Question

The logic operations performed by the given digital circuit is equivalent to:

• A. NAND
• B. NOR
• C. OR
• D. AND

Answer 1

The Correct Option is D

Step 1: Analyze the First NAND Gate

The output of the first (topmost) NAND gate is:

\[ \overline{A \cdot B}. \]

This output is also connected as an input to the second NAND gate (middle). Let this output be \( X_1 \):

\[ X_1 = \overline{A \cdot B}. \]

Step 2: Analyze the Second NAND Gate

The second NAND gate takes \( X_1 \) and the output of the third NAND gate (\( X_2 \)) as inputs. The output of the second NAND gate is:

\[ X_3 = \overline{X_1 \cdot X_2}. \]

From the first NAND gate, \( X_1 = \overline{A \cdot B} \), and from the third NAND gate, \( X_2 = \overline{A \cdot B} \). Substituting these values:

\[ X_3 = \overline{\overline{A \cdot B} \cdot \overline{A \cdot B}}. \]

Using the property of NAND gates:

\[ X_3 = A \cdot B. \]

Step 3: Analyze the NOT Gate

The NOT gate inverts the output of the second NAND gate. Let \( Y \) be the output of the NOT gate. Then:

\[ Y = \overline{X_3}. \]

Substituting \( X_3 = A \cdot B \):

\[ Y = A \cdot B. \]

Step 4: Simplify the Expression

Using Boolean algebra, the final output \( Y \) simplifies to:

\[ Y = A \cdot B. \]

This represents the logic for an AND gate.

Final Answer:

The given digital circuit is equivalent to an AND gate.