SchoolWork-LaTeX/数字逻辑及实验/平时作业/第一章作业-订正.tex

\documentclass[全部作业]{subfiles}

\fancyhead{}
\fancyhead[C]{\mysignature}
\begin{document}
\renewcommand{\bar}{\xoverline} % 这一行在编译时可以取消注释，注意一定要放在\begin{document}下面才有用
\chapter{逻辑代数基础}
% 思考题和习题  1、2、4、5、6、8、9
\begin{enumerate}

\item 运用基本定理证明下列等式。
\begin{enumerate}
    \item $AB+\bar{A}C+\bar{B}C = AB+C $
        \begin{proof}
        $$
        \begin{aligned}
        &\\
        &\begin{aligned}
        AB+\bar{A}C+\bar{B}C & = AB+(\bar{A}+\bar{B})C  \\
        & = AB+\overline{AB}C  \\
        & = AB+C  \\
        \end{aligned}
        \end{aligned}
        $$
        \end{proof}
    \item $BC+D+\bar{D}(\bar{B}+\bar{C})(DA+B)=B+D$
        \begin{proof}
        $$
        \begin{aligned}
        BC+D+\bar{D}(\bar{B}+\bar{C})(DA+B) & = BC+D+(\bar{B}+\bar{C})(DA+B)  \\
        & = BC+D+\overline{BC}(DA+B)  \\
        & = BC+D+DA+B  \\
        & = B+D  \\
        \end{aligned}
        $$
        \end{proof}
    \item $ABC+\bar{A}\bar{B}\bar{C}=\overline{A\bar{B}+B\bar{C}+C\bar{A}}$
        \begin{proof}
        $$
        \begin{aligned}
        ABC+\bar{A}\bar{B}\bar{C} & = (A+\bar{A})(A+\bar{B})(A+\bar{C})(B+\bar{A})(B+\bar{B})(B+\bar{C})(C+\bar{A})(C+\bar{B})(C+\bar{C})  \\
        & = (A+\bar{B})(A+\bar{C})(B+\bar{A})(B+\bar{C})(C+\bar{A})(C+\bar{B})  \\
        & = \overline{A\bar{B}+A\bar{C}+B\bar{A}+B\bar{C}+C\bar{A}+C\bar{B}} \\
        & \xlongequal{\textit{冗余律}} \overline{A\bar{B}+B\bar{C}+C\bar{A}}  \\
        \end{aligned}
        $$
        \end{proof}
    \item $AB+BC+CA=(A+B)(B+C)(C+A)$
        \begin{proof}
        $$
        \begin{aligned}
        (A+B)(B+C)(C+A) & = ABC+ABA+ACC+ACA+BBC+BBA+BCC+BCA  \\
        & = ABC+AB+AC+AC+BC+BA+BC+ABC  \\
        & = ABC+AB+AC+BC  \\
        & = AB+BC+CA  \\
        \end{aligned}
        $$
        \end{proof}
    \item $\bar{A}BC+AB+A\bar{C}=BC+A\bar{C}$
        \begin{proof}
        $$
        \begin{aligned}
        \bar{A}BC+AB+A\bar{C} & = B(\bar{A}C+A)+A\bar{C}  \\
        & = B(C+A)+A\bar{C}  \\
        & = BC+AB+A\bar{C}  \\
        & = BC+A\bar{C}  \\
        \end{aligned}
        $$
        \end{proof}
    \item $\overline{A\bar{B}+\bar{A}B}=(A+\bar{B})(\bar{A}+B)$
        \begin{proof}
        $$
        \begin{aligned}
        \overline{A\bar{B}+\bar{A}B} & = \overline{A\bar{B}}\ \overline{\bar{A}B}  \\
        & = (\bar{A}+B)(A+\bar{B})  \\
        & = (A+\bar{B})(\bar{A}+B)  \\
        \end{aligned}
        $$
        \end{proof}
    \item $\bar{A}\bar{B}+AB+BC=\bar{A}\bar{B}+AB+\bar{A}C$
        \begin{proof}
        $$
        \begin{aligned}
        \bar{A}\bar{B}+AB+BC & = \bar{A}\bar{B}+AB+BC(A+\bar{A})  \\
        & = \bar{A}\bar{B}+AB+BCA+BC\bar{A}  \\
        & = \bar{A}\bar{B}+AB+\bar{A}BC  \\
        & = \bar{A}(\bar{B}+BC)+AB  \\
        & = \bar{A}(\bar{B}+C)+AB  \\
        & = \bar{A}\bar{B}+AB+\bar{A}C  \\
        \end{aligned}
        $$
        \end{proof}
    \end{enumerate}

\item 用逻辑代数定理化简下列逻辑函数式。
    \begin{enumerate}
    \item $AB+\bar{A}B\bar{C}+BC$
        $$
        \begin{aligned}
        AB+\bar{A}B\bar{C}+BC & = B(A+\bar{A}\bar{C}+C)  \\
        & = B(A+\bar{C}+C)  \\
        & = B  \\
        \end{aligned}
        $$
    \item $\bar{A}\bar{B}\bar{C}+A\bar{B}\bar{C}+A\bar{B}C$
        $$
        \begin{aligned}
        \bar{A}\bar{B}\bar{C}+A\bar{B}\bar{C}+A\bar{B}C & = \bar{B}(\bar{A}\bar{C}+A\bar{C}+AC)  \\
        & = \bar{B}(\bar{C}+AC)  \\
        & = \bar{B}(\bar{C}+A)  \\
        \end{aligned}
        $$
    \item $ab(cd+\bar{c}d)$
        $$
        ab(cd+\bar{c}d)=abd
        $$
    \item $[x \overline{(xy)}][y \overline{(xy)}]$
        $$
        \begin{aligned}
        \relax[x \overline{(xy)}][y \overline{(xy)}] & = xy\overline{(xy)}\ \overline{(xy)}  \\
        & = xy\overline{(xy)}  \\
        & = 0  \\
        \end{aligned}
        $$
    \item $\overline{(a+b)}\ \overline{(\bar{a}+\bar{b})}$
        $$
        \begin{aligned}
        \overline{(a+b)}\ \overline{(\bar{a}+\bar{b})} & = \bar{a}\bar{b}ab  \\
        & = 0  \\
        \end{aligned}
        $$
    \item $\bar{a} \bar{b} \bar{c}+\bar{a} \bar{b}c+a \bar{b}\bar{c}+abc$
        $$
        \begin{aligned}
        \bar{a} \bar{b}\bar{c}+\bar{a} \bar{b}c+a \bar{b}\bar{c}+abc & = \bar{a} \bar{b}+a\bar{b}\bar{c}+abc  \\
        & = \bar{b}(\bar{a}+a\bar{c})+abc  \\
        & = \bar{b}(\bar{a}+\bar{c})+abc  \\
        & = \bar{b}\overline{ac}+bac  \\
        \end{aligned}
        $$
    \end{enumerate}
    \item[4.] 用卡诺图化简下列最小项表达式.\\
    $G=f(a,b,c)=\sum m(1,3,5,6,7)$\\
    \includesvg{1.4.G}\\
    $$
    G=f(a,b,c)=ab+c
    $$
    $H=f(w,x,y,z)=\sum m(0,2,8,10)$\\
    \includesvg{1.4.H}\\
    $$
    H=f(w,x,y,z)=\bar{x}\bar{z}
    $$

    \newpage
    $I=f(w,x,y,z)=\sum m(1,3,4,6,9,12,14,15)$\\
    \includesvg{1.4.I}\\
    $$
    I=f(w,x,y,z)=x\oplus z\oplus (wyz)
    $$
    
    \vspace{10em}
    $J=f(a,b,c)=\sum m(0,1,2,3,4,5,7)$\\
    \includesvg{1.4.J}\\
    $$
    J=f(a,b,c)=\sum M(6)=\bar{a}+\bar{b}+c
    $$

    \newpage
    $K=f(a,b,c,d)=\sum m(3,4,5,7,9,13,14,15)$\\
    \includesvg{1.4.K}\\
    $$
    K=f(a,b,c,d)=bd+\bar{a}b \bar{c}+\bar{a}cd+abc+\bar{ac}d
    $$
    
    \vspace{10em}
    $L=f(a,b,c,d)=\sum m(0,1,2,5,6,7,8,9,13,14)$\\
    \includesvg{1.4.L}\\
    $$
    L=f(a,b,c,d)=\bar{b}\bar{c}+\bar{c}d+\bar{a}bc+\bar{a}c \bar{d}+bc \bar{d}
    $$

    \newpage
    \item[5.] 用卡诺图化简下列最大项表达式。\\
    $H=f(a,b,c,d)=\prod M(2, 3, 4, 6, 7, 10, 11, 12) $\\
    \includesvg{1.5.H}\\
    $$
    H=f(a,b,c,d)=(a+\bar{c})(\bar{b}+c+d)(b+\bar{c})
    $$
    
    \vspace{10em}
    $F=f(u,v,w,x,y)=\prod M(0, 2, 8, 10, 16, 18, 24, 26)$\\
    \includesvg{1.5.F}\\
    $$
    F=f(u,v,w,x,y)=w+y
    $$

    \newpage
    \item[6.] 化简下列带任意项的逻辑函数。\\
    $V=f(a,b,c,d)=\sum m(2,3,4,5,13, 15)+\sum d(8,9,10, 11)$\\
    \includesvg{1.6.V}\\
    $$
    V=f(a,b,c,d)=ad+\bar{b}c+\bar{a}b \bar{c}
    $$
    
    \vspace{10em}
    $Y=f(u,v,w,x)=\sum m(1, 5, 7, 9, 13, 15)+\sum d(8, 10, 11, 14)$\\
    \includesvg{1.6.Y}\\
    $$
    Y=f(u,v,w,x)=x \overline{\bar{u}\bar{v}w}=x(u+v+\bar{w})
    $$

    \newpage
    $P=f(r,s,t,u)=\sum m(0, 2, 4, 8, 10, 14)+\sum d(5,6,7,12)$\\
    \includesvg{1.6.P}\\
    $$
    P=f(r,s,t,u)=\bar{u}
    $$
    
    \vspace{10em}
    $H=f(a,b,c,d,e)=\sum m(5,7,9,12,13,14,17,19,20,22,25,27,28,30)+\sum d(8,10,24,26)$\\
    \includesvg{1.6.H}\\
    $$
    H=f(a,b,c,d,e) = a \bar{c}e + ac \bar{e}+\bar{a}\bar{b}ce+bcd \bar{e}+b \bar{c}\bar{d}+\bar{a}bc \bar{d}
    $$

    \newpage
    $I=f(d,e,f,g,h)=\prod M(5,7,8,21,23,26,30)\cdot \prod D(10,14,24,28)$\\
    \includesvg{1.6.I}\\
    $$
    I=f(d,e,f,g,h)=(e+\bar{f}+\bar{h})(\bar{e}+\bar{g}+h)(d+\bar{e}+f+h)
    $$
    
    \vspace{10em}
    \item[8.] 将下列逻辑函数化简成与非形式最简式。\\
    $U=f(a,b,c,d)=\sum m(3,4,6,11,12,14)$\\
    \includesvg{1.8.U}\\
    $$
    U=f(a,b,c,d)=b \bar{d}+\bar{b}cd = \overline{\overline{b \bar{d}}\ \overline{\bar{b}cd}}
    $$

    \newpage
    $V=f(a,b,c,d)=\sum m (0,1,2,5,8,10,13)$\\
    \includesvg{1.8.V}\\
    $$
    \begin{aligned}
    &V=f(a,b,c,d)=(\overline{a \oplus b \oplus d}+\bar{a}\bar{b})\overline{cd}=\overline{a\oplus b\oplus d \overline{\bar{a}\bar{b}}}\overline{cd}=\overline{(\bar{a}b+a \bar{b}) \oplus d}\ \overline{cd} \\
    &=\overline{\overline{(\bar{a}b +a \bar{b})}d}\ \overline{(\bar{a}b+a \bar{b})\bar{d}}\ \overline{cd}=\overline{\overline{\bar{a}b} \overline{a \bar{b}} d} \ \overline{\overline{\overline{\bar{a}b} \overline{a \bar{b}}}\bar{d}}\ \overline{cd}\\
    \end{aligned}
    $$
    
    \vspace{10em}
    $W=f(a,b,c,d) = \sum m(3,5,7,10,11)$\\
    \includesvg{1.8.W}\\
    $$
    W=f(a,b,c,d)=\bar{a}cd+\bar{a}bd+a \bar{b}\bar{c}d+a \bar{b}c \bar{d} = \overline{\overline{\bar{a}cd}\ \overline{\bar{a}bd}\ \overline{a \bar{b}\bar{c}d}\ \overline{a \bar{b}c \bar{d}}}
    $$

    \newpage
    \item[9.] 将下列逻辑函数化简成或非形式最简式。\\
    $G=f(a,b,c,d)=\prod M(0,1,2,5,8,10,13) $\\
    \includesvg{1.9.G}\\
    $$
    G=f(a,b,c,d)=(b+d)(c+\bar{d}+a \bar{b}\bar{c})=\overline{\overline{b+d}\ \overline{c+\bar{d}+\overline{\bar{a}+b+c}}}
    $$
    
    \vspace{10em}
    $H=f(a,b,c,d)=\prod M(3,5,7,9,11)$\\
    \includesvg{1.9.H}\\
    $$
    H=f(a,b,c,d) = \bar{d}+\bar{a} \bar{b}\bar{c}+ab=\bar{d}+\overline{a+b+c}+\overline{\bar{a}+\bar{b}}
    $$
\end{enumerate}

\end{document}