81 lines
4.3 KiB
TeX
81 lines
4.3 KiB
TeX
\section{Rechteckspannung}
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Gegeben ist eine periodische Rechteckspannung mit der Periodendauer von $10\,\milli\second$.\\
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Berechnen Sie den Effektivwert, wenn der arithmetische Mittelwert gleich Null ist.\\
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\begin{align*}
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\begin{tikzpicture}[scale=0.5]
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\begin{scope}[>=latex,thick]
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\draw [->](0,-2) -- (10.75,-2) node [right] {$t\,[\milli\second]$};
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\draw [->](0,0) -- (0,5.5) node [above] {$u\,[\volt]$};
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\draw [<->,blue, very thick] (-.5,0)--(-.5,5) node at (0,2.5)[right]{$5\,\volt$};
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\draw [red,very thick](-0.5,0)--(0,0)--(0,5)--(7,5)--(7,0)--(10,0)
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--(10,5)--(10.5,5);
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\foreach \x in {0,1,...,10}
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\draw (\x,-2) -- (\x,-2.2) node[anchor=north] {$\x$};
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\foreach \y in {0,1,...,5}
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\draw (0,\y) -- (-0.2,\y);% node[anchor=east] {$\y$};
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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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\overline{u}&=\frac{1}{T}\cdot \int_{t=0}^{T}{u(t)\cdot dt}&\text{Arithmetischer Mittelwert}\\
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U&=U_{\textrm{eff}}=\sqrt{\frac{1}{T}\cdot \int_{t=0}^{T}{u^2(t)\cdot dt}}&\text{Effektivwert}&
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\end{align}
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\begin{align*}
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\intertext{Berechnung:}
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\intertext{a) Arithmetischer Mittelwert $\overline{u}=0$}
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&\text{$\Rightarrow$ Fläche ober- und unterhalb der Nulllinie muß gleich sein!}\\
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&\text{$\Rightarrow$ Wo ist die Nulllinie?}
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\end{align*}
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\begin{align*}
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\begin{tikzpicture}[scale=0.5]
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\begin{scope}[>=latex,thick]
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\draw [->](0,0) -- (10.75,0) node [right] {$t\,[\milli\second]$};
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\draw [->](0,0) -- (0,5.5) node [above] {$u\,[\volt]$};
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\fill [black!15!] (0,3.5)rectangle(7,5) (7,3.5)rectangle(10,0);
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\draw [red,very thick](-0.5,0)--(0,0)--(0,5)--(7,5)--(7,0)--(10,0)
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--(10,5)--(10.5,5);
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\draw [blue,very thick] (0,3.5)--(10,3.5) node [right]{NULL};
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\draw [->,blue,very thick] (3.5,3.5)--(3.5,5) node at(3.5,4.25)[right]{$u_1$};
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\draw [<-,blue,very thick] (8.5,0)--(8.5,3.5) node at(8.5,1.75)[right]{$u_2$};
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\foreach \x in {0,1,...,10}
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\draw (\x,0) -- (\x,-0.2) node[anchor=north] {$\x$};
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\foreach \y in {-3.5,-2.5,...,1.5}
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\draw (0,\y+3.5) -- (-0.2,\y+3.5) node[anchor=east] {$\y$};
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\end{scope}
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\end{tikzpicture}
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\end{align*}
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\enlargethispage{1cm}
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\begin{align*}
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&u_1+(-u_2)=5\,\volt\rightarrow u_2=-(5\,\volt-u_1)\\
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&\text{Fläche:}\\
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&\frac{1}{T}\int_{0}^{T}{u(t)\cdot dt}\stackrel{!}{=}0\\
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&\frac{1}{T}\cdot \left(\int_{0}^{7\,\milli\second}{u_1\cdot dt}-\int_{7\,\milli\second}^{10\,\milli\second}{u_2\cdot dt}\right)\stackrel{!}{=}0\\
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&u_1\cdot 7\,\milli\second-u_2\cdot 3\,\milli\second=0\\
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&u_1\cdot 7\,\milli\second-(5\,\volt-U_1)\cdot 3\,\milli\second=0\\
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&(7\,\milli\second+3\,\milli\second)\cdot u_1=15\,\volt\cdot \milli\second\\
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&u_1=\frac{15\,\volt\cdot \milli\second}{10\,\milli\second}=\uuline{1{,}5\,\volt}\\
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&u_2=-(5\,\volt-1{,}5\,\volt)=\uuline{-3{,}5\,\volt}
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\intertext{Alternativ mit Beträgen}
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&|u_1|+|u_2|=5\,\volt\rightarrow |u_2|=5\,\volt-|u_1|\\
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&\text{Fläche:}\\
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&\frac{1}{T}\int_{0}^{T}{u(t)\cdot dt}\stackrel{!}{=}0\\
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&\frac{1}{T}\cdot \left(\int_{0}^{7\,\milli\second}{u_1\cdot dt}-\int_{7\,\milli\second}^{10\,\milli\second}{u_2\cdot dt}\right)\stackrel{!}{=}0\\
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&u_1\cdot 7\,\milli\second-u_2\cdot 3\,\milli\second=0\\
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&u_1\cdot 7\,\milli\second-(5\,\volt-|u_1|)\cdot 3\,\milli\second=0\\
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&(7\,\milli\second+3\,\milli\second)\cdot |u_1|=15\,\volt\cdot \milli\second\\
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&|u_1|=\frac{15\,\volt\cdot \milli\second}{10\,\milli\second}=\uline{1{,}5\,\volt}\qquad
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|u_2|=5\,\volt-1{,}5\,\volt)=\uline{3{,}5\,\volt}\\
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&\text{Da $u_1$ positives Vorzeichen in der Skizze hat, muss $u_2$ ein negatives Vorzeichen erhalten.}\\
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\Rightarrow& u_1=\uuline{1{,}5\,\volt} \qquad u_2=\uuline{-3{,}5\,\volt}
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\intertext{b) Effektivwert}
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U^2&=U^2_{\textrm{eff}}=\frac{1}{T}\int{\left(u(t)\right)^2\cdot dt}\\
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&=\frac{1}{10\,\milli\second}\left(\int_{0}^{7\,\milli\second}{(1{,}5\,\volt)^2\cdot dt}+\int_{7\,\milli\second}^{10\,\milli\second}{(-3{,}5\,\volt)^2\cdot dt}\right)\\
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&=\frac{1}{10\,\milli\second}\cdot \left(2{,}25\,\volt^2\cdot \big[t\big]_{0}^{7\,\milli\second}+12{,}25\,\volt^2\cdot \big[t\big]_{7\,\milli\second}^{10\,\milli\second}\right)\\
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&=\frac{1}{10\,\milli\second}\cdot \left(2{,}25\,\volt^2\cdot 7\,\milli\second+12{,}25\,\volt^2\cdot (10\,\milli\second-7\,\milli\second)\right)=5{,}25\,\volt^2\\
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U&=\sqrt{5{,}25\,\volt^2}=\uuline{2{,}29\,\volt}
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\end{align*}
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\clearpage
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}{}%
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