Question

The logic circuit given below converts a binary code \(Y_1, Y_2, Y_3\) into _______.

 

• A. Excess-3 code
• B. Gray code
• C. BCD code
• D. Hamming code

Hint:

Gray code? Think XOR-based conversion: First bit unchanged, rest are XORs of adjacent bits.

Answer 1

The Correct Option is B

Step 1: Observe the Logic Circuit
The given circuit shows XOR gates arranged such that:
- \( Z_1 = Y_1 \)
- \( Z_2 = Y_1 \oplus Y_2 \)
- \( Z_3 = Y_2 \oplus Y_3 \)
Step 2: Gray Code Conversion Rule
Binary to Gray code conversion is done using:
- First bit remains the same: \( G_1 = B_1 \)
- Rest: \( G_i = B_i \oplus B_{i-1} \)
This matches exactly with the logic implemented in the circuit.
Step 3: Rule Out Other Codes
- Excess-3 → add 3 to BCD, needs binary adders.
- BCD → Binary Coded Decimal, no XOR used.
- Hamming code → error correction, uses parity bits.
Conclusion:
Option (2) is correct — the circuit converts binary to Gray code.