# Where am I wrong in terms of 16-QAM with union bounds?

The following is my analysis. (The analysis is based on analysis from ESE 471: Example Union Bound M=8 Box QAM) Please correct me if I am wrong.

Suppose each symbol has a distance of $$d$$ from its nearest neighbour.

4 corner points has 2 neighbours.

$$4 \times 2 Q(\frac{d} {\sqrt{2 N_0}})$$

Four inner points has 4 neighbours

$$4 \times 4 Q(\frac{d} {\sqrt{2 N_0}})$$

8 edge points has 3 neighbours

$$8 \times 3 Q(\frac{d} {\sqrt{2 N_0}})$$

So the final union bounds of $$P_e$$ will be the sum of these.

$$P_e \leq \frac{1} {16} (4 \times 2 Q(\frac{d} {\sqrt{2 N_0}}) + 4 \times 4 Q(\frac{d} {\sqrt{2 N_0}}) + 8 \times 3 Q(\frac{d} {\sqrt{2 N_0}})) = \frac{7} {2} Q(\frac{d} {\sqrt{2 N_0}})$$

The $$P_e$$ approximation based on the union bound is: $$P_e \leq \bar{v} Q \left( \frac{d}{\sqrt{2N_0}} \right)$$ where $$\bar{v}$$ is the average number of neighbors at miniminum distance $$d$$ over the constellation.
In 16-QAM, there are 4 points with 4 neighbors, 4 points with 2 neighbors, and 8 points with 3 neighbors, so that $$\bar{v} = \frac{16+8+24}{16} = 3$$ and then we have that $$P_e \leq 3 Q \left( \frac{d}{\sqrt{2N_0}} \right) .$$
• Keep in mind that there are different ways to calculate / approximate $P_e$. One way is to find an exact formula; this is often quite difficult. Another way is to find a simpler formula that either approximates or bounds the actual $P_e$. There are many ways to calculate bounds and approximations, which can be more or less accurate, and more or less tight. The approximation in that link is very simple to calculate, but it is not as accurate as the one given in my answer. None of them are exact.