Solved Examples on Minterms
Example 1: If there are three Boolean variables A = 0, B = 0 and C = 1 then find the minterms for the given values.
Solution:
Given the values of the Boolean variables
A = 0, B = 0, C = 1
The required minterm is given by = A’B’C
We complemented A and B as its value is 0.
Example 2: Simplify SOP and find 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 a K-map to simplify SOP
Example 3: Find the SOP for F(A, B, C, D) = ∑m (4, 5, 7, 8, 9, 12, 14)
Solution:
What is Minterm ?
Minterms are the fundamental part of Boolean algebra. Minterm is the product of N literals where each literal occurs exactly once. Minterm is represented by m. The output for the minterm functions is 1. This article explores the minterms in depth in addition to the two-variable, three variable and four variable minterm tables and K-maps. We will also solve some examples based on Minterms.