CHAPTER – 4 : COMPUTER ELECTRONICS
Introduction :
Early computer systems used electrical switches and when electrical switches were replaced by less mechanical devices such as vacuum tubes, than the transistor, the integrated circuit, the concept of switching on and off remained with computers but a representation of the on/off behavior of computers had to be made.
The term, digital, in computing and electronics applies to converting real world information to binary numeric form. The binary (base 2) number system represents two discreet values using two symbols or digits i.e., 0 and 1. The binary number system, where a zero symbolizes no electrical current (OFF) and one represents electrical current exists (ON).
All computer data (alpha-numeric, symbols and characters, audio, graphics and video) are represented or encoded using sequences of binary digits that are interpreted according to appropriate software. Computers are made up of electronic devices and electronic device can exist or either in ON or OFF state. For our convenience an ON State is represented by the code ‘1’ and OFF state is represented by the code‘0. They are called ‘bits’, an abbreviation for Binary Digit. The numbers represented by bits are known as binary numbers.
Basically number systems are classified in two types :
1)
Non – Positional Number System
2)
Positional Number System
Non-Positional Number System :
In this number system any number can be represented by arranging symbols in various positions. Each symbol represents a definite value irrespective of the position in which they appear. For example : Roman Number System
Positional Number System :
In a positional number system, a number is represented by a set of symbols. Each symbol represents a particular value, depending on its position. The actual number of symbols used in a position system depends on its base.
Base :
The number of digits or basic symbols used in a positional number system is known as the
BASE or Radix of the system.
1. Decimal Number System (Base 10) :
It is a positional number system with base 10 and thus it uses 10 symbols i.e., 0 to 9. Any number can be represented by arranging symbols in various positions. In the decimal system, each position represents a specific power of 10.
For example : The decimal number 654.52, written as (654.52)10 or 654.52(10) to specify base 10 is represented as follows :
|
Increasing Powers of 10 |
Decreasing Powers of 10 |
||||
|
|
Hundreds |
Tens |
Units |
One Tenth |
One Hundredth |
|
Weights |
102 |
101 |
100 |
10-1 |
10-2 |
|
Digits |
6 |
5 |
4 |
5 |
2 |
Thus, the expanded notation of
654.52(10) = 6 X 102 + 5 X 101 + 4 X 100 + 5 X 10-1 + 2 X 10-2
= 6 X 100 + 5 X 10 + 4 X 1 + 5 X 0.1 + 2 X 0.01
= 600 + 50 + 4 + 0.5 + 0.02
= 654.52
Note : In a positional number system all bits or digits to the left of the decimal or binary point have weights that are positive powers of base and those to the right have weights that are negative powers of base. The base is also called Radix and fractional point is called as Radix point.
2. Binary Number System (Base 2) :
The binary system is a positional system to the base 2. It uses two symbols 0 and 1. Each position represents specific power of 2.
For example : The binary number 1 1 0 1 . 1 1 written as 1 1 0 1 . 1 1 (2) or (1 1 0 1 . 1 1)2 to specify base 2, is represented as follows :
|
Increasing Powers of 2 |
Decreasing Powers of 2 |
|||||
|
Weights |
23 |
22 |
21 |
20 |
2-1 |
2-2 |
|
Digits |
1 |
1 |
0 |
1 |
1 |
1 |
Thus, the expanded notation of 1 1 0 1 . 1 1 (2)
= 1 X 23 + 1 X 22 + 0 X 21 + 1 X 20 + 1 X 2-1 + 1 X 2-0
3. Octal Number System (Base 8) :
The octal number system is a positional number system with base 8. It uses 8 symbols i.e., 0 to 7. In octal number system, each digit position corresponding to a power of 8.
For example : The octal number 43.12 written as (43.12)8 or 43.12(8) to specify base 8, is represented as follows :
|
Increasing Powers of 8 |
Decreasing Powers of 8 |
|||
|
Weights |
81 |
80 |
8-1 |
8-2 |
|
Digits |
4 |
3 |
1 |
2 |
Thus, the expanded notation of 43.12(8)
= 4 X 81 + 3 X 80 + 1 X 8-1 + 2 X 8-0
4. Hexadecimal Number System (Base 16) :
The hexadecimal number system is a positional number system to the base 16. It uses 16 symbols to represent any number. The first 10 symbols are represented by digits 0 to 9 and the remaining 6 symbols by the letters A to F, representing the decimal values 10 to 15 respectively. Each position represents a specific powers of 16.
For example, the hexadecimal number BA85.12 is written as (BA85.12)16 or BA85.12(16), is represented as follows :
|
Increasing Powers of 16 |
Decreasing Powers of 16 |
|||||
|
Weights |
163 |
162 |
161 |
160 |
16-1 |
16-2 |
|
Digits |
B |
A |
8 |
5 |
1 |
2 |
Thus, the expanded notation of BA85.12(16)
= B X 163 + A X 162 + 8 X 161 + 5 X 160 + 1 X 16-1 + 2 X 16-2
INTER CONVERSION OF NUMBER FROM ONE SYSTEM TO ANOTHER INTEGER PART
1.
Divide the integer part of the decimal number by the
base ‘b’ of the new system. The remainder will give the right most digit of the
integer part of the new number.
2.
Divide the quotient again by the base ‘b’. The remainder is the next
digit from right.
3.
Repeat step 2 until a zero quotient is obtained. Last
remainder is the left most digit of the new number.
FRACTIONAL PART
1.
Multiply the fractional part of the decimal number by the base ‘b’ of the new system. The
integer part of the product gives the left most digit of the fractional part of
the new number.
2.
Multiply the fractional part of the product by the base ‘b’. The integer part of the
resultant product is the next digit from left.
3.
Repeat step 2 until a zero fractional part or a
repeated fractional part or a non-terminating fractional part occurs.
1. Decimal to Binary Conversion :
1) 25.
625 (10) = ? (2)
|
2 |
2 5 |
|
|
Product |
Integer |
Fractional-Part |
|
2 |
1 2 |
1 |
0.625 X 2 |
1.250 |
1 |
0.250 |
|
2 |
6 |
0 |
0.250 X 2 |
0.500 |
0 |
0.500 |
|
2 |
3 |
0 |
0.500 X 2 |
1.000 |
1 |
0.000 |
|
2 |
1 |
1 |
|
|
|
|
|
2 |
0 |
1 |
Therefore,
25(10) = 1 1 0 0 1 Thus, the result is 25.625(10) |
& = |
0.625(10)
= 1 0 1 1 1 0 0 1 . 1 0 1(2) |
|
2. Binary to Decimal Conversion :
1) 1 1 . 1 0 1 1 (2) = ? (10)
1 1 . 1 0 1 1(2) = 1 X 21 + 1 X 20 + 1 X 2-1 + 0 X 2-2 + 1 X 2-3 + 1 X 2-4
= 1 X 2 + 1 X 1 + 1 X 0.5 + 0 X 0.25 + 1 X 0.125 + 1 X 0.0625
= 2 + 1 + 0.5 + 0 + 0.125 + 0.0625
= 3.6575(10)
3. Binary to Octal Conversion :
1) 0 1 1 1 0 1 (2) = ?(8)
|
Octal |
0 |
1 |
2 |
3 |
4 |
5 |
6 |
7 |
|
Binary |
000 |
001 |
010 |
011 |
100 |
101 |
110 |
111 |
Note : Place the 3 binary bits into one group from right side and write equivalent octal digit.
0 1 1 1 0 1
3 5
Thus, the result is 0 1 1 1 0 1 (2) = 3 5 (8)
4. Binary to Hexadecimal Conversion :
1) 01101101(2) = ?(16)
|
Hexa Decimal |
0 |
1 |
2 |
3 |
4 |
5 |
6 |
7 |
8 |
9 |
A |
B |
C |
D |
E |
F |
|
Binary |
0000 |
0001 |
0010 |
0011 |
0100 |
0101 |
0110 |
0111 |
1000 |
1001 |
1010 |
1011 |
1100 |
1101 |
1110 |
1111 |
Note : Place the 4 binary bits into one group from right side and write equivalent Hexadecimal digit.
0 1 1 0 1 1 0 1
6 D
Thus, the result is 0 1 1 0 1 1 0 1 (2) = 6 D (16)
5. Octal to Decimal Conversion :
1) 43.12(8) = ? (10)
43.12(8) = 4 X 81 + 3 X 80 + 1 X 8-1 + 2 X 8-2
= 4 X 8 + 3 X 1 + 1 X 0.125 + 2 X 0.015625
= 32 + 3 + 0.125 + 0.03125
= 35.15625 (10)
6. Octal to Binary Conversion :
|
Octal |
0 |
1 |
2 |
3 |
4 |
5 |
6 |
7 |
|
Binary |
000 |
001 |
010 |
011 |
100 |
101 |
110 |
111 |
Note : Place the 3 bit binary equivalent of each digit below the number 1) 67.35(8) = ?(2)
|
6 |
7 . |
3 5 |
|
110 |
111 . |
011 101 |
Thus the result is 67.35(8) = 110111.011101(2)
7. Octal to Hexadecimal Conversion :
(a)
First
Covert Octal to Decimal 43.12(8) = ? (10)
= 4 X 81 + 3 X 80 + 1 X 8-1 + 2 X 8-2
= 4 X 8 + 3 X 1 + 1 X 0.125 + 2 X 0.015625
= 32 + 3 + 0.125 + 0.031250
= 35.15625 (10)
(b) Then Convert Decimal to Hexadecimal
35.15625 (10) = ? (16)
|
16 |
3 5 |
|
|
Product |
Integer |
Fractional-Part |
|
16 |
2 |
3 |
0.15625 X 16 |
2.500 |
2 |
0.500 |
|
|
0 |
2 |
0.50000 X 16 |
8.000 |
0 |
0.000 |
Therefore, 35(10) = 2 3 & 0.15625(10) = 2 0
Thus, the result is 35.15625(10) = 2 3 . 2 0 (16)
Representation of Signed Numbers
In Binary System, we represent the sign of a number using an extra bit at the extreme right of the number and is called ‘sign’ bit. By convention the symbol ‘0’ is used to represent the (+) sign and ‘1’ to represent (-) sign.
In the case of binary numbers used in computers, the most significant bit represents the sign
and the remaining bits the magnitude of the number. For example a 6 bit binary equivalent of a decimal number -15 is represented by ( 1 0 1 1 1 1 )2 and + 15 is represented by ( 0 0 1 1 1 1 )2.
This method of representing is known as sign magnitude representation.
Representation of Signed numbers are :
1)
By sign magnitude representation
2)
By using 1’s complement
3)
By using 2’s complement
Sign and Magnitude Representation
To represent Positive numbers, the magnitude is represented in the true binary form, and a sign bit 0 is placed in front of the MSB.
For example + 47 represented as follows
:
|
0 |
1 |
0 |
1 |
1 |
1 |
1 |
Sign bit True Number
To represent negative number, the magnitude is represented into 2’s complement form, and a sign bit of 1 is placed in front of the MSB.
For example + 47 represented as follows :
|
1 |
0 |
1 |
0 |
0 |
0 |
1 |
Sign bit True Number
1’s Complement
The 1’s complement of a binary number is obtained by complementary all the bits i.e., by changing 1’s to 0’s.
Example : Find the 1’s complement of 1 0 1 1 0 0
Step1 : 1 0 1 1 0 0 given number
Step2 : 0 1 0 0 1 1 complement each bit to form 1’s complement.
Thus, 1’s complement of 1 0 1 1 0 0 = 0 1 0 0 1 1
2’s Complement
The 2’s complement of a binary number is obtained by adding 1 to 1’s complement of a binary number.
Example : Find the 2’s complement of 1 0 1 1 0 0
Step1 : 1 0 1 1 0 0 given number
Step2 : 0 1 0 0 1 1 complement each bit to form 1’s complement.
+ 1
Step3 : 0 1 0 1 0 0 is 2’s complement of original number.
Binary Arithmetic :
Binary Arithmetic is a fundamental code for all digital computers and most other digital systems. It can perform all arithmetic operations like addition, subtraction, multiplication and division.
Binary Addition :
It is performed in the same manner as decimal addition. The following table shows that four basic rules for binary addition.
|
A |
|
B |
Sum |
Carry |
|
0 |
+ |
0 |
0 |
0 |
|
0 |
+ |
1 |
1 |
0 |
|
1 |
+ |
0 |
1 |
0 |
|
1 |
+ |
1 |
0 |
1 |
Examples :
|
|
1 0 1 |
|
1 0 1 1 . 1 0 1 |
|
+ |
0 1 1 |
+ |
1 0 1 0 . 1 0 0 |
|
|
1 0 0 0 |
|
1 0 1 1 0 . 0 0 1 |
Binary Substraction :
Binary subtraction rules are as follows :
|
A |
|
B |
Sum |
Carry |
|
0 |
- |
0 |
0 |
0 |
|
1 |
- |
0 |
1 |
0 |
|
1 |
- |
1 |
0 |
0 |
|
0 |
- |
1 |
1 |
1 |
Examples :
|
|
1 0 1 |
|
1 0 0 1 0 1 0 |
|
- |
0 1 1 |
- |
1 0 1 0 1 |
|
|
0 1 0 |
|
0 1 1 0 1 0 1 |
Subtraction of Binary number using 1’s Complement :
Case 1: Substraction of Smaller number from Larger
number
1.
Determine 1’s Complement of smaller number
2.
Add 1’s compliment to the larger number
3. We get
carry. Addition end around carry to the above gives the result of subtraction.
Example :
Subtract 1001 – 1000 using 1’s Complement
1.
1’st complement of 1000 is 0111
2. Add 1’s
complement to the larger number 1 0 0 1
0 1 1 1
1 0 0 0 0
3. Add carry 1
to the above result 0 0 0 0
+ 1
0 0 0 1
Therefore, 1 0 0 1 – 1 0 0 0 = 0 0 0 1 (
9 – 8 = 1 )
Case 2: Substraction of Larger number from Smaller
number
1.
Determine 1’s Complement of larger number
2.
Add 1’s compliment to the smaller number
3. There will
be no carry. To get answer take the 1’s complement of the result and put
negative sign.
Example :
Subtract 111 – 101 using 1’s Complement
1.
1’s complement of larger number 111 is
000
2. Add 1’s
complement to the smaller number 1 0 1
+ 0 0 0
1 0 1
3. Find 1’s
complement of 101 and put negative sign i.e., -
0 1 0
Therefore, 1 0 1 – 1 1 1 = - 0 1 0 ( 5 – 7 = -2 )
Subtraction of
Binary number using 2’s Complement :
Case 1: Substraction of Smaller number from Larger
number
1.
Determine 2’s Complement of smaller number
2.
Add 2’s compliment to the larger number
3. We get
carry. Neglect the carry and the remaining gives the result of subtraction
Example :
Subtract 1001 – 1000 using 2’s Complement
1.
2’s complement of 1000 is
0 1 1 1
+ 1
1 0 0 0
2. Add 2’s
complement to the larger number 1 0 0 1
+ 1 0 0 0
1 0 0 0 1
3.
Neglect the carry i.e., 0 0 0 1
Therefore, 1 0 0 1 – 1 0 0 0 = 0 0 0 1 (
9 – 8 = 1 )
Case 2: Substraction of Larger number from Smaller
number
1.
Determine 2’s Complement of larger number
2.
Add 2’s compliment to the smaller number
3. There will
be no carry. To get answer take the 2’s complement of the result and put
negative sign.
Example :
Subtract 111 – 101 using 1’s Complement
1. 2’s
complement of larger number 111 is i.e., 0
0 0
+ 1
0 0 1
2. Add 2’s
complement to the smaller number 1 0 1
+ 0 0 1
1 1 0
3. Find 2’s
complement of 110 and put negative sign i.e., 0
0 1
+ 1
0 1 0
i.e., - 0 1 0
Therefore, 1 0 1 – 1 1 1 =
- 0 1 0 ( 5 – 7 = -2 )
Computer Codes :
Computer understands everything only in binary. Therefore,
when we input numbers, alphabets and other special symbols, they must be
represented in the binary format. There are three such coding standards. There
are BCD, ASCII and EBCDIC.
1.
BCD (Binary Coded Decimal
System :
In BCD each digit of a decimal number is independently
converted to 4-bit binary number. For example the decimal number 573 would be
represented in the 4 bit BCD code as
0 1 0 1 0 1 1 1 0 0 1 1
5 7 3
BCD Coding system is used to represent only decimal numbers,
4-bits are insufficient to represent the various characters used by computer.
Therefore 6-bit BCD Code was developed. In this code two more bits called as
‘zero position’ are added. It is possible to represent 64 code groups i.e., 10
decimal digits, 26 alphabets and 28 special characters using 6 bit code.
2.
ASCII (American Standard
Code for Information Interchange :
It is most widely used alphanumeric code for printers,
keyboards and terminals which are interface with computer to represent data.
This is 7 bit code and also it has
128
possible code groups i.e., alphabets, numbers, special characters and control
character (Enter Key, Escape Key, Space bar etc.,). A is represented in ASCII
as 1000001 whose decimal equivalent is 65.
3.
EBCDIC (Extended Binary
Coded Decimal Interchange Code) :
It is also an alphanumeric code used in IBM computers and
mainframe applications. It is an 8 bit code representing 256 different code
group. A is represented in EBCDIC code as 11000001.
Digital Logic :
The word
Logic is used to describe the circuits which can duplicate specific function of
decision making performed by the human mind. In the logic relevant to computers
we know that there can be only two states. This logic is thus called two state
logic or bivalent logic. Such a logical method was developed by ARISTOTAL for
getting at the truth. The method was subsequently developed by mathematician
DE-MORGAN and GEORGE BOOLE into a very powerful mathematical tool.
A
Gate is a simple electronic circuit or device that performs logical functions. It
has one or more inputs and output. Gates are called binary logic gates 1 and 0
are inputs and outputs.
Truth Table
Truth
Table is a table which shows all inputs and outputs possibilities of a logical
circuits or gate.
Types of Logic Gates
1.
AND Gate
2. OR Gate
3. NOT Gate
4. NAND Gate
5. NOR Gate
AND Gate
This is an electronic decision making element with one or
more inputs and single outputs. Its function is to implement the AND operation
(i.e., Logical Multiplication). The logical symbol for AND Gate is
A
Q
B
Two inputs AND Gate truth
table
|
Input A |
Input B |
Output Q=A.B |
|
0 |
0 |
0 |
|
0 |
1 |
0 |
|
1 |
0 |
0 |
|
1 |
1 |
1 |
Three inputs AND Gate truth
table
|
Input A |
Input B |
Input C |
Output Q=A.B.C |
|
0 |
0 |
0 |
0 |
|
0 |
0 |
1 |
0 |
|
0 |
1 |
0 |
0 |
|
0 |
1 |
1 |
0 |
|
1 |
0 |
0 |
0 |
|
1 |
0 |
1 |
0 |
|
1 |
1 |
0 |
0 |
|
1 |
1 |
1 |
1 |
AND Gate Rule
1.
Logic 1 when all inputs are 1
2. Logic 0 when any input is 0
OR Gate
This is an electronic decision making element with one or
more inputs and single output. Its function is to implement the OR operation
(i.e., Logical Addition). The logical symbol for OR Gate is
Two inputs OR
Gate truth table
|
Input A |
Input B |
Output Q=A+B |
|
0 |
0 |
0 |
|
0 |
1 |
1 |
|
1 |
0 |
1 |
|
1 |
1 |
1 |
Three inputs OR Gate truth
table
|
Input A |
Input B |
Input C |
Output Q=A+B+C |
|
0 |
0 |
0 |
0 |
|
0 |
0 |
1 |
1 |
|
0 |
1 |
0 |
1 |
|
0 |
1 |
1 |
1 |
|
1 |
0 |
0 |
1 |
|
1 |
0 |
1 |
1 |
|
1 |
1 |
0 |
1 |
|
1 |
1 |
1 |
1 |
OR Gate Rule
1.
Logic 1 when an inputs is 1
2. Logic 0 when all inputs are 0
NOT Gate
This is an electronic decision making element with one input
and one output. Its function is to implement the NOT operation (i.e., Inversion
or Logical Complement) . The logical symbol for NOT Gate is
Truth table of NOT Gate is
Two inputs OR
Gate truth table
|
Input A |
Output Q (Complement of A) |
|
0 |
1 |
|
1 |
0 |
NOT Gate Rule is
1.
Logic 1 when input is 0
2. Logic 0 when input is 1
NAND Gate
This is
combination of NOT and AND Gates. This is like AND Gate but with the output
complimented.
The logical symbol for NAND Gate is
Two inputs
NAND Gate truth table
|
Input A |
Input B |
A.B |
Output Q=NOT(A.B) |
|
0 |
0 |
0 |
1 |
|
0 |
1 |
0 |
1 |
|
1 |
0 |
0 |
1 |
|
1 |
1 |
1 |
0 |
NAND Gate rule is
1.
Logic 1 when any input is 0
2. Logic 0 when all inputs are 1
NOR Gate
This
is combination of NOT and OR Gates. This is like OR Gate but with the output
complimented.
The logical symbol for NOR Gate is
Two inputs NOR Gate truth
table
|
Input A |
Input B |
A+B |
Output Q=NOT(A+B) |
|
0 |
0 |
0 |
1 |
|
0 |
1 |
1 |
0 |
|
1 |
0 |
1 |
0 |
|
1 |
1 |
1 |
0 |
NOR Gate rule is
1.
Logic 1 when all inputs are 0
2. Logic 0 when any inputs is 1
Universal
Gates
NAND and
NOR gates are called universal gates because the other gates (i.e., AND, OR and
NOT) can realized these individual gates.
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