5
$\begingroup$

I'm trying to re-learn digital communication theory again and am creating a bit error probability simulation for various modulations. So far I've added BPSK, QPSK, and 8PSK and below is a plot that gets generated.

enter image description here

I noticed the BPSK and QPSK lines (orange) are on top of each other and the 8PSK line (green) is about what I would expect. The equations I am using are from the paper mentioned below. If this is really the case, why would it make sense to use BPSK when QPSK has the same performance?

Jianhua Lu, K. B. Letaief, J. C. -. Chuang and M. L. Liou, "M-PSK and M-QAM BER computation using signal-space concepts," in IEEE Transactions on Communications, vol. 47, no. 2, pp. 181-184, Feb. 1999, doi: 10.1109/26.752121.

Improve this question
$\endgroup$

3 Answers 3

Reset to default
1
$\begingroup$

Yes, they do have the same bit error rate (there are other ways to measure performance, though!)

All else being equal, there are few reasons to use BPSK instead of QPSK. Probably its main advantage in terms of system complexity is that it doesn't require a quadrature transceiver.

Improve this answer
$\endgroup$
5
  • 1
    $\begingroup$ What about spectral efficiency? We can send twice as much data in the same BW with QPSK, and at the same BER. $\endgroup$
    – Dan Boschen
    Dec 8, 2021 at 16:30
  • 2
    $\begingroup$ @DanBoschen Exactly; that's why I mentioned there are other ways to measure performance (besides BER), and why in almost all cases one should prefer QPSK. $\endgroup$
    – MBaz
    Dec 8, 2021 at 17:59
  • 2
    $\begingroup$ "...main advantage in terms of system complexity is that it doesn't require a quadrature receiver." @MBaz Do you have a reference for an actual recent BPSK implementation that uses squaring and bandpass-filtering the RF BPSK signal to remove the phase modulation and DC term, followed by a phase-lock loop (VCO etc) that tracks the resulting double-frequency signal to derive a local reference signal for the single RF mixer? I believe that most modern BPSK implementations use Costas loops (so no squaring of the RF signal) and are thus most of the way to a full-blown quadrature receiver anyway. $\endgroup$
    – Dilip Sarwate
    Dec 8, 2021 at 20:28
  • $\begingroup$ @DilipSarwate I'm not aware of any. I think the only modern context where one may find a non-quadrature system is in education. From my own experience, I have designed an undergrad course in which students design a full communication system using two computers' sound cards as front-ends. The flat region of the resulting channel is actually bandpass. In order to complete in one semester, corners need to be cut: in my case, that means using BPSK, and large-carrier modulation to avoid having to implement a frequency tracker. $\endgroup$
    – MBaz
    Dec 8, 2021 at 21:54
  • $\begingroup$ @DilipSarwate A paper describing this course is here $\endgroup$
    – MBaz
    Dec 8, 2021 at 21:54
3
$\begingroup$

BPSK and QPSK have the same BER performance under additive white Gaussian noise only if the symbols are appropriately mapped to bits (such as Gray coding) since they can be viewed / treated as two independent BPSK signals that can share the same frequency due to their orthogonality (quadrature carrier). This is not true with higher order waveforms beyond QPSK such as 16-QAM, 8-PSK etc where we find that a greater SNR is needed to achieve the same BER, with the benefit of better spectral efficiency (more data in the same occupied bandwidth), and we see in these cases we no longer have complete orthogonality between the I and Q components in the waveform.

The biggest and primary advantage of QPSK over BPSK is spectral efficiency: for the same BER performance and data rate we can use half the bandwidth! This is also intuitively explained that we are sending two BPSK signals independently in the same bandwidth.

A common confusion when considering error rates between the two different constellations in the presence of noise is Symbol Error Rate vs Bit Error Rate. In the QPSK constellation the decision locations are $1/\sqrt{2}$ closer (1/2 in power units), leading one to possibly make the mistake that the QPSK system is more susceptible to noise (including phase noise and additive white Gaussian noise) but once you consider that the QPSK system is transmitting 2 bits for each symbol, the Bit Error Rate (not the Symbol Error Rate) is the same since with Gray coding only 1 bit error results at the threshold of error condition. (under higher SNR conditions where the noise isn't strong enough statistically to cause the double errors to occur).

This is illustrated below starting with the constellations of BPSK and QPSK with the QPSK constellation properly Gray coded, and then with the effects of AWGN (additive white Gaussian Noise) and Phase Noise separately. Note how AWGN contains both phase noise and amplitude noise components! The drawings show how the Symbol Error Rate would be larger (the noise distributions are $1/sqrt{2}$ closer in magnitude to the decision boundaries), but as long as we appropriately code the constellation points (Gray coding so that only one bit changes between closest symbols) such that only one bit error results, the Bit Error Rate will be the same.

Constellations

AWGN

Phase Noise

To avoid any confusion, the plots above depict noise with the same standard deviation in all cases. The noise as AWGN extends to infinity with increasingly smaller distribution so there will be errors in all cases (as given by the standard waterfall curves of SNR vs error rate for this).

Improve this answer
$\endgroup$
0
1
$\begingroup$

You are analysing different modulation schemes under the assumption of perfect carrier recovery coherent demodulation. If your system has phase noise, BPSK will be superior to QPSK. Likewise QPSK will be more robust to phase noise than 8PSK.

Without implementing a carrier recovery loop, you can experiment the above by adding some phase noise into your system, on top of the white Gaussian noise.

Improve this answer
$\endgroup$
4
  • $\begingroup$ White Gaussian noise is half phase noise and half amplitude noise. The carrier tracking loop will track out the close in phase noise. I agree the higher order modulations above QPSK require better phase noise for same performance but don’t think that is the case with BPSK vs QPSK for the same arguments made with BER of BPSK vs QPSK: you can treat QPSK as two independent BPSK waveforms that can be sent on the same frequency— any phase noise leaking from I to Q is similar to the quadrature component of white Gaussian Noise - and is the noise we are already considering. Am I mistaken? $\endgroup$
    – Dan Boschen
    Dec 9, 2021 at 12:11
  • $\begingroup$ The statement "White Gaussian is haff phase noise and half amplitude noise" is mathematically incorrect (what does "half" mean?). Also, you can not decompose a phase noise into 2 independent I&Q components. It is this decomposition into independent components (of a complex AWGN) that gives you the identity in Eb/N0 for BPSK and QPSK (for a given BER). $\endgroup$
    – Ng Ph
    Dec 9, 2021 at 13:34
  • $\begingroup$ White Gaussian Noise is a complex waveform that has I and Q components and when added to a sinusoidal carrier will have equal AM and PM components (for the typical small angle approximations we would get with practical implementations involving phase noise). Half of the noise power is AM and half is PM. Phase Noise by definition is only PM and has no I component, only Q. $\endgroup$
    – Dan Boschen
    Dec 9, 2021 at 13:37
  • $\begingroup$ But I do see how if the phase noise is excessive, the BPSK system will be more sensitive (from that quadrature leakage), but we can make the same argument with the quadrature portion of AWGN. This explains the difference in Symbol Error Rate between QPSK and BPSK (BPSK has a 3 dB better SER), but since QPSK sends 2 bits per symbol we get the same BER. For this reason there is no difference in BPSK vs QPSK with regards to phase noise. Make sense? $\endgroup$
    – Dan Boschen
    Dec 9, 2021 at 13:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.