A Forward Error-Correction Example
© Copyright University of New Haven 1999
How many redundant bits are required for 2-bit correction of 8 different messages?
Let N = (message bits + redundant bits) be the required total number of bits.
3 message bits are required for 8 messages.
1 code is required for a correct message.
There are N ways to make a 1-bit change to a correct code (i.e. a 1-bit error).
There are ½N×(N-1) ways to make a 2-bit change to a correct code (i.e. a 2-bit error).
Total number of codes required per message = 1 + N + ½N×(N-1) = 1 + ½N + ½N².
Thus total number of codes required for all 8 messages = 8 × (1 + ½N + ½N²) = 8 + 4N +4N².
Tabulating some N values:
N 8 + 4N + 4N² 2N
4 8+16+64=88 16
7 8+28+196=232 128
8 8+32+256=296 256
9 8+36+324=368 512
Thus N=9 yields sufficient bits to create the required number of codes, giving 9-3 = 6 redundant bits required.
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