IN 1300 LEC Basic electronic 1 Basic electronic 1 .pdf

IN 1300
Digital Fundamental and
Computer System
FACULTY of INFORMATION
TECHNOLOGY
UNIVERSITY of MORATUWA
1

Important Notice
Please
Switch
Off
Mobile
Phones
2

Lesson 01
Digital Fundamental
by
Eng. H. D. A. Gunasekara
Lecturer
Department of Information Technology
University of Moratuwa
3

4
Course Organization
Lectures :Friday 8.15 to 10.15 ( Morning )
(At Lecture Hall- 1LH02 (1st Floor)
Textbook:
Logic and Computer Design Fundamentals
- 5th Edition
Final Grade Calculation
Final Grade = 70% (From final Exam)
+ 30% (CA Marks )
CA Marks Calculation =
60%  ( 03 – Assignment ) +
40%  ( Lab Practical & Tutorials).

5
Course Organization
This is my email address: harshanag@uom.lk.
Contact Number: 0716-130233 / 0773-606860
My office room : 1st Floor – Department of IT
Course Moodle website: http://moodle.itfac.mrt.ac.lk/
• Extra Lecture schedule , Subject details, etc. posted.
• A tentative set of lecture notes will be posted.
• Modifications may take place right up to, or during
lecture.
• Therefore you should download the class notes after a
lesson.

Which size of data storage device
has in your computer?
8 bits = 1 Byte
1024 Bytes = 1 KB
1024 KB = 1 MB
1024 MB = 1 GB
1024 GB = 1 TB
GB means Giga Byte
6
bit: A Bit is the smallest unit of data that a computer uses.
It can be used to represent two states of information, such as Yes or No.

Basic Data Types: Character Data
7
 Numeric
0 1 2 … 9
 Alphabetic
a b c …… z
 Special
# @ % ( $ &

Basic Data Types: Numeric data
8
 Integer
+ & - whole numbers
4251 , -582
 Real
All numbers including everything between
integers
0.23, 0, 5½, -2.3,
Least significant Num.
Most significant Num.

Data Representation in
Computers
How do computers represent data?
Modern computers are digital
 Recognize only two discrete states: on or off.
 Computers are electronic devices powered by
electricity, which has only two states, „1‟ or „0‟.
1 1 1 1 1
0 0 0 0 0
9

 Binary representation
 A number system that has just two unique
digits, 0 and 1
 The two digits represent the two on and off
states.
Binary
Digit (bit)
Electronic
Charge
Electronic
State
Data Representation in Computers
10

BCD (Binary Coded Decimal)
 4 bit code for numeric (0,1,2,3,4,5,…...9)
values only
 9 1001
11

ASCII (American Standard Code for Information
Interchange)
 7 bit code for all 128 characters
 A=1000001
EBCDIC (Extended BCD Interchange Code)
 8 bit ASCII
12

Computer Number Systems
 Decimal Numbers
 Binary Numbers
 Octal Numbers
 Hexadecimal Numbers
Decimal, b=10
a={0,1,2,3,4,5,6,7,8,9}
Binary, b=2
a={0,1}
Octal, b=8
a={0,1,2,3,4,5,6,7}
Hexadecimal, b=16
a={0,1,2,3,4,5,6,7,8,9,A,B,C,D,E,F} 15

Decimal Number System
 The prefix “deci-” stands for 10.
 The decimal number system is a Base 10
number system:
 There are 10 symbols that represent quantities:
0, 1, 2, 3, 4, 5, 6, 7, 8, 9
 Each place value in a decimal number is a power
of 10.
16

Decimal Number System
10
3
10
2
10
1
10
0
1000 100 10 1
1 x 1000 = 1000
4 x 100 = 400
9 x 10 = 90
2 x 1 = + 2
1492
Example : 149210
18

Binary Numbers System
The prefix “bi-” stands for 2
The binary number system is a Base 2 number
system:
 There are 2 symbols that represent quantities: 0,1.
 Each place value in a binary number is a power of 2.
8 4 2 1
2
3
2
2
2
1
2
0
19
Example : 10112
1 x 8 = 8
0 x 4 = 0
1 x 2 = 2
1 x 1 = + 1
11

Repeated Division Method
 Divide the number successively by 2.
 After each division record the remainder which is
either 1 or 0.
Example : 12310 becomes
123/2 = 61 r=1
61/2 = 30 r=1
30/2 = 15 r=0
15/2 = 7 r=1
7/2 = 3 r=1
3/2 = 1 r=1
1/2 = 0 r=1
The result is read from the last remainder upwards
12310 = 11110112
16510 =101001012
19810 =110001102
6410 = 10000002
21
1111011
2
Number Base Conversion
1.1 Decimal to Binary

 After each division record the remainder which is a
number in the range 0 to 7.
Example: 462910 becomes
4629/8 = 578 r= 5
578/8 = 72 r= 2
72/8 = 9 r= 0
9/8 = 1 r= 1
1/8 = 0 r= 1
462910 =110258
22
1.2 Decimal to Octal
1
1
0
2
5
8
16510 = 2458
19810 = 3068
6410 = 1008

 After each division record the remainder which is in the decimal
range 0 to 15, corresponding to the hexadecimal range 0 to F.
 Hexadecimal Range  0 ,1 ,2 ,3 ,4 ,5 ,6 ,7 ,8 ,9 ,A ,B ,C ,D ,E ,F
53241/16= 3327 r = 9 = 9
3327/16 = 207 r =15 = F
207/16 = 12 r =15 = F
12/16 = 0 r =12 = C
5324110 = CFF916
23
1.3 Decimal to Hexadecimal
C
F
F
9
16
16510 = A516
19810 = C616
6410 = 4016

24
 Take the left most none zero bit,
 Double (X 2) it and add it to the bit on its right.
 Now take this result, double it and add it to the next bit on the right.
 Continue in this way until the least significant bit has been added in.
2.1 Binary to Decimal
Example : 10101112 becomes
Therefore, 10101112 = 8710
1100101102 =40610
1110101002 =46810

 Take the left-most digit,
 Multiply it by eight and add it to the digit on its right.
 Then, multiply this subtotal by eight and add it to the next
digit on its right.
 The process ends when the left-most digit has been
added to the subtotal.
25
2.2 Octal to Decimal
Therefore, 64378 = 335910
17558 =100510
6258 = 40510

26
2.3 Hexadecimal to Decimal
Therefore, 1AC16 = 42810
1BC16 = 44410
2DE16 = 73410
The method is identical to the procedures for binary and
octal except that 16 is used as a multiplier.
Example: 1AC16 becomes

27
3.1.1 Binary to Octal
1100101102 =6268
1110101002 =7248
 Form the bits into groups of three starting at the binary
point and move leftwards.
 Replace each group of three bits with the corresponding
octal digit (0 to 7).
011 001 011 101
3 1 3 5
Therefore,110010111012 = 31358

28
3.1.2 Octal to Binary
 Each octal digit is simply replaced by its 3-bit binary
equivalent.
 It is important to remember that (say) 3 must be replaced
by 011 and not 11.
4 1 3 5 7
100 001 011 101 111
Therefore, 413578 = 1000010111011112

29
3.2.1 Binary to Hexadecimal
1100101102 =19616
1110101002 =1D416
Form the bits into groups of four bits starting at the decimal
point and move leftwards.
Replace each group of four bits with the corresponding
hexadecimal digit from 0 to 9, A, B, C, D, E, and F.
0110 0101 1101
6 5 D
Therefore, 110010111012 = 65D16

30
3.2.2 Hexadecimal to Binary
 Each hexadecimal digit is replaced by its 4-bit binary
equivalent.
Example : AB4C16 becomes
A B 4 C
1010 1011 0100 1100
Therefore, AB4C16 = 10101011010011002

31
Summary
How to remember
All of these conversions follow certain patterns that you need
to remember.
 When converting from decimal you always use divide.
 When converting to decimal you always multiply.
 Converting between hexadecimal and binary as well as
octal and binary is a bit easier to remember.
 Just remember that hexadecimal is 8421 and octal is 421.
 The only thing you need to know about converting
between hexadecimal and octal is that you must
always convert to binary first.

IN 1300 LEC Basic electronic 1 Basic electronic 1 .pdf

More Related Content

Similar to IN 1300 LEC Basic electronic 1 Basic electronic 1 .pdf

Recently uploaded

IN 1300 LEC Basic electronic 1 Basic electronic 1 .pdf