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This is something that bothers me from time to time and has always been hand-waved away with some plausible but non-rigorous explanation. For clarity, consider the QPSK case where we have the constellation points $(0,1), (1,0),(-1,0),(0,-1)$. What is the fundamental flaw in using the origin, and then three equidistant points separated by $\frac{2\pi}{3}$ radians? We would have the same minimum distance but with smaller average energy.

I understand the added annoyance in synchronization and differentiating between "no message" and "the zero message". But these should be solvable with preambles and whatnot. I'm curious about the flaws that are more than just annoyances, the ones that makes this constellation a fundamentally bad idea.

EDIT: I spoke without thinking, clearly the minimum distance would not be the same, it would be one in the zero-signal case and $\sqrt{2}$ in the standard case. The number of pairs at min. distance would be... uh... six...?

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    $\begingroup$ usually your QPSK signal multiplies a carrier signal. the way it is with the four points equidistant from the origin, it's only the phase of the carrier that gets changed. if you included the origin in the constellation, for that case the carrier would disappear. your receiver might react like it's "loss of carrier" or "disconnect from transmitter" instead of "no message". the reason we like FM or PM is that the amplitude stays constant and present, independent of the message being sent. $\endgroup$ – robert bristow-johnson Oct 23 '15 at 7:09
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    $\begingroup$ also, i think the QPSK constellation is normally (1,1), (-1,1), (-1,-1), (1,-1). $\endgroup$ – robert bristow-johnson Oct 23 '15 at 7:11
  • $\begingroup$ Isn't that 4-QAM? $\endgroup$ – Benjamin Lindqvist Oct 23 '15 at 7:24
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    $\begingroup$ QPSK and 4-QAM are the same thing: a symobl alphabet with constant magnitude and four phases. The phase offset is quite irrelevant for most applications as it has to be estimated at the receiver anyway. But I agree with Robert int that (1,1), (-1,1), (-1,-1), (1,-1) is the common QPSK/4-QAM alphabet. $\endgroup$ – Deve Oct 23 '15 at 7:47
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    $\begingroup$ In addition to the answers related to error rates, a constellation including zero will have a greater peak-to-average power ratio, which may increase the cost of solving other engineering problems (power supply filtering, etc.) $\endgroup$ – hotpaw2 Oct 23 '15 at 16:50
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As a complement to Deve's nice answer:

In regular QPSK, with symbol energy $E_s=1$, the distance between neighboring symbols is $d_{min}=\sqrt2$. For the same (average) symbol energy, the distance between neighbors in the proposed constellation is $d_{0,min}\approx1.15$.

If two symbols are at distance $d$, then the probability of the receiver making an error, if the noise variance is $\sigma_n^2$, is $$P(e)=Q\left(\sqrt{\frac{(d/2)^2}{\sigma_n^2}}\right).$$

In QPSK each symbol has two neighbors, while in the proposed constellation each symbol has only one neighbor. However, the quasi-exponential nature of the $Q$ function means that the distance is the most important consideration to evaluate the probability of error. This means that, as Deve shows, the proposed constellation will perform worse than QPSK.

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  • $\begingroup$ This is probably exactly what I'm looking for. Will mull it over a little bit before accepting... $\endgroup$ – Benjamin Lindqvist Oct 23 '15 at 16:09
  • $\begingroup$ @BenjaminLindqvist -- sure thing; let me know if you want a more detailed explanation. $\endgroup$ – MBaz Oct 23 '15 at 16:56
  • $\begingroup$ So, basically all it boils down to is the rapid decay of the Gaussian. Actually, I'm interested enough to go through all the numbers rigorously soon enough. Will post an update at that point. $\endgroup$ – Benjamin Lindqvist Oct 25 '15 at 8:14
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All implementation aspects aside, the constellation you propose performs worse than QPSK in an additve white gaussian noise (AWGN) channel.

I claim this based on simulations that I have run with Matlab calculating the symbol error rate (SER) as a function of signal-to-noise ratio (SNR). Here is the result: SER vs SNR

As you can see, for a given SNR, the proposed constellation has an higher SER than a QPSK constellation. Additionally, I expect the proposed scheme to perform even worse in terms of bit error rate (BER) as -- in contrast to QPSK -- Gray labelling is not possible. Some details concerning the simulation:

  • 1 million bits per SNR value
  • maximum likelihood decision
  • additive, Gaussian distributed noise
  • memoryless channel (no inter-symbol interference)

It might well be that I made a mistake writing the simulation script, so I uploaded it in case you would like to check my implementation.

This might not be the mathematically rigorous explanation that you have been hoping for. To answer your question theoretically, one would have to calculate the SER analytically. For that, one should first find the decision thresholds, which is the main problem (for me, at least). It's far from obvious, what the optimal decision thresholds would be. They are not straight lines as with QAM constellations but probably a kind of zero-centered disc with three "rays" emerging from its center. Calculation of the SER should probably involve a transformation of the problem to polar coordinates.

With that kind of irregular constellation it also doesn't help much to compare the minimum distance, i.e. the length of a straight line between two given constellation points. It's true, that for a given mean signal energy, the outer points of the proposed contellation have greater distance to the inner point than any two points of the QPSK constellation. However, the inner point of the proposed contellation is surrounded by three other points which decreases its decision area and consequently increases the symbol error probability for that point. It's not obvious what is better. According to the simulation, QPSK is better.

Update

The areas of the maximum likelihood decision can also be simulated. In the following image, each color corresponds to a distinct, decided symbol. The input to the decision device is a very noisy (0 dB SNR) signal with the proposed constellation. enter image description here

We can clearly see how the inner point is "squeezed" between the outer points. In my opinion, that is the reason why the proposed constellation performs worse than QPSK. Now that I see this plot, I find the SER calculation not so difficult anymore. With a little time it could be done.

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    $\begingroup$ Nice addition! That constellation pic really shows the decision boundaries nicely. $\endgroup$ – Peter K. Oct 23 '15 at 11:52
  • $\begingroup$ Great, great response! But there's a catch I think - we don't really NEED to use ML decoding. If you would expend less energy transmitting the zero-symbol, an optimal coding scheme would make sure to transmit that more often. This would expand the decision region containing the origin. $\endgroup$ – Benjamin Lindqvist Oct 23 '15 at 12:20
  • $\begingroup$ We don't need to use the ML detector, but as it is optimum in an AWGN channel, I think we should. Not sure if I get your (pre)coding idea but if you change the probability of the transmitted symbols to a non-uniform distribution you also change the mean energy and the entropy, i.e. the amount of information, of the source. A fair comparison with other constellations is not possible in that case. $\endgroup$ – Deve Oct 23 '15 at 12:42
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    $\begingroup$ @endolith Here you go: pastebin.com/Rtaa6xDh Enjoy ;) $\endgroup$ – Deve Oct 23 '15 at 16:53
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    $\begingroup$ @Deve FYI your answer was great, but the other answer gets down to the real crux - the rapid decay of the Gaussian. I accepted that one but upvoted yours. $\endgroup$ – Benjamin Lindqvist Oct 25 '15 at 8:15

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