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Is this correct? I thought in Manchester encoding the 0's had to be high-low transition, even if the initial state is assumed to be low and 1's low-high.

  • \$\begingroup\$ Manchester coding is a general idea of having at least one edge per code unit. Any particular implementation can choose how it encodes 1s and 0s, or even larger code units (trinary or quaternary), whether scrambling is applied before encoding, etc. The simplest implementation is a XOR of the bit stream with the transmit clock, eg. from legacy UARTs that have the transmit clock output. You can invert either one before XORing – as long as the receiver expects that. \$\endgroup\$ Jun 13 at 2:41
  • \$\begingroup\$ Does manchester encoding not normally go between a positive and a negative voltage? So starting from 0V would be in-between High and Low \$\endgroup\$
    – Password
    Jun 13 at 3:41
  • 1
    \$\begingroup\$ Manchester is the encoding - the data is enclosed in the change. Being it voltage, current, magnetism etc, the absolute value is immaterial as long as there is a change. \$\endgroup\$
    – Kartman
    Jun 13 at 4:58
  • \$\begingroup\$ 1) Don't you need to show the initial level before the first bit? 2) For differential Manchester encoding the first half bit is missing also (and you can cut the last half bit). \$\endgroup\$ Jun 13 at 11:49


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