Symbol Error Rate (SER) for 16-QAM

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Symbol Error Rate (SER) for 16-QAM

Posted by Krishna Pillai in Error Rate on 12 9th, 2007 | 16 responses

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.

Download free e-Book discussing theoretical and simulated error rates for the digital modulation schemes like BPSK, QPSK, 4-PAM, 16PSK and 16QAM. Further, Bit Error Rate with Gray coded mapping, bit error rate for BPSK over OFDM are also discussed.
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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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16 Responses to “Symbol Error Rate (SER) for 16-QAM”

mdkn45 says:

January 4, 2008 at 11:36 am

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

Reply
Krishna says:

January 5, 2008 at 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?

Reply
zuainroh says:

February 17, 2008 at 1:05 pm

zuainroh…

zuainroh…

Reply
rajesh says:

May 27, 2008 at 7:07 pm

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

Reply
Krishna Pillai says:

May 29, 2008 at 4:24 am

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

Reply
hwan says:

June 3, 2008 at 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’

Reply
Krishna Pillai says:

June 4, 2008 at 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.

Reply
Philips says:

October 30, 2008 at 9:15 pm

Hi friends,
I’m studying a project which is a simulated OFDM on PLC(PowerLine Communication) with M-files in MatLab. I have not imaged my work yet. But I’ve read some documents which is described about OFDM and PLC(PowerLine Communication) models. Could you help me! Thank you so much.

Reply
- Tesla says:
  
  April 14, 2009 at 8:16 pm
  
  me too
  
  anybody help us
  
  thanks
  
  Reply
  - Krishna Pillai says:
    
    April 17, 2009 at 5:51 am
    
    @Tesla: I do not know much about the channel model for power line communications (except that the channel is a wire, ofcourse).
    However, for OFDM aspect may I point to couple of articles which I have written in the past.
    (a) Understanding an OFDM transmission
    URI: http://www.dsplog.com/2008/02/03/understanding-an-ofdm-transmission/
    (b) Cylcic prefix in Orthogonal Frequency Division Multiplexing
    URI: http://www.dsplog.com/2008/02/17/cylcic-prefix-in-orthogonal-frequency-division-multiplexing/
    (c) Peak to Average Power Ratio for OFDM
    URI: http://www.dsplog.com/2008/02/24/peak-to-average-power-ratio-for-ofdm/
    
    Reply
Krishna Pillai says:

October 31, 2008 at 6:06 am

@Philips: I do not know much about the channel model for power line communications (except that the channel is a wire, ofcourse).
However, for OFDM aspect may I point to couple of articles which I have written in the past.
(a) Understanding an OFDM transmission
URI: http://www.dsplog.com/2008/02/03/understanding-an-ofdm-transmission/
(b) Cylcic prefix in Orthogonal Frequency Division Multiplexing
URI: http://www.dsplog.com/2008/02/17/cylcic-prefix-in-orthogonal-frequency-division-multiplexing/
(c) Peak to Average Power Ratio for OFDM
URI: http://www.dsplog.com/2008/02/24/peak-to-average-power-ratio-for-ofdm/

Reply
Melinda says:

March 21, 2009 at 5:55 am

Hi,

Why points on picture above are as Eb ?

On http://www.dsplog.com/2007/11/06/symbol-error-rate-for-4-qam/ – QPSK case is as Es?

Can You please explain this.

Best regards

Reply
- Krishna Pillai says:
  
  March 21, 2009 at 5:02 pm
  
  @Melinda: Hmm… yes. I just checked. All the equations and BER plots are using Es/N0. However, I missed correcting the constellation plot diagram. Yes, the constellation point diagram should have shown Es/N0 (instead of Eb/N0).
  
  Thanks for noticing and pointing that out. I will correct in future.
  
  Reply
Melinda says:

April 12, 2009 at 6:07 am

Hi,

I think that on figure is an error(horizontal axes)
It is not realy Es/No but Eb/No .
(see MATLAB bertool for example)

Am I right?

Best regards

Reply
- Krishna Pillai says:
  
  April 16, 2009 at 5:30 am
  
  @Melinda: No, its indeed Es/N0. Was wondering, what made you think otherwise.
  
  Reply
Melinda says:

April 12, 2009 at 6:11 am

Hi,
Sorry for this new comment but just to be clear, I mean on figure with SER curves

Reply

Symbol Error Rate (SER) for 16-QAM

Noise model

Computing the probability of error

Simulation model

Reference

16 Responses to “Symbol Error Rate (SER) for 16-QAM”

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