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Boolean Algebra

Question
CBSEENCO12011700

Derive a Canonical SOP expression for a Boolean function F, represented by the following truth table :

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

Solution

F(A,B,C) = A’B’C’ + A’BC + AB’C’ + ABC