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

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

where

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

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.

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1 Answer 1

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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).

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  • $\begingroup$ 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

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