# Digital Design Fifth Edition Exercise 1 of Chapter 1 , Exercises for Digital Systems Design. National University of Modern Languages

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Kingdom of Saudi Arabia

Ministry of Higher Education Majma’ah University College of Engineering EE & CEN

Logic Design (CE 207, CE 213) Chapter No. 1 – Part No. 1

Problem 1.1: List the Octal and Hexadecimal numbers from 16 to 32.

Sol:

Decimal (Base – 10):

16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32

Octal (Base – 8):

20, 21, 22, 23, 24, 25, 26, 27, 30, 31, 32, 33, 34, 35, 36, 37, 40

10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 1A, 1B, 1C, 1D, 1E, 1F, 20

Problem 1.2: What is the exact number of bytes in a system that contains?

a. 32 Kbytes

Sol:

As 1 Kilo (K) is referred as 210

32 Kbytes = 32 x 210 = 32 x 1024 = 32768 Bytes

b. 44 Mbytes

Sol:

As 1 Mega (M) is referred as 220

44 Mbytes= 64 x 220 = 32 x 1048576 = 67108864 Bytes

c. 6.4 Gbytes

Sol:

As 1 Giga (G) is referred as 230

6.4 Gbytes = 6.4 x 230 = 32 x 1073741824 = 6871947673.6 Bytes

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Problem 1.3: Convert the following numbers with the indicated base to decimal.

a. (4310)5

Sol:

(4310)5 = 4 x 53 + 3 x 52 + 1 x 51 + 0 x 50

(4310)5 = 4 x 125 + 3 x 25 + 1 x 5 + 0 x 1

(4310)5 = 500 + 75 + 5 + 0

(4310)5 = (580)10

b. (198)12

Sol:

(198)12 = 1 x 122 + 9 x121 + 8 x 120

(198)12 = 1 x 144 + 9 x12 + 8 x 1

(198)12 = 144 + 108 + 8

(198)12 = (260)10

c. (735)8

Sol:

(735)8 = 7 x 82 + 7 x 81 + 7 x 80

(735)8 = 7 x 64 + 7 x 8 + 7 x 1

(735)8 = 448 + 56 + 7

(735)8 = (511)10

d. (525)6

Sol:

(525)6 = 5 x 62 + 2 x 61 + 5 x 60

(525)6 = 5 x 36 + 2 x 6 + 5 x 1

(525)6 = 180 + 12 + 5

(525)6 = (197)10

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Problem 1.4: What is the largest number that can be expressed with 14 – bits? What are the equivalent decimal and hexadecimal numbers?

Sol:

Largest 14 – bits binary number = 11111111111111

Decimal equivalent number = 214 – 1 = 16384 – 1 = (16383)10

= 0011 1111 1111 1111

= (3FFF)16

Problem 1.7: Convert the Hexadecimal number 68BE to Binary and then convert it from Binary to Octal.

(68BE)16 = 6 8 B E

= 0110 1000 1011 1110

= (110100010111110)2

Binary to Octal:

(110100010111110)2 = 110 100 010 111 110

= 6 4 2 7 6

= (64276)8

Problem 1.8: Convert the decimal number 431 to binary number in two ways.

a. Convert directly to Binary:

Sol:

Theref ore (431)10 = (110101111)2

b. Convert first to Hexadecimal and then from Hexa decimal to Binary:

Therefore (431)10 = (1AF)16

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(1AF)16 = 1 A F

= 0001 1010 1111

= (10101111)2

Problem 1.9: Express the following numbers in decimals.

a. (10110.0101)2

Sol:

(10110.0101)2 = 1 x 24 + 0 x 23 + 1 x 22 + 1 x 21 + 0 x 20 + 0 x 2-1 + 1 x 2-2

+ 0 x 2-3 + 1 x 2-4

(10110.0101)2 = 1 x 16 + 0 + 1 x 4 + 1 x 2 + 0 + 0 + 1/4 + 0 + 1/16

(10110.0101)2 = 16 + 4 +2 + 0.25 + 0.0625

(10110.0101)2 = (22.3125)10

b. (16.5)16

Sol:

(16.5)16 = 1 x 161 + 6 x 160 + 5 x 16-1

(16.5)16 = 1 x 16 + 6 x 1 + 5 x 1/16

(16.5)16 = 16 + 6 + 0.3125

(16.5)16 = (22.3125)10

c. (26.24)8

Sol:

(26.24)8 = 2 x 81 + 6 x 80 + 2 x 8-1 + 4 x 8-2

(26.24)8 = 2 x 8 + 6 x 1 + 2 x 1/8 + 4 x 1/64

(26.24)8 = 16 + 6 + 0.25 + 0.0625

(26.24)8 = (22.3125)10

d. (FAFA.B)16

Sol:

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(FAFA.B)16 = 15 x 163 + 10 x 162 + 15 x 161 + 10 + 160 + 11 x 16-1

(FAFA.B)16 = 15 x 4096 + 10 x 256 + 15 x 16 + 10 + 1 + 11 x 1/16

(FAFA.B)16 = 61440 + 2560 + 240 + 10 + 0.6875

(FAFA.B)16 = (64250.6875)10

Problem 1.10:Convert the following numbers from Binary to Hexadecimal and then from Hexadecimal to Decimal.

A. (1.10010)2

(1.10010)2 = 0001.1001

(1.10010)2 = 1 . 9

(1.10010)2 = (1.9)16

(1.9)16 = 1 X 160 + 9 X 16-1 = 1 X 1 + 9 X 1/16 = 1 + 0.5625

(1.9)16 = (1.5625)10

B. (110.0101)2

(110.0101)2 = 0110.0101

(110.0101)2 = 6 . 5

(110.0101)2 = (6.5)16

(6.5)16 = 6 X 160 + 5 X 16-1 = 6 X 1 + 5 X 1/16 = 6 + 0.3125

(6.4)16 = (6.3125)10

Problem 1.13:Do the following conversions. a. Convert Decimal 27.315 to Binary

Sol: Convert 27 and 0.315 separately from Decimal to Binary:

0.315 x 2 = 0.63 0 0.63 x 2 = 1.26 1 0.26 x 2 = 0.52 0 0.52 x 2 = 1.04 1

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(27)10 = (11011)2 (0.315)10 = (0.0101)2 (27.315)10 = (11011.0101)2

b. Calculate the Binary equivalent of 2/3 up to eight places. Then convert from Binary to Decimal. How close is the result to 2/3?

Sol: Decimal to Binary: (2/3)10 = (0.66666666)10 0.66666666 x 2 = 1.33333333 1 0.33333333 x 2 = 0.66666666 0 0.66666666 x 2 = 1.33333333 1 0.33333333 x 2 = 0.66666666 0 0.66666666 x 2 = 1.33333333 1 0.33333333 x 2 = 0.66666666 0 0.66666666 x 2 = 1.33333333 1 0.33333333 x 2 = 0.66666666 0 (2/3)10 = (0.66666666)10 = (0.10101010)2

Binary to Decimal: (0.10101010)2 = 1 x 2-1 + 0 x 2-2 + 1 x 2-3 + 0 x 2-4 + 1 x 2-5 + 0 x 2-6 + 1 x 2-7

+ 0 x 2-8

(0.10101010)2 = 1/2 + 0 + 1/8 + 0 + 1/32 + 0 + 1/128 + 0

(0.10101010)2 = 0.5 + 0.125 + 0.03125 + 0.0078125

(0.10101010)2 = (0.6640625)10

c. Convert the Binary result in part (b) to Hexadecimal and then convert the result to Decimal. Is the answer the same?

Sol: Binary to Hexadecimal: (0.10101010)2 = 0.1010 1010 (0.10101010)2 = A A (0.10101010)2 = (0.AA)16 Hexadecimal to Decimal: (0.AA)16 = 10 x 16-1 + 10 x 16-2

(0.AA)16 = 10 x 1/16 + 10 x 1/256 (0.AA)16 = 10 x 0.0625 + 10 x 0.0039062 (0.AA)16 = 0.625 + 0.0390625 (0.AA)16 = (0.6640625)10

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