问答题
将下列逻辑函数式化为或非—或非形式。
(1) Y1(A,B,C)=A'B'C+A'BC'+AB'C'+ABC
(2) Y2(T,U,V)=TUV'+UV+U'V'
(3) Y3(A,B,C,D)=∑m(2,3,4,5,10,11,12,13)
(4) Y4(M,N,p,Q)=∑m(0,1,4,6,8,9,12,14)
【正确答案】(1) Y1(A,B,C)=A'B'C+A'BC'+AB'C'+ABC
=m1 +m2 +m4 +m7
=(m0 +m3 +m5 +m6)'
=(A'B'C'+A'BC+AB'C+ABC')'
=((A+B+C)'+(A+B'+C')'+(A'+B+C')'
+(A'+B'+C)')'
(2) Y2(T,U,V)=TUV'+UV+U'V'
=TUV'+T'UV+TUV+T'U'V'+TU'V'
=m0 +m3 +m4 +m6 +m7
=(m1 +m2 +m5)'
=(T'U'V+T'UV'+TU'V)'
=((丁+U+V')'+(T+U'+V)'+(T'+U+V')')'
如果要求得到最简的或非—或非形式,则可将上式化简为
Y2(T,U,V)=(U'V+T'UV')'
=((U+V')'+(T+U'+V)')'
(3) Y3(A,B,C,D)=∑m(2,3,4,5,10,11,12,13)
=(∑m(0,1,6,7,8,9,14,15))'
=(A'B'C'D'+A'B'C'D+A'BCD'+A'BCD
+AB'C'D'+AB'C'D+ABCD'+ABCD)'
=((A+B+C+D)'+(A+B+C+D')'
+(A+B'+C'+D)'+(A+B'+C'+D')'
+(A'+B+C+D)'+(A'+B+C+D')'
+(A'+B'+C'+D)'+(A'+B'+C'+D')')'
如果要求得到最简的或非—或非形式,则可将上式化简为
Y3(A,B,C,D)=(B'C'+BC)'
=((B+C)'+(B'+C')')'
(4) Y4(M,N,P,Q)=∑m(0,1,4,6,8。9,12,14)
=(∑m(2,3,5,7,10,11,13,15))'
=(M'N'PQ'+M'N'PQ+M'NP'Q+M'NPQ
+MN'PQ'+MN'PQ+MNP'Q+MNPQ)'
=((M+N+P'+Q)'+(M+N+P'+Q')'
+(M+N'+P+Q')'+(M+N'+P'+Q')
+(M'+N+P'+Q)'+(M'+N+P'+Q')'
+(M'+N'+P+Q')'+(M'+N'+P'+Q')')'
如果要求得到最简的或非—或非形式,则可将上式化简为
Y4(M,N,P,Q)=(N'P+NQ)'
=((N+P')'+(N'+Q')')'
【答案解析】