TIFR ECE 2011 | Question: 7

Question

Assume you are using a binary code error correcting code $C$. If the minimum Hamming distance between any two codewords of $C$ is $3$. Then

• A. We can correct and detect $2$ bit errors.
• B. We can correct $1$ bit errors and detect $2$ bit errors.
• C. We can correct $2$ bit errors and detect $1$ bit errors.
• D. We can correct $1$ bit errors and detect $1$ bit errors.
• E. None of the above.

Answer 1

B

Answer 2

1. Error Detection Capability ($t_d$)

To detect $t_d$ errors, the minimum Hamming distance must satisfy:

 

$$d_{\min} \ge t_d + 1$$

 

Substituting $d_{\min} = 3$:

 

$$3 \ge t_d + 1 \implies t_d \le 2$$

 

Result: The code can detect up to 2 bit errors.

2. Error Correction Capability ($t_c$)

To correct $t_c$ errors, the minimum Hamming distance must satisfy:

 

$$d_{\min} \ge 2t_c + 1$$

 

Substituting $d_{\min} = 3$:

 

$$3 \ge 2t_c + 1$$

$$2 \ge 2t_c$$

$$t_c \le 1$$

 

Result: The code can correct 1 bit error.

Conclusion

Combining these two maximum capabilities:

• The system can correct 1 bit error.

• The system can detect 2 bit errors.

This corresponds to Option B.

Correct Answer:

B. We can correct 1 bit errors and detect 2 bit errors.