1.1 decimal number system the decimal number system …

Dr. I. Manimehan, Department of Physics, MRGAC

NUMBER SYSTEM

A digital computer is a programmable machine that processes binary data, i.e., data

represented in binary number system. We have to understand computers and digital electronics

only with the help of binary number system. But we are used to decimal number system in our

everyday life for a long period of time and we take the rules for granted.

1.1 DECIMAL NUMBER SYSTEM

The decimal number system makes use of ten digits namely, 0, 1, 2, 3, 4, 5, 6, 7, 8 and 9.

Since ten basic symbols or digits are used, the decimal number system is said to have a base or

radix of ten. When we want to express numbers of value greater than 9, we use two or more

digits and the position of each digit within the number gives us the magnitude it represents. All

the digits in the decimal system are expressed in powers of 10 like 100, 101, 102 etc., for the

integer part and 10−1, 10−2, 10−3 etc., for the fractional part. The integer portion and fractional

portion in a decimal portion in a decimal number system are separated by a decimal point.

The quantities 100, 101, 102 etc., simply represent the units, tens, hundreds etc and the

quantities 10−1, 10−2, 10−3 etc., represent one tenth, one hundredth, one thousandth etc. The

quantities 102, 101, 100, 10−1, 10−2, 10−3 etc., are called weights.

i) 25 = 20 + 5

= 2 x 10 + 5 x 1

= 2 x101 + 5 x100

The digit 2 has a weight 10 and the digit 5 has a weight 1.

ii) 7694 = 7000 + 600 + 90 + 4

= 7x1000 + 6x100 + 9x10 + 4x1

= 7x103 + 6x102 + 9x101 + 4x100

The digit 7 has a weight 1000, digit 6 has a weight 100, the digit 9 has a weight 10 and the digit

4 has a weight 1.

1.2 BINARY NUMBER SYSTEM

A binary number system uses only two symbols or digits namely, 0 and 1. That is the

binary number system has a base or radix of 2. A binary 0 or 1 is often called a bit. All the bits

will have powers of 2 like 20, 21, 22, etc., for the integer portion and 2−1, 2−2, 2−3, etc., for the

fractional portion. Here we have a binary point in a binary number system.

A 4-bit binary word is called as a nibble. An 8-bit binary word is called as a byte. A 16-

bit binary word is simply called as a word and a 32-bit binary word is called as double word.


1.3 BINARY TO DECIMAL CONVERSION

A binary number can be converted into a decimal number by adding the products of each

bit and its weight. Let us take a few examples

i) 1 0 1 2

1 x20 = 1 x 1 =1

0 x21

= 0 x 2 = 0

1 x22 = 1 x 4 = 4

(101)2 = (5)10

ii) (1 0 0 1 1)2

1x20 = 1x1 = 1

1x21 = 1x2 = 2

0x22 = 0x4 = 0

0x23 = 0x8 = 0

1x 24 = 1x16 = 16

(1011)2 = (19)10

iii) (0. 1 0 1)2

1x2−1 = 1x0.5 = 0.5

0x2−2 = 0x0.25 = 0

1x2−3 = 1x0.125 = 0.125

(0.101)2 = (0.625)10

Example 1.1

Convert (1101011.1011)2 to its equivalent decimal number.

Solution:

For integer part,


(1 1 0 1 0 1 1)2

1 x 20 = 1x1 = 1

1 x 21 = 1x2 = 1

0 x 22 = 0x4 = 0

1x 23 = 1x8 = 8

0x 24 = 0x16 = 0

1x25 = 1x32 = 32

1x26 = 1x64 = 64

(1101011)2 = 107 10

For fraction part,

(.1 0 1 1)2

1x2−4 = 1x0.0625 = 0.0625

1x2−3 = 1x0.125 = 0.125

0x2−2 = 0x0.25 = 0

1x2−1 = 1x0.5 = 0.5

(.1 0 1 1)2 = (6875)10

(1101011.1011)2 = (107.6875)10


1.4 DECIMAL TO BINARY CONVERSION

A decimal like 19 can be converted into binary by repeatedly dividing the number by 2

and collecting the remainders (double dabble method)

2 19

9 - 1 (LSB)

4 - 1

2 - 0

1 - 0

0 - 1 (MSB)

(19)10 = (10011)2

Example 1.2

Convert (107.6875)10 to its equivalent binary number.

Solution:

For integer part, divide by 2 repeatedly and collect the remainders.

2 107

53 - 1 (LSB)

26 - 1

13 - 0

6 - 1

3 - 0

1 - 1

0 - 1 (MSB)

(107)10 = (110 1011)2


For the fractional part, multiply by 2 repeatedly and collect the carries.

0.6875x2 = 1.3750; carry 1 (MSB)

0.3750x2 = 0.7500; carry 0

0.7500x2 = 1.5000; carry 1

0.5000x2 = 1.0000; carry 1 (LSB)

Therefore, (107.6875)10 = (110 1011.1011)2

1.5 HEXADECIMAL NUMBER SYSTEM

The hexadecimal number system as a base 16. The basic digits are 0, 1, 2, 3, 4, 5, 6, 7,

8, 9, A, B, C, D, E, F. The base of the hexadecimal number system is 16 and the base of the

binary number system is 2. Since 24 = 16, it follows that any hex digit can be represented by a

group of four bit binary sequence.

The table (1.1) gives the decimal, hexadecimal and the four bit binary equivalence for the

decimal numbers 0 to 15.

Decimal

Hexadecimal

Binary

23 + 22 + 21 + 20

8 + 4 + 2 +1

0 0 0000

1 1 0001

2 2 0010

3 3 0011

4 4 0100

5 5 0101

6 6 0110

7 7 0111

8 8 1000

9 9 1001

10 A 1010

11 B 1011

12 C 1100

13 D 1101

14 E 1110

15 F 1111


1.6 HEXADECIMAL TO DECIMAL COVERSION

The hex to decimal conversion is similar to binary to decimal conversion, only the

weights are different. In this case, the weights used are 160, 161, 162 etc., for the integer part

and 16−1, 16−2, 16−3 etc., for the fractional part. In hexadecimal number system we have a

hexadecimal point.

Example 1.3

Convert the following hexadecimal numbers to decimal.

(a) (E9)H (b) (3FC.8)H (C) (FFFF)H

Solution:

(a) (E 9)H

9x160 = 9x1 = 9

14x161 = 14x16 = 224

(E9)H = (233)10

(b) (3FC.8)H

8x16−1 = 8x0.0625 = 0.5

12x 160 = 12x1 = 12

15x161 = 15x16 =240

3x 162= 3x 256 = 768

(3FC.8)H = (1020.5)10

(C) (F FF F)H

15x160 =15x1 = 15

15x161 = 15x16 = 240

15x 162 = 1 x256 = 3840

15x 163 = 15x4096 = 61440

(FFFF)H = (65535)10


1.7 DECIMAL TO HEXADECIMAL CONVERSION

To convert a decimal number to hex, we have to divide the decimal number by 16

repeatedly and collect the remainders from top to bottom (for the integer part). The remainders

also must be taken in hex.

Example 1.4

Convert the following decimal numbers to hex equivalents

(a) 1020 (b) 98.625

Solution:

(a) 16 1020

63 - 12 C (LSB)

3 - 15 F

0 - 3 3 (MSB)

(1020)10 = (3FC) H

(b) The hex number 980625 is separated as the integer part 98 and fractional part .625. The

integer 98 is converted into hex by repeated division by 16.

16 98

6 - 2

0 - 6

(98)10 = (62)H

The fraction 0.625 is converted into hex by multiplying 0.625 by 16 and collecting the carry.

0.625 x 16 = 10.000

The carry obtained is 10 which must be taken in hex as ‘A’.

Therefore, (98.625)10 = (62.A) H


1.8 HEXADECIMAL TO BINARY CONVERSION

To convert a decimal number to binary we have adopted a procedure of repeatedly

dividing the given decimal number by 2. Since the base of hexadecimal number system is 16

which is equal to 24, to convert a hexadecimal number to binary, all we have to do is replace

each hex digit with its equivalent 4-bit binary.

Table 1.1 gives the hex, decimal and the corresponding binary combination. Using the

table, we can write the binary equivalent of any hexadecimal number.

Example 1.5

Convert the following hexadecimal numbers to binary.

(a) (25)H (b) (3A.7)H (C) (CD.E8)H

Solution:

(a) (25)H = (0010 0101)2

(b) (3A.7)H = (0011 1010)2

(c) (CD.E8)H = (1100 1101.1110 1000)2

1.9 BINARY TO HEXADECIMAL CONVERSION

To convert a binary number to hex, we have to arrange the bits into group of 4 bits

starting from LSB (Least Significant Bit). If the final group has less than 4 bits, just include

zeros to make it a group of 4 bits. For example, 100101 2 into hex, arrange the bits as

10 0101 2. Now include two zeros for the first group at the front. The binary combination

now becomes 0010 0101 2. In last step replace each 4 bit binary group by its equivalent hex

digit. i.e, 0010=2 and 0101=5.

Therefore, 100101 2 = 0010 0101 2 = 25 H

The digit, whether it is an integer part or fractional part must be represented by a group of 4-bits.

For integer part, 4-bit groups are formed starting from the hexadecimal point and moving

towards left. For the fractional part, 4-bit groups are formed starting from the hexadecimal point

and moving towards right.

Example 1.6

Convert the following binary numbers to hexadecimal numbers.

(a) (101011)2 (b) (11110.111)2 (c) (111011011.111011)2

Solution:


(a) (101011)2 = (0010 1011)2

= ( 2 B )H

= (2B)H

(b) (11110.111)2 = (0001 1110.1110)2

= ( 1 E . E )H

= (1E.E)H

(c) (111011011.111011)2 = (0001 1101 1011.1110 1100)2

= ( 1 D B . E G )H

= (1DB.EG)H

1.10 OCTAL NUMBER SYSTEM

The octal number system has a base 8. The basic digits used are 0, 1, 2, 3, 4, 5, 6, 7. The

base of the octal number system is 8 and the base of the binary number system is 2. Since 28 =

8, it follows that any octal digit can be represented by a group of 3-bit binary sequence. The

following table (1.2) gives the decimal, octal and the three bit binary equivalence for the decimal

numbers 0 to 7.

Decimal

Octal

Binary

22 + 21 + 20

0 0 000

1 1 001

2 2 010

3 3 011

4 4 100

5 5 101

6 6 110

7 7 111

1.11 OCTAL TO DECIMAL CONVERSION

The octal to decimal conversion is similar to binary to decimal conversion, only the

weights are different. In this case, the weights used are 80, 81, 82, etc., for the integer part 8−1,

8−2, etc., for the fractional part. In octal number system, we have an octal point to separate the

integer and fractional parts.


Example 1.7

Convert the following octal numbers to decimal.

(a) (62)8 (b) (45.6)8

Solution:

(a) (6 2)8

2x80 = 2x1 = 2

6x 81 = 6x8 = 48

(62)8 = (50)10

(b) (4 5 .6)8

6x8−1 = 6x0.125 = 0.75

5x 80 = 5x1 = 5

4x81 = 4x8 = 32

(45.6)8 = (37.75)10

1.12 DECIMAL TO OCTAL CONVERSION

To convert a decimal number to octal, we have to divide the decimal number by 8

repeatedly and collect the remainders from top to bottom (for the integer part). The remainders

also must be taken in octal.

Example 1.8

Convert the following decimal numbers to octal equivalents.

(a) 109 (b) 98.625

Solution:

(a) 8 109

13 - 5 (LSB)

1 - 5

0 - 1 (MSB)

(109)10 = (155)8


(b) The octal number 98.625 is separated as the integer part 98 and fractional part .625. The

integer 98 is converted into octal by repeated division by 8.

8 98

12 - 2 (LSB)

1 - 4

0 - 1 (MSB)

(98)10 = (142)8

The fraction 0.625 is converted into octal by multiplying 0.625 by 8 and collecting the carry.

0.625x8 = 5.000; carry is 5.

Therefore, (98.625)10 = (142.5)8

1.13 OCTAL TO BINARY CONVERSION

To convert a decimal number to binary we have adopted a procedure of repeatedly

dividing the given decimal number by 2. Since the base of octal number system is 8 which is

equal to 23, to convert an octal number to binary, all we have to do is replace each octal digit

with its equivalent 3-bit binary. Using table 1.3, we can write the binary equivalent of any octal

number. We can see that for 3-bit binary combination, the weights required are only 22, 21 and

20.

Octal Binary Decimal

27 010 111 23

135 001 011 101 93

45.5 100 101 . 101 37.625

1.14 BINARY TO OCTAL CONVERSION

To convert a binary number to octal, we have to arrange the bits into group of 3 bits

starting from LSB (Least Significant Bit). If the final group has less than 3 bits, just include

zeros to make it a group of 3 bits. For example, to convert (10 101) into octal, arrange the bits as

(10 101). Now include a zero for the first group at the front. The binary combination now

becomes (010 101). In the last step, replace each 3-bit binary group by its equivalent octal digit.

i.e. 010 = 2 and 101 =5.

Therefore, 10 101 2 = 010 101 2 = 25 8


The digit, whether it is an integer part or fractional part must be represented by a group of 3-bits.

For integer part, 3-bit groups are formed starting from the octal point and moving towards left.

For the fractional part, 3-bit groups are formed starting from the octal point and moving towards

right.

Example 1.10

Convert the following binary numbers to hexadecimal numbers.

(a) (101011)2 (b) (11110.11)2 (c) (11011011.1111)2

Solution:

(a) (101011)2 = (101 011)2

= ( 5 3 )8

= (53)8

(b) (11110.11)2 = (011 110.110)2

= ( 3 6 . 6 )8

= (36.6)8

(c) (11011011.1111)2 = (011 011 011.111 100)2

= ( 3 3 3 . 7 4 )8

= (333.74)8


BINARY ARITHMETIC

2.1. 1’s AND 2’s COMPLEMENTS

There are two types of complements for binary number system, namely, 1’s complement

and 2’s complement. 1’s complement and 2’s complement can be used to perform subtraction

using adder circuits. Also they are used to represent negative numbers.

1’s Complement:

The 1’s complement of the binary digit (bit) is defined as 1 minus that bit.

i.e 1’s complement of 1 is 1 – 1=0

1’s complement of 0 is 1 – 0 =1

However, it is easy to remember that 1’s complement of 0 is 1 and 1’s complement of 1 is 0. In

other words, 1’s complement of any binary number is formed by simply changing all the 1’s to

0’s and all 0’s to 1’s.

Few examples are given below:

Binary Number 1’s Complement

1011 0100

110001 001110

100100 011011

11001110 00110001

10101101 01010010

2’s Complement:

2’s complement of a binary number is formed by adding 1 to the 1’s complement of that

number. A 1 is added to the least significant bit position. For example, let us find the 2’s

complement of 1010.

1’s complement of 1010 = 0 1 0 1 +

Add 1 = 1

2’s complement of 1010 = 0 1 1 0


A few more examples are given below

BOOLEAN ALGEBRA

For better understanding of digital systems one must have through knowledge in Boolean

algebra and De – morgan’s theorems.

Boolean algebra invented by George Boole, provides a simple mathematical technique

for the design, simplification and analysis of digital switching circuits, which are very complex

for visual analysis. The technique also gives a method of representing different switches and

logic function in mathematical form.

Boolean algebra will be useful in the following ways:

1. The logic functions may be expressed as algebraic equations offering more compactness

and clarity than the truth tables of logic gates.

2. Am algebraic equation can be obtained from a truth table by using the Boolean algebra.

3. A Boolean equation may be simplified, leading to an understanding of implementation of

a desired logic function.

Boolean algebra requires the use of a set of special symbols given below.

1. Symbols for constants 0 and 1

2. Symbols for variables A, B, C, X, Y, Z, W, Q, etc.,

3. Symbols for funtions; AND function (.)

OR function (+)

NOT function (-)

4. Symbols for equality (=)

Boolean postulates

The Boolean postulates originate from the three basic logic functions of AND, OR and

NOT, whose truth tables are shown below.

Binary Number 1’s Complement 2’s Complement

1011 0100 0101

110001 001110 001111

100100 011011 011100

11001110 00110001 00110010

10101101 01010010 01010011


AND

A B C

0 0 0

0 1 0

1 0 0

1 1 1

The postulates are expressed as algebraic equations, using the appropriate functions symbols as

given below.

Algebraic equations

AND function OR function NOT function

0.0 = 0 0+0 = 0 0 = 1

1.0 = 0 0+1 = 1 1 = 0

1.0 = 0 1+0 = 1 -

1.1 = 1 1+1 = 1 -

OR

A B C

0 0 0

0 1 1

1 0 1

1 1 1

NOT

A Y

0 1

1 0


Laws of Boolean Algebra:

i) Commutative Law

A + B = B + A

A . B = B . A

ii) Associative Law

A + (B + C) = (A + B) + C

A . (B . C) = (A .B) . C

iii) Distributive Law

A . (B + C) = AB + AC


DeMorgan’s Laws:

De morgan’s laws are helpful to see how an expression using AND operation can be

converted into equivalent expression using OR operation and vice versa.

Any logic expression can be inverted (or complement) using simply by inverting every

terms so that A becomes𝐴 , B becomes 𝐵 and every AND becomes OR and vice – versa.

This is stated mathematically by Demorgan’s laws as follows.

LAW I:

The complement of sum equals the product of complements.

The law reveals that a NOR gate 𝐴 + 𝐵 is equivalent to bubbled AND gate 𝐴 . 𝐵

𝐴 + 𝐵 = 𝐴 . 𝐵

LAW II:

The complement of product equals the sum of complements.

The law reveals that a NAND gate 𝐴. 𝐵 is equivalent to bubbled OR gate 𝐴 + 𝐵

𝐴. 𝐵 = 𝐴 + 𝐵

Proof:

Case (i): A = 0, B = 0

LHS; 𝐴. 𝐵 = 0.0 = 0 = 1

RHS; 𝐴 + 𝐵 = 0 + 0 = 1 + 1 = 1

Case (ii): A = 0, B = 1


LHS; 𝐴. 𝐵 = 0.1 = 0 = 1

RHS; 𝐴 + 𝐵 = 0 + 1 = 1 + 0 = 1

Case (iii): A = 1, B = 0

LHS; 𝐴. 𝐵 = 1.0 = 0 = 1

RHS; 𝐴 + 𝐵 = 1 + 0 = 0 + 1 = 1

Case (iv): A = 1, B = 1

LHS; 𝐴. 𝐵 = 1.1 = 1 = 0

RHS; 𝐴 + 𝐵 = 1 + 1 = 0 + 0 = 0

In all the four cases, 𝐴. 𝐵 = 𝐴 + 𝐵

Using Truth table

Duality principle:

1) A + 0 = A

2) A + 1 = 1

3) A + A= A

4) A + 𝐴 = 1

5) 𝐴 = A

6) 𝐴 . 0 =0

7) 𝐴 . 1 = A

8) 𝐴 . A = A

A B A.B 𝐴. 𝐵 𝐴 𝐵 𝐴 + 𝐵

0 0 0 1 1 1 1

0 1 0 1 1 0 1

1 0 0 1 0 1 1

1 1 1 0 0 0 0


9) 𝐴 . 𝐴 = 0

10) 𝐴 = A

A + AB = A

A . (1 + B) = A . 1 = A

A(A + B) = A . A + A . B

= A + AB

= A

Exercise: 1

Y = 𝐴 𝐵 + 𝐴 B

= 𝐴 (𝐵 . B)

= 𝐴 . 1

Y = 𝐴

𝐴 𝐵 𝐶

Y


Exercise: 2

𝐴 𝐵 C + A 𝐵 C + A B 𝐶 + A B C = AB + 𝐵 C

Solution:

LHS = 𝐴 𝐵 C + A 𝐵 C + A B 𝐶 + A B C

= 𝐵 C (𝐴 + 𝐴) + AB (𝐶 + C)

= 𝐵 C.1 + AB.1

= AB + 𝐵 C

LHS = RHS

Hence, it is proved.

𝑩 . 𝑪 𝐵 . 𝐶

𝐵 . 𝐶

C𝐵 . 𝐶

𝐵 . 𝐶

Y = A.B + 𝑩 . 𝑪


PROBLEMS:

F (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 + 𝐴 𝐵 𝐶 + 𝐴 B C

= A B (C + 𝐶 ) + 𝐴 𝐵 (C + 𝐶 ) + 𝐴 B C

= A B.1 + 𝐴 𝐵 .1 + 𝐴 B C

= A B + 𝐴 𝐵 + 𝐴 B C

= 𝐴 𝐵 + B (A + 𝐴 C)

= 𝐴 𝐵 + B (A + C)

= 𝐴 𝐵 + A B + B C

Exercises:

1) Show that 𝐴 B + 𝐵 C + 𝐶 A = A 𝐵 + B 𝐶 + C 𝐴

Solution:

SOP = Sum of Product

LHS = 𝐴 B + 𝐵 C + 𝐶 A

= 𝐴 B .1+ 𝐵 C.1 + 𝐶 A.1

= 𝐴 B.(C + 𝐶 )+ 𝐵 C.(A + 𝐴 ) + 𝐶 A.(B + 𝐵 )

= 𝐴 B C + 𝐴 B 𝐶 + 𝐵 C A + 𝐵 C 𝐴 +𝐶 A B + 𝐶 A 𝐵

= 𝐴 B C + 𝐴 B 𝐶 + A 𝐵 C + 𝐴 𝐵 C + A B 𝐶 + A 𝐵 𝐶

= A 𝐵 (C +𝐶 ) + B 𝐶 (𝐴 + A) + C 𝐴 (B + 𝐵 )

= A 𝐵 + B 𝐶 + C 𝐴 LHS = RHS

Hence it is proved.


POS:

F (A,B,C)= (A + B + C) . (𝐴 +𝐵 + 𝐶 )

2) Show that (𝐴 + B) (𝐵 + C) (𝐶 + A ) = (A + 𝐵 ) (B +𝐶 ) (C + 𝐴 )

LHS =(𝐴 + B) (𝐵 + C) (𝐶 + A )

= (𝐴 𝐵 + 𝐴 C + B 𝐵 + B C ) (𝐶 + A)

= (𝐴 𝐵 + 𝐴 C + B C ) (𝐶 + A)

= 𝐴 𝐵 𝐶 + 𝐴 C 𝐶 + B C 𝐶 + 𝐴 𝐵 . A + 𝐴 C . A + B C . A

= 𝐴 𝐵 𝐶 + A B C

RHS = (A + 𝐵 ) ( B + 𝐶 ) (C + 𝐴 )

3) Show that 𝐴 B C + A 𝐵 C + A B 𝐶 + A B C = A B +B C + C A

LHS =𝐴 B C + A 𝐵 C + A B 𝐶 + A B C

= 𝐴 B C + A 𝐵 C + A B (𝐶 + C)

= 𝐴 B C + A 𝐵 C + A B

= 𝐴 B C + A (𝐵 C + B) therefore, 𝐵 C + B = B + C

= 𝐴 B C + A ( B + C)

= 𝐴 B C + A B + A C

= B (𝐴 C + A ) + A C therefore 𝐴 C + A = A + C

= B (A + C) + A C

= A B +B C + C A

= RHS


Exercise: 3

F(A, B, C, D) = ABC + ABD + 𝐴 B 𝐶 + CD + B 𝐷

= ABC + ABD + B 𝐷 + 𝐴 B 𝐶 + CD

= ABC + B (AD + 𝐷 ) + 𝐴 B 𝐶 + CD AD + 𝐷 = A + D

= ABC + B (A + D) + 𝐴 B 𝐶 + CD

= ABC + BA + BD + 𝐴 B 𝐶 + CD

= AB (C + 1) + BD + 𝐴 B 𝐶 + CD

= AB + BD + 𝐴 B 𝐶 + CD

= B(A + 𝐴 𝐶 ) + BD + CD

= B(A + 𝐶 ) + BD + CD

= AB + B 𝐶 + BD + CD

= AB + B 𝐶 + BD (𝐶 + C) + CD

= AB + B 𝐶 + BD 𝐶 + BDC + CD

= AB + B 𝐶 (D + 1) + CD (B + 1)

= AB + B 𝐶 + CD

Logic Gates:

A digital circuit with one or more input pulse voltages but only one output pulse voltage

is called Gate. They are normally used to make logical decisions.

The most basic Gates are the AND gate, the OR gate and the NOT gate. By connecting

these gates in different ways, we can construct with the human brain. Since these circuits

simulate mental processes, the gates are often called logic gates.

The basic logic gates are:

1. The AND gate,

2. The OR gate,

3. The NOT gate.

The AND gate:

Consider the series electrical circuit shown in fig. The bulb will lit only if both switches

are closed. This circuit action is called the logical AND functions.


The AND gate is a logic circuit which gives an output of 1 state only when all the

inputs are in 1 state.

The logic symbol and the truth table (one which gives the summary of possible actions) for two

inputs A and B are given below.

The output is given by Y = A.B. Which is read as Y equals A and B. This is Boolean expression

for the output of AND gate.

The truth table shows that the output is 1 state only when both the inputs A and B are in the 1

state. In the expression Y = A.B the dot (.) indicates the AND functions.

The OR gate:

Consider the circuit shown in the fig. The bulb will be lit when weather or both of the

parallel switches are closed. This circuit action is called the logical OR functions.

Inputs Outputs

A B Y = A.B

0 0 0

0 1 0

1 0 0

1 1 1


The OR gate is a logic circuit which gives an output of ‘1’ state if one or more inputs

are in ‘1’ state. The logic symbol, Truth table and Boolean expression for two input OR gate

are given below.

The output is given by Y = A + B, which is read as Y equals A and B. This is Boolean

expression for the output of OR gate. The truth table shows that the output is 1 state when either

or both of the inputs A and B are in the 1 state when either or both of the inputs A and B are in

the state. In the expression for A + B (+ plus) indicates the OR function. So the plus sign is read

as OR.

The NOT gate:

The NOT gate has only one input and one output. Its function is to invert the imput so

that the output is always opposite. If the input is 1, the output is 0; if the input is 0, the output

will be 1. The NOT gate is commonly known as complementary circuit or an inverter. Thus the

NOT gate is an inverter circuit which is used to negative signal.

The logic symbol, the truth table and the Boolean expression of the NOT gate are given

below.

Inputs Outputs

A B Y = A + B

0 0 0

0 1 1

1 0 1

1 1 1


The output is in the 1 state when the input is in the 0 state and vice-versa. This result can be

stated as,

Y = 𝐴 in Boolean notation

𝐴 (A bar or not A) is the negation of A

The NOT logic gate for positive logic can be implemented using an NPN transistor as follows.

The NAND gates as universal building block:

The NAND logic gate is a negated AND logic gate i.e., when NOT gate follows an AND

gate, the combination is called the NAND gate.

The logic symbol the truth table and the Boolean expression for the output of the NAND

are given below.

The Boolean expression for the output is Y equals A dot B the whole bar i.e., not (A and

B). The bubble or complementing circle in the symbol indicates inversion of the AND.

The chip IC 7400 is a quad – two – input NAND gate.

Input Output

A Y = 𝐴

0 1

1 0

Inputs Outputs

A B Y = 𝐴. 𝐵

0 0 1

0 1 1

1 0 1

1 1 0


The NAND gate can be regarded as universal building block because it can be connected to other

NAND gates to generate any logic function. The NOT, the AND, the OR logic gates can be

fabricated with the use of NAND gates only.

NOT gate using NAND gates:

In a NAND gate, let the inputs be A and B

The output Y = 𝐴. 𝐵 . If the inputs are same, each equal to A, then

Y = 𝐴. 𝐴

Y = 𝐴

For the input A, the output is 𝐴 . Thus, the function of the NOT gate is performed.

AND gate using NAND gates:

The inputs A and B are fed to NAND gate (1) and its output is 𝐴𝐵 . This is used to feed two equal

inputs, 𝐴𝐵 each to the NAND gate (2); The second NAND gate works as a NOT gate.

The output Y = 𝐴. 𝐵

Y = A.B

Thus, the function of an AND gate is carried out by the use of two NAND gates.

OR gate using NAND gates:

The input A is fed to the NAND gate which acts as an inverter (NOT gate). The output is 𝐴

similarly, the input B is fed to the NAND, which acts as an inverter, to get the output 𝐵 . Now, 𝐴

and 𝐵 are fed as inputs to the NAND. After the NAND operation, the output becomes.

Y = 𝐴 . 𝐵

Using Demorgan’s law

Y= 𝐴 + 𝐵

Y = A + B

Thus, the OR operation is carries out with three NAND gates only.

1.1 decimal number system the decimal number system …

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