精华]数字逻辑(第二版)毛法尧课后题谜底ppt课件

1、第一章第一章 数制与码制数制与码制1.1 1.1 把以下各进制数写成按全展开的方式把以下各进制数写成按全展开的方式1 14517.2394517.23910 10 =4 =4103+5103+5102+1102+1101+7101+7100+2100+210-1+310-1+310-2 10-2 +9 +910-310-32 210110.010110110.01012 2 =1 =124+024+023+123+122+122+121+021+020+020+02-1 +12-1 +12-2 2-2 +0 +02-3+12-3+12-42-43 3325.744325.7448 8 =3 =

2、382+282+281+581+580+780+78-1+48-1+48-2+48-2+48-38-34 4785.4AF)16785.4AF)16 =7 =7162+8162+8161+5161+5160+4160+416-1+A16-1+A16-2+F16-2+F16-316-31.2 1.2 完成以下二进制表达式的运算完成以下二进制表达式的运算1 110111+101.101 10111+101.101 2 21100-111.0111100-111.0113 310.0110.011.01 1.01 4 41001.00011001.000111.10111.101 10111+ 10

3、1.101= 11100.101 11000.000- 00111.011= 10000.101 10.01 1.01 1001 0000 1001 10.1101 10.0111101 )10010001 1.5 1.5 如何判别一个二进制正整数如何判别一个二进制正整数B=b6b5b4b3b2b1b0B=b6b5b4b3b2b1b0能否被能否被4 41010整除？整除？解：解：b1b0b1b0同为同为0 0时能整除，否那么不能。时能整除，否那么不能。 1.6 写出以下各数的原码、反码和补码。 10.1011 20.0000 3-10110 解：0.1011原=0.1011反=0.1011补=

4、0.1011 0.0000原=0.0000反=0.0000反=0.0000 -10110原=110110 -10110反=101001 -10110反=101010 1.7 知N补=1.0110，求N原、N反和N 解： N原=1.1010 N反和=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 10000101-00110100000101-0011010 解10000101-0011010原=10010101 0000101-0011010=-0010101 0000101-0011010反=0000101反+-0011010反

5、 =00000101+11100101=11101010 0000101-0011010=-0010101 0000101-0011010反=0000101补+-0011010补 =00000101+11100110=11101011 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 20.010110-0.1001100.010110-0.100110 解20.010110-0.100110原=1.010000 0.010110-0.100110=-0.010000 0.010110-0.100110反=0.01

6、0110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=-0.010000 1.9 分别用“对9的补数“和对10的补数完成以下十进制数的运算 12550-123 解：12550-1239补=2550-01239补=25509补+-01239补 =02550+99876=02427 2550-123=+2427 2550-12310补

7、=2550-012310补=255010补+-012310补 =02550+99877=02427 2550-123=+2427 1.9 分别用“对9的补数“和对10的补数完成以下十进制数的运算 2537-846 解：2537-8469补=5379补+-8469补 =0537+9153=9690 537-846=-309 537-84610补=53710补+-84610补 =0537+9154=9691 537-846=-309 1.10 将以下8421BCD码转换成十进制数和二进制数 (1)011010000011 (2)01000101.1001 解：(1)0110100000118421

8、BCD=(683)D=(1010101011)2 (2)(01000101.1001)8421BCD=(45.9)D=(101101.1110)21.11 1.11 试用试用8421BCD8421BCD码、余码、余3 3码和格雷码分别表示以下各数码和格雷码分别表示以下各数(1)578)10 (1)578)10 (2)(1100110)2(2)(1100110)2 解：(578)10 =(010101111000)8421BCD =(100010101011)余3 =10010000102 =1101100011G 解：11001102 =1010101G =(102)10 =(00010000

9、0010)8421BCD =(010000110101)余3 1.12 将以下一组数按从小到大顺序排序 (11011001)2,(.6)8,(27)10,(3AF)16,(00111000)8421BCD (11011001)2=(217)10 (.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 ,(.6)8,(11011001)2 (3AF)16,)1111,1101,1100,0111,0100() 1 (CABBDF)1111,1101,1011,1

10、001,0111,0101,0011,0001()()()2(DDBABADBABABABAF第二章 逻辑代数根底时，即：时，为或11101101110010101001100001110110010101000010000100001010, 0)()()3(FABDCDCBADCBADCADACDBADCAAFCABACAAB)( ) 1 (2.2 用逻辑代数的公理、定理和规那么证明以下表达式CABACBCABACABACAAB）（证明：)(1)2(BABABAAB1AABABABAAB证明：CABCBACBAABCA) 3(CABACBAAABCA)(证明：CABACABCBACBACA

11、BCBACBAABCACBAABCCABCCABACACBBACACBBA)()()(证明：CACBBACBAABC)4(BABCBAABC)5(BABABCCBBACABABCBACBABCBAABC)(证明：CDBBDACACBBACADCA)()()6(CDBBDACADCBABCDACDBACDBADCBADCABCADCABDCBACDBADCBABCDACDBACADCBDBACDACACBBACADCACBBACADCA)()()(证明：2.4 求以下函数的反函数和对偶函数)()() 1 (CBCAFCBCAFCBCAF)()()()()()2(DCACBBAFDCACBBAFD

12、CACBBAF)()()()3(GFEDCBAFGFEDCBAFGFEDCBAFEDCBAFEDCBAFEDCBAF)4()()()()()5(CABBAFCABBACABBAFCABBAF11)2(BABABABABAABBAAFBABBABCDBBAF) 1 (DBADCADBADADCADBADBAABDCADBADABF)3(ADEACEDCEADACDCEDCAADCEDCACBAAF)()()()()4(CABCBBCAACF)5(BCCABCCBBACCBACBBACCBACBBCAACCABCBBCAACF)()()()()()(BCBACBCBACBBACCBACBBCAAC

13、CABCBBCAACF)()()()()()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 . 2mCBAFmC

14、BAFmCBAFDCBDCACDBDCBADCACDBDCADCBADCABCDBCDDACBADCCBCDBffF)()9 . 221)()()() 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(ABCD000111

15、100001111011101111111110001111000000001111ABCD0001111000011110DACDCDADBDCBAF),() 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 1

16、ABCD000111100001111011111111ABCD0001111000011110G是F的子集ABCCBACBACBACBABACBACBACBABACBABADCBAF)()()(),()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 ,

17、2 , 0(),(213. 21mDCBAF）（)10, 8 , 7 , 6 , 5 , 2 , 1 , 0(),(2mDCBAF)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）（DCABCDADCADB

18、F2BCDADCBACBAF33.1将以下函数化简，并用将以下函数化简，并用“与非、与非、 “或非或非 门画门画出逻辑电路图。出逻辑电路图。)7 , 3 , 2 , 0(),() 1 (mCBAF)6 , 3(),()2(MCBAFCBCADCABADCBAF),()3(CDBCAABDCBAF),()4(解1BCCABCCAFCBCAFCBCACBCAFF)()6 , 3(),()2(MCBAF解1ACCABACCABFBCACABFCBACBACBACBAFF)(CBCADCABADCBAF),()3(解:BACBCABACBCAFCBCADCABADCBAF),()3(解:ABCCBAF

19、CBACBACBACBAFF)(BACDBCABACDBCAABDCBAF),()4(解：ABF &ABFBABAF1ABF1CDBCABADCBAF),()4(解：DAABCACBFDABACACBDABACACBFF)()()()(CBABAABCBAF)(),() 1 (3.2 将以下函数化简，并用“与或非门画出逻辑电路图。)15,14,13,10, 9 ,78, 6 , 2 , 1 (),()2(mDCBAF解：BCACBAABCBABAABCBAF)(),() 1 (BCACABBCACABF)15,14,13,10, 9 ,78, 6 , 2 , 1 (),()2(mDCBAF解：

20、CBADCBDCADCBDCBADCBDCADCBDFABCABCZ Z0000000 00010011 10100100 00110111 11001000 01011010 01101101 11111111 13.3 解：由时间图得真值表如下：)7 , 6 , 3 , 1 (),(mCBAFBACAFBACABACAFF)(3.4解：CBAABCCBABCBCACCBBCACBCBCACBCBAF)()()()(3.5 1解：CBAFBCCABAG一位二进制数全减器：一位二进制数全减器：2全加器3. 6 解：BABAF1BABAF2BABABABAF3当A=B时等效F1=F2=F3=0X

21、YXYY Y3 3Y Y2 2Y Y1 1Y Y0 00000000000000101000100011010010001001111100110013.7 解(1)BYYBABAYABABY012302XY BYYABAABABAYABABY01230&AB3Y2Y1Y0Y11XYXYY Y4 4Y Y3 3Y Y2 2Y Y1 1Y Y0 0000000000000000101000010000110100100001000111111011110113.7 解(2)BYABYYAYABABY0123403XY &AB4Y2Y1Y0Y3Y1C C1 1 C C0 0 X XY Y00000

22、00 00010010 00100100 00110111 11001001 11011010 01101101 11111111 13.8 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 6F F5 5F F4 4F F3 3F F2 2F F1 1F F0 00000000000000000000000000001000100000101000001010010001000010000000100000

23、01100110001010100010101010001000010000000100000010101010010010100100101011001100011000000110000011101110011010100110101100010000100000001000000100110010100010101000101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 00000000011110001000101010

24、01000100101001100110101010001001010010101011111011001100101011101110101100010001010100110011010101010101111101110110101110011001010110111011010111011101010111111111111)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 , 4 , 0(1mZABCDDCBACADADCZ0A

25、BCDDCBACACBBAZ1)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 0111111111 1)15,12,10, 9 ,

26、 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 6.1：用两个4位二进制并行加法器实现两位十进制8421BCD码到余3码的转换高位低位A3A2A1A0B3B2B1B0ABAb a=b aBAb a=b ab74LS85(2)0 0A A7 7A A6 6A A5 50 0B B7 7B B6

27、6B B5 5A A4 4A A3 3A A2 2A A1 1B B4 4B B3 3B B2 2B B1 16.2：用两块4位数值比较器芯片实现两个7位二进制的比较6.3：用3-8线译码器74和必要逻辑门实现以下函数yxxyzyxFyxzyxFzxyyxzyxF),(),(),(321解：6106101),(mmmmmmzxyzyxzyxzxyyxzyxF7632107632102),(mmmmmmmmmmmmxyzzxyyzxzyxyzxzyxzyxzyxyxzyxF761076103),(mmmmmmmmxyzzxyzyxzyxyxxyzyxFA074LSY0A1A2G2AG1G2BY1Y2Y3Y4Y5Y6Y7&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