(2)Y2=(A+C+D) ′+A′B′CD′+AB′C′D,给定约束条件为: AB′CD′+AB′CD+ABC′D′+ABC′D+ABCD′+ABCD=0

解: Y2=A′C′D′(B+B′)+A′B′CD′+AB′C′D=A′BC′D′+A′B′C′D′+ A′B′CD′+AB′C′D

=m0+m2+m4+m9. 约束条件:

AB′CD′+AB′CD+ABC′D′+ABC′D+ABCD′+ABCD=0,即 d10+d11+d12+d13+d14+d15=0. 所以Y2可表示为:

Y2=∑m(0,2,4,9)+d(10,11,12,13,14,15)

用卡诺图法化简,得:Y2=AD+A′C′D′+A′B′D′.或Y2=AD+A′C′D′+B′CD′(见虚红线圈, 实红线圈去掉).

CD AB 00 01 11 10 1 1 00 01 1 11 × × × × 10 1 × ×

题2.23 将下列具有无关项的逻辑函数化为最简与或形式. (3) Y3(A,B,C,D)= ∑m(3,5,6,7,10)+d(0,1,2,4,8) 解:

CD AB 00 01 11 10

00 1 × × × 01 1 1 1 × 11 10 × 1

Y3=A′+B′D′

(4) Y4(A,B,C,D)= ∑m(2,3,7,8,11,14)+d(0,5,10,15) 解: CD AB 00 01 11 10

00 1 1 ×

01 1 ×

11 1 ×

10 1 1 ×

Y4= B′D′+CD+AC

附:1.

5

2.

6