Sum of Products (SOP) and Minterms

Start here You have a truth table, and you want a formula. Here’s the trick: look only at the rows where the output is 1. Write a small “detector” for each of those rows, then OR all the detectors together. That’s it. You’ve written a Sum of Products.

Rule: if the input is 1, write the letter; if it’s 0, write the letter with a bar.

The number is just the row’s binary value (011 = 3, so $m_{3}$).

Each minterm is 1 in exactly one row: its own row. That’s why it works as a “row detector”.

1. Find every row where $F = 1$.
2. Write the minterm for each of those rows.
3. OR them all together.

Example:

$F = \overline{A} B C + A \overline{B} \overline{C} + A \overline{B} C + A B \overline{C} + A B C$

Short form ($\Sigma$ notation): $F \left(A , B , C\right) =$ $\Sigma m \left(3 , 4 , 5 , 6 , 7\right)$, the list of rows where $F = 1$.

Take input $A = 1 , B = 0 , C = 1$ (row 5). Only the minterm $A \overline{B} C$ becomes 1 (1·1·1). Every other minterm has at least one 0 in it, so it’s 0. So $F = 0 + \ldots + 1 + \ldots + 0 =$ 1. Correct!

For a row that’s not in the list, no minterm turns on, so $F = 0$. Correct again.

• A truth table is the unique “signature” of a function, but it’s long.
• Many different expressions can give the same truth table (and different gate circuits).
• A canonical (standard) form, like the full SOP of minterms, gives one standard expression for each function.
• Every Boolean function can be written in SOP form.

SOP is a two-level circuit: a row of AND gates, then one OR gate.

In one line: For each row where $F = 1$, write a minterm (letter for 1, barred letter for 0) and OR them together. That’s the canonical Sum of Products, written short as $\Sigma m \left(\ldots \right)$. Exam words: literal, implicant, product term, minterm, maxterm, sum of products $\left(S O P\right) , \Sigma m$ notation, canonical form, two-level logic.