数字电路第二章例题

第二章例题信电学院马草原例化简函数CBACBAY解：BACCBACBACBAY)(例化简函数解：CBACBACBACBAYAABBACCABCCBAY)()(AABBAY或：并项例化简函数解：例化简函数解：)(FECDBABAYBAFECDBABAY)()(EFFEDABCDCDABYDCDABEFFEDABCDCDABY)(吸收例化简函数解：例化简函数解：CBCAABYCABCABABCBAABCBCAABY)(FEFEABCDY)(FEABCDFEABCDFEFEABCDFEFEABCDY)()(消去例化简函数解：BACBCBBAYCACBBABBCACBBABCACBACBACBACBBACCBACBAACBBABACBCBBAY)()()(配项解2：BACBCABACBBBCABACBCABCBACBACBABACBCBAACCBABACBCBBAY)()()(例化简函数BACBCBBAY配项例化简函数解：CACBBACABACBBACABACBCBBABACBCBBAYBACBCBBAY添加项例将Y=AB+BC展开成最小项表达式。解：BCAABCCABBCAACCABBCABY)()(例已知Y的真值表，要求画Y的卡诺图。表逻辑函数Y的真值表ABCY00000011010101101001101011001111例画出函数Y(A、B、C、D)=∑m(0,3,5,7,9,12,15)的卡诺图。1111AB＝11例已知Y＝AB＋ACD＋ABCD，画卡诺图。最后将剩下的填01+1ACD=1011ABCD=0111例用卡诺图化简逻辑函数Y(A、B、C、D)=∑m(0,1,2,3,4,5,6,7,8,10,11)不正确A例用卡诺图化简逻辑函数Y(A、B、C、D)=∑m(0,1,2,3,4,5,6,7,8,10,11)相邻BCA例化简图示逻辑函数。解：多余的圈ABCDCACBACDAY11223344例设ABCD是十进制数X的二进制编码，当X≥5时输出Y为1，求Y的最简与或表达式。XABCDY000000100010200100300110401000501011601101701111810001910011/1010×/1011×/1100×/1101×/1110×/1111×解：列真值表。画卡诺图并化简。充分利用无关项化简后得到的结果要简单得多。注意：当圈组后，圈内的无关项已自动取值为1，而圈外无关项自动取值为0。利用无关项化简结果为：Y＝A＋BD＋BC例化简逻辑函数Y(A、B、C、D)=∑m(1,2,5,6,9)+∑d(10,11,12,13,14,15)式中d表示无关项。卡诺图解：画函数的卡诺图并化简。结果为：Y＝CD＋CD=C⊕D2.3(3)：F＝(A+BC)(A+DE)F＝A(B+C)+A(D+E)F＝ABC+(A+BC)(A+C)F＝(A+B+C)·(A·(B+C)+AC)2.3(7)：2.4(1)：F＝ABC+A+B+CF＝ABC+ABC+ABC+B+C＝BC+ABC+B+C＝BC+ABC+BC＝BC+BC＝1111111110001111001BCA0001111001BCA2.4(4)：111001100001111001BCA0001111001BCAF＝AB+(AB+AB+AB)CF＝ABC+ABC+ABC+ABC+ABC＝ABC+(AB+AB+AB+AB)C＝ABC+(B+B)C＝ABC+C2.4(7)：F＝(A+BC)(A+DE)F＝(A+BC)(A+DE)＝ADE+ABC+BCDE＝ADE+ABC＝ADE·ABC＝(A+D+E)(A+B+C)2.4(10)：F＝XY+(X+Y)ZF＝XY+(X+Y)Z＝XY+(XY)Z＝XY+ZF＝AB+AB+CDF＝AB+AB+CD＝AB(C+C)(D+D)+AB(C+C)(D+D)+(A+A)(B+B)CD＝ABCD+ABCD+ABCD+ABCD+ABCD+ABCD+ABCD+ABCD+ABCD+ABCD+ABCD+ABCD)12,10,6,3,2,1,0(mS,13,14,15F(A,B,C)=Sm(0,1,2,4,5,7)2.10(1)化简ABC000111100111011110F＝B+AC+ACF(A,B,C,D)=Sm(0,2,3,5,7,8,10,11)+Sd(14,15)2.10(2)化简00001111000011110CDAB0dd101110110110F＝000+BC+BD+ABDF(A,B,C,D)=Sm(4,5,6,7,9,12)2.10(3)化简10001111000011110CDAB000010000001111F＝AB+BCD+ABCDF(A,B,C,D)=Sm(0,1,2,3,4,5,6,7,16,17,20,21)+Sd(24,25,27,28,30)2.10(4)化简00000101101000011110CDEAB11011110110011000011d00d0ddd0000000011111111F(A,B,C,D)=Sm(0,1,2,5,6,7,8,9,13,14)2.10(5)化简F＝AC+BC+BD+ABD0001111000011110CDAB1111111111F(A,B,C,D)=Sm(0,1,5,7,8,11,14)+Sd(3,9,15)2.10(6)化简0001111000011110CDAB1111111ddd