\ANG{
\EPSR{7cm}{elrr4551}{7cm}{}{
\PSFM{U}{$U$} \PSFM{U0}{$U_{0}$} \PSFM{t}{$t$} \PSFM{0}{0} 
\PSFM{P}{Parabeln}  
\PSFM{t1}{$t_{1}$} \PSFM{mt1}{$-t_{1}$} \PSFM{t2}{$t_{2}$} \PSFM{T}{$T$}  
}{
Nebenstehende Abbildung zeigt den zeitlichen Verlauf einer periodischen
Spannung $U(t)$. Dabei gilt
\[t_{2}=\alpha t_{1}\quad\text{mit}\quad \alpha\geqq 1\]
\begin{enumerate}
\item Berechne den Mittelwert $\BRA{U}$ und den Effektivwert $U_{\text{eff}}$
  von $U$, ausgedr"uckt durch $\alpha$ und $U_{0}$.
\item F"ur welches $\alpha$ ist $U_{\text{eff}}=0{,}6\cdot U_{0}$?
\item Untersuche, ob $\BRA{U}=U_{\text{eff}}$ sein kann.
\end{enumerate}
}
}
\LSG{
\begin{enumerate}
\item F"ur $-t_{1}\leqq t\leqq t_{1}$ lautet die Gleichung von $U$:
\[U(t)=U_{0}\left(1-\frac{t^{2}}{t_{1}^{2}}\right)\] 
Mit $T=t_{1}+t_{2}=t_{1}+\alpha t_{1}=(1+\alpha)t_{1}$ gilt:
\begin{align}
\BRA{U} &= \frac{1}{T}\int\limits_{-t_{1}}^{t_{2}}U(t)\,\RMd t=
\frac{2U_{0}}{T}\int\limits_{0}^{t_{1}}
  \left(1-\frac{t^{2}}{t_{1}^{2}}\right)\,\RMd t=
\frac{2U_{0}}{T}\left[
  t-\frac{t^{3}}{3t_{1}^{2}}\right]_{0}^{t_{1}}=\nonumber\\
  &= \frac{2U_{0}}{T}\left(t_{1}-\frac{t_{1}}{3}\right)=
     \frac{4U_{0}t_{1}}{3(1+\alpha)t_{1}}=\frac{4U_{0}}{3(1+\alpha)}\nonumber
\end{align}
\begin{align}
U_{\text{eff}}^{2} &= \BRA{U^{2}}= \frac{2}{T}\int\limits_{0}^{t_{1}}U(t)^{2}\,\RMd t=
\frac{2U_{0}^{2}}{T}\int\limits_{0}^{t_{1}}
  \left(1-\frac{t^{2}}{t_{1}^{2}}\right)^{2}\,\RMd t=\nonumber\\
  &= \frac{2U_{0}^{2}}{T}\int\limits_{0}^{t_{1}}\left(
     1-\frac{2t^{2}}{t_{1}^{2}}+\frac{t^{4}}{t_{1}^{4}}\right)\,\RMd t=
     \frac{2U_{0}^{2}}{T}\left[ 
       t-\frac{2t^{3}}{3t_{1}^{2}}+\frac{t^{5}}{5t_{1}^{4}}\right]_{0}^{t_{1}}=\nonumber\\
  &= \frac{2U_{0}^{2}}{T}\left(t_{1}-\frac{2t_{1}}{3}+\frac{t_{1}}{5}\right)=
     \frac{2U_{0}^{2}}{1+\alpha}\left(1-\frac23+\frac15\right)=
    \frac{16U_{0}^{2}}{15(1+\alpha)}\nonumber
\end{align}
\[U_{\text{eff}}=\frac{4U_{0}}{\sqrt{15(1+\alpha)}}\]
\item $\D \frac{4}{\sqrt{15(1+\alpha)}}=0{,}6=\frac35\QDQ
\frac{16}{15(1+\alpha)}=\frac{9}{25}\QDQ
1+\alpha=\frac{16\cdot 25}{15\cdot 9}=\frac{80}{27}$
\[\alpha=\frac{53}{27}\approx 1{,}96\]
\item \begin{align}
\BRA{U}=U_{\text{eff}}\quad&\GENAUDANN\quad
3(1+\alpha)=\sqrt{15(1+\alpha)}\nonumber\\
&\GENAUDANN\quad 9(1+\alpha)^{2}=15(1+\alpha)\nonumber\\
&\GENAUDANN\quad 9(1+\alpha)=15\nonumber\\
&\GENAUDANN\quad \alpha=\frac23<1\QDQ\text{nicht m"oglich}\nonumber
\end{align} 
\end{enumerate}
}
