Mumbai University > EXTC > Sem 8 > Satellite Communication and Networks

Marks: 5M

There will be east and west limits on the geostationary arc visible from any given earth station. This is referred to as the Limits of Visibility for that earth station. The figure below illustrates the limits of visibility for an earth station at the equator.

The Limits of Visibility depend on

Geographic coordinates of the earth station and

Antenna elevation.

For a geo stationary satellite to be visible from an earth station its elevation angle must be above some minimum value, which is at least 00.

Expression for Limits of Visibility:

Case 1:When earth station is at equator

Consider an Earth station at the equator with the antenna pointing either east or west along the horizontal, as shown in Fig. The limiting angle $γ_max$ is given by

In this case γ represents not only the central angle subtended by the earth station and satellite at the centre of the earth but also the longitudinal separation between the earth station and the satellite. Thus for this situation, an earth station could see satellites over a geostationary arc bounded by ±81.3° about the earth station longitude.

Hence the satellites which are in the limits of visibility of an earth station at the equator would be located between longitudes,$ \theta_e±\gamma=\theta_e±81.3^0$where le is the Earth station longitude.

In practice, to avoid reception of excessive noise from the earth, some finite minimum value of elevation is used.

$\because \sigma_min=90^0+E_{min}$

[A typical value is$ El_{min}=5^0]$ This is illustrated by dotted lines in the fig. above

Applying sine rule to the ∆OSE, we get

$\frac{sin {\gamma}}{d}$=$\frac{sin {S}}{r_e}$=$\frac{sin {\sigma_{min}}}{r_s}$

$\because s=sin^{-1}\Big(\frac{r_e}{r_s}sin\sigma_{min}\Big)\ = sin^{-1}\Big(\frac{r_e}{r_s}sin(90^0+El_{min})\Big)=sin^{-1}\Big(\frac{r_e}{r_s}cosEl \Big) $

For $El_{min}=5^0$

$\because S=sin^{-1}(\frac{6378}{42164}cos 5^0)=sin^{-1}(0.15069)=8.66^0$

Also, $\gamma=180^0-\sigma_{min}-S=180^0-90^0-El_{min}-S=90^0-El_{min}-S$

$\gamma=90^0-5-8.66=76.34^0$

Hence the satellites which are in the limits of visibility of an earth station at the equator would be located between longitudes $\theta_e±\gamma=\theta_e±76.3^0$

Case 2:

When earth station is at a non zero latitude (Not on equator)

The limits of visibility will also depend on the earth station latitude if the earth station is not at the equator. In such a case, the limits of visibility are evaluated as follows:

i. Calculate central angle γ as described above.

ii. Use the relation:

$cosγ ={ cosλ_e}{cos_β}$ where β is the difference between satellite & earth station longitude.

iii. Hence the satellites which are in the limits of visibility of an earth station at a latitude λe would be located between longitudes θe ± β = θe ± β, where θe is the Earth station longitude.

The maximum central angular separation between earth station and the sub satellite point is limited by:

Therefore maximum value of γ works out to be 81.30.

In practice to avoid the reception of excessive noise from earth, some finite minimum value of elevation is used, which will be denoting by Elmin. A typical value is 5°.