Computer Networks · TEE · papers
Q2(b)(i) · Block-code received words
2023-01-27 · AI-only Sem V
Exact historical variant. Current-syllabus questions solved; excluded topics and source limitations stated.
Q2(b)(i). For k=2,n=3, the encoding table is 00→000, 01→011, 10→101, 11→110. The sender sends dataword 01 as codeword 011. Explain the receiver outcome if it receives (a)111 and (b)000. [5 marks]
Answer
Given
Sent data01, encoded011; valid codewords000,011,101,110.
To find
Receiver outcome for111 and000.
Formula and why
A receiver first checks membership in the valid-codeword set, then decodes a valid word.
This even-parity code detects any single-bit error but cannot identify a unique bit to correct.
Solution
Step 1. Received111
111 has odd parity and is not a valid word. Relative to011, one bit differs. The receiver detects an error and rejects/requests retransmission; it cannot uniquely infer the original data.
Step 2. Received000
000 is a valid word and decodes to00. Relative to the actual sent011, two bits changed. Therefore this is an undetected error: the receiver accepts the wrong dataword00.