The output Y for the inputs A and B of circuit is given by Truth table of the shown circuit is :

Question

Aicharya · Accepted Answer

Output using DeMorgan's theorem can be written as,

$Y = ((A \cdot A \cdot B) \cdot (A \cdot B \cdot B)) = (A \cdot A \cdot B) + (A \cdot B \cdot B) = (A \cdot A \cdot B) + (A \cdot B \cdot B) = (A \cdot (A + B)) + ((A + B) \cdot B) = A \cdot A + A \cdot B + A \cdot B + B \cdot B = (A + B) \cdot (A + B)$

Which represents XOR gate.

Therefore, required truth table will be

$[\begin{matrix} A & B & Y \\ 0 & 0 & 0 \\ 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \end{matrix}]$

The output Y for the inputs A and B of circuit is given by Truth table of the shown circuit is :

Solution