课后习题答案-第2章-逻辑代数及其化简

1第2章逻辑代数及其化简2-1分别将十进制数29.625，127.175和378.425转换成二进制数。解答：(29.625)10=(1,1101.101)2(127.175)10=(111,1111.0010,1100,…)2(378.425)10=(1,0111,1010.0110,1100,…)22-2分别将二进制数101101.11010111和101011.101101转换成十进制数。解答：(101101.11010111)2=(45.83984375)10(101011.101101)2=(43.703125)102-3分别将二进制数100110.100111和101011101.1100111转换成十六进制数。解答：(100110.100111)2=(0010,0110.1001,1100)2=(26.9C)16(101011101.1100111)2=(1,0101,1101.1100,1110)2=(15D.CE)162-4分别将十六进制数3AD.6EBH和6C2B.4A7H转换成二进制数。解答：(3AD.6EB)16=(11,1010,1101.0110,1110,1011)2(6C2B.4A7)16=(110,1100,0010,1011.0100,1010,0111)22-5试用真值表法证明下列逻辑等式：(1)ABACBCABC++=+(2)ABABBCABABAC++=++(3)ABBCCAABBCCA++=++(4)ABABBCACABC+++=+(5)ABBCCDDAABCDABCD+++=+(6)ABABABCAB++=+证明：(1)ABACBCABC2真值表如下所示：ABCABACBCABC0000000111010000111110000101111101111111由真值表可知，逻辑等式成立。(2)ABABBCABABAC真值表如下所示：ABCABABBCABABAC0000000100010110111110011101111100011111由真值表可知，逻辑等式成立。(3)ABBCCAABBCCA真值表如下所示：ABCABBCCAABBCCA00000001110101101111310011101111101111100由真值表可知，逻辑等式成立。(4)ABABBCACABC真值表如下所示：ABCABABBCACABC0001100111010110111110000101001100011111由真值表可知，逻辑等式成立。(5)ABBCCDDAABCDABCD真值表如下所示：ABCDABBCCDDAABCDABCD0000110001000010000011000100000101000110004011100100000100100101000101100110000110100111000111111由真值表可知，逻辑等式成立。(6)ABABABCAB真值表如下所示：ABCABABABCAB0001100111010110111110011101111100011100由真值表可知，逻辑等式成立。2-6求下列各逻辑函数F的反函数F和对偶式F¢:(1)1FAABCAC=++(2)2()()()FABAABCABCABABC=++++++(3)3FABCDADB=+++(4)4FABBDCABBD=+++++5(5)()()5FABABBCBC=++(6)6FCDCDACDB=+++解答：(1)1FAABCAC1()()FAABCAC1'()()FAABCAC(2)2()()()FABAABCABCABABC=++++++2()()()FABAABCABCABABC2'()()()FABAABCABCABABC(3)3FABCDADB=+++3FABCDADB3'FABCDADB(4)4FABBDCABBD=+++++4()()()FABBDCABBD4'()()()FABBDCABBD(5)()()5FABABBCBC=++5()()()()FABABBCBC5'()()()()FABABBCBC(6)6FCDCDACDB=+++6()()()()FCDCDACDB6'()()()()FCDCDACDB62-7某逻辑电路有A、B、C共3个输入端，一个输出端F，当输入信号中有奇数个1时，输出F为1，否则输出为0，试列出此逻辑函数的真值表，写出其逻辑函数表达式，并画出逻辑电路图。解答：由题意可列出真值表如下：ABCF00000011010101101001101011001111由真值表可以得到函数表达式为：FABCABCABCABC逻辑电路如图T2-7所示：ABCABCABCABCF图T2-72-8设计一个3人表决电路，要求：当输入A、B、C中有半数以上人同意时，决议才能通过，但A有否决权，如A不同意，即使B、C都同意，决议也不能通过。解答：定义变量A、B、C，1代表同意，0代表不同意；F为结果，1代表通过，0代表不能通过。由题意可列出真值表如下：ABCF700000010010001101000101111011111由真值表可以得到函数表达式为FABCABCABC，化简可以得到FACAB。2-9试用代数公式法证明题2-5中的各等式。（1）ABACBCABC证明：()ABACBCABABCABABCABC（2）ABABBCABABAC证明：()ABABBCABBCABABBCACABABABAC（3）ABBCCAABBCCA证明：()()()()()()ABBCCAABBCBCCAABCAABBCCACAABBCABCABCABBCCACABCABABBCCA（4）ABABBCACABC证明：(1)ABABBCACABCACACBCABC8（5）ABBCCDDAABCDABCD证明：()()()()()()ABBCCDDAABBCCDDAABACBCCDCADAABCDABCD（6）ABABABCAB证明：()()ABABABCABABABCAABCABBAB2-10证明下列异或运算公式：(1)0AA?(2)1AA?(3)0AA?(4)1AA?(5)ABABA?(6)ABAB??解答：（1）0AA证明：000AAAAAA（2）1AA证明：11101011AAAAA（3）0AA证明：00010AAAAAA（4）1AA证明：91AAAAAAAAAAAA（5）ABABA证明：()()ABABABABABABABABABABABABA（6）ABAB证明：()()ABABABABABABABABABABABABABAB2-11用公式法化简下列逻辑函数为最简与或式：(1)1()FABABABABCD=+++(2)2FABCACABCAC=+++(3)3()()FABABABAB=++(4)4()()FAABABCC=+++(5)5()FABACDBCD=+++(6)6()()()FABAABCABCABABC=++++++解答：（1）1()FABABABABCD化简：1()()()()FABABABABCDAABABCDABABCDABABCDAB（2）2FABCACABCAC化简：2()()()()FABCACABCACABCCABCACABCABCACABCABCACABCACABCACABCABCACABCABACACABCABAABCAABC10（3）3()()FABABABAB化简：3()()()000FABABABABABABABABABABAB（4）4()()FAABABCC化简：4()()()()()0FAABABCCABABCABABC（5）5()FABACDBCD化简：5()()()()()()()()()()()()FABACDBCDABACDBCDAAACADABBCBDBCDACABBCADBDBCDACABADBDBCDACABADBCDABCACACDABABCABDABDACDADACABAD（6）6()()()FABAABCABCABABC化简：6()()()()FABAABCABCABABCAABCABCABABCACABCABABCABCABABBCABC2-12用卡诺图化简下列逻辑函数为最简与或式：(1)1(3,5,6,7)Fm=å(2)2(4,5,6,7,8,9,10,11,12,13)Fm=å(3)3(2,3,6,7,10,11,12,15)Fm=å(4)4(1,3,4,5,8,9,13,15)Fm=å11(5)5(1,3,4,6,7,9,11,12,14,15)Fm=å(6)6(0,2,4,7,8,9,12,13,14,15)Fm=å解答：（1）13,5,6,7Fm（）卡诺图：BCA000111100001010111由卡诺图可知：13,5,6,7FmACABBC（）（2）24,5,6,7,8,9,10,11,12,13Fm（）卡诺图：CDAB00011110000000011111111100101111由卡诺图可知：24,5,6,7,8,9,10,11,12,13FmABABAC（）（3）32,3,6,7,10,11,12,15Fm（）卡诺图：CDAB0001111000001101001111101010001112由卡诺图可知：32,3,6,7,10,11,12,15FmABCDACBCCD（）（4）4134,5,8,9,13,15Fm（，，）卡诺图：CDAB00011110000110011100110110101100由卡诺图可知：4134,5,8,9,13,15FmABDABCABDABC（，，）（5）5134,6,7,9,11,12,1415Fm（，，，）卡诺图：CDAB00011110000110011011111011100110由卡诺图可知：5134,6,7,9,11,12,1415FmBDBDCD（，，，）（6）6024,7,8,9,12,13,14,15Fm（，，）卡诺图：CDAB0001111000100101101011111113101100由卡诺图可知：6024,7,8,9,12,13,14,15FmABACCDABCBCD（，，）2-13对具有无关项0ABAC+=的下列逻辑函数进行化简：(1)1FACAB=+(2)2FACAB=+(3)3FABCABDABDABCD=+++(4)4FBCDABCDABCABD=+++（5）5FACDABCDABDABCD=+++(6)6FBCDABCDABCD=++解答：（1）1FACAB1FACABACABABACACBAC（2）2FACAB解：2FACABACABABACBC（3）3FABCABDABDABCD3FABCABDABDABCDABACABCABABCDABACABCBABC