# What is the effect of redundancy coding on the bit error rate?

If I have a channel with bit error rate BER = $$\10^{-3}\$$ after modulation, but this is too high so I decide to add repetition coding. With 2 code words and block length of 3 such that I have a minimum hamming distance of 3, giving the system ability to correct 1 error and detect 2 errors. How do I calculate the new bit error rate?

• You also need to know the statistics of the errors. Random single errors or bursts of errors? Are all redundant words sent consecutively or are they interleaved? – Elliot Alderson Mar 13 '19 at 0:22
• The errors are random. I'm not sure about the last part, which one makes it the most simple? – Joakim Hansen Mar 13 '19 at 0:29
• I am not sure I understand your coding. You can either correct one error or detect up to two errors, but you can’t do both, for that you would need a block length of four. – Edgar Brown Mar 13 '19 at 5:49
• Lets say correct one error then – Joakim Hansen Mar 13 '19 at 12:48

The error probability in the channel is $$\P_c = 10 ^ {-3}\$$. There will be a bit error after the decoder if a 3-bit codeword contains two or three errors.

The probability that the codeword has two errors after going through the channel is:

$$P_2 = \binom{3}{2} (1-P_c) P_c^2.$$

The term $$\\binom{3}{2} = 3\$$ is equal to the number of ways the channel can introduce two errors. If "c" stands for "correct" and "e" for "error", the ways to make two errors out of three bits are: "cee", "ece", and "eec".

The term $$\1-P_c\$$ is the probability that one bit is correct; the term $$\P_c^2\$$ is the probability that two bits are incorrect.

The probability that the codeword has three errors after going through the channel is

$$P_3 = P_c^3.$$

There is only one way to make three errors ("eee"), and no correct bits.

So the probability that the repetition decoder gives you an incorrect bit is

$$P_b = P_2 + P_3 \approx 3 \times 10 ^ {-6}.$$