Question

Consider the following logic circuit diagram. 

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

Hint:

Using Boolean algebra simplification techniques and Karnaugh maps can help verify logic expressions efficiently.

Answer 1

The Correct Option is A, B, C

Step 1: Identifying the Logic Gates
- The circuit consists of AND, OR, and NOT gates.
- Let's define intermediate expressions based on the given circuit.
Step 2: Deriving the Boolean Expression
1. First AND gate:
- Inputs: \( X \) and \( Y \)
- Output: \( X Y \)
2. NOT gates:
- \( X \) passes through a NOT gate, giving \( \overline{X} \).
- \( Y \) passes through a NOT gate, giving \( \overline{Y} \).
3. Second AND gate:
- Inputs: \( \overline{X} \) and \( Y \)
- Output: \( \overline{X} Y \)
4. Third AND gate:
- Inputs: \( X \) and \( \overline{Y} \)
- Output: \( X \overline{Y} \)
5. Final OR gate:
- Inputs: \( X Y \), \( \overline{X} Y \), and \( X \overline{Y} \).
- Output: \[ F = X Y + \overline{X} Y + X \overline{Y} \]
Step 3: Verifying the Options
- Option (A): \( \overline{X} Y \) is part of \( F \), so it is correct.
- Option (B): \( \overline{X} + \overline{Y} + X Y \) simplifies to the same Boolean function as \( F \), hence correct.
- Option (C): \( X Y + \overline{X} + X \overline{Y} \) is another valid form of \( F \). It simplifies to the same expression as \( F \), hence correct.
- Option (D): \( X + \overline{Y} \) does not match \( F \), so it is incorrect.

Thus, the correct answers are (A), (B), and (C).