Digital Design, Global Edition, 6th edition M Morris R Mano Solution Manual

Type:

Solution Manual

Resource:

Digital Design

Edition:

6th Edition

Author(s):

M. Morris Mano

.

1

SOLUTIONS MANUAL PART 1: CHAPTERS 1-5 Rev 06/05/2017

DIGITAL DESIGN WITH AN INTRODUCTION to the VERILOG HDL, VHDL, and SystemVerilog Sixth Edition Global Edition

M. MORRIS MANO Professor Emeritus California State University, Los Angeles

MICHAEL D. CILETTI Professor Emeritus University of Colorado, Colorado Springs Note: Solutions to problems requiring HDL code are presented in Verilog and VHDL

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CHAPTER 1 1.1 Octal and hexadecimal numbers from 5010 to 6410: Base 10 50 51 52 53 54 55 56 57 58 59 (decimal) Base 8 62 63 64 65 66 67 70 71 72 73 (octal) Base 16 32 33 34 35 36 37 38 39 3A 3B (hexadecimal)

60

61

62

63

64

74

75

76

77

100

3C 3D

3E

3F

40

Numbers from 1010 to 3010 in base 14: Base 10

11

12

13

14

15

16

17

18

19

20

21

22

23

24

25

26

27

28

29

30

Base 14

B

C

D

10

11

12

13

14

15

16

17

18

19

1A

1B

1C

1D

20

21

22

1.2

(a) 16K bytes = 16  210 = 16384 bytes (b) 32M bytes = 32  220 = 33554432 bytes (c) 2G bytes = 2  230 = 2147483648 bytes

1.3

(a) (1203)4 = 1× 43 + 2 × 42 + 3 × 40 = 9910 (b) (5243)6 = 5 × 63 + 2 × 62 + 4 × 61 + 3 × 60 = 117910 (c) (9922)14 = 9 × 143 + 9 × 142 + 2 × 141 + 2 × 140 = 2649010 (d) (248)9 = 2 × 92 + 4 × 91 + 8 × 90 = 20610

1.4

Maximum number with 12 bits in binary: 1111_1111_1111 Decimal Equivalent: = 211 × 210 × 29 × 28 × 27 × 26 × 25 × 24 × 23 × 22 × 21 × 20 = 4,09510 Octal Equivalent: 77778 Hexadecimal Equivalent: FFF16

1.5

Let b = base of the number system

6* b  7

 1* b  3 , 6b  7  5b  15, b  8 5 (b) 15  3  51, 15b  3b  51b ,so b  5 *3  5* b  1,so b  7 (a) 67 / 5  13,

67b / 5b  13b ,so

(c) 123  120  303, 123b  120b  303b



 



so b 2  2* b  3  b 2  2* b  3* b 2  3, so 2b 2  4b  3  3b 2  3, so b 2  4b,so b  4 1.6

 x  7 x  2  x 2 – 7  2 x  7 * 2  x 2  13x  22 Therefore: 7  2  b  3,so b  6 Also, 7 * 2  1410  226

1.7 ..

CA5E16 = 1100_1010_0101_11102 = 1_100_101_001_011_1102 = (145136)8

3 1.8

(a) Results of repeated division by 2 (quotients are followed by remainders in brackets): 25310 = 126(1); 63(0); 31(1); 15(1);7(1); 3(1); 1(1);0(1) Answer∶ 1111_110110 (b) Results of repeated division by 16 (quotients are followed by remainders in brackets): 25310 = 15(13); 0(15) Answer∶ FD16 = 1111_110110 Method (b) is faster since the number of steps for repeated division is much less. 〖

1.9

(a) (10101.101)2 = 16 + 4 + 1 + 0.5 + 0.125 = (21.625)10 (b) 64.816  6  16  4 

8  100.510 16

4 4  177.562510 (c) 261.448  2  64  6  8  1   8 64 (d) 51DE.C 16  5  163  162  13  16  14 

12   20,958.7510 16

1 (e) 110011.0012  32  16  2  1   51.12510 8 1.10

(a) 1.000112  0001.0001 10002  1.1816  1 

1000.112  1000.11002  8.C16  8 

1 8   1.0937510 16 256

12  8.7510 16

Reason: If we left shift 1.000112 by three places, we get 1000.112. Hence the value of 1000.112 is 23 = 8 times more. 1.11

Perform the following division in binary: 101010 ÷ 100. 1010.1 100 | 101010.0 100 101 100 100 100 0 Checking: 1010102 ÷ 1002 = 4210 ÷ 410 ≅ 10.510 = 1010.12

4 1.12

(a) 1101 and 110. 1101 + 110 10011  11012 + 1102 = 100112 Multiply 1101 and 110: 1101 × 110 0000 1101 1101 . 1001110  11012 × 1102 = 10011102 (b) D0h and 1Fh

D0 + 1F EF  D016 + 1F16 = EF16 Multiply D0 and 1F: D0 × 1F C30 D0 . 1930  D016 × 1F16 = 193016

1.13

(a) Integer portion of 45.125 = 45 45/2 = 22/2 11/2 5/2 2/2 1/2

Integer Quotient 22 11 5 2 1 0

Remainder 1 0 1 1 0 1

Coefficient a0 = 1 a1 = 0 a2 = 1 a3 = 1 a4 = 0 a5 = 1

4510 = 1011012 Fractional Portion of 45.125 = 0.125 Multiplication Integer 0.125 × 2 0 0.25 × 2 0 0.5 × 2 1 45.12510 = 101101.0012

Fraction .25 .5 0

Coefficient a–1 = 0 a–2 = 0 a–5 = 0