Probability of error of QPSK

Probability of error of QPSK :-

In order to understand the probability of error of a QPSK system, Let us consider a coherent QPSK system, the received signal x(t) is defined by   X(t)= Si(t) + w(t)  0≤ t ≤ T where i=1,2,3,4,  where w(t) is the sample function of a white Gaussian Noise process of Zero mean and power spectral density No/2.

Correspondingly, the observation vector x has 2 elements x1 and x2 defined by x_{1} = \int_{0}^{T}x(t)\phi_{1} (t)dt  and x_{2}= \int_{0}^{T}x(t)\phi _{2}(t)dt

x_{1}= \int_{0}^{T}x(t)\phi _{1}(t)dt

    = \sqrt{E}cos(2i-1)\frac{\pi }{4} + W_{1}

     = \pm \sqrt{\frac{E}{2}} + W_{1}

simillarly x_{2}= \int_{0}^{T}x(t)\phi _{2}(t)dt

                          = - \sqrt{E}sin(2i-1)\frac{\pi }{4} + W_{2}

                          = \mp \sqrt{\frac{E}{2}} +W_{2}

Thus the observable elements x1 and x2 are sample values of independent Gaussian random variables with mean values equal to \pm \sqrt{\frac{E}{2}} and \mp \sqrt{\frac{E}{2}}, respectively and with a common variance.

The decision rule is to simply to decide that

x1 falls inside the region z1 —–> S1 was transmitted.

x2 falls inside the region z2 —–> S2 was transmitted.

x3 falls inside the region z3 —–> S3 was transmitted.

x4 falls inside the region z4 —–> S4 was transmitted.

The regions and the signals were easily identified in the signal space diagram.

An erroneous decision will be made if for example signal S4(t) is transmitted . but the noise W(t) is such that the received signal point falls outside the region z4.

To calculate the  average probability of symbol error from the figure.

the coherent QPSK Receiver is in fact equivalent to two coherent binary PSK systems working in parallel and using two carriers that are in-phase and Quadrature-phase. x1 and x2 may be viewed as the individual outputs of the ‘2’ coherent BPSK systems may be characterized as follows.

The signal energy per bit = Eb/2 and the Noise power spectral density = No/2.

from the signal space representation , reference carriers are Φ1(t) and Φ2(t) all the signal vectors S1, S2, S3 and S4 are at 45o (or multiple of 45o) to these reference carriers. Suppose if S1 is transmitted. If the phase-shift of reference carrier is more than 45o (either Φ1 or Φ2)  it will be detected as some other symbol apart from S1.

The probability of error of BPSK system due to imperfect phase is given by 

P^{1}_{e BPSK}= \frac{1}{2}erfc\sqrt{\frac{E_{b}cos^{2}\phi }{N_{o}}} 

                 = \frac{1}{2}erfc\sqrt{\frac{E_{b}}{2N_{o}}} when Φ = 45o.

or  Pe of a coherent BPSK system is Pe

P_{e BPSK} = \frac{1}{2}erfc(\sqrt{\frac{E_{b}}{N_{o}}}) here Eb = Eb /2

there are two BPSK systems P^{'}_{e1} = P^{'}_{e2} = \frac{1}{2}erfc\sqrt{\frac{E_{b}}{2N_{o}}} the bit errors in the in-phase and Quadrature-phase channels of QPSK system are stastically independent.

The in-phase channel makes a decision on one of the two bits constituting a symbol.(di-bit) of the QPSK signal and the Q-channel takes care of the other bit. 

The Average probability of a correct decision resulting from the combined action of the two channels working together is 

P_{c} = (1-P^1_{e1})(1-P^1_{e2})

      =(1-P^{1}_{e1}) ^{2} as P^{1}_{e1} = P^{1}_{e2}.

      = 1+ P^{2}_{e1} - 2P_{e1}

Normally P_{e1} is very very small,P_{e1} <<1 —–> P^{2}_{e1} is negligible

P_{c}= 1-2P_{e BPSK}

P_{e} of coherent QPSK is P_{eQPSK} = 1-P_c{}

                                                               = 1-(1-2P_{e1})

                                                              = 2P_{e1}

P_{eQPSK}= 2X\frac{1}{2}erfc\sqrt{\frac{E_{b}}{2N_{o}}}

P_{eQPSK}= erfc\sqrt{\frac{E_{b}}{2N_{o}}}

here E_{b} is the energy per bit, in terms of symbol energy E_{s}=2E_{b} in QPSK 

P_{eQPSK}= erfc\sqrt{\frac{E_{s}}{4N_{o}}} where E_{s} is the energy per symbol in QPSK.

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Author: vikramarka

Completed M.Tech in Digital Electronics and Communication Systems and currently working as a faculty.