%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% % % % Corrigé du contrôle de Transmissions Numériques % % 2TR / PROMO 2005 / 3 décembre 2003 % % % % Auteur : Didier Barvaux % % % % % % Date: 14 novembre 2004 % % % % Document publié sous license GNU FDL (http://www.gnu.org/copyleft/fdl.html) % % % % --------------------------------------------------------------------------- % % % % Ceci est un source latex. Pour obtenir un fichier PDF à partir de ce % % source, il suffit de le compiler avec les commandes : % % % % $ latex transmissions_numeriques_2003.tex % % $ latex transmissions_numeriques_2003.tex (oui, une 2nde fois) % % $ dvipdf transmissions_numeriques_2003.dvi % % % % Un peu d'aide sur latex : http://www.maths.tcd.ie/~dwilkins/LaTeXPrimer/ % % % % Le répertoire 'diagrammes' contient les dessins insérés dans le document % % latex. Les fichiers *.dia peuvent être édités avec DIA, un programme pour % % le dessin de diagrammes structurés (http://www.lysator.liu.se/~alla/dia). % % Les fichiers *.tex ont été générés par DIA à partir des fichiers *.dia % % (export au format 'Macro TeX PSTricks'). % % % %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \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 } \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*} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \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 : \[ x_6(t) = x_5(t) * g_2(t) = \frac{1}{2} x_2(t) * g_2(t) = \frac{1}{2} x_1(t) * g_1(t) * g_2(t) \] %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \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*} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \subsection{En déduire la chaîne passe-bas équivalente avec le bruit additif $n_1(t)$.} \begin{figure}[H] \input{./diagrammes/question06.tex} \end{figure} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \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 $. \[ TEB = \frac{1}{2} Q\left[ \frac{1,0256V}{\sigma} \right] + \frac{1}{2} Q\left[ \frac{0,9744V}{\sigma} \right] \] %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \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}