F(A,B,C) = A — SOP and POS
exam standardexpanding a term with 2 missing variables
Answer
Σm(4,5,6,7) = ΠM(0,1,2,3)
Why this example is worth doing
A single literal that has to be expanded over three variables, producing four minterms. The interesting part is that the variable list is a parameter of the question, not a property of the expression: the same F = A is Σm(1) over one variable, Σm(2,3) over two and Σm(4,5,6,7) over three. The page therefore requires the variable order to be declared explicitly and shows how the minterm indices shift when a variable is added, which is where index lists copied between problems go wrong.
Try your own input in the SOP & POS canonical forms. Expand to canonical minterms and maxterms, and convert between the two.
How the answer is reached
Canonical sum of products
A · B′ · C′ + A · B′ · C + A · B · C′ + A · B · C
Canonical product of sums
(A + B + C) · (A + B + C′) · (A + B′ + C) · (A + B′ + C′)
Truth table
| # | A | B | C | F |
|---|---|---|---|---|
| 0 | 0 | 0 | 0 | 0 |
| 1 | 0 | 0 | 1 | 0 |
| 2 | 0 | 1 | 0 | 0 |
| 3 | 0 | 1 | 1 | 0 |
| 4 | 1 | 0 | 0 | 1 |
| 5 | 1 | 0 | 1 | 1 |
| 6 | 1 | 1 | 0 | 1 |
| 7 | 1 | 1 | 1 | 1 |