# ELEC5510_Satellite Communication Systems_2013 Semester 2_Tutorial 4 solution(1)

Tutorial 4 solutions

Q1:

a).

The receiver noise power N 2 is

228.6 10 70 10log 7.5 10 141.1

The carrier power at the input to the earth stations receiver is given by 123.4 The satellite receiver SNR

is given by

, where

is the propagations loss.

80.6 20 4 36 10 0.05 17.6 20 23.25 The contributions of the transponder noise to the total noise power at the

earth station is 123.4 23.25 146.65 b). 1 16.8

Q2:

a).

From figure 52 in lecture notes,

15.5 10 100 15.5 20 35.5 3548

b).

For P=100w, on a single radio link, the signal would be satisfactory 70% of time.

Q3:

a).

80 100 20 .

b).

Assuming a Rayleigh distribution, a fade exceeding 20dB occurs for 1% of time. This corresponds to about 36 second per hour. (Fig. 52 in lecture notes)

c).

To obtain an outage of 3.6 second per hour, or 0.1%, Fig. 52 indicates that a fade margin of 30dB is necessary. This can be achieved by increasing P r by 30-20=10dB.

d).

The reliability is 99.5% if the power is doubled.

e).

An improvement of 99% to 99.99% is achieved by using two-branch selection diversity. (Fig. 52 in lecture notes)

Q4:

a).

The coherence bandwidth is given by 1 1

15923.6

b).

The channel is frequency selective relative to the signal with bandwidth B r , since B>> B c .

c).

The ratio of B/B c is about 10. Hence, up to tenth order diversity is achieved by subdividing the channel bandwidth into 10 sub channels, each with bandwidth of 16 kHz.

However, increasing the diversity order above five is not very effective. We may therefore decide to choose the diversity order of 5 and increase the

bandwidth of each parallel channel to 32 kHz. Then the maximum possible symbol rate would be 32 ks/s and the maximum data rate of 64 kb/s.

d).

The transmit power reduction with a 3rd order selection diversity combiner is 12dB.

Q5:

a).

The maximum Doppler shift in Hz is

110 10 10

101.852

The bandwidth of the received carrier B c is

2 203.704

Since the Doppler shift can go both ways depending on the directions of reflected energy.

b).

The time it takes the vehicle to travel a half wavelength is / 4.909 Thus we can expect significant fades about every 4.0909ms or at a rate of 203.707Hz.

Q6:

a).

The symbol period is given by 1 1 /2 12000

0.5 Since T s is less than 5ms, the channel is frequency selective and equalizer is necessary.

b).

Δ 10 , thus the whole delay consists of 10 symbol delays and the module consists a delayed tap line with 11 taps:                     Coefficients a1, a2... a11 are Rayleigh distributed. Q7: c

Q8: c

Q9: b

Q10: a

Q11: c

Q12: a