Ch1.number systems

Dr. Mahmoud Abo_elfetouh

LOGIC DESIGN

• This course provides you with a basic understanding of what digital devices are, how they operate, and how they can be designed to perform useful functions.

• The course is intended to give you an understanding of Binary systems, Boolean algebra, digital design techniques, logic gates, logic minimization, standard combinational circuits, sequential circuits, flip-flops, synthesis of synchronous sequential circuits, and arithmetic circuits.

Course objectives

Contents

Total Practical Lecture Topic Week No.

5 2 3 Number Systems and Codes 1

5 2 3Number Systems and Codes 2

5 2 3Boolean Algebra and Logic Simplification

3

5 2 3Minimization Techniques-Karnaugh Map

4

5 2 3Minimization Techniques-Karnaugh Map

5

5 2 3 Logic Gates 6

5 2 3 Arithmetic Circuits-Adders7

Mid- Term Exam 8

ContentsTotal Practical Lecture Topic Week No.

5 2 3Arithmetic Circuits -Subtracter

9

5 2 3Combinational Circuits 10

5 2 3 Combinational Circuits 11

5 2 3 Flip-Flops 12

5 2 3Flip-Flops

13

5 2 3 Counters - Registers 14

5 2 3 Memory Devices 15

Final- Term Exam 16

Textbook

Logic and Computer Design Fundamentals, 4th Edition by M. Morris Mano and Charles R. Kime, Prentice Hall, 2008

Chapter 1

Number Systems

ITEC 1011 Introduction to Information Technologies

1. Number Systems

Chapt. 1

Location in course textbook


Common Number Systems

System Base SymbolsUsed by humans?

Used in computers?

Decimal 10 0, 1, … 9 Yes No

Binary 2 0, 1 No Yes

Octal 8 0, 1, … 7 No No

Hexa-decimal

16 0, 1, … 9,A, B, … F

No No


Quantities/Counting (1 of 3)

Decimal Binary OctalHexa-

decimal

0 0 0 0

1 1 1 1

2 10 2 2

3 11 3 3

4 100 4 4

5 101 5 5

6 110 6 6

7 111 7 7p. 33




decimal

8 1000 10 8

9 1001 11 9

10 1010 12 A

11 1011 13 B

12 1100 14 C

13 1101 15 D

14 1110 16 E

15 1111 17 F




decimal

16 10000 20 10

17 10001 21 11

18 10010 22 12

19 10011 23 13

20 10100 24 14

21 10101 25 15

22 10110 26 16

23 10111 27 17 Etc.


Conversion Among Bases

• The possibilities:

Hexadecimal

Decimal Octal

Binary

pp. 40-46


Quick Example

2510 = 110012 = 318 = 1916

Base


Decimal to Decimal (just for fun)

Hexadecimal

Decimal Octal

Binary

Next slide…


12510 => 5 x 100 = 52 x 101 = 201 x 102 = 100

125

Base

Weight


Binary to Decimal

Hexadecimal

Decimal Octal

Binary


Binary to Decimal

• Technique– Multiply each bit by 2n, where n is the “weight”

of the bit– The weight is the position of the bit, starting

from 0 on the right– Add the results


Example

1010112 => 1 x 20 = 11 x 21 =

20 x 22 =

01 x 23 =

80 x 24 =

01 x 25 =

32

4310

Bit “0”


Octal to Decimal

Hexadecimal

Decimal Octal

Binary


Octal to Decimal

• Technique– Multiply each bit by 8n, where n is the “weight”

of the bit– The weight is the position of the bit, starting



Example

7248 => 4 x 80 = 42 x 81 = 167 x 82 = 448

46810


Hexadecimal to Decimal

Hexadecimal

Decimal Octal

Binary


Hexadecimal to Decimal

• Technique– Multiply each bit by 16n, where n is the

“weight” of the bit– The weight is the position of the bit, starting



Example

ABC16 => C x 160 = 12 x 1 = 12 B x 161 = 11 x 16 = 176 A x 162 = 10 x 256 = 2560

274810


Decimal to Binary

Hexadecimal

Decimal Octal

Binary


Decimal to Binary

• Technique– Divide by two, keep track of the remainder– First remainder is bit 0 (LSB, least-significant

bit)– Second remainder is bit 1– Etc.


Example

12510 = ?22 125 62 12 31 02 15 12 7 12 3 12 1 12 0 1

12510 = 11111012


Decimal to Octal

Hexadecimal

Decimal Octal

Binary


Decimal to Octal

• Technique– Divide by 8– Keep track of the remainder


Example

123410 = ?8

8 1234 154 28 19 28 2 38 0 2

123410 = 23228


Decimal to Hexadecimal

Hexadecimal

Decimal Octal

Binary


Decimal to Hexadecimal

• Technique– Divide by 16– Keep track of the remainder


Example

123410 = ?16

123410 = 4D216

16 1234 77 216 4 13 = D16 0 4


Octal to Binary

Hexadecimal

Decimal Octal

Binary


Octal to Binary

• Technique– Convert each octal digit to a 3-bit equivalent

binary representation


Example

7058 = ?2

7 0 5

111 000 101

7058 = 1110001012


Hexadecimal to Binary

Hexadecimal

Decimal Octal

Binary


Hexadecimal to Binary

• Technique– Convert each hexadecimal digit to a 4-bit

equivalent binary representation


Example

10AF16 = ?2

1 0 A F

0001 0000 1010 1111

10AF16 = 00010000101011112


Binary to Octal

Hexadecimal

Decimal Octal

Binary


Binary to Octal

• Technique– Group bits in threes, starting on right– Convert to octal digits


Example

10110101112 = ?8

1 011 010 111

1 3 2 7

10110101112 = 13278


Binary to Hexadecimal

Hexadecimal

Decimal Octal

Binary


Binary to Hexadecimal

• Technique– Group bits in fours, starting on right– Convert to hexadecimal digits


Example

10101110112 = ?16

10 1011 1011

2 B B

10101110112 = 2BB16


Octal to Hexadecimal

Hexadecimal

Decimal Octal

Binary


Octal to Hexadecimal

• Technique– Use binary as an intermediary


Example

10768 = ?16

1 0 7 6

001 000 111 110

2 3 E

10768 = 23E16


Hexadecimal to Octal

Hexadecimal

Decimal Octal

Binary


Hexadecimal to Octal

• Technique– Use binary as an intermediary


Example

1F0C16 = ?8

1 F 0 C

0001 1111 0000 1100

1 7 4 1 4

1F0C16 = 174148


Exercise – Convert ...

Don’t use a calculator!


decimal

33

1110101

703

1AF

Skip answer Answer


Exercise – Convert …


decimal

33 100001 41 21

117 1110101 165 75

451 111000011 703 1C3

431 110101111 657 1AF

Answer


Common Powers (1 of 2)

• Base 10Power Preface Symbol

10-12 pico p

10-9 nano n

10-6 micro

10-3 milli m

103 kilo k

106 mega M

109 giga G

1012 tera T

Value

.000000000001

.000000001

.000001

.001

1000

1000000

1000000000

1000000000000


Common Powers (2 of 2)

• Base 2Power Preface Symbol

210 kilo k

220 mega M

230 Giga G

Value

1024

1048576

1073741824

• What is the value of “k”, “M”, and “G”?• In computing, particularly w.r.t. memory,

the base-2 interpretation generally applies


Example

/ 230 =

In the lab…1. Double click on My Computer2. Right click on C:3. Click on Properties


Exercise – Free Space

• Determine the “free space” on all drives on a machine in the lab

DriveFree space

Bytes GB

A:

C:

D:

E:

etc.


Review – multiplying powers

• For common bases, add powers

26 210 = 216 = 65,536

or…

26 210 = 64 210 = 64k

ab ac = ab+c


Fractions

• Decimal to decimal (just for fun)

pp. 46-50

3.14 => 4 x 10-2 = 0.041 x 10-1 = 0.1

3 x 100 = 3 3.14


Fractions

• Binary to decimal

pp. 46-50

10.1011 => 1 x 2-4 = 0.06251 x 2-3 = 0.1250 x 2-2 = 0.01 x 2-1 = 0.50 x 20 = 0.01 x 21 = 2.0 2.6875


Fractions

• Decimal to binary

p. 50

3.14579

.14579x 20.29158x 20.58316x 21.16632x 20.33264x 20.66528x 21.33056

etc.11.001001...


Fractions

• Octal to decimal

pp. 46-50

15.42 => 2 x 8-2 = 0.031254 x 8-1 = 0.55 x 80 = 5.01 x 81 = 8.0 13.53125


Fractions

• Decimal to octal

p. 50

3.14

.14x 81.12x 80.96x 87.68x 85.44x 83.52x 84.16

etc. 3.107534...


Fractions

• Hexadecimal to decimal

pp. 46-50

2B.84 => 4 x 16-2 = 0.015625 8 x 16-1 = 0.5

B x 160 = 11.02 x 161 = 32.0 43.515625


Fractions

• Decimal to Hexadecima

p. 50

3.1

.1x 161.6x 16 9.6x 16 9.6x 16 9.6x 16 9.6x 16 9.6

etc. 3.199999...


Exercise – Convert ...

Don’t use a calculator!


decimal

29.8

101.1101

3.07

C.82

Skip answer Answer


Exercise – Convert …


decimal

29.8 11101.110011… 35.63… 1D.CC…

5.8125 101.1101 5.64 5.D

3.109375 11.000111 3.07 3.1C

12.5078125 1100.10000010 14.404 C.82

Answer


Binary Addition (1 of 2)

• Two 1-bit values

pp. 36-38

A B A + B

0 0 0

0 1 1

1 0 1

1 1 10“two”


Binary Addition (2 of 2)

• Two n-bit values– Add individual bits– Propagate carries– E.g.,

10101 21+ 11001 + 25 101110 46

11


Multiplication (1 of 3)

• Decimal (just for fun)

pp. 39

35x 105 175 000 35 3675



• Binary, two 1-bit values

A B A B0 0 00 1 01 0 01 1 1



• Binary, two n-bit values– As with decimal values– E.g.,

1110 x 1011 1110 1110 0000 111010011010


Binary Subtraction (1 of 2)

• Two 1-bit values

pp. 36-38

A B A - B

0 0 0

0 1 1

1 0 1

1 1 0

Borrow 1



• Two n-bit values– Subtract individual bits– Propagate borrows– E.g.,

11001 25- 10101 - 21 00100 4

10

0



• Two n-bit values– Subtract individual bits– Propagate borrows– E.g.,

11001 25- 10101 - 21 00100 4

10 0 10001 - 10101 00100

10


Subtraction with Complements

• Complements are used for simplifying the subtraction operations.

• There are two types of complements for each base-r system: the r's complement and the (r — l)'s complement.

• 2's complement and 1's complement for binary numbers, and the 10's complement and 9's com plement for decimal numbers.


9's complement

• The 9's complement of a decimal number is obtained by subtracting each digit from 9.

• • The 9's complement of 546700 is:• 999999 - 546700 = 453299.• The 9's complement of 012398 is:• 999999 - 012398 = 987601.


Binary numbers, the 1's complement

• The 1's complement of a binary number is formed by changing 1's to 0's and 0's to 1's.

• Examples: • The 1's complement of 1011000 is• 0100111• The 1's complement of 0101101 is• 1010010


10's complement

• The 10's complement can be formed by leaving all least significant 0's un changed, subtracting the first nonzero least significant digit from 10, and subtracting all higher significant digits from 9.

• The 10's complement of 012398 is 987602.• The 10's complement of 246700 is 753300.


2's complement

• The 2's complement can be formed by leaving all least significant 0's and the first 1 unchanged, and replacing 1's with 0's and 0's with 1's in all other higher significant digits.

• The 2's complement of 1101100 is• 0010100• The 2's complement of 0110111 is• 1001001


Subtraction with Complements

• The subtraction of two n-digit unsigned numbers M — N in base r can be done as follows:

• Add M to the r's complement of N. • If M ≥ N, the sum will produce an end carry, r n,

which is discarded; what is left is the result M - N.• If M < N, the sum does not produce an end carry.

To obtain the answer in a familiar form, take the r's complement of the sum and place a negative sign in front.


Examples to illustrate the procedure

• Given the two binary numbers;• X = 1010100 and Y = 1000011, • perform the subtraction:• (a) X — Y • (b) Y — X • using 2's complements.


X — Y= 1010100 —1000011

• X = 1010100• 2's complement of Y = + 0111101• Sum = 10010001• Discard end carry = -10000000• Answer: X — Y = 0010001•


Y — X= 1000011 — 1010100

• Y = 1000011• 2's complement of X = + 0101100• Sum = 1101111• There is no end carry.• Answer:• Y - X = -(2's complement of 1101111) • = - 0010001


Thank you

Next topic


Binary Codes

• Binary codes are codes which are represented in binary system with modification from the original ones.

• Binary codes are classified as:– Weighted Binary Systems – Non Weighted Codes


Weighted Binary Systems

• Weighted binary codes are those which obey the positional weighting principles,

• Each position of the number represents a specific weight.

• The codes 8421, 2421, 5421, and 5211 are weighted binary codes.


Weighted Binary Systems


8421 Code/BCD Code

• The BCD (Binary Coded Decimal) is a straight assignment of the binary equivalent.

• It is possible to assign weights to the binary bits according to their positions.

• The weights in the BCD code are 8,4,2,1.• Example: The bit assignment 1001, can be seen by

its weights to represent the decimal 9 because:• 1x8+0x4+0x2+1x1 = 9• Ex. number 12 is represented in BCD as [0001 0010]


2421 Code

• 2421 Code This is a weighted code, its weights are 2, 4, 2 and 1.

• A decimal number is represented in 4-bit form and the total four bits weight is 2 + 4 + 2 + 1 = 9.

• Hence the 2421 code represents the decimal numbers from 0 to 9.


5211 Code

• 5211 Code This is a weighted code, its weights are 5, 2, 1 and 1.

• A decimal number is represented in 4-bit form and the total four bits weight is 5 + 2 + 1 + 1 = 9.

• Hence the 5211 code represents the decimal numbers from 0 to 9.


Reflective Code• Reflective Code A code is said to be

reflective when code for 9 is complement for the code for 0, and so is for 8 and 1 codes, 7 and 2, 6 and 3, 5 and 4.

• Codes 2421, 5211, and excess-3 are reflective, whereas the 8421 code is not.


Sequential Codes• Sequential Codes A code is said to be sequential

when two subsequent codes, seen as numbers in binary representation, differ by one.

• This greatly aids mathematical manipulation of data.

• The 8421 and Excess-3 codes are sequential, whereas the 2421 and 5211 codes are not.


Excess-3 Code• Excess-3 Code Excess-3 is a non weighted code

used to express decimal numbers. • The code derives its name from the fact that each

binary code is the corresponding 8421 code plus 0011(3).

• Example: 1000 of 8421 = 1011 in Excess-3


Error Detecting and Correction Codes

• For reliable transmission and storage of digital data, error detection and correction is required.


Error Detecting Codes

• When data is transmitted from one point to another there are chances that data may get corrupted.

• To detect these data errors, we use special codes, which are error detection codes.


Parity check

• In parity codes, every binary message is checked if they have even number of ones or even number of zeros.

• Based on this information an additional bit is appended to the original data.

• At the receiver side, once again parity is calculated and matched with the received parity, and if they match, data is ok, otherwise data is corrupt.

• There are two types of parity: Even parity and Odd Parity


Parity

• There are two types of parity:• Even parity: Checks if there is an even

number of ones; if so, parity bit is zero. When the number of ones is odd then parity bit is set to 1.

Even parity codexyz p

Messagexyz

000 0 000

001 1 001

011 0 011


Parity

• Odd Parity: Checks if there is an odd number of ones; if so, parity bit is zero. When number of ones is even then parity bit is set to 1.

• •

Odd parity codexyz p

Messagexyz

000 1 000

001 0 001

011 1 011


Alphanumeric Codes

• The binary codes that can be used to represent all the letters of the alphabet, numbers and mathematical symbols, punctuation marks, are known as alphanumeric codes or character codes.

• These codes enable us to interface the input-output devices like the keyboard, printers, video displays with the computer.


ASCII Code

• ASCII Code ASCII stands for American Standard Code for Information Interchange.

• It has become a world standard alphanumeric code for microcomputers and computers.

• It is a 7-bit code representing 27 = 128 different characters.

• These characters represent 26 upper case letters (A to Z), 26 lowercase letters (a to z), 10 numbers (0 to 9), 33 special characters and symbols and 33 control characters.

•


ASCII Code

• The 7-bit code is divided into two portions, The leftmost 3 bits portion is called zone bits and the 4-bit portion on the right is called numeric bits.

7-bit ASCII Character

100 0001 A

100 0010 B

011 0011 3


ASCII Code

• An 8-bit version of ASCII code is known as ASCII-8.

• The 8-bit version can represent a maximum of 256 characters.


EBCDIC Code

• EBCDIC Code EBCDIC stands for Extended Binary Coded Decimal Interchange.

• It is mainly used with large computer systems like mainframes.

• EBCDIC is an 8-bit code and thus accommodates up to 256 characters.

• An EBCDIC code is divided into two portions: 4 zone bits (on the left) and 4 numeric bits (on the right).

Ch1.number systems

Technology

information technologies

octal technique

binary technique

hexadecimal technique

logic minimization

arithmetic circuitsadders

common number systems

logic gates32