Chapter 4 solution to problems
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This document summarizes the solutions to two problems involving satellite communication systems. For the first problem, the document calculates the path loss, received power, noise power, and C/N ratio for a satellite system operating at 6.1 GHz. The C/N ratio at the transponder output was found to be 26.4 dB. The second problem performs similar calculations for a different satellite operating at 3.875 GHz, serving the contiguous US from a distance of 39,000 km. The gain of the satellite antenna was calculated to be 29.6 dB, and the C/N ratio for the receiving earth station was found to be 29.6 dB.
![Chapter 4 Solution to Problems
Question #1.
A C-band earth station has an antenna with a transmit gain of 54 dB. The transmitter output
power is set to 100 W at a frequency of 6.100 GHz. The signal is received by a satellite at a
distance of 37,500 km by an antenna with a gain of 26 dB. The signal is then routed to a
transponder with a noise temperature of 500 K, a bandwidth of 36 MHz, and a gain of 110 dB.
a. Calculate the path loss at 6.1 GHz. Wavelength is 0.04918 m.
Answer: Path loss = 20 log ( 4 R / ) = 20 log ( 4 37,500 103 / 0.04918) dB
Lp = 199.6 dB
b. Calculate the power at the output port (sometimes called the output waveguide flange) of the
satellite antenna, in dBW.
Answer: Uplink power budget gives
Pr = Pt + Gt + Gr - Lp dBW
= 20 + 54 + 26 – 199.6 = -99.6 dBW
c. Calculate the noise power at the transponder input, in dBW, in a bandwidth of 36 MHz.
Answer: N = k Ts BN = -228.6 + 27 + 75.6 = -126.0 dBW
d. Calculate the C/N ratio, in dB, in the transponder.
Answer: C/N = Pr – N = -99.6 + 126.0 = 26.4 dB
e. Calculate the carrier power, in dBW and in watts, at the transponder output.
Answer: The gain of the transponder is 110 dB. Output power is
Pt = Pr + G = -99.6 + 110 = 10.4 dBW or 101.04 = 11.0 W.
2. The satellite in Question #1 above serves the 48 contiguous states of the US. The antenna
on the satellite transmits at a frequency of 3875 MHz to an earth station at a distance of 39,000
km. The antenna has a 6o E-W beamwidth and a 3o N-S beamwidth. The receiving earth station
has an antenna with a gain of 53 dB and a system noise temperature of 100 K and is located at
4-2
the edge of the coverage zone of the satellite antenna. (Assume antenna gain is 3 dB lower than
in the center of the beam)
Ignore your result for transponder output power in Question 1 above. Assume the
transponder carrier power is 10 W at the input port of the transmit antenna on the satellite.
a. Calculate the gain of the satellite antenna in the direction of the receiving earth station.
[Use the approximate formula G = 33,000/(product of beamwidths).]
Answer: G = 33,000 / ( 6 x 3) = 1833 or 32.6 dB on axis.
Hence satellite antenna gain towards earth station is 32.6 – 3 = 29.6 dB.](https://image.slidesharecdn.com/chapter4solutiontoproblems-120612072838-phpapp02/85/Chapter-4-solution-to-problems-1-320.jpg)
Chapter 4 solution to problems
- 1. Chapter 4 Solution to Problems Question #1. A C-band earth station has an antenna with a transmit gain of 54 dB. The transmitter output power is set to 100 W at a frequency of 6.100 GHz. The signal is received by a satellite at a distance of 37,500 km by an antenna with a gain of 26 dB. The signal is then routed to a transponder with a noise temperature of 500 K, a bandwidth of 36 MHz, and a gain of 110 dB. a. Calculate the path loss at 6.1 GHz. Wavelength is 0.04918 m. Answer: Path loss = 20 log ( 4 R / ) = 20 log ( 4 37,500 103 / 0.04918) dB Lp = 199.6 dB b. Calculate the power at the output port (sometimes called the output waveguide flange) of the satellite antenna, in dBW. Answer: Uplink power budget gives Pr = Pt + Gt + Gr - Lp dBW = 20 + 54 + 26 – 199.6 = -99.6 dBW c. Calculate the noise power at the transponder input, in dBW, in a bandwidth of 36 MHz. Answer: N = k Ts BN = -228.6 + 27 + 75.6 = -126.0 dBW d. Calculate the C/N ratio, in dB, in the transponder. Answer: C/N = Pr – N = -99.6 + 126.0 = 26.4 dB e. Calculate the carrier power, in dBW and in watts, at the transponder output. Answer: The gain of the transponder is 110 dB. Output power is Pt = Pr + G = -99.6 + 110 = 10.4 dBW or 101.04 = 11.0 W. 2. The satellite in Question #1 above serves the 48 contiguous states of the US. The antenna on the satellite transmits at a frequency of 3875 MHz to an earth station at a distance of 39,000 km. The antenna has a 6o E-W beamwidth and a 3o N-S beamwidth. The receiving earth station has an antenna with a gain of 53 dB and a system noise temperature of 100 K and is located at 4-2 the edge of the coverage zone of the satellite antenna. (Assume antenna gain is 3 dB lower than in the center of the beam) Ignore your result for transponder output power in Question 1 above. Assume the transponder carrier power is 10 W at the input port of the transmit antenna on the satellite. a. Calculate the gain of the satellite antenna in the direction of the receiving earth station. [Use the approximate formula G = 33,000/(product of beamwidths).] Answer: G = 33,000 / ( 6 x 3) = 1833 or 32.6 dB on axis. Hence satellite antenna gain towards earth station is 32.6 – 3 = 29.6 dB.
- 2. b. Calculate the carrier power received by the earth station, in dBW. Answer: Calculate the path loss at 3.875GHz. Wavelength is 0.07742 m. Path loss = 20 log ( 4 R / ) = 20 log ( 4 39,000 103 / 0.07742) dB Lp = 196.0 dB Downlink power budget gives Pr = Pt + Gt + Gr - Lp dBW = 10 + 29.6 + 53 – 196.0 = -103.4 dBW c. Calculate the noise power of the earth station in 36 MHz bandwidth. Answer: N = k Ts BN = -228.6 + 20 + 75.6 = -133.0 dBW d. Hence find the C/N in dB for the earth station. Answer: C/N = Pr – N = -103.4 + 133.0 = 29.6 dB