Fundamentals of Computer 20CS11T.pdf

Vidya Vikas Educational Trust (R),
Vidya Vikas Polytechnic
27-128, Mysore - Bannur Road Alanahally,Alanahally Post, Mysuru, Karnataka 570028
Prepared by: Mr Thanmay J.S, H.O.D Mechanical Engineering VVETP, Mysore Page | 1
Department of Computer Science
Theory Notes
Subject : Fundamentals of Computers
Subject Code : 20CS11T
Semester : 1st
Semester
Name of the Student: …………………………………………….
Register Number: …………………………………………….

Course Details
Course Code 20CS11T Semester 01
Course Title Fundamentals of Computer Course Group Core
No. of Credit 4 Type of Course Lecture
Course Category PC Total contact Hours 52 Hours per Semester
Prerequisites Nil Teaching Scheme (L: P: T) = 4:0:0
CIE Marks 50 SEE Marks 50
Course Outcome:
CO-01: Apply the knowledge of number system and Boolean algebra in computer system
CO-02: Apply the knowledge of logic circuits for practical application
CO-03: Recognize the various hardware and software associated with computer
CO-04: Comprehend the functional units of a computer
CO-05: Represent simple problems in terms of algorithm and flowchart
Course Assessment:
Sl No Assessments Type Schedule Marks Conversion
1 CIE Assessment 1 Written Test -1 3rd
week 30 Average of 3
Test for 30
Marks
2 CIE Assessment 2 Written Test -2 7th
week 30
3 CIE Assessment 3 Written Test -3 13th
week 30
4 CIE Assessment 4 MCQ/Quiz 5th
week 20 Average of 3
Test for 20
Marks
5 CIE Assessment 5 Open book Test 9th
week 20
6 CIE Assessment 6 Student activity 11th
week 20
Total Internal Marks 50
Semester End Examination (SEE) [Written Exam] 100 50
Total 100
Students Activity:
1. Prepare a report on programming languages and their features
2. Prepare a report on open source and proprietary, system and application software
3. Prepare a report on recent viruses(computer)
4. Identify the logic circuits used in construction of memory and prepare a report
5. Identify the utilities of OS and prepare a report

Contents
Chapter 01: BASICS OF LOGIC DESIGN
1.1 Introduction to number system.
• Decimal
• Binary
• Octal
• Hexadecimal
(Characteristics of each number system)
1.2 Conversion from one number system to other
1.3 Complements of number systems and arithmetic operations
1.4 Computer codes (BCD, EBCDIC, ASCII Code, Gray code, Excess-3 code and Unicode)
1.5 Logic gates
1.6 Boolean algebra (rules, laws, De-Morgan Theorem, Boolean expressions and simplifications)
Chapter 02: LOGIC CIRCUITS
2.1 Combinational Circuits
• Characteristics
• Logic circuit design
• Block diagram, features &
• Applications of adders, subtractors and comparators
• Multiplexers, demultiplexers
• Encoders, decoders and code converters (7 segment)
2.2 Sequential Circuits
• Characteristics
• Types
i. Asynchronous
ii. Synchronous (clocked, unlocked)
• Flip flops
i. Types,
ii. circuit analysis
iii. truth table
• Applications of sequential circuits
i. Shift registers (types and application)
ii. Counters (classification and application)

Chapter 03: INTRODUCTION TO COMPUTER CONCEPTS
3.1 Introduction to computers
• Evolution of computer (abstract only)
• Generation of computers
• Classification of computer
• Applications
3.2 Components of computers
• Hardware (different types of hardware components)
• Software (System Software, Application Software, E-accessibility Software, Open source,
freeware and proprietary software)
• Peripherals (working of keyboard and laser printer)
3.3 Computer Network (Concept Only)
• Basics
• Categories
• Protocols (Application layer)
• Advantages.
3.4 Methods of data processing (concepts only)
• Single user programming
• Multi programming
• Real-time processing
• On-line processing
• Time sharing processing
• Distributed processing
3.5 Computer Security: Types of threats and source of threats
Chapter 04: INTRODUCTION TO COMPUTER ORGANIZATION & OPERATING SYSTEM
4.1 Introduction
• Overview of functional units of a computer
• Stored Program Concept
• Flynn's Classification of Computers
4.2 Memory Hierarchy
• Main memory
• Auxiliary memory
• Cache memory

4.3 Introduction to BIOS and UEFI
4.4 OS Concepts
• Overview
• Types (Batch Operating System, Multitasking/Time Sharing OS, Multiprocessing OS, Real Time
OS, Distributed OS, Network OS, Mobile OS)
• Services
Chapter 05: INTRODUCTION TO COMPUTER PROGRAMMING
5.1 Basics of programming
• Algorithms and Flowcharts
• Basics
• Decision making
• Iterative
(With sufficient examples)
5.2 Programming Languages
• Generation of languages
• General concepts of variables and constants

Chapter 01: BASICS OF LOGIC DESIGN
1.1 Introduction to number system.
Definition: A number system is defined as a system of writing to express numbers. It is the mathematical
notation for representing numbers of a given set by using digits or other symbols in a consistent manner. It
provides a unique representation of every number and represents the arithmetic and algebraic structure of
the figures. It also allows us to operate arithmetic operations like addition, subtraction, multiplication and
division.
The value of any digit in a number can be determined by:
• The digit
• Its position in the number
• The base of the number system
(𝑵𝒖𝒎𝒃𝒆𝒓 𝑫𝒊𝒈𝒊𝒕)(𝑺𝒚𝒔𝒕𝒆𝒎 𝒃𝒂𝒔𝒆 𝑵𝒖𝒎𝒃𝒆𝒓) = (𝑵)𝒃
Types of Number Systems
In the Number System, each number is represented by its base. If the base is 2 it is a binary number, if the
base is 8 it is an octal number, if the base is 10, then it is called decimal number system and if the base is
16, it is part of the hexadecimal number system. Based on the base value and the number of allowed digits,
number systems are of many types. The four common types of Number System are:
1) Decimal Number System (Base 10 Number System)
The decimal number system has a base of 10 because it uses ten digits from 0 to 9. In the decimal number
system, the positions successive to the left of the decimal point represent units, tens, hundreds, thousands and
so on. This system is expressed in decimal numbers. Every position shows a particular power of the base (10).
Example of Decimal Number System:
The decimal number 1457 consists of the digit 7 in the units position, 5 in the tens place, 4 in the hundreds
position, and 1 in the thousands place whose value can be written as:
(1×103
) + (4×102
) + (5×101
) + (7×100
)
(1×1000) + (4×100) + (5×10) + (7×1)
1000 + 400 + 50 + 7
1457

2) Binary Number System (Base 2 Number System)
The base 2 number system is also known as the Binary number system wherein, only two binary digits exist,
i.e., 0 and 1. Specifically, the usual base-2 is a radix of 2. The figures described under this system are known
as binary numbers which are the combination of 0 and 1. For example, 110101 is a binary number.
We can convert any system into binary and vice versa.
Example: Write (14)10 as a binary number.
2 14 Reminder
2 7 0
2 3 1
1 1 ∴ (14)10 = (1110)2
3) Octal Number System (Base 8 Number System)
In the octal number system, the base is 8 and it uses numbers from 0 to 7 to represent numbers. Octal numbers
are commonly used in computer applications. Converting an octal number to decimal is the same as decimal
conversion and is explained below using an example.
Example: Convert 2158 into decimal.
Method Verification (Converting to Octal)
2158 = 2 × 82
+ 1 × 81
+ 5 × 80
= 2 × 64 + 1 × 8 + 5 × 1
= 128 + 8 + 5
= (141)10
8 141 Reminder
8 17 5
(Start) 2 1
𝑨𝒏𝒔𝒘𝒆𝒓 = 215
4) Hexadecimal Number System (Base 16 Number System)
In the hexadecimal system, numbers are written or represented with base 16. In the hexadecimal system, the
numbers are first represented just like in the decimal system, i.e., from 0 to 9. Then, the numbers are
represented using the alphabet from A to F. The below-given table shows the representation of numbers in
the hexadecimal number system.
Hexadecimal 0 1 2 3 4 5 6 7 8 9 A B C D E F
Decimal 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
Characteristics of Number System:
• Each numeral system is characterized by its base and the base is never a digit.
• The numeral systems have a base or set of symbols that allow representing the different numeral quantities.
• They have a number or quantity that is formed by the juxtaposition of the different elements.
• Each element within the numeral system has a weighted value.

• The number 0 expresses or denotes the absence of a given quantity.
• It is a positional system.
• They are composed of digits.
Explanation:
The correct characteristic is that the 'The size of base is more than the number of digits’, that is
• In binary system, the digits can be 0 and 1 and the base is 2.
• In octal system, the digits are 0, 1, 2, 3, 4, 5, 6 and 7 and the base is 8.
• In decimal system, the digits are 0, 1, 2, 3, 4, 5, 6, 7, 8 and 9 and the base is 10.
1.2 Conversion from one number system to other
Decimal Conversion
a) Decimal to Binary: Example (456)10 = (𝐵𝑖𝑛𝑎𝑟𝑦)2
2 456 Remainder
2 228 (End) 0
2 114 0
2 57 0
2 28 1
2 14 0
2 7 0
2 3 1
(Start) 1 1
∴ (𝟒𝟓𝟔)𝟏𝟎 = (𝟏𝟏𝟏𝟎𝟎𝟏𝟎𝟎𝟎)𝟐
b) Decimal to Octal: Example (567)10 = (𝑂𝑐𝑡𝑎𝑙 𝑁𝑢𝑚𝑏𝑒𝑟)8
8 567 Remainder
8 70 (End) 7
8 8 6
(Start) 1 0
∴ (𝟓𝟔𝟕)𝟏𝟎 = (𝟏𝟎𝟔𝟕)𝟖
c) Decimal to Hexadecimal: Example (555)10 = (Hexadecimal)16
Before solving Hexadecimal, one should be familiar with Hexadecimal Conversion
Decimal 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
16 555 Remainder
16 34 11 = B
(Start) 2 2
∴ (𝟓𝟓𝟓)𝟏𝟎 = (𝟐𝟐𝐁)𝟏𝟔

Binary Conversion
a) Binary to Decimal: Example (1110011)2 = (Decimal)10
Place Value 7th
6th
5th
4th
3rd
2nd
1st
Binary Number 1 1 1 0 0 1 1
Power of 2 𝟐𝟔
𝟐𝟓
𝟐𝟒
𝟐𝟑
𝟐𝟐
𝟐𝟏
𝟐𝟎
𝟐𝟔
× 1 + 𝟐𝟓
× 1 + 𝟐𝟒
× 1 + 𝟐𝟑
× 0 + 𝟐𝟐
× 0 + 𝟐𝟏
× 1 + 𝟐𝟎
× 1
𝟔𝟒 × 1 + 𝟑𝟐 × 1 + 𝟏𝟔 × 1 + 𝟖 × 0 + 𝟒 × 0 + 𝟐 × 1 + 𝟏 × 1
𝟔𝟒 + 𝟑𝟐 + 𝟏𝟔 + 𝟐 + 𝟏 = 𝟏𝟏𝟓
∴ (𝟏𝟏𝟏𝟎𝟎𝟏𝟏)𝟐 = (𝟏𝟏𝟓)𝟏𝟎
b) Binary to Octal: Example (1110011)2 = (Decimal)10 = (Octal)8
Convert Binary to Decimal Conver Answer Decimal to Octal
𝟐𝟔
× 1 + 𝟐𝟓
× 1 + 𝟐𝟒
× 1 + 𝟐𝟑
× 0 + 𝟐𝟐
× 0
+ 𝟐𝟏
× 1 + 𝟐𝟎
× 1
𝟔𝟒 × 1 + 𝟑𝟐 × 1 + 𝟏𝟔 × 1 + 𝟖 × 0 + 𝟒 × 0
+ 𝟐 × 1 + 𝟏 × 1
𝟔𝟒 + 𝟑𝟐 + 𝟏𝟔 + 𝟐 + 𝟏 = 𝟏𝟏𝟓
∴ (𝟏𝟏𝟏𝟎𝟎𝟏𝟏)𝟐 = (𝟏𝟏𝟓)𝟏𝟎
8 115 Remainder
8 14 (End) 3
(Start) 1 6
∴ Answer for (𝟏𝟏𝟓)𝟏𝟎 = (𝟏𝟔𝟑)𝟖
𝐓𝐡𝐞 𝐅𝐢𝐧𝐚𝐥 𝐀𝐧𝐬𝐰𝐞𝐫 𝐢𝐬 (𝟏𝟏𝟏𝟎𝟎𝟏𝟏)𝟐 = (𝟏𝟏𝟓)𝟏𝟎 = (𝟏𝟔𝟑)𝟖
c) Binary to Hexadecimal: Example (1101010)2
Binary is the simplest kind of number system that uses only two digits of 0 and 1 (i.e., value of base 2).
Whereas Hexadecimal number is one of the number systems which has value is 16 and it has only 16 symbols
− 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 and A, B, C, D, E, F. Where A, B, C, D, E and F are single bit representations of
decimal value 10, 11, 12, 13, 14 and 15 respectively.
𝐹𝑖𝑟𝑠𝑡 𝑐𝑜𝑛𝑣𝑒𝑟𝑡 𝑡ℎ𝑖𝑠 𝑖𝑛𝑡𝑜 𝑑𝑒𝑐𝑖𝑚𝑎𝑙 𝑛𝑢𝑚𝑏𝑒𝑟: (1101010)2
= 1 × 26
+ 1 × 25
+ 0 × 24
+ 1 × 23
+ 0 × 22
+ 1 × 21
+ 0 × 20
= 64 + 32 + 0 + 8 + 0 + 2 + 0
= (106)10
Method 1 Verification
𝑇ℎ𝑒𝑛, 𝑐𝑜𝑛𝑣𝑒𝑟𝑡 𝑖𝑡 𝑖𝑛𝑡𝑜 ℎ𝑒𝑥𝑎𝑑𝑒𝑐𝑖𝑚𝑎𝑙 𝑛𝑢𝑚𝑏𝑒𝑟
= (106)10
= 6 × 161
+ 10 × 160
.
16 106 Remainder
16 (Start) 6 10 = A
(𝟔𝐀) 𝟏𝟔 𝐢𝐬 𝐭𝐡𝐞 𝐚𝐧𝐬𝐰𝐞𝐫

Octal Conversion
a) Octal to Decimal: Example (564)8 = (… . )10
Method 1 Verification
= 5 × 82
+ 6 × 81
+ 4 × 80
= 5 × 64 + 6 × 8 + 4 × 1 = (372)10
8 372 Remainder
8 46 4
8 (Start) 5 6
b) Octal to Binary: Example (564)8 = (… . )2
Octal to Decimal Conversion to Binary
= 5 × 82
+ 6 × 81
+ 4 × 80
= 5 × 64 + 6 × 8 + 4 × 1 = (372)10
2 372 Remainder
2 186 0
2 93 0
2 46 1
2 23 0
2 11 1
2 5 1
2 2 1
(Start) 1 0
𝐴𝑛𝑠𝑤𝑒𝑟 𝑓𝑜𝑟 (𝟓𝟔𝟒)𝟖𝐢𝐬 (𝟏𝟎𝟏𝟏𝟏𝟎𝟏𝟎𝟎)𝟐
c) Octal to Hexadecimal: Example (564)8 = (…. )16
Hexadecimal number is one of the number systems which has value is 16 and it has only 16 symbols − 0, 1,
2, 3, 4, 5, 6, 7, 8, 9 and A, B, C, D, E, F. Where A, B, C, D, E and F are single bit representations of decimal
value 10, 11, 12, 13, 14 and 15 respectively
Octal to Decimal Conversion to Hexadecimal
= 5 × 82
+ 6 × 81
+ 4 × 80
= 5 × 64 + 6 × 8 + 4 × 1 = (372)10
16 372 Remainder
16 23 4
(Start) 1 7
Hexadecimal Conversion
Decimal 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
a) Hexadecimal to Decimal: Example (1𝐷𝐴6)16 = (…. )10
Here,
1 = 1
D = 13
A = 10
6 = 6
Thus (1DA6)16 = (1 × 163) + (13 × 162) + (10 × 161) + (6 × 160)
(1DA6)16 = (1 × 4096) + (13 × 256) + (10 × 16) + (6 × 1)
= 4096 + 3328 + 160 + 6 = 7590
𝑇ℎ𝑒𝑟𝑒𝑓𝑜𝑟𝑒, (1𝐷𝐴6)16 = (7590)10

b) Hexadecimal to Binary: Example (𝐴6)16 = (…. )2
Hexadecimal to Decimal (𝐴6)16 = Conversion to Binary
(10 × 161
) + (6 × 160
)
(10 × 16) + (6 × 1)
= 160 + 6
= 166
𝑇ℎ𝑒𝑟𝑒𝑓𝑜𝑟𝑒, (𝐴6)16 = (166)10
2 166 Remainder
2 83 0
2 41 1
2 20 1
2 10 0
2 5 0
2 2 1
2 (Start) 1 0
𝑇ℎ𝑒𝑟𝑒𝑓𝑜𝑟𝑒, (𝐴6)16 = (10100110)2
c) Hexadecimal to Octal: Example (𝐴6)16 = (…. )2
Hexadecimal to Decimal (𝐴6)16 = Conversion to Octal
(10 × 161
) + (6 × 160
)
(10 × 16) + (6 × 1)
= 160 + 6
= 166
𝑇ℎ𝑒𝑟𝑒𝑓𝑜𝑟𝑒, (𝐴6)16 = (166)10
8 166 Remainder
8 20 6
(Start) 2 4
𝑇ℎ𝑒𝑟𝑒𝑓𝑜𝑟𝑒, (𝐴6)16 = (246)8
1.3 Complements of number systems and arithmetic operations
Complements are used in digital computers to simplify the subtraction operation and for logical manipulation.
There are two types of complements for each base‐r system: the radix complements and the diminished radix
complement. The first is referred to as the r’s complement and the second as the (r - 1)’s complement.
1.3.1: Binary numbers Complement:
1- One's (first) Complement:
1's complement= 𝑟𝑛
– 𝑁 − 1
where (𝑛): number of bits
N: binary number
r: system base
Simply the 1’s complement of binary number is the number we get by changing each bit
(0 to 1) and (1 to 0).
Example: the first complement of (101100)2
Solution:
binary number 101100
1’s complement 010011

2- The Two's (second) Complement:
2's complement = 𝑟𝑛
– 𝑁
Simply the 2's complement is equal to 1's complement added by one.
Example: find the 2's complement of (101101)2
Solution:
binary number 101101
1’s complement 010010
2’s complement 010010 + 1 = 010011
Binary Arithmetic Operations
a. Binary additions and subtractions are performed as same in decimal additions and subtractions. When
we perform binary additions, there will have two outputs: Sum (S) and Carry (C). There are four rules for
binary addition. These are given as following below,
Input A Input B Sum (S) =A+B Carry (C)
0 0 0 0
0 1 1 0
1 0 1 0
1 1 0 1
b. Binary subtraction. These are given as following below,
Input A Input B Subtract (S)= A-B Borrow (B)
0 0 0 0
0 1 0 1
1 0 1 0
1 1 0 0
Borrow 1 is required from next higher order bit to subtract 1 from 0. So, result became 0.
c. Binary multiplication. These are given as following below,
Input A Input B Multiply (M)
AxB
0 0 0
0 1 0
1 0 0
1 1 1
Whenever at least one input is 0, then multiplication is always 0.

d. Binary division: Dividend, Divisor, quotient, and remainder. These are given as following rules for
binary division,
Input A Input B Divide (D)
A/B
0 0 Not defined
0 1 0
1 0 Not defined
1 1 1
Whenever divisor is 0, then result is always not defined.
Examples for Asthmatic Operations
Example: Add the two binary numbers (001) and (100)
0 0 1
+ 1 0 0
1 0 1
Example: Add the two binary numbers (111) and (001)
Carry 1 1
1 1 1
+ 0 0 1
1 0 0 0
Example: subtract the binary number (100) from (101)
1 0 1
- 1 0 0
0 0 1
Example: subtract the binary number (1101) from (1110)
1 1 1 0
- 1 1 0 1
0 0 0 1
Example: Multiply the two binary numbers (111) 2𝑎𝑛𝑑 (101)2.
Example: divide the number (11011) on (101)

1.4 Computer codes (BCD, EBCDIC, ASCII Code, Gray code, Excess-3 code and Unicode)
Computer code refers to a set of instructions that a computer can understand and execute. It can be written in
a variety of programming languages, such as C++, Python, Java, and many others. These instructions tell the
computer what and how to do it, and they can be used to create software, websites, and other digital products.
Computers and digital circuits processes information in the binary format. Each character is assigned 7 or 8
bit binary code to indicate its character which may be numeric, alphabet or special symbol. Example - Binary
number 1000001 represents 65(decimal) in straight binary code, alphabet A in ASCII code and 41(decimal)
in BCD code.
Types of codes
a) BCD (Binary-Coded Decimal) code:
• Four-bit code that represents one of the ten decimal digits from 0 to 9.
• Example - (37)10 is represented as 0011 0111 using BCD code, rather than (100101)2 in straight binary
code.
• Thus, BCD code requires more bits than straight binary code.
• Still, it is suitable for input and output operations in digital systems.
Note: 1010, 1011, 1100, 1101, 1110, and 1111 are INVALID CODE in BCD code.
ASCII (American Standard Code Information Interchange) code:
• It is 7-bit or 8-bit alphanumeric code.
• 7-bit code is standard ASCII supports 127 characters.
• Standard ASCII series starts from 00h to 7Fh, where 00h-1Fh are used as control characters and 20h-
7Fh as graphics symbols.
• 8-bit code is extended ASCII supports 256 symbols where special graphics and math's symbols are
added.
• Extended ASCII series starts from 80h to FFh.
EBCDIC (Extended Binary Coded Decimal Interchange Code) code
• 8-bit alphanumeric code developed by IBM, supports 256 symbols.
• It was mainly used in IBM mainframe computers.
Gray code
• Differs from leading and following number by a single bit.
• Gray code for 2 is 0011 and for 3 is 0010.
• No weights are assigned to the bit positions.
• Extensively used in shaft encoders.
Excess-3 code
• 4-bit code is obtained by adding binary 0011 to the natural BCD code of the digit.
• Example - decimal 2 is coded as 0010 + 0011 = 0101 as Excess-3 code.
• It not weighted code.
• Its self-complimenting code, means 1's complement of the coded number yields 9's complement of the
number itself.
• Used in digital system for performing subtraction operations.

Unicode:
A standard for representing characters as integers. Unlike ASCII, which uses 7 or 8 bits for each character,
Unicode uses 16 bits, which means that it can represent more than 65,000 unique characters. This is a bit of
overkill for English and Western-European languages, but it is necessary for some other languages, such as
Greek, Chinese and Japanese. Many analysts believe that as the software industry becomes increasingly
global, Unicode will eventually supplant ASCII as the standard character coding format.

1.5 Logic gates
Computers and digital component use binary 0
and 1, where 0 is low voltage (0 volts) and 1 is
high voltage (+5 volts). Binary information is
carried by signals and manipulation of binary
information is done by logic circuits called as
gates. A circuit whose input and output signals
are two states on and off. A gate is logic circuit
with one or more input signals but only one output signal.
Basic types of Gates:
AND Gate
• Two or more input signals and one output signal.
• Output is high when both the inputs are high.
• Logic equation Y = A X B called as Boolean
equation.
• Where A and B are the inputs and Y is the output for
all standard symbols of gates shown below.
OR Gate
• Two or more input signals and one output signal.
• Output is low when both the inputs are low.
• Logic equation Y = A + B.
NOT Gate
• One input signal and one output signal, also called as
inverter.
• Output is always opposite state of the input.
• Logic equation Y = A
• Where is A is the complement of A.
NAND (Not AND) Gate
• Two or more input signals and one out signal.
• It has high output when at least one of the inputs is
zero or low.
• All input signals must be high to obtain low output.
• Logic equation Y = A X B
NOR (Not OR) Gate
• One or more input signals and one output signal.
• If one of the input is high then output is low
• Logic equation Y = A + B
XOR (Exclusive OR) Gate
• Two or more input signal and one output signal.
• Output is low when both the inputs are same.
• Logic equation Y = A XOR B
XNOR (Exclusive NOR) Gate
• Two or more input signal and one output signal.
• Is combination of XOR gate followed by invertor.
• Output is high when both the inputs are same and Logic equation is Y = A XOR B

1.6 Boolean algebra (rules, laws, De-Morgan Theorem, Boolean expressions and simplifications)
Boolean algebra is a branch of algebra that only uses ‘false’ and ‘true’ values for variables and is usually
denoted by 0 and 1. It is defined as a system of logic where variables are represented as whole numbers
between 0 and 1. The values that you take on the numbers are true or false, but not both at the same time. A
variable is either true or false, but never both true and false simultaneously.
The variables of Boolean algebra can take only one of two possible values, zero and one. Still, like any other
mathematical expression, Boolean expressions too can have an infinite number of variables, all of which
represent different individual inputs to the expression.
A complete understanding of the laws and theorems must be grasped to use Boolean algebra properly.
Boolean expressions can also be converted using logic gates like OR gate, AND gate, NOT gate, NOR gates,
XOR gates, XNOR gates, NAND gates, etc.
The three basic Boolean operations are:
1. OR gate returns ‘true’ or ‘1’ if either of the input variables is true.
A B A OR B = A + B
0 (False) 0 (False) 0 (False)
0 (False) 1 (True) 1 (True)
1 (True) 0 (False) 1 (True)
1 (True) 1 (True) 1 (True)
2. AND gate returns ‘true’ or ‘1’ only if all the input variables are true.
A B A AND B = A • B
0 (False) 0 (False) 0 (False)
0 (False) 1 (True) 0 (False)
1 (True) 0 (False) 0 (False)
1 (True) 1 (True) 1 (True)
3. NOT gate returns the complement value of the input variable.
A Ā
0 (False) 1 (True)
1 (True) 0 (False)
Basic Laws of Boolean Algebra
In all the cases given below A can either be 0 or 1
1. A × 0 = 0
2. A × 1 = A
3. A × A = A
4. A × A = 0
5. A + 0 = A
6. A + 1 = 1
7. A + A = 1
8. A + A = A

Some basic Boolean algebra laws that are used to simplify Boolean expressions are:
1. Idempotent Law
𝐴 × 𝐴 = 𝐴
𝐴 + 𝐴 = 𝐴
2. Associative Law
(𝐴 × 𝐵) × 𝐶 = 𝐴 × (𝐵 × 𝐶)
(𝐴 + 𝐵) + 𝐶 = 𝐴 + (𝐵 + 𝐶)
3. Commutative Law
𝐴 × 𝐵 = 𝐵 × 𝐴
𝐴 + 𝐵 = 𝐵 + 𝐴
4. Distributive Law
𝐴 × (𝐵 + 𝐶) = 𝐴 × 𝐵 + 𝐴 × 𝐶
𝐴 + (𝐵 × 𝐶) = (𝐴 + 𝐵) × (𝐴 + 𝐶)
5. Identity Law
𝐴 × 0 = 0 𝐴 × 1 = 𝐴
𝐴 + 1 = 1 𝐴 + 0 = 𝐴
6. Complement Law
A × A = 0
A + A = 1
7. Involution Law
(A )
̅̅̅̅̅̅ = 𝐴
8. DeMorgan’s Law
(𝐴 × 𝐵)
̅̅̅̅̅̅̅̅̅̅̅ = A + 𝐵
(𝐴 + 𝐵)
̅̅̅̅̅̅̅̅̅̅̅̅ = A × 𝐵
9. Absorption
𝐴 + (𝐴 × 𝐵) = 𝐴
𝐴 × (𝐴 + 𝐵) = 𝐴
(𝐴 × 𝐵) + (𝐴 × 𝐵) = 𝐴
(𝐴 + 𝐵) × (𝐴 + 𝐵) = 𝐴
𝐴 + (A × 𝐵) = 𝐴 + 𝐵
𝐴 × (A + 𝐵) = 𝐴 × 𝐵

Previous Years Question Papers with solutions
April / May 2021
1 a Convert the following: 6
I. Binary number 11011 to decimal number.
Place Value 5th
4th
3rd
2nd
1st
Binary Number 1 1 0 1 1
Power of 2 𝟐𝟒
𝟐𝟑
𝟐𝟐
𝟐𝟏
𝟐𝟎
(𝟐𝟒
× 1) + (𝟐𝟑
× 1) + (𝟐𝟐
× 0) + (𝟐𝟏
× 1) + (𝟐𝟎
× 1)
(𝟏𝟔 × 1) + (𝟖 × 𝟏) + (𝟒 × 0) + (𝟐 × 1) + (𝟏 × 1)
𝟏𝟔 + 𝟖 + 𝟎 + 𝟐 + 𝟏 = 𝟐𝟕
∴ (𝟏𝟏𝟏𝟎𝟎𝟏𝟏)𝟐 = (𝟐𝟕)𝟏𝟎
II. Decimal number 497 to Octal number.
8 497 Remainder
8 62 (End) 1
(Start) 7 6
∴ (𝟒𝟗𝟕)𝟏𝟎 = (𝟕𝟔𝟏)𝟖
b Express the decimal number (-49) in 8-bit binary form of 6
I. Sign-Magnitude form
II. 1st
complement form
III. 2nd
complement form
Start with (−49 ) = 45
2 49 Remainder
2 24 1
2 12 0
2 6 0
2 3 0
2 (Start) 1 1
∴ (𝟒𝟗)𝟏𝟎 = (𝟏𝟏𝟎 𝟎𝟎𝟏)𝟐 → 6 𝑏𝑖𝑡
6 𝑏𝑖𝑡 𝑡𝑜 8 𝑏𝑖𝑡 𝐶𝑜𝑛𝑣𝑒𝑟𝑠𝑖𝑜𝑛
∴ (𝟒𝟗)𝟏𝟎 = (𝟎𝟎𝟏𝟏 𝟎𝟎𝟎𝟏)𝟐 → 8 𝑏𝑖𝑡
To get the Negative integer number representation of (-49) replace all the bits
on 0 with 1’s and 1’s with 0
(𝟒𝟗)𝟏𝟎 = (𝟎𝟎𝟏𝟏 𝟎𝟎𝟎𝟏)𝟐 → (𝟏𝟏𝟎𝟎 𝟏𝟏𝟏𝟎)𝟐 = (−𝟒𝟗)𝟏𝟎
∴ (−𝟒𝟗)𝟏𝟎 = (𝟏𝟏𝟎𝟎 𝟏𝟏𝟏𝟎)𝟐
Adding 1 to (𝟏𝟏𝟎𝟎 𝟏𝟏𝟏𝟎)𝟐 gives us signed binary two’s component
(−𝟒𝟗)𝟏𝟎 = (𝟏𝟏𝟎𝟎 𝟏𝟏𝟏𝟎)𝟐 + (𝟏)𝟐 = (𝟏𝟏𝟎𝟎 𝟏𝟏𝟏𝟏)𝟐

c Write binary equivalent ASCII code for the words 8
CART
blue
Hint: ASCII code of A' is 65 in decimal, 'a' is 97 in decimal.
A B C D E F G H I J K L M N O P Q R
65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82
S T U V W X Y Z
83 84 85 86 87 88 89 90
a b c d e f g h i j k l m n
97 98 99 100 101 102 103 104 105 106 107 108 109 110
o p q r s t u v w x y z
111 112 113 114 115 116 117 118 119 120 121 122
𝐶𝐴𝑅𝑇 = 67 65 82 84 = ( 01000011 01000001 01010010 01010100)2
𝐵𝑙𝑢𝑒 = 98 108 117 101 = (01100010 01101100 01110101 01100101)2
2 a Explain universal gates with logic symbol, expressions, truth table. 8
Universal Gates: A universal gate is a gate which can implement any Boolean function without need to
use any other gate type. The NAND and NOR gates are universal gates.
In practice, this is advantageous since NAND and NOR gates are economical and easier to fabricate
and are the basic gates used in all IC digital logic families. In fact, an AND gate is typically implemented
as a NAND gate followed by an inverter not the other way around. Likewise, an OR gate is typically
implemented as a NOR gate followed by an inverter not the other way around.
NAND Gate: The NAND gate represents the complement of the AND operation. Its name is an
abbreviation of NOT AND. The graphic symbol for the NAND gate consists of an AND symbol with
a bubble on the output, denoting that a complement operation is performed on the output of the AND
gate. The truth table and the graphic symbol of NAND gate is shown in the figure.
The truth table clearly shows that the NAND operation is the complement of the AND.
NOR Gate: The NOR gate represents the complement of the OR operation. Its name is an abbreviation
of NOT OR. The graphic symbol for the NOR gate consists of an OR symbol with a bubble on the
output, denoting that a complement operation is performed on the output of the OR gate.
The truth table and the graphic symbol of NOR gate is shown in the figure.

b Develop a truth table for 3 input AND gate. 8
c Determine when the output of XOR and XNOR logic gates are high 4
The XNOR gate is the complement of the XOR gate. It is a hybrid gate. Simply, it is the combination
of the XOR gate and NOT gate. The output level of the XNOR gate is high only when both of its inputs
are the same, either 0 or 1. The symbol of the XNOR gate is the same as XOR, only complement sign
is added. Sometimes, the XNOR gate is also called the Equivalence gate.
XOR, which stands for Exclusive OR, and XNOR, which stands for Exclusive NOR. In an XOR gate,
the output is HIGH if one, and only one, of the inputs is HIGH. If both inputs are LOW or both are
LOW, the output is LOW.
Input A Input B Output
0 0 0
0 1 1
1 0 1
1 1 0
Another way to explain an XOR gate is as follows: The output is HIGH if the inputs are different; if the
inputs are the same, the output is LOW. The XOR gate has a lesser-known cousin called the XNOR
gate. An XNOR gate is an XOR gate whose output is inverted.
Input A Input B Output
0 0 1
0 1 0
1 0 0
1 1 1
As you can see, the only difference between these two symbols is that the XNOR has a circle on its
output to indicate that the output is inverted.

Previous Years Question Papers with solutions
Oct / Nov 2021
1 a Define number system. Explain the characteristics of number system.
Answer Page Number 6 to 8
b Convert the following:
i. Binary to Decimal: (𝟏𝟏𝟏𝟎𝟏)𝟐
Place Value 5th
4th
3rd
2nd
1st
Binary Number 1 1 1 0 1
Power of 2 24
23
22
21
20
(24
× 1) + (23
× 1) + (22
× 1) + (21
× 0) + (20
× 1)
(16 × 1) + (8 × 1) + (4 × 1) + (2 × 0) + (1 × 1)
16 + 8 + 4 + 0 + 1 = 29
∴ (111001)2 = (29)10
ii. Decimal to Binary: 45610
2 456 Remainder
2 228 (End) 0
2 114 0
2 57 0
2 28 1
2 14 0
2 7 0
2 3 1
(Start) 1 1
Answer
∴ (456)10 = (111001000)2
iii. Binary to Octal: (𝟏𝟎𝟎𝟎𝟏)𝟐
Convert Binary to Decimal Conver Answer Decimal to Octal
24
× 1 + 23
× 0 + 22
× 0 + 21
× 0 + 20
× 1
16 × 1 + 8 × 0 + 4 × 0 + 2 × 0 + 1 × 1
16 + 0 + 0 + 0 + 1 = 17
∴ (10001)2 = (17)10
8 17 Remainder
8 (Start) 2 (End) 1
∴ Answer for (17)10 = (21)8
The Final Answer is (1110011)2 = (17)10 = (21)8
iv. Decimal to Octal: (𝟓𝟔𝟕)𝟏𝟎
8 567 Remainder
8 70 (End) 7
8 8 6
(Start) 1 0
(567)10 = (1067)8
v. Hexadecimal to Decimal (𝟕𝑨𝑪)𝟏𝟔
Decimal 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
(7𝐴𝐶)16 = (7 × 162
) + (10 × 161
) + (12 × 160
)
(7𝐴𝐶)16 = 1792 + 160 + 12 = (1964)10

2 a Explain the following terms
(i) BCD, (ii) EBCDIC, (i) ASCII, (iv) UNICODE, (v) GRAY CODE
10
Answer Page Number 14 and 15 except Table chart
b Draw a circuit to realize the expressions using AND gates, OR gates and
Invertors and write truth table
𝑌 = (𝐴. 𝐵) + (𝐴. 𝐵)
There are three basic logic gates. The first and the basic one is NOT a gate. The
output due to this logic gate is nothing but the complement of the input signal. Let
us say our input signal is Y than the output signal X is given by 𝑋 = 𝑌
The next gate we are going to discuss is the OR gate. This gate has basically two
inputs and one output. The operation performed on the two input signals is such
that the output due to the gate is equal to the sum of the two input signals. Let us
say the two inputs are A and B. Then the output C is given by, 𝐶 = 𝐴 + 𝐵
The last basic gate is the AND gate. This gate also has two inputs and one output. The
operation performed on the two input signals is such that the output due to the gate is
equal to the product of the two input signals. Let us say the two inputs are A and B.
Then the output C is given by, 𝐶 = 𝐴 ⋅ 𝐵
The logic gate has basically its inputs equal to either one or zero. Let us say we are constructing a logic gate
for the above expression i.e., 𝑌 = (𝐴. 𝐵) + (𝐴. 𝐵). Let us say A and B are the two inputs and Y is the
output. Let us first determine the truth table of the above Boolean equation such that we take different
possibilities of inputs A and B
A 𝐴 B 𝐵 𝐴. 𝐵 𝐴. 𝐵 (𝐴. 𝐵) + (𝐴. 𝐵) (𝐴. 𝐵) + (𝐴. 𝐵)
0 1 0 1 1 X 0 = 0 0 X 1 = 0 0 + 0 = 0 1
0 1 1 0 1 X 1 = 1 0 X 0 = 0 1 + 0 = 1 0
1 0 0 1 0 X 0 = 0 1 X 1 = 1 0 + 1 = 1 0
1 0 1 0 0 X 1 = 0 1 X 0 = 0 0 + 0 = 0 1
Now from the symbols we know let us construct the logic gate having the input A and B and the output Y.

Assignments
April / May 2022
June / July 2023

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