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1、单击此处编辑母版标题样式,单击此处编辑母版文本样式,第二级,第三级,第四级,第五级,*,*,第,3,章习题解答,1,分析图,3-1,所示电路，写出电路输出,Y,1,和,Y,2,的逻辑函数表达式，列出真值表，说明它的逻辑功能。,=1,=1,&,&,&,A,B,C,Y,1,Y,2,Y,1,=A B C,A B C,Y,1,Y,2,0 0 0,0 0 1,0 1 0,0 1 1,1 0 0,1 0 1,1 1 0,1 1 1,0 0,1 0,1 0,0 1,1 0,0 1,0 1,1 1,Y,2,=(A B)C AB,当,A,、,B,、,C,中有奇数个,1,时，,Y,1,为真。,当,A,、,B,、,
2、C,中有两个或以上的,1,时，,Y,2,为真。,5,用与非门设计下列函数，允许反变量输入。,(2)F(A,B,C,D)=,M,(0,2,4,5,9,10,13,14),0,0,0,1,1,1,1,0,1,0,1,1,0,0,0,1,F=CD+ACD+ABD+ABC,=CD ACD ABD ABC,&,&,&,&,&,B,A,C,D,C,A,B,D,F,8,已知输入信号,A,，,B,，,C,，,D,的波形如下图所示，用或非门设计产生输出,F,波形的组合电路，允许反变量输入。,B,C,D,F,1,1,1,1,1,1,1,A,0,0,0,0,0,0,0,X,X,F=(A+C+D)(B+C)(A+D)
3、,=(A+C+D)+(B+C)+(A+D),1,1,1,1,C,D,C,A,B,D,F,A,10,分析图,3-7,所示由集成,8,选,1,数据选择器,CT74151,构成的电路，写出电路输出,F,1,的最简逻辑函数表达式，列出真值表。,CT74151,D,7,D,6,D,5,D,4,D,3,D,2,D,1,D,0,ST,A,0,A,1,A,2,1,A,B,C,F,1,AB,C,0,0,0,0,1,1,1,A B C,F,1,0 0 0,0 0 1,0 1 0,0 1 1,1 0 0,1 0 1,1 1 0,1 1 1,0,1,0,1,1,0,1,0,F,1,=AC+AC,0,0,0,1,0,0
4、,0,0,0,12,采用降维法用一片集成,8,选,1,数据选择器,CT74151,和必要的门电路实现下列逻辑函数。,(1)F(A,B,C,D)=,m,(0,2,8,10,11,13,14,15),1,1,1,1,1,1,1,AB,C,0,1,1,0,D,D,D,D,D,7,=1 D,6,=D D,5,=1 D,4,=D D,3,=0 D,2,=0 D,1,=D D,0,=D,CT74151,D,7,D,6,D,5,D,4,D,3,D,2,D,1,D,0,ST,A,0,A,1,A,2,1,A,B,C,F,D,1,15,用一片集成,3,线,8,线译码器,CT74138,和必要的门电路实现下列多输出
5、组合逻辑函数。,F,1,=ABC+A(B+C),F,2,=AB+AB,F,3,=(A+B)(A+C),F,4,=ABC+ABC,F,3,=(A+B)(A+C)=BC+AB+AC=m,2,+m,3,+m,4,+m,6,=m,2,m,3,m,4,m,6,=Y,2,Y,3,Y,4,Y,6,F,4,=ABC+ABC=m,0,+m,7,=m,0,m,7,=Y,0,Y,7,F,1,=ABC+A(B+C)=ABC+AB+AC=m,1,+m,2,+m,3,+m,7,=m,1,m,2,m,3,m,7,=Y,1,Y,2,Y,3,Y,7,F,2,=AB+AB=m,2,+m,3,+m,4,+m,5,=m,2,m,3,
6、m,4,m,5,=Y,2,Y,3,Y,4,Y,5,&,&,&,&,Y,0,Y,1,Y,2,Y,3,Y,4,Y,5,Y,6,Y,7,ST,A,A,2,A,1,A,0,1,A B C,ST,B,ST,C,CT74138,F,1,F,2,F,3,F,4,17,用一片集成,3,线,8,线译码器,CT74138,和必要的门电路设计一个运算电路，当,K=1,时，实现一位全加器；当,K=0,时，实现一位全减器。,CI,为全加器中低位向本位的进位或者全减器中低位向本位的借位，,CO,1,为全加器中本位向高位的进位，,CO,2,为全减器中本位向高位的借位。,6,4,0,2,CI A B,CO,1,F,1,0 0
7、 0,0 0 1,0 1 0,0 1 1,1 0 0,1 0 1,1 1 0,1 1 1,0 0,0 1,0 1,1 0,0 1,1 0,1 0,1 1,CO,2,F,2,0 0,0 1,1 1,0 0,1 1,1 1,0 0,1 0,序号,1,3,5,7,F=F,1,=F,2,=CIAB+CIAB+CIAB+CIAB=m,1,+m,2,+m,4,+m,7,=m,1,m,2,m,4,m,7,=Y,1,Y,2,Y,4,Y,7,CO,1,=CIAB+CIAB+CIAB+CIAB=m,3,+m,5,+m,6,+m,7,=m,3,m,5,m,6,m,7,=Y,3,Y,5,Y,6,Y,7,CO,2,=CIAB+CIAB+CIAB+CIAB=m,1,+m,4,+m,5,+m,7,=m,1,m,4,m,5,m,7,=Y,1,Y,4,Y,5,Y,7,CO=KCO,1,+KCO,2,Y,0,Y,1,Y,2,Y,3,Y,4,Y,5,Y,6,Y,7,ST,A,A,2,A,1,A,0,1,CI,A B,ST,B,ST,C,CT74138,&,F,&,&,&,&,1,K,1,CO,
数字逻辑电路习题解答3
