Average bit error and symbol energy

Question

How do you draw decision boundaries given this type of 16-qam constellation diagram, I know that there should be no vacant decision boundary but I can't seem to get it.Also, assuming that the correct decision boundary is drawn, how do you derive its total probability of error, if we let the SNR ratios of all symbols are equal?

Answer 1

The optimal decision regions are the Voronoi Regions. I dont know, if this is what you are looking after.

import numpy as np

points = np.array([(1,1), (1,-1), (-1,1), (-1,-1), (3,3), (3,0), (3,-3), (0,-3), (-3,-3), (-3,0), (-3,3), (0,3), (5,0), (0,5), (-5,0), (0,-5)])
from scipy.spatial import Voronoi, voronoi_plot_2d
vor = Voronoi(points)
voronoi_plot_2d(vor)
plt.axis('equal'); plt.xlim((-7,7)); plt.ylim((-7,7));

I actually, consider analytically calculating the error probability of this constellation a very hard task, since the regions are not very regular.

Answer 2

As explained in Maximilian Matthé's answer, the exact computation of the symbol error probability of this constellation (ITU-T V.29 modem standard) is quite complex. However, you can quite easily compute an approximation which becomes very good for relatively large signal to noise ratios (SNRs). This approximation is based on the union bound.

The symbol error probability is given by

$$P_S=\sum_{i=1}^MP[a_i]P[\text{error}|a_i]\tag{1}$$

where $M$ is the number of symbols, $P[a_i]$ is the probability that symbol $a_i$ is sent, and $P[\text{error}|a_i]$ is the error probability given that symbol $a_i$ was sent. This latter probability can be bound by the union bound:

$$P[\text{error}|a_i]\le\sum_{j=1\\j\neq i}^MQ\left(\frac{d_{ij}}{2\sigma}\right)\tag{2}$$

where $Q(\cdot)$ is the Q-function (assuming Gaussian noise), $d_{ij}$ is the distance between symbols $a_i$ and $a_j$, and $\sigma^2$ is the noise variance.

For large SNRs, the sum in $(2)$ is dominated by the term with the smallest value of $d_{ij}=d_{i,\text{min}}$, so we can write

$$P[\text{error}|a_i]\approx k_iQ\left(\frac{d_{i,\text{min}}}{2\sigma}\right)\tag{3}$$

where $k_i$ is the number of symbols with minimum distance $d_{i,\text{min}}$ from symbol $a_i$. Using this approximation in $(1)$, and assuming equally likely symbols (i.e., $P[a_i]=1/M$), we obtain the approximation

$$P_S\approx\frac{1}{M}\sum_{i=1}^{M}k_iQ\left(\frac{d_{i,\text{min}}}{2\sigma}\right)\tag{4}$$

Applying this formula to the given constellation works as follows. There are four groups of symbols with increasing distance from the origin. The first group (the one closest to the origin) has two nearest neighbors with distance $2$. The second group has one nearest neighbor with distance $2$. The third group has one nearest neighbor with distance $2\sqrt{3}-\sqrt{2}$, and the last group (furthest from the origin) has one nearest neighbor with distance $2$. This results in

$$\begin{align}P_S&\approx \frac14 2Q\left(\frac{1}{\sigma}\right)+\frac14 Q\left(\frac{1}{\sigma}\right)+\frac14 Q\left(\frac{2\sqrt{3}-\sqrt{2}}{2\sigma}\right)+\frac14 Q\left(\frac{1}{\sigma}\right)\\&=Q\left(\frac{1}{\sigma}\right)+\frac14 Q\left(\frac{2\sqrt{3}-\sqrt{2}}{2\sigma}\right)\tag{5}\end{align}$$