第四章 逻辑函数及其符号简化
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 )
F=A B C+A B C +AB C +ABC F= (A+B+C) ( A +B +C )
F=A BC+AB C+ABC +ABC ,F 的值为“1”: (3) F= (A+B+C) (A+B+C ) (A+B +C) (A+B +C )
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=1
3. 用真值表证明下列等式.
(1) A+BC = (A+B) (A+C)
(2) A BC+AB C+ABC = BCABC +ACABC +ABABC
(3) A B +B C +A C =ABC+A B C
(4) AB+BC+AC=(A+B)(B+C)(A+C)
(5) ABC+A +B +C =1
证:
( 1 )
( 2 )
( 3 )
( 4 )
( 5 )
4. 直接写出下列函数的对偶式F ′及反演式F 的函数表达式.
(1) F= [A B (C+D)][BC D +B (C +D)]
(2) F= ABC + (A +B C ) (A+C)
(3) F= AB+CD +BC +D +C E +D +E
(4) F=C +A B •A B +D
解:
(1) F`= [A +B+CD]+[(B+C +D ) •(B+C D]]
F = [A+B +C D ]+[(B +C+D)•(B +CD ]]
(2) F`= (A+B +C ) •[A •(B +C ) AC ]
F = (A +B +C ) •[A •(B +C ) +A C ]
(3) F`=C •(A +B ) +(A +B ) •D
F =C •(A +B ) +(A +B ) •D
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 = 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) A C +A B +BC+A C D =A +BC
证:
左=A BC + BC +A C D
=A + BC +A C D =A (1+C D ) + BC
=A +BC = 右边
(2) B C D +BC D+ACD+A B C D +A B CD+BC D +BCD=B C+BC +BD 证:
左 = (B C D +A B CD+ACDB )+(ABCD+BCD+BC D)+(BC D+BC D +A B C D ) =B C(D +A D+AD)+BD(AC+C+C )+BC (D+D +A D )
=B C+BC +BD
(3) AB ∙B +D ∙C D +BC +A B D +A +C D =1
证:
左 = (AB •B +D +C D) •BC + A (B D •A )+(C+D )
= [(A +B )(B +D )+C D]( B +C )+C+D
= [A B +B +A D +B D +C D][ B +C ]+C+D = [B +A D +C D][ B +C ]+C+D =B +B C +A D C +C D+C+D
=B +C +C+D =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)] = wy+[(x+uv)(x+z)] = wy+[x+xuv+xz+uvz] = wy+[x+uvz] = wy+x+uvz (5) A ⊕B ⊕C=A⊙B ⊙C
证:
左 = (A⊕B) ⊕C
=A ⊕B •C + (A⊕B) C
= (A⊙B)C+ (A B ) C
= A⊙B ⊙C
(6) A ⊕B ⊕C =A ⊙B ⊙C 证: 左 =A ⊕B •C +(A ⊕B ) C
= [(A⊕B)+C ]•[(A⊙B)+C]
= (A⊙B) C +[(A⊕B)C]
=A B C +ABC +A BC+AB C
右 = (A ⊙B ) ⊙C
= [(A ⊙B ) •C +A B •C ] = [(A B +AB) C +A B +AB •C ]
9. 证明 =A B C +ABC +A B C =A B C +ABC +(A⊕B)C =A B C +ABC +A BC+AB C
(1) 如果a b +a b = c,则a c +a c = b,反之亦成立
(2) 如果a b +ab = 0,则 ax +by = ax +by
证:
(1) a c + a c = a (a b +a b )+a (ab +a b)
= a (ab+a b )+a b
= ab+a b = b
(2) a b +ab = 0 说明a =b 或b =a
ax +by =ax +a y =ax •a y = (a +x )(a+y ) = ax +a y +x y = ax +a y
= ax +by
10. 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+BC D (2)F= AB +A B+BC
(3)F= AB +C +BD +A D +B +C
解:
(1) F=∑m (4, 11, 12, 15)
F =∑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)
F =∑m (0,1,6) F`=∑m (7,6,1)
(3) F= ∑m (1,5,6,7,8,913,14,15)
F = ∑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= (A⊙B)(A+B)+(A⊙B)AB
(2) F= (A⊕B)+A (B⊕C)
解:
•((1) F= (A⊙B) A+B+AB)
= (A B +AB)(A+B)
= AB+AB
= AB=∑m (3)
=ΠM (0,1,2)
(2) F= (A⊕B)+A (B C+BC )
= A B+AB +A B C+A B C
= A B+AB +A B C
= ∑m (1,2,3,4,5)
=ΠM (0,6,7)
12. 用公式法化简下列各式
(1) F= A+ABC +ABC+BC+B
解:
F= A(1+BC +BC)+B(C+1) = A+B
(2) F= AB C+A C D+AC
解:
F=AB +AC +C D
(3) F= (A+B)(A+B+C)(A +C)(B+C+D)
解:
F`= AB+ABC+A C+BCD = AB+A C+BCD = AB+A C
F``= F= (A+B)(A +C)
(4)
解: F=AB +A B •BC +B C
F= AB+A B +BC+B C
= AB+A C+B C
a) F=AB +B C +B (A C +A C )
解:
F=B C+AC
(5) F= (x+y+z+w ) (v+x) (v +y+z+w )
解:
F`= xyzw +vx+v yz w
= vx+v yz w +xyzw
= vx+v yz w
F``= F= (v+x) (v +y+z+w )
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= 0
14. 指出下列函数在什么组合时使F=1
(1) F=ΠM (4,5,6,7,8,9,12)
(2) F=ΠM (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=1
15. 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)
⑤F= (AC+B C)( B +AC+A C )
= A
B F=AC+B C
图P4.A16 ( 1 )
(2) F=
F=A B C +A B D+A C D
+AB D +ABCD+A BC D
(3) F=∑
(3) F=ΠM (0,3,6,7) (4) F=ΠM (0,1,2,3,4,6,12) 解: (1) F=ΠM (0,2,4,5,6) (2) F=ΠM (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 (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=ΠM (5,7,13,15) (5) F=ΠM (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)
F=A B C +AB C +ABD
+BCD + ACD
(4) F=
(5) F=
(6) F=
F=A B (7) F=
17. (1) F=∑(2) F=∑m (0,2,3,4,5,6,7,12,14,15) B C E
(3) F=∑m (0,1,4,5,12,13)
(4) F=ΠM (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)
F= AC+BC+D++ABD =
(2)
F=∑m (0,2,3,4,5,6,7,12,14,15)
( b ) 图P4.A17 ( 2 )
F=A C+BC+A D +BD +A B =A C •BC •A D •B D •AB
(3) F=∑m (0,1,4,5,12,13)
( b ) F=A 图P4.A17 ( 3 )
(4) F=
( b )
图P4.A17 ( 4 ) F = A B +ABD+ABC+B C D = A B •ABD •ABC •B C D
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 (0,2,8,10,14,15)
(3) F= AB +A C+B CD
(4) F= AB+B C+A C
解: 圈“0”格化简
(1) F=∑m (0,1,2,4,5)
( b ) F = AB+BC
F = (
A +B ) (B +C ) =A +B +B +C
(2) F=
( b )
图P4.A18 ( 2 ) F =A F= (A+D )(B +C)•(B+D ) •(A+B ) =A +D +B +C +B +D +A +B
(3) F= AB +A C+B CD
( b )
图P4.A18 ( 3 )
F =A C +AB F= (A+C) (A +B ) =A +C +A +B (4) F= AB+B C+A C
( b )
F =A C +B C
F = (A+C) (B+C) = 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 = AB +CD +A C+AC +BC +A 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
(1) F = A
B +CD +A C+AC +BC +A D
( b )
P4.A19 ( 1 )
F =A B F=A B C D +ABCD
(2) F=∑m (1,2,6,7,8,9,10,13,14,15)
( b )
P4.A19 ( 2 )
F =B CD+A B C +BC D +A C D F=B CD +A B C +B CD +A C D
(4) F=∑m (0,1,3,7,8,9,13,15,17,19,23,24,25,28,30)
( b )
=C +A+BD+D +C +ABCE
F = A C E +A B E +B C D +A D E +B C D +ABCE
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 = ABC +AB C +A B C D +AB C D , 变量A,B,C,D 不可能出现相同的取值. (4) F=A B C +ABC+A B C D , 约束条件A ⊕B = 0
解:化简为最简“与或”式 圈“1”格, 化简为最简“或与”式圈“0”格
F = F D
F = (A+C +D) •(B+C +D) •(A +B +C) ( B +C+D) (2) F=∑m (1,2,5,6,10,11,12,15) +∑m (3,7,8,14)
D +AC
(3) F= ABC +AB C +A B C D +AB C D , 变量A,B,C,D 不可能出现相同的取值.
图
(4) F=A B C +ABC+A B C D , 约束条件A ⊕B = 0
图
F= (B +C) •(A+C +D )
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= AB +BC +A 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= AB +BC +A C
( a ) ( b )
图P4.A21 ( 1 ) F = A•ABCD +B•ABCD +C•ABCD = A •ABCD •B •ABCD •C •ABCD (2) F
( b )
图P4.A21 ( 2 ) F= B
( b )
图P4.A21 ( 3 )
F= B•(AD +BC )+C•(BC +CD )+D(AD +CD
= B•BC •AD +CBC •CD +D•AD •CD =B •BC •AD •C BC •CD •D •AD •CD
(4) F=∑m (1,5,6,7,9,11,12,13,14)
( a ) ( b )
图P4.A21 ( 4 )
F= AB•ABCD +BC•ABCD +AD•ABCD +•C D
= AB∙ABCD +D•CD +BC•ABCD +AD•ABCD
=AB ∙ABCD ∙D ∙CD ∙BC ∙ABCD ∙AD ∙ABCD 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,即可得出F 的或非一或非表达式.
(1) F=∑m (0,6,7)
F =∑m (1,2,3,4,5) F`=∑m (6,5,4,3,2)
F ′= A•ABC +B•ABC =
F ″= F=A +A +B +C +B +A +B +C
图P4.A22 ( 1 ) ( b )
(2)F=∑m (0,1,2,3,4,6,7,8,9,11,15)
F =∑m (5,10,12,13,14)
F ′=∑m (10,5,3,2,1)
F ′
=
F ″ = F= D +A +D +B +C +C +A +D +B +C
图P4.A22 ( 2 ) ( b )
(3) F=∑m (0,4,5,7,11,12,13,15)
F =∑m (1,2,3,6,8,9,10,14)
F ′=∑m (14,13,12,9,7,6,5,1)
=
F ″= F=B +C +A +C +D +A +B +A +C +D +D +C +D
( b )
图P4.A22 ( 3 ) ( b )
23. 化简下列函数, 并用与非门组成电路.
F 1=AC +B C +A BC
(1) F 2=AC+A B C F 3=A+BC F 1=∑m (1,3,5,7,8,9,12,14) F 2=∑m (1,3,8,12,14) F 3=∑m (5,7,9,14)
(3) F 1=∑m (0,1,2,4,5,6,8,10,14,15)+ ∑m (3,7,11)
F 2=∑m (0,1,2,4,5,8,9,10,12,13,15)+ ∑m (3,7,11)
F 1 F 2 = A B C +AC=A B C •AC F 3 = A
BC+AC +AC=A BC •A C •AC
图P4.A23 ( 1 ) ( b )
图P4.A23 ( 2 ) ( a ) F 1 =A B D+A BD+AC D +AB C D+ABCD
=A B D •A BD •A C D •A B C D •ABC D
F 2 =A B D+AC D +ABCD =A B D •A C D •ABC D F 3 =A BD+AB C D+ABCD =A BD •A B C D •ABC D
图P4.A23 ( 2 ) ( b )
( 3 )
F 1=A F 2=C +D+B =C •D •B
图P4.A23 ( 3 ) ( b )