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I am reading the specs of a DAC (this one), and it says the resolution is 10 bit. Does it mean it can generate 1024 voltage levels at the sample rate which is 12 GSa/s? It sounds pretty surreal. Can it really do communications at 120 Gb/s or what would be the practical limitation?

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    \$\begingroup\$ The text at the linked page says "... DAC is able to sustain full-bandwidth streaming at 120 Gigabits per second". We don't even have to open a the data sheet. They didn't use the word "really", though. \$\endgroup\$
    – Kaz
    May 19, 2013 at 4:22
  • \$\begingroup\$ The trick is that the data comes in over 320 lines that are individually clocked much slower. The 320 lines come in as 5 groups of 64, into 5 muxes. Each mux is 8:1, so it reduces 64 to 8. So then we have 40 bits coming out of these muxes. They go to a 4:1 that picks 10 bits. \$\endgroup\$
    – Kaz
    May 19, 2013 at 4:33

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"Does it mean it can generate 1024 voltage levels at the sample rate which is 12 GSa/s?" : yes, although with limits on analog bandwidth. If you ask it to generate alternating 0 / 1023 levels, for a 6GHz signal, it will be attenuated a bit as the rated output bandwidth is only 5GHz.

"Can it really do communications at 120 Gb/s or what would be the practical limitation?" : yes, via a number of parallel digital links that operate at a more reasonable 375Mbit each.

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DAC resolution mean number of digital bits can be converted to analog signal. Yes, it means you can generate 2^10=1024 voltage levels like your thought.

The only limitation would be input rate i suppose.

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  • \$\begingroup\$ Please explain "Sample rate is really low". \$\endgroup\$
    – Anindo Ghosh
    May 19, 2013 at 5:39
  • \$\begingroup\$ I should have wrote like "sample rate time is low" or "sampling time is really fast". My bad. \$\endgroup\$
    – Hammers
    May 19, 2013 at 5:47
  • \$\begingroup\$ "Sampling time is very high," still doesn't make any sense... or, perhaps, even incorrect. "Sampling time" is a period, so for sampling rate to be high, "sampling time" has to be low. \$\endgroup\$
    – DrFriedParts
    May 19, 2013 at 10:16

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