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首页Floyd《数字电子技术第10版》答案解析教程

Floyd《数字电子技术第10版》答案解析教程

数字电子技术

Digital

Fundamentals

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《数字电子技术第十版》(Digital Fundamentals 10th Ed) 是由Thomas L. Floyd编写的经典教材，该书深入浅出地介绍了数字电路和系统的基础知识。这本书是电子工程和技术专业学生的必备参考书，它涵盖了数字逻辑、计算机体系结构、数字信号处理等多个核心主题，旨在帮助学生理解和应用这些概念于实际设计中。该教材第十版的配套在线教师手册提供了丰富的教学支持，包括详细解答和习题解析，这对于学习者来说是宝贵的补充资源。教师可以利用这些解答来指导课堂讨论，帮助学生解决在学习过程中遇到的问题。版权信息显示，这本书受到Pearson Education Inc.的保护，所有复制、存储、检索或任何形式的传播都需要事先获得出版商的许可，以确保尊重知识产权。作为一本权威教材，Floyd的《数字电子技术第十版》强调理论与实践的结合，不仅包含理论知识的讲解，还有大量的习题和实验设计，让学生通过动手实践加深对理论的理解。例如，章节可能涵盖了布尔代数、数字门电路（如与门、或门、非门等）、触发器、组合逻辑电路、时序逻辑电路的设计和分析，以及数字信号的量化和编码等关键概念。在课程学习中，教师可能会引导学生通过分析Floyd给出的解决方案来理解电路行为、电路性能优化以及如何解决实际问题。此外，书中还可能涉及模拟电子技术与数字电子技术的对比，以及现代数字系统如微处理器和FPGA的基础知识。《Digital Fundamentals 10th Ed》是一本全面且实用的数字电子技术教材，对于希望在这个领域深入学习的学生和教师来说，它不仅是理论知识的源泉，也是提高实践技能的重要工具。

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Chapter 2

1001

11(a)

17.

10101001

1101

0000

1101

1101(e)

1000

100

000

100(b)

10110110

1110

0000

1110

1101

1110(f)

100011

111

000

111

101

111(c)

110110

1001

0000

110

1001(d)

18.

(a)

100

= 010 (b)

0011

1001

= 0011 (c)

0100

1100

= 0011

Section 2-5 1’s and 2’s Complements of Binary Numbers

19. Zero is represented in 1’s complement as all 0’s (for +0) or all 1’s (for −0).

20. Zero is represented by all 0’s only in 2’s complement.

21.

(a) The 1’s complement of 101 is 010.

(b) The 1’s complement of 110 is 001.

(d) The 1’s complement of 11010111 is 00101000.

(e) The 1’s complement of 1110101 is 0001010.

(f) The 1’s complement of 00001 is 11110.

22. Take the 1’s complement and add 1:

(a) 01 + 1 = 10 (b) 000 + 1 = 001

(e) 00011 + 1 = 00100 (f) 01100 + 1 = 01101

(g) 01001111 + 1 = 01010000 (h) 11000010 + 1 = 11000011

http://angoothachaap.blogspot.com

Chapter 2

Section 2-6 Signed Numbers

23. (a) Magnitude of 29 = 0011101 (b) Magnitude of 85 = 1010101

+ 29 = 00011101 −85 = 11010101

= 1100100 (d) Magnitude of 123 = 1111011

+100 = 01100100 −123 = 11111011

24. (a) Magnitude of 34 = 0100010 (b) Magnitude of 57 = 0111001

−34 = 11011101 +57 = 00111001

−99 = 10011100 +115 = 01110011

25. (a) Magnitude of 12 = 1100 (b) Magnitude of 68 = 1000100

+12 = 00001100 −68 = 10111100

= 1100101 (d) Magnitude of 125 = 1111101

+101

= 01100101 −125 = 10000011

26. (a) 10011001 = −25 (b) 01110100 = +116 (c) 10111111 = −63

27. (a) 10011001 = −(01100110) = −102

(b) 01110100 = +(1110100) = +116

28. (a) 10011001 = −(1100111) = −103

(b) 01110100 = +(1110100) = +116

29. (a) 0111110000101011 → sign = 0

1.11110000101011 × 2

→ exponent = 127 + 14 + 141 = 10001101

Mantissa = 11110000101011000000000

01000110111110000101011000000000

(b) 100110000011000 → sign = 1

1.10000011000 × 2

→ exponent = 127 + 11 = 138 = 10001010

Mantissa = 11000001100000000000000

11000101011000001100000000000000

30. (a) 11000000101001001110001000000000

Sign = 1

Exponent = 10000001 = 129 − 127 = 2

Mantissa = 1.01001001110001 × 2

= 101.001001110001

−101.001001110001 = −

5.15258789

(b) 01100110010000111110100100000000

Sign = 0

Exponent = 11001100 = 204 − 127 = 77

Mantissa = 1.100001111101001

1.100001111101001 × 2

http://angoothachaap.blogspot.com

Chapter 2

Section 2-7 Arithmetic Operations with Signed Numbers

31. (a) 33 = 00100001 00100001

15 = 00001111 + 00001111

00110000

(c)

46 = 00101110 11010010

−46 = 11010010 + 00011001

25 = 00011001 11101011

(b) 56 = 00111000 00111000

27 = 00011011 + 11100101

−27 = 11100101 00011101

(d)

110

= 01101110 10010010

−110

= 10010010 + 10101100

84 = 01010100

100111110

−84 = 10101100

32. (a) 00010110 (b) 01110000

+ 00110011

+ 10101111

01001001

100011111

33. (a) 10001100 (b) 11011001

+ 00111001

+ 11100111

11000101

11000000

34. (a) 00110011 00110011

− 00010000

+ 11110000

00100011

(b)

01100101 01100101

− 11101000

+ 00011000

01111101

35. 01101010 01101010

× 11110001

× 00001111

01101010

100111110

01101010

1011100110

01101010

11000110110

Changing to 2’s complement with sign: 100111001010

36.

00011001

01000100

= 00000010

= 2, remainder of 18

Section 2-8 Hexadecimal Numbers

37.

(a) 38

= 0011 1000

(b) 59

= 0101 1001

= 1010 0001 0100

(d) 5C8

= 0101 1100 1000

(e) 4100

= 0100 0001 0000 0000

(f) FB17

= 1111 1011 0001 0111

(g) 8A9D

= 1000 1010 1001 1101

http://angoothachaap.blogspot.com

Chapter 2

38. (a) 1110 = E

(b) 10 = 2

(d) 1010 0110 = A6

(e) 0011 1111 0000 = 3F0

(f) 1001 1000 0010 = 982

39. (a) 23

= 2 × 16

+ 3 × 16

= 32 + 3 = 35

(b) 92

= 9 × 16

+ 2 × 16

= 144 + 2 = 146

= 1 × 16

+ 10 × 16

= 16 + 10 = 26

(d) 8D

= 8 × 16

+ 13 × 16

= 128 + 13 = 141

(e) F3

= 15 × 16

+ 3 × 16

= 240 + 3 = 243

(f) EB

= 14 × 16

+ 11 × 16

= 224 + 11 = 235

(g) 5C2

= 5 × 16

+ 12 × 16

+ 2 × 16

= 1280 + 192 + 2 = 1474

(h) 700

= 7 × 16

= 1792

40. (a)

= 0, remainder = 8

hexadecimal number = 8

(c)

= 2, remainder = 1 (LSD)

= 0, remainder = 2

hexadecimal number = 21

(e)

284

= 17, remainder = 12 = C

(LSD)

= 1, remainder = 1

= 0, remainder = 1

hexadecimal number = 11C

(g)

4019

= 251, remainder = 3 (LSD)

251

= 15, remainder = 11 = B

= 0, remainder = 15 = F

hexadecimal number = FB3

(b)

= 0, remainder = 14 = E

hexadecimal number = E

(d)

= 3, remainder = 4 (LSD)

= 0, remainder = 3

hexadecimal number = 34

(f)

2890

= 180, remainder = 10 = A

(LSD)

180

= 11, remainder = 4

= , remainder = 11 = B

hexadecimal number = B4A

(h)

6500

= 406, remainder = 4 (LSD)

406

= 25, remainder = 6

= 1, remainder = 9

= 0, remainder = 1

hexadecimal number = 1964

41. (a) 37

+ 29

= 60

(b) A0

+ 6B

= 10B

+ BB

= 1BA

http://angoothachaap.blogspot.com

Chapter 2

42. (a) 51

− 40

= 11

(b) C8

− 3A

= 8E

− 88

= 75

Section 2-9 Octal Numbers

43. (a) 12

= 1 × 8

+ 2 × 8

= 8 + 2 = 10

(b) 27

= 2 × 8

+ 7 × 8

= 16 + 7 = 23

= 5 × 8

+ 6 × 8

= 40 + 6 = 46

(d) 64

= 6 × 8

+ 4 × 8

= 48 + 4 = 52

(e) 103

= 1 × 8

+ 3 × 8

= 64 + 3 = 67

(f) 557

= 5 × 8

+ 5 × 8

+ 7 × 8

= 320 + 40 + 7 = 367

(g) 163

= 1 × 8

+ 6 × 8

+ 3 × 8

= 64 + 48 + 3 = 115

(h) 1024

= 1 × 8

+ 2 × 8

+ 4 × 8

= 512 + 16 + 4 = 532

(i) 7765

= 7 × 8

+ 7 × 8

+ 6 × 8

+ 5 × 8

= 3584 + 448 + 48 + 5 = 4085

44. (a)

= 1, remainder = 7 (LSD)

= 0, remainder =1

octal number = 17

(c)

= 5, remainder = 6 (LSD)

= 0, remainder = 5

octal number = 56

(e)

100

= 12, remainder = 4 (LSD)

= 1, remainder = 4

= 0, remainder = 1

octal number = 144

(g)

219

= 27, remainder = 3 (LSD)

= 3, remainder = 3

= 0, remainder = 3

octal number = 333

(b)

= 3, remainder = 3 (LSD)

= 0, remainder = 3

octal number = 33

(d)

= 8, remainder = 6 (LSD)

= 1, remainder = 0

= 0, remainder = 1

octal number = 106

(f)

142

= 17, remainder = 6 (LSD)

= 2, remainder = 1

= 0, remainder = 2

octal number = 216

(h)

435

= 54, remainder = 3 (LSD)

= 6, remainder = 6

= 0, remainder = 6

octal number = 663

http://angoothachaap.blogspot.com

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