数字逻辑第二版毛法尧课后题答案

1、第一章第一章 数制与码制数制与码制1.1 1.1 把下列各进制数写成按全展开的形式把下列各进制数写成按全展开的形式（1 1）（）（4517.2394517.239）1010 =4 =410103 3+5+510102 2+1+110101 1+7+710100 0+2+21010-1-1+3+31010-2 -2 +9 +91010-3-3（2 2）（）（10110.010110110.0101）2 2 =1 =12 24 4+0+02 23 3+1+12 22 2+1+12 21 1+0+02 20 0+0+02 2-1 -1 +1+12 2-2 -2 +0 +02 2-3-3+1+12 2

2、-4-4（3 3）（）（325.744325.744）8 8 =3 =38 82 2+2+28 81 1+5+58 80 0+7+78 8-1-1+4+48 8-2-2+4+48 8-3-3（4 4）（）（785.4AF)16785.4AF)16 =7 =716162 2+8+816161 1+5+516160 0+4+41616-1-1+A+A1616-2-2+F+F1616-3-31.2 1.2 完成下列二进制表达式的运算完成下列二进制表达式的运算（1 1）10111+101.101 10111+101.101 （2 2）1100-111.0111100-111.011（3 3）10.01

3、10.011.01 1.01 （4 4）1001.00011001.000111.10111.101 10111 10111+ 101.101+ 101.101= 11100.101= 11100.101 11000.000 11000.000- 00111.011- 00111.011= 10000.101= 10000.101 10.01 10.01 1.011.01 1001 1001 0000 0000 1001 1001 10.1101 10.1101 10.0110.0111101 )1001000111101 )10010001 1.5 1.5 如何判断一个二进制正整数如何判断一

4、个二进制正整数B=bB=b6 6b b5 5b b4 4b b3 3b b2 2b b1 1b b0 0能否被（能否被（4 4）1010整整除？除？解：解：b b1 1b b0 0同为同为0 0时能整除，否则不能。时能整除，否则不能。1.6 1.6 写出下列各数的原码、反码和补码。写出下列各数的原码、反码和补码。（1 1）0.1011 0.1011 （2 2）0.0000 0.0000 （3 3）-10110-10110解：解：0.10110.1011原原=0.1011=0.1011反反=0.1011=0.1011补补=0.1011=0.1011 0.0000 0.0000原原=0.0000=

5、0.0000反反=0.0000=0.0000反反=0.0000=0.0000 -10110 -10110原原=110110=110110 -10110 -10110反反=101001=101001 -10110 -10110反反=101010=101010 1.7 1.7 已知已知NN补补=1.0110=1.0110，求，求NN原原、NN反反和和N N解：解： NN原原=1.1010 N=1.1010 N反反和和=1.1001 N=-0.1010=1.1001 N=-0.10101.8 1.8 用原码、反码和补码完成如下运算用原码、反码和补码完成如下运算（1 1）0000101-0011010

6、0000101-0011010 解解（1 1）0000101-00110100000101-0011010原原=10010101=10010101 0000101-0011010=-0010101 0000101-0011010=-0010101 0000101-0011010 0000101-0011010反反=0000101=0000101反反+-0011010+-0011010反反 =00000101+11100101=11101010=00000101+11100101=11101010 0000101-0011010=-0010101 0000101-0011010=-0010101

7、 0000101-0011010 0000101-0011010反反=0000101=0000101补补+-0011010+-0011010补补 =00000101+11100110=11101011=00000101+11100110=11101011 0000101-0011010=-0010101 0000101-0011010=-00101011.8 1.8 用原码、反码和补码完成如下运算用原码、反码和补码完成如下运算 （2 2）0.010110-0.1001100.010110-0.100110解解（2 2）0.010110-0.1001100.010110-0.100110原原=1

8、.010000=1.010000 0.010110-0.100110=-0.010000 0.010110-0.100110=-0.010000 0.010110-0.100110 0.010110-0.100110反反=0.010110=0.010110反反+-0.100110+-0.100110反反 = 0.010110+1.011001=1.101111= 0.010110+1.011001=1.101111 0.010110-0.100110=-0.010000 0.010110-0.100110=-0.010000 0.010110-0.100110 0.010110-0.10011

9、0补补=0.010110=0.010110补补+-0.100110+-0.100110补补 = 0.010110+1.011010=1.110000= 0.010110+1.011010=1.110000 0.010110-0.100110=-0.010000 0.010110-0.100110=-0.0100001.9 1.9 分别用分别用“对对9 9的补数的补数“和和”对对1010的补数完成下列十进制的补数完成下列十进制数的运算数的运算（1 1）2550-1232550-123解解：（：（1 1）2550-1232550-1239 9补补=2550-0123=2550-01239 9补补=

10、2550=25509 9补补+-0123+-01239 9补补 =02550+99876=02427=02550+99876=02427 2550-123=+2427 2550-123=+2427 2550-123 2550-1231010补补=2550-0123=2550-01231010补补=2550=25501010补补+-0123+-01231010补补 =02550+99877=02427=02550+99877=02427 2550-123=+2427 2550-123=+24271.9 1.9 分别用分别用“对对9 9的补数的补数“和和”对对1010的补数完成下列十进制的补数完成

11、下列十进制数的运算数的运算 （2 2）537-846537-846解解：（：（2 2）537-846537-8469 9补补=537=5379 9补补+-846+-8469 9补补 =0537+9153=9690=0537+9153=9690 537-846=-309 537-846=-309 537-846 537-8461010补补=537=5371010补补+-846+-8461010补补 =0537+9154=9691=0537+9154=9691 537-846=-309 537-846=-3091.10 1.10 将下列将下列8421BCD8421BCD码转换成十进制数和二进制数码

12、转换成十进制数和二进制数(1)011010000011(1)011010000011(2)01000101.1001(2)01000101.1001解解：(1)(1)（011010000011011010000011）8421BCD8421BCD=(683)=(683)D D=(1010101011)=(1010101011)2 2 (2)(01000101.1001) (2)(01000101.1001)8421BCD8421BCD=(45.9)=(45.9)D D=(101101.1110)=(101101.1110)2 21.11 1.11 试用试用8421BCD8421BCD码、余码、

13、余3 3码和格雷码分别表示下列各数码和格雷码分别表示下列各数(1)578)(1)578)1010 (2)(1100110)(2)(1100110)2 2解解：( (578)578)1010 =(010101111000)=(010101111000)8421BCD8421BCD =(100010101011) =(100010101011)余余3 3 = =（10010000101001000010）2 2 = =（11011000111101100011）G G 解解：（11001101100110）2 2 = =（10101011010101）G G = =( (102)102)1010

14、=(000100000010)=(000100000010)8421BCD8421BCD =(010000110101) =(010000110101)余余3 3 1.12 1.12 将下列一组数按从小到大顺序排序将下列一组数按从小到大顺序排序(11011001)(11011001)2 2,(135.6),(135.6)8 8,(27),(27)1010,(3AF),(3AF)1616,(00111000),(00111000)8421BCD8421BCD(11011001)(11011001)2 2=(217)=(217)1010 (135.6) (135.6)8 8=(93.75)=(93

15、.75)1010 (3AF) (3AF)1616=(431)=(431)1010 (00111000) (00111000)8421BCD8421BCD=(38)=(38)1010按从小到大顺序排序为：按从小到大顺序排序为：(27)(27)10 10 , (00111000), (00111000)8421BCD 8421BCD ,(135.6),(135.6)8 8,(11011001),(11011001)2 2 (3AF) (3AF)1616, ,)1111,1101,1100,0111,0100() 1 (CABBDF)1111,1101,1011,1001,0111,0101,001

16、1,0001()()()2(DDBABADBABABABAF时，即：时，为或11101101110010101001100001110110010101000010000100001010, 0)()()3(FABDCDCBADCBADCADACDBADCAAFCABACAAB)( ) 1 (CABACBCABACABACAAB）（证明：)(1)2(BABABAAB1AABABABAAB证明：CABCBACBAABCA) 3(CABACBAAABCA)(证明：CABACABCBACBACABCBACBAABCACBAABCCABCCABACACBBACACBBA)()()(证明：CACBBAC

17、BAABC)4(BABCBAABC)5(BABABCCBBACABABCBACBABCBAABC)(证明：CDBBDACACBBACADCA)()()6(CDBBDACADCBABCDACDBACDBADCBADCABCADCABDCBACDBADCBABCDACDBACADCBDBACDACACBBACADCACBBACADCA)()()(证明：)()() 1 (CBCAFCBCAFCBCAF)()()()()()2(DCACBBAFDCACBBAFDCACBBAF)()()()3(GFEDCBAFGFEDCBAFGFEDCBAFEDCBAFEDCBAFEDCBAF)4()()()()()5

18、(CABBAFCABBACABBAFCABBAF11)2(BABABABABAABBAAFBABBABCDBBAF) 1 (DBADCADBADADCADBADBAABDCADBADABF)3(ADEACEDCEADACDCEDCAADCEDCACBAAF)()()()()4(CABCBBCAACF)5(BCCABCCBBACCBACBBACCBACBBCAACCABCBBCAACF)()()()()()(BCBACBCBACBBACCBACBBCAACCABCBBCAACF)()()()()()5(EDCADCEDCADCEDCAADCEDCACBAAF)()()()()()4()12, 8

19、 , 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)()

20、()() 1 (CBCABACABBAFCACBF) 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(ABCD000111100001111011101111111110001111000000001111ABCD0001111000011110DACDCDADB

21、DCBAF),() 1 (的关系、和、用卡诺图判断函数D)CBG(AD)CBF(A2.110000111111110000ABCD0001111000011110ABDDCACDDBDCBAG),() 1 (GF ) 1 (ABCCBAABCCBACAABCCBACACBAABCCBACACBCABAABCCBACACABABCCBACACBBAABCCBAACBCABDCBAG)()()()()()()()()()(),()2(1 11 11 11 1ABCD00011110000111101 11 11 11 11 11 11 11 1ABCD0001111000011110ABCCBAC

22、BACBACBABACBACBACBABACBABADCBAF)()()(),()2(DCACF1(2)aDBCAC0112. 2CBbDCACBFa时时）当（)10, 8 , 6 , 5 , 4 , 3 , 1 ()15,13, 7 , 2 , 0(),(113. 2dmDCBAF）（ABCD00011110000111101 1d d0 0d dd dd d1 10 0d d1 11 10 01 1d d0 0d dBDAF)15,13,10, 8 , 7 , 4 , 2 , 0(),(213. 21mDCBAF）（)10, 8 , 7 , 6 , 5 , 2 , 1 , 0(),(2mD

23、CBAF)7 , 4 , 3 , 2(),(3mDCBAF1 11 11 11 11 11 11 11 1ABABCDCD000001011111101000000101111110101 11 11 11 11 11 11 11 1ABABCDCD000001011111101000000101111110101 11 11 11 1ABABCDCD00000101111110100000010111111010ABDBCDADCBADBF12）（DCABCDADCADBF2BCDADCBACBAF33.1将下列函数化简，并用将下列函数化简，并用“与非与非”、 “或非或非” 门画门画出逻辑电

24、路图。出逻辑电路图。)7 , 3 , 2 , 0(),() 1 (mCBAF)6 , 3(),()2(MCBAFCBCADCABADCBAF),()3(CDBCAABDCBAF),()4(解解（1 1）BCCABCCAFCBCAFCBCACBCAFF)()6 , 3(),()2(MCBAF解解（1 1）ACCABACCABFBCACABFCBACBACBACBAFF)(CBCADCABADCBAF),()3(解解: :BACBCABACBCAFCBCADCABADCBAF),()3(解解: :ABCCBAFCBACBACBACBAFF)(BACDBCABACDBCAABDCBAF),()4(解

25、解：ABF &ABFBABAF1ABF1CDBCABADCBAF),()4(解解：DAABCACBFDABACACBDABACACBFF)()()()(CBABAABCBAF)(),() 1 (3.2 3.2 将下列函数化简，并用将下列函数化简，并用“与或非门与或非门”画出逻辑电路图。画出逻辑电路图。)15,14,13,10, 9 ,78, 6 , 2 , 1 (),()2(mDCBAF解解：BCACBAABCBABAABCBAF)(),() 1 (BCACABBCACABF)15,14,13,10, 9 ,78, 6 , 2 , 1 (),()2(mDCBAF解解：CBADCBDCA

26、DCBDCBADCBDCADCBDFABCABCZ Z0000000 00010011 10100100 00110111 11001000 01011010 01101101 11111111 13.3 3.3 解：由时间图得真值表如下：解：由时间图得真值表如下：)7 , 6 , 3 , 1 (),(mCBAFBACAFBACABACAFF)(3.43.4解：解：CBAABCCBABCBCACCBBCACBCBCACBCBAF)()()()(3.5 13.5 1）解：）解：CBAFBCCABAG一位二进制数全减器：一位二进制数全减器：2 2）全加器）全加器3. 6 3. 6 解：解：BABA

27、F1BABAF2BABABABAF3当当A=BA=B时等效时等效F F1 1=F=F2 2=F=F3 3=0=0XYXY Y Y3 3Y Y2 2Y Y1 1Y Y0 00000 000000000101 000100011010 010001001111 100110013.7 3.7 解解(1)(1)BYYBABAYABABY012302XY BYYABAABABAYABABY01230&AB3Y2Y1Y0Y11XYXY Y Y4 4Y Y3 3Y Y2 2Y Y1 1Y Y0 00000 00000000000101 00001000011010 01000010001111 1

28、1011110113.7 3.7 解解(2)(2)BYABYYAYABABY0123403XY &AB4Y2Y1Y0Y3Y1C C1 1 C C0 0 X XY Y0000000 00010010 00100100 00110111 11001001 11011010 01101101 11111111 13.8 3.8 解：真值表如下：解：真值表如下：XCXCXCXCF0101)7 , 6 , 4 , 3(),(01mXCCFY&11CXY0C&3.9 3.9 依题意得真值表如下：依题意得真值表如下：B B8 8B B4 4B B2 2B B1 1F F7 7F F6

29、 6F F5 5F F4 4F F3 3F F2 2F F1 1F F0 00 0 0 00 0 0 0 0 0 0 0 0 0 0 00 0 0 0 0 0 0 00 0 0 10 0 0 1 0 0 0 0 0 1 0 10 0 0 0 0 1 0 10 0 1 00 0 1 0 0 0 0 1 0 0 0 00 0 0 1 0 0 0 00 0 1 10 0 1 1 0 0 0 1 0 1 0 10 0 0 1 0 1 0 10 1 0 00 1 0 0 0 0 1 0 0 0 0 00 0 1 0 0 0 0 00 1 0 10 1 0 1 0 0 1 0 0 1 0 10 0 1 0

30、 0 1 0 10 1 1 00 1 1 0 0 0 1 1 0 0 0 00 0 1 1 0 0 0 00 1 1 10 1 1 1 0 0 1 1 0 1 0 10 0 1 1 0 1 0 11 0 0 01 0 0 0 0 1 0 0 0 0 0 00 1 0 0 0 0 0 01 0 0 11 0 0 1 0 1 0 0 0 1 0 10 1 0 0 0 1 0 1依真值表得依真值表得:07F86BF 45BF 24BF 03F12BF 01F10BF 3.10 3.10 依题意得真值表如下：依题意得真值表如下：y y1 1y y0 0 x x1 1x x0 0Z Z1 1 Z Z0

31、0000000001111000100010101001000100101001100110101010001001010010101011111011001100101011101110101100010001010100110011010101010101111101110110101110011001010110111011010111011101010111111111111)15,14,13,12,10, 9 , 8 , 5 , 4 , 0(1mZ)15,11,10, 7 , 6 , 5 , 3 , 2 , 1 , 0(0mZ)15,14,13,12,10, 9 , 8 , 5 ,

32、4 , 0(1mZABCDDCBACADADCZ0ABCDDCBACACBBAZ1)15,11,10, 7 , 6 , 5 , 3 , 2 , 1 , 0(0mZ3.113.11依题意得真值表如下：依题意得真值表如下：B B1 1B B0 0B B1 1B B0 0F F000000001 1000100010 0001000100 0001100111 1010001000 0010101011 1011001101 1011101110 0100010000 0100110011 1101010101 1101110110 0110011001 1110111010 0111011100

33、0111111111 1)15,12,10, 9 , 3 , 5 , 3 , 0(mF1234BBBBF=1=1=1=1=1=1=1=14B3B2B1B1F BCD 码 0 0 1 1 余 3 码 S4 S3 S2 S1 C4 C0 A4 A3 A2 A1 B4 B3 B2 B1 BCD 码 0 0 1 1 余 3 码 S4 S3 S2 S1 C4 C0 A4 A3 A2 A1 B4 B3 B2 B1 6.16.1：用两个：用两个4 4位二进制并行加法器实现两位十进制位二进制并行加法器实现两位十进制8421BCD8421BCD码到余码到余3 3码的转换码的转换高位高位低位低位A3A2A1A0B

34、3B2B1B0ABAb a=b aBAb a=b ab74LS85(2)0 0A A7 7A A6 6A A5 50 0B B7 7B B6 6B B5 5A A4 4A A3 3A A2 2A A1 1B B4 4B B3 3B B2 2B B1 16.26.2：用两块：用两块4 4位数值比较器芯片实现两个位数值比较器芯片实现两个7 7位二进制的比较位二进制的比较6.36.3：用：用3-83-8线译码器线译码器7413874138和必要逻辑门实现下列函数和必要逻辑门实现下列函数yxxyzyxFyxzyxFzxyyxzyxF),(),(),(321解：解：6106101),(mmmmmmzxy

35、zyxzyxzxyyxzyxF7632107632102),(mmmmmmmmmmmmxyzzxyyzxzyxyzxzyxzyxzyxyxzyxF761076103),(mmmmmmmmxyzzxyzyxzyxyxxyzyxFA074LS138Y0A1A2G2AG1G2BY1Y2Y3Y4Y5Y6Y7&F1xyz 100&F2F3变量多数表决电路）（全加器用四选一电路实现函数32) 1 (:4 . 61111iS) 1 (iiiiiiiiiiiiCBACBACBACBA：由全加器的真值表可得解：iiiCBA输出，输出输出端，、制端输入从四路选择器的选择控2YSY1i 12D 2D 2D 02D 1D1D 1D1D312110121130可得：iiiiCCCC1111iiiiiiiiiiiiiCBACBACBACBAC1A1AiAiB01D11D21D31D02D12D22D32DY2Y2iSiC 12D 2D 2D