274 lines
8.4 KiB
TeX
274 lines
8.4 KiB
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}\\
|
|
$$
|
|
G=f(a,b,c)=a \bar{b}+c
|
|
$$
|
|
$H=f(w,x,y,z)=\sum m(0,2,8,10)$\\
|
|
\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}\\
|
|
$$
|
|
J=f(a,b,c)=\sum M(6)=\bar{a}+b+c
|
|
$$
|
|
|
|
\newpage
|
|
$K=f(a,b,c,d)=\sum m(3,4,5,7,9,13,14,15)$\\
|
|
\includesvg{1.4.K}\\
|
|
$$
|
|
K=f(a,b,c,d)=bd+\bar{a}b \bar{c}+\bar{a}cd+abc
|
|
$$
|
|
|
|
\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}\\
|
|
$$
|
|
H=f(a,b,c,d)=(a+\bar{c})(\bar{b}+c+d)(\bar{a}+b+\bar{c})
|
|
$$
|
|
|
|
\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}\\
|
|
$$
|
|
V=f(a,b,c,d)=b \bar{c}+\bar{b}c
|
|
$$
|
|
|
|
\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} |