# How is the symbol error rate for M-QAM, 4QAM,16QAM and 32QAM derived?

How do you derive the theoretical symbol error rate as a function of $E_\mathrm{b}/N_0$ for 4QAM? I know that the result should be $Q\left(\sqrt{2E_\mathrm{b}/N_0}\right)$ but I am ĺooking for the derivation. Also, what are the symbol error rates vs $E_\mathrm{b}/N_0$ for 16QAM and 32QAM?

• Homework, or do you need it for a particular reason? – MSalters May 2 '14 at 16:16
• I am studying for an exam but I can't find this information in textbooks – user1930901 May 2 '14 at 16:18
• The theoretical symbol error rate for 4-QAM is not $Q(\sqrt{2E_b/N_0})$; that's the bit error rate. The $2$-bit 4-QAM symbol can have zero or one or two bit errors in it, and the probability that the symbol is in error is not the same as the probability that a bit is in error. – Dilip Sarwate May 2 '14 at 19:53

In $2^{2n}$-QAM with a square constellation, there are $4$ "corner" points and $4(2^n-2)$ "edge" points, and $(2^n-2)^2$ "interior" points. The conditional symbol error probabilities given that each type of point is transmitted, are \begin{align} P_e(\text{corner}) &= 2Q(x) - Q^2(x)\\ P_e(\text{edge}) &= 3Q(x) - 2Q^2(x)\\ P_e(\text{interior}) &= 4Q(x) - 4Q^2(x)\\ \end{align} where $Q(x)$ is the complementary cumulative probability distribution function of the standard Gaussian random variable. Combining these using the law of total probability (with the assumption that all $2^{2n}$ signals are equally likely) gives $$P_e\left(2^{2n}\text{-QAM}\right) = 4 \left[1 - 2^{-n}\right]Q(x) - 4\left[1 - 2^{-n}\right]^2Q^2(x)$$ For $4$-QAM, where $n = 1$, this reduces to $P_e(4\text{-QAM}) = 2Q(x) - Q^2(x)$. For more details, see, for example, pp. 147-153 of this ancient lecture note that I once wrote.