ABCD001011011101110100100100
00011110F7(A,B,C,D)=(A+B+C+D)(B+C)(A+C)(A+B+D)
（8）
F8(A,B,C,D)=000100011011110101?M(0,3,5,6,8,10,12,15)
101010
ABCD00011110F8(A,B,C,D)=(B+C+D)(A+C+D)(A+B+D)(A+B+C+D)=(A+B+C+D)(A+B+C+D)(A+B+C+D) 1-18
（1）F1=ABCD+ABCD+ABCD 约束条件 AB+AC=0
ABCD001001001100101000011110××××10××
F1=BD
（2）F2=ABD+ABD+BCD 约束条件 AB+AC=0
ABCD0001111000100101011011××××1010××
F2=BD+BD
（3）F3=BCD+BCD+ABCD ABCD000111100001×00111××1110××1001×0
F3=BD+AD+BD
（4）F4=ABCD+BCD+ABCD ABCD0001111000×11×011××0111××010×00×
F4?BC?AD
（5）F5=AC+BD+ABC+ABCD 约束条件BC+CD=0
约束条件BD+BD=0
约束条件ABCD+ABCD=0
ABCD001110111001110011000011110×111×
F5=D+BC+AC
（6）F6=AB+BC+CD 约束条件
?d(0,1,2,6)=0
ABCD00011111111000011110××110110××10
F6=B+CD
1-19
（1）F1(A,B,C,D)=邋m(0,1,3,5,10,15)+10d(2,4,9,11,14)
ABCD001011101110100011110×0×00×1
××F1(A,B,C,D)=AB+AC+AC
（2）F2(A,B,C,D)=邋m(0,1,5,7,8,11,14)+d(3,9,15)
ABCD0010010111011100010
00011110×1×1×F2(A,B,C,D)=BC+AD+CD+ABC
（3）F3(A,B,C,D)=邋m(2,6,9,10,13)+101101
d(0,1,4,5,8,11)
ABCD00011100000011110××××011××F3(A,B,C,D)=AD+CD+AB
（4）F4(A,B,C,D)=邋m(1,3,7,11,13)+1000d(5,9,10,12,14,15)
ABCD0000011111100011110×1×0××1××
F4(A,B,C,D)=D
（5）F5(A,B,C,D)=邋m(2,4,6,7,12,15)+d(0,1,3,8,9,11)
ABCD000111101100
00011110×××110011×××F5(A,B,C,D)=CD+CD+AD
（6）F6(A,B,C,D)=邋m(2,3,6,10,11,14)+101111
d(0,1,4,9,12,13)
ABCD000111100100011110×××00×××F6(A,B,C,D)=CD+BC
（7）F7(A,B,C,D)=邋m(3,5,6,7,10)+10d(0,1,2,4,8,15)
ABCD0001111100011110×××0100×101
×0×F7(A,B,C,D)=A+BD
（8）F8(A,B,C,D)=邋m(0,4,8,11,12,15)+d(2,3,6,7,13)
数字电路与逻辑设计课后习题答案蔡良伟(第三版)
