Question

The inputs to a digital circuit are as shown below. The output Y is

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

Hint:

To verify Boolean logic circuits, you can substitute test values (like \( A=1, B=1, C=1 \)) and trace them through the gates. For these inputs, the NAND gate gives 0, the NOT gate gives 0, and the OR gate outputs 0. Substituting these into the correct option (C) gives \( 0 + 0 + 0 = 0 \), which confirms the expression is correct.

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
The given logic circuit contains a NAND gate and a NOT gate feeding into an OR gate. We need to find the final Boolean expression for the output \( Y \) in terms of the inputs \( A \), \( B \), and \( C \).

Step 2: Key Formula or Approach:
We will find the output of each individual logic gate step-by-step and then use De Morgan's laws to simplify the final Boolean expression:
- NAND gate output for inputs \( A \) and \( B \): \( \overline{A \cdot B} \)
- NOT gate output for input \( C \): \( \bar{C} \)
- OR gate output: \( Y = \text{Input}_1 + \text{Input}_2 \)
- De Morgan's Law: \( \overline{A \cdot B} = \bar{A} + \bar{B} \)

Step 3: Detailed Explanation:
1. The top gate is a NAND gate with inputs \( A \) and \( B \). Its output is:
\[ X_1 = \overline{A \cdot B} \]
2. The bottom gate is a NOT gate with input \( C \). Its output is:
\[ X_2 = \bar{C} \]
3. These two intermediate outputs are passed as inputs to the final OR gate. The output \( Y \) is:
\[ Y = X_1 + X_2 = \overline{A \cdot B} + \bar{C} \]
4. Applying De Morgan's Law to the first term:
\[ \overline{A \cdot B} = \bar{A} + \bar{B} \]
Substituting this back into the expression for \( Y \):
\[ Y = \bar{A} + \bar{B} + \bar{C} \]

Step 4: Final Answer:
The Boolean expression for the output \( Y \) is \( \bar{A} + \bar{B} + \bar{C} \).