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Information Theory and Coding Basics

Five Hamming (7, 4) codewords are lined up and sent in the order of bit 1 of each of the five codewords, then bit 2 of each, and so on (interleaving of depth 5). The receiver restores the original order and then corrects a 1-bit error in each codeword. Which of the following is the maximum length of a burst error, concentrated in consecutive bits, that can always be corrected?

  1. A5 bits
  2. B7 bits
  3. C35 bits
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Answer: A. 5 bits

In the transmitted bit stream, bits of the same codeword appear every 5 bits. As a result, a burst error of 5 consecutive bits or fewer is split, after restoring the order, into at most one bit in each of the five codewords, and every one of them can be corrected by the Hamming code. With 6 bits, some codeword receives a 2-bit error and can no longer be corrected. Without interleaving, even a 2-bit burst error would fall into a single codeword.

  • A. Correct. With a depth of 5, a burst error of up to 5 bits is split into one bit per codeword.
  • B. 7 bits is the codeword length. A 6-bit burst error already puts a 2-bit error into some codeword.
  • C. 35 bits is the total number of bits sent. Each codeword can correct only 1 bit, so a burst that long cannot be corrected.
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