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%                                                                             %
% Corrigé du contrôle de Transmissions Numériques                             %
% 2TR / PROMO 2005 / 3 décembre 2003                                          %
%                                                                             %
% Auteur : Didier Barvaux <didier.barvaux@etu.enseeiht.fr>                    %
%                         <didier@barvaux.org>                                %
%                                                                             %
% Date: 14 novembre 2004                                                      %
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% Document publié sous license GNU FDL (http://www.gnu.org/copyleft/fdl.html) %
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\documentclass[11pt,a4paper,french,oneside]{article}
\usepackage[french]{babel}
\usepackage[latin1]{inputenc}
\usepackage[T1]{fontenc}
\usepackage{pstricks}
\usepackage{float}




\title{Corrigé du contrôle de Transmissions Numériques}
%\author{Didier Barvaux <didier@barvaux.org>}
\date{\today}

\begin{document}

\maketitle
Ce document est le corrigé du contrôle de Transmissions Numériques donné aux 2TR le 3 décembre 2003. L'épreuve a duré 2 heures (le double de temps aurait été nécessaire, mais bon...). Les documents étaient autorisés. Merci à M. Roviras pour son aide sur certaines questions.
\clearpage

\small
\tableofcontents
\normalsize

\clearpage



\section{Exercice 1.}




%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\subsection{Comment appelle-t-on un signal du type de $x_2(t)$ ?}

\[ x_1(t) = \sum_k{ a_k \hspace{3pt} \delta(t-kT_s) } \]

\begin{eqnarray*}
x_2(t) & = & x_1(t) * g_1(t)                                                    \\
       & = & \left[ \sum_k{ a_k \hspace{3pt} \delta(t-kT_s) } \right] * g_1(t)  \\
       & = & \sum_k{ a_k \hspace{3pt}  g_1(t-kT_s) }
\end{eqnarray*}

Comme $g_1(t)$ est un filtre de mise en forme rectangulaire égal à $1$ entre $0$ et $T_s$ et que $a_k = \pm V$ : \[ x_2(t) = NRZ\left(\pm V, T_s\right) \]

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






%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\subsection{Calculer la puissance du signal $x_2(t)$ ?}

\begin{eqnarray*}
P_{x_2} & = & \lim_{T \to +\infty} \frac{1}{T} \int_{[T]} |x_2(t)|^2\,dt          \\
        & = & \lim_{T \to +\infty} \frac{1}{T} \int_{[T]} V^2\,dt                 \\
        & = & \lim_{T \to +\infty} \frac{1}{T} \left( V^2 T \right)  \\
        & = & V^2
\end{eqnarray*}

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






%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\subsection{Calculer la puissance du signal $x_3(t)$ ?}

\begin{eqnarray*}
\frac{1}{T} \int_{[T]} |x_3(t)|^2\,dt & = & \frac{1}{T} \int_{[T]} |x_2(t) \cos(2\pi f_pt)|^2\,dt    \\
                                      & = & \frac{1}{T} \int_{[T]} |x_2(t)|^2 \cos^2(2\pi f_pt)\,dt  \\
                                      & = & \frac{1}{T} \int_{[T]} V^2 \cos^2(2\pi f_pt)\,dt         \\
                                      & = & \frac{V^2}{T} \int_{[T]} \cos^2(2\pi f_pt)\,dt
\end{eqnarray*}

\begin{eqnarray*}
P_{x_3} & = & \lim_{T \to \frac{1}{f_p}} \left[ \frac{1}{T} \int_{[T]} |x_3(t)|^2\,dt \right]           \\
        & = & \lim_{T \to \frac{1}{f_p}} \left[ \frac{V^2}{T} \int_{[T]} \cos^2(2\pi f_pt)\,dt \right]  \\
        & = & V^2f_p \hspace{3pt} \int_{[\frac{1}{f_p}]} \cos^2(2\pi f_pt)\,dt                          \\
        & = & V^2f_p \hspace{3pt} \frac{1}{2f_p}                                                        \\
        & = & \frac{V^2}{2}
\end{eqnarray*}

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\subsection{Montrer que du point de vue relation entrée/sortie la chaîne de transmission présentée dans l'énoncé est équivalente à la chaîne passe-bas équivalente présenté dans la question 4.}

\[ x_2(t) = x_1(t) * g_1(t) \]
\[ x_3(t) = x_2(t) \cos(2\pi f_pt) \]

Le canal est idéal donc : $ x_4(t) = x_3(t) * c(t) = x_3(t) = x_2(t) \cos(2\pi f_pt) $

\begin{eqnarray*}
x_5(t) & = & x_4(t) \cos(2\pi f_pt)                                   \\
       & = & x_2(t) \cos^2(2\pi f_pt)                                 \\
       & = & x_2(t) \frac{ \cos(4\pi f_pt) + 1 }{2}                   \\
       & = & \frac{1}{2} x_2(t) + \frac{1}{2} x_2(t) \cos(4\pi f_pt)
\end{eqnarray*}

$g_2(t)$ est un filtre passe-bas dont la fréquence de coupure est petite devant $f_p$, donc :
\[ x_6(t) = x_5(t) * g_2(t) = \frac{1}{2} x_2(t) * g_2(t) =  \frac{1}{2} x_1(t) * g_1(t) * g_2(t) \]

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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\subsection{Montrer que la DSP du bruit $n_1(t)$ est égale à $\frac{N_0}{4}$.}

\[ n_1(t) = n(t) \cos(2\pi f_pt) \]
donc : \[ N_1(f) = N(f) * TF\left[ \cos(2\pi f_pt) \right] \]
d'où : \begin{eqnarray*}
S_{n_1}(f) & = & S_n(f) * DSP\left[ \cos(2\pi f_pt) \right]                                        \\
           & = & S_n(f) * \left[ \frac{1}{4} \left( \delta(f-f_p) + \delta(f+f_p) \right) \right]  \\
           & = & \frac{1}{4} \left( S_n(f-f_p) + S_n(f+f_p) \right)                                \\
           & = & \frac{1}{4} \left( \frac{N_0}{2} + \frac{N_0}{2} \right)                          \\
           & = & \frac{N_0}{4}
\end{eqnarray*}

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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\subsection{En déduire la chaîne passe-bas équivalente avec le bruit additif $n_1(t)$.}

\begin{figure}[H]
\input{./diagrammes/question06.tex}
\end{figure}

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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\subsection{Calculer la réponse impulsionnelle de $g_2(t)$ de façon à avoir un filtre adapté en réception.}

\vspace{10pt}

Posons : $ p(t) = g_1(t) * \frac{1}{2} $ et $ h_e(t) = g_1(t) * \frac{1}{2} * g_2(t) $

\vspace{10pt}

On a un filtre adapté en réception si : $g_2(t) = \frac{k}{S_{n_1}(f)} p^*(t_0-t)$

\vspace{10pt}

D'où :
\begin{eqnarray*}
g_2(t) & = & \frac{k}{S_{n_1}(f)} p^*(t_0-t)                         \\
       & = & \frac{k}{S_{n_1}(f)} p(t_0-t)                           \\
       & = & \frac{4k}{N_0} \left( g_1(t_0-t) * \frac{1}{2} \right)  \\
       & = & \frac{2k}{N_0} g_1(t_0-t)
\end{eqnarray*}


\begin{figure}[H]
\centering
\input{./diagrammes/question07.tex}
\end{figure}

\vspace{10pt}

par suite : $ g_2(t) = \frac{2k}{N_0} g_1(t) $

\vspace{10pt}

$g_2(t)$ est donc égal à $g_1(t)$ à une constante près.

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




%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\subsection{La chaîne passe-bas équivalente vérifie-t-elle le 1er critère de Nyquist ?}

\begin{eqnarray*}
h_e(t) & = & g_1(t) * \frac{1}{2} * g_2(t)                                                           \\
       & = & \frac{1}{2} \left[ \prod_{T_s}(t-\frac{T_s}{2}) * \prod_{T_s}(t-\frac{T_s}{2}) \right]  \\
       & = & \frac{1}{2} T_s \bigwedge_{T_s}(t-\frac{T_s}{2})
\end{eqnarray*}


\begin{figure}[H]
\centering
\input{./diagrammes/question08.tex}
\end{figure}

\vspace{10pt}

$\exists t_0 = T_s$ tel que : $h_e(t_0) = 1$ et $h_e(t_0+kT_s) = 0 \hspace{3pt} (k \not= 0)$, la chaîne passe-bas équivalente vérifie donc Nyquist.

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





\subsection{Calcul du TEB sur canal idéal.}



%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\subsubsection{Donner les valeurs de $x_6(kT_s)$ aux instants de prise de décision.}

Prise de décision en $ t_0 = T_s $ : \[ x_6(kT_s) = \pm V \]

\[ P(\pm V) = \frac{1}{2} \]

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



%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\subsubsection{Calculer la variance du bruit aux instants de prise de décision.}

\begin{figure}[H]
\centering
\input{./diagrammes/question09-2.tex}
\end{figure}

\[ S_{n_2}(f) = S_{n_1}(f) |G_2(f)|^2 \]

\[  \Longrightarrow P_{n_2} = \frac{N_0}{4} \int |G_2(f)|^2\,df \]

d'après Perceval :
\begin{eqnarray*}
P_{n_2} & = & \frac{N_0}{4} \int |g_2(f)|^2\,df  \\
        & = & \frac{N_0}{4} T_s
\end{eqnarray*}

On sait que : $ P_{n_2} = \mu_{n_2}^2 + \sigma_{n_2}^2 $

\vspace{10pt}

Or $n_2(t)$ est un bruit à moyenne nulle, donc : $\mu_{n_2}^2 = 0$

\vspace{10pt}

d'où : \[ \sigma_{n_2}^2 = P_{n_2} = \frac{N_0}{4} T_s \]

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




%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\subsubsection{Dessiner le diagramme de l'\oe il reçu en l'absence de bruit.}

\begin{figure}[H]
\centering
\input{./diagrammes/question09-3.tex}
\end{figure}

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




%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\subsubsection{Donner la régle de décision MAP qui permet de distinguer les symboles reçus.}

$\pm V$ équiprobables $\Longrightarrow seuil = \frac{(+V)+(-V)}{2} = 0$

\vspace{10pt}

Règle MAP :
\[ si \hspace{3pt} x_6(t_0+kT_s) > 0 \Longrightarrow +V \]
\[ si \hspace{3pt} x_6(t_0+kT_s) < 0 \Longrightarrow -V \]
\[ si \hspace{3pt} x_6(t_0+kT_s) = 0 \Longrightarrow au \hspace{3pt} choix \]

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




%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\subsubsection{Calculer le TEB de la liaison.}

\[ TEB = Q\left[ \frac{(+V)-(-V)}{2\sigma_{n_2}} \right] = Q\left[ \frac{V}{\sigma_{n_2}} \right] \]

Or : $ \sigma_{n_2}^2 = \frac{N_0}{4} T_s $

\vspace{10pt}

donc : \[ TEB = Q\left[ \frac{2V}{\sqrt{N_0T_s}} \right] \]

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




\subsection{TEB sur canal idéal en fonction de $\frac{E_b}{N_0}$.}


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\subsubsection{Calculer l'énergie $E_b$ consommée pour émettre un bit dans le canal.}

\begin{figure}[H]
\centering
\input{./diagrammes/question10-1.tex}
\end{figure}

Énergie $E_s$ consommée pour émettre un symbole dans le canal : \[ E_s = P_{x_3} T_s = \frac{V^2}{2} T_s \]

D'où l'énergie consommée pour émettre un bit dans le canal : \[ E_b = \frac{E_s}{nombre \hspace{3pt} de \hspace{3pt} bits \hspace{3pt} par \hspace{3pt} symbole} \]

Or, $1$ symbole dure $T_s$, donc :

\[ E_b = E_s = \frac{V^2}{2} T_s \]

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





%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\subsubsection{Exprimer le TEB de la liaison en fonction de $\frac{E_b}{N_0}$.}

$ TEB = Q\left[ \frac{2V}{\sqrt{N_0T_s}} \right] $ et $ \frac{E_b}{N_0} = \frac{V^2T_s}{2N_0} $, donc :

\[ TEB = Q\left[ \frac{E_b}{N_0} \left( \frac{4 \sqrt{N_0}}{VT_s \sqrt{T_s}} \right) \right] \]

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




\subsection{Canal non idéal.}


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\subsubsection{A-t-on de l'ISI ?}

La réponse du canal de transmission vaut $ c(t) = \delta(t) - 0,4 \delta(t-T_s) $, donc on a de l'ISI.

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





%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\subsubsection{Donner les valeurs de $x_6(kT_s)$ aux instants de prise de décision ainsi que les probabilités respectives de ces différentes valeurs.}

Prise de décision en $t_0 = T_s$. Les valeurs sont : $\pm 1,4 V$ et $\pm 0.6 V$. Les quatre valeurs sont équiprobables.

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




\subsection{Égaliseur ZFE en présence d'un canal non idéal.}


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\subsubsection{Donner l'architecture de l'égaliseur ZFE avec 4 coefficients.}

cf. polycopié.

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





%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\subsubsection{Calculer les coefficients de l'égaliseur ZFE avec le canal de la question 1.11.}

\[ R . C = Y \]

$ R = \left( \begin{array}{cccc}
r(0) & 0    & 0    & 0    \\
r(1) & r(0) & 0    & 0    \\
r(2) & r(1) & r(0) & 0    \\
r(3) & r(2) & r(1) & r(0) \end{array} \right) $ avec $ r(k) = x_6(t_0+kT_s) $

\[ C = \left( \begin{array}{ccc}
c_0  \\
c_1  \\
c_2  \\
c_3  \end{array} \right) \]

On veut obtenir : $ Y = \left( \begin{array}{c}
1  \\
0  \\
0  \\
0  \end{array} \right) $

Un dirac en entrée donne : \[ r(0) = 1 \] \[ r(1) = -0,4 \] \[ r(k) = 0 \hspace{3pt} (k \ge 2) \]

\vspace{10pt}

Quels coefficients $c_0$, $c_1$, $c_2$ et $c_3$ ?

\[ \begin{array}{c}
\{ \hspace{3pt} y(0) = 1 = c_0 r(0)                                   \hspace{126pt}  \\
\{ \hspace{3pt} y(1) = 0 = c_0 r(1) + c_1 r(0)                        \hspace{84pt}   \\
\{ \hspace{3pt} y(2) = 0 = c_0 r(2) + c_1 r(1) + c_2 r(2)             \hspace{42pt}   \\
\{ \hspace{3pt} y(3) = 0 = c_0 r(3) + c_1 r(2) + c_2 r(1) + c_3 r(0)                  \end{array} \]

\[ \Longleftrightarrow \begin{array}{c}
\{ \hspace{3pt} c_0 = 1     \hspace{20pt}    \\
\{ \hspace{3pt} c_1 = 0,4   \hspace{12pt}    \\
\{ \hspace{3pt} c_2 = 0,16  \hspace{5pt}     \\
\{ \hspace{3pt} c_3 = 0,064                  \end{array} \]

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





%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\subsubsection{Donner les valeurs du signal , après l'égaliseur ZFE, aux instants de prise de décision.}

\begin{figure}[H]
\centering
\input{./diagrammes/question12-3.tex}
\end{figure}

\vspace{10pt}

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





\subsection{Calcul du TEB dans le cas du canal non idéal (sans ZFE).}


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\subsubsection{Donner la régle MAP de décision avec le canal de la question 1.11 ainsi que le seuil de décision utilisé.}

\begin{eqnarray*}
0 & \to & -1,4 V  \\
  & \to & -0,6 V
\end{eqnarray*}

\begin{eqnarray*}
1 & \to & +1,4 V  \\
  & \to & +0,6 V
\end{eqnarray*}

\[ seuil = 0 \]

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




%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\subsubsection{Calculer le TEB de la liaison en fonction de $\frac{E_b}{N_0}$.}

\begin{eqnarray*}
P(erreur) & = &   P(emettre \hspace{3pt} 0) P(erreur / 0)
                + P(emettre \hspace{3pt} 1) P(erreur / 1)  \\
          & = & \frac{1}{2} P(erreur / 0) + \frac{1}{2} P(erreur / 1)                                         \\
          & = &   \frac{1}{2} \left[ P(erreur / -1,4V) P(-1,4V / 0) + P(erreur / -0,6V) P(-0,6V / 0) \right]  \\
          &   & + \frac{1}{2} \left[ P(erreur / +1,4V) P(+1,4V / 0) + P(erreur / +0,6V) P(+0,6V / 0) \right]  \\
          & = &   \frac{1}{2} \left[ \frac{1}{2} P(erreur / -1,4V) + \frac{1}{2} P(erreur / -0,6V) \right]    \\
          &   & + \frac{1}{2} \left[ \frac{1}{2} P(erreur / +1,4V) + \frac{1}{2} P(erreur / +0,6V) \right]    \\
          & = & \frac{1}{2} \left[ Q\left( \frac{(+1,4V)-(-1,4V)}{2\sigma_{n_2}} \right)
                                 + Q\left( \frac{(+0,6V)-(-0,6V)}{2\sigma_{n_2}} \right) \right]              \\
          & = &   \frac{1}{2} Q\left[ \frac{1,4V}{\sigma_{n_2}} \right]
                + \frac{1}{2} Q\left[ \frac{0,6V}{\sigma_{n_2}} \right]
\end{eqnarray*}

Or : $ \sigma_{n_2} = \frac{\sqrt{N_0T_s}}{\sqrt{2}} $ et $ \frac{E_b}{N_0} = \frac{V^2T_s}{2N_0} $

\vspace{10pt}

\begin{eqnarray*}
TEB & = &   \frac{1}{2} Q\left[ \frac{1,4V\sqrt{2}}{\sqrt{N_0T_s}} \right]
          + \frac{1}{2} Q\left[ \frac{0,6V\sqrt{2}}{\sqrt{N_0T_s}} \right]                                     \\
    & = &   \frac{1}{2} Q\left[ \frac{1,4 E_b}{N_0} \left( \frac{2\sqrt{2N_0}}{VT_s\sqrt{Ts}} \right) \right]
          + \frac{1}{2} Q\left[ \frac{0,6 E_b}{N_0} \left( \frac{2\sqrt{2N_0}}{VT_s\sqrt{Ts}} \right) \right]
\end{eqnarray*}

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






%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\subsection{Calcul du TEB dans le cas d'un égaliseur ZFE avec un canal non idéal.}

La variance du bruit après l'égaliseur est égale à $ \sigma^2 $.

\[ TEB = \frac{1}{2} Q\left[ \frac{1,0256V}{\sigma} \right] + \frac{1}{2} Q\left[ \frac{0,9744V}{\sigma} \right] \]

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\subsection{Calcul du TEB dans le cas d'une erreur sur l'instant de décision.}


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\subsubsection{Donner les valeurs de $ x_6(kT_s + \frac{T_s}{4} $ aux instants de prise de décision ainsi que les probabilités respectives de ces différentes valeurs.}

Instant de décision en $ t_0 = kT_s + \frac{T_s}{4} $.

\begin{eqnarray*}
x_6(\frac{T_s}{4} + kT_s) & = & \pm 1,2V  \\
                          & = & \pm 0,7V  \\
                          & = & \pm 0,6V  \\
                          & = & \pm 0,1V
\end{eqnarray*}

Ces valeurs sont équiprobables.

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




%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\subsubsection{En déduire le TEB de la liaison.}

\begin{eqnarray*}
TEB & = & P(emettre \hspace{3pt} 0) P(erreur / 0) + P(emettre \hspace{3pt} 1) P(erreur / 1)  \\
    & = & \frac{1}{2} P(erreur / 0) + \frac{1}{2} P(erreur / 1)			 
\end{eqnarray*}

\begin{eqnarray*}
0 & \to & -1,2 V  \\
  & \to & -0,7 V  \\
  & \to & -0,6 V  \\
  & \to & -0,1 V
\end{eqnarray*}

\begin{eqnarray*}
1 & \to & +1,2 V  \\
  & \to & +0,7 V  \\
  & \to & +0,6 V  \\
  & \to & +0,1 V
\end{eqnarray*}

\begin{eqnarray*}
TEB & = &   \frac{1}{2} [ P(erreur/-1,2V) P(-1,2V/0) + P(erreur/-0,7V) P(-0,7V/0)      \\
    &   &               + P(erreur/-0,6V) P(-0,6V/0) + P(erreur/-0,1V) P(-0,1V/0) ]    \\
    &   & + \frac{1}{2} [ P(erreur/+1,2V) P(+1,2V/0) + P(erreur/+0,7V) P(+0,7V/0)      \\
    &   &               + P(erreur/+0,6V) P(+0,6V/0) + P(erreur/+0,1V) P(+0,1V/0) ]    \\
    & = &   \frac{1}{2} [ \frac{1}{4} P(erreur/-1,2V) + \frac{1}{4} P(erreur/-0,7V)    \\
	 &   &               + \frac{1}{4} P(erreur/-0,6V) + \frac{1}{4} P(erreur/-0,1V) ]  \\
    &   & + \frac{1}{2} [ \frac{1}{4} P(erreur/+1,2V) + \frac{1}{4} P(erreur/+0,7V)    \\
    &   &               + \frac{1}{4} P(erreur/+0,6V) + \frac{1}{4} P(erreur/+0,1V) ]  \\
	 & = &   \frac{1}{2} Q\left[ \frac{1,2V}{2\sigma_{n_2}} \right]
	       + \frac{1}{2} Q\left[ \frac{0,7V}{2\sigma_{n_2}} \right]
          + \frac{1}{2} Q\left[ \frac{0,6V}{2\sigma_{n_2}} \right]
          + \frac{1}{2} Q\left[ \frac{0,1V}{2\sigma_{n_2}} \right]
\end{eqnarray*}

Fin du contrôle.
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%





\end{document}

