# Why is 16 APSK arrangement 4 + 12?

Why does constellation arrangement affect the BER ? For example 4 + 12 vs 6 + 10 ? Is it because the correlation coefficient ?

• I recommend getting familiar with the union bound to the probability of error. It is particulary well explained in Heath, "intro to wireless communications, but you can find it in any decent book on digital comms.
– MBaz
May 14, 2022 at 14:00

The arrangement on constant circles for APSK leads to lower peak to average ratio in comparison to QAM which is a primary motivation for the arrangement. To then optimize the symbol error rate given additive white noise (AWGN) conditions, the symbols should be equally spaced. If we compare 4 + 12 to 6 + 10 for the given radii, we see that 4 + 12 results in a closer approximation to equal distance.

While AWGN will effect each constellation point equally, phase noise will have greater impact in terms of distance spread on the outer ring than the inner ring.

With reference to the figure below from the DVB-S2 Standard we can compute the symbol distances and normalize by the standard deviation of the waveform as a baseline to compare other options. Using the 4/5 Code Rate as an example, and with equally probable symbols the standard deviation is given by 4 out of 16 at magnitude R1 and 12 out of 16 at magnitude R2 leading to:

$$\sigma = \sqrt{\bigg(\frac{1}{4}\bigg)^2 + \bigg(\gamma \frac{3}{4}\bigg)^2}$$

$$= \sqrt{\frac{1}{16} + 2.75^2\frac{9}{16}} \approx 1.567$$

The distance between the symbol choices on the inner ring normalized by the standard deviation of the waveform is:

$$d_1 \approx 2 \sin(\pi/4) / 1.567 = 0.902$$

The closest distance, normalized by the standard deviation, between samples on ring1 and ring2 is:

$$d_2 \approx (\gamma-1)/1.567 = 1.75/1.567 = 1.117$$

The distance between the symbol choices on the outer ring normalized by the standard deviation of the waveform is:

$$d_3 = 2 \gamma \sin(\pi/12) / 1.567 = 0.908$$

Note the close match between distances on inner and outer ring and a reasonably similar distance from inner to outer ring. If the outer ring radius was reduced to match the distances closer between the two rings, the output samples would get closer together. Also note in the constellation the alignment of symbol choices on 45° line in each quadrant which also simplifies implementation. 