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?

  • $\begingroup$ 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 $\endgroup$ – user1930901 May 2 '14 at 16:18
  • $\begingroup$ 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. $\endgroup$ – 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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.