1. Net425:Satellite Communications
Networks and Communication Department
LAB 5

2. C-Band GEO Satellite - LinkBudget in Clear Air
TABLE 4.4a
C-band satellite parameters
Transponder saturated output power
20 W
Antenna gain, on axis
20 dB
Transponder bandwidth
36 MHz
Downlink frequency band
GHz
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3. Signal FM-TV analog signal FM-TV signal bandwidth 30 MHz
Minimum permitted overall C/N in receiver
9.5 dB
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4. Receiving C-band station
Downlink frequency
4.00GHz
Antenna gain, on axis
49.7 dB
Receiver IF bandwidth
27 MHz
Receiving system noise temperature
75 K
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Networks and Communication Department

5. Downlink power budget Pt = Satellite transponder output power, 20 W
13.0dBW
Bo = Transponder output backoff
-2.0 dB
Gt = Satellite antenna gain, on axis
20.dB
Gr = Earth station antenna gain
49.7 dB
Lp = Free space path loss at 4 GHz dB
Lant=Edge of beam loss for satellite antenna
-3.0 dB
La = clear air atmospheric loss
-0.2 dB
Lm = Other losses
-0.5 dB
Pr = Received power at earth station
-119.5dBw
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Networks and Communication Department

6. Figure 4.4 (p. 104) A satellite link.
Satellite Communications, 2/E by Timothy Pratt, Charles Bostian, & Jeremy Allnutt
Copyright © 2003 John Wiley & Sons. Inc. All rights reserved.

7. Ex 1
A GEO satellite at a distance of 36,000 km from a point on the earth’s surface Find The received carrier power in dB watts ? The rule : Pr = EIRP + Gr – LP – La – Lta – Lra dBW Where EIRP = effective isotropically radiated power Pt * Gt watt La = attenuation in the atmosphere Lta = losses associated with the transmitting antenna Lra= losses associated with the receiving antenna
Rules are in blue color , parameters are in green color
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8. Given Parameters Using the following parameters :
Transponder saturated output power which is Satellite transponder output power, (Pt) = 20 W
Convert in decibel : 10 log(20) = 13 dB
Antenna gain, on axis for Satellite transmitter which is Gt = Satellite antenna gain, on axis 20.dB
EIRP = pt+Gt = 13+20= 33 dB
Antenna gain, on axis, for Receiving C-band station Gr = Earth station antenna gain = dB
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9. Con. Given Parameters path loss Lp= (4πR/λ) 2
It is not a loss in the sense of power being
absorbed; it accounts for the way energy spreads
out as an electromagnetic wave travels away from
a transmitting source in three-dimensional space.
Using this parameter Downlink frequency (F) GHz
λ= c/f =3*108 / =0.075 m
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10. Con. Given Parameters
Using the parameter distance of 36,000 km ( R) Lp=(4*3.14* /0.075) 2 = watt
Convert in decibel : log( ) =196.5 dB
Constant Lant=Edge of beam loss for satellite antenna dB
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11. Sol. 1 La = clear air atmospheric loss -0.2 dB
Lm = Other losses -0.5 dB
Loss at transmitter Bo = Transponder output backoff dB
The equation will be
Pr = = dBW
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Networks and Communication Department

12. A noise power Pn
A receiving terminal with a system noise temperature TsK and a noise bandwidth Bn Hz has a noise power Pn referred to the output terminals of the antenna where
Pn = kTsBn watts
The receiving system noise power is usually written in decibel units as
N = k + Ts + Bn dBW
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13. Con. A noise power Pn where
k = Boltzmann’s constant = 1.39 × 10^-23 J/K
= dBW/K/Hz
Ts is the system noise temperature in dBK
Bn = noise bandwidth in which the noise
power is measured (the receiver ), in dBHz
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14. Con. A noise power Pn
Note that because we are working in units of power, all decibel conversions are made as 10 log 10(Ts) or 10 log 10 (Bn).
The rule : Pn = 10 log(kTsBn )=10 log(k)+10 log (T)+10 log (Bn)
Ex2: find the noise power Pn ?
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15. Downlink noise power budget in clear air
K = Boltzmann’s constant
-228.6dBW/K/Hz
Ts = System noise temperature, 75 K
18.8 dBK
Bn = Noise bandwidth, 27 MHz
74.3dBHz
N = Receiver noise power
dBW
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16. Sol.2 Using the following parameters :
k = Boltzmann’s constant = 1.39 × J/K
10 log (1.39 × 10^-23) = dBW/K/Hz
Ts = System noise temperature, 75 K log (75)=18.8 dBK
Bn = Noise bandwidth, 27 MHz log (27*10^6) =74.3dBHz
N= =-135.5dBW
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17. C/N ratio in receiver in clear air
The carrier-to-noise ratio, often written CNR or C/N, is a measure of the received carrier strength relative to the strength of the received noise. High C/N ratios provide better quality of reception, and generally higher communications accuracy and reliability, than low C/N ratios.
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18. Ex3 Find carrier-to-noise ratio ? C/N = Pr － N
= dBW － ( dBW)
= 16.0 dB
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