Is there any method of decoding a Manchester encoded signal (IEEE 802.3) that is independent of the bit period?

Suppose the data I want to send is $\rm 10110$. The Manchester code of this data will be $\rm 0110010110$ (according to IEEE standard). We can see that in the $\rm 0101$ part, we need to consider the $\rm 0 \rightarrow 1$ transition (to decode a $\rm 1$) and not the $\rm 1 \rightarrow 0$ transition (falsely decoding a $\rm 0$). To do this, we need to know the bit period a priori.

My question is, is there a way to do this without knowing the bit period a priori?


I think what you're really asking is whether you can decode a Manchester-encoded bit stream without being synchronized to a frame boundary. As you noted, a mid-symbol transition from $0$ to $1$ indicates one state (i.e. a transmitted $1$ for instance), while a transition from $1$ to $0$ would indicate the opposite state (a transmitted $0$). You seem to want to decode a Manchester-encoded bit stream without necessarily knowing where the symbol boundaries are.

Whether you will be able to do this effectively would depend upon the bit error rate in the stream of encoded bits that you observe. Notice that for each two-bit sequence of Manchester-encoded bits, only two of the four possible states correspond to valid observations:

$$ \begin{align} \text{received 01} &\to \text{1 transmitted} \\ \text{received 01} &\to \text{1 transmitted} \\ \text{received 00} &\to \text{invalid!} \\ \text{received 11} &\to \text{invalid!} \\ \end{align} $$

Therefore, one way to get at the encoded bits in a Manchester bit stream without knowing where the symbol boundaries are would be to try both cases. Since there are two channel bits per symbol, there are only two possible alignments of the encoded symbols on the Manchester-encoded bit stream. Therefore, one approach would be to try both and select the timing offset that gives you the smallest number of "invalid" observations as shown above. Unless you have a lot of bit errors and are very unlucky, the correct alignment should yield a high percentage of valid sequences according to the above criteria, while you would observe a lot of invalid symbols if you were offset by one bit in the stream.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.