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%Tapuscript : Laurent Mérat, Lycée Les Lombards, Troyes

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\begin{document}
\titre{Corection BTS A Métropole - 2005}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\exo{9} \begin{enumerate}
\item
	\begin{enumerate}
		\item $g'(t) = -2\cos t\sin t\times \sin^2 t+(1+\cos^2 t)\times 2\cos t \sin t = 2\cos t \sin t(-\sin^2 t+1+\cos^2 t)\\= 2\cos t \sin t\times 2\cos^2 t = 4 \sin t \cos^3 t$.
		\item $\cos^3 t$ et $\cos t$ sont de m\^eme signe donc :
$\begin{array}{|c|ccccc|}
\hline
t&0& &\frac{\pi}2& &\pi\\
\hline
4\sin t&0&+& &+&0\\
\hline
\cos^3 t& &+&0&-& \\
\hline
g'(t)&0&+&0&-&0\\
\hline
 & & &1& & \\
g(t)& &\nearrow& &\searrow& \\
 &0& & & &0\\
\hline
\end{array}$
	\end{enumerate}
\item
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		\item \psset{xunit=1.0cm,yunit=1.0cm}
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		\item $f$ est paire donc :
\begin{itemize}
\item[\textbullet] pour tout $n\in\N^*$, $b_n=0$
\item[\textbullet] $a_0=\dfrac21\displaystyle\int_0^{\frac12} f(t)dt = 2\displaystyle\int_0^{\tau} \dfrac12-\tau dt +2\displaystyle\int_{\tau}^{\frac12} -\tau dt = 
2\tau\left(\dfrac12-\tau \right)-2\left(\dfrac12-\tau \right)\tau =0$
\item[\textbullet]  soit $n\in\N^*$, $a_n = \dfrac41 \displaystyle\int_0^{\frac12} f(t)\cos\left(2\dfrac{2\pi}1t\right)dt =
4\displaystyle\int_0^{\tau} \left(\dfrac12-\tau\right)\cos(2n\pi t) dt +4\displaystyle\int_{\tau}^{\frac12} -\tau \cos(2n\pi t) dt\\ = 
4\left[ \left(\dfrac12-\tau\right)\dfrac{\sin(2n\pi t)}{2n\pi}\right]_0^{\tau}-4\left[\tau \dfrac{\sin(2n\pi t)}{2n\pi}\right]_{\tau}^{\frac12}
= 4\left(\dfrac12-\tau\right)\dfrac{\sin(2n\pi \tau)}{2n\pi}-0-0+4\tau \dfrac{\sin(2n\pi \tau)}{2n\pi}\\
= 4\times \dfrac12\times \dfrac{\sin(2n\pi \tau)}{2n\pi}=\dfrac{\sin(2n\pi \tau)}{n\pi}$
\end{itemize}
Ainsi, $ S(t)=\displaystyle\sum_{n=1}^{+\infty} \dfrac{1}{n\pi} \sin(2 n \pi \tau) \cos(2 n \pi t) $
	\end{enumerate}
\item
	\begin{enumerate}
		\item D'après la formule de Parseval, $E^2_h=\dfrac{\left(\dfrac{1}{\pi}\sin(2 \pi \tau)\right)^2+\left(\dfrac{1}{2\pi}\sin(4 \pi \tau)\right)^2}{2}\\
= \dfrac{\left(\dfrac{1}{\pi}\sin(2 \pi \tau)\right)^2+\left(\dfrac{2}{2\pi}\sin(2 \pi \tau)\cos(2 \pi \tau)\right)^2}{2}
= \dfrac{\sin^2(2 \pi \tau)+\sin^2(2 \pi \tau)\cos^2(2 \pi \tau)}{2\pi^2}\\
= \dfrac{(1+\cos^2(2 \pi \tau))\sin^2(2 \pi \tau)}{2\pi^2}$.
		\item Ainsi $E^2_h=\dfrac{1}{2\pi^2}\times[(1+\cos^2(2 \pi \tau))\sin^2(2 \pi \tau)]=\dfrac{1}{2\pi^2}~g(2\pi\tau)$.
	\end{enumerate}
\item Par conséquent, d'après \textbf{1.}, $E^2_h$ est maximal quand $2\pi\tau =\dfrac{\pi}2 \iff \tau =\dfrac14$
\end{enumerate}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%


\exo{11} \noindent {\textbf{Partie A : }} 
\begin{enumerate}
\item
	\begin{enumerate}
		\item On pose $(H)$ : $\dfrac{1}{200}y^{\prime}(t) + y(t) =0 \iff y^{\prime}(t) = -200 y(t)$\\
Les solutions de $(H)$ sont de la forme $y(t) = \lambda \text{e}^{-200t}$.\\
Soit $g$ une solution constante de $(1)$. On a $g^{\prime}(t)=0$ et $\dfrac{1}{200}\times 0 + g(t) = 146 \iff g(t) = 146$\\
Ainsi les solutions de  $(1)$ sont de la forme $y(t) = 146+\lambda \text{e}^{-200t}$.
		\item
 $\omega(0) = 150 \iff 146+\lambda \text{e}^{-200\times 0}=150 \iff 146+\lambda=150 \iff \lambda=4$\\
Ainsi $\omega(t) = 146 + 4\text{e}^{-200t}$ pour tout $t \in [ 0~;~ + \infty[$.
	\end{enumerate}
\item
	\begin{enumerate}
		\item $\displaystyle\lim_{t\to+\infty} \text{e}^{-200t} = 0$ donc $\omega_{\infty}=\displaystyle\lim_{t\to+\infty} \omega(t)=146$.\\
 Ainsi,   $\omega(0)-\omega_{\infty}=150-146=4 $ rad/s.

		\item $\begin{array}{|c|}
 \dfrac{\omega(t)-\omega_{\infty}}{\omega_{\infty}} \\
\end{array} = 0,01 \iff \begin{array}{|c|}
 \dfrac{146 + 4\text{e}^{-200t}-146}{146} \\
\end{array} = 0,01 \iff 4\text{e}^{-200t} = 1,46 \\\iff \text{e}^{-200t} = 0,365 \iff -200 t =\ln(0,365) \iff t=-\dfrac{\ln(0,365)}{200}\approx 0,005$
	\end{enumerate}
\end{enumerate}

\medskip

\noindent{ \large {\textbf{Partie B :}}} \begin{enumerate}
\item
	\begin{enumerate}
		\item Voir \textbf{3.d.} SVP.
		\item $\Gamma(p)=\Lp(K [U(t) - U(t - \tau)])=\dfrac{K}p-\dfrac{K\e^{-\tau p}}p$


	\end{enumerate}
\item $Lp\left(\dfrac{1}{200} f^{\prime}(t) + f(t)\right) =\Lp(\gamma(t)) \iff \dfrac1{200}(pF(p)-f(0^+))+F(p)=\Gamma(p) \\\iff \left(\dfrac{p}{200}+1\right)F(p)=\Gamma(p)
\iff \dfrac{p+200}{200}F(p)=\dfrac{K}p-\dfrac{K\e^{-\tau p}}p \iff F(p)=\dfrac{200K}{p(p+200)}-\dfrac{200K\e^{-\tau p}}{p(p+200)}$


\item
	\begin{enumerate}
		\item $\dfrac{200}{p(p + 200)}=\dfrac{a}{p}+\dfrac{b}{p + 200} \iff 200=a(p+200)+bp \iff 200 =(a+b)p+200 a \\\iff \Syst{a+b}{0}{200a}{200} \iff \Syst{a}{1}{b}{-1}$
		
Ainsi, $\dfrac{200}{p(p + 200)}=\dfrac{1}{p}-\dfrac{1}{p + 200}$
		\item $f(t)=\Lp^{-1}\left(\dfrac{200K}{p(p+200)}-\dfrac{200K\e^{-\tau p}}{p(p+200)}\right)
= K\Lp^{-1}\left(\dfrac{1}{p}-\dfrac{1}{p + 200}-\dfrac{\e^{-\tau p}}{p}+\dfrac{\e^{-\tau p}}{p + 200}\right)\\
=K\left(\U(t)-\e^{-200t}\U(t)-\U(t-\tau)+\e^{-200(t-\tau)}\U(t-\tau)\right)$.

Ainsi, sur $[0~;~\tau[$, $f(t)= K\left(1-\e^{-200t}\times1-0+\e^{-200(t-\tau)}\times0\right)=K (1 - \text{e}^{ -200t} )$

et sur $[\tau~;~+\infty[$, $f(t)= K\left(1-\e^{-200t}-1+\e^{-200(t-\tau)}\right)=  K\left(-\e^{-200t}+\e^{-200t+200\tau}\right)\\= K (\text{e}^{200\tau}-1) \text{e}^{ - 200t}$

		\item Sur $[0~;~\tau[$, $f'(t)=200K\e^{-200t} >0$ donc $f$ est strictement croissante.
		
Sur $[\tau~;~+\infty[$, $f(t)= -200 K (\text{e}^{200\tau}-1) \text{e}^{ - 200t} <0$ car $\tau \geqslant 0$ et $\text{e}^{200\tau}-1 \geq 0$. Ainsi $f$ est décroissante.

$f(0^+)=K (1 - \text{e}^{ -200\times 0} )=K(1-1)=0$ et $f(\tau^-)=K (1 - \text{e}^{ -200\tau} )$

$f(\tau^+)= K (\text{e}^{200\tau}-1) \text{e}^{ - 200\tau}=K (1 - \text{e}^{ -200\tau} )$  et comme $\displaystyle\lim_{t\to+\infty} \text{e}^{-200t} = 0$, $\displaystyle\lim_{t\to+\infty} f(t) = 0$
		\item
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	\end{enumerate}
\end{enumerate}
\end{document}

