Add first exercices
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\date{Dernière compilation~: \today{} à \currenttime}
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\usepackage{../../cours}
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\usepackage{enumitem}
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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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\begin{multicols}{4}
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\begin{enumerate}[label=\alph*)]
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\item $a \cdot b \cdot c \cdot d = 1$
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\item $a + b + c + d = 0$
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\item $a + b + c + d = 1$
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\item $a \cdot b \cdot c \cdot d = 0$
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\end{enumerate}
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\end{multicols}
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\subsection{Exercice 2}
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\begin{enumerate}[label=\alph*)]
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\item \begin{align*}
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F(a, b, c) &= a\bar{b} + abc + a\bar{c} \\
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F(0,1,1) &= 0 \\
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F(1,1,0) &= 1 \\
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F(1,0,0) &= 1 \\
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\end{align*}
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\item \begin{equation*}
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F(a, b) = \bar{a}\bar{b} + b = \bar{a} + b
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\end{equation*}
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\begin{center}
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\begin{tabular}{cc|c|c|c|c}
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\toprule
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$a$ & $b$ & $\bar{a}\bar{b}$ & $\bar{a}\bar{b} + b$ & $\bar{a}$ & $\bar{a} + b$ \\
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\midrule
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0 & 0 & 1 & 1 & 1 & 1 \\
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0 & 1 & 0 & 1 & 0 & 1 \\
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1 & 0 & 0 & 0 & 0 & 0 \\
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1 & 1 & 0 & 1 & 0 & 1 \\
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\midrule
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& & & $\uparrow$ & = & $\uparrow$ \\
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\bottomrule
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\end{tabular}
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\end{center}
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\item \begin{align*}
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F(a, b, c) &= a\bar{b} + ab + ac \\
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&= a (\bar{b} + b + c) \\
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&= a (1 + c) \\
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&= a \cdot 1 = a
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\end{align*}
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\item \begin{align*}
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F(a, b, c) &= \bar{a}\bar{b}c + a\bar{b}\bar{c} + a\bar{b}c + ab\bar{c} + abc \\
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&= \bar{a}\bar{b}c + a\bar{b}\bar{c} + ab\bar{c} + ac(b + \bar{b}) \\
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&= \bar{a}\bar{b}c + a\bar{b}\bar{c} + ab\bar{c} + ac \\
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&= \bar{a}\bar{b}c + a\bar{c} + ac \\
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&= \bar{a}\bar{b}c + a \\
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\end{align*}
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\end{enumerate}
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\subsection{Exercice 3}
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\begin{enumerate}[label=\alph*)]
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\item \begin{align*}
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F_1 &= (a + b) \cdot (\bar{a} + \bar{b}) \\
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\overline{F_1} &= \overline{(a + b) \cdot (\bar{a} + \bar{b})} \\
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&= \overline{(a + b)} + \overline{(\bar{a} + \bar{b})} \\
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&= \bar{a}\bar{b} + ab \\
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&= \overline{a \oplus b} \\
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&= \bar{a} \oplus b = a \oplus \bar{b} \\
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\end{align*}
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\item \begin{align*}
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F_2 &= a (c + d) + (\bar{a} + c) \cdot (\bar{b} + c + d) \\
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\overline{F_2} &= \overline{a (c + d) + (\bar{a} + c) \cdot (\bar{b} + c + d)} \\
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&= \overline{a (c + d)} \cdot \overline{(\bar{a} + c)} + \overline{(\bar{b} + c + d)} \\
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&= \overline{ac + ad} \cdot a\bar{c} + b\bar{c}\bar{d} \\
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&= \overline{ac} \cdot \overline{ad} \cdot a\bar{c} + b\bar{c}\bar{d} \\
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&= (\bar{a} + \bar{c}) \cdot (\bar{a} + \bar{d}) \cdot a\bar{c} + b\bar{c}\bar{d} \\
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% TODO: finish
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\end{align*}
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\end{enumerate}
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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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@ -72,22 +165,22 @@
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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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\begin{tabularx}{0.7\linewidth}{X|YYY|Y|Y}
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\toprule
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& D & A & B & C & H \\
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& A & B & C & D & 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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3 & 0 & 0 & 1 & 1 & 1 \\
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4 & 0 & 1 & 0 & 0 & 0 \\
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5 & 0 & 1 & 0 & 1 & 0 \\
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5 & 0 & 1 & 0 & 1 & 1 \\
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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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10 (2) & 1 & 0 & 1 & 0 & 0 \\
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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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@ -98,9 +191,10 @@
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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) = AB\bar{C}\bar{D} + \bar{A}B\bar{C}D + \bar{A}BCD + A\bar{B}CD + AB\bar{C}D
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\end{equation*}
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\begin{align*}
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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} \\
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&= \sum(3, 5, 6, 11, 13, 14)
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\end{align*}
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\subsection{Exercice 4}
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