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Symbol Error Rate (SER) for 16-QAM

Posted By Krishna Sankar On December 9, 2007 @ 12:25 pm In Modulation | 61 Comments

Given that we have went over the symbol error probability for 4-PAM [1] and symbol error rate for 4-QAM [2] , let us extend the understanding to find the symbol error probability for 16-QAM (16 Quadrature Amplitude Modulation). Consider a typical 16-QAM modulation scheme where the alphabets (Refer example 5-37 in [DIG-COMM-BARRY-LEE-MESSERSCHMITT] [3]).

$\alpha_{16QAM}=\left{\pm 1+\pm 1j,\ \pm 1+\pm 3j,\\\pm 3 + \pm 3j,\ \pm 3+\pm 1j \right}$ are used.

The average energy of the 16-QAM constellation is $\begin{eqnarray}E_{16QAM} &=&10\end{eqnarray}$ (here [4]). The 16-QAM constellation is as shown in the figure below

Figure: 16-QAM constellation

## Noise model

Assuming that the additive noise $n$ follows the Gaussian probability distribution function,

$p(x) = \frac{1}{\sqrt{2\pi\sigma^2}}e^{\frac{-(x-\mu)^2}{2\sigma^2}$ with $\mu=0$ and $\sigma^2 = \frac{N_0}{2}$.

## Computing the probability of error

Consider the symbol in the inside, for example $s_5$

The conditional probability distribution function (PDF) of $y$given $s_5$ was transmitted is:

$p(y|s_5) = \frac{1}{\sqrt{\pi N_0}}e^{\frac{-\left(y-\sqrt{\frac{E_s}{10}}\right)^2}{N_0}$.

As can be seen from the above figure, the symbol $s_5$ is decoded correctly only if $y$falls in the area in the black hashed region i.e.

$p(c|s_5) = p\left(\Re{y\le 0,\Re{y}>-2\sqrt{\frac{E_s}{10}}}|s_5\right)p\left(\Im{y>0,\Im{y}\le2\sqrt{\frac{E_s}{10}}}|s_5\right)$.

Using the equations from (symbol error probability of 4-PAM [1] as reference)

$p(c|s_5) = \left[1-erfc\left(\sqrt{\frac{E_s}{10N_o}}\right)\right]\left[1-erfc\left(\sqrt{\frac{E_s}{10N_o}}\right)\right]$.

The probability of $s_5$ being decoded incorrectly is,

$\begin{eqnarray}p(e|s_5) & = & 1-\left[1-erfc\left({\sqrt{\frac{E_s}{10N_0}}}\right)\right]^2\\ & \approx & 2erfc\left(\sqrt{\frac{E_s}{10N_0}}\right) \\ \end{eqnarray}$.

Consider the symbol in the corner, for example $s_3$

The conditional probability distribution function (PDF) of $y$given $s_3$ was transmitted is:

$p(y|s_3) = \frac{1}{\sqrt{\pi N_0}}e^{\frac{-\left(y-\sqrt{\frac{E_s}{10}}\right)^2}{N_0}$.

As can be seen from the above figure, the symbol $s_3$ is decoded correctly only if $y$falls in the area in the red hashed region i.e.

$p(c|s_3) = p\left(\Re{y}>2\sqrt{\frac{E_s}{10}}|s_3\right)p\left(\Im{y}>2\sqrt{\frac{E_s}{10}}|s_3\right)$.

Using the equations from (symbol error probability of 4-QAM [2] as reference)

$p(c|s_3) = \left[1-\frac{1}{2}erfc\left(\sqrt{\frac{E_s}{10N_o}}\right)\right]\left[1-\frac{1}{2}erfc\left(\sqrt{\frac{E_s}{10N_o}}\right)\right]$.

The probability of $s_3$ being decoded incorrectly is,

$\begin{eqnarray}p(e|s_3) & = & 1-\left[1-\frac{1}{2}erfc\left({\sqrt{\frac{E_s}{10N_0}}}\right)\right]^2\\ & \approx & erfc\left(\sqrt{\frac{E_s}{10N_0}}\right) \\ \end{eqnarray}$.

Consider the symbol which is not in the corner OR not in the inside, for example $s_{11}$

The conditional probability distribution function (PDF) of $y$given $s_{11}$ was transmitted is:

$p(y|s_{11}) = \frac{1}{\sqrt{\pi N_0}}e^{\frac{-\left(y-\sqrt{\frac{E_s}{10}}\right)^2}{N_0}$.

As can be seen from the above figure, the symbol $s_{11}$ is decoded correctly only if $y$falls in the area in the blue hashed region i.e.

$p(c|s_{11}) = p\left(\Re{y}>2\sqrt{\frac{E_s}{10}}|s_{11}\right)p\left(\Im{y}\le0,\Im{y}>-2\sqrt{\frac{E_s}{10}}|s_{11}\right)$.

Using the above two cases are reference,

$p(c|s_{11}) = \left[1-\frac{1}{2}erfc\left(\sqrt{\frac{E_s}{10N_o}}\right)\right]\left[1-erfc\left(\sqrt{\frac{E_s}{10N_o}}\right)\right]$.

The probability of $s_{11}$ being decoded incorrectly is,

$\begin{eqnarray}p(e|s_{11}) & = & 1-\left[1-\frac{1}{2}erfc\left(\sqrt{\frac{E_s}{10N_o}}\right)\right]\left[1-erfc\left(\sqrt{\frac{E_s}{10N_o}}\right)\right]\\ & \approx & \frac{3}{2}erfc\left(\sqrt{\frac{E_s}{10N_0}}\right) \\ \end{eqnarray}$.

Total probability of symbol error

Assuming that all the symbols are equally likely (4 in the middle, 4 in the corner and the rest 8), the total probability of symbol error is,

$\begin{eqnarray}\mathbf{P}_{16QAM} & \approx & \frac{4}{16}\cdot 2erfc\left(\sqrt{\frac{E_s}{10N_0}}\right) + \frac{4}{16}erfc\left(\sqrt{\frac{E_s}{10N_0}}\right) + \frac{8}{16} \cdot \frac{3}{2}erfc\left(\sqrt{\frac{E_s}{10N_0}}\right)\\ & \approx & \frac{3}{2}erfc\left(\sqrt{\frac{E_s}{10N_0}}\right) \end{eqnarray}$.

## Simulation model

Simple Matlab/Octave code for generating 16QAM constellation, transmission through AWGN channel and computing the simulated symbol error rate.

Figure: Symbol Error Rate curve for 16QAM modulation

Observations

1. Can observe that for low $\frac{E_s}{N_0}$ values, the theoretical results seem to be ‘pessimistic’ ‘optimistic’ compared to the simulated results. This is because for the approximated theoretical equation, the $erfc^2()$ term was ignored. However, this approximation is valid only when the $erfc^2()$ term is small, which need not be necessarily true for low $\frac{E_s}{N_0}$ values.

## Reference

Hope this helps.

Krishna

URL to article: http://www.dsplog.com/2007/12/09/symbol-error-rate-for-16-qam/

URLs in this post:

[1] symbol error probability for 4-PAM: http://www.dsplog.com/2007/10/07/symbol-error-rate-for-pam/

[2] symbol error rate for 4-QAM: http://www.dsplog.com/2007/11/06/symbol-error-rate-for-4-qam/