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- \section {Klirrfaktor}
- Am Eingang liegt die Spannung\\
- $u_e(t)=2\,\volt+3\,\volt\cdot \sin(2\pi\cdot 50\,\power{\second}{-1}\cdot t) + 4\,\volt\cdot \sin(2\pi\cdot 100\,\power{\second}{-1}\cdot t)$
- \begin{align*}
- \begin{tikzpicture}[very thick,scale=2]
- \begin{scope}[>=latex,xshift=0cm,yshift=1cm]%Spule -
- \draw (0,0)--(.3,0) (.7,0)--(1,0)node at (.5,.0667) [above] {$L$};
- \fill (.3,-0.0667)rectangle(.7,0.0667);
- % \draw [->,blue] (.3,-.2)--(.7,-.2)node at(.5,-.2)[below]{\footnotesize$U_{L}$};
- \end{scope}
- \begin{scope}[>=latex,very thick,xshift=1cm,yshift=0cm,rotate=90]%Kondensator |
- \draw (0,0)--(.475,0) (.475,-.125)--(.475,.125) (.525,-.125)--(.525,.125) (.525,0)--(1,0)node at (.5,-.133) [right] {$C$};
- % \draw [<-,blue] (.3,.2)--(.7,.2) node at (.5,.2)[left]{\footnotesize$U_{C1}$};
- \end{scope}
- \draw(0,0)--(1.5,0)(1,1)--(1.5,1);
- \fill(0,0)circle(.05cm)(0,1)circle(.05cm)(1.5,0)circle(.05cm)(1.5,1)circle(.05cm);
- \draw[->,blue](0,.9)--(0,.1)node at(0,.5)[left]{$u_e$};
- \draw[->,blue](1.5,.9)--(1.5,.1)node at(1.5,.5)[right]{$u_a$};
- \draw[->,red](1.5,1.1)--(1.3,1.1)node at(1.2,1.1)[above]{$i_a=0$};
- \end{tikzpicture}
- \end{align*}
- \renewcommand{\labelenumi}{\alph{enumi})}
- \begin{enumerate}
- \item Berechnen Sie den Klirrfaktor $k_a$ der Ausgangsspannung $u_a$
- \item Berechnen Sie den Effektivwert $U_a$
- \end{enumerate}
- $L=100\,\milli\henry$, $C=250\,\micro\farad$
- \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}}\\
- k&=\sqrt{\frac{A^2_2+A^2_3+\cdots+A^2_n}{A^2_1+A^2_2+\cdots+A^2_n}}\\
- A_0&\quad\text{Gleichanteil}\notag\\
- A_1&\quad\text{Grundschwingung}\notag\\
- A_2\cdots A_n&\quad\text{Oberwellen}\notag
- \end{align}
- \clearpage
- Berechnung:
- \begin{align*}
- \intertext{a) Klirrfaktor am Ausgang\ $k_a$}
- U_{a0}&=U_{e0}=2\,\volt\qquad\text{Gleichanteil ist ohne Einfluß auf den Klirrfaktor}\\[\baselineskip]
- \frac{\uline{U}_a}{\uline{U}_e}&=\frac{jX_C}{jX_L+jX_C}=\frac{X_C}{X_L+X_C}\qquad\text{Wechselspannung}\\[\baselineskip]
- \text{Für }&50\,\hertz\text{ Grundwelle}\\
- X_{C1}&=-\frac{1}{2\pi f\cdot C}=\frac{-1}{2\pi\cdot 50\,\cancel{\frac{1}{\second}}\cdot 250\cdot \power{10}{-6}\,\frac{\ampere\cancel{\second}}{\volt}}=-12{,}73\,\ohm\\
- X_{L1}&=2\pi f\cdot L=2\pi\cdot 50\,\cancel{\frac{1}{\second}}\cdot 0{,}1\cdot \,\frac{\volt\cancel{\second}}{\ampere}=31{,}42\,\ohm\\
- U_{a1}&=\frac{-12{,}73\,\ohm}{31{,}42\,\ohm-12{,}73\,\ohm}\cdot U_{e1}=-0{,}6811\cdot U_{e1}\\
- U_{a1}&=|-0{,}6811\cdot U_{e1}|=0{,}6811\cdot \frac{3\,\volt}{\sqrt{2}}=1{,}445\,\volt\quad \text{Effektivwert}\\
- U^2_{a1}&=2{,}088\,\volt^2\\[\baselineskip]
- \text{Für }&100\,\hertz\text{ 1. Oberwelle}\\
- X_{C2}&=\frac{1}{2}\cdot X_{C1}=-6{,}365\,\ohm\\
- X_{L2}&=2X_{L1}=62{,}84\,\ohm\\
- U_{a2}&=\left|\frac{-6{,}365\,\ohm}{62{,}84\,\ohm-6{,}365\,\ohm}\right|\cdot \frac{4\,\volt}{\sqrt{2}}=0{,}1127\cdot \frac{4\,\volt}{\sqrt{2}}=0{,}3188\,\volt\quad \text{Effektivwert}\\
- U^2_{a2}&=0{,}1016\,\volt^2\\[\baselineskip]
- k_a&=\sqrt{\frac{U^2_{a2}}{U^2_{a1}+U^2_{a2}}}=\sqrt{\frac{0{,}1016\,\volt^2}{2{,}088\,\volt^2+0{,}1016\,\volt^2}}
- =\uuline{0{,}215}=\uuline{21{,}5\%}
- \intertext{b) Effektivwert $U_a$}
- U_a&=\sqrt{U^2_{a0}+U^2_{a1}+U^2_{a2}}=\sqrt{(2\,\volt)^2+2{,}088\,\volt^2+0{,}1016\,\volt^2}=\uuline{2{,}49\,\volt}
- \end{align*}
- \clearpage
- }{}%
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