Chapter 2 Problem Solutions 2.1. Decimal 32 33 34 35

8/14/2019 Chapter 2 Problem Solutions 2.1. Decimal 32 33 34 35

1/34

- 2.1 -

Chapter 2Problem Solutions

2.1.

Decimal Binary Ternary Octal Hexadecimal

32 100000 1012 40 20

33 100001 1020 41 21

34 100010 1021 42 22

35 100011 1022 43 23

36 100100 1100 44 24

37 100101 1101 45 25

38 100110 1102 46 26

39 100111 1110 47 27

40 101000 1111 50 28

41 101001 1112 51 29

42 101010 1120 52 2A

43 101011 1121 53 2B

44 101100 1122 54 2C

45 101101 1200 55 2D

46 101110 1201 56 2E

47 101111 1202 57 2F

48 110000 1210 60 30

49 110001 1211 61 31

50 110010 1212 62 32


2/34

- 2.2 -

2.2.

Decimal Quaternary Quinary Duodecimal

0 0 0 0

1 1 1 1

2 2 2 23 3 3 3

4 10 4 4

5 11 10 5

6 12 11 6

7 13 12 7

8 20 13 8

9 21 14 9

10 22 20 A

11 23 21 B

12 30 22 10

13 31 23 11

14 32 24 12

15 33 30 13

16 100 31 14

17 101 32 15

18 102 33 16

19 103 34 17

20 110 40 18

21 111 41 19

22 112 42 1A

23 113 43 1B

24 120 44 20

25 121 100 21

26 122 101 22

27 123 102 23

28 130 103 24

29 131 104 25

30 132 110 26

31 133 111 27


3/34

- 2.3 -

2.3. (a) 1 1 (b) 1111 (c) 11 11 11101 1001 1100.01

+ 1001_____ + 111_____ + 101.11________

10110 10000 10010.00

(d) 111 11 (e) 111 (f) 1 11

11.011 111.01 0.1110+ 10.111_______ + 11.10_______ +0.1011______

110.010 1010.11 1.1001

2.4. (a) 00 (b) 001 (c) 0110 11/1/01 1/1/0/01.1 1/0/0/1/.0/0

- 110____ - 1011.0_______ - 11.01_______111 1110.1 101.11

(d) 0 0 (e) 01 0 (f) 0 11/01/.01 1/0/0.11/01 1011/.0/0

- 10.10______ - 11.1010________ - 10.11_______10.11 1.0011 1000.01

2.5. (a) 10111 (b) 11011

110_____ 1011_____

00000 1101110111 11011

10111________ 00000

10001010 11011_________100101001

(c) 1010 (d) 101.1

1.01____ 11.01_____

10 10 1 011000 0 00 00

1010_______ 101 1

1100.10 1011_________10001.111

2.6. (a) 10.1 ________ (b) 110011____________

1010)11001.0 10101)10000101111-1010____ - 10101______

101 0 11000-101 0 _____ -10101_____

0 11111

-10101_____10101

-10101_____

0


4/34

- 2.4 -

2.6. (continued)

(c) 110.101 ___________ (d) 101 1.01____________

110)100111.110 101.1 )111101.1 11- 110 ____ -1011____

111 10001

-110 ___ - 1011_____11 1 110 1

-11 0 ____ -101 1_____

110 1 0 11-110 ___ -1 0 11______

0 0

2.7. (a) 1 1 (b) 11 1 (c) 1202012.0 220.12 2/0/1/02

+ 1102.1_______ + 121.20_______ -12121_____

10121.1 1112.02 211

(d) 012 1 (e) 120.21 (f) 122021/2/0/02/.12 122______ 21.2_____

- 2121.20________ 1011 12 10211 1

2110.22 10111 2 1220212021_________ 102111_________

100221.02 1122000.1

(g) 210.2 _________ (h) 1 2.02___________

1012)221200.1 212.1 )11111.2 12-2101 ____ - 2121_____

1110 1220 2

-1012 ____ -1201 2______210 1 12 0 12

-210 1 _____ -12 0 12_______

0 0

2.8. (a) Quaternary

a+b b

0 1 2 3

a

0 0 1 2 3

1 1 2 3 1:0

2 2 3 1:0 1:1

3 3 1:0 1:1 1:2

Notation: x:y denotes carry digit x and sum digit y.


5/34

- 2.5 -

2.8. (continued)

a-b b

0 1 2 3

a

0 0 1:3 1:2 1:1

1 1 0 1:3 1:2

2 2 1 0 1:3

3 3 2 1 0

Notation: x:y denotes borrow digit x and difference digit y.

ab b

0 1 2 3

a

0 0 0 0 0

1 0 1 2 3

2 0 2 1:0 1:2

3 0 3 1:2 2:1

Notation: x:y denotes carry digit x and product digit y.

(b) Octal

a+b b0 1 2 3 4 5 6 7

a

0 0 1 2 3 4 5 6 7

1 1 2 3 4 5 6 7 1:0

2 2 3 4 5 6 7 1:0 1:1

3 3 4 5 6 7 1:0 1:1 1:2

4 4 5 6 7 1:0 1:1 1:2 1:3

5 5 6 7 1:0 1:1 1:2 1:3 1:46 6 7 1:0 1:1 1:2 1:3 1:4 1:5

7 7 1:0 1:1 1:2 1:3 1:4 1:5 1:6



6/34

- 2.6 -

2.8. (continued)

a-b b

0 1 2 3 4 5 6 7

a

0 0 1:7 1:6 1:5 1:4 1:3 1:2 1:1

1 1 0 1:7 1:6 1:5 1:4 1:3 1:2

2 2 1 0 1:7 1:6 1:5 1:4 1:3

3 3 2 1 0 1:7 1:6 1:5 1:4

4 4 3 2 1 0 1:7 1:6 1:5

5 5 4 3 2 1 0 1:7 1:6

6 6 5 4 3 2 1 0 1:7

7 7 6 5 4 3 2 1 0


ab b

0 1 2 3 4 5 6 7

a

0 0 0 0 0 0 0 0 0

1 0 1 2 3 4 5 6 7

2 0 2 4 6 1:0 1:2 1:4 1:6

3 0 3 6 1:1 1:4 1:7 2:2 2:5

4 0 4 1:0 1:4 2:0 2:4 3:0 3:4

5 0 5 1:2 1:7 2:4 3:1 3:6 4:3

6 0 6 1:4 2:2 3:0 3:6 4:4 5:2

7 0 7 1:6 2:5 3:4 4:3 5:2 6:1



7/34

- 2.7 -

2.8. (continued)

(c) Hexadecimal

a+b b

0 1 2 3 4 5 6 7 8 9 A B C D E F

a

0 0 1 2 3 4 5 6 7 8 9 A B C D E F

1 1 2 3 4 5 6 7 8 9 A B C D E F 1:0

2 2 3 4 5 6 7 8 9 A B C D E F 1:0 1:1

3 3 4 5 6 7 8 9 A B C D E F 1:0 1:1 1:2

4 4 5 6 7 8 9 A B C D E F 1:0 1:1 1:2 1:3

5 5 6 7 8 9 A B C D E F 1:0 1:1 1:2 1:3 1:4

6 6 7 8 9 A B C D E F 1:0 1:1 1:2 1:3 1:4 1:5

7 7 8 9 A B C D E F 1:0 1:1 1:2 1:3 1:4 1:5 1:6

8 8 9 A B C D E F 1:0 1:1 1:2 1:3 1:4 1:5 1:6 1:7

9 9 A B C D E F 1:0 1:1 1:2 1:3 1:4 1:5 1:6 1:7 1:8

A A B C D E F 1:0 1:1 1:2 1:3 1:4 1:5 1:6 1:7 1:8 1:9

B B C D E F 1:0 1:1 1:2 1:3 1:4 1:5 1:6 1:7 1:8 1:9 1:A

C C D E F 1:0 1:1 1:2 1:3 1:4 1:5 1:6 1:7 1:8 1:9 1:A 1:B

D D E F 1:0 1:1 1:2 1:3 1:4 1:5 1:6 1:7 1:8 1:9 1:A 1:B 1:C

E E F 1:0 1:1 1:2 1:3 1:4 1:5 1:6 1:7 1:8 1:9 1:A 1:B 1:C 1:D

F F 1:0 1:1 1:2 1:3 1:4 1:5 1:6 1:7 1:8 1:9 1:A 1:B 1:C 1:D 1:E



8/34

- 2.8 -

2.8. (continued)

a-b b

0 1 2 3 4 5 6 7 8 9 A B C D E F

a

0 0 1:F 1:E 1:D 1:C 1:B 1:A 1:9 1:8 1:7 1:6 1:5 1:4 1:3 1:2 1:1

1 1 0 1:F 1:E 1:D 1:C 1:B 1:A 1:9 1:8 1:7 1:6 1:5 1:4 1:3 1:2

2 2 1 0 1:F 1:E 1:D 1:C 1:B 1:A 1:9 1:8 1:7 1:6 1:5 1:4 1:3

3 3 2 1 0 1:F 1:E 1:D 1:C 1:B 1:A 1:9 1:8 1:7 1:6 1:5 1:4

4 4 3 2 1 0 1:F 1:E 1:D 1:C 1:B 1:A 1:9 1:8 1:7 1:6 1:5

5 5 4 3 2 1 0 1:F 1:E 1:D 1:C 1:B 1:A 1:9 1:8 1:7 1:6

6 6 5 4 3 2 1 0 1:F 1:E 1:D 1:C 1:B 1:A 1:9 1:8 1:7

7 7 6 5 4 3 2 1 0 1:F 1:E 1:D 1:C 1:B 1:A 1:9 1:8

8 8 7 6 5 4 3 2 1 0 1:F 1:E 1:D 1:C 1:B 1:A 1:9

9 9 8 7 6 5 4 3 2 1 0 1:F 1:E 1:D 1:C 1:B 1:A

A A 9 8 7 6 5 4 3 2 1 0 1:F 1:E 1:D 1:C 1:B

B B A 9 8 7 6 5 4 3 2 1 0 1:F 1:E 1:D 1:C

C C B A 9 8 7 6 5 4 3 2 1 0 1:F 1:E 1:D

D D C B A 9 8 7 6 5 4 3 2 1 0 1:F 1:E

E E D C B A 9 8 7 6 5 4 3 2 1 0 1:F

F F E D C B A 9 8 7 6 5 4 3 2 1 0



9/34

- 2.9 -

2.8. (continued)

ab b

0 1 2 3 4 5 6 7 8 9 A B C D E F

a

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

1 0 1 2 3 4 5 6 7 8 9 A B C D E F

2 0 2 4 6 8 A C E 1:0 1:2 1:4 1:6 1:8 1:A 1:C 1:E

3 0 3 6 9 C F 1:2 1:5 1:8 1:B 1:E 2:1 2:4 2:7 2:A 2:D

4 0 4 8 C 1:0 1:4 1:8 1:C 2:0 2:4 2:8 2:C 3:0 3:4 3:8 3:C

5 0 5 A F 1:4 1:9 1:E 2:3 2:8 2:D 3:2 3:7 3:C 4:1 4:6 4:B

6 0 6 C 1:2 1:8 1:E 2:4 2:A 3:0 3:6 3:C 4:2 4:8 4:E 5:4 5:A

7 0 7 E 1:5 1:C 2:3 2:A 3:1 3:8 3:F 4:6 4:D 5:4 5:B 6:2 6:9

8 0 8 1:0 1:8 2:0 2:8 3:0 3:8 4:0 4:8 5:0 5:8 6:0 6:8 7:0 7:8

9 0 9 1:2 1:B 2:4 2:D 3:6 3:F 4:8 5:1 5:A 6:3 6:C 7:5 7:E 8:7

A 0 A 1:4 1:E 2:8 3:2 3:C 4:6 5:0 5:A 6:4 6:E 7:8 8:2 8:C 9:6

B 0 B 1:6 2:1 2:C 3:7 4:2 4:D 5:8 6:3 6:E 7:9 8:4 8:F 9:A A:5

C 0 C 1:8 2:4 3:0 3:C 4:8 5:4 6:0 6:C 7:8 8:4 9:0 9:C A:8 B:4

D 0 D 1:A 2:7 3:4 4:1 4:E 5:B 6:8 7:5 8:2 8:F 9:C A:9 B:6 C:3

E 0 E 1:C 2:A 3:8 4:6 5:4 6:2 7:0 7:E 8:C 9:A A:8 B:6 C:4 D:2

F 0 F 1:E 2:D 3:C 4:B 5:A 6:9 7:8 8:7 9:6 A:5 B:4 C:3 D:2 E:1


2.9. (a) 11 1 (b) 0233 (c) 1111 1

31213(4)

1/3/0/0/12(4)

466735(8)

+ 23102(4)_________

- 33321(4)_________

+ 375627(8)__________

120321(4)

30031(4)

1064564(8)

(d) 6775 (e) 1 1 (f) C5 AF47/0/0/6/05

(8)8C9F65

(16)D/6/2B/0/5/3

(16)

-356742(8)_________ +374B27(16)__________ -47E3C89(16)___________

321643(8)

C3EA8C(16)

8E473CA(16)


10/34

- 2.10 -

2.10. (a) 101101.1(2)

= 1105+ 010

4+ 110

3+ 110

2+ 010

1

+ 1100+ 110

-1

/ 125+ 02

4+ 12

3+ 12

2+ 02

1

+ 120+ 12

-1

= 32 + 0 + 8 + 4 + 0 + 1 + 0.5

= 45.5(10)

(b) 110111.101(2)

= 1105+ 110

4+ 010

3+ 110

2+ 110

1

+ 1100+ 110

-1+ 010

-2+ 110

-3

/ 125+ 12

4+ 02

3+ 12

2+ 12

1

+ 120+ 12

-1+ 02

-2+ 12

-3

= 32 + 16 + 0 + 4 + 2 + 1 + 0.5 + 0

+ 0.125

= 55.625(10)

(c) 2110(3)

= 2103

+ 1102+ 110

1+ 010

0

/ 233+ 13

2+ 13

1+13

0

= 54 + 9 + 3 + 0

= 66(10)

(d) 12021.1(3)

= 1104+ 210

3+ 010

2+ 210

1+ 110

0

+ 110-1

/ 134 + 233 + 032 + 231 + 130 + 13-1

= 81 + 54 + 0 + 6 + 1 + 0.333...

= 142.333...(10)

(e) 362(8)

= 3102

+ 6101+ 210

0

/ 382+ 68

1+ 28

0

= 192 + 48 + 2

= 242(10)

(f) 1475.2(8) = 1103+ 410

2+ 710

1+ 510

0+ 210

-1

/ 183+ 48

2+ 78

1+ 58

0+ 28

-1

= 512 + 256 + 56 + 5 + 0.25

= 829.25(10)


11/34

- 2.11 -

2.10. (continued)

(g) 2C3(16)

= 2102

+ C101+ 310

0

/ 2162+ 1216

1+ 316

0

= 512 + 192 + 3

= 707(10)

(h) AD.E(16)

= A101+ D10

0+ E10

-1

/ 10161+ 1316

0+ 1416

-1

= 160 + 13 + 0.875

= 173.875(10)

2.11. (a) 42(10)

= 4101

+ 2100

/ 10010101+ 101010

0

= 101000 + 10

= 101010(2)

(b) 78.5(10)

= 7101+ 810

0+ 510

-1

/ 11110101+ 10001010

0+ 1011010

-1

= 1000110 + 1000 + 0.1

= 1001110.1(2)

(c) 201(3)

= 2102

+ 0101+ 110

0

/ 10112 + 0111 + 1110

= 10010 + 0 + 1

= 10011(2)

(d) 21.2(3)

= 2101

+ 1100+ 210

-1

/ 10111+ 111

0+ 1011

-1

= 110 + 1 + 0.101010...

= 111.101010...(2)

(e) 204(8) = 2102

+ 0101+ 410

0

/ 1010002+ 01000

1+ 1001000

0

= 10000000 + 0 + 100

= 10000100(2)


12/34

- 2.12 -

2.11. (continued)

(f) 56.3(8)

= 5101

+ 6100+ 310

-1

/ 10110001+ 1101000

0+ 111000

-1

= 101000 + 110 + 0.011

= 101110.011(2)

2.12. (a) 11010(2)

= 1104+ 110

3+ 010

2+ 110

1+ 010

0

/ 124+ 12

3+ 02

2+ 12

1+ 02

0

= 1121 + 122 + 011 + 12 + 01

= 121 + 22 + 0 + 2 + 0

= 222(3)

(b) 73.2

(8)

= 7101

+ 3100+ 210

-1

/ 21221+ 1022

0+ 222

-1

= 2002 + 10 + 0.020202...

= 2012.020202...(3)

(c) 75(10)

= 7101

+ 5100

/ 211011+ 12101

0

= 2121 + 12

= 2210(3)

(d) 3D(16)

= 3101 + D100

/ 101211+ 111121

0

= 1210 + 111

= 2021(3)


13/34

- 2.13 -

2.13. Using the polynomial method of converting 225(b)

to

decimal:

225(b)

= 2102+ 210

1+ 510

0

/ (2b2+ 2b

1+ 5b

0)(10)

= 89(10)

Solving the quadradic 2b2+2b-84=0 for b:

b = +6 or -7

Since the base of a number system is defined as the number

of symbols in the system, the base must be positive.

b = 6

2.14. (a) Remainders Integers

1632=81 1(10)

/1(2)

0.752=1.50 1(10)

/1(2)

812=40 1(10)

/1(2)

0.502=1.00 1(10)

/1(2)

402=20 0(10)

/0(2)

202=10 0(10)

/0(2)

102= 5 0(10)

/0(2)

52= 2 1(10)

/1(2)

22= 1 0(10)

/0(2)

12= 0 1

(10)

/1

(2) 163.75

(10)/ 10100011.11

(2)

Remainders Integers

2022=101 0(10)

/0(2)

0.92=1.8 1(10)

/1(2)

1012= 50 1(10)

/1(2)

0.82=1.6 1(10)

/1(2)

502= 25 0(10)

/0(2)

0.62=1.2 1(10)

/1(2)

252= 12 1(10)

/1(2)

0.22=0.4 0(10)

/0(2)

122= 6 0(10)

/0(2)

0.42=0.8 0(10)

/0(2)

62= 3 0(10)/0(2)

32= 1 1(10)

/1(2)

12= 0 1(10)

/1(2)

202.9(10)

/ 11001010.111001100...(2)


14/34

- 2.14 -

2.14. (continued)

(b) Remainders Integers

1633=54 1(10)

/1(3)

0.753=2.25 2(10)

/2(3)

543=18 0(10)

/0(3)

0.253=0.75 0(10)

/0(3)

183= 6 0(10)/0(3)

63= 2 0(10)

/0(3)

23= 0 2(10)

/2(3)

163.75(10)

/ 20001.2020...(3)

Remainders Integers

2023=67 1(10)

/1(3)

0.93=2.7 2(10)

/2(3)

673=22 1(10)

/1(3)

0.73=2.1 2(10)

/2(3)

223= 7 1

(10)

/1

(3)

0.13=0.3 0

(10)

/0

(3)73= 2 1

(10)/1(3)

0.33=0.9 0(10)

/0(3)

23= 0 2(10)

/2(3)

202.9(10)

/ 21111.22002200...(3)

(c) Remainders Integers

1638=20 3(10)

/3(8)

0.758=6.00 6(10)

/6(8)

208= 2 4(10)

/4(8)

28= 0 2(10)

/2(8)

163.75(10)

/ 243.6(8)

Remainders Integers

2028=25 2(10)

/2(8)

0.98=7.2 7(10)

/7(8)

258= 3 1(10)

/1(8)

0.28=1.6 1(10)

/1(8)

38= 0 3(10)

/3(8)

0.68=4.8 4(10)

/4(8)

0.88=6.4 6(10)

/6(8)

0.48=3.2 3(10)

/3(8)

202.9(10)

/ 312.714631463...

(3)

(d) Remainders Integers

16316=10 3(10)

/3(16)

0.7516=12.00 12(10)

/C(16)

1016= 0 10(10)

/A(16)

163.75(10)

/ A3.C(16)


15/34

- 2.15 -

2.14. (continued)

Remainders Integers

20216=12 10(10)

/A(16)

0.916=14.4 14(10)

/E(16)

1216= 0 12(10)

/C(16)

0.416= 6.4 6(10)

/6(16)

202.9(10)/ CA.E66...(16)

2.15. (a) Remainders

1110001011=1001011 1(2)

/1(3)

100101111= 11001 0(2)

/0(3)

1100111= 1000 1(2)

/1(3)

100011= 10 10(2)

/2(3)

1011= 0 10(2)

/2(3)

Integers

0.110111=10.0111 10(2)

/2(3)

0.011111= 1.0101 1(2)

/1(3)

0.010111= 0.1111 0(2)

/0(3)

0.111111=10.1101 10(2)

/2(3)

11100010.1101(2)

/ 22101.21022102...(3)

(b) Remainders

111000101000=11100 10(2)/2(8)

111001000= 11 100(2)

/4(8)

111000= 0 11(2)

/3(8)

Integers

0.11011000=110.1000 110(2)

/6(8)

0.10001000=100.0000 100(2)

/4(8)

11100010.1101(2)

/ 342.64(8)

(c) Remainders

111000101010=10110 110(2)

/6(10)

101101010= 10 10(2)

/2(10)

101010= 0 10(2)

/2(10)


16/34

- 2.16 -

2.15. (continued)

Integers

0.11011010=1000.0010 1000(2)

/8(10)

0.00101010= 1.0100 1(2)

/1(10)

0.01001010= 10.1000 10(2)/2(10)

0.10001010= 101.0000 101(2)

/5(10)

11100010.1101(2)

/ 226.8125(10)

(d) Remainders

1110001010000=1110 10(2)

/2(16)

111010000= 0 1110(2)

/E(16)

Integers

0.110110000=1101.0000 1101(2)/D(16) 11100010.1101

(2)/ E2.D

(16)

2.16. (a) Remainders Integers

101122=1202 1(3)

/1(2)

0.12=0.2 0(3)

/0(2)

12022= 212 1(3)

/1(2)

0.22=1.1 1(3)

/1(2)

2122= 102 1(3)

/1(2)

1022= 12 1(3)

/1(2)

122= 2 1(3)/1(2)

22= 1 0(3)

/0(2)

12= 0 1(3)

/1(2)

10112.1(3)

/ 1011111.0101...(2)

(b) Remainders Integers

1011222=102 21(3)

/7(8)

0.122= 2.2 2(3)

/2(8)

10222= 1 10(3)

/3(8)

0.222=12.1 12(3)

/5(8)

122= 0 1(3)

/1(8)

10112.1(3)

/ 137.2525...(2)

(c) Remainders Integers

10112101=100 12(3)

/5(10)

0.1101=10.1 10(3)

/3(10)

100101= 0 100(3)

/9(10)

10112.1(3)

/ 95.33...(10)


17/34

- 2.17 -

2.16. (continued)

(d) Remainders Integers

10112121=12 120(3)

/F(16)

0.1121=12.1 12(3)

/5(16)

12121= 0 12(3)

/5(16)

10112.1(3)/ 5F.55...(16)

2.17. (a) 7 7 1 . 1 7 2(8)

8 8 8 8 8 8

000111111001.001111010(2)

9 9 9 9 9

1 F 9 . 3 D(16)

(b) 1 2 1 3 . 4

(8)

8 8 8 8 8

001010001011.1000

(2)

9 9 9 9

2 8 B . 8(16)

(c) 2 7 4 2 . 2 2 6(8)

8 8 8 8 8 8 8

010111100010.010010110(2)

9 9 9 9 9

5 E 2 . 4 B(16)

(d) 3 4 4 6 . 5(8)

8 8 8 8 8

011100100110.1010

(2)

9 9 9 9

7 2 6 . A(16)

2.18. (a) 3 7 . 5(8)

9 9 9

011111.101(2)


18/34

- 2.18 -

2.18. (continued)

(b) 4 5 . 1(8)

9 9 9

100101.001

(2)(c) 6 1 . 3

(8)9 9 9

110001.011(2)

(d) 7 2 4 . 0 6(8)

9 9 9 9 9

111010100.000110(2)

2.19. (a) 1 C . 3(16)

9 9 9

00011100.0011(2)

(b) F 2 . C(16)

9 9 9

11110010.1100(2)

(c) 4 5 0 . B(16) 9 9 9 9

010001010000.1011(2)

(d) 8 E A . 5 9(16)

9 9 9 9 9

100011101010.01011001(2)


19/34

- 2.19 -

2.20. Algorithm to convert between base-3 and base-9:

1. Form blocks of 2 digits of the base-3 number starting

at the radix point and working both right and left,

adding leading and trailing 0's if necessary.

2. Replace each block of 2 digits by its equivalent digit

in base-9.

The conversion for 21021.112(3)

is

021021.1120(3)

9 9 9 9 9

2 3 7 . 4 6(9)

2.21. (a) Let N=be the r's-complement of N. Therefore,

N= = rn - N

The r's-complement of N=is

rn- N

== r

n- (r

n- N) = (r

n- r

n) + N = N

(b) Let N-be the (r-1)'s-complement of N. Therefore,

N-= r

n- r

-m- N

The (r-1)'s-complement of N-is

rn- r

-m- N-= r

n- r

-m- (r

n- r

-m- N)

= (rn- r

n) + (r

-m- r

-m) + N

= N

2.22. 1's-complements:

(a) 108- 10

-0- 10111011 = 01000100

(b) 109- 10

-0- 101110100 = 010001011

(c) 106- 10

-0- 101100 = 010011

(d) 107- 10

-0- 0110101 = 1001010

(e) 103- 10

-2- 010.11 = 101.00

(f) 105- 10

-3- 11011.100 = 00100.011

(g) 106- 10

-3- 100101.101 = 011010.010


20/34

- 2.20 -

2.22. (continued)

(h) 107- 10

-3- 1010110.110 = 0101001.001

2's-complements:

(a) 10

8

- 10111011 = 01000101

(b) 109- 101110100 = 010001100

(c) 106- 101100 = 010100

(d) 107- 0110101 = 1001011

(e) 103- 010.11 = 101.01

(f) 105- 11011.100 = 00100.100

(g) 10

6

- 100101.101 = 011010.011

(h) 107- 1010110.110 = 0101001.010

2.23. 9's-complements:

(a) 106- 10

-0- 285302 = 714697

(b) 105- 10

-0- 39040 = 60959

(c) 106- 10

-0- 059637 = 940362

(d) 106- 10

-0- 610500 = 389499

(e) 104- 10

-2- 7142.89 = 2857.10

(f) 104- 10

-4- 5263.4580 = 4736.5419

(g) 104- 10

-3- 0283.609 = 9716.390

(h) 103- 10

-4- 134.5620 = 865.4379

10's-complements:

(a) 106- 285302 = 714698

(b) 105- 39040 = 60960

(c) 106- 059637 = 940363

(d) 106- 610500 = 389500


21/34

- 2.21 -

2.23. (continued)

(e) 104- 7142.89 = 2857.11

(f) 104- 5263.4580 = 4736.5420

(g) 10

4

- 0283.609 = 9716.391

(h) 103- 134.5620 = 865.4380

2.24. r's-complements:

(a) 104- 0120.21

(3)= 2102.02

(3)

(b) 103- 101.120

(3)= 121.110

(3)

(c) 103- 241.03

(5)= 203.42

(5)

(d) 103- 031.240

(5)= 413.210

(5)

(e) 103- 407.270

(8)= 370.510

(8)

(f) 104- 0156.0037

(8)= 7621.7741

(8)

(g) 103- 83D.9F

(16)= 7C2.61

(16)

(h) 105- 0070C.B6E

(16)= FF8F3.492

(16)

(r-1)'s-complements:

(a) 104- 10

-2- 0120.21

(3)= 2102.01

(3)

(b) 103- 10

-3- 101.120

(3)= 121.102

(3)

(c) 103- 10

-2- 241.03

(5)= 203.41

(5)

(d) 103- 10

-3- 031.240

(5)= 413.204

(5)

(e) 103- 10

-3- 407.270

(8)= 370.507

(8)

(f) 104- 10

-4- 0156.0037

(8)= 7621.7740

(8)

(g) 103- 10

-2- 83D.9F

(16)= 7C2.60

(16)

(h) 105- 10

-3- 0070C.B6E

(16)= FF8F3.491

(16)


22/34

- 2.22 -

2.25. Using 1's-complements:

(a) N1= 1101101 6 N

1= 1101101

N2= 0110110 6 + N

-2= + 1001001_________

1 0110110

+ 1_______0110111

(b) N1= 010100 6 N

1= 010100

N2= 110000 6 + N

-2= + 001111

______

100011

(c) N1= 10010.00111 6 N

1= 10010.00111

N2= 00101.11000 6 + N

-2= + 11010.00111_____________

1 01100.01110

+ 1___________01100.01111

(d) N1= 10110.0100 6 N

1= 10110.0100

N2= 01011.1101 6 + N

-2= + 10100.0010____________

1 01010.0110+ 1__________

01010.0111

(e) N1= 01101.1011 6 N

1= 01101.1011

N2 = 10110.1100 6 + N-2 = + 01001.0011__________

10110.1110

(f) N1= 00101.100100 6 N

1= 00101.100100

N2= 11010.010011 6 + N

-2= + 00101.101100

____________

01011.010000

Using 2's-complements:

(a) N1= 1101101 6 N

1= 1101101

N2 = 0110110 6 + N=

2 = + 1001010_________

1/ 0110111

(b) N1= 010100 6 N

1= 010100

N2= 110000 6 + N

=

2= + 010000

______

100100


23/34

- 2.23 -

2.25. (continued)

(c) N1= 10010.00111 6 N

1= 10010.00111

N2= 00101.11000 6 + N

=

2= + 11010.01000_____________

1/ 01100.01111

(d) N1= 10110.0100 6 N

1= 10110.0100

N2= 01011.1101 6 + N

=

2= + 10100.0011____________

1/ 01010.0111

(e) N1= 01101.1011 6 N

1= 01101.1011

N2= 10110.1100 6 + N

=

2= + 01001.0100

__________

10110.1111

(f) N1= 00101.100100 6 N

1= 00101.100100

N2 = 11010.0100116 + N=2 = + 00101.101101____________

01011.010001


(a) N1= 0

s10110 6 N

1= 0

s10110

N2= 0

s01101 6 + N

-2= + 1

s10010

_________

1 0s01000

+ 1_______

0s01001

(b) N1= 0

s010111 6 N

1= 0

s010111

N2= 0

s110100 6 + N

-2= + 1

s001011

________

1s100010

(c) N1= 0

s110.1001 6 N

1= 0

s110.1001

N2= 0

s011.0100 6 + N

-2= + 1

s100.1011

____________

1 0s011.0100

+ 1__________

0s011.0101

(d) N1= 0

s10101.1 6 N

1= 0

s10101.1

N2= 0

s10101.1 6 + N

-2= + 1

s01010.0

_________

1s11111.1


24/34

- 2.24 -

2.26. (continued)

(e) N1= 0

s101.11000 6 N

1= 0

s101.11000

N2= 0

s010.01011 6 + N

-2= + 1

s101.10100

_____________

1 0s011.01100

+ 1___________0s011.01101

(f) N1= 0

s010111.10101 6 N

1= 0

s010111.10101

N2= 0

s111010.11000 6 + N

-2= + 1

s000101.00111

______________

1s011100.11100


(a) N1= 0

s10110 6 N

1= 0

s10110

N

2

= 0

s

01101 6 + N=

2

= + 1

s

10011

_________1/ 0

s01001

(b) N1= 0

s010111 6 N

1= 0

s010111

N2= 0

s110100 6 + N

=

2= + 1

s001100

________

1s100011

(c) N1= 0

s110.1001 6 N

1= 0

s110.1001

N2= 0

s011.0100 6 + N

=

2= + 1

s100.1100

____________

1/ 0s011.0101

(d) N1= 0

s10101.1 6 N

1= 0

s10101.1

N2= 0

s10101.1 6 + N

=

2= + 1

s01010.1

___________

1/ 0s00000.0

(e) N1= 0

s101.11000 6 N

1= 0

s101.11000

N2= 0

s010.01011 6 + N

=

2= + 1

s101.10101

_____________

1/ 0s011.01101

(f) N1= 0

s010111.10101 6 N

1= 0

s010111.10101

N2= 0

s111010.11000 6 + N=

2= + 1

s000101.01000

______________

1s011100.11101


25/34

- 2.25 -


(a) N1= 7842 6 N

1= 7842

N2= 3791 6 + N

-2

= + 6208______

1 4050

+ 1____4051

(b) N1= 265 6 N

1= 265

N2= 894 6 + N

-2

= + 105___

370

(c) N1= 508.3 6 N

1= 508.3

N2= 094.7 6 + N

-2= + 905.2_______

1 413.5

+ 1_____413.6

(d) N1= 073.68 6 N

1= 0.73.68

N2= 538.90 6 + N

-2= + 461.09

______

534.77

(e) N1= 427.08 6 N

1= 427.08

N2= 089.30 6 + N

-2= + 910.69________

1 337.77+ 1______

337.78

(f) N1= 0804.20 6 N

1= 0804.20

N2= 3621.47 6 + N

-2= + 6378.52

_______

7182.72


(a) N1= 7842 6 N

1= 7842

N2 = 3791 6 + N=

2 = + 6209______

1/ 4051

(b) N1= 265 6 N

1= 265

N2= 894 6 + N

=

2= + 106

___

371


26/34

- 2.26 -

2.27. (continued)

(c) N1= 508.3 6 N

1= 508.3

N2= 094.7 6 + N

=

2= + 905.3_______

1/ 413.6

(d) N1= 073.68 6 N

1= 073.68

N2= 538.90 6 + N

=

2= + 461.10

______

534.78

(e) N1= 427.08 6 N

1= 427.08

N2= 089.30 6 + N

=

2= + 910.70________

1/ 337.78

(f) N1= 0804.20 6 N

1= 0804.20

N2 = 3621.476 + N=2 = + 6378.53_______

7182.73


(a) N1= 0

s546 6 N

1= 0

s546

N2= 0

s232 6 + N

-2= + 1

s767

_______

1 0s313

+ 1_____

0s314

(b) N1= 0

s384 6 N

1= 0

s384

N2= 0

s726 6 + N

-2= + 1

s273

_____

1s657

(c) N1= 0

s326.4 6 N

1= 0

s326.4

N2= 0

s087.2 6 + N

-2= + 1

s912.7

_________

1 0s239.1

+ 1_______

0s239.2

(d) N1= 0

s076.23 6 N

1= 0

s076.23

N2= 0

s209.40 6 + N

-2= + 1

s790.59

________

1s866.82


27/34

- 2.27 -

2.28. (continued)

(e) N1= 0

s406.9 6 N

1= 0

s406.9

N2= 0

s406.9 6 + N

-2= + 1

s593.0

_______

1

s

999.9

(f) N1= 0

s063.40 6 N

1= 0

s063.40

N2= 0

s240.36 6 + N

-2= + 1

s759.63

________

1s823.03


(a) N1= 0

s546 6 N

1= 0

s546

N2= 0

s232 6 + N

=

2= + 1

s768

_______

1/ 0

s

314

(b) N1= 0

s384 6 N

1= 0

s384

N2= 0

s726 6 + N

=

2= + 1

s274

_____

1s658

(c) N1= 0

s326.4 6 N

1= 0

s326.4

N2= 0

s087.2 6 + N

=

2= + 1

s912.8

_________

1/ 0s239.2

(d) N1= 0

s076.23 6 N

1= 0

s076.23

N2= 0

s209.40 6 + N

=2= + 1

s790.60

________

1s866.83

(e) N1= 0

s406.9 6 N

1= 0

s406.9

N2= 0

s406.9 6 + N

=

2= + 1

s593.1

_________

1/ 0s000.0

(f) N1= 0

s063.40 6 N

1= 0

s063.40

N2= 0

s240.36 6 + N

=

2= + 1

s759.64

________

1s823.04


28/34

- 2.28 -

2.29. A=0s1000110 A

==1s0111010

B=1s1010011 B

==0s0101101

(a) A+B (b) A-B = A+B=

A = 0

s

1000110 A = 0

s

1000110

+ B = + 1s1010011

___________+ B= = + 0

s0101101

_________

1/ 0s0011001 0

s1110011

(c) B-A = B+A=

(d) -A-B = A=+B=

B = 1s1010011 A

== 1

s0111010

+ A== + 1

s0111010

___________+ B== + 0

s0101101

_________

1/ 1s0001101 1

s1100111

2.30. A=0s601.7 A

=

=1s398.3B=1

s754.2 B

==0s245.8

(a) A+B (b) A-B = A+B=

A = 0s601.7 A = 0

s601.7

+ B = + 1s754.2

_________+ B=

= + 0s245.8

_______

1/ 0s355.9 0

s847.5

(c) B-A = B+A=

(d) -A-B = A=+B=

B = 1s754.2 A

== 1

s398.3

+ A== + 1

s398.3

_________+ B=

= + 0s245.8

_______

1/ 1s152.5 1

s644.1

2.31. A=0s1010110 A

-=1s0101001

B=1s1101100 B

-=0s0010011

(a) A+B (b) A-B = A+B-

A = 0s1010110 A = 0

s1010110

+ B = + 1s1101100___________ + B-= + 0s0010011_________

1 0s1000010 0

s1101001

+ 1_________

0s1000011


29/34

- 2.29 -

2.31. (continued)

(c) B-A = B+A-

(d) -A-B = A-+B-

B = 1s1101100 A

-= 1

s0101001

+ A-= + 1

s0101001

___________+ B-= + 0

s0010011

_________

1 1s0010101 1s0111100+ 1_________

1s0010110

2.32. A=0s418.5 A

-=1s581.4

B=1s693.0 B

-=0s306.9

(a) A+B (b) A-B = A+B-

A = 0s418.5 A = 0

s418.5

+ B = + 1s693.0_________ + B

=

= + 0s306.9_______1 0

s111.5 0

s725.4

+ 1_______

0s111.6

(c) B-A = B+A-

(d) -A-B = A-+B-

B = 1s693.0 A

-= 1

s581.4

+ A-= + 1

s581.4

_________+ B-

= + 0s306.9

_______

1 1s274.4 1

s888.3

+ 1_______

0s274.5

2.33. (a) 100001010011 (b) 011001000010 (c) 101110000110

8 5 3 8 5 3 8 5 3

(d) 100100010000010101000 (e) 100100101000110

8 5 3 8 5 3

2.34. (a) 8 9 5 8 3

(b) 5 6 2 5 0

10001001010110000011

(c) 7 2 6 1


30/34


31/34

- 2.31 -

2.37. Assume two weights exceed 4. Then, only two weights, say,

wi1

and wi2, are less than or equal to 4. With these two

weights it is necessary to encode the decimal digits 0, 1,

2, 3, and 4 by assigning binary digits 0's and 1's to the

ai's in the formula a1wi1+a2wi2. Since there are only four

ways of making this assignment, the five decimal digits can

not be represented. Thus, no more than one weight can

exceed 4.

2.38. Divide the weights of the code into two groups: those which

are used in the representation of the decimal digit 0 and

those which are not used in this representation. Let the

sum of the weights in the first group be Ewi and the sum

of the weights in the second group be Ewj. Clearly,

Ewi=0. By the definition of a self-complementing code,

those weights which are not used to represent the decimal

digit 0 must be used to represent the decimal digit 9.

Hence, Ewj=9. Combining these results, Ew

i+Ew

j=9.

2.39. Assume three weights are negative and only one weight is

positive. There can exist at most 9 non-negative

combinations of the four weights. Therefore it is

impossible to represent the ten non-negative digits which

must be coded. Hence, at most two weights can be negative.

2.40. (a) 011100101101100110000

9 6 0

(b) 101100001010111011001

X + Y

(c) 1000011110111111001001100101

C o d e


32/34

- 2.32 -

2.41. (a) 101100101011100000110011

2 8 3

which is represented by B2B833 in hexadecimal.

(b) 010110101011110110110001

Z = 1

which is represented by 5ABDB1 in hexadecimal.

(c) 01000010011010010111010011110011

B i t s

which is represented by 426974F3 in hexadecimal.

2.42. (a)

Decimaldigit

7 6 5 4 3 2 1 Position

2 4 2 p3

1 p2

p1Format

0 0 0 0 0 0 0 0

1 0 0 0 0 1 1 1

2 0 0 1 1 0 0 1

3 0 0 1 1 1 1 0

4 0 1 0 1 0 1 0

5 1 0 1 0 1 0 1

6 1 1 0 0 0 0 1

7 1 1 0 0 1 1 0

8 1 1 1 1 0 0 0

9 1 1 1 1 1 1 1


33/34

- 2.33 -

2.42. (continued)

(b)

Decimal

digit

9 8 7 6 5 4 3 2 1 Position

b

5

p

4

b

4

b

3

b

2

p

3

b

1

p

2

p

1

Format

0 1 1 1 0 0 1 0 1 0

1 0 0 0 0 1 1 1 1 0

2 0 0 0 1 0 1 1 0 1

3 0 0 0 1 1 0 0 1 1

4 0 0 1 0 0 1 1 0 0

5 0 0 1 0 1 0 0 1 0

6 0 0 1 1 0 0 0 0 1

7 1 1 0 0 0 0 1 1 0

8 1 1 0 0 1 1 0 0 0

9 1 1 0 1 0 1 0 1 1

2.43.

12 11 10 9 8 7 6 5 4 3 2 1 Position

b8

b7

b6

b5

p4

b4

b3

b2

p3

b1

p2

p1Format

(a) 1 1 1 0 1 0 0 1 0 1 1 1

(b) 0 1 0 1 0 1 0 0 1 0 0 1

(c) 1 0 0 1 0 0 1 0 0 1 0 0

2.44. (a) 7 6 5 4 3 2 1 Position

0 0 1 1 0 0 0 Received code group

c*

3

c*

2

c*

1

=001 which indicates position 1 in error.

Transmitted code group: 0011001

(b) 7 6 5 4 3 2 1 Position1 1 1 1 0 0 0 Received code group

c*3c*2c*1=000 which indicates no error.



34/34

2.44. (continued)

(c) 7 6 5 4 3 2 1 Position

1 1 0 1 1 0 0 Received code group

c*3c*2c*1=110 which indicates position 6 in error.


2.45. (a) Position: 11 10 9 8 7 6 5 4 3 2 1

Format: poverall

b6b5p4b4b3b2p3b1p2p1

Code group: 1 1 0 1 0 1 0 1 1 1 1

(b) If errors occur in bit positions 2 and 9, the received

code group is 11110101101. Using the received code

group, the overall parity bit is correct. The binary

check number is c*4c*3c*2c*1=1011. Since c*4c*3c*2c*10000 and

the overall parity bit is correct, a double error has

occurred.

2.46. Upon comparing each pair of code groups, the minimum

distance of this code is found to be three. Therefore, it

can be used for double-error detection or single-error

correction.

Chapter 2 Problem Solutions 2.1. Decimal 32 33 34 35

Documents

Transcript of Chapter 2 Problem Solutions 2.1. Decimal 32 33 34 35