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I have a simple S/PDIF protocol question that I can't find the answer to anywhere. I'm trying to understand the protocol, and everywhere I look says the same thing: The length of a preamble pattern is 8 bits, but the allocated length in the word frame is only 4 bits. Here's an example from an ST application note:

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Everywhere else I have looked says the same thing too. How do I reconcile this information? Is the preamble slot 4 or 8 bits?

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The preamble length is 4 unencoded bits as the whole data frame is 32 unencoded bits.

The preamble length is 8 biphase Manchester encoded time slots as the whole data frame is 64 time slots.

As the preamble violates the bit encoding it can only be represented as 8 encoded symbols and not unencoded bits.

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  • \$\begingroup\$ That makes perfect sense, thank you. So you can scan the bitstream for a violation and you know it’ll be a sync point (preamble)? \$\endgroup\$
    – ezra_vdj
    Commented Jul 31, 2023 at 6:34

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