Boolean Algebra is a kind of algebra that is used to analyze and simplify the logic circuits. Boolean algebra is frequently used in digital electronics. It is also known as Binary Algebra of Logical Algebra.
It was developed by English mathematician George Boole in the in 1854.
It uses only two variables 1 to represent High and 0 to represent Low.
It uses three logical operators namely AND, OR, NOT . It represents the relationships between variables.
The variables that can take only two values i.e True and False where True denotes to 1 and False Denotes to 0.
A Boolean variable is a variable that can contain only two types of values such as 0 and 1 where 0 denotes to FALSE and 1 denotes to TRUE. Boolean variables are used in Boolean Functions or expressions.
Boolean algebra contains three Boolean operators to express boolean functions. It is also known as Logical operators. The Boolean Operators are AND , OR, NOT that are used to performs logical operations.
This operator operates on two or more operands. It is used to perform logical Multiplication in Boolean Algebra. It is denoted by symbol dot( . )
| A | B | Y(A.B) |
| 0 | 0 | 0 |
| 0 | 1 | 0 |
| 1 | 0 | 0 |
| 1 | 1 | 1 |
This operator operates on two or more operands. It is used to perform logical Addition in Boolean Algebra. It is denoted by symbol Plus( +).
| A | B | Y(A+B) |
| 0 | 0 | 0 |
| 0 | 1 | 1 |
| 1 | 0 | 1 |
| 1 | 1 | 1 |
It is an unary operator. It requires one operand to operate . This operator is also known as inverter or complement. It returns 1 for 0 and vice versa(0 for 1).
It is denoted by symbol tilde(~) , Bar( -) or single inverted comma (‘).
| A | A’ |
| 0 | 1 |
| 1 | 0 |
A Truth table is a table that shows all the Possibilities of a Logic Circuit.
A Truth Table is a tabular form that shows all the possible values of logical variables with possible outcomes for the given condition.
Example
A Literal is a variable or it’s complement such as A, B or A’ etc. used in Boolean Expressions. The number of literals in a Boolean function is minimized by Algebraic manipulation to prepare a Circuit.
A Product term of all the variables in the Sum-Of-Product form in the expression.
Example AB+A’B+AB’
Rules for Obtaining Minterms
| A | B | Minterm | Designator |
| 0 | 0 | A’B’ | 0 |
| 0 | 1 | A’B | 1 |
| 1 | 0 | AB’ | 2 |
| 1 | 1 | AB | 3 |
| A | B | C | Minterm | Designator |
| 0 | 0 | 0 | A’B’C’ | 0 |
| 0 | 0 | 1 | A’B’C | 1 |
| 0 | 1 | 0 | A’BC’ | 2 |
| 0 | 1 | 1 | A’BC | 3 |
| 1 | 0 | 0 | AB’C’ | 4 |
| 1 | 0 | 1 | AB’C | 5 |
| 1 | 1 | 0 | ABC’ | 6 |
| 1 | 1 | 1 | ABC | 7 |
| A | B | C | D | Minterm | Designator |
| 0 | 0 | 0 | 0 | A’B’C’D’ | 0 |
| 0 | 0 | 0 | 1 | A’B’C’D | 1 |
| 0 | 0 | 1 | 0 | A’B’CD’ | 2 |
| 0 | 0 | 1 | 1 | A’B’CD | 3 |
| 0 | 1 | 0 | 0 | A’BC’D’ | 4 |
| 0 | 1 | 0 | 1 | A’BC’D | 5 |
| 0 | 1 | 1 | 0 | A’BCD’ | 6 |
| 0 | 1 | 1 | 1 | A’BCD | 7 |
| 1 | 0 | 0 | 0 | A’BCD | 8 |
| 1 | 0 | 0 | 1 | AB’C’D | 9 |
| 1 | 0 | 1 | 0 | AB’CD’ | 10 |
| 1 | 0 | 1 | 1 | AB’CD | 11 |
| 1 | 1 | 0 | 0 | ABC’D’ | 12 |
| 1 | 1 | 0 | 1 | ABC’D | 13 |
| 1 | 1 | 1 | 0 | ABCD’ | 14 |
| 1 | 1 | 1 | 1 | ABCD | 15 |
Q. Find Minterm.
A+B
=A.1+B.1
=A(B+B’)+B(A+A’) Inverse law[A+A’=1]
=AB+AB’+AB+A’B Idempotent Law[A+A=A]
=AB+AB’+A’B
Q.2 A+AB
=A(B+B’)+AB
=AB+AB’+AB
=AB+AB’
Shorthand Minterm Notation
Q1. Find the minterm of AB’C’D
Solution: AB’C’D
= 1 0 0 1
=9
m9
A sum term of all the variables in the Product of Sum form in the expression. In Maxterm all complements(Barred) variable is represented by 1 and unbarred number is represented by 0.
example (A+B+C’)(A+B’+C)
| A | B | Maxterm | Designator |
| 0 | 0 | A+B | 0 |
| 0 | 1 | A+B’ | 1 |
| 1 | 0 | A’+B | 2 |
| 1 | 1 | A’+B’ | 3 |
| A | B | C | Maxterm | Designator |
| 0 | 0 | 0 | A+B+C | 0 |
| 0 | 0 | 1 | A+B+C’ | 1 |
| 0 | 1 | 0 | A+B’+C | 2 |
| 0 | 1 | 1 | A+B’+C’ | 3 |
| 1 | 0 | 0 | A’+B+C | 4 |
| 1 | 0 | 1 | A’+B+C’ | 5 |
| 1 | 1 | 0 | A’+B’+C | 6 |
| 1 | 1 | 1 | A’+B’+C’ | 7 |
| A | B | C | D | Maxterm | Designator |
| 0 | 0 | 0 | 0 | A+B+C+D | 0 |
| 0 | 0 | 0 | 1 | A+B+C+D’ | 1 |
| 0 | 0 | 1 | 0 | A+B+C’+D | 2 |
| 0 | 0 | 1 | 1 | A+B+C’D’ | 3 |
| 0 | 1 | 0 | 0 | A+B’+C+D | 4 |
| 0 | 1 | 0 | 1 | A+B’+C+D’ | 5 |
| 0 | 1 | 1 | 0 | A+B’+C’+D | 6 |
| 0 | 1 | 1 | 1 | A+B’+C’+D’ | 7 |
| 1 | 0 | 0 | 0 | A’+B+C+D | 8 |
| 1 | 0 | 0 | 1 | A’+B+C+D’ | 9 |
| 1 | 0 | 1 | 0 | A’+B+C’+D | 10 |
| 1 | 0 | 1 | 1 | A’+B+C’+D’ | 11 |
| 1 | 1 | 0 | 0 | A’+B’+C+D | 12 |
| 1 | 1 | 0 | 1 | A’+B’+C+D’ | 13 |
| 1 | 1 | 1 | 0 | A’+B’+C’+D | 14 |
| 1 | 1 | 1 | 1 | A’+B’+C’+D’ | 15 |
Q. Find out the designator of the maxterms A’+B’+C
1+1+0
m(6)
A Boolean expression composed completely either of minterms or maxterms refers to canonical expression. A canonical expression provides two ways to represents itself.
a. Sum of Product(S-O-P)
b. Product of Sum (P – O -S)
A Boolean Expression when composed of purely minterms is called Sum of Product(S-O-P) expression.
example F=A’BC +AB’C+ABC’
A Boolean Expression when composed of purely maxterms is called Product of Sum (P – O -S) expression.
example F=(A’ +B+C ) ( A+ B’+C) ( A+B+C’)
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