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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?

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  • $\begingroup$ Homework, or do you need it for a particular reason? $\endgroup$
    – MSalters
    May 2 '14 at 16:16
  • $\begingroup$ I am studying for an exam but I can't find this information in textbooks $\endgroup$ May 2 '14 at 16:18
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    $\begingroup$ 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. $\endgroup$ May 2 '14 at 19:53
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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 $P_e(4\text{-QAM}) = 2Q(x) - Q^2(x)$.


(This added part is for those who like to cross their eyes and dot their teas)

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 $$P_e(\text{interior}) = 4Q(x) - 4Q^2(x).$$

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