\documentclass[a4paper,french,12pt]{article} \title{Logique Programmable --- Prérequis --- Exercices} \author{} \date{Dernière compilation~: \today{} à \currenttime} \usepackage{../../cours} \usepackage{enumitem} \begin{document} \maketitle \section{Simplification des fonctions logiques} \subsection{Exercice 1} \begin{multicols}{4} \begin{enumerate}[label=\alph*)] \item $a \cdot b \cdot c \cdot d = 1$ \item $a + b + c + d = 0$ \item $a + b + c + d = 1$ \item $a \cdot b \cdot c \cdot d = 0$ \end{enumerate} \end{multicols} \subsection{Exercice 2} \begin{enumerate}[label=\alph*)] \item \begin{align*} F(a, b, c) &= a\bar{b} + abc + a\bar{c} \\ F(0,1,1) &= 0 \\ F(1,1,0) &= 1 \\ F(1,0,0) &= 1 \\ \end{align*} \item \begin{equation*} F(a, b) = \bar{a}\bar{b} + b = \bar{a} + b \end{equation*} \begin{center} \begin{tabular}{cc|c|c|c|c} \toprule $a$ & $b$ & $\bar{a}\bar{b}$ & $\bar{a}\bar{b} + b$ & $\bar{a}$ & $\bar{a} + b$ \\ \midrule 0 & 0 & 1 & 1 & 1 & 1 \\ 0 & 1 & 0 & 1 & 0 & 1 \\ 1 & 0 & 0 & 0 & 0 & 0 \\ 1 & 1 & 0 & 1 & 0 & 1 \\ \midrule & & & $\uparrow$ & = & $\uparrow$ \\ \bottomrule \end{tabular} \end{center} \item \begin{align*} F(a, b, c) &= a\bar{b} + ab + ac \\ &= a (\bar{b} + b + c) \\ &= a (1 + c) \\ &= a \cdot 1 = a \end{align*} \item \begin{align*} F(a, b, c) &= \bar{a}\bar{b}c + a\bar{b}\bar{c} + a\bar{b}c + ab\bar{c} + abc \\ &= \bar{a}\bar{b}c + a\bar{b}\bar{c} + ab\bar{c} + ac(b + \bar{b}) \\ &= \bar{a}\bar{b}c + a\bar{b}\bar{c} + ab\bar{c} + ac \\ &= \bar{a}\bar{b}c + a\bar{c} + ac \\ &= \bar{a}\bar{b}c + a \\ \end{align*} \end{enumerate} \subsection{Exercice 3} \begin{enumerate}[label=\alph*)] \item \begin{align*} F_1 &= (a + b) \cdot (\bar{a} + \bar{b}) \\ \overline{F_1} &= \overline{(a + b) \cdot (\bar{a} + \bar{b})} \\ &= \overline{(a + b)} + \overline{(\bar{a} + \bar{b})} \\ &= \bar{a}\bar{b} + ab \\ &= \overline{a \oplus b} \\ &= \bar{a} \oplus b = a \oplus \bar{b} \\ \end{align*} \item \begin{align*} F_2 &= a (c + d) + (\bar{a} + c) \cdot (\bar{b} + c + d) \\ \overline{F_2} &= \overline{a (c + d) + (\bar{a} + c) \cdot (\bar{b} + c + d)} \\ &= \overline{a (c + d)} \cdot \overline{(\bar{a} + c)} + \overline{(\bar{b} + c + d)} \\ &= \overline{ac + ad} \cdot a\bar{c} + b\bar{c}\bar{d} \\ &= \overline{ac} \cdot \overline{ad} \cdot a\bar{c} + b\bar{c}\bar{d} \\ &= (\bar{a} + \bar{c}) \cdot (\bar{a} + \bar{d}) \cdot a\bar{c} + b\bar{c}\bar{d} \\ &= (\bar{a} + \bar{a}\bar{d} + \bar{a}\bar{c} + \bar{c}\bar{d}) \cdot a\bar{c} + b\bar{c}\bar{d} \\ &= a\bar{c}\bar{d} + b\bar{c}\bar{d} \\ &= (a + b) \cdot \bar{c}\bar{d} \\ \end{align*} \end{enumerate} \subsection{Exercice 4} \subsection{Exercice 5} \subsection{Exercice 6} \subsection{Exercice 7} \section{Circuits de logique combinatoire} \subsection{Exercice 1} \includegraphics[width=0.6\linewidth]{./img/2.1.png} \subsection{Exercice 2} a/ Élaborer l'équation logique $F(S_1,S_0,C,a,b)$ du circuit ci-dessous. \includegraphics[width=\linewidth]{./img/2.2.png} \begin{equation*} F = \bar{c} + \overline{S_0}(ab + \bar{a}\bar{b}S_1) \end{equation*} b/ Compléter le tableau suivant pour C = 0~: $F = \bar{c}$ \begin{tabularx}{0.8\linewidth}{cccY} \toprule C & $S_1$ & $S_0$ & $F(a,b)$ \\ \midrule 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 1 \\ 0 & 1 & 0 & 1 \\ 0 & 1 & 1 & 1 \\ \bottomrule \end{tabularx} c/ Compléter le tableau suivant pour C = 1~: $F = \overline{S_0}(ab + \bar{a}\bar{b}S_1)$ \begin{tabularx}{0.8\linewidth}{cccY} \toprule C & $S_1$ & $S_0$ & $F(a,b)$ \\ \midrule 1 & 0 & 0 & $ab$ \\ 1 & 0 & 1 & 0 \\ 1 & 1 & 0 & $ab + \bar{a}\bar{b}$ \\ 1 & 1 & 1 & 0 \\ \bottomrule \end{tabularx} \subsection{Exercice 3} Soit le schéma de $H(A,B,C,D)$ utilisant un multiplexeur à 3 entrées d'adresse. \includegraphics[width=0.6\linewidth]{./img/2.3.png} a/ Donner la table de vérité de la fonction $H$. \begin{tabularx}{0.7\linewidth}{X|YYY|Y|Y} \toprule & A & B & C & D & H \\ \midrule 0 & 0 & 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 & 1 & 0 \\ 2 & 0 & 0 & 1 & 0 & 0 \\ 3 & 0 & 0 & 1 & 1 & 1 \\ 4 & 0 & 1 & 0 & 0 & 0 \\ 5 & 0 & 1 & 0 & 1 & 1 \\ 6 & 0 & 1 & 1 & 0 & 1 \\ 7 & 0 & 1 & 1 & 1 & 0 \\ \midrule 8 (0) & 1 & 0 & 0 & 0 & 0 \\ 9 (1) & 1 & 0 & 0 & 1 & 0 \\ 10 (2) & 1 & 0 & 1 & 0 & 0 \\ 11 (3) & 1 & 0 & 1 & 1 & 1 \\ 12 (4) & 1 & 1 & 0 & 0 & 0 \\ 13 (5) & 1 & 1 & 0 & 1 & 1 \\ 14 (6) & 1 & 1 & 1 & 0 & 1 \\ 15 (7) & 1 & 1 & 1 & 1 & 0 \\ \bottomrule \end{tabularx} b/ Exprimer $H(A,B,C,D)$ sous la forme disjonctive. \begin{align*} H(A,B,C,D) &= \bar{A}\bar{B}CD + \bar{A}B\bar{C}D + \bar{A}BC\bar{D} + A\bar{B}CD + AB\bar{C}D + ABC\bar{D} \\ &= \sum(3, 5, 6, 11, 13, 14) \end{align*} \subsection{Exercice 4} \end{document}