ERROR PROBABILITY IN BPSK | what is error probability of bpsk formula with example

what is error probability of bpsk formula with example ?
(U.P. Tech., Sem. Examination, 2003-04) (10 marks)
We have studied Binary PSK (BPSK) earlier in this chapter. Let us study the detection of BPSK signal using matched filter receiver. We have seen that phase of the carrier is shifted by 180° for two symbols. These two symbols (bits) are represented by two signals as follows:
Binary ‘1’  Þ   x1(t) =  cos (2pfct)                                …(8.118)
and Binary ‘0’ Þ  x2(t) =        cos (2pfct)                             …(8.119)
Here P is normalized power of the carrier; and P = , where A is amplitude of the carrier. From the above two equations we can write,
x2(t) = -x2(t)                                                     …(8.120)
We know that probability of error of the matched filter is given as
And for a matched filter detection in presence of white gaussian noise,
Here we know that x(t) = x1(t) – x2(t). For PSK, x2(t) = – x1(t)
Thus,                           x(t) = x1(t) – [- x2(t)] = 2x1(t)
Hence, equation (8.122) becomes,
As in the previous subsection, it can be shown very easily that,
In the above equation, second integration will be zero since it is integration of cosine wave over one bit period. This we have proved in last subsection. Hence, above equation becomes,
Thus, energy E = Power (P) x bit duration (T).
Substituting the above result in equation (8.123) we get,
Substituting this result in equation (8.121) we get,
On simplification of above equation, we obtain
Error probability in PSK : Pe=                                               …(8.126)

When somewhat higher data rates are required in a band-limited channel than can be achieved with FSK, phase-shift keying (PSK) is often used.

This is the expression which gives error probability of BPSK using matched filter detection. This expression shows that the probability of error depends upon the energy contents of the signal ‘E’. It does not depend upon the shape of the signal. As the energy increases, value of erfc function decreases and the probability of error also reduces. The expression in equation (8.126) can also be expressed in terms of the Q function as under:
Pe = Q
This is because the relation between erfc and Q function is given by
Q(x) =   and
NOTE Let us compare equations (7.126) and (7.117). The `erfc’ function is a monotonic decreasing function. Therefore, the value of erfc  is less than erfc Hence, the probability of error with BPSK technique is less than that with the ASK technique. Therefore, the BPSK system is superior in performance as compared to the ASK system.
EXAMPLE 8.4. A bandpass data transmission scheme uses a PSK signaling scheme with x2(t) = A cos (2p fct) and x1(t) = – A cos (2pfct), 0 ≤ t ≤ Tb, Tb = 0.2 m sec and fc = 5 fb. The carrier amplitude at the receiver input is 1 millivolt and power spectral density (psd) of the additive white gaussian noise at the input is 10-11 watt/Hz. Assume that an ideal correlation receiver is used. Calculate probability of error, Pe, of the receiver.
Solution: The amplitude of the carrier is,
A = 1 x 10-3 v
psd of white noise is,  watt/Hz
Bit duration Tb = 0.2 x 10-3 sec
The normalized power of the carrier in 1W load is,
Ps =  A2
Bit energy is, Eb = PS Tb (i.e., Power x Bit duration)
Eb =  A2Tb = x (1 x 10-3)2 x 0.2 x 10-3 = 1 x 10-10
Error probability of PSK is given by
Pe =
or                                             Pe =    Here E = Eb = bit energy
EXAMPLE 8.5. Determine the bit error probability (BEP) for a BPSK system having a bit rate of 1 Mbits/s. The receiver receives the waveforms s1(t) = A cos ct and s2(t) = – A cos ct. The received signals are coherently detected using a matched filter. If A = 10 mV and single sided noise power spectral density is N0 = 10-11 W/Hz. Assume that the signal power and energy per bit are normalized.
Solution : We know that                                A =   = 10 x 10-3 V
But, T = one bit period  =   = 1 μs
Therefore,                    Eb =  =  = 5 x 10-11 Joules
This is the energy per bit.
Hence, we have  = 2.24
Therefore, bit error probability will be
Pe =
=  x 0.00041 = 2.05 x 10-4     Ans.
EXAMPLE 8.6. It is desired to generate a random signal x(t), with autocorrelation function Rxx(T) = , by passing a white noise Ƞ(t), with power spectral density Sxx(f) = Ƞ/2 watts/Hz, through a LTI system. Obtain an expression for the transfer function H(f) of the LTI system.
(Gate Examination  1933)
Solution : Given                     Rxx(t) =  ,
Then, power spectral density of signal x(t),
An output, power spectral density
… [using equation (i)]
or                                 |H(f)|2 =
or                                 H(f) =
Example 8.7. Binary data is transmitted at a rate of 106 bits/sec over a chennel having a BW of 3 MHz. Assume that the noise power spectral density (psd) at the receiver is  W/Hz. Determine the average carrier amplitude required at the received at the receiver input for coherent PSK and DPSK signalling schemes to maintain Pe < 10-4.
Solution: We have
Pe(PSK) =  = 10-4
This means that
erfc  ≤ 2 x 10-4
or                                               ≤ 2.6
or                                               ≥ 6.9
or                                             Eb ≥ 6.9 x N0 ≥ 6.9 x 2 x 10-10
or                                             Eb ≥ 6.9 x 2 x 10-10
or                                             Eb =   ≥ 13.8 x 10-10
But, we have
T –
Therefore, using equation (i), we have
or                                             A ≥ 5.26 x 10-2 ≥ 52.6 mV.
or                                             A ≥ 52.6 mV               Ans.

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