THE ISLAMIC UNIVERSITY OF GAZA

ENGINEERING FACULTY DEPARTMENT OF

COMPUTER ENGINEERING DIGITAL LOGIC

DESIGN DISCUSSION – ECOM 2012

Chapter (1)

Eng. Mai Z. Alyazji

September, 2016

DIGITAL LOGIC DESIGN ECOM 2012 ENG. Mai Z. Alyazji

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1.1 List the octal and hexadecimal numbers from 16 to 32. Using A and B for

the last two digits, list the numbers from 8 to 28 in base 12.

Answer:

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

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

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

Decimal 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24

25 26 27 28

Base 12 8 9 A B 10 11 12 13 14 15 16 17 18 19 1A 1B 20

21 22 23 24

1.2 What is the exact number of bytes in a system that contains (a) 32K

bytes, (b) 64M bytes, and (c) 6.4G bytes?

Answer:

(a) 32 * 210 = 32768 bytes

(b) 64 * 220 = 67108864 bytes

(c) 6.4 * 230 = 6871947674 bytes

1.3 Convert the following numbers with the indicated bases to decimal:

(a) (4310) 5 (b) (198) 12

(c) (435) 8 (d) (345) 6

Answer:

(a) 4 * 53 + 3 * 52 + 1 * 51 + 0 * 50 = (580)10

(b) 1* 122 + 9 * 121 + 8 * 120 = (260)10

(c) 4 * 82 + 3 * 81 + 5 * 80 = (285)10

(d) 3 * 62 + 4 * 61 + 5 * 60 = (137)10

DIGITAL LOGIC DESIGN ECOM 2012 ENG. Mai Z. Alyazji

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1.4 What is the largest binary number that can be expressed with 16 bits?

What are the equivalent decimal and hexadecimal numbers?

Answer:

The largest number of any k-digits binary number is that with all digits

being 1, such that the largest number of 16-digit binary number is

(1111 1111 1111 1111)2 = (65535)10 = (FFFF)16

To calculate the largest number of k bits we can simply use the formula

2k -1

In this question 216 -1 = 65536-1=65535

1.5 Determine the base of the numbers in each case for the following

operations to be correct:

(a) 14/2 = 5 (b) 54/4 = 13 (c) 24 + 17 = 40.

Answer:

(a) (1 * b1 + 4 * b0 1) / 2 * b0 1 = 5 * b0 1

(b + 4) / 2 = 5

b/2 + 4/2 = 5 b = 6

(b) (5 * b1 + 4 * b0 1) / 4 * b0 1 = 1 * b1 + 3 * b0 1

(5b + 4) / 4 = 1b + 3

5b/4 + 4/4 = 1b + 3 b = 8

(c) (2 *b + 4) + (b + 7) = 4b b = 11

1.6 The solutions to the quadratic equation x2 - 11x + 22 = 0 are x = 3 and x

= 6. What is the base of the numbers?

Answer:

Notice that the solutions to the equation x2 - 11x + 22 in decimal system

are 8.37 and 2.62, which means that this equation is in another system,

we are asked to get the base of it.

DIGITAL LOGIC DESIGN ECOM 2012 ENG. Mai Z. Alyazji

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431/2

215 1

107 1

53 1

26 1

13 0

6 1

3 0

1 1

0 1

(x – 3) (x - 6) = x2 – 9x + 18

(x2 - 11x + 22)b = (x2 – 9x + 18)10

(11)b = (9)10

(1 * b1 + 1 * b0 1) = 9

b = 8

1.7 Convert the hexadecimal number 64CD to binary, and then convert it

from binary to octal.

Answer:

(64CD)16 = (0110 0100 1100 1101)2

(110 010011 001 101)2 = (62315)8

1.8 Convert the decimal number 431 to binary in two ways: (a) convert

directly to binary; (b) convert first to hexadecimal and then from

hexadecimal to binary. Which method is faster? Answer:

(a)

Integer Remainder

431 = (110101111)2

DIGITAL LOGIC DESIGN ECOM 2012 ENG. Mai Z. Alyazji

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(b)

Integer Remainder

431/16

26 F

1 A

0 1

431 = (1AF)16 = (110101111)2

The second method is faster than the first one.

1.9 Convert Express the following numbers in decimal:

(a) (10110.0101) 2 (b) (16.5) 16

(c) (26.24) 8

Answer:

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

+ 1 * 2-2 + 0 * 2-3 + 1 * 2-4 = 22.3125

(b) (16.5) 16 = 1 * 161 + 6 * 160 + 5 * 16-1 = 22.3125

(c) (26.24) 8 = 2 * 81 + 6 * 80 + 2 * 8-1 + 4 * 8-2 = 22.3125

1.10 Convert the following binary numbers to hexadecimal and to decimal:

(a) 1.10010, (b) 110.010. Explain why the decimal answer in (b) is 4

times that in (a).

Answer:

(a) 1.100102 = 0001.10012 = 1.916 = 1 + 9/16 = 1.56310

(b) 110.0102 = 0110.01002 = 6.416 = 6 + 4/16 = 6.2510

Reason: 110.0102 is the same as 1.100102 shifted to the left by two places.

العدد القديم فضع دصبح العدد الجدية اليمين يهما بتحرك خانة لجكل منصف العدد القديعدد الجديد صبح الي يسارة الهبتحرك خانة لج امكل

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1.11 Perform the following division in binary: 111011 ÷ 101. Answer:

1011.11

101 111011.00 101

01001 101

1001 101

1000 101

0110 101

001

1.12 Add and multiply the following numbers without converting them to

decimal.

(a) Binary numbers 1011 and 101.

(b) Hexadecimal numbers 2E and 34.

Answer:

a) Addition Multiplication 1

1001 +

0101

1001 ×

0101

1110

1001 00000

100100

101101

DIGITAL LOGIC DESIGN ECOM 2012 ENG. Mai Z. Alyazji

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2 27/2 0.315*2

13 1 0.36 0

6 1 0.26 1

3 0 0.52 0

1 1 0.04 1

0 1 0.08 0 0.16 0 0.32 0

b) Addition Multiplication 1

2E 2E + ×

34 34

62 38 + 80

B8

2A0 +

1.13 Do the following conversion problems:

(a) Convert decimal 27.315 to binary.

Answer:

600

958

a) Integer Remainder Fraction Integer

7.31510 = 11011.01012

1.14 Obtain the 1’s and 2’s complements of the following binary numbers:

(a) 00010000 (b) 00000000

(c) 11011010 (d) 10101010

Answer:

00010000 00000000 11011010 10101010 1st complement 11101111 11111111 00100101 01010101 2nd complement 11110000 00000000 00100110 01010110

DIGITAL LOGIC DESIGN ECOM 2012 ENG. Mai Z. Alyazji

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1.15 Find the 9’s and the 10’s complement of the following decimal numbers:

(a) 25,478,036 (b) 63, 325, 600

(c) 25,000,000 (d) 00,000,000.

Answer:

25478036 63325600 25000000 00000000 9’s complement 74521963 36674399 74999999 99999999 10’s complement 74521964 36674400 75000000 00000000

1.16 (a) Find the 16’s complement of C3DF.

(b) Convert C3DF to binary.

(c) Find the 2’s complement of the result in (b).

(d) Convert the answer in (c) to hexadecimal and compare with the

answer in (a).

Answer:

a) FFFF – C3DF + 1 = 3C21

b) 1100 0011 1101 1111

c) 0011 1100 0010 0001

d) 3C21

1.17 Perform subtraction on the given unsigned numbers using the 10’s

complement of the subtrahend. Where the result should be negative,

find its 10’s complement and affix a minus sign. Verify your answers.

(a) 4,637 - 2,579 (b) 125 - 1,800

Answer:

a) 10’s complement of 2579 = 9999 - 2579 + 1 = 7421

4637 + 7421 = 12058 here, the result should be positive; we discard

the 1.

result = + 2058

b) 10’s complement of 1800 = 8200

DIGITAL LOGIC DESIGN ECOM 2012 ENG. Mai Z. Alyazji

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125 + 8200 = 8325 here, the result should be negative; we find its

10’s complement and affix a minus sign.

result= - 1675

1.18 Perform subtraction on the given unsigned binary numbers using the

2’s complement of the subtrahend. Where the result should be negative,

find its 2’s complement and affix a minus sign.

(a) 10011 – 10010 (b) 100010 – 100110

Answer:

a) 10011 + 01110 = 100001

Result = + 00001, as we discard 1

b) 100010 + 011010 = 111100

Result= - 000100, as the result should be negative.

1.22 Convert decimal 6,514 to both BCD and ASCII codes. For ASCII, an even

parity bit is to be appended at the left.

Answer:

6514 = ( 0110 0101 0001 0100 )BCD

= ( 0 0110110 0110101 0110001 0110100)ASCII

1.23 Represent the unsigned decimal numbers 791 and 658 in BCD, and then

show the steps necessary to form their sum. Answer:

1 1 1

0111 1001 0001 +

0110 0101 1000

1 1 1 1 1 1 1

1101 1110 1001

0110 0110 +

= (1449)BCD

0001 0100 0100 1001

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