1第二章逻辑代数习题解答2.1.1用真值表证明下列恒等式(3)ABABAB(A⊕B)=AB+AB解:真值表如下ABABABABABAB+AB0001011011000010100001100111由最右边2栏可知,AB与AB+AB的真值表完全相同。2.1.3用逻辑代数定律证明下列等式(3)()AABCACDCDEACDE解:()AABCACDCDE(1)ABCACDCDEAACDCDEACDCDEACDE2.1.4用代数法化简下列各式(3)()ABCBC解:()ABCBC()()ABCBCABACBBBCCBC(1)ABCABBABC(6)()()()()ABABABAB解:()()()()ABABABAB()()ABABABAB2BABABABBABAB(9)ABCDABDBCDABCBDBC解:ABCDABDBCDABCBDBC()()()()()ABCDDABDBCDCBACADCDBACADBACDABBCBD2.1.7画出实现下列逻辑表达式的逻辑电路图,限使用非门和二输入与非门(1)LABAC(2)()LDAC3(3)()()LABCD2.2.2已知函数L(A,B,C,D)的卡诺图如图所示,试写出函数L的最简与或表达式解:(,,,)LABCDBCDBCDBCDABD2.2.3用卡诺图化简下列个式(1)ABCDABCDABADABC4解:ABCDABCDABADABC()()()()()ABCDABCDABCCDDADBBCCABCDDABCDABCDABCDABCDABCDABCDABCD(6)(,,,)(0,2,4,6,9,13)(1,3,5,7,11,15)LABCDmd解:LAD(7)(,,,)(0,13,14,15)(1,2,3,9,10,11)LABCDmd解:LADACAB52.2.4已知逻辑函数LABBCCA,试用真值表,卡诺图和逻辑图(限用非门和与非门)表示解:1由逻辑函数写出真值表ABCL000000110101011110011011110111102由真值表画出卡诺图3由卡诺图,得逻辑表达式LABBCAC用摩根定理将与或化为与非表达式LABBCACABBCAC4由已知函数的与非-与非表达式画出逻辑图