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I've been looking into the way that 10 Gbps Ethernet encodes its data and I'm a little confused by the nomenclature.

I have gathered, e.g. from here http://www.ikn.no/download/Whitepaper-10G-Ethernet-10-08.pdf, that Pulse Amplitude Modulation is used to pack more data into each "symbol" transmitted down the twisted pairs of a cable. Specifically, 10GBASE-T uses 16 of these levels, its key advantage over slower schemes.

However, my understanding is that two of these channels (or is it one channel and two consecutive symbols?) are combined into the 128 bit DSQ128 constellation, as shown in this diagram from the above reference. PAM16 DSQ128 encoding constellation

This is done to increase the separation in phase space between adjacent encoding values, boosting the SNR by a factor of 3dB.

However, it seems to me that by chucking out every other point in the constellation, we've now limited ourselves to 8 levels for each symbol. So, we use 2x PAM16 symbols and end up with 128 possible messages.

My question is, why bother with using PAM16 and then this "constellation" arrangement when we could get 128 possible messages out of two symbols by simply using PAM8? This has the same spacing between adjacent symbols as DSQ128 does. The only difference I can see is that DSQ128 has half of its voltages offset by 1 unit compared with a PAM 8 scheme.

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You're talking about 10GBASE-T - most other 10G flavors use straight 64b/66b (10GBASE-R), but that doesn't work on 500 MHz Cat 6A.

With 10GBASE-T, two consecutive PAM-16 levels (on each lane) represent one two-dimensional symbol. Not all possible 256 symbols are used, but they are selected from 128 maximally spaced combinations (DSQ128). Of these 7 "raw" information bits, 3 are uncoded data, 4 are used for low-density parity check (LDPC). Check Clause 55.1.3 for details.

Since that selection effectively doubles the signal spacings, the signal-to-noise ratio increases by a factor of two = 3 dB. That's the same as with two PAM-8 symbols, but those would convey only six bits of information, not seven.

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    Thanks for the answer Zac, I hadn't even realised there was a variety of 10G flavours. Thanks for the link too. I'm a bit clearer now, but I still don't see the difference between this 2D symbol and two 1D symbols at PAM8 in terms of bitrate... Commented Jun 23, 2017 at 20:07
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    The 2D code is a clever trick to increase the symbol distance and thus increase the SNR by 3 dB or so (you do need what you can get with twisted pair). Commented Jun 23, 2017 at 23:03
  • @user253751 A fair amount of error correction is required since the channel is pretty noisy with PAM-16 at 800 MBd, even with that 3 dB extra margin. Commented Feb 4, 2023 at 7:36
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    @CharlieB I think the answer is that the first symbol in the pair must be one of the blue dots (it can't be a red dot) and the second must be a red dot (it can't be a blue dot). So you're using the temporal separation/synchronization of the two symbols to double the noise margin for both of them, which gives you 3dB extra noise margin over a single 7-bit symbol (where any of those 128 codes can appear at any time). You could do the same with PAM8 by using a shifted version of it for every other symbol, but it would still only give you 6 bits (64 possible values) per symbol. Commented Jan 7 at 15:38

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