DIGITAL DESIGNDIGITAL DESIGNTHIRD EDITIONTHIRD EDITION

M. MORRIS MANOM. MORRIS MANO

CHAPTER 1 : BINARY CHAPTER 1 : BINARY SYSTEMS PROBLEMS SYSTEMS PROBLEMS

1.1-) List the octal and the hexadecimal numbers 1.1-) List the octal and the hexadecimal numbers

from 16 to 32. Using A and B for the last two from 16 to 32. Using A and B for the last two

digits, list the numbers from 10 to 26 in base 12digits, list the numbers from 10 to 26 in base 12 . .

Octal :

16 = 8¹ x 2 + 8º x 0 => (16)10 = (20)8

32 = 8¹ x 4 + 8º x 0 => (32)10 = (40)8

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

Hexadecimal :

16 = 16¹ x 1 + 16º x 0 => (16)10 = (10)16

32 = 16¹ x 2 + 16º x 0 => (32)10 = (20)8

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

Base-12 :

10 = 12º x A => (10)10 = (A)12

26 = 12¹ x 2 + 12º x 2 => (26)10 = (22)12

A, B, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 1A, 1B, 20, 21, 22

1.2-) What is the exact number of bytes in a 1.2-) What is the exact number of bytes in a

system that contains (a) 32K byte, (b)64M bytes, system that contains (a) 32K byte, (b)64M bytes,

and (c)6.4G byteand (c)6.4G byte ??

(a) 32K byte:

1K = 2¹º = 1,024

32K = 32 x 2¹º = 32 x 1,024 = 32,768

32K byte = 32,768 byte

(b) 64M byte:

1M = 2²º = 1,048,576

64M = 64 x 2²º = 64 x 1,048,576 = 67,108,864

64M byte = 67,108,864 byte

(c) 6.4G byte:

1G = 2³º = 1,073,741,824

6.4G = 6.4 x 2³º = 6.4 x 1,073,741,824 = 6,871,747,674

6.4G byte = 6,871,747,674 byte

1.3-) What is the largest binary number that can 1.3-) What is the largest binary number that can

be expressed with 12 bits? What is the equivalent be expressed with 12 bits? What is the equivalent

decimal and hexadecimal ?decimal and hexadecimal ?

Binary:

(111111111111)2

Decimal:

(111111111111)2 = 1x 2º+ 1 x 2¹ + 1 x 2² +…..+ 1 x 2¹¹ + 1 x 2¹²

(111111111111)2 = 4,095

Hexadecimal:

(1111 1111 1111)2

F F F

(FFF)16=

1.4-) Convert the following numbers with the 1.4-) Convert the following numbers with the

indicated bases to decimal : (4310)indicated bases to decimal : (4310)5 5 , and (198), and (198)1212 . .

(4310)5 = 0 x 5º + 1 x 5¹ + 3 x 5² + 4 x 5³ = 0 + 5 + 75 + 500

(4310)5 = 580

(198)12 = 8 x 12º + 9 x 12¹ + 1 x 12² = 8 + 108 + 144

(198)12 = 260

1.5-) Determine the base of the numbers in each 1.5-) Determine the base of the numbers in each

case for the following operations to be correct :case for the following operations to be correct :

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

(a) (14)a / (2)a = (5)a

(4 x aº + 1 x a¹) / (2 x aº) = 5 x aº

(4 + a) / 2 = 5

4 + a = 10

a = 6

(b) (54)b / (4)b = (13)b

(4 x bº + 5 x b¹) / (4 x bº) = 3 x bº + 1 x b¹

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

4 + 5b = 12 + 4b

b = 8

(4 x cº + 2 x c¹) + (7 x cº + 1 x c¹) = 4 x c¹

4 + 2c + 7 + c = 4c

c = 11

(c) (24)c + (17)c = (40)c

1.6-) The solution to the quadratic equation x1.6-) The solution to the quadratic equation x²² - -

11x + 22 = 0 is x=3 and x=6. What is the base of 11x + 22 = 0 is x=3 and x=6. What is the base of

the numbers?the numbers?

x² - 11x + 22 = (x – 3) . (x – 6)

x² - 11x + 22 = x² - (6 + 3)x + (6.3)

(11)a = (6)a + (3)a

1 + a = 6 + 3

a = 8

1.7-) Express the following numbers in decimal : 1.7-) Express the following numbers in decimal :

(10110.0101)(10110.0101)2 2 , (16.5), (16.5)16 16 , (26.24), (26.24)88 ..

( 1 0 1 1 0 . 0 1 0 1 )2

4 3 2 1 0 -1 -2 -3 -4

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

(10110.0101)2 = 22.3125

= 2¹ + 2² + (2^4) +( 2^-2) + (2^-4)

( 1 6 . 5 )16

1 0 -1

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

(16.5)16 = 22.3125

( 2 6 . 2 4 )8

1 0 -1 -2

(26.24)8 = 6 + 16 + (2/8) + (4/64)

(26.24)8 = 22.3125

= 6 x 16º + 1 x 16¹ + 5 x (16^-1)

= 6 x 8º + 2 x 8¹ + 2 x (8^-1) + 4 x (8^-2)

1.8-) Convert the following binary numbers to 1.8-) Convert the following binary numbers to

hexadecimal and to decimal : (a) 1.11010 , (b) hexadecimal and to decimal : (a) 1.11010 , (b)

1110.10 . Explain why the decimal answer in (b) is 1110.10 . Explain why the decimal answer in (b) is

8 times that of (a) .8 times that of (a) .

(a) ( 1 . 1101 0 )2 = ( 1 . D )16 = 1 x 16º + D x (16^-1)

1 D 0 0 -1

1.9-) Convert the hexadecimal number 68BE to 1.9-) Convert the hexadecimal number 68BE to

binary and then from binary convert it to octal .binary and then from binary convert it to octal .

(68BE) 16

Binary form:

(0110 1000 1011 1110)2=(0110100010111110)2

6 8 B E

Octal form:

(0 110 100 010 111 110)2

0 6 4 2 7 6

=(064276)8

(a)(a) 1.10-) Convert the decimal number 345 to 1.10-) Convert the decimal number 345 to

binary in two ways :binary in two ways :

Convert directly to binary; Convert directly to binary;

Convert first to hexadecimal, then from Convert first to hexadecimal, then from

hexadecimal to binary. Which method is hexadecimal to binary. Which method is

faster ?faster ?

(345)10

Number Divided by 2

Remainder

345 345/2=172 1

172 172/2=86 0

86 86/2=43 0

43 43/2=21 1

21 21/2=10 1

10 10/2=5 0

5 5/2=2 1

2 2/2=1 1

Method 1:

Method 2:

Number Divided by 16

Remainder

345 345/16=21 9

21 21/16=1 5

(345)10=(159)16 (1 101 1001)2

1.11-) Do the following conversion problems :1.11-) Do the following conversion problems :

(a) Convert decimal 34.4375 to binary .(a) Convert decimal 34.4375 to binary .

(b) Calculate the binary equivalent of 1/3 (b) Calculate the binary equivalent of 1/3

out to 8 places.out to 8 places.

Then convert from binary to decimal. How Then convert from binary to decimal. How

close is the close is the

result to 1/3 ?result to 1/3 ?

(c) Convert the binary result in (b) into (c) Convert the binary result in (b) into

hexadecimal. Then hexadecimal. Then

convert the result to decimal . Is the answer convert the result to decimal . Is the answer

the same ?the same ?

(a) 34.4375

34 0.4375

34:2=17 r=0

17:2=8 r=1

8:2=4 r=0

4:2=2 r=0

2:2=1 r=0

34=(100010)2

0.4375*2=0.875 r=0

0.875*2=1.75 r=1

0.75*2=1.5 r=1

0.5*2=1.0 r=1

0*2=0 r=0

0.4375=(0.01110)2

34.4375=(100010.01110)234.4375=(100010.01110)2

(b) 1/3=0.3333…

0.33333*2=0.66666 r=0 0.66666*2=1.33332 r=1 0.33332*2=0.66664 r=0 0.66664*2=1.33328 r=1

. .

.0.3333…=(0.010101….)= 0+ ¼ + 0 +

1/8 + 0 + 1/32 +… =~0.33333…

(c)

0.010101010…=0.0101 0101 0101

(0.555..)16=5/16 +5/256 +5/4096 +…

=~0.33203

1.12-) Add and multiply the following numbers 1.12-) Add and multiply the following numbers

without without

converting them to decimal.converting them to decimal.

(a) Binary numbers 1011 and 101 .(a) Binary numbers 1011 and 101 .

(b) Hexadecimal numbers 2E and 34 .(b) Hexadecimal numbers 2E and 34 .

(a) 1011 (11) (a) 1011 (11) 1011(11)1011(11)

101 (5) 101(5) 101 (5) 101(5) +__________ x_____ +__________ x_____

10000(16) 1011 10000(16) 1011 0000 0000 + 1011 + 1011 _________ _________

110111 (55) 110111 (55)

(b) 2E (46) 2E 34 (52) 34 +____ x____ 62 (98) B8 8A +____ 958(2392)

1.13-) Perform the following division in binary : 1.13-) Perform the following division in binary :

1011111 1011111 ÷÷ 101 . 101 . (1011111)2=95

(101)2=5

95/5=19 (10011)2

1011111 101101 10011000111 101

0101

101

0000

1.14-) Find the 9’s- and the 10’s-complement of 1.14-) Find the 9’s- and the 10’s-complement of

the following decimal numbers :the following decimal numbers :

(a) 98127634 (b) 72049900 (c) 10000000 (d) (a) 98127634 (b) 72049900 (c) 10000000 (d)

00000000 .00000000 .

9’s comlements :

(a)99999999-98127634=01872365

(b)99999999-72049900=27950099

(c)99999999-10000000=89999999

(d)99999999-0000000=99999999

10’s complements

(a)100000000- 98127634= 01872366(b)100000000-72049900=27950100(c)100000000-10000000=90000000(d)100000000-0000000=00000000

1.15-) (a) Find the 16’s-complement of AF3B .1.15-) (a) Find the 16’s-complement of AF3B .

(b) Convert AF3B to binary . (b) Convert AF3B to binary .

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

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

compare with the answer in (a) compare with the answer in (a)

(a)(a)16^5-AF3B=50C516^5-AF3B=50C5

(b)(b) (AF3B)16=1010 1111 0011 1011(AF3B)16=1010 1111 0011 1011

(c)1010111100111011 (c)1010111100111011 01010000110001010101000011000101

(d)0101 0000 1100 0101= 50C5(d)0101 0000 1100 0101= 50C5

1.16-) Obtain the 1’s and 2’S complements of the 1.16-) Obtain the 1’s and 2’S complements of the

following binary numbers :following binary numbers :

(a)11101010 (b)01111110 (c)00000001 (a)11101010 (b)01111110 (c)00000001

(d)10000000 (e)00000000(d)10000000 (e)00000000

1’s complements:1’s complements:(a) 00010101 (b)10000001 (c)11111110 (d)01111111 (e)11111111(a) 00010101 (b)10000001 (c)11111110 (d)01111111 (e)11111111

2’s complement :2’s complement :(a) 00010110 (b)10000010 (c)11111111 (d)10000000 (e)00000000(a) 00010110 (b)10000010 (c)11111111 (d)10000000 (e)00000000

1.17-) Perform subtraction on the following 1.17-) Perform subtraction on the following

unsigned numbers using the 2’s-complement of unsigned numbers using the 2’s-complement of

the subtrahend. Where the result shoud be the subtrahend. Where the result shoud be

negative, 10’s complement it and affix a minus negative, 10’s complement it and affix a minus

sign. Verify your answers .sign. Verify your answers .

(a) 7188-3049 (b)150-2100 (c)2997-7992 (d)1321-(a) 7188-3049 (b)150-2100 (c)2997-7992 (d)1321-

375375

(a)7188+6951=4139 (a)7188+6951=4139 One carry out so One carry out so answer is correct. answer is correct.

(b)150+7900=8050 (b)150+7900=8050 correct answer=-1950correct answer=-1950

(c)2997+2008=5005 (c)2997+2008=5005 correct answer=-4995correct answer=-4995

(d)1321+9625=0946 (d)1321+9625=0946 One carry out so One carry out so answer is correct.answer is correct.

1.18-) Perform subtraction on the following 1.18-) Perform subtraction on the following

unsigned binary numbers using the 2’s-unsigned binary numbers using the 2’s-

complement of the subtrahend. Where the result complement of the subtrahend. Where the result

should be negative, 2’s complement it and affix a should be negative, 2’s complement it and affix a

minus sign .minus sign .

(a)11011-11001 (b)110100-10101 (c)1011-110000 (a)11011-11001 (b)110100-10101 (c)1011-110000

(d)101010-101011(d)101010-101011

(a)(a)11011+00111=11011+00111=00010(27-25=2)00010(27-25=2)

(b)(b)110100+01011=110100+01011=011111(52-21=31)011111(52-21=31)

(c)1011+010000=011011 (c)1011+010000=011011 -100101(11-48=-37)-100101(11-48=-37)

(d)101010+010101=111111(d)101010+010101=111111-000001(42-43=-1)-000001(42-43=-1)

1.19-) The following decimal numbers are shown 1.19-) The following decimal numbers are shown

in sign- magnitude form : +9826 and +801. in sign- magnitude form : +9826 and +801.

Convert them to signed 10’s-complement form Convert them to signed 10’s-complement form

and perform the following operations : (Note that and perform the following operations : (Note that

the sum is +10627 and requires six digits). the sum is +10627 and requires six digits).

(a) (+9826)+(+801) (b)(+9826)+(-801) (a) (+9826)+(+801) (b)(+9826)+(-801)

(c)(-9826)+(+801) (d)(-9826)+(-801)(c)(-9826)+(+801) (d)(-9826)+(-801)

(a)(a)009826+00801=010627 009826+00801=010627

(b)009826+999199=09025(b)009826+999199=09025

(c)990174+000801=990975(c)990174+000801=990975-09025-09025

(d)990174+999199=989373(d)990174+999199=989373-10627-10627

1.20-) Convert decimal +61 and +27 to binary 1.20-) Convert decimal +61 and +27 to binary

using the signed-2’s complement representation using the signed-2’s complement representation

and enough digits to accomodate the numbers. and enough digits to accomodate the numbers.

Then perform the binary equivalent of (+27) + (- Then perform the binary equivalent of (+27) + (-

61) , (- 27) + (+61) and (-27) + (- 61) .61) , (- 27) + (+61) and (-27) + (- 61) .

Convert the answers back to ecimal and verify that Convert the answers back to ecimal and verify that

they are correctthey are correct . .

+61=0111101 -61=1000011+27=0011011 -27=1100101

(a)27+(-61)=0011011+1000011=1011110

(b)-27+(+61)=1100101+0111101=0100010

(c)-27+(-61)= 1100101+1000011=0101000(overflow)

11100101+11000011=10101000

1.21-) Convert decimal 9126 to both BCD and 1.21-) Convert decimal 9126 to both BCD and

ASCII codes. For ASCII, an odd parity bit is to be ASCII codes. For ASCII, an odd parity bit is to be

appended at the left .appended at the left .

BCD: 1001 0001 0010 0110

ASCII: 10111001 00110001 00110010 10110110

1.22-) Represent the unsigned decimal numbers 1.22-) Represent the unsigned decimal numbers

965 and 672 in BCD and then show the steps 965 and 672 in BCD and then show the steps

necessary to form their sum .necessary to form their sum .

965= 1001 0110 0101965= 1001 0110 0101672= 0110 0111 0010672= 0110 0111 0010 +___ ___ ____ +___ ___ ____ 1 0000 1101 0111 1 0000 1101 0111 +0110 +0110 +0110 +0110 +_________________+_________________0001 0110 0011 0111 0001 0110 0011 0111 (1637)10 (1637)10

1.23-) Formulate a weighted binary code for the 1.23-) Formulate a weighted binary code for the

decimal digits using weights 6, 3, 1, 1 .decimal digits using weights 6, 3, 1, 1 .

6 3 1 1 Decimal

0 0 0 0 0

0 0 0 1 1

0 0 1 1 2

0 1 0 0 3

0 1 1 0 4(0101)

0 1 1 1 5

1 0 0 0 6

1 0 0 1 7(1010)

1 0 1 1 8

1 1 0 0 9

1.24-) Represent decimal number 6027 in 1.24-) Represent decimal number 6027 in

(a) BCD, (b) excess-3 code, and (c) 2421 (a) BCD, (b) excess-3 code, and (c) 2421

code .code .

(a)(a)60276027 BCD : 0110 0000 0010 0111 BCD : 0110 0000 0010 0111

(b)(b)excess3: 1001 0011 0101 1010excess3: 1001 0011 0101 1010

(c)(c)(c)0110 0000 0010 1101(c)0110 0000 0010 1101

1.25-) Find the 9’s complement of 6027 and 1.25-) Find the 9’s complement of 6027 and

express it in 2421 code. Show that the result is express it in 2421 code. Show that the result is

the 1’s complement of the answer to (c) in the 1’s complement of the answer to (c) in

Problem 1.24 . This demonstrates that the 2421 Problem 1.24 . This demonstrates that the 2421

code is self-complementing .code is self-complementing .

9’s complement of 6027 is 39729’s complement of 6027 is 3972

6027 as 2421 code is 6027 as 2421 code is 0110 0000 0010 1101 0110 0000 0010 1101

3972 as 2421 code is 3972 as 2421 code is 0011 1111 1101 00100011 1111 1101 0010

1.26-) Assign a binary code in some orderly 1.26-) Assign a binary code in some orderly

manner to the 51 playing cards. Use the minimum manner to the 51 playing cards. Use the minimum

number of bits.number of bits.

2^4 =16 2^4 =16

2^5 =322^5 =32 2^6=64 2^6=64 6 bits are necessary. 6 bits are necessary.

1.27-) Write the expresion “G. Boole” in ASCII 1.27-) Write the expresion “G. Boole” in ASCII

using an eight-bit code. Include the period and using an eight-bit code. Include the period and

the space. Treat the leftmost bit of each character the space. Treat the leftmost bit of each character

as a parity bit. Each 8-bit code shouls have even as a parity bit. Each 8-bit code shouls have even

parity.parity.

G . B O O L EG . B O O L E(01000111)(00101110) (01000010) (01101111) (01101111) (01101100) (01100101)(01000111)(00101110) (01000010) (01101111) (01101111) (01101100) (01100101)

1.28-) Decode the following ASCII code : 1001010 1.28-) Decode the following ASCII code : 1001010

1100001 1100001

1101110 1100101 0100000 1000100 1101111 1101110 1100101 0100000 1000100 1101111

1100101 .1100101 .

Jane Jane Doe Doe

1.29-) The following is a string of ASCII characters 1.29-) The following is a string of ASCII characters

whose bit patterns have benn converted into whose bit patterns have benn converted into

hexadecimal for compactness : 4A EF 68 6E 20 hexadecimal for compactness : 4A EF 68 6E 20

C4 EF E5 . Of the 8 bits in each pair of digits, the C4 EF E5 . Of the 8 bits in each pair of digits, the

leftmost is a parity bit. The remaining bits are the leftmost is a parity bit. The remaining bits are the

ASCII code.ASCII code.

01001010 11101111 01101000 01101110 00100000 11000100 11101111 1110010101001010 11101111 01101000 01101110 00100000 11000100 11101111 11100101

J O H N (space) D O EJ O H N (space) D O E

1.30-) How many printing characters are there in 1.30-) How many printing characters are there in

ASCII ?ASCII ?

How many of them are special characters (not How many of them are special characters (not

letters or numerals) ?letters or numerals) ?

94 characters 94 characters

62 of them are numbers and letters.62 of them are numbers and letters.

32 of them are special characters.32 of them are special characters.

1.31-) What bit must be complemented to change 1.31-) What bit must be complemented to change

an ASCII letter from capital to lowercase, and vice an ASCII letter from capital to lowercase, and vice

versa ? versa ?

Cevap: Bir ASCII karakteri büyük harften küçük harfe çevirmek için sağdan 6. bit 0 iken 1 yapılır. Küçükten büyüğe çevrilecekse 1 iken 0 yapılır.

1.32-) The state of a 12-bit register is 1.32-) The state of a 12-bit register is

100010010111 . What is its content if it 100010010111 . What is its content if it

represents represents

(a) three decimal digits in BCD?(a) three decimal digits in BCD?

(b) three decimal digits in the excess-3(b) three decimal digits in the excess-3 code?code?

(c) three decimal digits in 84-2-1 code?(c) three decimal digits in 84-2-1 code?

(d) binary number?(d) binary number?

Three Decimal Digits in BCD:

1000 1001 0111 1000 1001 0111

Three Decimal Digits in Exces-3 Code:

1000 1001 01111000 1001 0111

(8-3) (9-3) (7-3)(8-3) (9-3) (7-3)

Three Decimal Digits in the 8-4-2-1 Code::

1000 1001 0111 1000 1001 0111

88 9 7 9 7

Binary Code:

100010010111100010010111

897897

564564

897897

2^11+2^7+2^4+2^2+2+1=21992^11+2^7+2^4+2^2+2+1=2199

1.33-) List the ASCII code for the 10 decimal digits 1.33-) List the ASCII code for the 10 decimal digits

with an even parity bit in the leftmos position.with an even parity bit in the leftmos position.

00110000

10110001

10110010

00110011

10110100

00110101

00110110

10110111

10111000

00111001

1.34-) Assume a 3-input AND gate with output F 1.34-) Assume a 3-input AND gate with output F

and a 3-input OR gate with output G. Inputs are and a 3-input OR gate with output G. Inputs are

A, B, and C . Show the signals (by means of a A, B, and C . Show the signals (by means of a

timing diagram) of the outputs F and G as timing diagram) of the outputs F and G as

functions of three inputs ABC. Use all possible functions of three inputs ABC. Use all possible

combinations of ABC.combinations of ABC.

F : A , B , C

F : A , BX , CX

AX , B , CX

AX , BX , C

NOT: X’ler HIGH ya da LOW olabilir