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用卡诺图法化简函数

画卡诺图 填最小项 圈图,写答案F=BD+CD+A'C'D'

Y的标上1 红色; 约束条件标上x,绿色 再圈图 蓝色,写出最简与或式Y=AD+A'C'D'+A'B'D' ===================================== Y'=A'D+BCD'+AD' Y=(A+D')(B'+C'+D)(A'+D) 验证: =(AB'+AC'+AD+B'D'+C'D'+0)(A'+D) =0+0+0+A'B'D'+A'C'D'+0+AB'D+...

Y(A,B,C,D)=∑m(0,1,4,5,7,8,13,15) =∑m(0,1)+∑m(0,4)+∑m(0,8)+∑m(5,7,13,15) =A'B'C'+A'C'D'+B'C'D'+BD

F=AC'+B'C Y=A'B'+AC+AD

与单输出的有点不同,请阅读http://www.doc88.com/p-950219968373.html

f=cd+c'd'

F=A’B’C’+AB’CD+AB’+AD’+AB’C =∑m(0,1)+m11+∑m(8,9,10,11)+ ∑m(8,10,12,14)+ ∑m(10,11) =∑m(0,1,8,9,10,11,12,14) =∑m(0,1,8,9)+ ∑m(8,9,10,11)+ ∑m(8,10,12,14) =B’C’+A’B’+AD’

Y=AB'+AD+BC+CD'

y=b'cd+bc'+a'c'd+ab'd 每一项4元变量都对应ABCD,1或0,有非 ' 号的为0,缺少对元的用x代替,表示0或1两项 那么以上四项分别对应:x011 x10x 0x01 10x1 x011, x=1,即1011,十进制即11;x=0,即0011,十进制即是3; x10x有2个x,则对应4个最小项:...

Y=ABC+ABD+C’D’+AB’C+ACD’ =∑m(0,4,8,10,11,12,13,14,15) =∑m(0,4,8,12)+∑m(12,13,14,15)+∑m(10,11,14,15) =C’D’+AB+AC Y=ABC+ABD+C’D’+AB’C+ACD’ =AB(C’D’)’+C’D’+(ABC+AB’C)+ACD’ =AB+C’D’+AC+ACD’ =AB+C’D’+AC(1+D’) =AB+C’D’+AC

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