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.

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

Four inner points has 4 neighbours

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

8 edge points has 3 neighbours

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

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

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


1 Answer 1


(See https://dsp.stackexchange.com/a/16254/11256)

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 $$ P_e \leq 3 Q \left( \frac{d}{\sqrt{2N_0}} \right) .$$

  • $\begingroup$ But the link dsp.stackexchange.com/questions/45748/… said it is 15 instead of 3 $\endgroup$
    – kile
    Commented Dec 29, 2023 at 23:07
  • $\begingroup$ 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. $\endgroup$
    – MBaz
    Commented Dec 29, 2023 at 23:14

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