108 lines
3.1 KiB
TeX
108 lines
3.1 KiB
TeX
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\documentclass[a4paper,french,12pt]{article}
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\title{Logique Programmable --- Exercices}
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\author{Catherine MARECHAL --- \href{mailto:catherine.marechal@efrei.fr}{\nolinkurl{catherine.marechal@efrei.fr}}}
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\date{Dernière compilation~: \today{} à \currenttime}
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\usepackage{../../cours}
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\begin{document}
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\maketitle
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\section{Simplification des fonctions logiques}
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\subsection{Exercice 1}
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\subsection{Exercice 2}
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\subsection{Exercice 3}
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\subsection{Exercice 4}
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\subsection{Exercice 5}
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\subsection{Exercice 6}
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\subsection{Exercice 7}
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\section{Circuits de logique combinatoire}
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\subsection{Exercice 1}
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\includegraphics[width=0.6\linewidth]{./img/2.1.png}
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\subsection{Exercice 2}
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a/ Élaborer l'équation logique $F(S_1,S_0,C,a,b)$ du circuit ci-dessous.
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\includegraphics[width=\linewidth]{./img/2.2.png}
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\begin{equation*}
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F = \bar{c} + \overline{S_0}(ab + \bar{a}\bar{b}S_1)
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\end{equation*}
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b/ Compléter le tableau suivant pour C = 0~:
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$F = \bar{c}$
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\begin{tabularx}{0.8\linewidth}{cccY}
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\toprule
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C & $S_1$ & $S_0$ & $F(a,b)$ \\
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\midrule
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0 & 0 & 0 & 1 \\
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0 & 0 & 1 & 1 \\
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0 & 1 & 0 & 1 \\
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0 & 1 & 1 & 1 \\
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\bottomrule
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\end{tabularx}
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c/ Compléter le tableau suivant pour C = 1~:
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$F = \overline{S_0}(ab + \bar{a}\bar{b}S_1)$
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\begin{tabularx}{0.8\linewidth}{cccY}
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\toprule
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C & $S_1$ & $S_0$ & $F(a,b)$ \\
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\midrule
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1 & 0 & 0 & $ab$ \\
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1 & 0 & 1 & 0 \\
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1 & 1 & 0 & $ab + \bar{a}\bar{b}$ \\
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1 & 1 & 1 & 0 \\
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\bottomrule
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\end{tabularx}
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\subsection{Exercice 3}
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Soit le schéma de $H(A,B,C,D)$ utilisant un multiplexeur à 3 entrées d'adresse.
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\includegraphics[width=0.6\linewidth]{./img/2.3.png}
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a/ Donner la table de vérité de la fonction $H$.
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\begin{tabularx}{0.7\linewidth}{X|Y|YYY|Y}
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\toprule
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& D & C & B & A & H \\
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\midrule
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0 & 0 & 0 & 0 & 0 & 0 \\
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1 & 0 & 0 & 0 & 1 & 0 \\
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2 & 0 & 0 & 1 & 0 & 0 \\
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3 & 0 & 0 & 1 & 1 & 0 \\
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4 & 0 & 1 & 0 & 0 & 0 \\
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5 & 0 & 1 & 0 & 1 & 0 \\
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6 & 0 & 1 & 1 & 0 & 1 \\
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7 & 0 & 1 & 1 & 1 & 0 \\
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\midrule
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8 (0) & 1 & 0 & 0 & 0 & 0 \\
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9 (1) & 1 & 0 & 0 & 1 & 0 \\
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10 (2) & 1 & 0 & 1 & 0 & 1 \\
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11 (3) & 1 & 0 & 1 & 1 & 1 \\
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12 (4) & 1 & 1 & 0 & 0 & 0 \\
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13 (5) & 1 & 1 & 0 & 1 & 1 \\
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14 (6) & 1 & 1 & 1 & 0 & 1 \\
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15 (7) & 1 & 1 & 1 & 1 & 0 \\
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\bottomrule
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\end{tabularx}
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b/ Exprimer $H(A,B,C,D)$ sous la forme disjonctive.
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\begin{equation*}
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H(A,B,C,D) = \bar{A}BC\bar{D} + \bar{A}B\bar{C}D + AB\bar{C}D + A\bar{B}CD + \bar{A}BCD
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\end{equation*}
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\subsection{Exercice 4}
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\end{document}
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