逻辑代数习题作业及答案

第一章习题1.1.数字信号与模拟信号的主要区别是什么？1.2.(1)(172)10=(?)2(2)(0.8123)10=(?)2(3)(10101101.0101)2=(?)10(4)(3625)10=(?)8=(?)16(5)(0.172)8=(?)16=(?)2(6)(4CA)16=(?)2=(?)10解：(1)(172)10=(10101100)2(2)(0.8123)10=(011001)2(3)(10101101.0101)2=(173.3125)10(4)(3625)10=(7051)8=(E29)16(5)(0.172)8=(03D0)16=(0.00111101)2(6)(4CA)16=(10011001010)2=(1226)101.3.1.))((CABABCA2.ACCBBAACCBBA解：1.ABCBCA+BCA+BA+C(A+B)(A+C)0000000000100010010001000111111110001111101011111100111111111111(2).ABCABCBACBCAACCBBABACBACACCBBA00011100000000001110001101010101010101100101110000111001100011100100111010101001010111000101010011111000000000001.4.求下列函数的对偶式和反函数式(1)DCBADCABZ)((1(2)CBADDCBAZ2解：(1)Z1的对偶式为:DABCDACBZ)('1Z1的反函数式为：DCABDCABZ1(2)Z2的对偶式为：)('2CBDADBCAZZ2的反函数式为：))((2CBDADCBAZ1.5.用逻辑代数证明下列等式(a)BABAA(b)ACABCABCBAABC(c)ECDAEDCCDACBAA解：(a)BABAA由交换律))((CABABCA，得BABAAABAA))(((b)ACABCABCBAABCACABBCACBCACBCBBCACABCBAABC)()()((c)ECDAEDCCDACBAAECDAECDCDAEDCCDAAEDCCDACBAA_____)(1.6.试从以下真值表写出LABCLABCL00001001001110100101110001101111解：ABCCBACBACBACBAfL),,(1.7.解：BABABABABABAfL))((),(ABBA0000111011101.8.(1)CABDACDABDCBAZ),,,(1(2)BCCBDADCAACDDCBAZ),,,(2解：（1）)14,13,12,10(1Z（2）)10,8,2,0(2Z1.9.用代数法将下列函数化简成为最简“与—或”表达式(1)CDBACDBCDA(2)EFBEFBABDCAABDAAD(3))()(DACBADACBA(4)))()((EDEDCCBBCBA(5))(ABCAB(6)CBBCCBAABCA(7)BABABABA(8)CEAEAECAEBECEBA)((9)CABCBCDABCDADA)()((10)CBBA1.10.利用与非门实现下列函数(a)L=AB+AC(b)_____________)(CADL(c)____________________))((DCBAL解：(a)L=AB+AC_______________________ACABL(b)_____________)(CADL≥1≥1≥111ABL_____________________________)(CADCADL(c)____________________))((DCBAL__________________________________________))((DCBADCBAL1.11.试用2输入与非门和反相器设计一个4位的奇偶校验器，即当4位数中有奇数个1时输出为0，否则输出为1。解：DCBAL_______________________BABABABAL1.12.用卡诺图法将下列函数化简成为最简“与-或”(1)ABBABAZ(2)CBACBA(3)CABCBCDABCDADAZ)()((4)EACEAECAEBECEBAZ)((5))15,14,11,10,8,7,6,5,2,0(),,,(mDCBAZ(6))13,12,11,10,9,8,7,6,5,3,2,1(),,,(mDCBAZ(7)CBCBCACADCBAZ),,,((8)DCACBADCDCAABDABCDCBAZ),,,((9))14,11,10,9,8,6,4,3,2,1,0(),,,(mDCBAZ(10)DBDCACBDBADBADCBAZ)()(),,,(解：(1)BAZ1(2)12ZBA01011110BCA000101111111011111100ABCD000101111110100000000000000111100ABCDE00001101111111001000101011010111111111111100000000000000000000(3)CBZ3(4)EAEBAZ4(5)BDADBBCBAZ5(6)BACACADCZ6(7)CBBACAZ7或CBBACA(8)DAZ8(9)DADCBZ9(10)CADABZ10BCA000101111111001011100ABCD00010111111010111111111110000000ABCD00010111111010111111111001001100ABCD00010111111010110101111011110000ABCD00010111111010111111110111100000ABCD000101111110101111111111110000