COMPUTER

FUNDAMENTALS
UMME HASUNAT TOAFIA & ANKON GOPAL BANIK

According to the syllabus of first semester of department of CSE Gono Bishwabidyalay

COMPUTER FUNDAMENTALS

BOOK REFERENCE- 01. COMPUTER FUNDAMENTALS- M. LUTFAR RAHMAN & M.

ALAMGIR HOSSAIN.

02. COMPUTER FUNDAMENTALS AND PROGRAMMING IN C-

ANITA GOLE & AJAY MITTAL.

IT WAS IMPOSSIBLE TO COMPLITE THIS WORK WITHOUT THE HELP OF OUR

RESPECTIVE TEACHER, HASNAT PARVEZ.

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COMPUTER FUNDAMENTALS

Chapter-1: Introduction to computer.

Lesson -1: Basic Organization.
Question-1: What is data processing circle?

Answer-1: Different programs are used to solve different programs. The first step of solving a
problem by a computer is to develop a computer program and then store it in the memory. The
computer then executes the instructions in the program. The instruction generally direct the
computer to perform three basic functions over and over again; these are input processing and
output. Collectively these functions constitute the data processing cycle.

Keyboard System Monitor

(Input Unit) Unit (Output Unit)

Fig (A)

Input Processing Output Unit

Fig (B)

Fig (A): Solving a problem with a computer.

Fig (B): Equivalent block diagram.

Input: Input devices feed the facts on data to be processed.

Processing: The control and storing of data, numerical comparisons and arithmetic operations
and performed on the input data to produce the results.

Output: The computer feeds the data processed the data or information to the output devices.

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Question-2: Discuss about components of a computer system.

Answer-2: A computer is used to process data and a data processing system consists of more
than just machine. A computer system must contain:

 Hardware.
 Software.
 Human ware.
 Operational producers.

Hardware: Hardware refers to machine or physical equipment that perform the basic functions
of the data processing cycles.

Software: Programs are referred to as software.

Human ware: Human refers to the person who design program and operate a computer.

Operational producers: These functions generally including obtaining, preparing and entering
data into the computer, processing job, initiating new programs and changing or deleting old
one etc.

Question-3: What is Software?

Answer-3: A program is sequence of instructions, which directs a computer to perform certain

functions. A computer must have access to stored programs and input devices for data, it must
make decision, arithmetically manipulate data, and output results in the correct sequence.
Programs are referred to as software.

Question-4: Define system and application software?

Answer-4: Software is generally categorized as:

 System Software.
 Application Software.

System Software: System consists of program that helps the use of computer. These program
such standard task as organizing and maintaining data files, translating program writing in
program languages to a language acceptable to the computer, scheduling jobs through the
computer, as well as aiding in other areas of general operations of all, this system software is
the most important one is known as operating system.

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COMPUTER FUNDAMENTALS

Application Software: Application software includes programs to perform user applications, a

computer program for preparing result of a public examinations is an example of application
software. Application programs are either purchased or written by the computer users for
specific applications.

Question-5: What are the importance and limitations of computer?

Answer-5: The following are some important criteria, which may be used to justify to use of
computers

Volume of data: Computer is suited to handing large volume of data.

Accuracy: Computer ensure high degree of accuracy and their consistency is reliable.

Repetitiveness: Processing cycles that repeat time again and again are ideally suited to
computers. Once a computer is programed, it goes on automatically performing the as many
time as required.

Complexity: Computers perform complex calculations. A computer provides the required

answers after running the program for complex calculations.

Speed: Computers work at very high speeds and this enables them to respond quickly to a given
situation.

Common data: One item of data can be used for different computer producers.

Besides the advantages of using a computer, there are few limitations of it. These are:

 Computers cannot think.

 Computers cannot do anything without human instructions.
 Computers cannot make any adjustment as human being can do.

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Lesson -2: Types of Computer.

Question-1: Explain about microcomputer?

Answer-1: Microcomputers (also known as personal computers-PCs) are microprocessor based

small laptop or desktop or notebook systems with varying capability. Personal digital assistants
(PDAs) are very small portable computers. PDAs are also known as palmtop computers. The
brain of microcomputer is microprocessor; it is a silicon chip containing necessary circuits to
perform arithmetic/logic operations and to control input/output operations.

A microcomputer system is formed by adding input/output capability and memory to the

microprocessor.

Question-2: Explain about minicomputer?

Answer-2: A minicomputer system performs the basic arithmetic and logic functions and
supports some of the programming languages used with large computer systems. They are
physically smaller, less expensive and small storage capacity compared to mainframe.
Minicomputers are ideally suited for processing task that do not required access to huge volume
of stored data. Use of minicomputers are gradually being demised with the rapid development
of microcomputers.

Question-3: Explain about mainframe computer?

Answer-3: A larger computer generally consists of modules mounted on a chassis (or main
frame) and is known as a mainframe main frame computer. They vary in size frame those
slightly large than a minicomputer to supercomputers. Mainframe computer systems offer
substantial advantages over minicomputers or microcomputers. Some of this are: greater
processing speed, greater storage capacity, a large verity of input and output devices and
support for a number of high speed storage devices, multiprogramming and time sharing.

Question-4: Explain about super computer?

Answer-4: A very large and powerful mainframe computer is called super computer. The
astronomical cost of super computers has limited their developments to only a few hundred

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worldwide. The Cary X-MP is an example of a super computer. Such super computers are applied
to the solutions of very complex and sophisticated scientific problems and for national security
purpose of some advanced nations.

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Chapter-2: Number system and codes.

Lesson -1: Number System.
Question-1: What are the base of binary, decimal, octal and hexadecimal numbers?

Question-2: What down digits for binary, decimal, octal and hexadecimal numbers?

NUMBER SYSTEM

Name of Numbers/Digits Base Example

Number System
Binary 0, 1 2 (1011)2
Octal 0, 1, 2, 3, 4, 5, 6, 7 8 (127)8
Decimal 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 10 (139)10
Hexadecimal 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, A, B, C, D, E, F 16 (3B5)16

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Lesson -2: Conversion of Numbers

Decimal to Binary:

(99.375)10 = (?)2
.375
2 99 × 2
2 49 - 1 .750
2 24 - 1 0 × 2
2 12 – 0 .50
1
2 6-0 × 2
3 1 00
2 1 -- 0
1
2 0-1

(99.375)10 = (1100011.011)2

Decimal to Octal:

(469.046875)10 = (?)8

8 469
8 58 - 5 .046875
8 7-2 × 8
8 0–7 0 .375
× 8
(469.046875)10 = (725.03)8 3 00

Decimal to Hexadecimal:

(850.850)10 = (?)16 .850

× 16
16 850
D(13) .60
16 53 - 2 × 16
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16 3-5
09 .60
16 0–3 × 16
9 .60
× 16
9
(850.850)10 = (352.D999)16 .60
Binary to Decimal:

(11011.101)2 = (?)10

(11011)2 = 1×24+1×23+0×22+1×21+1×20=27 = (27)10

(0.101)2 = 1×2-1+0×2-2+1×2-3 = ½+0× 1/8 = .5+.125 = (.625)10

(11011.101)2 = (25.625)10

Octal to Decimal:

(123.540)8 = (?)10

(123.540)8 = 1×82+2×82+3×80+5×8-1+4×8-2+0×8-3

=64+16+3+5/8+4/64+0

= 83+0.625+0.0625

= (83.6875)10

(123.540)8 = (83.6875)10

Hexadecimal to Decimal:

(B5D.48)16 = (?)10

(B5D.48)16 = B×162+5×161+D×160+4×16-1+8×16-2

=11×256+80+13+.25+.03125

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= (2909.28125)10

(B5D.48)16 = (2909.28125)10

Octal to Binary:

0 1 2 3 4 5 6 7
000 001 010 011 100 101 110 111

Example: (127.73)8 = (?)2

(127.73)8 = 001 010 111 . 111 011

= (1010111.111011)2

(127.73)8 = (1010111.111011)2

Binary to Octal:

(1111011.1111011)2 = (?)8

(1111011.1111011)2 = 1111011.1111011

= 001 111 011 . 111 101 100

= (173.758)8

(1111011.1111011)2 = (173.758)8

Hexadecimal to Binary:

0 0000
1 0001
2 0010
3 0011
4 0100
5 0101

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6 0110
7 0111
8 1000
9 1001
A 1010
B 1011
C 1100
D 1101
E 1110
F 1111

Example: (12A.5)16 = (?)2

(12A.5)16 = 0001 0010 1010 . 0101

= (100101010.0101)2

(12A.5)16 = (100101010.0101)2

Binary to Hexadecimal:

(1111011.1111011)2 = (?)16

(1111011.1111011)2 = 111 1011 . 1111 011

= 0111 1011 . 1111 0110

= (7B.F6)16

(1111011.1111011)2 = (7B.F6)16

Hexadecimal to Octal:

(12A) 16 = (?)8

(12A) 16 = 0001 0010 1010

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= (100 101 010)2

= (452)8

(12A) 16 = (458)8

Octal to Hexadecimal:

(127)8 = (?)16

(127)8 = 001 010 111

= (0101 0111)2

(127)8 = (57)16

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Lesson -3: Binary Arithmetic

Binary addition:

0+0 = 0

0+1 = 1

1+0 = 1

1+1 = 0 Carry 1

1+1+1 = 1 Carry 1

Example: (11011.001) 2 + (1011.11) 2

11011.001
+ 1011.110
100110.111

The addition result is = 100110.111

Binary Subtraction:

0-0 = 0

1-0 = 1

1-1 = 0

0-1 = 11 Carry 1

Example: (11011.001) 2 - (1011.11) 2

11011.001
- 1011.110
01111.011

The subtraction result is = 1111.011

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Binary Multiplication:

0×0 = 0

1×0 = 0

0×1 = 0

1×1 = 1

Example: (101011) 2 × (1011) 2

101011
× 1011
101011
1010110
00000000
101011000
111011001

The multiplication result is = 111011001

Binary Division:

0÷0 = Undefined

1÷0 = Undefined

0÷1 = 0

1÷1 = 1

Example 1:

(101011) 2 ÷ (11) 2

11) 101011 (11011

11
100
11
101
11
101

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11
10

The division result is = 11011

Example 2:

(111001) 2 ÷ (10) 2

10) 111001 (110

10
11
10
10
10
01
The division result is = 110

Complements:

111100001111
000011110000 1’s Complement
+ 1
000011110001 2’s Complement

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Chapter-3: Digital Circuits

Lession-1: Logic function and logic gates
Question-1: Draw symbol and truth table of primary logic gates.

Answer-1: The common use of logic elements is to act as switches. Although they have no
moving part, they open to pass on a pulse of electricity of close to shut it off. This is the way they
are known as gates. The primary gates are OR, AND and NOT

OR GATE- An OR gate has an output 1 if any of its inputs are 1. The symbol and truth table for a
two-input OR gate are shown in Figure. Here output, Y = A+B, Where + denotes OR operation.

Truth Table

A B C
0 0 0
0 1 1
1 0 1
1 1 1
(a)

A Y=A+B

(b)

Figure: Two-input OR gate.(a) Truth Table, (b) symbol.

AND GATE- An AND gate has an output 1 if all of its inputs are 1. The symbol and truth table for
a two-input AND gate are shown in Figure. Here output, Y=A.B , Where denotes AND operation.

Truth Table

A B C
0 0 0
0 1 0
1 0 0
1 1 1
(a)

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A Y= A.B

(b)

Figure: Two-input AND gate.(a) Truth Table,(b) symbol

NOT GATE- A Not gate has 1 input and 1 output. It has the effect of reversing the input signal
and is also called an inverter. The symbol and truth table for a NOT gate are shown in Figure.
Here output Y= A. Where – indicates NOT operator.

Truth Table

A B
0 1
1 0
(a)

A Y= A

(b)

Figure: NOT gate (a) Truth Table, (b) symbol.

Question-2: Explain the operator of following gate.

(a) NOR, NAND, XOR.

Answer-2: NOR: A NOR gate has the same effect as OR gate followed by a NOT gate. Hence the
output will be the opposite of OR gate.

A Y = (A+B)
B
(a)
A B Y
0 0 1
0 1 0
1 0 0
1 1 0

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Figure: NOR gate (a) symbol, (b) Truth Table.

NAND: A NAND gate has the same effect as AND gate followed by a NAND gate. Hence the
output will be the opposite of AND gate.

A Y = A.B
B
(a)

A B Y
0 0 1
0 1 1
1 0 1
1 1 0
(b)

Figure: NAND gate (a) symbol, (b) Truth Table.

EXOR: It is widely used in digital circuits. Here output Y = A B, where denotes EXOR
expression.

A Y=A B

(a)

A B Y
0 0 0
0 1 1
1 0 1
1 1 0
(b)

Figure: EXOR gate (a) symbol, (b) Truth Table.

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Lession-2: Boolean Algebra and Logic Simplification

Question-1: DeMorgan’s theorem for two and three variables.

Answer-1: DeMorgan’s theorems for two variables are-

A+B=A.B

A.B=A+B

The proof of the DeMorgan’s theorems for two variables using truth table follows-

A B A+B A.B A+B A B A.B A+B A.B

0 0 0 0 1 1 1 1 1 1
0 1 1 0 0 1 0 0 1 1
1 0 1 0 0 0 1 0 1 1
1 1 1 1 0 0 0 0 0 0
The validity of the theorems can be observed from the table.

The theorems are valid for any number of variables. For three variables, the theorems are-

A+B+C=A.B.C

A.B.C=A+B+C

A B C A B C A + B +C A + B +C A + B +C ABC ABC A.B.C

0 0 0 1 1 1 0 1 1 0 1 1
0 0 1 1 1 0 1 0 1 0 1 0
0 1 0 1 0 1 1 0 1 0 1 0
0 1 1 1 0 0 1 0 1 0 1 0
1 0 0 0 1 1 1 0 1 0 1 0
1 0 1 0 1 0 1 0 1 0 1 0
1 1 0 0 0 1 1 0 1 0 1 0
1 1 1 0 0 0 1 0 0 1 0 0

Question-2: Example of logic simplification.

1. X = ABC + ABC
2. X = (A + B) (A + B)
3. X = BC + BC + BC

Answer-2: 1. X = ABC + ABC

= AB(C + C)

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= AB as, C + C = 1

2. X = (A + B) (A + B)

= AA + AB + AB + BB

= 0 + AB + AB + 0 as, AA + 0

= AB + AB

3.X = BC + BC + BC

= B(C + C) + BC

= B + BC as, C + C = 1

Question-3: Universality of NAND gate.

Answer-3:

Fig: Realization of basic logic gates by NAND gets only.(a) NOT, (b) AND, (c) OR

Question-4: Universality of NOR gate.

Answer-4:

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Fig: Realization of basic logic gates by NOR gets only.(a) NOT, (b) AND, (c) OR

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Lession-2: Karnaugh Map and Logic Simplfication

Question-1: K-map for two, three and four variable.

Answer-1: Two variable:

Fig: K-map for two variables

Three variables:

Fig: K-map for three variables

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Four variables:

Fig: K-map for four variables

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Chapter-4: Microcomputer System

Lession-3: Bus Organized Architecture
Question-1: Define address bus.

Answer-1: The address bus is unidirectional, information flows over it only in on direction, from
the CPU to the memory or I/O devices. The CPU generates addresses on the lines of the address
bus. Each of the address corresponds to one memory location or one I/O device. When the CPU
wants to communicate with (READ or WRITE), certain memory location or I/O device, it places
the appropriate address on the address output. These address is then decoded to select the
desired memory location or I/O device. This decoding process usually requires address decoder
circuits.

Question-2: Define data bus.

Answer-2: The data bus is bidirectional and data can flow to and from the CPU through it. The
data bus can be either input or output depending on whether the CPU performs a READ or a
WRITE operation. During READ operation the data bus receives data that have been placed on
the data bus by the memory or I/O device selected by the address. During WRITE operation the
data bus acts as the output and places data on the data bus which are sent to selected memory
location or I/O device.

Question-3: Define data bus.

Answer-3: The control bus consists of a set signal that is used to synchronize the activities of
separate microcomputer elements. Some of this control signals are sent by the CPU to the other
components to tell them the type of operation in progress. The I/O devices can send control
signals to the CPU. Read/write, Reset, interrupt are examples of control signals used in
microcomputer.

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Chapter-5: Input and Output Devices

Lession-1: Input/Output Operations
Question-1: What are the different I/O methods.

Answer-1: There are three basic methods by which data can be read from or written to a
peripheral device and RAM. These are-

 Programmed I/O
 Interrupt I/O, and
 Direct memory access

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Lession-2: Input Devices

Question-1: What is optical mark reader.

Answer-1: An optical mark reader(OMR) optically reads marks on carefully printed forms.
Optical mark forms are relatively expensive, as they must be printed with exact tolerances. The
most popular use of such devices is optical readers for scoring examinations in educational
institutions.

Question-2: What is optical character reader.

Answer-2: Optical character recognition(OCR) devices can convert data from source document
to a machine-recognizable form. Applications of optical scanning include billing, insurance
premium notices, invoices etc. An OCR device can reliably read and interpret script or
handwriting.

Question-3: What is bar code reader.

Answer-3: A bar code reader is a device for scanning or reading a bar code. Bar code is printed
code that consists of parallel bars of different width and spacing. The application most
commonly observed is the coding on food and other goods that is read at the checkout counter
and translated into a line of a print on the bill showing product and cost.

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Lession-3: Outputput Devices

Question-1: Compare thermal, inkjet and laser printer.

Answer-1: Comparison of thermal, inkjet and laser printer is given bellow-

Printer Technology Advantages Disadvantages Typical speed

type
Thermal Temperature sensitive, Quiet, high quality Special paper Several dozens
paper changes color color output. required, to several
when treated, expensive, slow. hundred
characters are formed characters per
by selectively heating second.
print head.
Inkjet Electrostatically charged Quiet, prints color, Relatively slow, Several
ink drops hit paper. less expensive, fast. clogged jets, hundred
lower dpi. characters per
second.
Laser Laser beam directed to a Quiet, excellent High cost. Up to several
drum. “etching” spots quality, very high hundred pages
attract toner, which is speed, color or per minute.
then transferred to black and white.
paper.

Question-2: Advantage and dis advantage of liquid crystal display.

Answer-2: The principal advantages of LCDs are-

 Low power consumption

 Low cost
 Small size

The major disadvantages of LCDs are-

 LCDs do not emit light, as a result, the image has very little contrast.
 Screen is susceptible to glare, so the optimum viewing angle is narrow.
 LCDs have less color capability.
 Resolution is not as good as that of a CRT.

Question-3: Advantage of laser printers.

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Answer-3: The major advantages of laser printers are-

 Very high speed

 Low noise level
 Low maintenance requirement
 Very high image quality
 Excellent graphics capabilities
 A variety of type size and styles

Question-4: What is plotters?

Answer-4: An inexpansible portable plotter is capable of generating multicolor plots from data
stored on magnetic tape or disk. Plotters with multicolor capabilities generally use a writing
mechanism containing several pens, each capable of producing a different color. Some devices
for automated drafting are equipped with plotting surfaces larger than 10 square feet and very
costly. Plotters are available for general(such as designing, mapping, or plotting schematics) or
more specialized(such as three-dimensional data presentation, structural analysis, contouring,
or business chart) applications.

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Chapter-7: Memory Organization

Lession-1: Memory Basics
Question-1: What is access time?

Answer-1: Performance of a memory device is primarily determined by the rate at which

information can be red from or written into the memory. A convenient performance measure is
the average time required to read a fixed amount of information from the memory. This is
termed as access time.

Question-2: Explain about cycle time and data transferred rate.

Answer-2: The minimum time between two consecutive memory access operations called the
cycle time. It is generally convenient to assume that cycle time is the time needed to complete
any read or write operation in the memory. The maximum amount of information that can be
transferred to or from the memory every second is data transferred rate.

Question-3: Describe memory hierarchy with a diagram.

Answer-3: A variety of factors should be considered to explore the hierarchy of memories. The
major components are: nature of connection between memory and processor, cost per bit,
access time and storage capacity. Bellow figure shows these factors in hierarchical form.

It can be observed from the figure that the cost per bit increase with fall of access time and
increasing closeness with the CPU(microprocessor).

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Lession-2: Main Memory

Question-1: Write down difference between SRAM and DRAM.

Answer-1: Difference between SRAM and DRAM is given bellow-

SRAM DRAM
SRAM stands for Static Random Access DRAM stands for Dynamic Random Access
Memory. Memory.
High speed RAM. Relatively slower.
SRAM costly. Relatively low cost.
Does not require refreshing. Require refreshing periodically.
Generally used for cache. Used for main memory.
Control complexity is less. Relatively high control complexity.

Question-2: Write down difference between RAM and ROM.

Answer-2: Difference between RAM and ROM is given bellow-

RAM ROM
RAM stands for Random Access Memory. ROM stands for Read Only Memory
User can read or write into it. User can only read from it.
For any type of processing, user must store Manufacturing stores instruction into the
programmed data into RAM. ROM permanently.
Temporary or volatile memory. Permanent or non-volatile memory.

Question-3: What is cache memory?

Answer-3: Cache memories high-speed buffers for holding recently accessed data in memory. By
adding a cache memory between the fast device and slower memory system, an apparently fast
memory system can be provided.

CPU Cache Main memory

Fig: Cache and main memory

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Lession-3: Main Memory

Question-1: Write down comparison between primary memory and secondary memory.

Answer-1: Comparison between primary memory and secondary memory is given bellow-

Characteristics Primary memory Secondary memory

Location with Inside/Outside and directly accessible Outside and indirectly accessible by
respect to the by CPU. CPU.
CPU
Cost Most expensive. Less expensive then primary storage.
Capacity Lower as compared to secondary Several thousand time higher as
memory. compared to primary memory.
Average access In billionths of a second. In millionths of a second.
time
Data processing Yes. No, data must be moved into
directly from primary memory.
storage
Means of storing Semiconductor chips. Magnetic disk, tape and optical disk.
information

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QUESITION OF PREVIOUS YEAR

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