# 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? Commented May 2, 2014 at 16:16
• I am studying for an exam but I can't find this information in textbooks Commented May 2, 2014 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. Commented May 2, 2014 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 (or for other constellation points even farther away than the nearest neighbor). 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 $$P_e(\text{interior}) = 4Q(x) - 4Q^2(x).$$