Question

From the truth table below, determine SOP and POS expression.

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

Answer 1

From the truth table:

Output X = 1 for:

001, 011, 110

SOP Expression 

X = A′B′C + A′BC + ABC′

X = Σm(1, 3, 6)​

Simplified SOP:

X = A′C + ABC′

POS Expression

Output X = 0 for:

000, 010, 100, 101, 111

Therefore,

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

X = ΠM(0, 2, 4, 5, 7)​

Simplified POS:

X = (A + C)(A′ + B)(A′ + C′)