131 lines
7.5 KiB
TeX
131 lines
7.5 KiB
TeX
\section{Ortskurve}
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\renewcommand{\labelenumi}{\alph{enumi})}
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\begin{enumerate}
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\setlength{\itemsep}{-0.2\baselineskip}
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\item Zeichnen Sie maßstäblich die Ortskurve für das Spannungsverhältnis $U_2/U_1$ in Abhängigkeit von der Frequenz $f$.
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\item Geben Sie die Grenzfrequenz der Schaltung an.
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\item Wie groß ist die Dämpfung des Vierpols für die Frequenz $f=1{,}2\,\kilo\hertz$ (falls in Vorlesung behandelt: in $\deci\bel$)
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\end{enumerate}
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$\qquad \,\, R=16\,\kilo\ohm$; $C=12\,\nano\farad$; $200\,\hertz\leq f \leq 1{,}2\,\kilo\hertz$
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\begin{align*}
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\begin{tikzpicture}[very thick,scale=2]
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\begin{scope}[>=latex,very thick,xshift=0cm,yshift=1cm]%Widerstand - nach EN 60617
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\draw (0,0)--(.3,0) (.3,-0.0667)rectangle(.7,0.0667) (.7,0)--(1,0)node at (.5,.0667) [above] {$R$};
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\end{scope}
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\begin{scope}[>=latex,very thick,xshift=1cm,yshift=0cm,rotate=90]%Kondensator |
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\draw (0,0)--(.475,0) (.475,-.125)--(.475,.125) (.525,-.125)--(.525,.125) (.525,0)--(1,0)node at (.5,-.133) [right] {$C$};
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\end{scope}
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\begin{scope}[>=latex,very thick,xshift=0cm]%Knotenpunkte
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\draw (-.5,0)--(0,0) (-.5,1)--(0,1) (-.5,0)--(1.5,0) (.8,1)--(1.5,1);
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\draw [->,blue] (-.5,.9)--(-.5,.1)node at(-.5,.5)[right]{$\underline{U}_1$}; \fill (-.5,0)circle(.025) (-.5,1)circle(.025);
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\draw [->,red] (-.4,1.1)--(-.1,1.1) node at (-.25,1.1)[above]{$\underline{I}_1$};
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\draw [->,blue] (1.5,.9)--(1.5,.1)node at(1.5,.5)[right]{$\underline{U}_2$}; \fill (1.5,0)circle(.025) (1.5,1)circle(.025);
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\draw [->,red] (1.4,1.1)--(1.1,1.1) node at (1.25,1.1)[above]{$\underline{I}_2=0$};
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\end{scope}
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\end{tikzpicture}
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\end{align*}
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\ifthenelse{\equal{\toPrint}{Lösung}}{%
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%\begin{align}
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%\intertext{Formeln:}
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%\end{align}
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Berechnung: (Platzbedarf in x: $5\,\centi\metre$; in y: $15\,\centi\metre$)\\[0.5\baselineskip]
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a) Spannungsverhältnis $U_2/U_1$ in Abhängigkeit von der Frequenz $f$
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\begin{align*}
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\uline{I}_1&=\uline{U}_1 \cdot \uline{Y} = \uline{U}_1 \cdot \frac{j\omega\cdot C}{1+j\omega\cdot C\cdot R}=\frac{\uline{U}_1}{R+\frac{1}{j\omega\cdot C}} \\
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\uline{U}_2&= \uline{I}_1\cdot \frac{1}{j\omega\cdot C}= \frac{\uline{U}_1}{(R+\frac{1}{j\omega\cdot C})\cdot j\omega\cdot C}=
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\frac{\uline{U}_1}{1+j\omega\cdot R\cdot C}\\
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\frac{\uline{U}_2}{\uline{U}_1}&=\frac{1}{1+j\omega\cdot R \cdot C}
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=\frac{1-j\omega RC}{1+(\omega RC)^2}\\[0.5\baselineskip]
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&\begin{tabular}{|l|l|l|}
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\hline
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% after \\: \hline or \cline{col1-col2} \cline{col3-col4} ...
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$f/Hz$&$2\pi\cdot f\cdot R\cdot C$&$\uline{U}_2/\uline{U}_1$\\
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\hline
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200 & 0{,}241&0{,}945-j0{,}228\\
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828{,}9&1&0{,}5-j0{,}5\\
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1200&1{,}448&0{,}323-j0{,}468\\
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\hline
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\end{tabular}
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\intertext{b) Grenzfrequenz}
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\omega_g\cdot R\cdot C&=1\\
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f_g&=\frac{1}{2\pi\cdot R\cdot C}=\uuline{828{,}9\,\hertz}
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\intertext{c) Dämpfung}
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f&=1{,}2\,\kilo\hertz\qquad a=20\cdot \lg\frac{\uline{U}_2}{\uline{U}_1}\\
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\frac{\uline{U}_2}{\uline{U}_1}&=0{,}323-j0{,}468=0{,}569\cdot e^{-j50{,}56\,\degree}\\
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a&=20\cdot \lg(0{,}569)=\uuline{-4{,}9\,\deci\bel}
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\end{align*}
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\begin{align*}
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\begin{tikzpicture}[very thick,scale=1]
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\begin{scope}[>=latex,very thick,xshift=0cm,yshift=0cm]
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\draw[ultra thin,black!50!](0,-7.5)grid [step=.5cm](5,7.5);
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\draw[thin,->](0,0)--(5.5,0)node[right]{$\Re \, \left\{\frac{\uline{U}_2}{\uline{U}_1}\right\}$};
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\draw[thin,->](0,-8)--(0,8)node[above]{$\Im \, \left\{\frac{\uline{U}_2}{\uline{U}_1}\right\}$};
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\foreach \y in {1.5,1,...,-1.5}
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\draw(0,5*\y)--(-.1,5*\y)node[left]{$\y$};
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\draw[red,thin](5,-7.5)--(5,7.5);
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\draw[red](4.9,5)--(5.1,5)node [right]{$f_g=828{,}9\,\hertz$};
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\draw[red](4.9,-5)--(5.1,-5)node [right]{$f_g=828{,}9\,\hertz$};
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\draw[red](4.9,1.205)--(5.1,1.205)node [right]{$f=200\,\hertz$};
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\draw[red](4.9,-1.205)--(5.1,-1.205)node [right] {$f=200\,\hertz$};
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\draw[red](4.9,7.24)--(5.1,7.24)node [right] {$f=1200\,\hertz$};
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\draw[red](4.9,-7.24)--(5.1,-7.24)node [right] {$f=1200\,\hertz$};
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\draw node at(6.125,6.2){$(1+j\omega RC)$};
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\foreach \x in {.5,1}
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\draw(5*\x,.1)--(5*\x,-.1)node at (5*\x+.1,-.25){$\x$};
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\fill[blue](2.5,0)circle(.05cm);
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\draw[blue](0:2.5cm)+(180:2.5cm)arc(180:360:2.5cm);%Mittelpunkt+Start arc Start:End:Radius
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\draw[blue!50!red,thin](0,0)--(5,-1.205)(0,0)--(5,-5)(0,0)--(5,-7.24);
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\filldraw[blue!50!red](4.73,-1.14)circle(.05)node [below left]{$200\,\hertz$};
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\filldraw[blue!50!red](2.5,-2.5)circle(.05)node [below]{$f_g$};
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\filldraw[blue!50!red](1.62,-2.35)circle(.05)node [below left]{$1{,}2\,\kilo\hertz$};
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\draw [blue] node at(4,-2.5){$\frac{\underline{U}_2}{\underline{U}_1}$};
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\end{scope}
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\end{tikzpicture}
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\end{align*}
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% (Siehe Merksätze zu Inversion im Script, hier graphisch veranschaulicht:)
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%\begin{align*}
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% \begin{tikzpicture}[very thick,scale=1.25]
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% \begin{scope}[>=latex,very thick,xshift=0cm,yshift=0cm]
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% \draw(0,-1)--(0,1)node[above]{$\Im$};
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% \draw(0,0)--(2,0);
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% \draw[red](1,-1)--(1,1);
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% \draw[red,->](0,0)--(1,.75)node at(1.25,.75){$\uline{Y}_p$};
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% \end{scope}
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% \begin{scope}[>=latex,very thick,xshift=3cm,yshift=0cm]
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% \draw(0,-1)--(0,1)node[above]{$\Im$};
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% \draw(0,0)--(2,0);
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% \draw[blue](.5,0)circle(.5cm);
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% \draw[blue,->](0,0)--(.7,-.5cm)node at(1.25,-.5){$\uline{Z}_p$};
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% \end{scope}
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% \begin{scope}[>=latex,very thick,xshift=6cm,yshift=0cm]
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% \draw(0,-1)--(0,1);
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% \draw(-.5,0)--(2,0);
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% \draw[blue!50!red](-.5,-1)--(-.5,1)node[above]{$\Im$ neu};
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% \draw[blue](.5,0)circle(.5cm);
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% \draw[green!50!black,->](-.5,0)--(.7,-.5cm)node at(1,-.5){$\uline{Z}$}; \end{scope}
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% \begin{scope}[>=latex,very thick,xshift=9cm,yshift=0cm]
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% \draw(0,-1)--(0,1);
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% \draw[blue!50!red](-.5,-1)--(-.5,1)node[above]{$\Im$ neu};
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% \draw(-.5,0)--(2.5,0);
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% \draw[blue](1.25,0)circle(1cm);
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% \draw[blue!50!red,->](-.5,0)--(2.1,.5)node at(2.5,.5){$\uline{Y}$};
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% \end{scope}
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% \end{tikzpicture}
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%\end{align*}
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%Leitwertsortskuve der Parallelschaltung von $R_2$ und $C$ ergibt eine Gerade die im Abstand $G_2$ parallel zur imaginären Achse liegt.\\
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%$G_2=20\,\milli\siemens\,\widehat{=}\, 4\,\centi\metre$\\
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%Grenzen eintragen für $\uline{Y}_P=5\,\milli\siemens\,\widehat{=}\, 1\,\centi\metre (f_u)\quad 25\,\milli\siemens\,\widehat{=}\, 5\,\centi\metre (f_o)\quad\text{ auch für $\uline{Y}^*_P$}$\\
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%Inversion von $\uline{Y}_P$ liefert die Widerstandsortskurve $\uline{Z}_P$ ein Kreis durch den Ursprung.\\
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%$R_2=50\,\ohm\,\widehat{=}\, 5\,\centi\metre \quad\text{Kreis um P(2.5,0) $r=2{,}5\,\centi\metre$}$\\
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%Der Serienwiderstand $R_1$ wird durch verschieben der Koordinaten hinzugefügt.\\
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%Punktweise Inversion von $\uline{Z}$ ein Kreis der jetzt nicht mehr durch den Ursprung geht, liefert $\uline{Y}$, wieder einen Kreis.\\
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%$\uline{Y}\quad\text{Kreis um P(6.67,0) $r=5\,\centi\metre$, da $\uline{Y}_{min}$ und $\uline{Y}_{max}$ bekannt sind}$\\
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%\begin{align*}
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%R_2&=50\,\ohm\,\widehat{=}\, 5\,\centi\metre \quad\text{Kreis um P(2.5,0) $r=2{,}5\,\centi\metre$}\\
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%G_2&=20\,\milli\siemens\,\widehat{=}\, 4\,\centi\metre\\
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%\uline{Y}_P&=5\,\milli\siemens\,\widehat{=}\, 1\,\centi\metre (f_u)\quad 25\,\milli\siemens\,\widehat{=}\, 5\,\centi\metre (f_o)\quad\text{zeichnen, auch $\uline{Y}^*_P$}\\
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%\uline{Y}&\quad\text{Kreis um P(6.67,0) $r=5\,\centi\metre$, da $\uline{Y}_{min}$ und $\uline{Y}_{max}$ bekannt sind}\\
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%\end{align*}
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\clearpage
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}{}%
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