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Symbol Error Rate (SER) for QPSK (4-QAM) modulation

Posted by Krishna Pillai
November 6, 2007

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Given that we have discussed symbol error rate probability for a 4-PAM modulation, let us know focus on finding the symbol error probability for a QPSK (4-QAM) modulation scheme.

Background

Consider that the alphabets used for a QPSK (4-QAM) is (Refer example 5-35 in [DIG-COMM-BARRY-LEE-MESSERSCHMITT]).

4QAM constellation

Figure: Constellation plot for QPSK (4-QAM) constellation

The scaling factor of is for normalizing the average energy of the transmitted symbols to 1, assuming that all the constellation points are equally likely.

Noise model

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

with and .

Computing the probability of error

Consider the symbol

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

.

probability density function for 4QAM

Figure: Probability density function for QPSK (4QAM) modulation

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

.

Probability of real component of greater than 0, given was transmitted is (i.e. area outside the red region)

, where

the complementary error function, .

Similarly, probability of imaginary component of greater than 0, given was transmitted is (i.e. area outside the blue region).

.

The probability of being decoded correctly is,

.

Total symbol error probability

The symbol will be in error, it atleast one of the symbol is decoded incorrectly. The probability of symbol error is,

.

For higher values of , the second term in the equation becomes negligible and the probability of error can be approximated as,

.

Simulation Model

Simple Matlab/Octave script for generating QPSK transmission, adding white Gaussian noise and decoding the received symbol for various values.

Click here to download: Matlab/Octave script for computing the symbol error rate for QPSK modulation

Figure: Symbol Error Rate for QPSK (4QAM) modulation

Observations

1. Can see good agreement between the simulated and theoretical plots for 4-QAM modulation

2. When compared with 4-PAM modulation, the 4-QAM modulation requires only around 2dB lower for achieving a symbol error rate of .

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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    • Comments
    Comment by Ansar on June 11, 2008 @ 4:05 am

    Its nice article and thanx for post. I have a situation. Let we assume 4QAM modulation but only 3 symbols are transmitted and this fact is known to receiver (let there is no s2 symbol.. and this quadrant is empty). How the deceision region will look like and how to calculate BER for this so called 3QAM. Thanks…

    Comment by Krishna Pillai on June 12, 2008 @ 5:36 am

    @ansar: Let me try to answer”
    If s2 is not present, the area occupied by s2 can be used by both s3 and s1. The decision boundary can be the line drawn at an angle of 45degree from the origin (passing through the quadrant occupied by s2).

    For s3, the decision region will be the quadrant occupied by s3 plus the upper half the triangular area of the quadrant occupied by s2. Similarly for s1, the decision region will be the quadrant occupied by s1 plus lower half of the triangular area of the quadrant occupied by s2.

    Symbol error probability of s0 remains the same.

    Though I have not done the math (with integration of area). may I try to find an intuitive answer.
    Perror for s0 = erc(sqrt(Es/2No)) (remains same)
    Perror for s1 = erc(sqrt(Es/2No)) - (1/2)erc(sqrt(Es/2No)) = (1/2)erc(sqrt(Es/2No))
    Similiarly Perror for s3 = (1/2)erc(sqrt(Es/2No))

    Total symbol error probability = 1/3*erc(sqrt(Es/2No)) + (1/3)*(1/2)erc(sqrt(Es/2No)) + (1/3)*(1/2)erc(sqrt(Es/2No))
    = (2/3)*erc(sqrt(Es/2No))

    Do you agree? What do you think?

    Just as a final note: If one decides not to use s2, then a better way for defining the constellation points might be to place all of the constellation equidistant at maybe 120 degrees away.

    Thanks for this nice question. Kindly share your thoughts.

    Comment by Richard Mayo on June 27, 2008 @ 8:29 pm

    Very useful, thank you. Does the same analysis hold if the noise is only phase noise, or would the effect be reduced by 1/(sqrt(2)) ?

    Comment by Krishna Pillai on June 30, 2008 @ 5:16 am

    @Richard Mayo: Thanks. Sorry, I have not studied the BER vs phase noise. However, from a quick browsing, I found the article: Specifying Local Oscillator Phase Noise Performance: How Good is Good Enough? Gilmore, R.P. 1991
    Hope that reference helps.

    Kindly do let know, if you find some further references. Will be useful for the community.Thanks.

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