**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 x_{1}(t) = *A* cos (_{c} + W) *t*

and Binary ‘1’ Þ x_{2}(t) = *A* cos (_{c} – W) *t *…(8.127)

The general expression for probability of error P* _{e}* 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) = x

_{1}(t) – x

_{2}(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

_{0}T increases.

As

_{0}T > > 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 P

*is given by*

_{e}**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

E

*= . T. P. (x*

_{b}_{1}) + . T. P(x

_{2}) =

We have

P

*= Q*

_{e}__<__10

^{-4}

From Q-table, Q[3.71] = 10

^{-4}.

Therefore,

or A

_{c}=

i.e. A

_{c}, =

**2.35 mV**

**i.e. Minimum value of A**

*, = 2.35 mV.*

_{c}⸫ 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 10**

^{6}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 P**

*assuming A*_{e}^{2}/N_{0}= 8000.**Solution:**Here, we have

P

_{e}

But, E

*=*

_{b}Therefore, we write

P

_{e}

Also, T =

Hence, P

*=*

_{e}P

*= = 6.34 x 10*

_{e}^{-4}

**Ans.**

**EXAMPLE 8.11. For an FSK system, the following data are observed. Transmitted binary data rate = 2.5 x 10**

^{6}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 P

*= erfc T = T*

_{e}_{b}=

**equation**

Also, psd = = 10

^{-20}W/Hz/

**equation**

P

*= erfc (2.23)*

_{e}or P

*= x 1.84 x 10*

_{e}^{-3}= 0.92 x 10

^{-3}(from erfc table)

**Ans.**