ET2_Uebung_BEI/ET2_L_B19_A3.tex

\section {Effektivwert und Klirrfaktor}
\begin{minipage}[c]{.7\textwidth}
Bild 1 zeigt einen Teil aus dem Ersatzschaltbild eines Transformators, aus dem hervorgeht,
dass sich der Leerlaufstrom $i_0(t)$ zusammensetzt aus dem (verzerrten) Magnetisierungsstrom
$i_\mu(t)$ und dem Strom $i_{Fe}(t)$, der die Eisenverluste repräsentiert.
\end{minipage}
\begin{minipage}[c]{.3\textwidth}
\begin{align*}
	\begin{tikzpicture}[scale=1.5]
		\begin{scope}[>=latex,very thick,xshift=0cm,yshift=0cm,rotate=90]%Spule |
    		\draw (0,0)--(.3,0) (.7,0)--(1,0)node at(.5,-.0667) [right] {$L_{1h}$};
			\fill (.3,-0.0667)rectangle(.7,0.0667);
			\draw [<-,red] (.75,.1)--(.95,.1)node at(.85,.1)[left]{\footnotesize$i_{\mu}(t)$};
        \end{scope}
		\begin{scope}[>=latex,very thick,xshift=1cm,yshift=0cm,rotate=90] 			
			\draw (0,0)--(.3,0) (.3,-0.0667)rectangle(.7,0.0667) (.7,0)--(1,0)node at (.5,-.0667) [right] {$R_{Fe}$};
			\draw [<-,red] (.75,-.1)--(.95,-.1)node at(.85,-.1)[right]{\footnotesize$i_{Fe}(t)$}; 	
		\end{scope}
		\begin{scope}[>=latex,very thick,xshift=0cm,yshift=0cm]%Fehlstellen Eckverbindungen.
            \draw(0,.1)--(0,0)--(1,0)--(1,.1)(0,.9)--(0,1)--(1,1)--(.9,1)(.5,1)--(.5,1.5)(.5,0)--(.5,-.5);
            \fill (0.5,0)circle(0.025cm)(.5,1)circle(0.025cm);
            \draw [<-,red] (.6,-.35)--(.6,-.15)node at(.6,-.25)[right]{\footnotesize$i_{0}(t)$};
		\end{scope}	
\draw node at (.5,-1){Bild 1};
	\end{tikzpicture}
\end{align*}
\end{minipage}
Bild 2 zeigt die zeitlichen Verläufe von $i_\mu(t)$ und $i_{Fe}(t)$, welche durch folgende Fourier-Reihen approximiert werden können:\\
$i_\mu(t)=10\,\milli\ampere\cdot \cos(\omega t) + 2,88\,\milli\ampere\cdot \cos(3\omega t)$;\\ $i_{Fe}(t)=-4\,\milli\ampere\cdot \sin(\omega t)$\\[\baselineskip]
Der resultierende, in Bild 3 dargestellte Leerlaufstrom ist die Summe:\\
$i_0(t)= i_\mu(t)+i_{Fe}(t)$
\begin{align*}
%	\begin{tikzpicture}[scale=1.5]
%		\begin{scope}[>=latex,very thick,xshift=0cm,yshift=0cm,rotate=90]%Spule |
%    		\draw (0,0)--(.3,0) (.7,0)--(1,0)node at(.5,-.0667) [right] {$L_{1h}$};
%			\fill (.3,-0.0667)rectangle(.7,0.0667);
%			\draw [<-,red] (.75,.1)--(.95,.1)node at(.85,.1)[left]{\footnotesize$i_{\mu}(t)$};
%        \end{scope}
%		\begin{scope}[>=latex,very thick,xshift=1cm,yshift=0cm,rotate=90] 			
%			\draw (0,0)--(.3,0) (.3,-0.0667)rectangle(.7,0.0667) (.7,0)--(1,0)node at (.5,-.0667) [right] {$R_{Fe}$};
%			\draw [<-,red] (.75,-.1)--(.95,-.1)node at(.85,-.1)[right]{\footnotesize$i_{Fe}(t)$}; 		 
%		\end{scope}
%		\begin{scope}[>=latex,very thick,xshift=0cm,yshift=0cm]%Fehlstellen Eckverbindungen.
%            \draw(0,.1)--(0,0)--(1,0)--(1,.1)(0,.9)--(0,1)--(1,1)--(.9,1)(.5,1)--(.5,1.5)(.5,0)--(.5,-.5);
%            \fill (0.5,0)circle(0.025cm)(.5,1)circle(0.025cm);
%            \draw [<-,red] (.6,-.35)--(.6,-.15)node at(.6,-.25)[right]{\footnotesize$i_{0}(t)$};
%		\end{scope}	
%\draw node at (.5,-1){Bild 1};
%	\end{tikzpicture}
\begin{tikzpicture}[scale=1.25]
\begin{scope}[>=latex, xshift=0cm, yshift=0]
\foreach \ii in {5} {	% Enter Number of Decades in x
\foreach \jj in {2.5} {	% Enter Number of Decades in y
		\foreach \i in {1,2,...,\ii} {
			\foreach \j in {0,1,2,...,\jj} {			
\draw[black!50!, step=0.5] (0,-.5) grid (\ii,\jj); % Draw Sub Linear grid					
}}% End Log Grid
\draw[black!80!] (0,-.5) grid (\ii,\jj); % Draw Linear grid
\draw [->,thick] (0,-.5)--(0,\jj+.25) node (yaxis) [above] {$i\,[\milli\ampere]$};
\draw [->,thick] (0,1)--(\ii+.25,1) node (xaxis) [right] {$\omega t$}; % Draw axes
\draw node at (0,-.5)[below]{$0$};
\draw node at (2,-.5)[below]{$\pi$};
\draw node at (4,-.5)[below]{$2\pi$};
\foreach \y in {-10,0,10}% y Axis Label:
\node [anchor=east] at(0,\y/10+1){$\y$};
}}
\end{scope}
\begin{scope}[>=latex, xshift=0cm, yshift=1cm]
\draw[color=red,thick,domain=0:5,smooth,samples=100]  plot[id=iomega]  function{1*cos(.5*3.14*x)+.288*cos(1.5*3.14*x)};
\draw[color=blue,thick,domain=0:5,smooth,samples=100]  plot[id=iFe]  function{-.4*sin(.5*3.14*x)};
\draw[red] node at (.825,1.25) {{\footnotesize $i_{\mu}(t)$}};
\draw[blue] node at (2.5,.75) {{\footnotesize $i_{Fe}(t)$}};
\end{scope}
\draw node at (2.25,-1)[below]{Bild 2};
\end{tikzpicture}
\begin{tikzpicture}[scale=1.25]
\begin{scope}[>=latex, xshift=0cm, yshift=0]
\foreach \ii in {5} {	% Enter Number of Decades in x
\foreach \jj in {2.5} {	% Enter Number of Decades in y
		\foreach \i in {1,2,...,\ii} {
			\foreach \j in {0,1,2,...,\jj} {			
\draw[black!50!, step=0.5] (0,-.5) grid (\ii,\jj); % Draw Sub Linear grid					
}}% End Log Grid
\draw[black!80!] (0,-.5) grid (\ii,\jj); % Draw Linear grid
\draw [->,thick] (0,-.5)--(0,\jj+.25) node (yaxis) [above] {$i\,[\milli\ampere]$};
\draw [->,thick] (0,1)--(\ii+.25,1) node (xaxis) [right] {$\omega t$}; % Draw axes
\draw node at (0,-.5)[below]{$0$};
\draw node at (2,-.5)[below]{$\pi$};
\draw node at (4,-.5)[below]{$2\pi$};
\foreach \y in {-10,0,10}% y Axis Label:
\node [anchor=east] at(0,\y/10+1){$\y$};
}}
\end{scope}
\begin{scope}[>=latex, xshift=0cm, yshift=1cm]
\draw[color=red!50!blue,thick,domain=0:5,smooth,samples=100]  plot[id=iomega]  function{1*cos(.5*3.14*x)+.288*cos(1.5*3.14*x)-.4*sin(.5*3.14*x)};
\draw[red!50!blue] node at (.825,1.25) {{\footnotesize $i_0(t)$}};
\end{scope}
\draw node at (2.25,-1)[below]{Bild 3};
\end{tikzpicture}
\end{align*}
\renewcommand{\labelenumi}{\alph{enumi})}
\begin{enumerate}
\item Berechnen Sie Effektivwert und Klirrfaktor von $i_\mu(t)$
\item Berechnen Sie Effektivwert und Klirrfaktor von $i_0(t)$
\end{enumerate}
\ifthenelse{\equal{\toPrint}{Lösung}}{%
\begin{align}
\intertext{Formeln:}
k&=\frac{\sqrt{\sum \limits_{n=2}^{\infty}A^2_n}}{\sqrt{\sum \limits_{n=1}^{\infty}A^2_n}}=\frac{\text{Effektivwert der Oberschwingungen}}{\text{Effektivwert des Gesamtsignals}}
\end{align}
\clearpage
Berechnung:
\begin{align*}
\intertext{a) Effektivwert $I_{\mu}$ und Klirrfaktor $k_{\mu}$}
I_{\mu}&=\sqrt{I_{\mu,\omega}^{\phantom{\mu}2}+I_{\mu,3\omega}^{\phantom{\mu}2}}
=\sqrt{\left(\frac{\widehat{i}_{\mu,\omega}}{\sqrt{2}}\right)^2
+\left(\frac{\widehat{i}_{\mu,3\omega}}{\sqrt{2}}\right)^2}
=\frac{1}{\sqrt{2}}\cdot \sqrt{(10\,\milli\ampere)^2+(2{,}88\,\milli\ampere)^2}=\uuline{7{,}36\,\milli\ampere}\\
%I_{\mu}&=\sqrt{I_{\mu,\omega}^{\phantom{\mu}2}+I_{\mu,3\omega}^{\phantom{\mu}2}}=\frac{1}{\sqrt{2}}\cdot \sqrt{\widehat{i}_{\mu,\omega}^{\phantom{\mu}2}+\widehat{i}_{\mu,3\omega}^{\phantom{\mu}2}}=\frac{1}{\sqrt{2}}\cdot \sqrt{(10\,\milli\ampere)^2+(2{,}88\,\milli\ampere)^2}=\uuline{7{,}36\,\milli\ampere}\\
k_{\mu}&=\frac{I_{\mu,3\omega}}{I_{\mu}}
=\frac{\widehat{i}_{\mu,3\omega}\cancel{/\sqrt{2}}}{\sqrt{\widehat{i}_{\mu,\omega}^{\phantom{\mu}2}
+\widehat{i}_{\mu,3\omega}^{\phantom{\mu}2}}\cancel{/\sqrt{2}}}
=\frac{2{,}88\,\milli\ampere}{\sqrt{(10\,\milli\ampere)^2+(2{,}88\,\milli\ampere)^2}}=\uuline{0{,}277}=\uuline{27{,}7\%}\\
\end{align*}
\begin{minipage}[c]{.8\textwidth}
\begin{align*}
\intertext{b) Effektivwert $I_0$ und Klirrfaktor $k_0$}
\widehat{i}_{0,3\omega}&=\widehat{i}_{\mu,3\omega}=2{,}88\,\milli\ampere\\
\intertext{Nulldurchgang von $\widehat{i}_{Fe}$ bei den Spitzenwerten}\\
\widehat{i}_{0,\omega}&=\sqrt{\widehat{i}_{\mu,\omega}^{\phantom{\mu}2}+\widehat{i}_{Fe,\omega}^{\phantom{Fe}2}}\\
&=\sqrt{(10\,\milli\ampere)^2+(4\,\milli\ampere)^2}=10{,}77\,\milli\ampere\\[\baselineskip]
I_0&=\frac{1}{\sqrt{2}}\cdot \sqrt{\widehat{i}_{0,\omega}^{\phantom{\mu}2}+\widehat{i}_{0,3\omega}^{\phantom{\mu}2}}=\frac{1}{\sqrt{2}}\cdot \sqrt{(10{,}77\,\milli\ampere)^2+(2{,}88\,\milli\ampere)^2}=\uuline{7{,}88\,\milli\ampere}\\
k_0&=\frac{\widehat{i}_{0,3\omega}\cancel{/\sqrt{2}}}{\sqrt{\widehat{i}_{0,\omega}^{\phantom{\mu}2}+\widehat{i}_{0,3\omega}^{\phantom{\mu}2}}\cancel{/\sqrt{2}}}
=\frac{2{,}88\,\milli\ampere}{\sqrt{(10{,}77\,\milli\ampere)^2+(2{,}88\,\milli\ampere)^2}}=\uuline{0{,}258}=\uuline{25{,}8\%}
\end{align*}
\end{minipage}
\begin{minipage}[c]{.2\textwidth}
\begin{tikzpicture}[scale=1.2]
    \begin{scope}[>=latex, xshift=2cm, yshift=5]
        \draw[dashed](0,-2)rectangle(1,0);
        \draw[->,blue](0,0)--(1,0)node[right]{$\widehat{i}_{Fe,\omega}$};
        \draw[->,red](0,0)--(0,-2)node[below]{$\widehat{i}_{\mu,\omega}$};
        \draw[->,red!50!blue](0,0)--(1,-2)node[below right]{$\widehat{i}_{0,\omega}$};
    \end{scope}
\end{tikzpicture}
\end{minipage}\\[\baselineskip]

\uline{Nicht gefragt}\\[\baselineskip]
Typische Klirrfaktoren:\\
Rechteckschwingung $33\%$\\
Sprache noch verständlich $10\%$\\
Max. HiFi Verstärker $1\%$\\
Guter HiFi Verstärker $0{,}1\%$\\

Weiteres unter \url{http://de.wikipedia.org/wiki/Klirrfaktor}
\clearpage
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