Question

For the logic gate shown in the diagram, find the output \(Y\) for the given inputs \(A\) and \(B\).

• A. \(A\cdot \overline{B}\)
• B. \(\overline{A} + \overline{B}\)
• C. \(A + B\)
• D. \(\overline{A\cdot B}\)

Answer 1

The Correct Option is B

Concept: In digital electronics: • An AND gate gives output \(A\cdot B\). • A NOR gate gives output \( \overline{A+B} \). • De Morgan's Theorem states: \[ \overline{A+B} = \overline{A}\cdot\overline{B}, \qquad \overline{AB} = \overline{A} + \overline{B} \] Using these relations, complex logic circuits can be simplified.
Step 1:Identify the outputs of the AND gates. From the circuit, \[ P = A \cdot B \] \[ Q = A \cdot B \] Both gates are AND gates producing the same output.
Step 2:Determine the final gate operation. The outputs \(P\) and \(Q\) are fed into a NOR gate. \[ Y = \overline{P + Q} \] Substitute \(P\) and \(Q\): \[ Y = \overline{(A\cdot B) + (A\cdot B)} \]
Step 3:Simplify the Boolean expression. \[ Y = \overline{A\cdot B} \] Using De Morgan's theorem: \[ \overline{A\cdot B} = \overline{A} + \overline{B} \] \[ \boxed{Y = \overline{A} + \overline{B}} \]