# 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? May 2 '14 at 16:16
• I am studying for an exam but I can't find this information in textbooks 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. 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 $$n = 1$$ and all the constellation points are corner points, this reduces to $$P_e(4\text{-QAM}) = 2Q(x) - Q^2(x)$$.
So where do the above formulas come from? Well, for corner points, a symbol error occurs if the transmitted corner point is mistaken for either of its two nearest (edge) neighbors. These independent events have probability $$Q(x)$$ each, and using $$P(A\cup B) = P(A) + P(B) - P(A\cap B),$$, we get $$P_e(\text{corner}) = 2Q(x) - Q^2(x).$$ An edge point has three nearest neighbors, two of which are either corner points or edge points and one of which is an interior point. Now we use $$\require{cancel}P(A\cup B\cup C)= P(A)+P(B)+P(C)-P(A\cap B)-\cancelto{0}{P(A\cap C)}-P(B\cap C) + \cancelto{o}{P(A\cap B\cap C)}$$ and independence of $$A,B$$ and of $$B,C$$ to get $$P_e(\text{edge}) = 3Q(x) - 2Q^2(x).$$ Finally, an interior point has four nearest neighbors, the events "symbol error in the I direction" and "symbol error in the Q direction" are independent events of probability $$2Q(x)$$ each, and so $$P_e(\text{interior}) = 4Q(x) - 4Q^2(x).$$