# Chapter 4 Solution to Problems

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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 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 m. Answer: Path loss = 20 log ( 4 π R / λ) = 20 log ( 4 π 37, / ) db L p = 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 P r = P t + G t + G r - L p dbw = = dbw c. Calculate the noise power at the transponder input, in dbw, in a bandwidth of 36 MHz. Answer: N = k T s B N = = dbw d. Calculate the C/N ratio, in db, in the transponder. Answer: C/N = P r N = = 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 P t = P r + G = = 10.4 dbw or = 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 6 o E-W beamwidth and a 3 o 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 4-1

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 = 29.6 db. b. Calculate the carrier power received by the earth station, in dbw. Answer: Calculate the path loss at 3.875GHz. Wavelength is m. Path loss = 20 log ( 4 π R / λ) = 20 log ( 4 π 39, / ) db L p = db Downlink power budget gives P r = P t + G t + G r - L p dbw = = dbw c. Calculate the noise power of the earth station in 36 MHz bandwidth. Answer: N = k T s B N = = dbw d. Hence find the C/N in db for the earth station. Answer: C/N = P r N = = 29.6 db 3. A 14/11 GHz satellite communication link has a transponder with a bandwidth of 52 MHz which is operated at an output power level of 20W. The satellite transmit antenna gain at 11 GHz is 30 db towards a particular earth station. Path loss to this station is 206 db, including clear air atmospheric loss. The transponder is used in FDMA mode to send 500 BPSK voice channels with half rate FEC coding. Each coded BPSK signal has a symbol rate of 50 kbps and requires a receiver with a noise bandwidth of 50 khz per channel. The earth stations used to receive the voice signals 4-2

3 have antennas with a gain of 40 db (1m diameter) and a receiver with T system = 150K in clear air, and IF noise bandwidth 50 khz. a. Calculate the power transmitted by the satellite in one voice channel. Answer: In FDMA, the output power of the transmitter is divided equally between the channels. For P t = 20 W and 500 channels, power per channels is 20 / 500 = 40 mw/ch. b. Calculate the C/N in clear air for an earth station receiving one BPSK voice signal. Answer: Each channel receiver has a noise bandwidth of 50 khz or 47 dbhz. Path loss at 11GHz is db, including atmospheric loss.. Downlink power budget for one FDMA channel gives P r = P t + G t + G r - L p dbw = = dbw The noise power at the input to the receiver is N = k T s B N = = dbw Hence C/N = P r N = = 9.8 db. c. What is the margin over a coded BPSK threshold of 6 db? Answer: Margin is receiver C/N minimum permitted C/N, in db Margin = = 3.8 db. 4. Geostationary satellites use L, C, Ku and Ka bands. The path length from an earth station to the GEO satellite is 38,500 km. For this range, calculate the path loss in decibels for the following frequencies: Note: Round all results to nearest 0.1 db. a. 1.6 GHz, 1.5 GHz Wavelengths are: 1.6 GHz, λ = m; 1.5 GHz, λ = m. Answer: Path loss = 20 log ( 4 π R / λ) For 1.6 GHz, L p = 20 log ( 4 π 38, / ) = db For 1.5 GHz, L p = 20 log ( 4 π 38, / 0.200) = db 4-3

4 Path loss at frequency f 2 can be found from path loss at frequency f 1 by scaling: L p (f 2 ) = L p (f 1 ) + 20 log (f 2 / f 1 ). Using the result for 1.6 GHz, L p = db: b. 6.2 GHz, 4.0 GHz Answer: At 6.2 GHz, L p = db + 20 log (6.2 / 1.6) = db At 4.0 GHz, L p = db + 20 log (4.0 / 1.6) = db c GHz, 12.0 GHz Answer: At GHz, L p = db + 20 log (14.2 / 1.6) = db At 12.0 GHz, L p = db + 20 log (12.0 / 1.6) = db d GHz 20.0 GHz Answer: At 30.0 GHz, L p = db + 20 log (30 / 1.6) = db At 20.0 GHz, L p = db + 20 log (20 / 1.6) = db Note: All commercial satellite systems have path losses that fall within the above range, excepting any in the vhf and uhf bands, and above 40 GHz. 5. Low earth orbit satellites use mainly L band, with ranges varying from 1000 km to 2,500 km. Calculate the maximum and minimum path loss from earth to a satellite, in db, for the uplink frequency of 1.6 GHz, and the downlink frequency of 1.5 GHz. Answer: Wavelengths are: 1.6 GHz, λ = m; 1.5 GHz, λ = m. Path loss = 20 log ( 4 π R / λ) For 1.6 GHz, Maximum L p = 20 log ( 4 π 2, / ) = 64.5 db For 1.5 GHz, L p = 20 log ( 4 π 38, / 0.200) = db 6. A geostationary satellite carries a transponder with a 20 watt transmitter at 4 GHz. The transmitter is operated at an output power of 10 watts and drives an antenna with a gain of 30 db. An earth station is at the center of the coverage zone of the satellite, at a range of 38,500 km. Using decibels for all calculations, find: 4-4

6 Path loss L p = 20 log (4 π R / λ) = 10 log (4 π / 0.120) = db Downlink power budget gives P r = P t + G t + G r - L p - losses dbw = = dbw b. The noise power of the earth station receiver for a noise temperature of 260K and an RF channel bandwidth of 20 khz. Answer: The noise power at the input to the receiver is N = k T s B N = = dbw c. The C/N ratio in db for the LEO signal at the receiver output. Answer: C/N = P r N = = 8.1 db. 8. A satellite in GEO orbit is a distance of 39,000 km from an earth station. The required flux density at the satellite to saturate one transponder at a frequency of 14.3 GHz is dbw/m 2. The earth station has a transmitting antenna with a gain of 52 db at 14.3 GHz. Find: a. The EIRP of the earth station Answer: EIRP = P t + G t = P t + 52 dbw Flux density is given by F = 20 log [ EIRP / (4 π R 2 ) ] dbw/m 2 Hence for R = 39,000 km, f = 14.3 GHz, λ = m F = = EIRP - 10 log (4 π) - 20 log (39, ) dbw / m = EIRP dbw / m 2 EIRP = = 72.8 dbw b. The output power of the earth station transmitter. Answer: EIRP = P t + G t = 72.8 dbw. Hence P t = = 20.8 dbw. 4-6

7 9. A 12 GHz earth station receiving system has an antenna with a noise temperature of 50K, a LNA with a noise temperature of 100 K and a gain of 40 db, and a mixer with a noise temperature of 1000 K. Find the system noise temperature. Answer: System noise temperature is calculated from T s = T antenna + T LNA + T mixer / G LNA +. Hence for G LNA = 40 db = 10,000 as a ratio T s = / 10,000 = K 10. A geostationary satellite carries a C-band transponder which transmits 20 watts into an antenna with an on-axis gain of 30 db. An earth station is in the center of the antenna beam from the satellite, at a distance of 38,000 km. For a frequency of 4.0 GHz: a. Calculate the incident flux density at the earth station in watts per square meter and in dbw/m 2. Answer: Flux density is given by F = 20 log [ EIRP / (4 π R 2 ) ] dbw/m 2 Hence for R = 38,000 km, f = 4.0 GHz, λ = m, EIRP = = 43.0 dbw F = log (4 π) - 20 log (38, ) dbw / m 2 = = dbw / m 2 b. The earth station has an antenna with a circular aperture 2 m in diameter and an aperture efficiency of 65%. Calculate the received power level in watts and in dbw at the antenna output port. Answer: The effective area of the antenna is A eff = η A π r 2 = 0.65 π 1 = m or 3.1 dbm 2 For an incident flux density of dbw / m 2 or W/m 2 P r = = W or dbw or P r = = dbw 4-7

9 c. Calculate the on-axis gain of the antenna in db. Answer: At f = 20 GHz, λ = m: Antenna gain for a circular aperture is given by G = η A (π D / λ) 2 G = 10 log ( 0.65 (π 2 / ) 2 ) = 50.6 db d. Calculate the free space path loss between the satellite and the earth station. and the power received, Pr, at the earth station using the link equation: Answer: At a frequency of 20.0 GHz, λ = m. Path loss = 20 log ( 4 π R / λ) = 20 log ( 4 π 38, / 0.015) db L p = db Downlink power budget gives P r = P t + G t + G r - L p dbw = = dbw The answer is the same as in Question 10 for the 4 GHz satellite because the incident flux density on the earth s surface is the same, and the antenna effective area is the same. Note to instructor: Questions 12, 13 and 14 are based on problems from Take Home exams. A significant length of time is needed to complete the solutions. 12. This sequence of questions requires you to design a communication link through a geostationary satellite to meet a C/N and link margin specification. Use these constants: Boltzmann's constant k = dbw/k/hz Path length to satellite = 38,500 km Satellite: Geostationary at 73 o W longitude. 24 C band transponders, 28 Ku band transponders 3.2 kw RF power output Antenna gain, on axis, C-band and Ku-band (transmit and receive) = 31 db Receive system noise temperature (C-band and Ku-band) = 500 K Transponder saturated output power: C-band = 40 W Transponder bandwidth: C-band = 36 MHz 4-9

10 Transponder saturated output power: Ku-band = 80 W Transponder bandwidth: Ku-band = 54 MHz Signals: FM-TV analog signal to be received in a bandwidth of 27 MHz Multiplexed digital TV signals transmitted as QPSK with symbol rate 27 Msps using half rate FEC with coding gain 5.5 db Minimum permitted C/N overall = 9.5 db Question #12.1 Design a transmitting earth station to provide a clear air C/N of 26 db in a C-band transponder at a frequency of GHz. Use an uplink antenna with a diameter of 9 m and an aperture efficiency of 68%, and find the uplink transmitter power required to achieve the required C/N. The uplink station is located on the 2 db contour of the satellite footprint. Allow 0.5 db for clear air atmospheric attenuation and other losses. Answer: An uplink power budget and noise budget are required to find C/N at the satellite transponder input. Calculate the transmitting earth station antenna gain and path loss at GHz first. G t = η A (π D / λ) 2 = 10 log ( 0.68 (π 9 / ) 2 ) = 53.8 db L p = 20 log ( 4 π R / λ) = 20 log ( 4 π 38, / ) = db Uplink power budget P t G t G r 31 db 2 db off axis loss L p Atmospheric loss and other losses Receiver power P r Transponder input noise power budget k Boltzmann s constant TBD 53.9 db 29.0 db db -0.5 db P t dbw dbw/k/hz T s 500 K 27.0 dbk B N 27 MHz (FM-TV) 74.3 dbhz N dbw 4-10

11 The C/N ratio at the transponder input is C/N = C N = P t db For C/N = 26 db we require P t = C/N 9.6 = 16.4 dbw or 43.7 W Question #12.2 Design a C-band receiving earth station to provide an overall clear air C/N of 13 db in a 27 MHz IF noise bandwidth at a carrier frequency of 4.06 GHz. The antenna noise temperature is 20 K and the LNA noise temperature is 55 K. You may assume a high gain LNA and ignore the noise generated in other parts of the receiver. The C-band satellite transponder is operated with 1 db output backoff. Clear air atmospheric attenuation on the downlink and other losses total 0.5 db. Determine the diameter of the receiving antenna, assuming an aperture efficiency of 65%. The receiving terminal is located on the 3 db contour of the satellite footprint. Reminder: Overall C/N includes the effect of noise radiated by the satellite transponder. Answer: Set the receiving earth station antenna gain as G r db. Path loss at the downlink frequency of 4.06 GHz is L p = log (4.06/6.285) = db Transponder output power is P t saturated 1 db = 16 1 = 15.0 dbw Downlink power budget P t G t 31 db 3 db off axis loss G r L p Atmospheric loss and other losses Receiver power P r Earth station receiver input noise power budget k Boltzmann s constant 15.0 dbw 28.0 db TBD db -0.5 db G r dbw dbw/k/hz T s 20 K + 55 K = 75 K 18.8 dbk B N 27 MHz (FM-TV) 74.3 dbhz N dbw The downlink C/N ratio at the receiver input is (C/N) dn = C N = G r = Gr db 4-11

15 Uplink noise power budget: k Boltzmann s constant dbw/k/hz T s 500 K 27.0 dbk B N 27 MHz (27 Msps QPSK) 74.3 dbhz N dbw The C/N ratio at the transponder input is C/N = C N = P t db For C/N = 30 db we require P t = C/N 3.8 = 26.2 dbw or 417 W Question #12.5 Design a Ku-band receiving earth station to provide an overall clear air C/N of 17 db in a 27 MHz IF noise bandwidth at a carrier frequency of GHz. The antenna noise temperature is 30 K and the LNA noise temperature is 110 K. You may assume a high gain LNA and ignore the noise generated in other parts of the receiver. Determine the diameter of the receiving antenna. The receiving terminal is located on the 3 db contour of the satellite footprint, and clear air attenuation on the path and other losses total 0.8 db. Answer: Set the receiving earth station antenna gain as G r db. Path loss at the downlink frequency of GHz is L p = log (11.45/14.15) = db Transponder output power is P t saturated = 80 W or 19.0 dbw No back off is quoted. Downlink power budget P t G t 31 db 3 db off axis loss G r L p Atmospheric loss and other losses Receiver power P r Earth station receiver input noise power budget 19.0 dbw 28.0 db TBD db -0.8 db G r dbw 4-15

20 i. Calculate the noise power at the input to the satellite receiver in a noise bandwidth of 20 MHz. Hence find the uplink transmitter power required to achieve a C/N of 28 db in the satellite transponder. Answer: Transponder input noise power budget k Boltzmann s constant dbw/k/hz T s 500 K 27.0 dbk B N 20 MHz 73.0 dbhz N dbw The uplink C/N ratio is required to be 28.0 db. Hence the earth station transmitter power is given by P t where P t dbw = db P t = dbw = 19.8 dbw or 95.5 W. j. The gain of the satellite transponder must be set to amplify the received signal at the transponder input to an output level of 180 watts. Calculate the gain of the transponder in decibels. (Ignore the change in frequency in the transponder.) When designing RF equipment, a common rule to avoid oscillation is to make the amplification at any given frequency no higher than 60 db. How would you design a bent-pipe DBS-TV transponder to provide the end to end gain that you calculated? Answer: The power at the input to the transponder is P t dbw = dbw P r = dbw, P t = 22.6 dbw G xp = P t P r = = db. [Gain is P out / P in for any device, giving G = P out dbw P in dbw] RF and IF amplifiers should not be specified with G > 60 db because of the risk of oscillation. In this transponder we need a gain of db. This should be distributed through the amplifer,. Typically, transponders are built with excess gain and include an RF or IF attenuater. So a typical arrangement would set these gains: RF input (17 GHz) 30 db IF amplifier (~ 1 GHz) 60 db RF amplifier (12 GHz) (LPA + HPA) 60 db Attenuator (controlled from earth) -6.8 db 4-20

22 Satellite Transponder Maximum output power P t 20 W Transponder bandwidth 2 MHz Transponder input noise temperature T s 500 K Handheld Transceiver Parameters Transmitter output power 1.0 W Antenna gain (transmit and receive) G 0 db Receiver system noise temperature T s 300 K Receiver system noise bandwidth B n 10 khz Hub Station Parameters Maximum transmit power P t 100 W Receiver system noise temperature (clear air) T s 250 K Antenna gain at 29 GHz (transmit) G t 54 db Antenna gain at 19 GHz (receive) G r 52 db Constants: Boltzmann s constant k = 1.38 x J/K = dbw/k/hz Question #14.1 Preliminary calculations a. Calculate the path loss, in db, for a 2000 km path at 1.6, 2.5, 19, and 29 GHz. Answer: Path loss is calculated from L p = 20 log ( 4 π R / λ). For a worst case analysis, we must take the longest path length of 2000 km. At 1.6 GHz, λ = m, L p = 20 log ( 4 π / ) = db At 2.5 GHz, λ = m, L p = 20 log ( 4 π / ) = db At 19 GHz, λ = m, L p = 20 log ( 4 π / ) = db At 29 GHz, λ = m, L p = 20 log ( 4 π / ) = db b. Calculate the noise power, in dbw, for the receiver in the transponder and for the receivers at the hub station and the handheld unit, in a single voice channel bandwidth of 10 khz. (Note: use the bandwidth of one speech channel, 10 khz for all 4-22

23 the calculations, not 2 MHz.) Answer: For the transponder, Ts = 500 K N xp = k T s B N = = dbw For the hub station, T s = 250 K N H = k T s B N = = dbw For the handheld satellite phone, T s = 300 K N H = k T s B N = = dbw c. The satellite has broad coverage antennas at L-band and Ka-band with half-power beamwidths of 120 degrees. Estimate the gain, in db, of the antennas at each frequency. 2 Answer: The gain of an antenna with a beamwidth of θ 3 db is G = 33,000 / θ 3 db This result is independent of frequency, so all antennas have the same gain: G = 33,000 / = 2.29 or 3.6 db. Question #14.2 C/N ratios Use the values you obtained in Q#1 above, for path loss, antenna gain, and noise power in this question. Calculate C/N values for stations located at the edge of the coverage zone of the satellite, where the satellite antenna gain is 3 db below its maximum value, and the range to the satellite is 2000 km. Take care to use the correct path loss and receiver noise power values for each frequency. Give your answers in decibels. a. Calculate the C/N in the satellite transponder for the signal transmitted by one handheld transceiver located at the edge of the coverage zone (satellite antenna gain 3 db below maximum) and at maximum range from the satellite (2000 km). Answer: The EIRP of the satellite phone is 0 dbw (1 W transmitter, 0dB antenna gain) P r = EIRP + G r L p = = dbw The noise power at the input to the transponder is N xp = dbw Hence (C/N) up = -0.3 db 4-23

29 Answer: There are 60 satellites in the system at a cost of \$40 M each, so the in-orbit part of the system has a capital cost of \$2,400 M. Operating costs are \$100 M per year, total \$1000 M over ten years. Total cost over ten years is \$3,400 M. This cost, plus a return on investment of 27%, or \$918 M, for a grand total of \$4,318 M, is required to make the system viable. c. Calculate the cost per minute per voice circuit assuming that each satellite can be loaded to an average of 20% of its capacity over its lifetime. Answer: The total cost of \$4,318 M over 10 years must be recovered by selling satellite channels. With a 20% load factor (typical for GEO satellites), there are the equivalent of 200 telephone circuits available for 365 days per year on each of 50 satellites. There are 525,600 minutes in a year, so we can sell ,600 circuit minutes in the 10 year period. Hence cost per minute = \$ 4,318,000,000 / = \$ d. Write two paragraphs discussing the cost of the system and the cost of a voice circuit. What price per minute would you set for a satellite voice circuit? Would you expect customers to be willing to pay this amount for a satellite telephone connection? How does the cost compare to terrestrial cellular telephone charges? Answer: The cost per channel to use the satellite looks attractive, compared to a terrestrial cellular call costing \$0.20 per minute (typical year 2000 charge). It should be possible to charge a customer more than \$0.082 for the call via satellite and make additional profit. However, there are a number of difficulties with the LEO satellite system that are not obvious in the above analysis. If the satellites are in polar orbits, most of them are over the poles or the oceans and cannot be used. For example, if this system used polar orbits, only two satellites would be visible from the United States at any time, giving the entire population of the US a capacity of only 2000 telephone circuits. To reach a utilization factor of 20%, customers must be found in countries other than the United States. If each US customer were to use the two visible satellites with a 50% load factor for 10 minutes each day, during the hours 6 am 10 pm, the available circuit minutes = 960,000, so there could be a maximum of 96,000 US customers. To recover the system cost of \$431,800,000 per year, each customer must spend \$4,497 on satellite telephone 4-29

30 calls each year. Put this way, it is clear that a large number of overseas customers must be recruited to make the system pay. One further factor that makes satellite telephones less attractive than cellular phones is that the weak signals will not penetrate buildings. You have to go outside to make and receive calls, - not pleasant in the middle of winter. The difficulty for the satellite system owner is that the above calculation is misleading. Customers cannot be signed up until the system is operational, and then the customer base will grow only slowly. The owner of the system must pay interest on money borrowed to build the system (\$2,400 M in this example) while revenue is low. This has crippled several of the LEO satellite telephone systems built in the late 1990s. 4-30

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