Majority logic decoding

Majority logic decoding is a method to decode Repetition codes, based on assumption that the largest occurance of a symbol, was transmitted message. It bases the decision on the frequency of occurance of any symbol in the given received code vector.

Theory

If we have a binary alphabet made of $0,1$ respectively, then we use $(n,1)$ repetition code, we have the input bit mapped to the codeword as a string of $n$ duplicated input bits. Failed to parse (unknown function "\begin{array}"): {\displaystyle 0 \leftarrow \begin{array}[0 & 0 & 0 & 0 & 0 & ... & 0_{n^{th zero}}]\end{array},\\ 1 \leftarrow \begin{array}[1 & 1 & 1 & 1 & 1 & ... & 1_{n^{th zero}}]\end{array} } We generally choose $n=2*t+1$ an odd number.

The repetition codes, can correct upto $[n/2]$ errors. Also, decoding errors occur, when the more than, these specified errors occur. so the probability of error for a repetition code is given using, Failed to parse (unknown function "\begin{array}"): {\displaystyle P_e = \sum_{t=\frac{n+1}{2}}^{n} \begin{array} n\\ t\\ \end{array} P_e^(t)(1-P_e)^(n-t)}

Algorithm

Assumptions

You have a $(n,1)$ code word, where $n=2t+1$ an odd number.

• Calulate the $d_{H}$ Hamming weight of the Repetition code.
• if $d_{H}\leq t$, decode code word to be all 0's
• if $d_{H}\geq t+1$, decode code word to be all 1's

Example

If you had a $(n,1)$ code, with R=[1 0 1 1 0], then you would decode it as,

• $n=5,t=2$, $d_{H}=3$, so R'=[1 1 1 1 1]
• Hence the transmitted message bit, was 1.

References

Rice University, http://cnx.rice.edu/content/m0071/latest/ UT, Arlington, EE5364, http://www.uta.edu/faculty/prabhu/classes/ee5364/