SATELLITE COMMUNICATIONS I
1. Calculate the length of the path to a geostationary satellite from an earth station
where the angle of elevation is 30 degrees.
GIVEN:
=30
Geostationary satellite
REQD:
Length of the path (d)
SOLN:
Recall:
d km = [( R+h) R cos ]Rsin
Where:
R = 6371 km is the average earth radius or
6378 km, equatorial earth radius
h = 35 786 km, is the height of the geostationary satellite
is the angle of elevation
Solve for the length of the path (d).
d km = [(6378+35 786)2( 6378 )2 cos 2(30 ) ]( 6378 ) sin(30 )
d km =38 608.88 km
Therefore,
d km =38 611.64 km
2. Calculate the gain of a 3 m paraboloid antenna operating at a frequency of 12 GHz.
Assume an aperture frequency of 0.55.
GIVEN:
D=3m
f = 12 GHz
= 0.55
REQD:
G
SOLN:
Recall:
G=
( )
Where:
is the aperture efficiency
D is the diameter of the parabolic antenna
c
f
c = 3x
,is the wavelength and
10
m/s is the speed of light
Solve for the G.
G=
( )
G=(0.55)
(3 m)
8
3 x 10 m/s
12 x 109 Hz
G=78 167.27
In terms of dB,
GdB=10 logG
G dB=10 log ( 78 167.27)
GdB=48.93 dB
Therefore,
GdB=48.93 dB
3. Calculate the angle of declination for an antenna using a polar mount at latitude of
45 degrees.
GIVEN:
LA = 45
REQD:
Angle of declination (
SOLN:
Recall:
=arctan
RsinLA
h+ R(1cosLA)
Where:
R = 6371 km is the average earth radius or
6378 km, equatorial earth radius
h = 35 786 km, is the height of the geostationary satellite
LA is the earth station latitude
Solve for the
.
=arctan
( 6378 km) sin ( 45 )
35786 km+(6378 km)(1cos ( 45 ))
=6.8299
Therefore,
=6.82
4. A satellite downlink at 12 GHz operates with a transmit power of 6 W and an antenna
gain of 42.2 dB. Calculate the EIRP in dBw.
GIVEN:
f = 12 GHz
Pt =6 W
Gt =42.2 dB
REQD:
EIRP in dBw
SOLN:
Recall:
EIRP dBw =10logP t Gt
Where:
Solve for the
Pt
is the radiated power
Gt
is the transmit gain
EIRPdBw .
EIRPdBw =10 log (
EIRP dBw
6 W 10
1W
42.2
10
= 49.98 dBw
Therefore,
EIRP dBw
= 49.98 dBw
5. An LNA is connected to a receiver which has noise figure of 12 dB. The gain of LNA is
40 dB, and its noise temperature is 120 K. Calculate the overall noise temperature
referred to the LNA input.
GIVEN:
NF = 12 dB
A = 40 dB
T 1 =120 K
REQD:
T overall
SOLN:
Recall:
T e=T 0 ( F1)
Where:
T0
= 17C +273 K, is the noise temperature at room
temperature
Solve for the
is the noise ratio
T2 .
12
10
T 2 =(290 K )(10 1)
T 2 =4306.19 K
Recall:
T overall=T 1 +
T2
A
Where:
A
Solve for the
T overall=120 K +
is the gain of the amplifier
T overall .
4306.19 K
10
40
10
Therefore,
T e=120.43 K
6. A satellite transmitter operates at 4 GHz with a transmitter power of 7 W and an
antenna gain of 40 dBi. The receiver has an antenna gain of 30 dB i, and the path
length is 40,000 km. Calculate the signal strength at the receiver.
GIVEN:
f = 4 GHz
Pt = 7 W
Gt = 40 dBi
Gr = 30 dBi
d = 40 0000 km
REQD:
Pr (dBm)
SOLN:
Recall:
2
P r=
Pt G t G r
16 2 d 2
Where:
Pt
is the transmitted power
Gt
is the transmit gain
Gr
is the receive gain
d is the distance from the source
is the wavelength
Solve for the
P r=
Pr .
3 x 108 m 2
40
30
s
10
(7W )(10 )(10 10 )
4 x 10 9 Hz
16 2(40 000 000 m)2
Pr=1.56 pW
In terms of dBm,
Pr ( dBm ) =10 log
P
( 1mW
)
Pr ( dBm ) =10 log
pW
( 1.56
1 mW )
Pr ( dBm ) =88.07 dBm
Therefore,
Pr ( dBm ) =88.07 dBm
7. A typical TVRO installation for use with C-band satellite (downlink at approximately 4
GHz) has a diameter of about 3 meters and an efficiency of about 55%. Calculate its
gain and beamwidth.
GIVEN:
D=3m
f = 4 GHz
= 0.55
REQD:
G and
SOLN:
Recall:
2
G=
( D )
And
=70
( D ) degrees
Where:
is the aperture efficiency
D is the diameter of the parabolic antenna
c
f
c = 3x
,is the wavelength and
108 m/s is the speed of light
Solve for the G.
G=
( )
G=(0.55)
(3 m)
3 x 108 m/s
4 x 109 Hz
G=8685.25
In terms of dB,
GdB=10 logG
GdB=10 log ( 8685.25)
GdB=39.39 dB
Solve for the
3 x 10 m/ s
4 x 109 Hz
=70
degrees
3m
=(0.55)
(3 m)
8
3 x 10 m/ s
4 x 109 Hz
=1.75 degrees
Therefore,
GdB=39.39 dB
and =1.75 degrees
8. A receiver has a noise figure of 1.5 dB. Find its equivalent noise temperature.
GIVEN:
NF = 1.5 dB
REQD:
Te
SOLN:
Recall:
T e=T 0 ( F1)
Where:
T0
= 17C +273 K, is the noise temperature at room
temperature
F
Solve for the
is the noise ratio
Te .
1.5
T e=(290 K )(10 10 1)
T e=119.64 K
Therefore,
T e=119.64 K
9. A receiving antenna with a gain of 40 dB i looks at a sky with a noise temperature of
15 K. The loss between the antenna and the LNA input, due to the feed line, is 0.4 dB,
and the LNA has a noise temperature of 40 K, Calculate G/T.
GIVEN:
GdB =40 dBi
i
T sky =15 K
L = 0.4 dB
T e=40 K
REQD:
G/T
SOLN:
Recall:
G
( dB )=G dBi10 log ( T a +T e ) L
T
Where:
T a=
290 ( L1 )+T sky
L
, is the effective noise temperature of
antenna and feed line referenced to receiver antenna input in
Kelvin
T sky is the effective sky temperature in Kelvin
L is the loss in feed line and antenna as a ratio of input to
output power
GdBi is the gain of the receiving antenna in dBi
T e is the equivalent noise temperature of the amplifier
First solve for the
Ta .
290 K (10
T =
0.4
10
10
T e=39.2 K
Solve for
G
( dB ) .
T
1 +15 K
0.4
10
G
( dB )=40 dBi10 log ( 39.2 K + 40 K )0.4 dB
T
G
( dB )=20.6 dB
T
Therefore,
G
( dB )=20.6 dB
T
10. In satellite communication system, for a total transmit power of 1000 watts,
determine the energy per bit for a transmission rate of 50 Mbps expressed in dB w.
GIVEN:
Pt = 1000 W
f b = 50 Mbps
REQD:
Eb
N0
SOLN:
Recall:
E b Pt
=
N0 f b
Where:
Pt
is the transmitted power
f b is the bit rate
Solve for the
Te .
Eb
1000 W
=
N 0 50 x 106 bps
Eb
=20 W /bps
N0
6
Eb
20 x 10 W /bps
=10 log (
)
N0
1W
In terms of dBw,
Eb
=46.99 dBw /bps
N0
Therefore,
Eb
=46.99 dBw /bps
N0
11. Assuming a bandwidth of 20 MHz, a coherent BPSK transmitter operates at a bit rate
of 20 Mbps with a carrier-to-noise (C/N) of 8.8 dB. Determine Eb/No.
GIVEN:
C/N
= 8.8 dB
f b = 20 Mbps
REQD:
Eb
N0
SOLN:
Recall:
Eb C
B
= +10 log ( )
N0 N
fb
Where:
C
N
is the carrier-to-noise ratio
B is the bandwidth
f b is the bit rate
Solve for the
Eb
N0 .
Eb
20 x 106 Hz
=8.8 dB+10 log (
)
N0
20 x 10 6 bps
Eb
=8.8 dB
N0
Therefore,
Eb
=8.8 dB
N0
