I've read in a couple sources that in a channel consisting of only additive white Gaussian noise (AWGN), at low SNR, QPSK31 is actually worse than BPSK31.

http://www.arrl.org/psk31-spec, probably copied verbatim from http://aintel.bi.ehu.es/psk31theory.html:

By doing simulations in a computer, and tests on the bench with a noise generator, it has been found that when the bit error-rate is less than 1% with BPSK, it is much better than 1% with QPSK and error-reduction, but when the BER is worse than 1% on BPSK, the QPSK mode is actually worse than BPSK. Therefore, if we are dealing with radio paths where the signal is just simply very noisy, there is actually no advantage to QPSK at all!


Computer simulation with BPSK in white noise shows that when the SNR is good, the error-correction system does win, reducing the low error rate to very low levels, but at the SNR levels that are acceptable in live amateur contacts, it's better to transmit the raw data slowly in the narrowest bandwidth. It also takes up less band space of course!

The reasoning seems to be that the switch from BPSK to QPSK degrades the SNR:

Suppose I devise an error-correcting system that doubles the number of transmitted bits. If I wanted to keep the traffic throughput the same, I would need to double the bit rate. But with BPSK that means doubling the bandwidth, so I lose 3dB of signal-to-noise ratio and get more errors. The error-correction system will have to work twice as hard just to break even! It is no longer obvious that error-correction wins. It is interesting to note that with FSK, where the bandwidth is already much wider than the information content, you can double the bit-rate without doubling the bandwidth, and error-correction does work.

I'm skeptical. Doubling the symbol rate in FSK halves the duration of the symbols, and thus halves Eb/N0. Switching from BPSK to QPSK while holding the symbol rate the same splits the transmitter between two orthogonal systems, so again halving Eb/N0. I don't see any reason error correction would work for FSK but not QPSK.

But then, the channel decoding in QPSK31 combines two bits on the air into one bit received. So this is a doubling of energy per bit, for a net effect of no change in Eb/N0.

What's the basis for this claim that BPSK31 is worse than QPSK31 in an AWGN channel? Is there any formal proof or published empirical results to back it up?


1 Answer 1


Your reasoning is correct: in principle, given identical bit rate and signal power, the BER for BPSK and QPSK are identical. The Wikipedia article on PSK puts it nicely. The author there points out that you can think of a QPSK signal as two BPSK signals in the same channel, since they are orthogonal (one is built from sines, the other from cosines). You double the bit rate and double the power, and as you point out, it comes out the same in the end.

However, there are some assumptions made in this calculation that don't necessarily hold in the real world. In order to make the calculation tractable, you have to assume (1) at the receiver you know the frequency and phase of the carrier precisely, and (2) you know precisely where the boundaries between bits occur. Neither assumption is true in the real world.

Perhaps bit synchronization in noisy conditions is harder for QPSK than for BPSK. Maybe since BPSK's phase changes at the boundaries are on average bigger than QPSK's, you get better boundary estimates for BPSK. I'm sure there are other effects as well. The only way to be really sure is to run a simulation of a realistic demodulator, as the PSK31 author did.

I would think that the effects of FEC would be a separate issue. I didn't see an apples-to-apples comparison in the PSK31 paper you linked to, and I don't know enough about the performance of FEC as a function of SNR to comment anyway. The comparison that I saw was between QPSK with error correction and BPSK without. Assuming for the sake of argument that FEC improves BER up to a certain point, then collapses (seems reasonable), you might see the behavior described in the PSK31 paper.

This is all a bit speculative, but I hope I've made clear that the devil is in the details. Theory makes use of assumptions that sometimes are violated in the real world - otherwise you can't do the calculations - but in this case, I think that a thorough simulation is really the gold standard.

  • $\begingroup$ Sorry but this doesn't really answer the question. Best I can tell, about 20 years ago the author wrote a paper published in QST and everyone has been copying these two paragraphs since. I've never seen a BER curve, or any quantitative results at all. There's already a lot of speculation and assumption. I'm looking for something with some rigor. $\endgroup$ Commented Jul 27, 2017 at 11:52
  • $\begingroup$ My point is there is no rigor to be had, beyond the standard results I point to above. If you want to see a BER curve, look at the Wikipedia article on PSK, here: en.wikipedia.org/wiki/Phase-shift_keying#Bit_error_rate_3. You'll see that the curves for BPSK and QPSK are right on top of each other. The QST paper shows simulation and bench experiments that show different results, and I believe the simulations and experiments over the "rigorous" results, for the reasons I give above. $\endgroup$ Commented Jul 28, 2017 at 17:29
  • 1
    $\begingroup$ @RodneyPrice Please moderate your tone. Stack Exchange is for getting other people to answer your questions, to a greater or lesser degree depending on the particular sort of question. If you feel a question is the result of a lack of adequate research, then downvote it, and if answering further would require more effort, then just say that's all you've got. $\endgroup$
    – Kevin Reid AG6YO
    Commented Jul 28, 2017 at 19:03
  • 1
    $\begingroup$ @PhilFrost-W8II has reasonably said that this answer does not contain the specific thing he asked for, a formal proof or a citation of a published result. There's no sharp rules, but everyone should be starting from the perspective of collaborating to create a useful document — a question & its best answer. Incomplete answers can help, but it's best to write something like "This doesn't answer entirely because I would have to actually [build a simulation or whatever], but I can at least say…” $\endgroup$
    – Kevin Reid AG6YO
    Commented Jul 28, 2017 at 20:26
  • 1
    $\begingroup$ @RodneyPrice I apologize. Attacking you was not my intention. Just trying to make it clear why I'm not accepting the answer. $\endgroup$ Commented Jul 28, 2017 at 20:32

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .