I have a large number of samples which represent Manchester encoded bit streams as audio signals. The frequency at which they are encoded is the primary frequency component when it is high, and there is a consistent amount of white noise in the background.

I have manually decoded these streams, but I was wondering if I could use some sort of machine learning technique to learn the encoding schemes. This would save a great deal of time manually recognizing these schemes. The difficulty is that different signals are encoded differently.

Is it possible to build a model which can learn to decode more than one encoding scheme? How robust would such a model be, and what sort of techniques would I want to employ? Independent Component Analysis (ICA) seems like could be useful for isolating the frequency I care about, but how would I learn the encoding scheme?


1 Answer 1


I suggest the use of Hidden Markov Models, with two possible states: (1) high levels and (0) low levels.

This technique might be helpful to decode your signal. Probably you would need a specific HMM for each codification.

If noise is an issue an FIR filter with a Blackman-Harris window function would allow you to isolate the frequency you're concerned with.

  • $\begingroup$ would this work on a manchester encoded signal where the value is encoded in the state transitions? $\endgroup$ Jun 24, 2014 at 22:21
  • $\begingroup$ It depends on the Manchester codification but I would say so. Nonetheless, previous to a HMM training, I'd suggest to use a zero-crossing algorithm to detect flanks of the signal. With this, you could detect the minimum time a change occurs which can give you a hint on the clock speed. $\endgroup$
    – adesantos
    Jun 25, 2014 at 7:11
  • $\begingroup$ Why would I need clock speed? Manchester encoding is self clocking. Timing should be unimportant. $\endgroup$ Jun 25, 2014 at 14:21
  • $\begingroup$ I though it could be helpful to know the clock speed in order to know how fast are the transitions between low/high values. $\endgroup$
    – adesantos
    Jun 25, 2014 at 14:39

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