Question

Given the following truth table, derive a sum of product (SOP) and Product of Sum (POS) form of Boolean expression from it :

A	B	C	G(A,B,C)
0	0	0	0
0	0	1	1
0	1	0	1
0	1	1	0
1	0	0	0
1	0	1	1
1	1	0	0
1	1	1	1

Answer 1

The desired Canonical Sum-of-Product form is as following;

G = ∑(1, 2, 5, 7) = A’B’C + A’BC’ + AB’C + ABC 

The desired Canonical Product-of-Sum form is as following;

G = π(0, 3, 4, 6) = (A + B + C)(A + B’ + C’)(A’ + B + C)(A’ + B’ + C)