ERROR PROBABILITY IN BFSK | Binary data is transmitted over a microwave link at the rate of 1 Mbps and

By   July 28, 2020

ERROR PROBABILITY IN BFSK
Earlier, in this chapter, we studied binary FSK transmission. Two different carrier frequencies are used to transmit two binary levels. As discussed earlier, these two signals are as follows:
Binary ‘1’  Þ   x1(t) = A cos (c + W) t
and Binary ‘1’     Þ     x2(t) = A cos (c – W) t                                    …(8.127)
The general expression for probability of error Pe is given by
equation
The detector which maximises the ratio  =
=  is the matched filter detector.
Now, output SNR [i.e., (SNR)0], 2 can be obtained as
equation
By Parseval’s theorem, we have
equation
Where                          P(t) = x1(t) – x2(t)
⸫                                                           equation
equation
equation
Integrating and substituting limits, we get
equation
equation
If we assume that offset angular frequency ‘W’ is quite small in comparison with the carrier angular frequency, then the last three terms of the equation (8.128) are of the following form:
The ratio approaches 0 as 0T increases.
As 0T > > 1, the above expression simplifies as under:
equation
for orthogonal tone spacing, WT = np and  the above expression reduces to
equation
⸫         We get
Hence, the probability of error Pe is given by
equation
equation
The energy per bit is given by
equation
equation
EXAMPLE 8.8. Binary data is transmitted over a microwave link at the rate of 1 Mbps and psd at the input of the receiver is 10-7 W/Hz. Determine the average carrier power required to maintain an average probability of bit error ≤ 10-4 for coherent FSK. What will be the channel bandwidth?
Solution: For FSK, we have
Eb = . T. P. (x1) + . T. P(x2) =
We have
Pe = Q  < 10-4
From Q-table, Q[3.71] = 10-4.
Therefore,
or                     Ac =
i.e.                   Ac, = 2.35 mV
            i.e. Minimum value of Ac, = 2.35 mV.
⸫         Minimum average carrier power  = 2.76 watts.
Channel Bandwidth    =  =   = 1 MHz.                       Ans.
EXAMPLE 8.9. Binary data are transmitted over a microwave link at the rate of 106 bits/sec. Given Ƞ/2 = 10-7 W/Hz, A = 2.3 V. Determine the average probability of bit error
            (i)         for coherent FSK receiver
            (ii)        for non-coherent FSK receiver.
Solution: For FSK, we have
EQUATION
eB = .
(i) For coherent FSK receiver, we have
equation
(ii) For non-coherent FSK reciever, we have
equation
EXAMPLE 8.10. Binary data is transmitted over a bandpass channel at a rate of 300 bps using non-coherent FSK signalling scheme with tone frequency 1070 and 1270 Hz. Calculate the probability of error Pe assuming A2/N0 = 8000.
Solution: Here, we have
Pe
But,                                         Eb =
Therefore, we write
Pe
Also,                                        T =
Hence,                         Pe =
Pe =  = 6.34 x 10-4   Ans.
 
EXAMPLE 8.11. For an FSK system, the following data are observed. Transmitted binary data rate = 2.5 x 106 bits/sec power spectral density (psd) of noise = 10-20 watts/Hz. Amplitude of received signal = 1μV. Determine the average probability of symbol error assuming coherent detection.
Solution: Here, we have Pe = erfc                   T = Tb =
equation
Also,                            psd =  = 10-20 W/Hz/
equation
Pe =  erfc (2.23)
or                     Pe =  x 1.84 x 10-3 = 0.92 x 10-3 (from erfc table)     Ans.

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