Explore BrainMass

Analysis of link budget DTH QPSK Satellite link

This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!

A direct-to-home satellite television system uses a geostationary satellite in orbit above the equator at a longitude of 109 degrees west.

The parameters of the satellite transmission are as follows:
Transmit power 160 dBW
Transmit antenna gain (edge of coverage) 31 dB
Distance to earth 38,500 km
Transmission frequency 12.5 GHz
Atmospheric and other losses 1.0 dB

The receiving earth station has the following parameters.
Receiving antenna gain 34 dB
Receive system noise bandwidth 30 MHz
System noise temperature in clear air 90 K

a. Calculate the path loss Lp for the link. Remember to put R into meters.

b. Calculate the power received Pr, by a receiving terminal (earth station) at the edge of the satellite transmit beam (edge of coverage area).

c. Calculate the noise power, N, in the receiver, in dBW.

d. Hence find the CNR in the receiver in clear air conditions.

e. The transmissions from the satellite use QPSK modulation and the earth station receiver has an implementation margin of 0.2 dB. Find the average BER at the receiver output in clear air conditions.

f. Rain causes the CNR to fall by 4 dB. Recalculate the BER. If the BER has risen above 10^-6, suggest one way in which it could be reduced to an acceptable level (say < 10^-8).

© BrainMass Inc. brainmass.com March 21, 2019, 9:07 pm ad1c9bdddf

Solution Preview

(a) Path loss in this example is the sum of the free space loss FSL and other losses over the path, i.e. Atmospheric losses LA = 1.0 dB

The free space loss is given in terms of the operating wavelength of the transmission (Lamda) and the distance of the GEO satellite downlink to Earth (r) by (1) where Lamda and r measured in m.

FSL = 20*log10{4pi*r/Lamda} (1)

Putting in values:

r = 38,500 km = 3.85 x 10^7 m

Lamda = c/f where c is the speed of light, c = 3 x 108 m/s, f is the frequency f = 12.5 GHz

Lamda = (3 x 10^8)/(12.5 x 10^9) = 0.024 m

Therefore using (1)

FSL = 20*log10{(4pi x 3.85 x 10^7)/0.024} = 206.1 dB

So total path loss in downlink will be

LP = FSL + LA = 206.1 + 1.0 dB = 207.1 dB

(b) Power received (Pr) at edge of the satellite coverage area can be deduced by performing the downlink power link budget and thus the power received is given by (2) as the dB sum of the power transmitted {Pt + Gt}, the gain of the receive site antenna Gr minus the total path loss LP. Note that Pt is the actual dB power transmitted and Gt is the gain of the transmit antenna in the direction of edge coverage.

Pr = {Pt + Gt} + Gr - LP (2)

Transmit power is given as Pt = 160 W so we first need to convert this to a dBW equivalent. This can be achieved using identity (3).

Pr(dBW) = 10*log10{Pr/1 W} (3)

Pr(dBW) = 10*log10{160/1} = 22 dBW

We are given the transmit gain in the direction of beam edge coverage as Gt = 31 dB, the receive site antenna gain pointing at the satellite as ...

Solution Summary

A direct-to-home (QPSK) satellite television system uses a geostationary satellite in orbit above the equator. Some typical link parameters are determined including path loss over the link to Earth, the power received at the edge of the transmit beam, the noise power in the ground receiver, the CNR in clear sky conditions and the BER in rain conditions from known inputs