数字逻辑(第二版)毛法尧课后题答案

第一章数制与码制1.1把下列各进制数写成按全展开的形式（1）（4517.239）10=4×103+5×102+1×101+7×100+2×10-1+3×10-2+9×10-3（2）（10110.0101）2=1×24+0×23+1×22+1×21+0×20+0×2-1+1×2-2+0×2-3+1×2-4（3）（325.744）8=3×82+2×81+5×80+7×8-1+4×8-2+4×8-3（4）（785.4AF)16=7×162+8×161+5×160+4×16-1+A×16-2+F×16-31.2完成下列二进制表达式的运算（1）10111+101.101（2）1100-111.011（3）10.01×1.01（4）1001.0001÷11.10110111+101.101=11100.10111000.000-00111.011=10000.10110.01×1.0110010000100110.110110.0111101)100100011.5如何判断一个二进制正整数B=b6b5b4b3b2b1b0能否被（4）10整除？解：b1b0同为0时能整除，否则不能。1.6写出下列各数的原码、反码和补码。（1）0.1011（2）0.0000（3）-10110解：[0.1011]原=[0.1011]反=[0.1011]补=0.1011[0.0000]原=[0.0000]反=[0.0000]反=0.0000[-10110]原=110110[-10110]反=101001[-10110]反=1010101.7已知[N]补=1.0110，求[N]原、[N]反和N解：[N]原=1.1010[N]反和=1.1001N=-0.10101.8用原码、反码和补码完成如下运算（1）0000101-0011010解（1）[0000101-0011010]原=10010101∴0000101-0011010=-0010101[0000101-0011010]反=[0000101]反+[-0011010]反=00000101+11100101=11101010∴0000101-0011010=-0010101[0000101-0011010]反=[0000101]补+[-0011010]补=00000101+11100110=11101011∴0000101-0011010=-00101011.8用原码、反码和补码完成如下运算（2）0.010110-0.100110解（2）[0.010110-0.100110]原=1.010000∴0.010110-0.100110=-0.010000[0.010110-0.100110]反=[0.010110]反+[-0.100110]反=0.010110+1.011001=1.101111∴0.010110-0.100110=-0.010000[0.010110-0.100110]补=[0.010110]补+[-0.100110]补=0.010110+1.011010=1.110000∴0.010110-0.100110=-0.0100001.9分别用“对9的补数“和”对10的补数完成下列十进制数的运算（1）2550-123解：（1）[2550-123]9补=[2550-0123]9补=[2550]9补+[-0123]9补=02550+99876=02427∴2550-123=+2427[2550-123]10补=[2550-0123]10补=[2550]10补+[-0123]10补=02550+99877=02427∴2550-123=+24271.9分别用“对9的补数“和”对10的补数完成下列十进制数的运算（2）537-846解：（2）[537-846]9补=[537]9补+[-846]9补=0537+9153=9690∴537-846=-309[537-846]10补=[537]10补+[-846]10补=0537+9154=9691∴537-846=-3091.10将下列8421BCD码转换成十进制数和二进制数(1)011010000011(2)01000101.1001解：(1)（011010000011）8421BCD=(683)D=(1010101011)2(2)(01000101.1001)8421BCD=(45.9)D=(101101.1110)21.11试用8421BCD码、余3码和格雷码分别表示下列各数(1)578)10(2)(1100110)2解：(578)10=(010101111000)8421BCD=(100010101011)余3=（1001000010）2=（1101100011）G解：（1100110）2=（1010101）G=(102)10=(000100000010)8421BCD=(010000110101)余31.12将下列一组数按从小到大顺序排序(11011001)2,(135.6)8,(27)10,(3AF)16,(00111000)8421BCD(11011001)2=(217)10(135.6)8=(93.75)10(3AF)16=(431)10(00111000)8421BCD=(38)10∴按从小到大顺序排序为：(27)10,(00111000)8421BCD,(135.6)8,(11011001)2(3AF)16,2.1分别指出变量（A，B，C，D）在何种取值时，下列函数的值为1？)1111,1101,1100,0111,0100()1(CABBDF)1111,1101,1011,1001,0111,0101,0011,0001()())(()2(DDBABADBABABABAF第二章逻辑代数基础时，，，，，，，，，，，，即：时，为或11101101110010101001100001110110010101000010000100001010,0)()()3(FABDCDCBADCBADCADACDBADCAAFCABACAAB)()1(2.2用逻辑代数的公理、定理和规则证明下列表达式CABACBCABACABACAAB））（（证明：)(1)2(BABABAAB1AABABABAAB证明：CABCBACBAABCA)3(CABACBAAABCA)(证明：CABACABCBACBACABCBACBAABCACBAABCCABCCABACACBBACACBBA))(())()((证明：CACBBACBAABC)4(BABCBAABC)5(BABABCCBBACABABCBACBABCBAABC))((证明：CDBBDACACBBACADCA))()(()6(CDBBDACADCBABCDACDBACDBADCBADCABCADCABDCBACDBADCBABCDACDBACADCBDBACDACACBBACADCACBBACADCA)())()((证明：2.4求下列函数的反函数和对偶函数))(('))(()1(CBCAFCBCAFCBCAF))()(('))()(()()2(DCACBBAFDCACBBAFDCACBBAF]))([(']))([(])([()3(GFEDCBAFGFEDCBAFGFEDCBAF2.5答：（1）正确（2）不正确，A=0时，B可以不等于C（3）不正确，A=1时，B可以不等于C（4）正确EDCBAFEDCBAFEDCBAF')4()('))(()())(()5(CABBAFCABBACABBAFCABBAF2.6用代数法化简成最简“与或”表达式11)2(BABABABABAABBAAFBABBABCDBBAF)1(DBADCADBADADCADBADBAABDCADBADABF)3(ADEACEDCEADACDCEDCAADCEDCACBAAF))(())(())()(()4(CABCBBCAACF)5(BCCABCCBBACCBACBBACCBACBBCAACCABCBBCAACF)())(())()(())()((BCBACBCBACBBACCBACBBCAACCABCBBCAACF)())()(())()(()5(EDCADCEDCADCEDCAADCEDCACBAAF)())(())(())()(()4()12,8,4,3,2,1,0())((),,,()3()13,12,9,8,7,6,5,4(),,,()2()7,6,5,4,3())(()(),,()1(7.2mDCABADCBBCADCBAFmDCBBCDCABBADCBAFmCBACABACABACBAF)7,6,5,3(),,()3()6,5,3,0(),,,()2()4,2,1,0(),,()1(8.2mCBAFmCBAFmCBAFDCBDCACDBDCBADCACDBDCADCBADCABCDBCDDACBADCCBCDBffF))(()9.221))()(())(()1(CBCABACABBAFABC000111100000011101CACBF)1()(BACFFABCF最简或与表达式：与”表达式或”表达式和最简“或并写出最简“与，用卡诺图化简下列函数2.10CBACDCABADCBAF),,,()2(ABCD00011110000111101110111111111000CBACBADCBAF),,,()2())((CBACBAFFBCACBAF最简或与表达式：BADDBCBADCBDDBCDCBAF))((),,,()3(DBF)3(ABCD000111100001111011101111111110001111000000001111ABCD0001111000011110DACDCDADBDCBAF),,,()1(的关系、、、和、、、用卡诺图判断函数D)CBG(AD)CBF(A2.110000111111110000ABCD0001111000011110ABDDCACDDBDCBAG),,,()1(GF)1(ABCCBAABCCBACAABCCBACACBAABCCBACACBCABAABCCBACACABABCCBACACBBAABCCBAACBCABDCBAG)())(())()(())()(())()()(()(),,,()2(1111ABCD000111100001111011111111ABCD0001111000011110G是F的子集ABCCBACBACBACBABACBACBACBABACBABADCBAF))(()()(),,,()2(DCACF1(2)aDBCAC0112.2CBbDCACBFa时时）当（)10,8,6,5,4,3,1()15,13,7,2,0(),,,(113.2dmDCBAF）（ABCD00011110000111101d0ddd10d1101d0dBDAF)15,13,10,8,7,4,2,0(),,,(213.21mDCBAF）（)10,8,7,6,5,2,