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Consider the modelling of the Signal-to-Noise Ratio (SNR) for a satellite (sat) communication system linked to an earth

Posted: Fri May 20, 2022 11:32 am
by answerhappygod
Consider the modelling of the Signal-to-Noise Ratio (SNR) for a
satellite (sat) communication system linked to an earth station
(et) with the following parameters: 1. Radius of Earth (𝑅𝑒) in km =
6400 km 2. Radius of satellite above earth’s atmosphere (π‘Ÿ) in km=
Input variable 3. Central angle (𝛽) in degree = Output 4. Longitude
of satellite in degrees (π‘™π‘œπ‘›π‘”π‘ π‘Žπ‘‘) = 68.5 Β°E 5. Longitude of Earth
Station in degrees (π‘™π‘œπ‘›π‘”π‘’π‘‘) = Input variable 6. Frequency of
transmission in GHz (𝐹𝐺𝐻𝑧) = Input variable 7. Gain of transmitting
antenna in dB (𝐺𝑑 ) = 15 dB* 8. Gain of receiving antenna in dB (𝐺𝑑
) = 15 dB 9. Stefan-Boltzmann constant (𝐾) = 1.38 Γ— 10βˆ’23 𝐽/𝐾 10.
Noise temperature of receiving system in Kelvin (𝑇) = 190 Kelvin
11. Bandwidth of the receiver system in Hz (𝐡)= 75 Γ— 106 Hz 12.
Received Signal Power in dB (π‘ƒπ‘Ÿ) = Output 13. Transmitted Signal
Power in dBm (𝑃𝑑) = Input variable 14. Path Loss in dB (𝐿) = Output
15. Signal to Noise Ratio (𝑆𝑁𝑅) = Output *dB is the decibel unit In
order to calculate the SNR during signal transmission from space to
earth, a parameter which gives a statistic of the amount of signal
present during transmission, we employ: 𝑆𝑁𝑅 = π‘ƒπ‘Ÿ βˆ’ 𝑃𝑛 [𝑑𝐡] (1)
Where π‘ƒπ‘Ÿ is the received power level in dBm and 𝑃𝑛 is the noise
power in dB. Mathematically, π‘ƒπ‘Ÿ = 𝑃𝑑 + πΊπ‘Ÿ + 𝐺𝑑 βˆ’ 𝐿 [π‘‘π΅π‘š] (2) And,
𝑃𝑛 = 10πΏπ‘œπ‘”10(𝐾𝑇𝐡) [𝑑𝐡] (3) The path loss, 𝐿, is the amount of
transmitted signal lost due to the effects of travelling through
space and earth’s atmosphere. The path loss in dB is given as: 𝐿 =
92.44 + 20πΏπ‘œπ‘”10(π·π‘‘π‘œπ‘‘π‘Žπ‘™) + 20πΏπ‘œπ‘”10(𝐹𝐺𝐻𝑧) [𝑑𝐡] (4) π·π‘‘π‘œπ‘‘π‘Žπ‘™ is referred
to as the slant range, a parameter representing the actual distance
between the earth station and the satellite given by: π·π‘‘π‘œπ‘‘π‘Žπ‘™ = βˆšπ‘…π‘’
2 + π‘Ÿ 2 βˆ’ 2π‘Ÿπ‘…π‘’π‘π‘œπ‘ π›½ [π‘˜π‘š] (5) Where, 𝛽 = πΆπ‘œπ‘ βˆ’1[𝑆𝑖𝑛(π‘™π‘œπ‘›π‘”π‘ π‘Žπ‘‘)𝑆𝑖𝑛
(π‘™π‘œπ‘›π‘”π‘’π‘‘) + πΆπ‘œπ‘ (π‘™π‘œπ‘›π‘”π‘ π‘Žπ‘‘) πΆπ‘œπ‘ (π‘™π‘œπ‘›π‘”π‘’π‘‘)πΆπ‘œπ‘ (πœ”)] [π‘‘π‘’π‘”π‘Ÿπ‘’π‘’π‘ ] 3 For, πœ” =
π‘™π‘œπ‘›π‘”π‘ π‘Žπ‘‘ Β± π‘™π‘œπ‘›π‘”π‘’π‘‘* *If π‘™π‘œπ‘›π‘”π‘ π‘Žπ‘‘ and π‘™π‘œπ‘›π‘”π‘’π‘‘ belong to the same
longitude i.e. both Β°E or both Β°W, use the negative sign If π‘™π‘œπ‘›π‘”π‘ π‘Žπ‘‘
and π‘™π‘œπ‘›π‘”π‘’π‘‘ belong to the different longitude i.e . Β°E and Β°W or
vice versa, use the positive sign Your Tasks in this Laboratory
Using windows forms and writing in C++, design an application that
can be used to estimate the SNR of a typical earth-satellite
communication system using the given parameters as provided. For
this purpose, please note that the summary and interpretation of
the required parameters are given to help your understanding. For
the proposed application, please note that items 2, 5, 6 and 13 are
to be your input variables, while items 3, 12, 14 and 15 are the
output variables. It will an added advantage to you if your
designed GUI shows at attempt of creativity, innovation and
elegance. A REPORT must be written and submitted with respect to
the GUI application being built in this laboratory. A. Thereafter,
using your completed GUI, provide answers for the following
parameters in this use-case scenarios. Note that π‘™π‘œπ‘›π‘”π‘ π‘Žπ‘‘ is no
longer a constant as previously designated: 1. For Low Earth Orbit
(LEO) satellite in positional orbit at π‘™π‘œπ‘›π‘”π‘ π‘Žπ‘‘ = 15Β°W: π‘Ÿ = 1500 π‘˜π‘š,
π‘™π‘œπ‘›π‘”π‘’π‘‘ = 10Β°W, 𝐹𝐺𝐻𝑧 = 6 𝐺𝐻𝑧, 𝑃𝑑 = 120 π‘‘π΅π‘š [5 Points] 2. For Medium
Earth Orbit (MEO) satellite in positional orbit at π‘™π‘œπ‘›π‘”π‘ π‘Žπ‘‘ = 30Β°E:
π‘Ÿ = 10500 π‘˜π‘š,π‘™π‘œπ‘›π‘”π‘’π‘‘ = 40Β°E, 𝐹𝐺𝐻𝑧 = 12 𝐺𝐻𝑧, 𝑃𝑑 = 75 π‘‘π΅π‘š [5 Points]
3. For Geostationary Earth Orbit (GEO) satellite in positional
orbit at π‘™π‘œπ‘›π‘”π‘ π‘Žπ‘‘ = 69Β°E: π‘Ÿ = 36000 π‘˜π‘š,π‘™π‘œπ‘›π‘”π‘’π‘‘ = 59Β°W, 𝐹𝐺𝐻𝑧 = 40 𝐺𝐻𝑧,
𝑃𝑑 = 65 π‘‘π΅π‘š [5 Points] B. What are your observations when 𝑃𝑑 = 0
π‘‘π΅π‘š in all cases of LEO,MEO and GEO as given above? Explain your
results. [5 Points]