King Fahd University of Petroleum and Minerals
Department of Electrical Engineering
EE 418 Introduction to Satellite Communication
Section 01 (071)
Homework Assignment # 2 (Due Sunday, 25 November 2007)
As this homework set is relatively simple, you are asked to work on it alone.
No cooperation between students is allowed in solving this homework
assignment
1.
A satellite receiver has an overall System Temperature of TS = 97.45 K, and the
following other specifications:
TIn = 35 K
TRF = 50 K
TM = 250 K
TIF = Unknown
GRF = 1000
GM = 0.125
GRF = 10
a) Find the noise temperature of the IF stage (TIF),
b) If the bandwidth of the signal we are interested in is 20 MHz, find the noise power
that would be accompanying this signal at i) the output of the antenna, ii) the
output of the RF stage, iii) the output of the mixer, and iv) the output of the IF
stage.
a) We know that the System Noise Temperature TS is given by
T S  T In T RF 
TM
T IF

G RF G RF G M
So,
97.45  35  50 
T IF
250

1000 (1000)(0.125)
T IF  1525 K
b) The noise power that would accompany the signal at
i)
the output of the antenna is a function of TIn only
Pat output of antenna  k  T In   B N
 1.39*1023   35  20*106 
ii)
the output of the RF stage is a function of TIn , TRF , and GRF
Pat output of RF  k  T In T RF  G RF   B N
 1.39*1023   35  50 1000    20*106 
iii)
the output of the Mixer stage is a function of TIn , TRF , TM , GM and GRF


Pat output of M  k  T In T RF  G RF T M  G M  B N


 1.39*1023   35  50 1000   250 0.125  20*106 
iv)
the output of the IF stage is a function of TIn , TRF , TM , TIF , GIF , GM
and GRF


Pat output of IF  k  T In T RF  G RF T M  G M T IF G IF  B N


 1.39*1023   35  50 1000   250  0.125  1525 10  20*10 6 
Note that this could have also been obtained using
Pat output of IF  k  T S G RF G M G IF   B N
2.
A circular antenna on a GEO satellite is transmitting signals in the frequency range
10.2 GHz to 12.5 GHz. Determine the minimum diameter of the antenna in each case
(consider the worst situation in each case) that will be able to provide:
a) a minimum 3-dB coverage area on Earth with a circular shape around the subsatellite point with radius of 500 km,
b) a minimum antenna gain of 56.80 dB given that the antenna reflector has an
efficiency of 0.7
a) The altitude (not radius) of a GEO is (42,164 – 6,378 = 35,786 km)
So, a circular coverage area of radius 500 km will represent an angle of
3dB
 500 
 tan 1 
  0.800
2
 35, 786 
So,
 500 
  1.600
 35, 786 
3dB  2 tan 1 
Now, consider the two frequency extremes knowing that the 3 dB beamwidth in
  

degrees is given by 3dB  75   or D  75 
:
D 
 3dB 
3*108
 0.0294 
f C  10.2 GHz   
 0.0294 m  D  75 
  1.379 m
9
10.2*10
 1.600 
3*108
 0.024 
f C  12.5 GHz   
 0.024 m  D  75 
  1.125 m
9
12.5*10
 1.600 
For an antenna to work for all of our frequency range and provide coverage
to our area with radius of 500 km, it must have a radius of 1.125 m to do so.
If the larger diameter was used, it will provide coverage for low frequencies
around 10.2 GHz, but when frequency is increased, coverage area will
become smaller than 500 km in radius.
a) Consider the two frequency extremes knowing that the gain is given by
2
 D
G  A     56.8 dB  478, 630
 
or
D
 G  478, 630
:

 A 
0.7
3*108
0.0294 478,630
f C  10.2 GHz   
 0.0294 m  D 
 7.738 m
9
10.2*10

0.7
3*108
0.024 478,630
f C  12.5 GHz   
 0.024 m  D 
 6.317 m
9
12.5*10

0.7
For an antenna to work for all of our frequency range and provide the
minimum required gain, it must have a radius of 7.738 m to do so. If the
smaller diameter was used, it will provide the required gain at the high
frequency around 12.5 GHz, but when frequency is reduced, the gain will
become smaller than 56.8 dB.
3.
Solve Problem 3 on page 93 of your textbook (Remember that the sun provides 1.39
kW/m2 of power. Part of this is absorbed by solar cells depending on their efficiency
and that spinner satellites have an effective area of the satellites illuminated by the
sun that is equal to Height * Width = Height * Circumference / π).
a) For the 3-axis stabilized satellite,
Total Power Output by SCs  Efficincy * Effective SC Area * Sun Radiation per 1 m2
So,
Effective SC Area 
Total Power Output by SCs
Efficincy * Sun Radiation per 1 m2
Therefore,
Effective SC Area 
4 kW
0.15 * 1.39
kW
m2
 19.18 m 2
If the width of the solar sail is 2 m, the length of would be 9.59 m.
b) For the spinner satellite,
Total Power Output by SCs  Efficincy * Effective SC Area * Sun Radiation per 1 m2
So,
Effective SC Area  Height * Diameter =

Height * Circumference

Total Power Output by SCs
Efficincy * Sun Radiation per 1 m2
Therefore,
Effective SC Area 
4 kW
kW
0.18 * 1.39 2
m
 15.98 m 2
If the Diameter of the satellite is 3.5 m, its height would be 4.56 m.
4.
For a satellite downlink with the following specifications
Satellite Output Power
15.00
W
Satellite Output Backoff
2.5
dB
Satellite Antenna Gain
Antenna Efficiency
0.85
Diameter
1.20
m
3
dB
Edge of Beam Loss for Sat. Antenna
Signal Bandwidth
Carrier Signal Frequency
Minimum Permitted C/N ratio at
receiver
Earth Station Antenna Gain
22
11.3
16
Antenna Efficiency
0.6
Diameter
0.85
MHz
GHz
dB
m
Tin = 25 K
Receiver Noise Specifications
TRF = 35 K
GRF = 100
TM = 100 K
GM = 0.1
TIF = 200 K
GIF = 20
Maximum Sat.-Earth Station Distance
40000
km
Clear Air Atmospheric Loss
2
dB
Rain Loss
8
dB
Other Losses
1
dB
a) Fill in the table given below with the proper results obtained from the table above
both in linear units and in dB units (Include a sheet that shows all of your work).
Remember that losses represent gains that are smaller than zero or negative dBs.
Carrier Results
Effective Satellite Transmitted Power
W
dBW
Satellite Antenna Gain
Linear
dB
Earth Station Antenna Gain
Linear
dB
Free Space Path Loss at Edge of Beam
Linear
dB
All Other Losses
Linear
dB
Earth Station Received Carrier Power
Linear
dB
Boltzmann's Constant
J/K
dB(W/K/Hz)
System Noise Temperature
K
dBK
Noise Bandwidth
MHz
dBHz
Earth Station Received Noise Power
W
dBW
Linear
dB
Noise Results
C/N Ratio
C/N Ratio at Earth Station Antenna Output
b) Determine if this system will satisfies the minimum system carrier to noise ratio
(C/N) of not. Determine the (C/N) margin in dBs (remember that the (C/N)
margin is positive if the system performs better than minimum (C/N) permitted
and negative if does not).
c) If you were able to modify the satellite antenna diameter only, what would be the
new satellite antenna diameter that will give a (C/N) margin of +3 dB?
d) If you were able to modify the satellite transmitted power, how much would the
satellite transmit in W for the downlink to have (C/N) margin of +1 dB?
e) Statistics indicate that the coverage area of this satellite experiences rain
attenuation according to table below. If the (C/N) margin obtained in (b) is
completely allocated to rain attenuation, how many hours a year will this
downlink provide a (C/N) ratio below the minimum allowed (called outage time).
Use interpolation between two points to find an accurate outage time.
Attenuation (dB)
1 dB
2 dB
3 dB
4 dB
5 dB
6 dB
7 dB
8 dB
9 dB
10 dB
11 dB
12 dB
13 dB
14 dB
15 dB
16 dB
Duration
(Hours a year)
350
300
250
210
160
130
100
85
70
55
30
22
15
10
7
3
a) To fill in the table, first consider the quantities given in the first table. Assume that these
quantities are represented by the variables shown below:
b)
Satellite Output Power
Satellite Output Backoff
Satellite Antenna Gain
W
Pbackoff
dB
Antenna Efficiency
S
Diameter
DS
m
Ledge
dB
Edge of Beam Loss for Sat. Antenna
Signal Bandwidth
Carrier Signal Frequency
Minimum Permitted C/N ratio at
receiver
Earth Station Antenna Gain
Pmax
BWSignal
fc
MHz
GHz
(C/N)min
dB
Antenna Efficiency
ES
Diameter
DES
m
Tin = 25 K
Receiver Noise Specifications
TRF = 35 K
GRF = 100
TM = 100 K
GM = 0.1
TIF = 200 K
GIF = 20
Maximum Sat.-Earth Station Distance
Dist
km
Clear Air Atmospheric Loss
LA
dB
Rain Loss
LR
dB
Other Losses
LO
dB
Now, the requested table can be filled easily using the following formulas (either the linear of the
dB value is found in each case, the other one can be obtained easily using the 10 log (.) or 10( . /10)
Carrier Results
PT  PMax (dB)
Effective Satellite Transmitted Power
Satellite Antenna Gain
Earth Station Antenna Gain
Free Space Path Loss at Edge of Beam
All Other Losses
W
 DS 
G S  S  

 c /fC 
 D 
G ES  ES   ES 
 c /fC 

Dist 
L P   4

 c /fC 
PBackoff (dB)
dBW
2
Linear
dB
Linear
dB
Linear
dB
2
2
Linear
L A  LR  LO
dB
PR  PT (dB)
G S (dB)
G ES (dB)
L P  L A
Earth Station Received Carrier Power
Linear
 L R  LO
dB
Noise Results
Boltzmann's Constant
1.39*10-23
J/K
dB(W/K/Hz)
K
dBK
MHz
dBHz
W
dBW
T S  T In  T RF

TM
G RF

T IF
G RF G M
System Noise Temperature
Noise Bandwidth
BWSignal
Earth Station Received Noise Power
-23
PN = (1.39*10 )Ts BWSignal
C/N Ratio
C/N Ratio at Earth Station Antenna Output
Linear
(C / N ) 
PR (dB)  PN (dB)
dB
b) If (C/N) is greater than (C/N)min , the system is working fine, otherwise it is not working
properly.
c) The satellite antenna needs to be modified so that the C/N margin is only 3 dB. Weather
the (C/N) in dB was positive or negative, the satellite antenna gain will be changed to
make such that the new (C/N) is greater than (C/N)min by only 3 dB. That is,
The new gain of the antenna GS,new = GS,old – [(C/N) – ((C/N)min + 3)] (All in dB)
Once the new gain is obtained, use D S 
c
fC
G S ,new
S
to obtain the new diameter
d) The same concept as in (c) is applied here to make the C/N margin only 3 dB. Weather
the (C/N) in dB was positive or negative, the satellite needs to transmit an amount of
power in dB that is equal to
The new transmitted power
PT,new = PT,old – [(C/N) – ((C/N)min + 3)] (All in dB)
e) We will have to first obtain the total loss allocated to rain in dB. This can be obtained by
adding using the following formula
LR,max = LR (8 dB as given in problem) + [(C/N) – (C/N)min]
(All in dB)
Once this is found, use interpolation to find the amount of time (using the last table) by
interpolation in the following form (for illustration, let LR,max = 12.35 dB)
TOutage  22
15  22

L R ,max  12
13  12

12.35  12
 0.35
13  12
Solving for Toutage gives the required value.