INTERNATIONAL INSTITUTE OF TECHNOLOGY & MANAGEMENT, MURTHAL SONEPAT

E-NOTES , Subject : DIGITAL ELECTRONICS, Course: Diploma , Branch : Electrical Engineering ,

Sem-4th , Chapter Name: NUMBER SYTEM AND LOGIC GATES
( Prepared By: Mrs. Gurpreet Kaur , Assistant Professor , EE)

NUMBER SYSTEM AND CODES

INTRODUCTION:-
 INTERNATIONAL INSTITUTE OF TECHNOLOGY & MANAGEMENT, MURTHAL SONEPAT
E-NOTES , Subject : DIGITAL ELECTRONICS, Course: Diploma , Branch : Electrical Engineering ,
Sem-4th , Chapter Name: NUMBER SYTEM AND LOGIC GATES
( Prepared By: Mrs. Gurpreet Kaur , Assistant Professor , EE)

 The term digital refers to a process that is achieved by using discrete unit.
 In number system there are different symbols and each symbol has an absolute value and also has place
value.

RADIX OR BASE:-
The radix or base of a number system is defined as the number of different digits which can occur in
each position in the number system.

RADIX POINT :-
The generalized form of a decimal point is known as radix point. In any positional number system the
radix point divides the integer and fractional part.

Nr = [ Integer part . Fractional part ]

↑
Radix point
NUMBER SYSTEM:-
In general a number in a system having base or radix ‘ r ’ can be written as

an an-1 an-2 …………… a0 . a -1 a -2.................a - m

This will be interpreted as

Y = an x rn + an-1 x rn-1 + an-2 x rn-2 + ……… + a0 x r0 + a-1 x r -1 + a-2 x r -2 +…..........+a -m x r –m

where Y = value of the entire number

an = the value of the nth digit

r = radix

TYPES OF NUMBER SYSTEM:-

There are four types of number systems. They are

1. Decimal number system

2. Binary number system
3. Octal number system
4. Hexadecimal number system

DECIMAL NUMBER SYSTEM:-

 The decimal number system contain ten unique symbols 0,1,2,3,4,5,6,7,8 and 9.
 In decimal system 10 symbols are involved, so the base or radix is 10.
 It is a positional weighted system.
 The value attached to the symbol depends on its location with respect to the decimal point.
 INTERNATIONAL INSTITUTE OF TECHNOLOGY & MANAGEMENT, MURTHAL SONEPAT
E-NOTES , Subject : DIGITAL ELECTRONICS, Course: Diploma , Branch : Electrical Engineering ,
Sem-4th , Chapter Name: NUMBER SYTEM AND LOGIC GATES
( Prepared By: Mrs. Gurpreet Kaur , Assistant Professor , EE)

In general,

is given by
dn dn-1 dn-2 …………… d0 . d -1 d -2............................... d - m

(dn x 10n) + (dn-1 x 10n-1) + (dn-2 x 10n-2) + … + ( d0 x 100) + ( d-1 x 10 -1) + (d-2 x 10 -2) +…+(d -m x 10 –m)

For example:-

9256.26 = 9 x 1000 + 2 x 100 + 5 x 10 + 6 x 1 + 2 x (1/10) + 6 x ( 1/100)

= 9 x 103 + 2 x 102 + 5 x 101 + 6 x 100 + 2 x 10-1 + 6 x 10-2

BINARY NUMBER SYSTEM:-

 The binary number system is a positional weighted system.
 The base or radix of this number system is 2.
 It has two independent symbols.
 The symbols used are 0 and 1.
 A binary digit is called a bit.
 The binary point separates the integer and fraction parts.

In general,

dn dn-1 dn-2 …………… d0 . d -1 d -2.................d – k

is given by
(dn x 2n) + (dn-1 x 2n-1) + (dn-2 x 2n-2) + ….+ ( d0 x 20) + ( d-1 x 2 -1) + (d-2 x 2 -2) +….+(d -k x 2 –k)

OCTAL NUMBER SYSTEM:-

 It is also a positional weighted system.
 Its base or radix is 8.
 It has 8 independent symbols 0,1,2,3,4,5,6 and 7.
 Its base 8 = 23 , every 3- bit group of binary can be represented by an octal digit.

HEXADECIMAL NUMBER SYSTEM:-

 The hexadecimal number system is a positional weighted system.
 The base or radix of this number system is 16.
 The symbols used are 0,1,2,3,4,5,6,7,8,9,A,B,C,D,E and F
 The base 16 = 24 , every 4 – bit group of binary can be represented by an hexadecimal digit.

CONVERSION FROM ONE NUMBER SYSTEM TO ANOTHER :-

1. BINARY NUMBER SYSTEM:-
(a) Binary to decimal conversion:-
In this method, each binary digit of the number is multiplied by its positional weight and the product terms
are added to obtain decimal number.
For example:

(i) Convert (10101)2 to decimal.

Solution :
(Positional weight) 24 23 22 21 20
Binary number 10101
= (1 x 24) + (0 x 23) + ( 1x 22 ) + ( 0 x 21 ) + (1 x 20)
= 16 + 0+ 4+ 0+ 1
= (21)10

(ii) Convert (111.101)2 to decimal.

Solution:
(111.101)2 = (1 x 22) + (1 x 21) + ( 1x 20 ) + ( 1 x 2 -1 ) + (0 x 2 -2) + (1 x 2 -3)
= 4+ 2+ 1 + 0.5 + 0 + 0.125
= (7.625)10
(b)Binary to Octal conversion:-
For conversion binary to octal the binary numbers are divided into groups of 3 bits each, starting at the
binary point and proceeding towards left and right.

Octal Binary Octal Binary

0 000 4 100
 1 001 5 101

2 010 6 110
3 011 7 111

For example:

(i) Convert (101111010110.110110011)2 into octal.

Solution :

Group of 3 bits are 101 111 010 110 . 110 110 011
Convert each group into octal = 5 7 2 6 . 6 6 3
The result is (5726.663)8
(ii) Convert (10101111001.0111)2 into octal.
Solution :
Binary number 10 101 111 001 . 011 1
Group of 3 bits are = 010 101 111 001 . 011 100
Convert each group into octal = 2 5 7 1 . 3 4
The result is (2571.34)8
(c) Binary to Hexadecimal conversion:-
For conversion binary to hexadecimal number the binary numbers starting from the binary point, groups are
made of 4 bits each, on either side of the binary point.
 Hexadecimal Binary Hexadecimal Binary

0 0000 8 1000

1 0001 9 1001

2 0010 A 1010

3 0011 B 1011

4 0100 C 1100

5 0101 D 1101

6 0110 E 1110
7 0111 F 1111

For example:
(i) Convert (1011011011)2 into hexadecimal.
Solution:
Given Binary number 10 1101 1011
Group of 4 bits are 0010 1101 1011
Convert each group into hex = 2 D B
The result is (2DB)16
(ii) Convert (01011111011.011111)2 into hexadecimal.
Solution:
Given Binary number 010 1111 1011 . 0111 11
Group of 3 bits are = 0010 1111 1011 . 0111 1100
Convert each group into octal = 2 F B . 7 C
The result is (2FB.7C)16
2. DECIMAL NUMBER SYSTEM:-

(a) Decimal to binary conversion:-

In the conversion the integer number are converted to the desired base using successive division by the
base or radix.
For example:
(i) Convert (52)10 into binary.
Solution:
Divide the given decimal number successively by 2 read the integer part remainder upwards to get
equivalent binary number. Multiply the fraction part by 2. Keep the integer in the product as it is and multiply
the new fraction in the product by 2. The process is continued and the integer are read in the products from
top to bottom.

2 I 52
2 l 26 ―0
2 l 13 ―0
2l6 ―1
 2l3 ―0
2l1 ―1
0 ―1
Result of (52)10 is (110100)2
(ii) Convert (105.15)10 into binary.
Solution:
Integer part Fraction part
2 I 105 0.15 x 2 = 0.30
2l 52 ― 1 0.30 x 2 = 0.60
2l 26 ― 0 0.60 x 2 = 1.20
2l 13 ― 0 0.20 x 2 = 0.40
2l 6 ― 1 0.40 x 2 = 0.80
2l3 ― 0 0.80 x 2 = 1.60
2l1 ― 1
0 ― 1
Result of (105.15)10 is (1101001.001001)2

(b) Decimal to octal conversion:-

To convert the given decimal integer number to octal, successively divide the given number by 8 till the
quotient is 0. To convert the given decimal fractions to octal successively multiply the decimal fraction and
the subsequent decimal fractions by 8 till the product is 0 or till the required accuracy is obtained.

For example:
(i) Convert (378.93)10 into octal.
Solution:
8 I 378 0.93 x 8 = 7.44
8 l 47 ― 2 0.44 x 8 = 3.52
8l5 ― 7 0.52 x 8 = 4.16
0 ― 5 0.16 x 8 = 1.28

Result of (378.93)10 is (572.7341)8

(c) Decimal to hexadecimal conversion:-

The decimal to hexadecimal conversion is same as octal.

For example:
(i) Convert (2598.675)10 into hexadecimal.
Solution:
Remainder
Decimal Hex Hex
16 I 2598 0.675 x 16 = 10.8 A
16 l 162 ― 6 6 0.800 x 16 = 12.8 C
16 l 10 ― 2 2 0.800 x 16 = 12.8 C
0 ― 10 A 0.800 x 16 = 12.8 C
Result of (2598.675)10 is (A26.ACCC)16

3. OCTAL NUMBER SYSTEM:-

(a) Octal to binary conversion:-
To convert a given a octal number to binary, replace each octal digit by its 3- bit binary equivalent.

For example:
Convert (367.52)8 into binary.
Solution:
Given Octal number is 3 6 7 . 5 2
Convert each group octal = 011 110 111 . 101 010
to binary

Result of (367.52)8 is (011110111.101010)2

(b) Octal to decimal conversion:-

For conversion octal to decimal number, multiply each digit in the octal number by the weight of its position
and add all the product terms

For example: -
Convert (4057.06) 8 to decimal
Solution:
(4057.06) 8 = 4 x 83 + 0 x 82 + 5 x 81 + 7 x 80 + 0 x 8 – 1 + 6 x 8- 2
= 2048 + 0 + 40 + 7 + 0 +0.0937
= (2095. 0937)10

Result is (2095.0937)10
(c) Octal to hexadecimal conversion:-
For conversion of octal to Hexadecimal, first convert the given octal number to binary and then binary
number to hexadecimal.

For example :-
Convert (756.603)8 to hexadecimal.
Solution :-
Given octal no. 7 5 6 . 6 0 3
Convert each octal digit to binary = 111 101 110 . 110 000 011
Group of 4bits are = 0001 1110 1110 . 1100 0001 1000
Convert 4 bits group to hex. = 1 E E . C 1 8

Result is (1EE.C18)16

(4) HEXADECIMAL NUMBER SYSTEM :-

(a) Hexadecimal to binary conversion:-
For conversion of hexadecimal to binary, replace hexadecimal digit by its 4 bit binary group.

For example:
Convert (3A9E.B0D)16 into binary.
 Solution:
Given Hexadecimal number is 3 A 9 E . B 0 D
Convert each hexadecimal = 0011 1010 1001 1110 . 1011 0000 1101
digit to 4 bit binary

Result of (3A9E.B0D)8 is (0011101010011110.101100001101)2

(b) Hexadecimal to decimal conversion:-
For conversion of hexadecimal to decimal, multiply each digit in the hexadecimal number by its position
weight and add all those product terms.

For example: -
Convert (A0F9.0EB)16 to decimal

Solution:
(A0F9.0EB)16 = (10 x 163 )+(0 x 162 )+(15 x 161 ) +( 9 x 160 ) +(0 x 16 – 1) +(14 x 16- 2) +(11 x 16-3)
= 40960 + 0 + 240 + 9 + 0 +0.0546 + 0.0026
= (41209.0572)10

Result is (41209.0572)10
(c) Hexadecimal to Octal conversion:-
For conversion of hexadecimal to octal, first convert the given hexadecimal number to binary and then
binary number to octal.

For example :-
Convert (B9F.AE)16 to octal.

Solution :-
Given hexadecimal no.is B 9 F . A E
Convert each hex. digit to binary = 1011 1001 1111 . 1010 1110
Group of 3 bits are = 101 110 011 111 . 101 011 100
Convert 3 bits group to octal. = 5 6 3 7 . 5 3 4

Result is (5637.534)8

BINARY ARITHEMATIC OPERATION :-

1. BINARY ADDITION:-
The binary addition rules are as follows
0 + 0 = 0 ; 0 + 1 = 1 ; 1 + 0 = 1 ; 1 + 1 = 10 , i.e 0 with a carry of 1

For example :-

Add (100101)2 and (1101111)2.

Solution :-

100101
+ 1101111
10010100
Result is (10010100)2
2. BINARY SUBTRACTION:-
The binary subtraction rules are as follows
0 - 0 = 0 ; 1 - 1 = 0 ; 1 - 0 = 1 ; 0 - 1 = 1 , with a borrow of 1
For example :-
Substract (111.111)2 from (1010.01)2.
Solution :-

1010.010
- 111 .111
0 0 1 0 .0 1 1
Result is (0010.011)2
3. BINARY MULTIPLICATION:-
The binary multiplication rules are as follows
0x0=0;1x1=1;1x0=0;0x1=0
For example :-

Multiply (1101)2 by (110)2.

Solution :-

1101
x 110
0000
1101
+ 1101
1001110
Result is (1001110)2
4. BINARY DIVISION:-
The binary division is very simple and similar to decimal number system. The division by ‘0’ is meaningless.
So we have only 2 rules
0÷1=0
1÷1=1
For example :-
Divide (10110)2 by (110)2.

Solution :-

110 ) 101101 ( 111.1

- 110
1010
110
1001
110
110
 110
000
Result is (111.1)2
1’s COMPLEMENT REPRESENTATION :-
The 1’s complement of a binary number is obtained by changing each 0 to 1 and each 1 to 0.

For example :-
Find (1100)2 1’s complement.
Solution :-
Given 1 1 0 0
1’s complement is 0 0 1 1

Result is (0011)2

2’s COMPLEMENT REPRESENTATION :-

The 2’s complement of a binary number is a binary number which is obtained by adding 1 to the 1’s
complement of a number i.e.
2’s complement = 1’s complement + 1
For example :-
Find (1010)2 2’s complement.
Solution :-
Given 1 0 1 0
1’s complement is 0 1 0 1
+ 1
2’s complement 0 1 1 0
Result is (0110)2
SIGNED NUMBER :-
In sign – magnitude form, additional bit called the sign bit is placed in front of the number. If the sign bit
is 0, the number is positive. If it is a 1, the number is negative.

For example:-
0 1 0 1 0 0 1 = +41
↑
Sign bit

1 1 0 1 0 0 1 = -41
↑
Sign bit

SUBSTRACTION USING COMPLEMENT METHOD :-

1’s COMPLEMENT:-
In 1’s complement subtraction, add the 1’s complement of subtrahend to the minuend. If there is a carry out,
then the carry is added to the LSB. This is called end around carry. If the MSB is 0, the result is positive. If the
MSB is 1, the result is negative and is in its 1‘s complement form. Then take its 1’s complement to get the
magnitude in binary.
For example:-
Subtract (10000)2 from (11010)2 using 1’s complement.
Solution:-
11010 11010 = 26
-10000 => + 0 1 1 1 1 (1’s complement) = - 16
Carry → 101001 + 10
+ 1
01010 = +10

Result is +10
2’s COMPLEMENT:-
In 2’s complement subtraction, add the 2’s complement of subtrahend to the minuend. If there is a carry out,
ignore it. If the MSB is 0, the result is positive. If the MSB is 1, the result is negative and is in its 2‘s
complement form. Then take its 2’s complement to get the magnitude in binary.

For example:-
Subtract (1010100)2 from (1010100)2 using 2’s complement.
Solution:-
1010100 1010100 = 84
- 1010100 => + 0 1 0 1 1 0 0 (2’s complement) = - 84_
1 0 0 0 0 0 0 0 ( Ignore the carry) 0
= 0 (result = 0)
Hence MSB is 0. The answer is positive. So it is +0000000 = 0
LOGIC GATES:- LOGIC GATES
 Logic gates are the fundamental building blocks of digital systems.
 There are 3 basic types of gates AND, OR and NOT.
 Logic gates are electronic circuits because they are made up of a number of electronic devices and
components.
 Inputs and outputs of logic gates can occur only in 2 levels. These two levels are termed HIGH and
LOW, or TRUE and FALSE, or ON and OFF or simply 1 and 0.
 The table which lists all the possible combinations of input variables and the corresponding outputs is
called a truth table.

LEVEL LOGIC:-

A logic in which the voltage levels represents logic 1 and logic 0. Level logic may be positive or negative logic.
Positive Logic:-
A positive logic system is the one in which the higher of the two voltage levels represents the logic 1 and the
lower of the two voltages level represents the logic 0.
Negative Logic:-
A negative logic system is the one in which the lower of the two voltage levels represents the logic 1 and the
higher of the two voltages level represents the logic 0.
DIFFERENT TYPES OF LOGIC GATES:-
NOT GATE (INVERTER):-
 A NOT gate, also called and inverter, has only one input and one output.
 It is a device whose output is always the complement of its input.
 The output of a NOT gate is the logic 1 state when its input is in logic 0 state and the logic 0 state when
its inputs is in logic 1 state.

IC No. :- 7404

Logic Symbol Truth table

INPUT OUTPUT
A A
0 1
1 0
Timing Diagram
1 0 0 1

0 1 1 0
AND GATE:-
 An AND gate has two or more inputs but only one output.
 The output is logic 1 state only when each one of its inputs is at logic 1 state.
 The output is logic 0 state even if one of its inputs is at logic 0 state.
IC No.:- 7408
Logic Symbol Truth Table

OUTPUT
A B Q=A . B

Timing Diagram 0 0 0

0 0 1 1 0 1 0
1 0 0
A
1 1 1
0 1 0 1

0 0 0 1

OR GATE:-
 An OR gate may have two or more inputs but only one output.
 The output is logic 1 state, even if one of its input is in logic 1 state.
 The output is logic 0 state, only when each one of its inputs is in logic state.

IC No.:- 7432
Logic Symbol Truth Table
INPUT OUTPUT
A B Q=A + B
0 0 0
0 1 1
1 0 1

Timing Diagram 1 1 1

0 0 1 1

0 1 0 1

0 1 1 1
Q
NAND GATE:-
 NAND gate is a combination of an AND gate and a NOT gate.
 The output is logic 0 when each of the input is logic 1 and for any other combination of inputs, the
output is logic 1.
IC No.:- 7400 two input NAND gate
7410 three input NAND gate
7420 four input NAND gate
7430 eight input NAND gate

Logic Symbol Truth Table

INPUT OUTPUT
A B Q= A.B

0 0 1
0 1 1
1 0 1
Timing Diagram
1 1 0
0 0 1 1

0 1 0 1

1 1 1 0

NOR GATE:-
 NOR gate is a combination of an OR gate and a NOT gate.
 The output is logic 1, only when each one of its input is logic 0 and for any other combination of inputs,
the output is a logic 0 level.

IC No.:- 7402 two input NOR gate

7427 three input NOR gate
7425 four input NOR gate

Logic Symbol Truth Table

INPUT OUTPUT
A B Q= A + B

0 0 1
0 1 0
1 0 0
1 1 0
Timing Diagram

0 0 1 1

0 1 0 1

1 0 0 0

EXCLUSIVE – OR (X-OR) GATE:-

 An X-OR gate is a two input, one output logic circuit.
 The output is logic 1 when one and only one of its two inputs is logic 1. When both the inputs is logic 0
or when both the inputs is logic 1, the output is logic 0.

IC No.:- 7486

Logic Symbol Truth Table

INPUT OUTPUT
A B Q=A
B
INPUTS are A and B 0 0 0
OUTPUT is Q = A B 0 1 1
1 0 1
=AB+AB
1 1 0
Timing Diagram

0 0 1 1

0 1 0 1

0 1 1 0

EXCLUSIVE – NOR (X-NOR) GATE:-

 An X-NOR gate is the combination of an X-OR gate and a NOT gate.
 An X-NOR gate is a two input, one output logic circuit.
 The output is logic 1 only when both the inputs are logic 0 or when both the inputs is 1.
 The output is logic 0 when one of the inputs is logic 0 and other is 1.
IC No.:- 74266

Logic Symbol
INPUT OUTPUT
A B OUT =A XNOR B

0 0 1
0 1 0
1 0 0
OUT =A B + A B
1 1 1
= A XNOR B

Timing Diagram

0 0 1 1
 A

0 1 0 1
B

1 0 0 1
OUT

UNIVERSAL GATES:-
There are 3 basic gates AND, OR and NOT, there are two universal gates NAND and NOR, each of which can
realize logic circuits single handedly. The NAND and NOR gates are called universal building blocks. Both
NAND and NOR gates can perform all logic functions i.e. AND, OR, NOT, EXOR and EXNOR.
NAND GATE:-
a) Inverter from NAND gate

Input =A
Output Q = A

b) AND gate from NAND gate

Input s are A and B

Output Q = A.B

c) OR gate from NAND gate

Inputs are A and B

Output Q = A+B
 d) NOR gate from NAND gate

Inputs are A and B

Output Q = A+B

e) EX-OR gate from NAND gate

Inputs are A and B

Output Q = A B + AB

f) EX-NOR gate From NAND gate

Inputs are A and B

Output Q = A B + A B

NOR GATE:-
a) Inverter from NOR gate
Input =A
Output Q = A
b) AND gate from NOR gate
Input s are A and B
Output Q = A.B
c) OR gate from NOR gate

Inputs are A and B

Output Q = A+B

d) NAND gate from NOR gate

Inputs are A and B

Output Q = A.B

e) EX-OR gate from NOR gate

Inputs are A and B

Output Q = A B + AB

f) EX-NOR gate From NOR gate

Inputs are A and B

Output Q = A B + A B