数字电子技术（佘新平 著） 华中科技大学出版社 课后答案-

1、第一章数制与编码1.1 自测练习1.1.1、模拟量数字量1.1.2、(b1.1.3、(c1.1.4、(a是数字量,(b(c(d是模拟量1.2 自测练习1.2.1. 21.2.2. 比特bit1.2.3. 101.2.4. 二进制1.2.5. 十进制1.2.6. (a1.2.7. (b 1.2.8. (c 1.2.9. (b 1.2.10. (b 1.2.11. (b 1.2.12. (a 1.2.13. (c 1.2.14. (c 1.2.15. (c 1.2.16. 1001001 1.2.17. 111.2.18. 110010 1.2.19. 1101 1.2.20. 8进制1.2.21

2、. (a1.2.22. 0,1,2,3,4,5,6,71.2.23. 十六进制1.2.24. 0,1,2,3,4,5,6,7,8,9,A,B,C,D,E,F 1.2.25. (b1.3 自测练习1.3.1. 1221.3.2. 675.521.3.3. 011111110.011.3.4. 521.3.5. 1BD.A81.3.6. 1110101111.11101.3.7. 38551.3.8. 28.3751.3.9. 100010.111.3.10. 135.6251.3.11. 570.11.3.12. 120.51.3.13. 2659.A1.4 自测练习1.4.1. BCD Bin

3、ary coded decimal 二十进制码1.4.2. (a1.4.3. (b1.4.4. 8421BCD码,4221BCD码,5421BCD 1.4.5. (a1.4.6. 011001111001.10001.4.7. 111111101.4.8. 101010001.4.9. 111111011.4.10. 61.051.4.11. 01011001.01110101 1.4.12. 余3码1.4.13. XS31.4.14. XS31.4.15. 1000.10111.4.16. 1001100000111.4.17. 521.4.18. 110101.4.19. 0101111.

4、4.20. (b1.4.21. ASCII1.4.22. (a1.4.23. ASCII American Standard Code for Information Interchange美国信息交换标准码EBCDIC Extended Binary Coded Decimal Interchange Code 扩展二-十进制交换吗1.4.24. 10010111.4.25. ASCII1.4.26. (b1.4.27. (b1.4.28. 110111011.4.29. -1111.4.30. +231.4.31. -231.4.32. -861.5 自测练习1.5.1 略1.5.2 11

5、0111011.5.3 010001011.5.4 11100110 补码形式1.5.5 011111011.5.6 10001000 补码形式1.5.7 11100010 补码形式习题1.1 (a(d是数字量,(b(c是模拟量,用数字表时(e是数字量,用模拟表时(e是模拟量1.2 (a7, (b31, (c127, (d511, (e40951.3 (a,(b,(c(d522104108×+×+320410910×+×+×26108108×+×+321102105100×+×+×+21+1.4

6、 (a, (b, (c(d3212121×+×+984+12+12+×××4311212121×+×+×+212×64212+12+12+12+1××××1212+×2220110327.15310210710110510.=×+×+×+×+×3210-1-221011.0112+02+12+12+02+12=××××××210-18437.448

7、+38+78+48=××××10-1-2163A.1C316+A16+116+C16=××××,1.6 (a11110, (b100110,(c110010, (d10111.7 (a1001010110000, (b10010111111.8 110102 = 2610, 1011.0112 = 11.37510,57.6438 = 71.81835937510,76.EB16 = 118. 91796875101.9 1101010010012 = 65118 = D4916,0.100112 = 0.468

8、= 0.9816,1011111.011012 = 137.328 = 5F.68161.10 168 = 1410,1728 = 12210,61.538 = 49.671875,126.748 = 86.9375101.11 2A16 = 4210 = 1010102 = 528,B2F16 = 286310 = 1011001011112 = 54578,D3.E16= 211.87510 = 11010011.11102 = 323.78,1C3.F916 = 451.9726562510 =111000011.111110012 = 703.76281.12 (aE, (b2E, (

9、c1B3, (d3491.13 (a22, (b110, (c1053, (d20631.14 (a4094, (b1386, (c492821.15 (a23, (b440, (c27771.16 198610 = 111110000102 = 00011001100001108421BCD,67.31110 = 1000011.010012 =01100111.0011000100018421BCD, 1.183410 = 1.0010112 = 0001.00011000001101008421BCD ,0.904710 = 0.1110012 = 0000.10010000010001

10、118421BCD1.17 1310 = 000100118421BCD = 01000110XS3 = 1011Gray, 6.2510 = 0110.001001018421BCD =1001.01011000 XS3 = 0101.01Gray,0.12510 = 0000.0001001001018421BCD =0.010*XS3 = 0.001 Gray1.18 101102 = 11101 Gray,0101102 = 011101 Gray1.19 110110112 = 0010000110018421BCD,45610 = 0100010101108421BCD,1748=

11、0010011101008421BCD,2DA16 = 0111001100008421BCD,101100112421BCD = 010*BCD,11000011XS3 = 100100008421BCD1.20 0.0000原= 0.0000反= 0.0000补,0.1001原= 0.1001反= 0.1001补,11001原= 10110反= 10111补1.21 010100原= 010100补,101011原= 110101补,110010原= 101110补,100001原= 111111补1.22 1310 = 00001101补,11010 = 01101110补,-2510

12、= 11100111补,-90 =10100110补1.23 01110000补= 11210,00011111补= 3110,11011001补= -3910,11001000补= -56101.24 1000011 1000001 1010101 1010100 1001001 1001111 1001110 0100001 0100000 1001000 1101001 1100111 1101000 0100000 1010110 1101111 1101100 1110100 1100001 1100111 11001011.25 0100010 1011000 0100000 01

13、11101 0100000 0110010 0110101 0101111 1011001 01000101.26 BEN SMITH1.27 00000110 100001101.28 01110110 10001110第二章逻辑门2.1 自测练习2.1.1. (b 2.1.2. 162.1.3. 32, 6 2.1.4. 与2.1.5. (d 2.1.6. 162.1.7. 32, 6 2.1.8. 或2.1.9. 非2.1.10. 12.2 自测练习2.2.1. FAB=.2.2.2. (b2.2.3. 高2.2.4. 322.2.5. 16,52.2.6. 12.2.7. 串联2.2.

14、8. (d2.2.9. 不相同2.2.10. 高2.2.11. 相同2.2.12. (a2.2.13. (c2.2.14. 奇2.3 自测练习2.3.1. OC,上拉电阻2.3.2. 0,1,高阻2.3.3. (b2.3.4. (c2.3.5. ,FAB=. 高阻2.3.6. 不能2.4 自测练习2.4.1. TTL,CMOS2.4.2. Transisitor Transistor Logic2.4.3. Complementary Metal Oxide Semicoductor2.4.4. 高级肖特基TTL,低功耗和高级低功耗肖特基TTL 2.4.5. 高,强,小2.4.6. (c2.4

15、.7. (b2.4.8. (c2.4.9. 大2.4.10. 强2.4.11. (a2.4.12. (a2.4.13. (b2.4.14. 高级肖特基TTL2.4.15. (c习题2.1 与,或,与2.2 与门,或门,与门2.3 (aF=A+B, F=AB (bF=A+B+C, F=ABC (cF=A+B+C+D, F=ABCD 2.4 (a0 (b1 (c0 (d02.5 (a0 (b0 (c1 (d02.6 (a1 (b1 (c1 (d12.7 (a4 (b8 (c16 (d322.8 (a3 (b4 (c5 (d6ABCF0 0 0 0 1 1 0 1 0 1 0 1 1 0 1 0 0

16、 1 1 0 11111112.9 (a(b ABCDF1110 1 1 1 0 1 0 0 00 1 0 1 1 0 1 1 0 1 0 1 1 1 0 1 0 0 0 0 1 00 1 1 1 0 1 0 1 1 0 1 1 0 1 1 0 0 1 1 1 0 111111112.10 YABAC=+ 2.11ABCY0 0 0 0 1 0 0 1 0 1 0 1 1 1 1 0 0 0 1 0 1 11111112.122.13F1 = A(B+C, F2=A+BCA B C F1 F2 0 0 0 0 0 0 0 1 0 0 0 1 0 00 0 1 1 0 1 1 0 1 1 1 1

17、 0 0 0 1 1 1 0 1 1 111112.142.15 (a0 (b1 (c1 (d02.16 (a1 (b0 (c0 (d12.17 (a0 (b02.182.19 CDEF.2.20 CY ABDF.2.21 102.22 402.23 当TTL反相器的输出为3V,输出是高电平,红灯亮。当TTL反相器的输出为0.2 V时,输出是低电平,绿灯亮。2.24 当TTL反相器输出高电平时三极管会导通,LED灯会点亮;当TTL反相器输出低电平时三极管不会导通,LED灯不会点亮。3.1自测练习答案1. 逻辑函数2. 逻辑表达式、真值表、逻辑电路图、卡诺图和波形图3.表3-1 真值表ABC11

18、11111 1 1 0 0 1 1 0 1 1 1 1 0 1 1 1 1 04.ACD+BAC+5. (略3.2自测练习答案1. 与、或、非2. 代入规则、反演规则、对偶规则3. a和c4. a d5. a6. 、7. 、8. F3.3自测练习答案1. A2. AD3.4.5.6.7.8.9.3.4自测练习答案1. 标准与或表达式、标准或与表达式2. 1、3.4. 最大项5. 4,5,6,7,12,13,14,156.7.8.9.A B CF0 0 00 0 10 1 00 1 11 0 0 1 0 1 1 1 0 1 1 1 0111110.11.12. C3.5自测练习答案1. 12.3

19、. 格雷码4.、*F5. m66. M17.(A=AF*=8.9.+10.第三章练习答案3.1、(3.2、(a1,0,0 (b1,1,1 (c0,1,03.3.略3.4.(aCBC+(bF3.5(a(b3.6 提示: 列出真值表可知: (1不正确, (2不正确, (3 正确, (4 正确3.7(a(bABC (cF(d(e(fAB(g(hF (iF=(j3.83.9 (a(b3.10 函数Y和函数Z互补, 即:Y 3.11、111 1 03.12 3.133.14 3.15 3.163.17 AB习题4.1写出图所示电路的逻辑表达式,并说明电路实现哪种逻辑门的功能。ABABA习题4.1图解:该电路实现异或门的功能4.2分析图所示电路,写出输出函数F。习题4.2图解:4.3已知图示电路及输入A、B的波形,试画出相应的输出波形F,不计门的延迟.习题4.3图解:4.4由与非门构成的某表决电路如图