

Impedanz
\end{minipage}
\begin{minipage}{7.8cm}
\center

Admittanz
\end{minipage}
\begin{center}
verschwindet,
\end{center}
\begin{minipage}{7.8cm}
\center

Scheinwiderstand
\end{minipage}
\begin{minipage}{7.8cm}
\center

Scheinleitwert
\end{minipage}
\begin{center}
erreicht Minimum.
\end{center}

\begin{minipage}{7.0cm}
\[\underline U = \textrm{const.} \ \Rightarrow \ I \sim \frac 1 Z\]
Der Strom hat im spannungsgespeisten Resonanzkreis bei Resonanz Höchstwert.

\end{minipage}
\qquad
\begin{minipage}{7.0cm}
\[\underline I = \textrm{const.} \ \Rightarrow \ U \sim \frac 1 Y\]
Die Spannung hat im stromgespeisten Resonanzkreis bei Resonanz Höchstwert.
\end{minipage}

\section{Normierte Größen}

\[\boxed{\quad \varrho = \frac{|\,\text{Zwischen }L\text{ und }C\text{ pendelnde Blindleistung}\,|}{\text{Wirkleistung}}\vphantom{\Bigg|}\quad } \qquad \text{Güte}\]

\bigskip

\begin{minipage}{7.8cm}
\center

Reihenresonanzkreis
\begin{align*}
\varrho_r &= \frac{I^2 \cdot \omega_0 \cdot L_r\vphantom{C_p}}{I^2 \cdot R_r\vphantom{C_p}} = \frac{\omega_0 \cdot L_r}{R_r}\\
 &= \frac 1 {\omega_0 \cdot C_r \cdot R_r\vphantom{C_p}} = \frac{1}{R_r}\cdot\sqrt{\frac{L_r\vphantom{C_p}}{C_r\vphantom{C_p}}}
\end{align*}
\end{minipage}
\begin{minipage}{7.8cm}
\center

Parallelresonanzkreis
\begin{align*}
\varrho_p &= \frac{U^2 \cdot \omega_0 \cdot C_p}{U^2 \cdot R_p} = \frac{\omega_0 \cdot C_p}{R_p}\\
 &= \frac 1 {\omega_0 \cdot L_p \cot G_p} = \frac{1}{G_p}\cdot\sqrt{\frac{C_p}{L_p}}
\end{align*}
\end{minipage}

\subsection{Barkhausen-Verstimmung}

\[\boxed{\quad \vphantom{\\Bigg|} v = \frac{\omega}{\omega_0} - \frac{\omega_0}{\omega} = \Omega - \frac{1}{\Omega}\quad} \qquad \Omega = \frac{\omega}{\omega_0} \]

\subsubsection*{Näherung}

\[v \approx \left\{\begin{array}{ll}-\frac 1 {\Omega} & \Omega \ll 1\\ 2(\Omega -1) & |\Omega -1| \ll 1\quad *\vphantom{\Big|}\\\Omega & \Omega \gg 1\end{array}\right.\]

$*)$ Näherung für $\omega \approx \omega_0$:

\qquad mit $\omega = \omega_0 + \Delta \omega$
\begin{align*}
v &= \frac{\omega_0+\Delta\omega}{\omega_0} - \frac{\omega_0}{\omega_0+\Delta\omega} = 1+\frac{\Delta\omega}{\omega_0} - \frac{1}{1+\frac{\Delta\omega}{\omega_0}}, && \frac{1}{1+ \frac{\Delta\omega}{\omega_0}} \approx 1-\frac{\Delta\omega}{\omega_0} \text{ für }\left|\frac{\Delta\omega}{\omega_0}\right|\ll 1\\
&= 1 + \frac{\Delta\omega}{\omega_0} - 1+\frac{\Delta\omega}{\omega_0} = 2\cdot\frac{\omega-\omega_0}{\omega_0} = 2\cdot\left(\frac{\omega}{\omega_0}-1\right)\\
&=  2(\Omega -1)
\end{align*}

\subsection{45°--Frequenzen}

\begin{minipage}{7.8cm}
\center

Reihenresonanzkreis
\[\varphi_{z} = \pm \dfrac{\pi}{4}\]
\[\varrho_r\vphantom{\varrho_p} \cdot v_{45} = \pm 1\]
\[v_{45} = \pm \frac{1}{\varrho_r\vphantom{\varrho_p}}\]
\end{minipage}
\begin{minipage}{7.8cm}
\center

Parallelresonanzkreis
\[\varphi_{y} = \pm \dfrac{\pi}{4}\]
\[\varrho_p \cdot v_{45} = \pm 1\]
\[v_{45} = \pm \frac{1}{\varrho_p}\]
\end{minipage}

\bigskip

\[\omega_{\pm 45} = \omega_0\cdot\left(\sqrt{1+\left(\frac{1}{2\varrho}\right)} \pm \frac{1}{2\varrho}\right) \approx \omega_0 \cdot \left(1\pm \frac{1}{2\varrho}\right) \quad \text{für } \varrho \gg 1\]

\bigskip

\begin{minipage}{7.8cm}
\[Z(\omega_{45}) = \sqrt{2} \cdot R_r\vphantom{Y_p}\]
\end{minipage}
\begin{minipage}{7.8cm}
\[Y(\omega_{45}) = \sqrt{2} \cdot Y_p\]
\end{minipage}

\subsection{Bandbreite}
absolut:
\begin{align*}
b_{\omega} &= \omega_{+45}-\omega_{-45} & b_f &= f_{+45^{\circ}} - f_{-45^{\circ}} = \frac{b_{\omega}}{2\pi}\\
b_{\omega} &= \frac{\omega_0}{\varrho} & b_f &= \frac{f_0}{\varrho}
\end{align*}
relativ:
\[b = \frac{b_{\omega}}{\omega_0} = \frac{b_f}{f_0} = \frac{1}{\varrho}\]

\section{Strom-- und Spannungsverläufe (Reihenschaltung)}


\begin{minipage}{5cm}
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\end{minipage}
\begin{minipage}{10.8cm}
\[\underline I = \frac{\underline U}{\underline Z} = \frac{U}{R + j\omega L + \frac{1}{j\omega C}} = \frac{U}{R(1+j\varrho v)}\]
\[\boxed{ \quad v = \frac{\omega}{\omega_0} - \frac{\omega_0}{\omega} = \Omega - \frac{1}{\Omega} \quad} \qquad \boxed{\quad \Omega = \frac{\omega}{\omega_0}\quad}\]

\bigskip
\end{minipage}

\bigskip

\[\boxed{\quad I = \frac{U}{R\cdot \sqrt{1+(\varrho v)^2}}\quad} \hspace{2cm} \varphi_I = -\arctan (\varrho v)\]


\begin{align*}
\frac{\underline U_L}{U} &= \frac{j\omega L}{R(1+j\varrho v)} = j \frac{\omega_0 L \frac{\omega}{\omega_0}}{R(1+j\varrho v)} = j\varrho \frac{\Omega}{1+j\varrho v} = \frac{1}{1-\Omega^2 - j \frac{1}{\varrho \Omega}}\\
\frac{\underline U_C}{U} &= \frac{1}{j\omega C \cdot R(1+j\varrho v)} = \frac{\frac{\omega}{\omega_0}}{j\omega_0 C \cdot R(1+j\varrho v)} = \frac{\varrho}{j} \frac{\frac{1}{\Omega}}{1+j\varrho v} = \frac{1}{1-\Omega^2 - j \frac{\Omega}{\varrho}}
\end{align*}

\[\boxed{\quad \frac{U_L}{U} = \varrho \cdot \frac{\Omega}{\sqrt{1+(\varrho v)^2}}\quad} \hspace{2cm} \varphi_{U_L} = \frac{\pi}{2} -\arctan (\varrho v)\]

\bigskip

\[\left.\frac{U_L}{U}\right|_{\text{max}} = \frac{\varrho}{\sqrt{1-\frac{1}{4\varrho^2}}} \approx \varrho \left(1+\frac{1}{8\varrho^2}\right) \quad \text{bei}\quad \Omega_{L_{\text{max}}} = \frac{1}{\sqrt{1-\frac{1}{2\varrho^2}}} \approx 1 + \frac{1}{4\varrho^2}\]

\newpage

\[\boxed{\quad \frac{U_C}{U} = \varrho \cdot \frac{\frac{1}{\Omega}}{\sqrt{1+(\varrho v)^2}}\quad} \hspace{2cm} \varphi_{U_L} = -\frac{\pi}{2} -\arctan (\varrho v)\]

\bigskip

\[\left.\frac{U_C}{U}\right|_{\text{max}} = \frac{\varrho}{\sqrt{1-\frac{1}{4\varrho^2}}} \approx \varrho \left(1+\frac{1}{8\varrho^2}\right) \quad \text{bei}\quad \Omega_{C_{\text{max}}} = \frac{1}{\sqrt{1-\frac{1}{2\varrho^2}}} \approx 1 + \frac{1}{4\varrho^2}\]


\vspace{1cm}

\begin{minipage}{7cm}
\center

Ortskurven der Teilspannungen, $\varrho = 3$

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\begin{minipage}{8cm}
\center

Effektivwerte der Teilspannungen

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\put(71,125){$\underline U_L$}
\put(63,72){$\underline I$}
\put(160,35){$\Omega$}
\end{picture}
\end{minipage}

\chapter{Lineare Zweitore}
\section{Strom-Spannungsbeziehungen  (Zweitorgleichungen)}
\subsection{Allgemeines}

\begin{minipage}{7.8cm}
\center

\begin{picture}(140,50)
\multiput(50,0)(0,35){2}{\line(1,0){35}}
\multiput(50,0)(35,0){2}{\line(0,1){35}}
\multiput(40,5)(0,25){2}{\line(1,0){10}}
\multiput(85,5)(0,25){2}{\line(1,0){10}}
\multiput(39,5)(0,25){2}{\circle{2}}
\multiput(96,5)(0,25){2}{\circle{2}}
\put(15,30){\vector(1,0){20}}
\put(39,27){\vector(0,-1){19}}
\put(20,34){$\underline I_1$}
\put(24,15){$\underline U_1$}
\put(100,30){\vector(1,0){20}}
\put(96,27){\vector(0,-1){19}}
\put(105,34){$\underline I_2$}
\put(99,15){$\underline U_2$}
\end{picture}

\bigskip

\emph{Kettenpfeilsystem}

\smallskip

$\Rightarrow$ bei Kettenschaltung verwendet

\end{minipage}
\begin{minipage}{7.8cm}
\center

\begin{picture}(140,50)
\multiput(50,0)(0,35){2}{\line(1,0){35}}
\multiput(50,0)(35,0){2}{\line(0,1){35}}
\multiput(40,5)(0,25){2}{\line(1,0){10}}
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\put(39,27){\vector(0,-1){19}}
\put(20,34){$\underline I_1$}
\put(24,15){$\underline U_1$}
\put(120,30){\vector(-1,0){20}}
\put(96,27){\vector(0,-1){19}}
\put(105,34){$\underline I_2$}
\put(99,15){$\underline U_2$}
\end{picture}

\bigskip

\emph{symmetrisches Pfeilsystem}

\smallskip

$\Rightarrow$ nachfolgend verwendet
                    
\end{minipage}

\bigskip

\begin{description}
\item[Beispiele:] elektronische Bauelemente (Transistoren), Verstärker, Filter, Leitung
\item[Beschreibung:] Von den 4 Klemmengrößen können zwei vorgegeben werden (unabhängige Variable), die anderen beiden ergeben sich daraus.

$\Rightarrow$ 6 Möglichkeiten
\end{description}


\subsection{Widerstandsform der Zweitorgleichungen}

\[
\left.\begin{array}{ll}
\underline U_1 &= \underline Z_{11} \cdot \underline I_1 + \underline Z_{12} \cdot \underline I_2  \vphantom{\Big|}\\ 
\underline U_2 &= \underline Z_{21} \cdot \underline I_1 + \underline Z_{22} \cdot \underline I_2  \vphantom{\Big|}
\end{array}\right\}
\qquad (\underline U) = (\underline Z) \cdot (\underline I)
\]

\begin{description}
\item[$(\underline Z)$:] Impedanzmatrix

\item[$Z_{ij} (i,j = 1,2)$:] $Z$-- oder Widerstandsparameter des Zweipols
\end{description}

\begin{center}
\begin{tabular}{|c|c|}
\hline
Eingangsimpedanz & Transimpedanz rückwärts\\
$\underline Z_{11} = \left.\dfrac{\underline U_1}{\underline I_1}\right|_{\underline I_2 = 0} \vphantom{\dfrac{\dfrac{a}{a}}{\dfrac{a}{a}}}$ & $\underline Z_{12} = \left.\dfrac{\underline U_1}{\underline I_2}\right|_{\underline I_1 = 0}$\\
\hline
Transimpedanz vorwärts & Innenimpedanz vorwärts\\
$\underline Z_{21} = \left.\dfrac{\underline U_2}{\underline I_1}\right|_{\underline I_2 = 0} \vphantom{\dfrac{\dfrac{a}{a}}{\dfrac{a}{a}}}$ & $\underline Z_{22} = \left.\dfrac{\underline U_2}{\underline I_2}\right|_{\underline I_1 = 0}$\\
\hline
Bei Leerlauf am Ausgang ($\underline I_2 = 0 \vphantom{\Big|}$) &
Bei Leerlauf am Eingang ($\underline I_1 = 0$)\\
\hline
\end{tabular}
\end{center}

\subsection*{Bestimmung der Widerstandsparameter}

\subsubsection{Durch (Gesamt-)Netzwerkanalyse}

\begin{center}
\begin{picture}(220,75)
\put(0,27){$\underline I_1\Big\uparrow$}
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}
\put(150,60){\line(1,0){20}}
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\put(190,0){\line(0,1){60}}
\put(190,30){\circle{20}}
\put(203,27){$\Big\uparrow \underline I_2$}
\put(49,55){\vector(0,-1){50}}
\put(52,27){$\underline U_1$}
\put(156,27){$\underline U_2$}
\put(171,55){\vector(0,-1){50}}
\put(75,69){$\underline Z_1$}
\put(115,69){$\underline Z_3$}
\put(80,27){$\underline Z_2$}
\put(120,27){$\underline Z_4$}
\end{picture}
\end{center}
\begin{align*}
\underline U_1 &= \left(\underline Z_1 + \frac{1}{\frac{1}{\underline Z_2} + \frac{1}{\underline Z_3 + \underline Z_4}}\right) \cdot \underline I_1 + \frac{\underline Z_4}{\underline Z_2 + \underline Z_3 + \underline Z_4} \cdot \underline I_2\\
\\
\underline U_2 &= \frac{\underline Z_2 + \underline Z_4}{\underline Z_2 + \underline Z_3 + \underline Z_4} \cdot \underline I_1  + \frac{1}{\frac{1}{\underline Z_4} + \frac{1}{\underline Z_2 + \underline Z_3}} \cdot \underline I_2
\end{align*}

Durch Koeffizientenvergleich:

\begin{align*}
\underline Z_{11} &= \underline Z_1 + \frac{\underline Z_2 \cdot (\underline Z_3 + \underline Z_4)}{\underline Z_2 + \underline Z_3 + \underline Z_4} &
\underline Z_{12} &= \frac{\underline Z_4 \cdot \underline Z_2}{\underline Z_2 + \underline Z_3 + \underline Z_4}\\ \\
\underline Z_{21} &= \frac{\underline Z_2 \cdot \underline Z_4}{\underline Z_2 + \underline Z_3 + \underline Z_4} & \underline Z_{22} &= \frac{\underline Z_4 \cdot (\underline Z_2 + \underline Z_3)}{\underline Z_2 + \underline Z_3 + \underline Z_4}
\end{align*}

\subsubsection{Aus den Definitionen der Zweitorparameter}

\begin{enumerate}
\item $\underline I_2 = 0$: \qquad $\underline Z_{11} = \left.\dfrac{\underline U_1}{\underline I_1}\right|_{\underline I_2 = 0},\ \underline Z_{21} = \left.\dfrac{\underline U_2}{\underline I_1}\right|_{\underline I_2=0}$
\qquad
\raisebox{-20pt}{  %  naja
\begin{picture}(170,40)
\put(1,17){$\underline I_1\Big\uparrow$}
\multiput(30,5)(0,25){2}{\line(0,1){5}}
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\put(30,20){\circle{20}}
\put(20,20){\line(1,0){20}}
\multiput(45,5)(0,30){2}{\circle{2}}
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\multiput(65,0)(0,40){2}{\line(1,0){40}}
\put(45,32){\vector(0,-1){24}}
\put(48,17){$\underline U_1$}
\multiput(105,5)(0,30){2}{\line(1,0){19}}
\multiput(125,5)(0,30){2}{\circle{2}}
\put(125,32){\vector(0,-1){24}}
\put(129,17){$\underline U_2$}
\end{picture}
}
\item $\underline I_1 = 0$: \qquad $\underline Z_{12} = \left.\dfrac{\underline U_1}{\underline I_2}\right|_{\underline I_1 = 0},\ \underline Z_{22} = \left.\dfrac{\underline U_2}{\underline I_2}\right|_{\underline I_1=0}$
\qquad
\raisebox{-20pt}{  
\begin{picture}(170,40)
\put(150,17){$\Big\uparrow\underline I_2$}
\multiput(140,5)(0,25){2}{\line(0,1){5}}
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\put(140,20){\circle{20}}
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\put(45,32){\vector(0,-1){24}}
\put(29,17){$\underline U_1$}
\multiput(105,5)(0,30){2}{\line(1,0){19}}
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\put(125,32){\vector(0,-1){24}}
\put(110,17){$\underline U_2$}
\end{picture}
}
\end{enumerate}

\paragraph{Beispiel:} gesucht: $\underline Y_{12}$

\begin{minipage}{7cm}
\begin{picture}(180,90)
\put(5,27){$\underline U_1 = 0$}
\put(45,0){\line(0,1){60}}
\multiput(45,0)(0,60){2}{\line(1,0){9}}
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\put(56,0){\line(1,0){78}}
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\put(115,60){\line(1,0){19}}
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\put(155,0){\line(0,1){60}}
\put(155,30){\circle{20}}
\put(165,27){$\Big\downarrow \underline U_2$}
\put(80,69){$\underline Z_1$}
\put(123,27){$\underline Z_2$}
\put(47,65){\vector(1,0){18}}
\put(50,69){$\underline I_1$}
\end{picture}
\end{minipage}
\begin{minipage}{8cm}
\vspace{0.75cm}

\[\underline Y_{12} = \left.\frac{\underline I_1}{\underline U_2}\right|_{\underline U_1 = 0} = -\frac{\underline U_2}{\underline Z_1 \cdot \underline U_2} = -\frac{1}{\underline Z_1} \]
\end{minipage}

\subsection{Umrechnung von Zweitorparametern}

Zweitorparameter können ineinander umgerechnet werden. 
\begin{description}
\item[Beispiel 1:] $\underline Z$- $\to \underline Y$-Parameter

\smallskip

gegeben: $(\underline Z) = \left(\begin{array}{cc}\underline Z_{11} & \underline Z_{12}\\ \underline Z_{21} & \underline Z_{22}\end{array}\right)$ \qquad 
gesucht: $(\underline Y) = \left(\begin{array}{cc}\underline Y_{11} & \underline Y_{12}\\ \underline Y_{21} & \underline Y_{22}\end{array}\right)$ \qquad 

\smallskip

\[(\underline U) = (\underline Z) \cdot (\underline I), \qquad (\underline I) = (\underline Z) \cdot (\underline U)\]
\[(\underline Y) = (\underline Z)^{-1} = \frac{1}{\det(\underline Z)} \cdot \left(\begin{array}{rr} \underline Z_{22} & -\underline Z_{12} \\ -\underline Z_{21} & \underline Z_{11}\end{array}\right), \qquad \det(\underline Z) = \underline Z_{11} \cdot \underline Z_{22} - \underline Z_{12} \cdot \underline Z_{21}\]

\item[Beispiel 2:] $\underline A_{11}$ ist aus den $\underline Z$-Parametern zu berechnen.

\[\underline A_{11} = \left.\frac{\underline U_1}{\underline U_2}\right|_{I_2 = 0} = \frac{\underline Z_{11}}{\underline Z_{21}} \qquad \begin{array}{l}
\underline U_1 = \underline Z_{11} \cdot \underline I_1 + \underline Z_{12} \cdot \underline I_2 = \underline Z_{11} \cdot \underline I_1\\
\underline U_2 = \underline Z_{21} \cdot \underline I_1 + \underline Z_{22} \cdot \underline I_2 = \underline Z_{21} \cdot \underline I_1
\end{array} \]
\end{description}

\section{Zusammenschaltung von Zweitoren}

\begin{minipage}[t]{7.8cm}
\center 

\begin{picture}(130,90)
\multiput(0,0)(0,45){2}{%
 \multiput(50,0)(0,35){2}{\line(1,0){35}}
 \multiput(50,0)(35,0){2}{\line(0,1){35}}
 \multiput(41,5)(0,25){2}{\line(1,0){9}}
 \multiput(40,5)(0,25){2}{\circle{2}}
 \put(40,26){\vector(0,-1){17}}
}
\put(22,15){$\scriptstyle \underline U_1^{(2)}$}
\put(22,60){$\scriptstyle \underline U_1^{(1)}$}
\put(0,37){$\underline U_1$}
\multiput(15,5)(0,70){2}{\circle{2}}
\multiput(16,5)(0,70){2}{\line(1,0){23}}
\put(15,71){\vector(0,-1){62}}
\put(40,31){\line(0,1){18}}
\put(20,78){\vector(1,0){15}}
\put(23,83){$\underline I_1$}
\put(36,47.5){\vector(0,-1){15}}
\put(25,38){$\scriptstyle \underline I_1$}
\multiput(85,0)(0,45){2}{
 \multiput(0,5)(0,25){2}{\line(1,0){9}}
 \multiput(10,5)(0,25){2}{\circle{2}}
 \put(10,26){\vector(0,-1){17}}
}
\put(95,31){\line(0,1){18}}
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\put(99,47.5){\vector(0,-1){15}}
\put(115,78){\vector(-1,0){15}}
\put(103,83){$\underline I_2$}
\put(98,15){$\scriptstyle \underline U_2^{(2)}$}
\put(98,60){$\scriptstyle \underline U_2^{(1)}$}
\put(103,38){$\scriptstyle \underline I_2$}
\put(123,37){$\underline U_2$}
\put(60,14.5){$(2)$}
\put(60,59.5){$(1)$}
\end{picture}

\begin{align*}
(\underline U) &= \left(\underline U^{(1)}\right) + \left(\underline U^{(2)}\right)\\
&= \left(\underline Z^{(1)}\right) \cdot \underbrace{\left(I^{(1)}\right)}_{(\underline I)} + \left(\underline Z^{(2)}\right) \cdot \underbrace{\left(I^{(2)}\right)}_{(\underline I)}\\
&= \left(\left(\underline Z^{(1)}\right) + \left(\underline Z^{(2)}\right)\right) \cdot \left(\underline I\right)
\end{align*}

\end{minipage}
\begin{minipage}[t]{7.8cm}
\center

\begin{picture}(130,90)
\multiput(0,0)(0,45){2}{%
 \multiput(50,0)(0,35){2}{\line(1,0){35}}
 \multiput(50,0)(35,0){2}{\line(0,1){35}}
 \multiput(41,5)(0,25){2}{\line(1,0){9}}
 \multiput(40,5)(0,25){2}{\circle{2}}
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\put(-1,15){$\scriptstyle \underline U_1^{(2)}$}
\put(-1,60){$\scriptstyle \underline U_1^{(1)}$}
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\put(36,83){$\scriptstyle \underline I_1^{(1)}$}

\put(33,70){\vector(0,-1){15}}
\put(36,60){$\scriptstyle \underline I_1^{(2)}$}
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\put(35,5){\circle*{2}}

\put(30,30){\line(0,1){45}}
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\put(30,75){\circle*{2}}

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 \multiput(0,5)(0,25){2}{\line(1,0){9}}
 \multiput(10,5)(0,25){2}{\circle{2}}
}

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\put(96,30){\line(1,0){9}}
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\put(88,60){$\scriptstyle \underline I_2^{(2)}$}
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\put(123,37){$\underline U_2$}
\put(60,14.5){$(2)$}
\put(60,59.5){$(1)$}
\multiput(124,27)(0,23){2}{$\scriptstyle =$}
\end{picture}

\begin{align*}
(\underline I) &= \left(\underline I^{(1)}\right) + \left(\underline I^{(2)}\right)\\
&= \left(\left(Y^{(1)}\right)+\left(Y^{(2)}\right)\right)\cdot (\underline U)
\end{align*}
\end{minipage}

\section{Klassifikation von Zweitoren}

\begin{center}  % *hust*
Zweitore

$\swarrow \hspace{2.2cm} \searrow$

\hspace{0.7cm}Umkehrbare \hspace{3cm} nicht Umkehrbar

$\swarrow \hspace{2cm} \searrow$ \hspace{5.3cm}

 symmetrisch \qquad unsymmetrisch\mbox{\hspace{5.2cm}} 
\end{center}

\subsection{Umkehrbare Zweitore}
\begin{center}
\begin{picture}(150,60)
\put(1,17){$\underline I_0\Big\uparrow$}
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\multiput(125,5)(0,30){2}{\circle{2}}
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\put(129,17){$\underline U_2$}
\put(45,38){\vector(1,0){15}}
\put(41,42){$\scriptstyle \underline I_1 = \underline I_0$}
\put(76,17){$(\underline Z)$}
\put(157,17){$=$}
\end{picture}
\begin{picture}(170,60)
\put(150,17){$\Big\uparrow\underline I_0$}
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\put(107,42){$\scriptstyle \underline I_2 = \underline I_0$}
\put(76,17){$(\underline Z)$}
\end{picture}
\end{center}

\[\left.\frac{\underline U_2}{\underline I_1}\right|_{I_2 = 0} = \boxed{\vphantom{\Big|}\underline Z_{21} \qquad = \qquad \underline Z_{12}} = \left.\frac{\underline U_1}{\underline I_2}\right|_{I_1=0} \]

für $\underline Y$-Parameter:
\[-\frac{\underline Y_{21}}{\det(\underline Y)} = -\frac{\underline Y_{12}}{\det(\underline Y)}\]
\[\boxed{\vphantom{\Big|} \underline Y_{21} = \underline Y_{12}}\]

\begin{center}
\begin{picture}(150,40)
\put(0,17){$\underline U_0\Big\downarrow$}
\multiput(30,5)(0,25){2}{\line(0,1){5}}
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\multiput(125,5)(0,30){2}{\circle{2}}
\put(125,34){\line(0,-1){28}}
\put(129,17){$\Big\uparrow\; \underline I_2$}
\put(76,17){$(\underline Z)$}
\put(157,17){$=$}
\end{picture}
\begin{picture}(170,40)
\put(150,17){$\Big\downarrow\underline U_0$}
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\end{picture}
\end{center}
für $\underline A$-Parameter: \qquad $\boxed{\det(\underline A) = -1 \vphantom{\Big|}}$

\bigskip

Ein $RLCM$--Zweitor ist immer umkehrbar. Ein nicht umkehrbares Zweitor enthält stets gesteuerte Quellen. Ein Zweitor mit gesteuerten Quellen \emph{kann} umkehrbar sein.

\subsubsection*{Beispiel}

\begin{minipage}{7cm}
\begin{picture}(170,70)
\multiput(29,56)(107,0){2}{\vector(0,-1){52}}
\put(10,64){$\underline I_1$}
\put(5,60){\vector(1,0){20}}
\put(148,64){$\underline I_2$}
\put(160,60){\vector(-1,0){20}}
\put(12,27){$\underline U_1$}
\put(139,27){$\underline U_2$}
\multiput(29,0)(0,60){2}{\circle{2}}
\multiput(30,0)(0,60){2}{\line(1,0){40}}
\multiput(70,0)(0,40){2}{\line(0,1){20}}
\multiput(0,0)(25,0){2}{
\multiput(70,20)(-10,10){2}{\line(1,1){10}}
\multiput(70,20)(10,10){2}{\line(-1,1){10}}
}
\put(60,30){\line(1,0){20}}
\put(95,0){\line(0,1){60}}
\multiput(95,0)(0,60){2}{\line(1,0){40}}
\multiput(136,0)(0,60){2}{\circle{2}}
\put(57,40){\vector(0,-1){20}}
\put(37,27){$b\, \underline I_2$}
\put(108,40){\vector(0,-1){20}}
\put(111,27){$a\, \underline U_1$}
\end{picture}
\end{minipage}
\begin{minipage}{8cm}
Wann ist dieses Zweitor umkehrbar?
\[\begin{array}{l} \underline U_2 = a \cdot \underline U_1\\ \underline I_1 = b \cdot \underline I_2\end{array} \Rightarrow \text{inverse Hybriddarstellung}\]
\[\Rightarrow \underline U_1 = \frac 1 a \cdot \underline U_2, \qquad \underline I_1 = -b \cdot (-\underline I_2)\]
\end{minipage}

\bigskip

$\underline A$-Beschreibung:
\[\det(\underline A) = \left(\begin{array}{rr}\frac 1 a & 0 \\ 0 & -b\end{array}\right) = - \frac b a = -1 \quad \Leftrightarrow \quad b = a\]
$\Rightarrow$ idealer Transformator

\subsection{Symmetrische Zweitore}

Ein umkehrbares Zweitor heißt symmetrisch, wenn die Leerlauf-Eingangswiderstände an den Klemmenpaaren gleich sind.

\begin{center}
\begin{picture}(150,45)
\put(42,20){\makebox(0,0)[r]{$\underline Z_1 = \underline Z_{11} \Rightarrow$}}
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\multiput(65,0)(40,0){2}{\line(0,1){40}}
\multiput(65,0)(0,40){2}{\line(1,0){40}}
\multiput(105,5)(0,30){2}{\line(1,0){19}}
\multiput(125,5)(0,30){2}{\circle{2}}
\end{picture}
\begin{picture}(170,45)
\put(130,20){\makebox(0,0)[l]{$\Leftarrow \underline{Z}_2 = \underline Z_{22}$}}
\multiput(45,5)(0,30){2}{\circle{2}}
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\multiput(65,0)(40,0){2}{\line(0,1){40}}
\multiput(65,0)(0,40){2}{\line(1,0){40}}
\multiput(105,5)(0,30){2}{\line(1,0){19}}
\multiput(125,5)(0,30){2}{\circle{2}}
\end{picture}
\end{center}

\[\underline Z_{11} = \underline Z_{22}  \quad \longrightarrow \quad \underline Y_{11} = \underline Y_{22} , \quad \underline A_{11} = \underline A_{22} \qquad \text{(Symmetriebedingungen)}\]

\section{Ersatzschaltungen}
\subsection{Problem}

\begin{minipage}{5.9cm}
\begin{picture}(160,90)
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\multiput(0,0)(110,0){2}{\line(1,0){35}}
\put(0,80){\line(1,0){145}}
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\put(35,45){\line(1,0){75}}
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\multiput(0,0)(55,0){2}{
\multiput(0,0)(0,25){2}{
\multiput(35,5)(11,0){2}{\line(1,0){9}}
\put(45,5){\circle{2}}}}
\put(72.5,17.5){\makebox(0,0)[c]{ZT}}
\put(72.5,62.5){\makebox(0,0)[c]{Netzwerk}}
\end{picture}
\end{minipage}
\begin{minipage}{9cm}
\begin{description}
\item[gegeben:] Zweitor, das in ein Netzwerk eingeschaltet ist (z.B.~Transistor in einer Schaltung)
\item[gesucht:] Ströme und Spannungen in der Gesamtschaltung
\end{description}
\end{minipage}

\bigskip

Zur Netzwerkanalyse muß das Zweitor durch Netzwerkelemente ersetzt werden. Das leistet die Ersatzschaltung. Sie modelliert das Klemmenverhalten des Zweitores durch ein Netzwerk.

\subsection{Allgemeine Zweipole}
Aus jeder Darstellungsform der Zweitorgleichungen läßt sich eine Ersatzschaltung ableiten.

\subsubsection*{Prinzip}

\begin{enumerate}
\item Eine Spannungsbilanz entspricht einer Serienschaltung (Maschensatz)
\item Eine Strombilanz entspricht einer Parallelschaltung (Knotensatz)
\end{enumerate}

\subsection*{Beispiele}

\subsubsection{Y-Darstellung}

\begin{minipage}{8cm}
\quad\quad \begin{picture}(200,80)
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\multiput(95,0)(0,40){2}{\line(0,1){20}}
\multiput(95,20)(-10,10){2}{\line(1,1){10}}
\multiput(95,20)(10,10){2}{\line(-1,1){10}}
\put(85,30){\line(1,0){20}}
\put(38,30){\makebox(0,0)[r]{$\underline Y_{11}$}}
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\put(3,67){\makebox(0,0)[c]{$\underline I_1$}}
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\put(53,50){\makebox(0,0)[l]{$\scriptstyle \underline Y_{11} \cdot \underline U_1$}}
\put(50,65){\vector(1,0){20}}
\put(60,72){\makebox(0,0)[c]{$\scriptstyle\underline Y_{12} \cdot \underline U_2$}}
\put(83,40){\vector(0,-1){20}}
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\put(133,30){\makebox(0,0)[l]{$\scriptstyle\underline Y_{21} \cdot \underline U_1$}}
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\multiput(170,0)(0,40){2}{\line(0,1){20}}
\multiput(165,20)(10,0){2}{\line(0,1){20}}
\multiput(165,20)(0,20){2}{\line(1,0){10}}
\put(165,56){\vector(0,-1){12}}
\put(163,50){\makebox(0,0)[r]{$\scriptstyle \underline Y_{22} \cdot \underline U_2$}}
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\put(222,60){\vector(-1,0){18}}
\put(216,67){\makebox(0,0)[c]{$\underline I_2$}}
\put(204,30){\makebox(0,0)[l]{$\underline U_2$}}
\put(201,56){\vector(0,-1){52}}
\put(177,30){\makebox(0,0)[l]{$\underline Y_{22}$}}
\end{picture}
\end{minipage}
\begin{minipage}{7.8cm}
\begin{align*}
\underline I_1 &= \underline Y_{11} \cdot \underline U_{1} + \underline Y_{12} \cdot \underline U_2\\
\underline I_2 &= \underline Y_{21} \cdot \underline U_{1} + \underline Y_{22} \cdot \underline U_2
\end{align*}
\end{minipage}

\subsubsection{Hybrid-Darstellung}


\begin{minipage}{8cm}
\quad\quad \begin{picture}(200,80)
\put(0,30){\makebox(0,0)[l]{$\underline U_1$}}
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\multiput(15,60)(50,0){2}{\line(1,0){30}}
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\multiput(45,55)(20,0){2}{\line(0,1){10}}
\put(56,72){\makebox(0,0)[c]{$\underline H_{11}$}}
\qbezier(40,54)(55,47)(70,54)
\put(70,54){\vector(2,1){0}}
\put(55,44){\makebox(0,0)[c]{$\scriptstyle\underline H_{11}\cdot \underline I_1$}}
\put(95,0){\line(0,1){60}}
\multiput(95,20)(-10,10){2}{\line(1,1){10}}
\multiput(95,20)(10,10){2}{\line(-1,1){10}}
\qbezier(85,15)(77,30)(85,45)
\put(85.1,15){\vector(1,-2){0}}
\put(79,30){\makebox(0,0)[r]{$\scriptstyle\underline H_{12}\cdot \underline U_2$}}
\put(-7,60){\vector(1,0){18}}
\put(3,67){\makebox(0,0)[c]{$\underline I_1$}}
\multiput(120,0)(0,40){2}{\line(0,1){20}}
\multiput(120,20)(-10,10){2}{\line(1,1){10}}
\multiput(120,20)(10,10){2}{\line(-1,1){10}}
\put(110,30){\line(1,0){20}}
\put(132,40){\vector(0,-1){20}}
\put(133,30){\makebox(0,0)[l]{$\scriptstyle\underline H_{21} \cdot \underline I_1$}}
\multiput(120,0)(0,60){2}{\line(1,0){80}}
\multiput(170,0)(0,60){2}{\circle*{2}}
\multiput(170,0)(0,40){2}{\line(0,1){20}}
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\put(165,56){\vector(0,-1){12}}
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\put(222,60){\vector(-1,0){18}}
\put(216,67){\makebox(0,0)[c]{$\underline I_2$}}
\put(204,30){\makebox(0,0)[l]{$\underline U_2$}}
\put(201,56){\vector(0,-1){52}}
\put(177,30){\makebox(0,0)[l]{$\underline H_{22}$}}
\end{picture}
\end{minipage}
\begin{minipage}{7.8cm}
\begin{align*}
\underline U_1 &= \underline H_{11} \cdot \underline I_{1} + \underline H_{12} \cdot \underline I_2\\
\underline I_2 &= \underline H_{21} \cdot \underline I_{1} + \underline H_{22} \cdot \underline U_2
\end{align*}
\end{minipage}

\subsection{Zweitore mit durchgehender Masse (Dreipole)}
Widerstandsparameter ($Z$-Darstellung)

\begin{center}
\begin{picture}(200,80)
\put(29,30){\makebox(0,0)[r]{$\underline I_1 \Big\uparrow$}}
\multiput(40,0)(0,40){2}{\line(0,1){20}}
\put(40,30){\circle{20}}
\put(30,30){\line(1,0){20}}
\multiput(40,0)(0,60){2}{\line(1,0){14}}
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\put(56,0){\line(1,0){148}}
\put(56,60){\line(1,0){14}}
\multiput(70,55)(0,10){2}{\line(1,0){20}}
\multiput(70,55)(20,0){2}{\line(0,1){10}}
\put(81,72){\makebox(0,0)[c]{$\underline Z_1$}}
\put(90,60){\line(1,0){40}}
\multiput(130,55)(0,10){2}{\line(1,0){20}}
\multiput(130,55)(20,0){2}{\line(0,1){10}}
\put(141,72){\makebox(0,0)[c]{$\underline Z_3$}}
\put(150,60){\line(1,0){54}}
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\multiput(170,60)(10,10){2}{\line(1,-1){10}}
\multiput(205,0)(0,60){2}{\circle{2}}
\multiput(206,0)(0,60){2}{\line(1,0){14}}
\multiput(220,0)(0,40){2}{\line(0,1){20}}
\put(220,30){\circle{20}}
\put(210,30){\line(1,0){20}}
\put(231,30){\makebox(0,0)[l]{$\Big\uparrow \underline I_2$}}
\multiput(110,0)(0,60){2}{\circle*{2}}
\multiput(110,0)(0,40){2}{\line(0,1){20}}
\multiput(105,20)(10,0){2}{\line(0,1){20}}
\multiput(105,20)(0,20){2}{\line(1,0){10}}
\put(117,30){\makebox(0,0)[l]{$\underline Z_2$}}
\put(55,56){\vector(0,-1){52}}
\put(57,30){\makebox(0,0)[l]{$\underline U_1$}}
\put(86,30){\circle{27}}
\put(84,16){\vector(-1,0){0}}
\put(86,30){\makebox(0,0)[c]{$\underline I_1$}}

\put(161,30){\circle{27}}
\put(163,16){\vector(1,0){0}}
\put(161,30){\makebox(0,0)[c]{$\underline I_2$}}

\put(190,73){\vector(-1,0){20}}
\put(181,79){\makebox(0,0)[c]{$\scriptstyle \underline Z_4 \cdot \underline I_1$}}
\put(205,56){\vector(0,-1){52}}
\put(203,30){\makebox(0,0)[r]{$\underline U_2$}}
\end{picture}
\end{center}

\bigskip

\begin{minipage}{7.9cm}
\centering
Analyseergebnis 
\begin{align*}
\underline U_1 &= (\underline Z_1 + \underline Z_2)\underline I_1 + \underline Z_2 \cdot \underline I_2\\
\underline U_2 &= (\underline Z_2 + \underline Z_4) \underline I_1 + (\underline Z_2 + \underline Z_3)\underline I_2
\end{align*}
\end{minipage}
\begin{minipage}{7.9cm}
\centering
$Z$-Darstellung
\begin{align*}
\underline U_1 &= \underline Z_{11}\cdot \underline I_1 + \underline Z_{12}\cdot \underline I_2\\
\underline U_2 &= \underline Z_{21} \cdot \underline I_1 + \underline Z_{22} \cdot \underline I_2
\end{align*}
\end{minipage}

\bigskip

Koeffizientenvergleich

\begin{minipage}{7cm}
\begin{align*}
\underline Z_1 + \underline Z_2 &= \underline Z_{11}  & \underline Z_2 &= \underline Z_{12}\\
\underline Z_2 + \underline Z_4 &= \underline Z_{21} & \underline Z_2 + \underline Z_3 &=  \underline Z_{22}
\end{align*}
\end{minipage}
\begin{minipage}{7cm}
\begin{empheq}[innerbox=\fbox,left={\Rightarrow}]{align*}
\underline Z_1 &= \underline Z_{11} - \underline Z_{12} & \underline Z_2 &= \underline Z_{12}\\
\underline Z_3 &= \underline Z_{22} - \underline Z_{12} & \underline Z_4 &= \underline Z_{21} - \underline Z_{12}
\end{empheq}
\end{minipage}

\bigskip


Ersatzschaltung
\begin{center}
\begin{picture}(200,100)
\multiput(19,15)(0,60){2}{\circle{2}}
\put(19,71){\vector(0,-1){52}}
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\put(0,75){\vector(1,0){15}}
\put(7,82){\makebox(0,0)[c]{$\underline I_1$}}
\put(20,15){\line(1,0){150}}
\put(20,75){\line(1,0){15}}
\multiput(35,70)(0,10){2}{\line(1,0){20}}
\multiput(35,70)(20,0){2}{\line(0,1){10}}
\put(55,75){\line(1,0){40}}
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\multiput(75,15)(0,40){2}{\line(0,1){20}}
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\multiput(70,35)(0,20){2}{\line(1,0){10}}
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\multiput(95,70)(20,0){2}{\line(0,1){10}}
\put(115,75){\line(1,0){55}}
\multiput(171,15)(0,60){2}{\circle{2}}
\multiput(135,75)(10,-10){2}{\line(1,1){10}}
\multiput(135,75)(10,10){2}{\line(1,-1){10}}
\put(46,85){\makebox(0,0)[c]{$\scriptstyle \underline Z_{11}\!-\!\underline Z_{12}$}}
\put(106,85){\makebox(0,0)[c]{$\scriptstyle \underline Z_{22}\!-\!\underline Z_{12}$}}
\put(82,45){\makebox(0,0)[l]{$\scriptstyle \underline Z_{12}$}}
\multiput(0,-4)(100,0){2}{
\multiput(25,0)(0,4){25}{\line(0,1){2}}}
\multiput(0,-4)(0,100){2}{
\multiput(25,0)(4,0){25}{\line(1,0){2}}
}
\put(33,3){umkehrbarers ZT}
\put(155,88){\vector(-1,0){20}}
\put(152,95){\makebox(0,0)[c]{$\scriptstyle (\underline Z_{21}\!-\!\underline Z_{12})\underline I_1$}}

\put(171,71){\vector(0,-1){52}}
\put(174,45){\makebox(0,0)[l]{$\underline U_2$}}
\put(190,75){\vector(-1,0){15}}
\put(185,82){\makebox(0,0)[c]{$\underline I_2$}}

\end{picture}
\end{center}

\chapter{Transformator (Übertrager)}
\section{Transformatorgleichungen und Ersatzschaltung}

\begin{minipage}{6cm}
\begin{picture}(155,100)
\multiput(50,30)(0,60){2}{\line(1,0){60}}
\multiput(50,30)(60,0){2}{\line(0,1){60}}
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\multiput(65,45)(0,30){2}{\line(1,0){30}}
\thinlines
\put(25,30){
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\qbezier(32.5,26)(21.5,26)(21.5,24)
\qbezier(21.5,24)(21.5,22)(25,22)}
\qbezier(32.5,39.8)(25,40)(25,40)}
\multiput(40,50)(0,20){2}{\line(1,0){10}}
\multiput(39,50)(0,20){2}{\circle{2}}
\put(23,57){$u_1$}
\put(20,70){\vector(1,0){15}}
\put(09,67){$i_1$}
\put(70,30){
\multiput(-0.2,0)(0,4.6){4}{
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\qbezier(22,22)(22,26)(32.5,26)}
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\qbezier(32.5,39.8)(40,40)(40,40)
}
\multiput(110,50)(0,20){2}{\line(1,0){10}}
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\put(124,57){$u_2$}
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\put(138,70){\vector(-1,0){13}}
\put(142,67){$i_2$}
\put(71,63){$w_1$}
\put(78,51){$w_2$}
\put(52,74){$L_1$}
\put(96,74){$L_2$}
\put(52,40){$R_1$}
\put(96,40){$R_2$}
\end{picture}

\qquad\quad$\!$\begin{picture}(100,50)  % haha
\put(20,0){
\multiput(0,0)(52,0){2}{\multiput(5,5)(0,35){2}{\circle{2}}}
\multiput(0,0)(0,35){2}{\multiput(6,5)(35,0){2}{\line(1,0){15}}}
\multiput(0,0)(20,0){2}{\multiput(21,5)(0,27.5){2}{\line(0,1){7.5}}}
\multiput(21,12.5)(0,5){4}{\qbezier(0,0)(5,0)(5,2.5)\qbezier(5,2.5)(5,5)(0,5)}
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\put(26,47){$M$}
\put(08,20){$L_1$}
\put(43,20){$L_2$}}
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\put(10,20){$u_1$}
\put(81,20){$u_2$}
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\put(5,43){$i_1$}
\put(103,40){\vector(-1,0){20}}
\put(89,43){$i_2$}
\multiput(37,34)(28,0){2}{\circle*{3}}
\end{picture}



\end{minipage}
\begin{minipage}{9.8cm}
\begin{empheq}[box=\fbox]{align*}
u_1 &= i_1 \cdot R_1 + L_1 \frac{di_1}{dt} + M\frac{di_2}{dt}\vphantom{\int\limits_{.}}\\
u_2 &= i_2 \cdot R_2 + L_2\frac{di_2}{dt} + M\frac{di_1}{dt} \vphantom{\int\limits^{.}}
\end{empheq}
Kenngröße: $\text{\emph{ü}} = \sqrt{\dfrac{L_1}{L_2}} = \dfrac{w_1}{w_2}$ wenn $R_{m1} = R_{m2}$

\bigskip

$M = k \sqrt{L_1\cdot L_2}$ \hfill $k$: Koppelfaktor

\bigskip

$k = \sqrt{1-\sigma},\ \sigma = 1 - k^2 $ \hfill $\sigma$: Streufaktor
\end{minipage}

\subsubsection*{Transformation in das Komplexe}
(harmonische Zeitverläufe, eingeschwungener Zustand)


\begin{empheq}[innerbox=\fbox,right=\qquad Z\text{-Darstellung}]{align*}
\underline U_1 &= (\underbrace{R_1 + j\omega L_1}_{\underline Z_{11}}) \underline I_1 + \underbrace{j\omega M}_{\underline Z_{12}} \underline I_2\vphantom{\int\limits_{.}}\\
\underline U_2 &=  \underbrace{j\omega M}_{\underline Z_{21}} \underline I_1 +  (\underbrace{R_2 + j\omega L_2}_{\underline Z_{22}}) \underline I_2
\end{empheq}

\bigskip

\subsubsection*{Allgemeine Ersatzschaltung eines Transformators}
\begin{minipage}{9cm}
\begin{picture}(250,90)
\put(5,30){\makebox(0,0)[l]{$\underline U_1$}}
\multiput(19,0)(0,60){2}{\circle{2}}
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\put(0,60){\vector(1,0){16}}
\put(7,67){\makebox(0,0)[c]{$\underline I_1$}}
\multiput(20,60)(40,0){3}{\line(1,0){20}}
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\begin{minipage}{6.8cm}
\bigskip 

$R_{Fe}$ modelliert die Eisenverluste $\to$ Hysteresekurve durch Ellipse approximiert.
\end{minipage}

\pagebreak

\subsubsection*{Nachteile:}
\begin{enumerate}
\item Durchgehende Masse
\item Nur für $k < \text{\emph{ü}} < \frac 1 k$ positive Ersatzelemente
\end{enumerate}
$\to$ Behebung durch 7.2

\section{Vereinfachte Ersatzschaltung}

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\raisebox{-25pt}{$\Rightarrow$}
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\bigskip
\bigskip

\[\underline U_2 = \frac 1 {\alpha} \cdot {\underline U_2}^* \qquad \underline I_2 = \alpha \cdot {\underline I_2}^* \qquad \alpha = \frac{{\underline U_2}^*}{\underline U_2}\]
\begin{align*}
\underline U_1 &= (R_1 + j\omega L_1) \underline I_1 + j\omega M \cdot \alpha \cdot {\underline I_2}^*\\
\frac 1 {\alpha} {\underline U_2}^* &= j\omega M \cdot \underline I_1 + (R_2 + j\omega L_2)\cdot \alpha\cdot {\underline I_2}^*\\
{\underline U_2}^* &= j\omega M \cdot \alpha \cdot \underline I_1 + (R_2 + j\omega L_2) \cdot \alpha^2 \cdot {\underline I_2}^* 
\end{align*}

\subsubsection*{Durch geeignete Wahl von $\alpha$ unterschiedliche Ersatzschaltbilder}

\subsubsection{Symmetrisches Ersatzschaltbild}

\[\boxed{\vphantom{\int}\quad L_1 - \alpha M = \alpha^2 \cdot L_2 - \alpha \cdot M \quad} \quad \Rightarrow \alpha = \sqrt{\frac{L_1}{L_2}} = \text{\emph{ü}}\]
\begin{align*}
L_1 - \alpha \cdot M &= L_1 -  \sqrt{\frac{L_1}{L_2}} \cdot k \cdot \sqrt{L_1 \cdot L_2} = (1-k) \cdot L_1 = (1-\underbrace{\sqrt{1-\sigma}}_{\approx 1 - \frac{\sigma}{2}})\cdot L_1\\
 &\approx \frac{\sigma}{2}\cdot L_1
\end{align*}

\subsubsection{Nur eine Längsinduktivität}

\[\boxed{\vphantom{\int}\quad \alpha^2 \cdot L_2 - \alpha M = 0 \quad} \quad \Rightarrow \alpha = \frac M {L_2} = \frac{k\cdot\sqrt{L_1\cdot L_2}}{L_2} = k \cdot \sqrt{\frac{L_1}{L_2}} = k\cdot \text{\emph{ü}}\]
\[L_1 - \alpha \cdot M = L_1 - k \sqrt{\frac{L_1}{L_2}} \cdot k \cdot \sqrt{L_1 \cdot L_2} = L_1 - (1-k^2) = \sigma L_1\]

\newpage

\section{Leistungsübertrager}

Bestimmung der Elemente der Ersatzschaltung

\subsection{Leerlaufversuch}

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\bigskip

\begin{description}
\item[Messung:] $U_{1B} = 230\,\mathrm V$, $U_2 = 2040\,\mathrm V$, $I_1 = I_{1L} = 0,\!83\,\mathrm{A}$, $P_{1L} = 118\,\mathrm W$

\bigskip  % ab hier wird kräftig gerundet, aber so hat es der meister nun 
          % mal an die Tafel gepinselt

Bei Leerlauf: $\displaystyle P_{1L} \approx P_{Fe} \approx \frac{{U_{1B}}^2}{R_{Fe}} \quad \to \quad R_{Fe} \approx \frac{{U_{1B}}^2}{P_{1L}} = 450\,\Omega$

\item[Scheinleistung:] $S_{1L} = U \cdot I = 200\,\mathrm{VA}$

\item[Blindleistung:] $Q_{1L} = \displaystyle \sqrt{{S_{1L}}^2 - {P_{1L}}^2} = 162\,\mathrm{var} \approx \frac{{U_{1B}}^2}{X_M} = \frac{{U_{1B}}^2}{\omega L_1}$

$\displaystyle\Rightarrow L_1 = \frac{{U_{1B}}^2}{\omega Q_{1L}} = 1,\!1\,\mathrm H$

\item[Spannungsübersetzung:] $\displaystyle\text{\emph ü} = \frac{U_{1B}}{U_2} = 0,\!11 = \frac 1 {8,\!9}$
\end{description}

$\Rightarrow$ Im Leerlaufversuch werden die Elemente des Querzweiges bestimmt.

\subsection{Kurzschlußversuch}

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\bigskip

\begin{description}
\item[Messung:] $I_{1B} = 15\,\mathrm A$, $U_{1K} = 14,\!5\,\mathrm{V}$, $P = 107\,\mathrm{W}$, $I_2 = 1,\!66\,\mathrm{A}$

\bigskip

$\displaystyle P_{1K} \approx P_{Cu}  = {I_{1K}}^2 \cdot (R_1 + {R_2}^*) \quad \to \quad R_1 + {R_2}^* = \frac{P_{1K}}{{I_{1K}}^2} = \frac{P_{1K}}{{I_{1K}}^2} = 0,\!5\,\Omega$

Bei gleichem Wickelraum gilt $R_1 =  {R_2}^* = 0,\!25\,\Omega$ (gemessen: $0,\!29\,\Omega$)

\smallskip

$R_2 = \frac{{R_2}^*}{\text{\emph ü}^2} = 18,\!9\,\Omega$ (gemessen: $14,\!5\,\Omega$)

\item[Stromübersetzung:] $\displaystyle\text{\emph ü} = \frac{I_2}{I_{1B}} = 0,\!11 = \frac{1}{9,\!0}$

\item[Scheinleistung:] $S_{1K} = U_{1K} \cdot I_{1B} = 14,\!5\,\mathrm V \cdot 15\,\mathrm A = 218\,\mathrm{VA}$

\item[Blindleistung:] $Q_{1K} = \sqrt{{S_{1K}}^2 - {P_{1K}}^2} = 190\,\mathrm{var}$

\bigskip

$\displaystyle Q_{1K} \approx {I_{1B}}^2 \cdot X_{\sigma} = {I_{1B}}^2 \cdot \omega \sigma L_1 \quad \to \quad X_{\sigma} = \frac{Q_{1K}}{{I_{1B}}^2} = 0,\!84\,\Omega$

\smallskip

$\displaystyle\sigma L_1 = \frac{Q_{1K}}{\omega \cdot {I_{1B}}^2} = 2,\!7\,\mathrm{mH}$
\end{description}

Im Kurzschlußversuch werden die Elemente des Längszweiges bestimmt.

\smallskip

$L_1 = 1,\!1\,\mathrm H$, $L_2 = \dfrac{L_1}{\text{\emph ü}^2} = 83\,\mathrm H$, $M = \dfrac{L_1}{\text{\emph ü}} = 9,\!6\,\mathrm H$, 

$\sigma L_1 = 2,\!7\,\mathrm{mH} \quad \to \quad \sigma = 2,\!5\cdot 10^{-3} \to k = \sqrt{1-\sigma} = 0,\!9988$

\section{Signalübertrager: Frequenzverhalten}
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\begin{description}
\item[Annahme:] $R_1 = {R_2}^* = (k \cdot \text{\emph ü})^2 \cdot R_2$ \quad (Anpassung)
\item[gesucht:] Frequenzgänge und Ortskurve von $\dfrac{\underline U_2}{\underline U_G}$
\end{description}

\begin{align*}
\underline G(j\omega) &= \frac{\underline U_2}{\underline U_G} = \frac{1}{k \text{\emph ü}} \frac{{\underline U_2}^*}{\underline U_G} = \frac{1}{k \text{\emph ü}} \frac{R_1 \parallel j\omega k^2 L_1}{R_1 + j\omega \sigma L_1 + R_1\parallel j\omega k^2 L_1} = \frac{1}{k \text{\emph ü}} \frac{1}{2 + \frac{\sigma}{k^2} + \frac{j\omega \sigma L_1}{R_1} + \frac{R_1}{j\omega k^2 L_1}}\\
 &= \frac{1}{k \text{\emph ü}} \cdot \frac{1}{A(1+j\varrho v)} = \frac{1}{k \text{\emph ü}} \cdot \frac{1}{A\left(1+j\varrho \left(\frac{\omega}{\omega_0} - \frac{\omega_0}{\omega}\right)\right)}
\end{align*}

\subsubsection*{Koeffizientenvergleich}
\begin{align*}
\text{\textcircled{\raisebox{-0.95pt}{1}}}\ \omega^0\!: &\quad A = 2 + \frac{\sigma}{k^2} & \text{\textcircled{\raisebox{-0.95pt}{2}}}\  \omega\!: &\quad \frac{\sigma L_1}{R_1} = \frac{A \varrho}{\omega_0} &
\text{\textcircled{\raisebox{-0.95pt}{3}}}\ \frac{1}{\omega}\!: &\quad \frac{R_1}{k^2 L_1} = A \cdot \varrho \cdot \omega_0
\end{align*}

$\displaystyle\text{\textcircled{\raisebox{-0.95pt}{2}}} \cdot \text{\textcircled{\raisebox{-0.95pt}{3}}}\!: \quad A^2 \cdot \varrho^2 = \frac{\sigma}{k^2} = \frac{\sigma}{1-\sigma} \quad \to \quad \varrho^2 = \frac{1}{A^2}\frac{\sigma}{1-\sigma} \quad \Rightarrow\quad \varrho = \frac{\sqrt{\sigma}}{2}$

$\displaystyle \frac{\text{\textcircled{\raisebox{-0.95pt}{3}}}}{\text{\textcircled{\raisebox{-0.95pt}{2}}}}\!: \quad {\omega_0}^2 = \frac{{R_1}^2}{k^2 \sigma {L_1}^2} = \frac{{R_1}^2}{(1-\sigma) \sigma {L_1}^2} \approx \frac{R_1}{\sigma {L_1}^2}$

\subsubsection*{Ortskurve}

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\subsection{Grenzfrequenzen}

$\omega_u$ und $\omega_{ob}$ liegen weit auseinander. $\rightarrow$ Näherung.

\bigskip

\begin{center}
\begin{tabular}{c|c}
untere Grenzfrequenz $\omega_u$ & obere Grenzfrequenz $\omega_{ob}$ \\
\hline
$\vphantom{\Bigg|}2 \approx \dfrac{R_1}{\omega_u k^2 L_1}$ & $2 \approx \omega_{ob} \dfrac{\sigma L_1}{R_1}$ \\
$\vphantom{\Bigg|}\omega_u  = \dfrac{R_1}{2L_1}$ & $\omega_{ob} = \dfrac{2 R_1}{\sigma L_1}$\\
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bestimmt durch Querglied & bestimmt durch Längsglied
\\
\end{tabular}
\end{center}

\bigskip
\[\omega_{ob} \cdot \omega_u = \frac{{R_1}^2}{\sigma {L_1}^2}\ = {\omega_0}^2 \qquad\qquad \frac{\omega_{ob}}{\omega_{u}} = \frac{4}{\sigma}\]

\chapter{Periodische Signale und Netzwerke bei periodischer Erregung}

\section{Periodische Signale}

\subsection{Definition: periodisches Signal}

\bigskip

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\begin{minipage}{8cm}
Für alle $t_1 < t < t_1 +T$ gilt
\[\boxed{\quad y(t+kT) = y(t), k = 0,\, \pm 1,\, \pm 2,\, \ldots \quad \vphantom{\Big|}}\]

$T$: Periodendauer, $f = \frac 1 T,\ \omega = 2\pi f$
\end{minipage}

\paragraph{Bezeichnung:} $\displaystyle \int\limits_{t_1}^{t_1+T}y(t)\,dt = \int\limits_0^T y(t)\,dt = \int\limits_{(T)}y(t)\,dt$, unabhängig von $t_1$.

\subsection{Signalkenngrößen}

\subsubsection{(arithmetischer) Mittelwert, Gleichanteil}
\begin{minipage}{7cm}
\[\boxed{\quad Y_0 = \overline{y(t)} = \phantom{\int\limits^(}\frac 1 T \int\limits_{(T)} y(t) \, dt \quad}\]
\end{minipage}
\begin{minipage}{8.5cm}
\textbf{Messung:}
\begin{enumerate}
\setlength{\itemsep}{0ex}
\renewcommand{\labelenumi}{(\alph{enumi})}
\item Multimeter mit Einstellung "`DC"'
\item Tiefpassfilter
\end{enumerate}
\end{minipage}

\subsubsection{Signalleistung}
\begin{minipage}{5cm}
\centering
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\begin{minipage}{10.8cm}
\[p(t) = i(t) \cdot u(t) = \frac{u^2(t)}{R} = i^2(t) \cdot R\]
mittlere Leistung:
\[P = \frac 1 T \int\limits_{(T)}p(t)\,dt = \overline{p(t)} = \frac 1 R \frac 1 T \int\limits_{(T)} u^2(t)\,dt = R \frac 1 T \int\limits_{(T)} i^2(t)\,dt\]
\end{minipage}

Verallgemeinerung:
\[\boxed{\quad\vphantom{\int\limits^(} \frac 1 T \int\limits_{(T)} y^2(t)\,dt = \overline{y^2(t)} = P\quad} \qquad \text{Def.: Signalleistung}\]

quadratischer Mittelwert (mean squared value): $[P] = \left\{\begin{array}{ll}V^2 & \text{ wenn } y(t) = u(t)\\ A^2 & \text{ wenn } y(t) = i(t)\end{array}\right.$

\subsubsection*{Messung:}
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\end{center}


\subsubsection{Effektivwert (Root mean square value, RMS)}
Effektivwert = Wert einer leistungsäquivalenten Größe

\[i^2 \cdot R = R \cdot \frac 1 T \int\limits_{(T)} i^2(t)\,dt \qquad \Rightarrow \qquad i^2 = \frac 1 T \int\limits_{(T)} i^2(t)\,dt\]
\[\boxed{\quad Y = \sqrt{\overline{y^2(t)}} = \sqrt{\frac 1 T \int\limits_{(T)} i^2(t)\,dt}   \quad} \qquad \text{Def.: Effektivwert}\]

\paragraph*{Messung:} (True RMS-value measurement)
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\end{center}

\subsubsection{Gleichrichtwert}

\begin{minipage}{6cm}
\[Y_{Gl} = \overline{\left|y(t)\right|} = \frac 1 T \int\limits_{(T)} |y(t)|\,dt \]
\end{minipage}
\begin{minipage}{9.8cm}
\center
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\smallskip

Meßgerät mit Meßgleichrichter
\end{minipage}


\section{Spektraldarstellung periodischer Signale}
\subsection{Fourierentwicklung}

\[y(t) = Y_0 + \sum\limits_{n=1}^{N}\widehat{Y}_n \cdot \cos (n \omega_0 t + \varphi_n) = Y_0 + \sum\limits_{n=1}^N A_n \cos n\omega t + B_n \cos n \omega t + e(t)\]

\[\boxed{\quad \widehat{Y}_n \cdot e^{j\varphi_n} = A_n - jB_n\quad}\]

\subsubsection*{Approximationsproblem:}

Fehler: $\displaystyle E_N  = \frac 1 T \int\limits_{(T)} e^2(t) \, dt \quad \Rightarrow \quad \min(A_n,\,b_n,\,Y_0,\,n=1,\,2,\, \ldots ,\, N)$

\newpage

\[Y_0 = \frac 1 T \int\limits_{(T)} y(t)\,dt,\ A_n = \frac 2 T \int\limits_{(T)} y(t) \cos n \omega_0 t\,dt,\  B_n = \frac 2 T \int\limits_{(T)} y(t) \sin n \omega_0 t\,dt\]

\subsection{Amplituden- und Phasenspektrum}


\bigskip

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Amplitudenspektrum
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Phasenspektrum
\end{minipage}

\subsection{Kenngrößen periodischer Signale im Frequenzbereich}

\subsubsection{Mittelwert}
\[Y_0 = \frac 1 T \int\limits_{(T)} y(t)\,dt\]

\subsubsection{Signalleistung}

\begin{align*}
P &= \frac 1 T \int\limits_{(T)} y^2(t)\,dt = \frac 1 T \int\limits_{(T)}\left(Y_0 + \sum\limits_{n=1}^{\infty} \widehat{Y}_n \cos n(\omega_0 t + \varphi_n)\right) \cdot \left(Y_0 + \sum\limits_{m=1}^{\infty} \widehat{Y}_m \cos m(\omega_0 t + \varphi_m)\right)\,dt\\
 &= \frac 1 T \int\limits_{(T)} {Y_0}^2 + Y_0 \cdot \sum\limits_{n=1}^{\infty} \widehat{Y}_n \cos(n\omega_0 t + \varphi_n) + 
 \hfill \sum\limits_{n=1}^{\infty}\sum\limits_{m=1}^{\infty}\widehat Y_n \widehat Y_m \cos(n\omega_0 t + \varphi_n) \cdot \cos (m\omega_0 t +\varphi_m))\,dt 
\end{align*}
\[\frac 1 T \int\limits_{(T)} \cos (n\omega_0 t + \varphi_n)\,dt = 0, \quad 
\frac 1 T \int\limits_{(T)} \cos (n\omega_0 t + \varphi_n) \cdot \cos(m\omega_0 t + \varphi_m)\,dt = \left\{\begin{array}{ll}0 & n \ne m\\\frac 1 2 & n = m\end{array}\right.\]


\[P = {Y_0}^2 + \sum\limits_{n=1}^{\infty} \frac{{\widehat Y_n}^2}{2} = {Y_0}^2 + \sum\limits_{n=1}^{\infty} {Y_n}^2\]

\bigskip

Signalleistung = Gleichleistung + $\sum$ Leistung der Harmonischen

\subsubsection{Effektivwert}
\[Y = \sqrt{P} = \sqrt{\overline{y^2(t)}} = \sqrt{{Y_0}^2 + \sum\limits_{n=1}^{\infty} {Y_n}^2}\]

\subsubsection{Klirrfaktor}
\[k = \dfrac{\text{Effektivwert der Summe der Oberwellen}}{\text{Effektivwert}}\]
\[k \approx \sqrt{\frac{\sum\limits_{n=2}^{\infty}{Y_n}^2}{\sum\limits_{n=1}^{\infty}{Y_n}^2}} = \sqrt{\frac{Y^2 - {Y_1}^2}{Y^2}} \qquad Y^2 = {Y_n}^2 = \sum\limits_{n=1}^{\infty}{Y_n}^2 - {Y_1}^2\]

\section{Reaktion von Netzwerken auf periodische Signale}
\subsection{Grundprinzip}
Die Reaktionen des Netzwerkes auf die einzelnen Harmonischen überlagern sich bei linearen Netzwerken ohne gegenseitige Beeinflussung.

\subsubsection*{Rechenverfahren}
\begin{enumerate}
\item Zerlegung des Signals in Harmonische (Fourier-Analyse)
\item Berechnung der Netzwerkreaktion auf jede Harmonische
\item Addition der Teilreaktionen aus 2.
\end{enumerate}
\subsubsection{Beispiel}

\begin{minipage}{6cm}
\centering
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\centering

\bigskip
\bigskip

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\begin{enumerate}
\item Fourierreihe für $u_1(t)$
\[u_1(t) = \widehat U_1 \left(\frac 1 2 + \frac 2{\pi} \left(\cos \omega_1 t - \frac{\cos 3\omega_1 t}{3}+\frac{\cos 5\omega_1 t}{5} - \cdots + \cdots \right)\right)   \]
\item Berechnung der Netzwerkreaktion auf jede Harmonische
\begin{enumerate}
\item Übertragungsfaktor des Filters
\[\underline G (j\omega) = \frac{\underline U_2}{\underline U_1} = \frac{1}{1+j\omega\tau} = \frac{1}{\sqrt{1+(\omega\tau)^2}}\cdot e^{j\varphi}, \qquad \tau = RC\]
\[\varphi(\omega) = - \arctan \omega \tau\]
\item Berechnung der Teilreaktionen
\end{enumerate}
\end{enumerate}
\[
\begin{array}{c@{\ }c@{\ }c@{\ }c@{\ }c@{\ }c@{\ }c}  %% Achtung, nächste 10
                                                      %% Zeilen bitte übersehen!
u_1(t) & = & \dfrac{\widehat U_1}{2} & + & \dfrac{\widehat U_1 \cdot 2}{\pi } \cos \omega_1 t & - & \dfrac{\widehat U_1 \cdot 2}{3\pi } \cos 3 \omega_1 t \qquad + \quad  \cdots\\[0.5ex]
       &   &  
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&&
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\\[0.5ex]
u_2(t) & = & \dfrac{\widehat U_1}{2} & + &  \dfrac{\widehat U_1 \cdot 2}{\pi\sqrt{1 + (\omega_1\tau)^2}} \cdot \cos(\omega_1 t + \varphi(\omega_1)) & - & \dfrac{\widehat U_1 \cdot 2}{3\pi\sqrt{1 + (3\omega_1\tau)^2}} \cdot \cos(3\omega_1 t + \varphi(3\omega_1))
\end{array}
\]

\chapter{Schaltvorgänge}

\section{Zustandgleichungen}

Die Zustandsgleichungen eines Netzwerkes sind ein System von gewöhnlichen Differentialgleichungen erster Ordnung. Die Variablen in den Differentialgleichungen (Zustandsvariablen) sind die Ströme durch die Induktivitäten und die Spannungen an den Kapazitäten des Netzwerkes. Es gibt so viele Zustandsgleichungen, wie das Netzwerk unabhängige Energiespeicher ($L$, $C$) enthält.

\subsection[Algorithmus zur Ableitung der Zustandgleichungen aus einem Netzwerk]{Algorithmus zur Ableitung der Zustandgleichungen\\aus einem Netzwerk}

\begin{minipage}{8.5cm}
\begin{description}
\item[gegeben:] Netzwerk
\item[gesucht:] DGLn für alle Ströme durch die Induktivitäten und Spannungen an den Kapazitäten (Zustandsgleichungen)
\end{description}
\end{minipage}
\hfill
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\subsubsection*{1. Schritt}
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Ersetzen aller
\begin{itemize}
\item Induktivitäten durch Stromquellen und
\item Kapazitäten durch Spannunsgquellen
\end{itemize}
\textbf{Ergebnis:} Gleichstromnetzwerk
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\subsubsection*{2. Schritt}
\begin{minipage}{8.5cm}
Berechnung der Spannungen an den Induktivitäten und Ströme durch die Kapazitäten durch Gleichstromanalyse

\bigskip

\textbf{Ergebnis:} Algebraisches Gleichungssystem
\end{minipage}
\hspace{0.5cm}
\begin{minipage}{6.5cm}
\begin{align}
u_L &= -i_L\cdot R_L + u_C\tag{1}\\
i_C &= -i_L - \frac{u_C}{R_C} + i\tag{2}
\end{align}
\end{minipage}

\subsubsection*{3. Schritt}
\begin{minipage}{8.5cm}
Einführung der $u$-$i$-Relationen $i_c = C\frac{du_C}{dt}$ und $u_L = L\frac{di_L}{dt}$ für alle $L$ und $C$ auf den linken Seiten der Gleichungen
\bigskip

\textbf{Ergebnis:} DGL-System
\end{minipage}
\hspace{0.5cm}
\begin{minipage}{6.5cm}
\begin{align}
L\frac{di_L}{dt} &= -i_L\cdot R_L + u_C\tag{1}\\
C\frac{du_C}{dt} &= -i_L - \frac{u_C}{R_C} + i\tag{2}
\end{align}
\end{minipage}

\subsection{Anwendungsbeispiel: Transistorschalter mit induktiver Last}
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\bigskip

\[\text{gegeben: } i(t) = \left\{\begin{array}{ll}I_0 & 0\leq t \leq T\\0 & \text{sonst}\end{array}\right. = I_0(s(t) - s(t-T)), \qquad \text{gesucht: } u(t)\]

$I(s) = \dfrac{I_0}{s}\left(1-e^{-sT}\right)$, \quad $u_L = u_R = i_R \cdot R,\quad i_R = i-i_L \quad \Rightarrow u_L = (i-i_R)\cdot R$

\bigskip

$L\dfrac{di_L}{dt} = (i-i_L)\cdot R \quad \Rightarrow \quad L\dfrac{di_L}{dt} + i_L\cdot R = i \cdot R$, \qquad mit $\tau = \frac{L}{R}:$

\bigskip

$\tau \dfrac{di_L}{dt} + i_L = i$ \qquad \text{mit }$i_L(0) = 0$ \qquad \Big|$\mathcal{L}$

\bigskip

$\tau(I_L \cdot s - i_L(0)) + I_L = I = \frac{I_0}{s}\left(1-e^{-sT}\right)$

\bigskip

$(1-s\tau) \cdot I_L = \frac{I_0}{s}\left(1-e^{-sT}\right)$

\bigskip

$I_L(s) = \frac{I_0}{s(1+s\tau)} - \frac{I_0}{s(1+s\tau)}\cdot e^{-sT} \qquad \Big|\mathcal{L}^{-1}$

\bigskip

$i_L(t) = I_0\left(1-e^{-\frac{t}{\tau}}\right) - I_0\left(1-e^{-\frac{t-T}{\tau}}\right)\cdot s(t-T)$

\bigskip

\[i_L(t) = \left\{\begin{array}{ll}0 & t < 0\\ I_0\left(1-e^{-\frac{t}{\tau}}\right) & 0 < t < T\\ i_0\left(e^{\frac{T}{\tau} - 1}\right)\cdot e^{-\frac{t}{\tau}} & T \leq t \end{array}\right.
\qquad u_L(t) = \left\{\begin{array}{ll}0 & t < 0\\ \frac{L\cdot I_0}{\tau}\cdot e^{-\frac{t}{\tau}} & 0 \leq t < T\\ -\frac{L\cdot I_0}{\tau}\left(e^{\frac{T}{\tau}} - 1\right) \cdot e^{-\frac{t}{\tau}} & T \leq t  \end{array}\right.
\] 
$u(t) = U_0 - u_L(t)$





\end{document}