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Say I'm using binary phase-shift keying (BPSK) to transmit some data at some rate. Now say that I want to double the data rate, without increasing my transmitter power. I could either continue to use BPSK but double the rate, or I could send two bits per symbol at the same rate by switching to QPSK.

Either will double the data rate, but what tradeoffs are being made? If I understand correctly, QPSK will use the same spectrum as BPSK at the same symbol rate, provide twice the bit rate, but the bit error rate increases as the transmitter power is now divided between two bits per symbol. But how does doubling the bit rate, but continuing to use BPSK change things? I expect it will use more spectrum; what of the error rate?

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    \$\begingroup\$ More spectrum means more noise into your receiver therefore your signal to noise ratio reduces. \$\endgroup\$
    – Andy aka
    May 14, 2013 at 15:12
  • \$\begingroup\$ @Andyaka by how much? Besides the obvious change in (de)modulator complexity, and given that QPSK requires less spectrum, is there any advantage to BPSK? \$\endgroup\$
    – Phil Frost
    May 14, 2013 at 21:34
  • \$\begingroup\$ Phil, I'd have to look it up so instead I'm being lazy and watching the answers roll-in here. Sorry \$\endgroup\$
    – Andy aka
    May 14, 2013 at 21:40

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Assuming reasonable SNR (>10 dB), QPSK will double your data rate, but the error rate is similar since the two phases of QPSK are independent. Specifically BER at the same Eb/No is the same. Since there are 3x the number bits, Eb is 3 dB worse.

In general use pi/4-QPSK or OQPSK instead of QPSK.

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