Free Essay

Satllite

In: Science

Submitted By wangchao
Words 740
Pages 3
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Faculty of Engineering
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Subject: Satellite Communication Systems
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Assignment Number: Project One/Part A Date Submitted: 26/10/2012
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Assignment Title: Project One
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Student Name(s) and Number(s) Tutorial Group:
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Chao Wang 11275955
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Declaration of Originality:
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The work contained in this assignment, other than that specifically attributed to another source, is that of the author(s). It is recognised that, should this declaration be found to be false, disciplinary action could be taken and the assignments of all students involved will be given zero marks. In the statement below, I have indicated the extent to which I have collaborated with other students, whom I have named.
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Signature
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Chao Wang
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Statement of Collaboration:
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Preparation
My student ID: 11275955
The month I was born: March
So phi=mod (11275955+3, 360), phi=38°

TASK 1
From the wiki, I find the satellite which is the closest to the phi is called Paksat-1E. It is geosynchronous and communications satellite built and owned by the Boeing Company, leased to the SUPARCO as PakSat. ( longitude ØSS= 38.8° )
Then the cities I choose is England (359.5°E, 53.5°N), the second is, Australia (15°E, 47.5°N)

TASK 2
ØE1= 359.5°, ØSS= 38°, λE1= 53.5°,ØE2=15°,λE2=47.5°
Assume the ELmin = 5°
Ϭmin = π/2 + ELmin = 95° b = π - Ϭmin - sin−1{ R sin(Ϭmin)/ aGso }= 180°- 95° - sin−1{6371 x sin(95°)/ 42164} = 76.34°
England
B1 = cos−1[cos(b) cos(λE1)] = cos−1[cos(76.34°)cos(53.5°) ]= 81.92°
So the look angle for England is [ØE1- B1, ØE1+B1] = [ 277.58° E, 81.92° E ]
Australia
B2= cos−1[cos(b) cos(λE2)] = cos−1[cos(76.34°)cos(47.5°) ] = 80.82°
So the look angle for Australia is [ØE2- B2, ØE2+B2] = [-65.82E, 95.82E ] = [294.18E, 95.82E ] = [65.82W, 81.92° E ]

TASK 2
For the England:
Ls= (hR-h0)/ sin (EL1) =( 4- 0 )/ sin(35.48°)= 6.892 km
The horizontal projection of this slant path length is
LG= Ls cos(EL2) = 6.892 x cos (35.48°) = 5.612 km
For ran 0.1% of the time, the reduction factor is
Rp = R0.01 = 180/(180+LG)= 0.96976
So the effective path length may be determined:
L= LsRp = 6.892 x 0.96976 = 6.684
The attenuation A can now be calculated:
A = á L dB

From the Optus C1 satellite, it is at 14GHz
So I define the function for linear table:
Y = a X + b
From the table, to calculate the ah, use the 12GHz and 15GHz values to calculate
12 = a * 0.0188 + b
15 = a * 0.0367 + b a = 167.6 b = 8.85
So at the 14GHz, 14 = 167.6 * ah + 8.85 ah = 0.0307

Use the same function, to calculate av,
12 = a * 0.0168 + b
15 = a * 0.0335 + b a= 179.64 b = 8.98
At 14 GHz, 14 = 179.64 * av + 8.98 av = 0.0279

Follow above, to calculate bh
12 = a * 1.217 + b
15 = a * 1.154 + b a= -47.62 b= 69.65
At 14 GHz, 14 = -47.62 * bh + 69.65 bh= 1.169

To calculate bv :
12 = a * 1.2 + b
15 = a * 1.128 + b a= -41.67 b=62
At 14 GHz, 14 = -41.67* bv + 62 bv= 1.152

For horizontal polarization: á = ah * db/km 0.20148 db/km
So A =á L = 0.20148 * 6.684 = 1.3467 dB

For vertical polarization: á = av * db/km = 0.17816 db/km
So A =á L = 0.17816 * 6.684 = 1.1908 dB

For the Australia:
Ls= (hR-h0)/ sin (EL2) =( 4- 0 )/ sin(35.39°)= 6.907 km
The horizontal projection of this slant path length is LG= Ls cos(EL2) = 6.907 x cos (35.39°) = 5.631 km
For ran 0.1% of the time, the reduction factor is Rp = R0.01 = 180/ (180+LG) = 0.96967
So the effective path length may be determined:
L= LsRp = 6.907 x 0.96967 = 6.697
Use the same ah, bh, av, bv values as above

For horizontal polarization: á = ah * db/km = 0.20148 db/km
So A =á L = 0.20148 * 6.697 = 1.3493 dB

Similarly, for vertical polarization: á = av * db/km = 0.17816 db/km
So A =á L = 0.17816 * 6.697 = 1.1931 dB…...

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