I'm working a Viterbi equalizer for a BPSK signal which uses Fibonacci coding. In the more theoretical descriptions of the Viterbi algorithm I've read, there's a term for the probability of a particular observation from a given state. In the more practical descriptions I've seen this term does not exist. I assume it's because the probability of observing a 0 or a 1 are considered equally probable.

This isn't true with Fibonacci coding. I've read (but not seen the proof) that the probability of a 1-bit is ${1\over 2} - (1-\sqrt{1/5}) \approx 0.2764$.

Can this be used to advantage if the objective is minimizing the word error rate (words being delimited by two consecutive 1-bits)? Or would this make things worse, perhaps by making the detection of the word boundaries less likely to be detected, thus increasing the probability of word errors?

And if so, within the context of channel equalization or decoding, the Viterbi algorithm is used to find the minimum path metric through the trellis, with the path metric being the hamming or euclidean distance. More generally though, the Viterbi algorithm is used to find the maximum path through the trellis, using the probability of each transition. How to reconcile these two contradictory concepts?

  • $\begingroup$ If I understand well "Fibonacci coding" is source coding ?? Viterbi algorithm is usually used with channel coding. I mean the output of Fibonnaci coding needs to pass over a channel coder and the probability of 1-bit is not necessarily kept as at the output of Fibonacci code. $\endgroup$
    – AlexTP
    Commented Jul 20, 2017 at 13:56
  • $\begingroup$ In my case there is no other channel coding that would alter the bit probabilities. Regardless of the usual usage of the Viterbi algorithm in FEC, why stop there? Viterbi is used for all kinds of hidden Markov models. $\endgroup$
    – Phil Frost
    Commented Jul 20, 2017 at 15:22
  • $\begingroup$ ok sorry did not read the question well $\endgroup$
    – AlexTP
    Commented Jul 20, 2017 at 20:01


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Browse other questions tagged or ask your own question.