Module 10 Assignment Solutions

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Module 10 Assignment SOLUTIONS 525.440 Satellite Communications Module #10 Propagation There are four problems in this assessment. The problem set is due on Day 7 of Module #10. 1. A Direct-To-Home (DTH) Ku-band downlink has the following design specifications: Clear-sky T sys = 100 K Clear-sky CNR = 11 dB Performance spec: 99% of a year with CNR>10 dB (link is at performance for CNR>10 dB). Availability spec: 99.9% of a year with CNR > 6 dB (link is available for CNR > 6 dB). The climate in which the DTH service operates has the following rain attenuation statistics: Annual Percentage Time Rain Attenuation Fade Level 10% 0 dB 1% 1 dB 0.1% 3 dB 0.01% 10 dB a) Determine the reduction in CNR due to rain attenuation for the four percentage times listed. Assume the effective temperature of the rain is 280 K. T sys = 100 K. Assume T A,clearsky =10K, so T rcvr =90K (ok if you just assumed T sys = T rcvr .) T rain ’ = 280 (1 – 1/L rain ) T sys,rain = T rain ’ + T A,clearsky /L rain + T rcvr CNR cs = 11 dB CNR rain = CNR clearsky – L rain – 10log(T sys,rain /100) Performance Level à CNR > 10 dB Availability Level à CNR > 6 dB Annual % Time Lrain CNR rain Performance? Availability? 10% 0 dB 11 dB OK OK 1% 1 dB 8.1 dB NO OK 0.1% 3 dB 4.3 dB NO NO 0.01% 10 dB -4.4 dB NO NO

2. Looking at a case similar to that of the Basic Link Design Project, assume you have a 0.35m antenna pointed at the Sun for emergency command (uplink) reception from Earth. The range to the Sun is 50 AU. Frequency of operation is X-band (7.2 GHz). a. Estimate the antenna noise temperature with the Sun in the antenna beam. You can assume only the 3-dB beamwidth needs to be considered, that the antenna noise temperature when the Sun is not in the 3-dB beamwidth is 3 K, and that the antenna is lossless. b. Repeat part (a), using the same assumptions, for the case when the Sun is in beam for reception on a 2m High Gain Antenna (also at 50 AU range). Solution: a. The 3-dB beamwidth of the antenna is ࠵? = 70 ࠵? ࠵? = 70 3࠵?8 7.2࠵?9 0.35 = 8.3° The angular extent of the Sun at 50 AU is ࠵? !"# = 2 tan $% ࠵?࠵?࠵?࠵? 50࠵?࠵? = 2 tan $% 6.96࠵?5 ࠵?࠵? 7.48࠵?9 ࠵?࠵? = 1.066E-2° This is an extraordinarily small angle compared to the 3-dB beamwidth of the antenna. The peak gain is essentially constant across this area. For this problem, the only difference between the no-Sun and Sun case is in this small area. ࠵? &#’,!"# = ࠵? &#’,#)$!"# + B 1 4࠵? D ࠵?(࠵?, ࠵?) (40,000 ࠵? − 3࠵?) * !"# + $* !"# + ࠵?࠵?O The peak gain of the antenna is ࠵? ,- = ࠵? Q ࠵?࠵? ࠵? R + = 0.5 Q ࠵?0.35 4.17E-2 R + = 348 = 25.4 ࠵?࠵?࠵? The integral becomes (1/(4π))(348)(39,997)(the solid angle of the Sun at 50AU). This cone solid angle is 2π(1-cos( ࠵? Sun at 50AU /2)) = 2.718E-8 ster. So with the Sun in-beam, the MGA antenna noise temperature rises from 3 K to (3K + 0.03K) = 3.03 K. For part b, the peak antenna gain is 11,351 = 40.6 dBi, with a 3-dB beamwidth of 1.46° (still much greater than the Sun’s disk angular extent). The corresponding increase in antenna noise temperature is (1/(4π))(11,351)(39997) = 0.98 K, so the overall antenna noise temperature rises just under 1 K to 3.98 K. Part b is analogous to what occurred during the Pluto dark-side occultation observation using the radio on-board New Horizons. The APL-built, low-power radio was designed to include a fine microwave phase and amplitude recorder, to measure how the radio signal from Earth “winked” out as the spacecraft flew behind the planet (as viewed from Earth). During this time the high gain antenna was pointed to the unlit side of Pluto, and the measured antenna noise temperature rose to a maximum of 30 K (due to the microwave thermal emission characteristics of Pluto). At the end of the occultation period, the Sun emerged and a further brief increase in antenna noise temperature of about 2 K was observed while the Sun was in the HGA beam (New Horizons uses a 2.1m HGA at X-band, and was about 33 AU from the Sun at Pluto encounter).

3. The G/T of a large aperture ground station can be measured using the Y-factor technique. In this method, the ground station antenna is pointed first to a hot source (e.g., the Sun, or the black hole at the center of the Milky Way Cygnus X-1 (no kidding)), and resulting noise level is measured at some point in the receive chain. The antenna is then pointed at cold sky (basically any high-elevation direction in the sky that doesn’t have a radio source within it, like the galactic plane, Jupiter, a black hole like Cygnus X-1 or and Sagitarrius A* at the center of the galaxy, supernova remnants like Cassiopeia A, it’s a real mine field!), and the noise level measured again at the same point in the system. The ratio of the power measurements (N hot /N cold ) is called the Y-factor (usually measured in dB). If you know the noise temperature of the hot source and the cold source (and we can get specific information on the Sun’s radio flux, or noise temperature, across frequency on daily basis, then you can determine the antenna clear-sky G/T. The expression for determining G/T is: G/T = 8 π k (Y – 1) / ( l 2 f (f)) where k is Boltzmann’s constant (in W/Hz/K), Y is the ratio form of the Y-factor calculated in the measurement, and f (f) = radiation flux density of the hot sky source (Sun, radio star, etc.) a) You as the ground station operator need to measure the X-band (8.0 GHz) G/T for your 9.3m antenna. You perform a Y-factor measurement by pointing the antenna at the Sun and then at cold sky. You measure a Y-factor of 23.0 dB. You find from the interwebs that the solar flux density at X-band on this day is 226 solar flux units (1 SFU = 1E-22 Watts/m 2 ). What is your estimate of the 9.3m dish’s G/T in dB/K? b) If you assume the dish is 50% efficient, what then is the corresponding clear-sky system noise temperature?

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4. An advanced satellite digital video broadcast system is planned at Ka-band (f down = 20 GHz). Assume that the coding scheme used requires a 3 dB CNR at the ground receiver to meet acceptable performance standards. Assume that the clear sky received CNR is 16 dB, that the receiver system noise temperature is 150 K (of which 50 K is clear-sky antenna noise temperature), and that the physical temperature along any rain attenuation path is 280 K. Assume a 30° elevation angle and that the ground has a negligible contribution to the antenna noise temperature. a) Estimate the rain margin for this service. That is, how much rain loss do we need to see on the link before we fall below acceptable performance. CNR clearsky = 16 dB T sys,clearsky = 150 K T A,clearsky = 50 K T phys,rain = 280 K Elevation = 30° T sys,rain = 50/L rain + 280 (1-1/L rain ) + 100 CNR rain = CNR clearsky – L rain – 10log(T sys,rain /100) 13 dB = L rain + 10log(T sys,rain /100) 20 = (L rain ) (T sys,rain /100) = (L rain ) (1/(2L rain ) + 2.8 (1-1/L rain ) + 1) = (1/2 + 2.8L rain – 2.8 + 1) 20 = 2.8L rain – 2.3 L rain = 7.96 = 9.0 dB

b) Estimate the availability for this service in Orlando, FL and in Phoenix, AZ, using the ITU calculator attached to this module. Using the latitude and longitude for the two cities, the elevation angle and transmit frequency, and finding the availability corresponding to 9 dB attenuation: Orlando availability = 99.70% Phoenix availability = 99.96%