Symbol Error Rate (SER) for 16-QAM

Posted by Krishna Pillai

December 9, 2007

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Given that we have went over the symbol error probability for 4-PAM and symbol error rate for 4-QAM , 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]).

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

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

Constellation plot for 16QAM modulation

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 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 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.

Click here to download : Matlab/Octave script for simulating 16QAM symbol error rate

Symbol Error Rate curve for 16QAM modulation

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

[DIG-COMM-BARRY-LEE-MESSERSCHMITT] Digital Communication: Third Edition, by John R. Barry, Edward A. Lee, David G. Messerschmitt

Hope this helps.

Krishna

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Error Rate

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Comments

Comment by mdkn45 on January 4, 2008 @ 11:36 am

hi, it’s very useful. But i want to bit error. What can i do?

Comment by Krishna on January 5, 2008 @ 6:54 am

@mdkn45: Typically, we can assume that the bits assigned to constellation points are grey coded (i.e. bit groups allocated to adjacent symbols differ by only one bit). Which means that each symbol error causes only 1 out of 4 bits to be incorrectly detected. Given so, the bit error probability is 1/4th the symbol error probability.

Additionally, since we are transmitting 4 bits in a symbol,
the each symbol carries energy of four bits i.e Es/No = 4*Eb/No.

Combining, the bit error probability comes to
BER_{16QAM} = (3/8)*erfc(sqrt(4Eb/10No)).

Agree?

Trackback by zuainroh on February 17, 2008 @ 1:05 pm

zuainroh…

zuainroh…

Comment by rajesh on May 27, 2008 @ 7:07 pm

hi,
if i have 16 distinct points, what are the conditions on them to form 16 qam ???

Comment by Krishna Pillai on May 29, 2008 @ 4:24 am

@rajesh: Well, for square QAM, you need to arrange them into {+/-1+/-j, +/-3,+/-3j,+/-3,+/-1j,+/-1,+/-3j} constellation points.

Comment by hwan on June 3, 2008 @ 3:07 pm

what is difference between Eb/No to SNR/bit.

many communication book show us case of SNR/bit.
i.e)John G.proakis ‘communication systems engineering’

Comment by Krishna Pillai on June 4, 2008 @ 5:05 am

@hwan:
As you maybe aware, depending on the order of the modulation, constellation symbol might carry k bits, where k=log2(M). Typically k=1 for BPSK, 2 for QPSK, 4 for 16QAM etc.

The term SNR/bit refers to the energy required for transmitting a bit. So if the signal (constellation symbol) to noise ratio is Es/No, as each constellation carries k bits, then Es/No = k*Eb/No
Then term SNR/bit is equal to Eb/No.

Hope this helps.

Symbol Error Rate (SER) for 16-QAM

Noise model

Computing the probability of error

Simulation model

Reference

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Symbol Error Rate (SER) for 16-QAM

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