Minterm vs Maxterm

Minterms and Maxterms are important parts of Boolean algebra. Minterm is the product of N distinct literals where each literal occurs exactly once. The output of the minterm functions is 1. Minterm is represented by m. Maxterm is the sum of N distinct literals where each literals occurs exactly once. The output of the maxterm functions is 0. Maxterm is represented by M. In this article we will learn about minterms and maxterms, their difference, why we use minterms and maxterms along with the solved examples.

What is Minterm?

Minterm is the product of various different literals in which each literal occurs exactly once. The output result of the minterm functions is 1. It is represented by m. To represent a function, we perform a sum of minterms also called the Sum Of Products (SOP).

Example of SOP:

A'B + AC + BC

Table for Two-Variable Minterm

Minterms for Values

Minterms for values is the minterms obtained by the values of the Boolean variable.

Steps for Obtaining Minterms from Values

1. If the value of the Boolean variable is 1 then we will take the variable without complementing it.
2. If the value of the Boolean variable is 0 then we will take the variable by complementing it.

Minterm Example:

If there are four Boolean variables A, B, C, D with the values A = 1, B = 0, C = 0 and D = 1. Find the minterms for values given.

Solution:

Given the values of the Boolean variables:

A = 1, B = 0, C = 0 and D = 1

The required minterm is given by = AB'C'D

We complemented B and C as its value is 0.

What is Maxterm?

Maxterm is the sum of various different literals in which each literal occurs exactly once. The output result of the maxterm functions is 0. It is represented by M. To represent a function, we perform product of maxterms also called Product of Sum (POS).

Example of POS:

(A+B). (C+D)

Table for Two Variable Maxterm

Maxterms for Values

Maxterms for values is the maxterms obtained by the values of the Boolean variable.

Steps for obtaining maxterms from values:

1. If the value of the Boolean variable is 0 then we will take the variable without complementing it.
2. If the value of the Boolean variable is 1 then we will take the variable by complementing it.

Maxterm Example:

If there are four Boolean variables A, B, C, D with the values A = 1, B = 0, C = 0 and D = 1. Find the maxterm for values given.

Solution:

Given the values of the Boolean variables:

A = 1, B = 0, C = 0 and D = 1

The required maxterm is given by = A'+ B + C + D'

We complemented A and D as its value is 1 and B and C are 0 therefore there are no change.

Minterm vs Maxterm

Below mentioned is the image of Four Variable K-Map.

Why we use Minterms and Maxterms ?

Minterms and Maxterms are used to express the Boolean functions in its canonical forms. The minterms and maxterms are the basic building blocks for representing the Boolean functions in the canonical forms. The minterms are the terms which are used when we require active high output. The maxterms are the terms which are used when we require active low output.

Solved Examples on Minterms and Maxterms

Example 1: For the Boolean variables A = 0, B = 1 and C = 0 obtain the minterm and maxterm for the variables.

Solution:

Given the values of Boolean variables as:

A = 0, B = 1 and C = 0

The required minterm for above values = A'BC'

The required maxterm for above values = A + B' + C

Logic gate diagram for above minterm and maxterm

Example 2: Simplify the POS form and obtain the result in POS form only.

F(A, B, C, D) = (A + B + C + D)(A + B' + C + D')(A + B' + C' + D)(A + B' + C' + D')(A' + B + C + D)(A' + B + C' + D')(A' + B' + C' + D')

Solution:

We draw K-Map to simplify the given POS

The simplified POS form F(A, B, C, D) = (B + C + D)(A + B' + D' )(A + B' + C')(A' + C' + D')

Logic gate diagram for above maxterm:

Example 3: Simplify the SOP form and obtain the result in SOP only.

F(A, B, C, D) = A'BC'D' + A'BC'D + A'BCD + AB'C'D' + AB'C'D + ABC'D' + ABCD'

Solution:

We draw K-Map to simplify the given SOP

The Simplified SOP form is F(A, B, C, D) = ABD' + AB'C' + A'BC' + A'BD

Logic gate diagram for above minterm: