In an FM line of sight link , the frequency is 6 GHz and the beamwidth = 27 MHz

Question

In an FM line of sight link , the frequency is 6 GHz and the beamwidth = 27 MHz the transmitting and

receiving antennas have a gain of 33 dB , the receiving antenna is 40 Km apart from the transmitting

station :

i- find the required transmitter power in order to secure a C/N of 40 dB given that the noise temperature is

700°K and the receiver noise figure=9 dB, you need to account for a 3dB cable loss and other losses on the

receiving side.

ii- find the antenna heights assuming smooth earth in between.

iii- given that the FM thershold C/N = 10dB , find the probability of fading assuming normal terrain and weather.

Explanation / Answer

Given data:

In Fm link has frequency f=6GHz,

Beam width=27MHz

Distance between transmitter and receiver(d)=40 Km

Gain of transmitter and receiver=Gr=Gt=33 db

(i)carrier to noise ratio(C/N)=40 db

Noise temperature(Tant)=700degrees

Noise figure(f)=9db

(i)

We have Boltzman’s constant in db isKTdb=228.6

We have the formula that,

C/No=(EIRP)dbw+Grdb-Tdb-Lsdb+228.6

Tdb=Tant+Te

Here,Te is noise temperature

Te=(F-1)T

Here T is constant, at room temperature,T=273+17=290deg

Fdb=10logF

9=10logF

F=7.94….(1)

By substituting the value,

Te=(7.94-1)290deg

=2012.6deg

Now,

Tdb=10log(Tant+Te)

=10log(700+2012.6)

=10log (2712.6 deg)

=34.33db

Now,

(EIRP)dbw=Wtdb+Gtdb

Now ,

C/No=(EIRP)dbw+Grdb-Tdb-Lsdb+228.6

40db= Wtdb+33 db+33 db-34.33db-3db+228.6db

Therefore, the transmitted power is given as,

Wtdb=217.27 db

(ii)the height of the antenna at the smooth earth given as,

Let us auume both the antennas have same heights.

Then we have the equation as,

D=4.12(square root(hr)+ square root(hr))km

Here, we assume,hr=ht

40km=4.12(2square root(hr))km

hr=ht=94.26 meters.

(iii)

Now,

C/N=10 db

The probability of fading is given as,

We have the formula as

C/N=Wrdb-KTdb-Tdb

10db=217.27db-228.6db-Tdb

Tdb=21.33db

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