数字电路与数字电子技术课后答案

1、第四章 逻辑函数及其符号简化1 列出下述问题的真值表，并写出逻辑表达式：(1) 有A、B、C三个输入信号，如果三个输入信号中出现奇数个1时，输出信号F=1，其余情况下，输出F= 0.(2) 有A、B、C三个输入信号，当三个输入信号不一致时，输出信号F=1，其余情况下，输出为0.(3) 列出输入三变量表决器的真值表.解: ( 1 )A B C F0 0 0 00 0 1 10 1 0 10 1 1 01 0 0 11 0 1 01 1 0 01 1 1 1 F=C+B+A+ABC( 2 )A B C F0 0 0 00 0 1 10 1 0 10 1 1 11 0 0 11 0 1 11 1 0

2、 11 1 1 1 F= (A+B+C) ( +)( 3 )A B C F0 0 0 00 0 1 00 1 0 00 1 1 11 0 0 01 0 1 11 1 0 11 1 1 1 F=BC+AC+AB+ABC2. 对下列函数指出变量取哪些组值时,F的值为“1”：(1) F= AB+(2) F= AB+C(3) F= (A+B+C) (A+B+) (A+C) (A+)解:(1) AB = 00或AB=11时F=1(2) ABC110或111,或001,或011时F=1(3) ABC = 100或101或110或111时F=13. 用真值表证明下列等式.(1) A+BC = (A+B) (

3、A+C)(2) BC+AC+AB= BC+AC+AB(3) =ABC+(4) AB+BC+AC=(A+B)(B+C)(A+C)(5) ABC+=1证: ( 1 ) A B C A+BC (A+B)(A+C)0 0 0 0 00 0 1 0 00 1 0 0 00 1 1 1 11 0 0 1 11 0 1 1 11 1 0 1 11 1 1 1 1( 2 )A B C ABC + ABC + ABC BCABC + ACA B C + ABABC0 0 0 0 00 0 1 0 00 1 0 0 00 1 1 1 11 0 0 0 01 0 1 1 11 1 0 1 11 1 1 0 0( 3

4、 )A B C AB + BC + AC ABC + A B C0 0 0 1 10 0 1 0 00 1 0 0 00 1 1 0 01 0 0 0 01 0 1 0 01 1 0 0 01 1 1 1 1A B C AB+BC+AC (A+B)(B+C)(A+C)0 0 0 0 00 0 1 0 00 1 0 0 00 1 1 1 11 0 0 0 01 0 1 1 11 1 0 1 11 1 1 1 1( 4 )( 5 ) A B C ABC + A + B + C0 0 0 1 0 0 1 1 0 1 0 1 0 1 1 1 1 0 0 1 1 0 1 1 1 1 0 1 1 1 1

5、1 4. 直接写出下列函数的对偶式F及反演式的函数表达式.(1) F= B (C+D)B+B (+D)(2) F= A+ (+) (A+C)(3) F= AB+(4) F=解:(1) F= +B+CD+(B+)B+D= A+(+C+D)+C(2) F= (A+)= (+)(3) F=+=+5. 若已知x+y = x+z，问y = z吗?为什么?解:y不一定等于z,因为若x=1时,若y=0,z=1,或y=1,z=0,则x+y = x+z = 1,逻辑或的特点,有一个为1则为1。6. 若已知xy = xz，问y = z吗?为什么?解:y不一定等于z，因为若x = 0时，不论取何值则xy = xz

6、= 0,逻辑与的特点,有一个为0则输出为0。7. 若已知x+y = x+z Xy = xz 问y = z吗? 为什么?解:y等于z。因为若x = 0时，0+y = 0+z，y = z，所以xy = xz = 0，若x = 1时, x+y = x+z = 1，而xy = xz式中y = z要同时满足二个式子y必须等于z。8.用公式法证明下列个等式(1) +BC+=+BC证:左=+ BC + =+ BC +=(1+) + BC =+BC = 右边(2) C+BD+ACD+B+CD+B+BCD=C+B+BD证: 左= (C+CD+ACD)+(ABCD+BCD+BD)+(BD+B+B)=C(+D+AD

7、)+BD(AC+C+)+B(D+)=C+B+BD(3) +=1证: 左= (+D)+ ()+(C+)= (+)(+)+D( +)+C+= +D +C+= +D +C+=+D+C+=+C+=1(4) x+wy+uvz= (x+u+w) (x+u+y) (x+v+w) (x+v+y) (x+z+w) (x+z+y)证:对等式右边求对偶,设右边=F,则F= xuw+xuy+xvw+xvy+xzw+xzy= xu (w+y)+xv (w+y) +xz (w+y)= (w+y) (xu+xv+xz) F= F= wy+(x+u)(x+v) (x+z)= wy +(x+xu+xv+uv) (x+z)= w

8、y+(x+uv)(x+z)= wy+x+xuv+xz+uvz= wy+x+uvz= wy+x+uvz(5) ABC=ABC证: 左= (AB)C=+ (AB) = (AB)C+ ()= ABC(6) =证: 左= (AB)+(AB)+C= (AB) +(AB)C=+AB+BC+AC 右= ()= ()+= (+AB) +=+AB+=+AB+(AB)C=+AB+BC+AC9.证明(1) 如果a+b = c，则a+c = b，反之亦成立(2) 如果+ab = 0，则 = a+b证:(1) a+ c = a ()+(a+b)= a (ab+)+b= ab+b = b(2) +ab = 0 说明a =

9、或b = (+)(a+)= a+= a+= a+b10.写出下列各式F和它们的对偶式，反演式的最小项表达式(1)F= ABCD+ACD+B(2)F= A+B+BC(3)F= +解:(1) F=m=m (0,1,2,3,5,6,7,8,9,10,13,14)F=m (15,14,13,12,10,9,8,7,6,5,2,1)(2) F=m (2,3,4,5,7)=m (0,1,6)F=m (7,6,1)(3) F= m (1,5,6,7,8,913,14,15)= m (0,1,3,4,10,11,12)F= m (15,13,12,11,5,4,3)11.将下列函数表示成最大项之积(1) F=

10、 (AB)(A+B)+(AB)AB(2) F= (AB)+(BC)解:(1) F= (AB)A+B+AB)= (+AB)(A+B)= AB+AB= AB=m (3)=M (0,1,2)(2) F= (AB)+(C+B)= B+A+C+B= B+A+C= m (1,2,3,4,5)=M (0,6,7)12. 用公式法化简下列各式(1) F= A+AB+ABC+BC+B解:F= A(1+B+BC)+B(C+1) = A+B(2) F= AC+D+A解:F=A+A+D(3) F= (A+B)(A+B+C)(+C)(B+C+D)解:F= AB+ABC+C+BCD = AB+C+BCD = AB+CF=

11、 F= (A+B)(+C)(4) F=解:F= AB+BC+ = AB+C+a) F=解:F=C+AC(5) F= (x+y+z+) (v+x) (+y+z+)解:F= xyz+vx+yz = vx+yz+xyz = vx+yzF= F= (v+x) (+y+z+)13.指出下列函数在什么输入组合时使F=0(1) F=m (0,1,2,3,7)(2) F=m (7,8,9,10,11)解:(1) F在输入组合为4,5,6时使F= 0(2) F在输入组合为0,1,2,3,8,10,11,13,14,15时使F= 014.指出下列函数在什么组合时使F=1(1) F=M (4,5,6,7,8,9,1

12、2)(2) F=M (0,2,4,6)解:(1) F在输入组合为0,1,2,3,8,10,11,13,14,15时使F=1;(2) F 在输入组合为1,3,5,7时使F=115.变化如下函数成另一种标准形式(1) F=m (1,3,7)(2) F=m (0,2,6,11,13,14)(3) F=M (0,3,6,7)(4) F=M (0,1,2,3,4,6,12)解:(1) F=M (0,2,4,5,6)(2) F=M (1,3,4,5,7,8,9,10,12,15)(3) F=m (1,2,4,5)(4) F=m (5,7,8,9,10,11,13,14,15)16.用图解法化简下列各函数(

13、1) 化简题12中(1),(3),(5)(2) F=m (0,1,3,5,6,8,10,15)(3) F=m (4,5,6,8,10,13,14,15)(4) F=M(5,7,13,15)(5) F=M (1,3,9,10,11,14,15)(6) F=m (0,2,4,9,11,14,15,16,17,19,23,25,29,31)(7) F=m (0,2,4,5,7,9,13,14,15,16,18,20,21,23,25,29,30,31)解:(1) 化简题12中(1),(3),(5)ABCD 00 01 11 1000011110( b )0 1 0 00 1 0 00 1 1 10

14、1 1 1ABC 00 01 11 1001F = A + B( a )0 1 1 10 1 1 1 =+AF= (A+B) (+C)F= (AC+C)( +AC+)= AC+C+ACABC 00 01 11 1001( C )0 0 0 01 0 1 1 F=AC+C 图P4.A16 ( 1 )(2) F= m (0,1,3,5,6,8,10,15)ABCD 00 01 11 1000011110图P4.A16 ( 2 )1 11 1 1 1 1 1 F=+D+D +A+ABCD+BC(3) F=m (4,5,6,8,9,10,13,14,15)ABCD 00 01 11 100001111

15、0图P4.A16 ( 3 ) 1 1 1 1 1 1 1 1 1 F=B+A+ABD +BC+ AC(4) F=M(5,7,13,15)ABCD 00 01 11 1000011110图P4.A16 ( 4 ) 0 0 0 0 = BDF=+(5) F=M (1,3,9,10,11,14,15)ABCD 00 01 11 1000011110图P4.A16 ( 5 ) 0 00 0 0 0 0 = AC+D F = (+)(B+)(6) F=m (0,2,4,9,11,14,15, 16,17,19,23,25,29,31)ABCD 000 001 011 010 110 111 101 10

16、000011110图P4.A16 ( 6 )1 1 1 1 1 1 1 1 1 1 1 11 1F=+BCD+BE+ABE+ACDE+A+AE(7) F=m (0,2,4,5,7,9,13,14,15,16,18,20,21,23,25,29,30,31) ABCD 000 001 011 010 110 111 101 10000011110图P4.A16 ( 7 )1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1F= ACE+BE+BCD+C+17. 将下列各函数化简成与非一与非表达式,并用与非门实现(1) F=m (0,1,3,4,6,7,10,11,13,14,1

17、5)(2) F=m (0,2,3,4,5,6,7,12,14,15)(3) F=m (0,1,4,5,12,13)(4) F=M(4,5,6,7,9,10,11,12)解: 圈“1”格化简(1) F=m (0,1,3,4,6,7,10,11,13,14,15)ABCD 00 01 11 1000011110 ( a )1 1 1 1 1 1 1 1 1 1 1 ( b )图P4.A17 (1)F= AC+BC+D+ABD =(2) F=m (0,2,3,4,5,6,7,12,14,15)ABCD 00 01 11 1000011110( a )1 1 11 1 1 1 1 1 1 ( b )图

18、P4.A17 ( 2 )F=C+BC+B+B =(3) F=m (0,1,4,5,12,13) ABCD 00 01 11 1000011110( a )1 1 11 1 1 ( b )F=+B=图P4.A17 ( 3 )ABCD 00 01 11 1000011110( a )1 0 0 1 1 0 1 01 0 1 0 1 0 1 0(4) F=M(4,5,6,7,9,10,11,12) ( b ) 图P4.A17 ( 4 )F = +ABD+ABC+= 18. 将下列各函数化简成或非一或非表达式并用或非门实现(1) F=m (0,1,2,4,5)(2) F=m (0,2,8,10,14,

19、15)(3) F= A+C+CD(4) F= AB+C+C解: 圈“0”格化简(1) F=m (0,1,2,4,5) ABC 00 01 11 1001( a )1 1 0 11 0 0 1 ( b )图P4.A18 ( 1 )= AB+BCF = (+) (+) =ABCD 00 01 11 1000011110( a )1 0 0 1 0 0 0 00 0 1 0 1 0 1 1(2) F=m (0,2,8,10,14,15) ( b ) 图P4.A18 ( 2 )=D+B+D+BF= (A+)(+C)B+)A+) =(3) F= A+C+CDABCD 00 01 11 100001111

20、0( a )0 0 0 1 0 0 0 11 1 0 1 1 1 0 1 ( b ) 图P4.A18 ( 3 )=+ABF= (A+C) +) =(4) F= AB+C+CABC 00 01 11 1001( a )0 0 1 01 1 1 1 ( b )图P4.A18 ( 4 )=+F = (A+C) (B+C) = 19. 将下列各函数化简为与或非表达式,并用与或非门实现.(1) F = A+C+C+A+B+D(2) F=m (1,2,6,7,8,9,10,13,14,15)(3) F=m (0,1,3,7,8,9,13,15,17,19,23,24,25,28,30)解: 圈“0”格化简

21、(1) F = A+C+C+A+B+DABCD 00 01 11 1000011110( a )0 1 1 1 1 1 1 11 1 0 1 1 1 1 1 ( b ) 图P4.A19 ( 1 )=+ABCDF=(2) F=m (1,2,6,7,8,9,10,13,14,15)ABCD 00 01 11 1000011110 ( a )0 0 0 1 1 0 1 10 1 1 0 1 1 1 1 ( b ) 图P4.A19 ( 2 )=CD+B+B+F=(4) F=m (0,1,3,7,8,9,13,15,17,19,23,24,25,28,30)ABCD 000 001 011 010 11

22、0 111 101 10000011110( a )1 0 1 1 1 1 0 01 0 1 1 1 0 0 11 1 1 0 0 0 1 10 0 0 0 0 1 1 1 ( b ) 图P4.A19 ( 3 )=C+A+BD+D+C+ABCEF = 20.用卡诺图将下列含有无关项的逻辑函数化简为最简“与或”式和最简“或与”式。(1) F = m (0,1,5,7,8,11,14)+ m (3,9,15)(2) F = m (1,2,5,6,10,11,12,15)+ m (3,7,8,14)(3) F = AB+A+C+AC,变量A,B,C,D不可能出现相同的取值.(4) F=+ABC+C,

23、约束条件AB = 0解:化简为最简“与或”式圈“1”格, 化简为最简“或与”式圈“0”格(1) F = m (0,1,5,7,8,11,14)+ m (3,9,15)ABCD 00 01 11 1000011110( a )1 1 1 1 xx 1 x 1 1 ABCD 00 01 11 1000011110( b ) 0 0 0 x x 0 0 0 图P4.A20 ( 1 )F = +D+CD+ABC = C+C+AB+BF = (A+D) B+D) +C) ( +C+D)(2) F=m (1,2,5,6,10,11,12,15) +m (3,7,8,14)ABCD 00 01 11 100

24、0011110( a ) 1 x 1 1 x x 1 1 1 1 x 1ABCD 00 01 11 1000011110( b )0 0 x 0 0x x 图P4.A20 ( 2 )F = C+D+A =+ADF= (A+C+D) +C+)ABCD 00 01 11 1000011110 ( a )x 1 1 1 1 x 1 1ABCD 00 01 11 1000011110( b )x 0 0 0 0 0 x 0 0 0 (3) F= AB+A+C+AC,变量A,B,C,D不可能出现相同的取值.图P4.A20 ( 3 )F= A+=+CD+BCF= (A+C)(+)(+)(4) F=+ABC

25、+C,约束条件AB = 0ABCD 00 01 11 1000011110( a )1 x x 1 x x x 1 x 1 x 1 1ABCD 00 01 11 1000011110( b ) x 0 x x 0 x0 x x x x 图P4.A20 ( 4 )F= AC+= B+CDF= (+C) A+)21. 在输入只有原变量条件下，用最少与非门实现下列函数。(1) F= A+B+C(2) F=m (1,3,4,5,6,7,9,12,13)(3) F=m (1,2,4,5,10,12)(4) F=m (1,5,6,7,9,11,12,13,14)解:可利用禁止原理(1) F= A+B+CA

26、BC 00 01 11 10010 1 1 11 1 1 1 ( a ) ( b )图P4.A21 ( 1 )F = A+B+C= （2） F=m (1,3,4,5,6,7,9,12,13)ABCD 00 01 11 1000011110( a )0 1 1 0 1 1 1 11 1 0 0 1 1 0 0 ( b ) 图P4.A21 ( 2 )F= B+D=(3) F=m (1,2,4,5,10,12)ABCD 00 01 11 1000011110( a )0 1 1 0 1 1 0 00 0 0 0 1 0 0 1 ( b )图P4.A21 ( 3 )F= B）+C）+D（ = B+C+

27、D =(4) F=m (1,5,6,7,9,11,12,13,14)ABCD 00 01 11 10000111100 0 1 0 1 1 1 10 1 0 1 0 1 1 0 ( a ) ( b )图P4.A21 ( 4 )F= AB+BC+AD+D = AB+D+BC+AD =22. 输入只有原变量条件下,用或非门实现下列函数(1) F=m (0,6,7)(2) F=m (0,1,2,3,4,6,7,8,9,11,15)(3) F=m (0,4,5,7,11,12,13,15)解:输入只有原变量条件下用或非门实现逻辑函数时,应先求出F的对偶式F,将F化为与非一与非表达式,再求一次对偶F=F

28、,即可得出F的或非一或非表达式.(1)F=m (0,6,7)=m (1,2,3,4,5)F=m (6,5,4,3,2) ABC 00 01 11 1001图P4.A22 ( 1 ) ( a )0 1 1 10 1 0 1F= A+B=F= F=图P4.A22 ( 1 ) ( b )(2)F=m (0,1,2,3,4,6,7,8,9,11,15)=m (5,10,12,13,14)F=m (10,5,3,2,1)ABCD 00 01 11 1000011110图P4.A22 ( 2 ) ( a )0 0 0 0 1 1 0 01 0 0 0 1 0 0 1F= D+C= D+C= F= F= +图P4.