ET2_Uebung_BEI/ET2_L_B12_A5.tex

\section{Phasenanschnitt}
Berechnen Sie den Effektivwert dieser sinusförmigen Spannung mit Phasenanschnitt.\\
\begin{align*}
	\begin{tikzpicture}[scale=1]
		\begin{scope}[>=latex,thick]
		 	\draw [ultra thin] (0,-2)grid(6,2); 			
			\draw [->](0,0) -- (6.5,0) node [right] {$t\,[\milli\second]$};
			\draw [->](0,-2.25) -- (0,2.25) node [above] {$u\,[\volt]$};
            \draw[color=red,very thick,domain=0.6:2,smooth,samples=100](0,0)--(0.6,0)--(0.6,1.618)
            plot[id=sina]function{2*sin(.5*3.14*x)};
			\draw[color=red,very thick,domain=0.6:2,smooth,samples=100]plot[id=sinb]function{2*sin(0.5*3.14*x)};
			\draw[color=red,very thick,domain=2.6:4,smooth,samples=100]                 (2,0)--(2.6,0)-- (2.6,-1.618) plot[id=sinc] function{2*(sin(0.5*3.14*x))};
			\draw[color=red,very thick,domain=4.6:6,smooth,samples=100] (4,0)--(4.6,0)--(4.6,1.618)plot[id=sind]function{2*sin(0.5*3.14*x)};
			\foreach \x in {10,20,30}
				\draw(\x/5,0)--(\x/5,-0.2) node[anchor=north]{$\x$};
				\draw (3/5,0)--(3/5,-0.2)node[anchor=north]{$3$};
			\foreach \y in {-400,-200,...,400}
				\draw (0,\y/200)--(-0.2,\y/200)node[anchor=east]{$\y$};
		\end{scope}
	\end{tikzpicture}
\end{align*}
\ifthenelse{\equal{\toPrint}{Lösung}}{%
\begin{align}
\intertext{Formeln:}
U&=U_{\textrm{eff}}=\sqrt{\frac{1}{T}\cdot \int_{t=0}^{T}{u^2(t)\cdot dt}}&\text{Effektivwert}&\\
\sin^2\alpha&=\frac{1}{2}(1-\cos 2\alpha)
\end{align}
\begin{align*}
\intertext{Berechnung:}
\end{align*}
\begin{align*}
	\begin{tikzpicture}[scale=1]
		\begin{scope}[>=latex,thick]
		 	\draw [ultra thin] (0,-2)grid(6,2); 			
			\draw [->](0,0) -- (6.5,0) node [right] {$t\,[\milli\second]$};
			\draw [->](0,-2.25) -- (0,2.25) node [above] {$u\,[\volt]$};
			\draw[color=red,very thick,domain=0.6:2,smooth,samples=100]			 (0,0)--(0.6,0)--(0.6,1.618) plot[id=sin12a53] function{2*sin(0.5*3.14*x)};
					\draw[color=red,very thick,domain=2.6:4,smooth,samples=100] (2,0)--(2.6,0)--(2.6,-1.618)  plot[id=sin12a54] function{2*(sin(0.5*3.14*x))};								 
			\draw[color=red,very thick,domain=4.6:6,smooth,samples=100] (4,0)--(4.6,0)--(4.6,1.618) plot[id=sin12a55] function{2*sin(0.5*3.14*x)};
			\draw[color=blue,very thick,dashed, domain=0:4,smooth,samples=100] plot[id=sin12a56] function{2*sin(0.5*3.14*x)};			
			\foreach \x in {10,20,30}
				\draw (\x/5,0) -- (\x/5,-0.2) node[anchor=north] {$\x$};
					\draw (3/5,0) -- (3/5,-0.2)node[anchor=north] {$3$};
			\foreach \y in {-400,-200,0,200,400}
				\draw (0,\y/200) -- (-0.2,\y/200) node[anchor=east] {$\y$};
		\end{scope}
	\end{tikzpicture}
\end{align*}
\begin{align*}
\intertext{Periodendauer $T=20\,\milli\second$, da Symmetrie in einer Periode. Betrachtung nur einer Sinus-Halbwelle mit $\frac{1}{2}\cdot T=10\,\milli\second$}
\omega&=2\pi f=\frac{2\pi}{T}=\frac{2\pi}{20\,\milli\second}=314\,\frac{1}{\second}\\
u(t)&=
\begin{cases}
0&\text{ für }t=0\ldots 3\,\milli\second\\
400\,\volt\cdot \sin(\omega t)=400\,\volt\cdot \sin(314\,\frac{1}{\second}\cdot t) &\text{ für }t=3\,\milli\second\ldots 10\,\milli\second\\
\end{cases}\\[\baselineskip]
U^2&=\frac{1}{T/2}\int_{3\,\milli\second}^{10\,\milli\second}{(400\,\volt\cdot \sin(\omega t))^2\cdot dt}\\
&=\frac{1}{T/2}\cdot (400\,\volt)^2\int_{3\,\milli\second}^{10\,\milli\second}{\sin^2(\omega t)\cdot dt}\\[\baselineskip]
&\text{mit }\sin^2\alpha=\frac{1}{2}(1-\cos 2\alpha)\\[\baselineskip]
U^2&=\frac{(400\,\volt)^2}{10\,\milli\second}\cdot\frac{1}{2}\cdot  \left(\int_{3\,\milli\second}^{10\,\milli\second}{1\, dt} -\int_{3\,\milli\second}^{10\,\milli\second}{\cos(2 \omega t)\cdot dt}\right)\\
&=\frac{(400\,\volt)^2}{10\,\milli\second}\cdot \frac{1}{2}\cdot \left(\Big[t\Big]_{3\,\milli\second}^{10\,\milli\second}-\left[\sin(2 \omega t)\cdot \frac{1}{2\omega} \right]_{3\,\milli\second}^{10\,\milli\second}\right)\\
&=\frac{(400\,\volt)^2}{20\,\milli\second}\cdot \left(7\,\milli\second-\frac{1}{2\cdot \omega}\cdot \underbrace{\left(sin(\cancel{2}\cdot \frac{2\pi}{\cancel{20\,\milli\second}}\cdot \cancel{10\,\milli\second}\right)}_{sin(2\pi)=0}-sin\left(2\cdot \frac{\pi}{20\,\milli\second}\cdot 3\,\milli\second\right)\right)\\
&=\frac{(400\,\volt)^2}{20\,\milli\second}\cdot \left(7\,\milli\second-\frac{20\,\milli\second}{4\pi}\cdot \underbrace{(-sin(0{,}6\pi))}_{-0{,}951}\right)\\
&=\frac{(400\,\volt)^2}{20\,\milli\second}\cdot 8{,}51\,\milli\second=68109\,\square\volt\\[\baselineskip]
U&=400\,\volt\cdot \sqrt{\frac{8{,}51\,\milli\second}{20\,\milli\second}}=\uuline{261\,\volt}\\
\intertext{Zum Vergleich: Sinus ohne Phasenanschnitt hätte einen Effektivwert von}
U_{(sin)}&=\frac{400\,\volt}{\sqrt{2}}=283\,\volt
\end{align*}
\clearpage
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