If the probability of bit error for a square M-ary QAM is

$P_M = (1-(1-P_\sqrt{M})^2)$


$P_\sqrt{M} = 2(1-\frac{1}{\sqrt{M}})Q(\sqrt{\frac{3E_s}{(M-1)N_0}})$

and $E_s$ is the average symbol energy, can I assume that $E_s=10A^2$?. $2A$ is the minimum distance between two adjacent symbols.


1 Answer 1


A symbol with coordinates $(x,y)$ has energy $x^2+y^2$. In 16-QAM, minimum distance $2A$ implies that the values of $x$ and $y$ are restricted to the set $\lbrace\pm A,\pm3A\rbrace$, and in consequence the possible symbol energies are $2A^2$, $10A^2$ and $18A^2$.

Furthermore, there are 4 symbols with energy $2A^2$, 8 symbols with energy $10A^2$, and 4 symbols with energy $18A^2$. Calculating the averagey symbol energy, we conclude that $E_s=160A^2/16=10A^2$.

Note that a similar procedure can be used to calculate the average symbol energy of any quadrature modulation, even if they're not square (or even rectangular).

  • $\begingroup$ If a value for $P_M$ is given and if I want to find the value of $A$ can I just equate the value I get for $E_S$ as follows and find A? $E_S = 10A^2$ $\endgroup$ Sep 22, 2015 at 1:13
  • $\begingroup$ Yes, assuming you also know $M$ and $N_0$. $\endgroup$
    – MBaz
    Sep 22, 2015 at 1:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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