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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
    Commented May 2, 2014 at 16:16
  • $\begingroup$ I am studying for an exam but I can't find this information in textbooks $\endgroup$ Commented May 2, 2014 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. $\endgroup$ Commented May 2, 2014 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 (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 $$P_e(\text{interior}) = 4Q(x) - 4Q^2(x).$$

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  • $\begingroup$ For the corner points, there are 2 nearest neighbours and one farther (sqrt(2) times far as the nearest one). In your answer you disregard the probability that the corner point may end up at this farther point, but it can, which is why for an accurate answer i believe this should be factored in as well $\endgroup$ Commented Aug 18, 2023 at 6:56
  • $\begingroup$ @AnirudhRoy See revised wording. The result that I have stated is correct; if you feel you are right and something needs to be corrected, please write your own answer with what you believe are the correct formulas. $\endgroup$ Commented Aug 18, 2023 at 21:24

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