Question

Examine the given truth table and then write both SOP & POS boolean expressions describing the output.

A	B	C	Output
0	0	0	1
0	0	1	0
0	1	0	1
0	1	1	0
1	0	0	0
1	0	1	1
1	1	0	1
1	1	1	0

Answer 1

For the given truth table:

Output = 1 for:

000, 010, 101, 110

SOP Expression

F = A′B′C′ + A′BC′ + AB′C + ABC′ 

F = Σm(0, 2, 5, 6)​

Simplified SOP:

F = A′C′ + BC′ + AB′C

POS Expression

Output = 0 for:

001, 011, 100, 111 

F = (A + B + C′)(A + B′ + C′)(A′ + B + C)(A′ + B′ + C′)

F = ΠM(1, 3, 4, 7)

Simplified POS:

F = (A + C′)(B′ + C′)(A′ + B + C)