Q
I did some digging and found a couple of diagrams that explain the difference between QPSK and 8PSK
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Representing 2 bits of data vs 3 bits of data is the difference and hence the 33% more throughput. (depending on your symbol rate)
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THanks. Those diagrams are helpful for showing the data bit structure, which explains the part that I was having a hard time explaining.
But before this gets more confused, again, comparing the improvement you get from 8PSK to QPSK, it's not twice, as in previous message, not 30%, and not 33%, it's 50%, ie a factor 3/2=1.5 . You could say that QPSK is 33% less than 8PSK, but 8PSK is 50% more than QPSK. Not a big deal, but it's nice to get them right.
The actual equation for calculating the bitrate from a given SR/FEC is:
Bitrate= SR * (188/204) * N * FEC
Where 188/204 is a factor related to the number of actual useful payload data bytes vs total data bytes in a packet.
N= number of bits per symbol (ie 2 for QPSK and 3 for 8PSK).
and SR and FEC are obvious.
For example, for a 30000 SR with 3/4 FEC, you get:
QPSK bitrate= 30000*2*(188/204)*(3/4) = 41471
8PSK bitrate= 30000*3*(188/204)*(3/4) = 62206
Ie, the 8PSK bitrate is 62206/41471=1.5 times the QPSK bitrate.
OR... you could say that
the QPSK bitrate is 33% LESS than 8PSK, but it isn't really proper to say that 8PSK is anything other than 50% more than QPSK, or a factor of 1.5.
Anyway, this is an interesting concept since it seems intuitive that 8 phase angles would give twice the throughput of 4 phase angles, but I still think that the fundamental reason for the difference is because those 2 or 4 pairs of phase angles cannot be used at the same time, which is the reason for the complicated diagrams producing the 2 or 3 data bits.