习题2
2-1 试用列真值表的方法证明下列等式成立。 (1) A+BC=(A+B)(A+C) (2) A AB A B +=+ (3) 0A A ⊕= (4) 1A A ⊕=
(5) ()A B C AB AC ⊕+=⊕ (6) 1A B A B A B ⊕==⊕?
解:(1)设1F A BC =+ 2()()F A B A C =++
(2) 1F A AB =+ 2F A B
=+
(3) 10F A =⊕ 2F A =
(4) 11F A =⊕ 2F A =
(5) 1()F A B C =⊕+ 2F A B
A C =⊕
(6) 1F A B =⊕ 2F A B
= 31F A B =⊕?
2-2 分别用反演规则和对偶规则求出下列函数的反函数式和对偶式 。 (1) [()]F AB C D E B =++ (2) ()()F AB A C C DE =+++
(3) F A B C D E =++++ (4) ()0F A B C ABC =++= (5) F A B =⊕
解:(1)[()]F A B C D E B =+?++
'[()]F A B C D E B =+?+?+
(2) ()[()]F A B AC C D E =+?++
'()[()]F A B A C C D E =+?++
(3) ()F A B C D E =?+++
'F A B C D E =????
(4) ()1F A B C A B C =??+++=
'()1F A B C A B C =??+++=
(5) F A B =
'F AB AB =+
2-3 用公式法证明下列各等式。 (1) ()AB A C B C D AB A C D +++=++ (2) ()()BC D D B C AD B B D ++++=+ (3) AC AB BC ACD A BC +++=+ (4) AB BC C A AB BC CA ++=++ (5) A B C A B C ⊕⊕= (6) A B A B ⊕=⊕
(7) ()()A CD ACD A C A D +=⊕⊕ 解:(1)
()C B C D AB AC BC BCD AB AC BC D AB AC D ++=+++=+++=++=左边=AB+A 右边
(2)
()()()B C AD B BC D BC AD B BC D AD B B D ++=+++=+++=+=左边=BC+D+D 右边
(3) C AB BC ACD C AB BC A C ACD A BC +++=++++=+=左边=A A 右边 (4) BC CA AB BC AC CA BC AB AC BC AB ++=+++++=++=左边=AB 右边 (5) C A B C A B C ⊕⊕=⊕== 左边=A B 右边 (6) B AB AB A B ⊕=+=⊕=左边=A 右边 (7)
()()()()D AACD AACD ACD AC AC AD AD A C A D +++=++=⊕⊕=左边=AC 右边
2-4对于图P2-4(a )所示的每一个电路:
(1) 写出电路的输出函数表达式,列出完整的真值表。
(2) 若将图(b )所示的波形加到图(a )所示的电路的输入端,试分别画出12,F F 的输
出波形。
解:(1)1F A BB C =++
2F A B C =⊕⊕
(2)
A B C
1F 2
F
2-5 已知逻辑函数的真值表分别如表P2-5(a ),(b ),(c )所示。 (1) 试分别写出各逻辑函数的最小项之和表达式,最大项之积表达式。 (2) 分别求出各逻辑函数的最简与或式,最简或与式。 (a)
(b)
(c)
解:(1)(a)最小项之和的表达式为:1F ABC ABC ABC =++
最大项之积表达式为:1()()()()()F A B C A B C A B C A B C A B C =++++++++++ (b) 最小项之和的表达式为:1F ABC ABC ABC ABC =+++
最大项之积表达式为: 1()()()()F A B C A B C A B C A B C =++++++++ (c) 最小项之和的表达式为: 1F ABC ABC ABC ABC =+++
最大项之积表达式为: 1()()()()F A B C A B C A B C A B C =++++++++ (2) (a) 最简与或式:1F AB A C =+ 最简或与式:1()F A B C =+ (b) 最简与或式:1F AC BC =+ 最简或与式:1()()F A C B C =++ (c) 最简与或式:1F BC AC =+
最简或与式:1()()F B C A C =++ 2-6 对于图P2-6所示的每一个电路: (1) 试写出未经化简的逻辑函数表达式。 (2) 写出各函数的最小项之和表达式。
解:(1)(a )1F AB AB C =++ (b) 2F A B B C A C =+++++ (c) 3F A B B A AC =++++ (d) 4F A BC D A B C =++++ (2)各式的最小项表达式: (a )1F C = (b )2F AC = (c )3F A B C =++ (d )4F ABC =
2-7 用代数法化简下列逻辑函数,求出最简与或式。 (1) F AB B AB =++ (2) F ABC A B C =+++
(3) F ABC AB =+
(4) F ABCD ABD ACD =++ (5) ()()F AB A CD AD BC A B =+++ (6) ()()F AC CD AB BC B AD CE =++++ (7) F AC ABC ACD CD =+++
(8) ()()()F A B C A B C A B C =++++++ (9) ()()F BC ABCE B AD AD B AD AD =+++++
(10) ()F AC ACD ABEF B D E BCDE ABEF =+++⊕++ 解:(1)F A B =+ (2) 1F = (3) 1F = (4) F AD = (5) 0F = (6) F ABCDE = (7) F A CD =+ (8) F A BC =+ (9) ()F BC A D =+⊕
(10) ()F AC AD B D E AEF =++⊕+
2-8 判断图P2-8中个卡诺图的圈法是否正确。如有错请改正,并写出最简与或表达式。
解:各卡诺图均有错误,改正后的各卡诺图圈法如图解2-5所示。
(a)
(b)
(c )正确图示为:将原图的10对应的一列4个1圈起来,其他不变。 (d )正确图示为:将00行和10行的最后两个1圈起来,其他不变。
(e )正确图示为:将00行的中间两个1和10行的中间两个1圈起来,再将11行的4个1圈起来,其他不见。
(f )正确图示为:去掉00列的叉的圈,把11列的两个1和10列的两个叉圈起来,其他不变。
(g )正确图示为:把00列的两个1和10列的连个1圈起来,把00行的两个叉,1和10行的两个叉,1圈起来,然后把10行的四个圈起来,其他不变。
2-9 用卡诺图化简法将下列函数化简为最简与或式,并画出全部由与非门组成的逻辑电路图。
(1) (,,)(0,1,2,5,7)F A B C m =
∑
(2) (,,,)(2,3,6,7,8,10,12,14)F A B C D m =∑ (3) (,,,)(2,3,4,5,8,9,14,15)F A B C D m =
∑
(4) (,,,,)(0,4,18,19,22,23,25,29)F A B C D E m =∑
(5) (,,,)(0,1,2,3,6,8,10,11,12)F A B C D m =
∏
(6) F AB ABD A C BCD =+++
(7) F ACD BC BD AB A C BC BD =++++++
解:首先写出画出卡诺图,,圈“1”格,每个卡诺图圈对应写出一个与项,从而求出最简与
或式。得到各式的最简与或式与逻辑电路图分别如下:
=++
(1)最简式:F AB AC AC
逻辑电路图:
=+
(2)F A C AD
逻辑电路图:
=+++
(3)F ABC ABC ABC ABC
逻辑电路图:
=++
(4)F ABDE ABDE ABD
逻辑电路图:
=+++(5)F ABC ABC ACD BD 逻辑电路图:
=+(6)F AB A C
逻辑电路图:
=++(7)F B C AD 逻辑电路图:
2-10 用卡诺图化简法将下列函数化简为最简或与式,并画出全部由或非门组成的逻辑电路图。
(1) (,,,)(0,2,5,7,8,10,13,15)
F A B C D m
=∑
(2) (,,,)(0,2,3,7,8,10,11,13,15)
F A B C D M
=∏
(3) (,,,,)(0,1,3,4,5,7,10,14,19,23,26,27,30,31)
F A B C D E M
=∏
(4) ()
F AB AB AB AB C
=+++
(5) ()()()()
F A B A B C A C B C D
=++++++
解:将各逻辑函数分别填入卡诺图,圈“0”格,每个卡诺图对应写一个或项,从而求得最简或与式,进而求得或非式。
(1)最简或与式:
1
()()
F B D B D B D B D BD BD
=++=+++=+
逻辑电路图:
(2)最简或与式:
2()()()F B D C D A B D B D C D A B D BD CD ABD =++++=++++++=++
逻辑电路图:
(3) 最简或与式:3()()()()F A B D B D E A B E A D E =++++++++ 逻辑电路图:
(4) 最简或与式:4()()F A C B C A C B C AC BC =++=+++=+ 逻辑电路图:
(5) 最简或与式:5()()F A B A C A B A C AB AC =++=+++=+ 逻辑电路图:
2-11 已知试求1F ABD C =+,2()()()F B C A B D C D =++++,试求:
(1)12a F F F =?之最简与或式和最简与非-非与式。 (2)12b F F F =+之最简或与式和之最简或非-或非式。 (3)12c F F F =⊕之最简与或非式。
解:两函数之间的与,或,异或运算可由两个函数的卡诺图运算(即两个卡诺图中相应的方格作与,或,异或运算)来实现,分别求出a F ,b F ,c F 之卡诺图,分别如下图所示:
a F 图
CD AB 00 10 11 10 00 01 11
10
b F 图
CD AB 00 10 11 10 00 01 11
10
c F 图
CD AB 00 10 11 10 00 01 11
10
解:(1)求出各函数的表达式为:
a 12F F F ABC BCD =?=?
12()()b F F F A C D B C D =+=+++++ 12c F F F BCD ABD BCD =⊕=++
2-12 设有三个输入变量A,B,C ,试按下述逻辑问题列出真值表,并写出它们各自的最小项表达式,最大项表达式。
(1)当A+B=C 时,输出b F 为1,其余情况为0。
(2)当A B B C ⊕=⊕时,输出C F 为1,其余情况下为0。 解:a F ,b F ,c F 随A,B,C 变化的真值表如下图所示:
(1)m(0,3,5,7)(1,2,4,6)b F M ==∑∏
(2) m(0,2,5,7)(1,3,4,6)c F M =
=∑∏
2-13 将下列具有无关项的逻辑函数化简为与或表达式。 (1)(,,,)(0,1,4,7,9,10,13)(2,5,8,12,15)F A B C D M d =?∏∏
(2) (,,,)(1,3,6,8,11,14)(2,4,5,13,15)F A B C D m d =
+∑∑
(3)(,,,)(0,2,4,5,10,12,15)(8,14)F A B C D m d =
+∑∑F BCD BCD =+
(2) F ABD BCD ACD ABCD =+++ (3)F CD BD AD ABC ABC =++++
2-14 将下列具有约束条件的逻辑函数化简为最简或与表达式:
(1)
F ABC ABC ABCD ABCD ?=+++
?
?
??变量ABCD不可能出现相同的值
(2)
()
F A B CD ABC ACD AB CD
?=⊕++
?
?
+=
??
解:(1)F AC BD
=+ (2) F C
=