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Repetition Codes

Module by: Don Johnson. E-mail the author

Summary: Explains the repetition code for error correction.

Perhaps the simplest error correcting code is the repetition code.

Figure 1: The upper portion depicts the result of directly modulating the bit stream bn b n into a transmitted signal xt x t using a baseband BPSK signal set. R ' R ' is the datarate produced by the source coder. If that bit stream passes through a (3,1) channel coder to yield the bit stream cl c l , the resulting transmitted signal requires a bit interval TT three times smaller than the uncoded version. This reduction in the bit interval means that the transmitted energy/bit decreases by a factor of three, which results in an increased error probability in the receiver.
Repetition Code
Repetition Code (sys32.png)
Here, the transmitter sends the data bit several times, an odd number of times in fact. Because the error probability p e p e is always less than 12 1 2 , we know that more of the bits should be correct rather than in error. Simple majority voting of the received bits (hence the reason for the odd number) determines the transmitted bit more accurately than sending it alone. For example, let's consider the three-fold repetition code: for every bit bn b n emerging from the source coder, the channel coder produces three. Thus, the bit stream emerging from the channel coder cl c l has a data rate three times higher than that of the original bit stream bn b n . The coding table illustrates when errors can be corrected and when they can't by the majority-vote decoder.

Table 1: Coding Table
In this example, the transmitter encodes 0 0 as 000 000. The channel creates an error (changing a 0 0 into a 1 1) that with probability p e p e . The first column lists all possible received datawords and the second the probability of each dataword being received. The last column shows the results of the majority-vote decoder. When the decoder produces 00, it successfully corrected the errors introduced by the channel (if there were any; the top row corresponds to the case in which no errors occurred). The error probability of the decoders is the sum of the probabilities when the decoder produces 1 1.
Code Probability Bit
000 1 p e 3 1 p e 3 0
001 p e 1 p e 2 p e 1 p e 2 0
010 p e 1 p e 2 p e 1 p e 2 0
011 p e 2(1 p e ) p e 2 1 p e 1
100 p e 1 p e 2 p e 1 p e 2 0
101 p e 2(1 p e ) p e 2 1 p e 1
110 p e 2(1 p e ) p e 2 1 p e 1
111 p e 3 p e 3 1

Thus, if one bit of the three bits is received in error, the receiver can correct the error; if more than one error occurs, the channel decoder announces the bit is 1 instead of transmitted value of 0. Using this repetition code, the probability of b^ n0 b^ n 0 equals 3 p e 2×(1 p e )+ p e 3 3 p e 2 1 p e p e 3 . This probability of a decoding error is always less than p e p e , the uncoded value, so long as p e <12 p e 1 2 .

Exercise 1

Demonstrate mathematically that this claim is indeed true. Is 3 p e 2×(1 p e )+ p e 3 p e 3 p e 2 1 p e p e 3 p e ?

Solution

This question is equivalent to 3 p e ×(1 p e )+ p e 21 3 p e 1 p e p e 2 1 or 2 p e 23 p e +10 2 p e 2 3 p e 1 0 . Because this is an upward-going parabola, we need only check where its roots are. Using the quadratic formula, we find that they are located at 12 1 2 and 11. Consequently in the range 0 p e 12 0 p e 1 2 the error rate produced by coding is smaller.

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