( b )
图P4.A19 ( 1 )
= +ABCD
F=
(2) F=∑m (1,2,6,7,8,9,10,13,14,15)
( 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)
( b )
图P4.A19 ( 3 )
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) BC+A C+AB = BC +AC +AB
(3) =ABC+
(4) AB+BC+AC=(A+B)(B+C)(A+C)
(7) F=∑m (0,2,4,5,7,9,13,14,15,16,18,20,21,23,25,29,30,31)
F= ACE+B E+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 (0,1,3,4,6,7,10,11,13,14,15)
(2) F=∑m (0,2,3,4,5,6,7,12,14,15)
= ( +AB)(A+B)
= AB+AB
= AB=∑m (3)
=ΠM(0,1,2)
(2)F= (A⊕B)+ ( C+B )
= B+A + C+ B
= B+A + 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+AB +ABC+BC+B
解:
F= A(1+B +BC)+B(C+1) = A+B
(2) F= A C+ 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+ C
F``= F= (A+B)( +C)
(4)F=
解:
F= AB+ +BC+
= AB+ C+
a)F=
解:
F= C+AC
= + +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)]
= 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)
= (A⊙B)C+ ( )
= C( + D+AD)+BD(AC+C+ )+B (D+ + )
= C+B +BD
(3) + + =1
证:
左= ( + D) + ( )+(C+ )
= [( + )( + )+ D]( + )+C+
= [ + + + + D][ + ]+C+
= [ + + D][Hale Waihona Puke + ]+C+
= + + + D+C+
(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)
(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)
(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)
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。
(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)
( b )
图P4.A17 (1)
F= AC+BC+ D+ +ABD =
(2) F=∑m (0,2,3,4,5,6,7,12,14,15)
= A⊙B⊙C
(6) = ⊙ ⊙
证:
左=
= [(A⊕B)+ ] (A⊙B)+C]
= (A⊙B) +[(A⊕B)C]
= +AB + BC+A C
右= ( ⊙ )⊙
= [( ⊙ ) + ]
= [( +AB) + ]
= +AB +
= +AB +(A⊕B)C
= +AB + BC+A C
9.证明
(1)如果a + b = c，则a + c = b，反之亦成立
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 +B D+ACD+ B + CD+B +BCD= C+B +BD
证:
左= ( C + CD+ACD )+(ABCD+BCD+B D)+(B D+B + B )
(5) ABC+ + + =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 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]]
①③
= +A
F= (A+B) ( +C)
⑤F= (AC+ C)( +AC+ )
= A C+ C+AC
F=AC+ C
图P4.A16 ( 1 )
(2) F=∑m (0,1,3,5,6,8,10,15)
F= + D+ D
+A +ABCD+ BC
(3) F=∑m (4,5,6,8,9,10,13,14,15)
= C +A +B D+ D + C +ABCE
F =
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。