### 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)= S_{i}(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 N_{o}/2.

Correspondingly, the observation vector x has 2 elements x_{1} and x_{2} defined by and

simillarly

Thus the observable elements x_{1} and x_{2} are sample values of independent Gaussian random variables with mean values equal to and , respectively and with a common variance.

The decision rule is to simply to decide that

x_{1} falls inside the region z_{1} —–> S_{1} was transmitted.

x_{2} falls inside the region z_{2} —–> S_{2} was transmitted.

x_{3} falls inside the region z_{3} —–> S_{3} was transmitted.

x_{4} falls inside the region z_{4} —–> S_{4} 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 S_{4}(t) is transmitted . but the noise W(t) is such that the received signal point falls outside the region z_{4}.

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. x_{1} and x_{2} may be viewed as the individual outputs of the ‘2’ coherent BPSK systems may be characterized as follows.

The signal energy per bit = E_{b}/2 and the Noise power spectral density = N_{o}/2.

from the signal space representation , reference carriers are Φ_{1}(t) and Φ_{2}(t) all the signal vectors S_{1}, S_{2}, S_{3} and S_{4} are at 45^{o} (or multiple of 45^{o}) to these reference carriers. Suppose if S_{1} is transmitted. If the phase-shift of reference carrier is more than 45^{o} (either Φ_{1} or Φ_{2}) it will be detected as some other symbol apart from S_{1}.

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

when Φ = 45^{o}.

or P_{e} of a coherent BPSK system is P^{‘}_{e}

here E_{b} = E_{b} /2

there are two BPSK systems 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

.

Normally is very very small, <<1 —–> is negligible

of coherent QPSK is

here is the energy per bit, in terms of symbol energy in QPSK

where is the energy per symbol in QPSK.