Symbol Error Rate (SER) for 16-QAM

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]).

are used.

The average energy of the 16-QAM constellation is (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 follows the Gaussian probability distribution function,

with and .

Computing the probability of error

Consider the symbol in the inside, for example

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

.

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

.

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

.

The probability of being decoded incorrectly is,

.

Consider the symbol in the corner, for example

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

.

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

.

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

.

The probability of being decoded incorrectly is,

.

Consider the symbol which is not in the corner OR not in the inside, for example

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

.

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

.

Using the above two cases are reference,

.

The probability of being decoded incorrectly is,

.

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,

.

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 values, the theoretical results seem to be ‘pessimistic’ ‘optimistic’ compared to the simulated results. This is because for the approximated theoretical equation, the term was ignored. However, this approximation is valid only when the term is small, which need not be necessarily true for low 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

61 thoughts on “Symbol Error Rate (SER) for 16-QAM

  1. hi Krishna please i have a project on modelling simulation of OFDM on power line communication i need m-file
    please help me

  2. hi… Im doing my project in 2D channel estimation for ofdm system. I have a doubt whether its possible to plot ( no of bits vs SNR plot) ., (SNR transitted vs SNR recieved).. please do help me

  3. hi, i want to ask about 16 QAM matlab code under this web. why the mapping is a symbol for each bit? for N=256, your mapping result 256 symbol.
    i read in some reference, 16 QAM has 1 symbol for every 4 bit. it means for N=256, will craete 64 symbol.
    can you explain it? thanks

  4. hi,
    i need a code for 64 QAM modulation, also the BER and FER for this modulation, where the input is a sequence of ‘0’ and ‘1’.
    thank you

  5. hi!
    I have simulated 4-QAM, 16QAM under AWGN and 4QAM under rayleigh fading and got good results. but when I simulate the 16QAM under rayleigh fading the decoder (MLD) hardly decodes even up to 50dB (BER=0.08102). Can anyone tell me what I’m doing wrong? because I’m only adding coefficient (y = h*x+n)
    Thanks

  6. hi…….
    i am doing project in mimo signal detection using combined qrd-m and dfe detection technique for simple and efficient detection Can u please help me out with it ,thank u.

  7. Hi!
    I am doing a project on digital hardware implementation of D64-QAM for voice band data transmission. Can u please help me out with it .

  8. You have given the matlab code for simulating the SER of 16 QAM in AWGN channel. Can you provide some hints towards SER of 16 QAM in Rayleigh fading channel with diversity combining, i.e. how to incorporate the changes in your Matlab code given for AWGN case.

      1. Thanks very much Krishna for your guidelines. I have still another query.

        Regarding dual-hop Rayleigh channel (means between Tx and Rx, there exists another transceiver node), Can you provide some hints what would be the Equalization term if h1 and h2 are channel parameters in dual-hop case.

          1. No, Sir. I am doing the equalization after two-hop by dividing the received signal by (h1.*h2). Which one is correct? after each hop or after two-hops.

          2. @soumendra: Well, I believe the idea behind using relay communication is to improve the reliability of the link. So, the intermediete node might be doing something. Couple of ways are:
            a) decode the data, and then re-transmit.
            b) amplify the signal + noise, and then re-transmit.
            c) more ?

            Depending on your system design, you may chose the appropriate approach.

          3. Yes, Sir…. I have adopted the strategy (b) amplify the signal + noise, and then re-transmit….So, the equalization term after two-hop would be dividing the received signal by (G*h1.*h2) where G is the amplification factor. Am I right?

          4. Other way could be doing equalization after each hop, i.e.
            y1 = h1x + n1; y1=y1/h1;

            y2 = h2*g*y1+n2 = h2*g*(x+(n1/h1)) + n2 = h2*g*x + g*(h2/h1)*n1 + n2;

            y2=y2/(g*h2);

            Which one would be appropriate ? doing equalization after each hop or after two-hop.

          5. @soumendra: I have a feeling that your proposal i.e. doing equalization at each hop might be better. However, one needs to do the math to confirm. Please try simulating it and let us know.
            As an added thought, a third approach will be to decode x at y1 (after equalization) and then re-transmit.

  9. hi
    i wish i find any help in the topic of using ofdm in power line communication
    if any one get any results by using matlab codes i wish i can have these results
    thanks

  10. 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

    1. @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.

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

  12. 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.

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

  13. @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.

  14. 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’

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

  16. Pingback: zuainroh
  17. @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?

Leave a Reply

Your email address will not be published. Required fields are marked *