1.Digital Logic Design (Chap 01, Topic 1,2 - Binary Systems).pdf

Course Teacher :
Colonel S M Saiful Islam, SUP, psc
CSE (BUET), MBA (IBA), MDS
ITIL® (Expert), Prince2® (Practitioner), CDCP®, ISO 27001 Lead Auditor®
Course Title :
CSE-1201: Digital Logic Design
Credit Hour: 3.0
Department of Computer Science & Engineering
Bangladesh University of Professionals
Chapter 01
BINARY SYSTEMS

2
Chapter 01: Binary Systems
Learning Objective
▪ Learn various binary systems that are suitable for representing
information in digital components.
▪ Understand the binary number system and how it functions.
▪ Explore binary codes and their use in representing decimal and
alphanumeric information.
▪ Gain an initial understanding of binary logic from an intuitive
perspective.
10-Nov-24

3
Overview
1. Digital Systems
2. Binary Numbers
3. Number Base Conversions
4. Octal and Hexadecimal Numbers
5. Complements
6. Signed Binary Numbers
7. Binary Codes
10-Nov-24

Topic- 1: Introduction to Number Systems and
Digital Components (Binary, Octal,
Hexadecimal Systems)
10-Nov-24 Chapter 01: Binary Systems 4

5
Analog Vs Digital System
10-Nov-24
Analog System
• The physical quantities or signals may
vary continuously over a specified range
Digital System
• The physical quantities or signals can
assume only discrete values
• Greater accuracy
Analog Signal
Digital Signal

6
Digital Systems
10-Nov-24
• Digital systems are designed to store, process, and communicate
information in digital form.
• They are found in a wide range of applications, including process control,
communication systems, digital instruments, and consumer products. The
digital computer, more commonly called the computer, is an example of a
typical digital system.
• They are involved in our business transactions, communications,
transportation, medical treatment and entertainment. In industrial world
they are heavily employed in design, manufacturing, distribution and
sales.

7
Digital Systems (contd…)
10-Nov-24
• Takes a set of discrete
information inputs and
discrete internal information
(system state) and generates
a set of discrete information
outputs
Discrete
Information
Processing
System
System
State
Discrete
Inputs
Discrete
Outputs

8
10-Nov-24
Discrete
Information
Processing
System
System
State
Discrete
Inputs
Discrete
Outputs
• No state present
• Combinational Logic System
• Output = Function (Input)
• State present
• State updated at discrete times
• Synchronous Sequential System
• State updated at any time
• Asynchronous Sequential System
• State = Function (State, Input)
• Output = Function (State)
or Function (State, Input)

9
10-Nov-24
A Digital Counter (e. g., odometer):
Synchronous or Asynchronous?
1 3
0 0 5 6 4
Count Up
Reset
Inputs: Count Up, Reset
Outputs: Visual Display
State: "Value" of stored digits

10
10-Nov-24
Inputs: Keyboard, mouse,
modem, microphone
Outputs: CRT, LCD,
modem, speakers
Memory
Control
unit Datapath
Input/Output
CPU

11
10-Nov-24
• An information variable represented by physical quantity.
• For digital systems, the variable takes on discrete values.
• Two level, or binary values are the most prevalent values in digital
systems.
• Binary values are represented abstractly by:
• digits 0 and 1
• words (symbols) False (F) and True (T)
• words (symbols) Low (L) and High (H)
• and words On and Off.
• Binary values are represented by values or ranges of values of physical
quantities

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10-Nov-24
Analog
Asynchronous
Synchronous
Time
Continuous in value
& time
Discrete in value &
continuous in time
Discrete in value &
time
Digital
Signal

13
10-Nov-24
Binary Digital Signal
• An information variable represented by physical quantity.
• For digital systems, the variable takes on discrete values.
• Two level, or binary values are the most prevalent values
• Binary values are abstractly represented by:
• Digits 0 and 1
• Words (symbols) False (F) and True (T)
• Words (symbols) Low (L) and High (H)
• And words On and Off
• Binary values are represented by values
or ranges of values of physical quantities
Binary Digital Signal

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Number Systems
10-Nov-24
• Integers are normally written using positional number system, in
which each digit represents the coefficient in a power series
N = 𝑎𝑛−1 𝑟𝑛
− 1 + 𝑎𝑛−2 𝑟𝑛
− 2 + ⋯ + 𝑎2 𝑟2
+ 𝑎1 𝑟1
+ 𝑎0
Where,
n is the number of digit
r is the radix or base
𝑎𝑖 is the coefficient 0 ≤ 𝑎𝑖 ≤ r
• There are four systems of arithmetic which are often used in digital circuit

15
Number Systems
10-Nov-24
N = 𝑎𝑛−1 𝑟𝑛
− 1 + 𝑎𝑛−2 𝑟𝑛
− 2 + ⋯ + 𝑎2 𝑟2
+ 𝑎1 𝑟1
+ 𝑎0
0 ≤ 𝑎𝑖 ≤ r
2073 = 2 𝑥 103 + 0 𝑥102 + 7𝑥101 + 3
233 = 2 𝑥 82 + 3𝑥81 + 3
2A9 = 2 𝑥 162 + 𝐴𝑥161 + 9

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Number Systems (contd…)
10-Nov-24
• Decimal: it has a base (r = 10) and coefficients (a) are in the range 0 to 9
• Binary: it has a base (r = 2) and coefficients (a) are all either 0 or 1
• Octal: it has a base (r = 8) and coefficients (a) are in the range 0 to 7
• Hexadecimal: it has a base (r = 16) and coefficients (a) are in the range
{0, 1, 2, 3, 4, 5, 6, 7, 8, 9, A, B, C, D, E, F}

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10-Nov-24
General Decimal Binary
Radix (Base) r 10 2
Digits 0 => r - 1 0 => 9 0 => 1
0
1
2
3
Powers of 4
Radix 5
-1
-2
-3
-4
-5
r0
r1
r2
r3
r4
r5
r -1
r -2
r -3
r -4
r -5
1
10
100
1000
10,000
100,000
0.1
0.01
0.001
0.0001
0.00001
1
2
4
8
16
32
0.5
0.25
0.125
0.0625
0.03125

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10-Nov-24
Commonly Occurring Bases
Name Radix Digits
Binary 2 0,1
Octal 8 0,1,2,3,4,5,6,7
Decimal 10 0,1,2,3,4,5,6,7,8,9
Hexadecimal 16 0,1,2,3,4,5,6,7,8,9,A,B,C,D,E,F

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10-Nov-24
Decimal
(Base 10)
Binary
(Base 2)
Octal
(Base 8)
Hexadecimal
(Base 16)
00 00000 00 00
01 00001 01 01
02 00010 02 02
03 00011 03 03
04 00100 04 04
05 00101 05 05
06 00110 06 06
07 00111 07 07
08 01000 10 08
09 01001 11 09
10 01010 12 0A
11 01011 13 0B
12 01100 14 0C
13 01101 15 0D
14 01110 16 0E
15 01111 17 0F
16 10000 20 10

20
10-Nov-24
210 (1024) is Kilo, denoted "K"
220 (1,048,576) is Mega, denoted "M"
230 is Giga, denoted "G"
Exponent Value Exponent Value
0 1 11 2,048
1 2 12 4,096
2 4 13 8,192
3 8 14 16,384
4 16 15 32,768
5 32 16 65,536
6 64 17 131,072
7 128 18 262,144
19 524,288
20 1,048,576
21 2,097,152
8 256
9 512
10 1024
240 is Tera, denoted "G"
250 is Peta, denoted “P"

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10-Nov-24
(73925)10 = 7*104 + 3*103 + 9*102 + 2*101 + 5*100
(a4 a3 a2 a1 a0)10 = a4 *104 + a3*103 + a2*102 + a1 *101 + a0*100
(a4 a3 a2 a1 a0)r = a4 *r4 + a3*r3 + a2*r2 + a1 *r1 + a0*r0
(an an-1 ………a2 a1 a0)r = an *rn + an-1*rn-1 + …… + a2*r2 + a1 *r1 + a0*r0
(73925.613)10 = 7*104 + 3*103 + 9*102 + 2*101 + 5*100 + 6*10-1 + 1*10-2 + 3*10-3
(an an-1 ………a2 a1 a0.a-1 a-2 ......a-m)r
= an *rn + an-1*rn-1 + …… + a2*r2 + a1 *r1 + a0*r0 + a-1*r-1 + a-2 *r-2 + ………… +
a-m*r-m

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Binary Arithmetic
10-Nov-24
• Binary arithmetic is essential in all digital computers and in many other
types of digital systems.
• To understand digital systems, one must know the basics of
• Binary Addition
• Binary Subtraction
• Binary Multiplication
• Binary Division

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Binary Arithmetic (contd…)
10-Nov-24
Binary Addition
Carry

24
10-Nov-24
Binary Subtraction
Borrow a “Base” when needed

25
10-Nov-24
Binary Multiplication
Bit by Bit

26
10-Nov-24
Binary Division
101 110111
-101
111
-101
0101
-101
0
1011
Division in binary follows the same procedure as division in decimal.

27
Exercise
10-Nov-24
• Perform the following binary additions:
a) 1101 + 1010
b) 10111 + 01101
• Perform the following binary subtractions:
a) 1101 – 0100
b) 1001 – 0111
• Perform the indicated binary operations:
a) 110 × 111
b) 1100 ÷ 011

Topic- 2: Conversions Between Number
Systems, Complements and Codes
10-Nov-24 Chapter 01: Binary Systems 28

29
Conversion Between Bases
10-Nov-24
1) Convert the Integer Part:
Repeatedly divide the number by the new radix and save the remainders. The digits for
the new radix are the remainders in reverse order of their computation. If the new radix is
> 10, then convert all remainders > 10 to digits A, B, …
2) Convert the Fractional Part:
Repeatedly multiply the fraction by the new radix and save the integer digits that result.
The digits for the new radix are the integer digits in order of their computation. If the new
radix is > 10, then convert all integers > 10 to digits A, B, …
3) Join the two results with a radix point

30
Conversion Between Bases (Binary-to-Decimal)
10-Nov-24
(a) Integer (b) Fraction
110102 =1*24 + 1*23 + 0*22 + 1*21 + 0*20
= 16 + 8 + 0 + 2 + 0
= (26)10
1011102 = 1*25 + 0*24 + 1*23 + 1*22 + 1*21+ 0*20
= 32 + 8 + 4 + 2
= (46)10

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10-Nov-24
25 ÷ 2 = 12 +1
12 ÷ 2 = 6 +0
6 ÷ 2 = 3 +0
3 ÷ 2 = 1 +1
1 ÷ 2 = 0 +1
2510 = 110012
0.625 × 2 = 1.25 = 0.25 + 1
0.25 × 2 = 0.5 = 0.5 + 0
0.5 × 2 = 1.0 = 0.0 + 1
0.62510 = 0.1012
Conversion Between Bases (Decimal-to-Binary)

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Conversion Between Bases (Decimal-to-Octal)
10-Nov-24
3928610 = 1145668 0.34210 = 0.25708

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10-Nov-24
1408 = 9610
1725.438 = 981.54687510
Conversion Between Bases (Octal-to-Decimal)

34
10-Nov-24
Conversion Between Bases (Decimal-to-Hexadecimal)

35
10-Nov-24
54.D216 = 84.820312510
Conversion Between Bases (Hexadecimal-to-Decimal)

36
10-Nov-24
Works both ways (Binary to Octal & Octal to Binary)
10110.012 = 26.28
Conversion Between Bases (Binary to/from Octal)

37
10-Nov-24
10110.012 = 16.416
Conversion Between Bases (Binary to/from Hexadecimal)

38
10-Nov-24
26.28 = 16.416
Conversion Between Bases (Octal to/from Hexadecimal)

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10-Nov-24
Conversion of Binary to Octal & Hexadecimal
(10110001101011 . 111100000110)2 = (26153.7406)8
0 1 0 1 1 0 0 0 1 1 0 1 0 1 1 . 1 1 1 1 0 0 0 0 0 1 1 0
2 6 1 5 3 . 7 4 0 6
(10110001101011 . 111100000110)2 = (2C6B.F06)16
0 0 1 0 1 1 0 0 0 1 1 0 1 0 1 1 . 1 1 1 1 0 0 0 0 0 1 1 0
2 C 6 B F 0 6

40
10-Nov-24
Conversion of Hexadecimal-to-Binary-to-Octal
Hexa 2 C 6 B F 0 6
Binary 0 0 1 0 1 1 0 0 0 1 1 0 1 0 1 1 . 1 1 1 1 0 0 0 0 0 1 1 0
Octal 2 6 1 5 3 7 4 0 6
(2C6B.F06)16 = (26153.7406)8

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Exercise
10-Nov-24
• Convert the decimal number 250.5 to base 3, base 4, base 7, base 8,
and base 16.
• Convert the following decimal numbers to binary: 12.0625, 104, 673.23,
and 1998.
• Convert the following binary numbers to decimal: 10.10001, 101110.0101,
1110101.0101, 1110101.110.
• Convert the following numbers from the given base to the bases indicated:
a) Decimal 225.225 to binary, octal, and hexadecimal
b) Binary 11010111.110 to decimal, octal, and hexadecimal
c) Octal 623.77 to decimal, binary, and hexadecimal
d) Hexadecimal 2AC5.D to decimal, octal, and binary

42
Complements
10-Nov-24
• Complements are used in digital computers for simplifying the subtraction
operation and for logical manipulations.
• There are two types of complements for each base-r system:
• r’s complement
• (r-1)’s complement

43
Complements
10-Nov-24
r's complement of (N)r = rn - N
• 10's of (52520)10 = 105 – 52520 = 47480
• 10's of (0.3267)10 = 1 - 0.3267 = 0.6733
• 10's of (25.639)10 = 102 - 25.639 = 74.631
• 2's of (101100)2 = 26 – 101100
= 100000 – 101100
= (010100)2
• 2's of (0.0110)2 = 1 - 0.0110
= 0.1010

44
Complements
10-Nov-24
(r-1)'s complement of (N)r = rn -1- N
rn -1 = (r-1) (r-1) ……… (r-1) (r-1) (r-1) ….. up to n i.e. 105 -1 = 99999 ( five digits of 9)
• 9's of (52520)10 = 99999 – 52520 = 47479
• 9's of (0.3267)10 = 0.9999 - 0.3267 = 0.6732
• 9's of (25.639)10 = 99.999 - 25.639 = 74.630
• 1's of (101100)2 = 111111 – 101100
= (010011)2
• 1's of (0.0110)2 = 0.1111 - 0.0110
= 0.1001

45
Subtraction with Complements
10-Nov-24
Subtraction with r's complement
• Make the minuend and subtrahend equal in digits
• Add the minuend to the r's complement of the subtrahend
• If and end-carry occurs, ignore it
• If there is no end carry, make the r's complement of the result and place a
negative sign in front.

46
Subtraction with 10's complement
10-Nov-24
Example : Subtract 72532 - 3250
Minuend M = 7 2 5 3 2 M = 7 2 5 3 2 M = 7 2 5 3 2
Subtrahend N = 3 2 5 0 N = 0 3 2 5 0 10's complement of N = + 9 6 7 5 0
end carry 1 6 9 2 8 2
Answer = 69282
Minuend M = 3 2 5 0 M = 0 3 2 5 0 M = 0 3 2 5 0
Subtrahend N = 7 2 5 3 2 N = 7 2 5 3 2 10's complement of N = + 2 7 4 6 8
end carry 0 3 0 7 1 8
10's complement of 30718 = 69282
Answer = ‒ 69282

47
10-Nov-24
Example : Subtract 1010100 – 1000100
Minuend M = 1010100 M = 1010100
Subtrahend N = 1000100 2’s complement of N = + 0111100
end carry 1 0010000
Answer = 10000
Minuend M = 1000100 M = 1000100
end carry 0 1110000
2’s complement of 1110000 = 10000
Answer = - 10000

48
Subtraction with (r-1)'s complement
10-Nov-24
• Make the minuend and subtrahend equal in digits
• Add the minuend to the (r-1)'s complement of the subtrahend
• If and end-carry occurs, add 1 with the result
• If there is no end carry, make the (r-1)'s complement of the result and
place a negative sign in front.

49
10-Nov-24
Example: Subtract 3250 - 72532
Minuend M = 3 2 5 0 M = 0 3 2 5 0 M = 0 3 2 5 0
Subtrahend N = 7 2 5 3 2 N = 7 2 5 3 2 9’s complement of N = + 2 7 4 6 7
end carry 0 3 0 7 1 7
9’s complement of 30717 = 69282
Answer = - 69282
Minuend M = 7 2 5 3 2 M = 7 2 5 3 2 M = 7 2 5 3 2
Subtrahend N = 3 2 5 0 N = 0 3 2 5 0 9’s complement of N = + 9 6 7 4 9
end carry 1 6 9 2 8 1
+ 1
6 9 2 8 2
Answer = 69282

50
10-Nov-24
Minuend M = 1010100 M = 1010100
end carry 1 0001111
+ 1
0010000
Answer = 10000
Minuend M = 1000100 M = 1000100
end carry 0 1101111
1’s complement of 1101111 = 0010000
Answer = - 10000

51
Subtraction with 10's complement in Base-11 number system
10-Nov-24
Example : Subtract (72532)11 – (3250)11
10’s complement of (03250)11 is = A A A A A – 0 3 2 5 0 = A 7 8 5 A
Minuend M = 7 2 5 3 2 M = 7 2 5 3 2 M = 7 2 5 3 2
Subtrahend N = 3 2 5 0 N = 0 3 2 5 0 9’s complement of N = + A 7 8 5 A
end carry 1 6 A 2 9 1
+ 1
6 A 2 9 2
Answer = 69282

52
Complements
10-Nov-24
Additional Study Notes :
• Prove that “(r-1)’s complement of N is defined as rn – r-m–N”, where N is positive
number, r is base, n is number of digits, m is number of digits for fraction part.
Page -10-11
• Prove that if minuend (M) is smaller than the subtrahend (N), while both M and
N are positive, subtraction is obtained by making complement of the summation
of M and complement of N.
Page-12

53
Exercise
10-Nov-24
• Determine the 1’s complement of each binary number:
a) 00011010
b) 11110111
c) 10001101
a) 00010110
b) 11111100
c) 10010001

54
Exercise
10-Nov-24
a) 00011010
b) 11110111
c) 10001101
a) 00010110
b) 11111100
c) 10010001

55
Binary Codes
10-Nov-24
• When number, alphabets or words are represented by a specific group of
symbols i.e., they are encoded
• The group of symbols used to encode them are called codes. The digital
data is represented, stored and transmitted as a group of binary digits (bits)
• Group of bits binary code Numeric and alphanumeric code

56
Classification of Binary Codes
10-Nov-24

58
Binary Numbers & Binary Codes
10-Nov-24
• Do NOT mix up conversion of a decimal number to a binary
number with coding a decimal number with a BINARY CODE.
• 1310 = 11012 (This is conversion)
• 13  0001|0011 (This is coding)

59
Binary Codes
10-Nov-24
• Given n binary digits (called bits), a binary code
is a mapping from a set of represented
elements to a subset of the 2n binary numbers.
• Example: A binary code for the seven colors of
the rainbow
• Code 100 is not used
Binary Number
000
001
010
011
101
110
111
Color
Red
Orange
Yellow
Green
Blue
Indigo
Violet
Non-Numeric Binary Codes

60
Number of bits required
10-Nov-24
• Given M elements to be represented by a binary code, the minimum
number of bits, n, needed, satisfies the following relationships:
2n > M > 2(n – 1)
n = log2 M where x , called the ceiling function, is the integer greater
than or equal to x.
• Example: How many bits are required to represent decimal digits with a
binary code?

61
Binary Codes & Decimal Digits
10-Nov-24
Decimal 8,4,2,1 Excess3 8,4,-2,-1 Gray
0 0000 0011 0000 0000
1 0001 0100 0111 0100
2 0010 0101 0110 0101
3 0011 0110 0101 0111
4 0100 0111 0100 0110
5 0101 1000 1011 0010
6 0110 1001 1010 0011
7 0111 1010 1001 0001
8 1000 1011 1000 1001
9 1001 1100 1111 1000
There are over 8,000 ways that you can chose 10 elements from the 16
binary numbers of 4 bits. A few are useful:
2,4,2,1
0000
0001
0010
0011
1011
1100
1101
1110
1111
0100

62
Binary Coded Decimal (BCD)
10-Nov-24
• Binary coded decimal (BCD) is a way to express each of the decimal digits
with a binary code. There are only 10 code groups in the BCD system, so it
is very easy to convert between decimal and BCD.
• The 8421 code is a type of BCD code that indicates the binary weights of
the four bits (23, 22 , 21 , 20 ). The 8421 code is the predominant BCD code, and it
is referred to BCD, it is always meant to be the 8421 BCD code.
Decimal/ BCD conversion

63
Binary Coded Decimal (BCD)
10-Nov-24
• Invalid Codes
• With 4 bits 16 numbers (0000 through 1111) can be
represented but that, in the 8421 code only 10 of these are
used. The six combinations that are not used- 1010, 1011,
1100, 1101, 1110, and 1111 —are invalid in the 8421 BCD
code.

64
Conversion (Decimal to BCD)
10-Nov-24
• Convert each of the following decimal numbers to BCD:
• (a) 35 (b) 98 (c) 170 (d) 2469
Example: Convert the decimal number 9673 to BCD.

65
10-Nov-24
• Convert each of the following BCD codes to decimal:
(a) 10000110 (b) 001101010001 (c) 1001010001110000
Example: Convert the BCD code 10000010001001110110 to decimal.
Conversion (BCD to Decimal)

66
BCD Addition
10-Nov-24
• BCD is a numerical code and can be used in arithmetic operations.
Addition is the most important operation because the other three
operations (subtraction, multiplication, and division) can be accomplished
by the use of addition. Here is how to add two BCD numbers:
• Step 1: Add the two BCD numbers, using the rules for binary addition.
• Step 2: If a 4-bit sum is equal to or less than 9, it is a valid BCD number.
• Step 3: If a 4-bit sum is greater than 9, or if a carry out of the 4-bit group
is generated, it is an invalid result. Add 6 (0110) to the 4-bit sum in order
to skip the six invalid states and return the code to 8421. If a carry results
when 6 is added, simply add the carry to the next 4-bit group

67
BCD Addition
10-Nov-24
• Add the following BCD numbers:
• (a) 0011 + 0100
• (b) 00100011 + 00010101

68
BCD Addition
10-Nov-24
• (c) 10000110 + 00010011
• (d) 010001010000 + 010000010111
Note that in each case the sum in any 4-bit column does not exceed 9, and the
results are valid BCD numbers.
Example: Add the BCD numbers: 1001000001000011 + 0000100100100101.

69
BCD Addition
10-Nov-24
• (a) 1001 + 0100
• (b) 1001 + 1001

70
BCD Addition
10-Nov-24
• (c) 00010110 + 00010101
• (d) 01100111 + 01010011
Example : Add the BCD numbers: 01001000 + 00110100.

71
Exercise
10-Nov-24
• What is the binary weight of each 1 in the following BCD numbers?
(a)0010 (b) 1000 (c) 0001 (d) 0100
• Convert the following decimal numbers to BCD:
(a)6 (b) 15 (c) 273 (d) 849
• What decimal numbers are represented by each BCD code?
(a) 10001001 (b) 001001111000 (c) 000101010111
• In BCD addition, when is a 4-bit sum invalid?

72
Gray Code
10-Nov-24
• The Gray code is unweighted and is not an arithmetic code; that is, there are
no specific weights assigned to the bit positions.
• The important feature of the Gray code is that it exhibits only a single bit
change from one code word to the next in sequence.
• This property is important in many applications, such as shaft position
encoders, where error susceptibility increases with the number of bit changes
between adjacent numbers in a sequence

73
Gray Code
10-Nov-24
• What special property does the
Gray code have in relation to
adjacent decimal digits?
Decimal 8,4,2,1 Gray
0 0000 0000
1 0001 0100
2 0010 0101
3 0011 0111
4 0100 0110
5 0101 0010
6 0110 0011
7 0111 0001
8 1000 1001
9 1001 1000

74
Gray Code
10-Nov-24
• Going from decimal 3 to decimal 4, the Gray code changes from 0010 to 0110, while the binary code
changes from 0011 to 0100, a change of three bits. The only bit change in the Gray code is in the third
bit from the right: the other bits remain the same.

75
Conversion (Binary to Gray Code)
10-Nov-24
• Step- 1: The most significant bit (left-most) in the Gray code is the same
as the corresponding MSB in the binary number.
• Step- 2: Going from left to right, add each adjacent pair of binary code
bits to get the next Gray code bit. Discard carries.

76
Conversion (Binary to Gray Code)
10-Nov-24
• Convert the binary number 11000110 to Gray code.
Example: Convert binary 101101 to Gray code.

77
Conversion (Gray Code to Binary )
10-Nov-24
• Step- 1: The most significant bit (left-most) in the binary code is the same
as the corresponding bit in the Gray code.
• Step- 2: Add each binary code bit generated to the Gray code bit in the
next adjacent position. Discard carries.

78
Conversion (Gray Code to Binary )
10-Nov-24
• Convert the Gray code 10101111 to binary.
Example: Convert Gray code 100111 to binary.

79
Alphanumeric Codes
10-Nov-24
• In order to communicate, not only numbers, but also letters and other
symbols are needed.
• Alphanumeric codes are codes that represent numbers and alphabetic
characters (letters). Most such codes, however, also represent other
characters such as symbols and various instructions necessary for conveying
information.
• ASCII is a universally accepted alphanumeric code used in most computers
and other electronic equipment.

80
ASCII Character Codes
10-Nov-24
• American Standard Code for Information Interchange
• This code is a popular code used to represent information sent as
character-based data. It uses 7-bits to represent:
• 94 Graphic printing characters.
• 34 Non-printing characters
• Some non-printing characters are used for text format (e.g. BS =
Backspace, CR = carriage return)
• Other non-printing characters are used for record marking and flow
control (e.g. STX and ETX start and end text areas).

81
ASCII Character Codes
10-Nov-24
ASCII has some interesting properties:
• Digits 0 to 9 span Hexadecimal values 30₁₆ to 39₁₆.
• Upper case A - Z span 41₁₆ to 5A₁₆.
• Lower case a - z span 61₁₆ to 7A₁₆.
• Lower to upper case translation (and vice versa)
occurs by flipping bit 6.
• Delete (DEL) is all bits set, a carryover from when
punched paper tape was used to store messages.
Punching all holes in a row erased a mistake!

82
Unicode
10-Nov-24
A system to encode all characters across the world's languages by assigning
each a unique numeric value (code point) in the universal character set (UCS).
• Supports 100,000+ Characters: Includes text, symbols, and multilingual
scripts.
• Compatibility with ASCII: The first 128 Unicode characters match ASCII,
making ASCII a small subset of Unicode.
• Flexible Encoding: Uses "code points" mapped to sequences of bytes for
different characters.
• Comprehensive Standard: Covers rules for normalization, decomposition,
and supports bidirectional scripts (e.g., Arabic, Hebrew).

83
Error Detection Code
10-Nov-24
Message P ( Odd) P (even)
0000 1 0
0001 0 1
0010 0 1
0011 1 0
0100 0 1
0101 1 0
0110 1 0
0111 0 1
1000 0 1
1001 1 0
1010 1 0
1011 0 1
1100 1 0
1101 0 1
1110 0 1
1111 1 0
• Any external noise introduced into a physical communication
media changes the bit values from 0 to 1 or vice versa.
• Redundancy (e.g. extra information), in the form of extra bits,
can be incorporated into binary code words to detect and
correct errors.
• A simple form of redundancy is Parity i.e. an extra bit
included with the message to make the total number of 1’s
either odd or even.
• Parity can detect all single-bit errors and some multiple-bit
errors.
• A code word has even parity if the number of 1’s in the code
word is even.
• A code word has odd parity if the number of 1’s in the code
word is odd.
• An error is detected if the checked parity does not correspond
to the adopted one

84
Detecting an Error
10-Nov-24
• A parity bit helps detect a single-bit error (or any odd number of errors), but it cannot
detect two errors in one group.
• We transmit the BCD code 0101. The total code transmitted, including the even parity bit,
is :
Now let’s assume that an error occurs in the third bit from the left (the 1 becomes a 0)
When this code is received, the parity check circuitry determines that there is only a single 1, when
there should be an even number of 1s. Because an even number of 1s does not appear in the code
when it is received, an error is indicated.

85
Example
10-Nov-24
• Assign the proper even parity bit to the following code groups:
• (a) 1010 (b) 111000 (c) 101101 (d) 1000111001001 (e) 101101011111
Make the parity bit either 1 or 0 as necessary to make the total number of 1s
even. The parity bit will be the left-most bit (color).

86
Example
10-Nov-24
• An odd parity system receives the following code groups: 10110, 11010,
110011, 110101110100, and 1100010101010. Determine which groups, if
any, are in error.
Since odd parity is required, any group with an even number of 1s is
incorrect. The following groups are in error: 110011 and 1100010101010.
Example:
The following ASCII character is received by an odd parity system:
00110111. Is it correct?

87
Hamming Code
10-Nov-24
Hamming code is used to detect and correct single-bit errors in
transmitted data.
• Redundancy Bits:
• 4 redundancy bits are added to a 7-bit group of data bits.
• These bits are placed at positions 2ⁿ (n = 0, 1, 2, 3) within
the data.
• Error Correction Process:
• Redundancy bits are interspersed within the data bits
during transmission.
• After transmission, redundancy bits are removed to retrieve
the original data.

88
Exercise
10-Nov-24
• .Which odd-parity code is in error?
(a) 1011 (b) 1110 (c) 0101 (d) 1000
• Which even-parity code is in error?
(a) 11000110 (b) 00101000 (c) 10101010 (d) 11111011
• Add an even parity bit to the end of each of the following codes.
(a) 1010100 (b) 0100000 (c) 1110111 (d) 1000110

For Your Attention
Thank You
+8801769021807 Chairman.cse@bup.edu.bd

1.Digital Logic Design (Chap 01, Topic 1,2 - Binary Systems).pdf

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