Question

The logic performed by the circuit shown in the figure is equivalent to: 

• A. \( AND \)
• B. \( NAND \)
• C. \( OR \)
• D. \( NOR \)

Hint:

De Morgan’s laws simplify logic circuits: \( \overline{A + B} = A \cdot B \) and \( \overline{A \cdot B} = A + B \). These transformations can turn complex circuits into simpler ones.

Answer 1

The Correct Option is A

The circuit executes the following logical operations:

1. Two initial NOT gates receive inputs \( a \) and \( b \), producing their respective complements \( \bar{a} \) and \( \bar{b} \).
[ \text{Output of the first NOT gate: } \bar{a}, \quad \text{Output of the second NOT gate: } \bar{b}. ]
2. An OR gate then combines these complements:
[ \text{Output of the OR gate: } \bar{a} + \bar{b}. ]
3. A subsequent NOT gate complements the OR gate's output, resulting in the final output:
[ Y = \overline{\bar{a} + \bar{b}}. ]

Applying De Morgan's law simplifies this to:

[ Y = a \cdot b. ]

Consequently, the circuit functions as an AND gate.

Truth Table:

[ \begin{array}{|c|c|c|} \hline A & B & Y = A \cdot B \\ \hline 0 & 0 & 0 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \\ 1 & 1 & 1 \\ \hline \end{array} ]

The circuit's output corresponds to the truth table of an AND gate.

Final Answer:

[ \boxed{\text{AND Gate}} ]