P
Philip Newman
Guest
FINAL SOLUTION:
Question: A goonhilly aerial locaed at 5.17W and 50.05N transmites 500W of
14-14.5GHZ microwave power in a beamwidth of 2.6mR. if the aerial of a
geostationary satellite was 1m in diameter and was located at 30W, what
would be the carrier to noise ratio? Make reasonable assumptions.
Answer:
Let y-axis correspond to longitude of Earth Station, therefore the
co-ordinates are (0, rCos(theta, rSin(theta))
let Phi be the longitude of the satellite (w.r.t Earth Station)
Here, Phi = (30-5.17) = 25 (roughly)
Co-ordinates are [(r+h)sin(Phi), (r+h)Cos(Phi), 0]
Now r = 6378km and r+h = 42612 (depending on G values, mass of earth etc
taken)
therefore h = 35786km
ES^2 = (X1-X2)^2 + (Y1-Y2)^2 + (Z1-Z2)^2
Therefore the range to satellite is:
ES^2 = [0 - 42164Sin(25)]^2 + [65378*0.64279 - 42164*0.9063]^2 +
[6378*0.76604]^2
which gives ES = 38.796km. Problem solved
But what of the rest of the question??
Power received by the Satellite = Prx * Area of Antenna/Area of beam (W)
= [500*(pi/4)*1^2] /[pi/4(3.88*10^5 * 2.6*10^3)] = 4.8*10^-8
remember that pi*r^2 = pi*(D/2)^2 = pi/4 * D^2 and D = 1;
SO, the noise power at the satellite = k(Tae+Tamp)B where B = bandwidth
Assume the Noise Figure for the amp is 1.5dB
Therefore the noise power = 1.38*10^-23*(290+120)*5*10^8 = 3.5pW.
So the CNR = 41dB
Hope this helps
Phil
"Richard Henry" <rphenry@home.com> wrote in message
news:45Vjc.2058$fE.724@fed1read02...
Question: A goonhilly aerial locaed at 5.17W and 50.05N transmites 500W of
14-14.5GHZ microwave power in a beamwidth of 2.6mR. if the aerial of a
geostationary satellite was 1m in diameter and was located at 30W, what
would be the carrier to noise ratio? Make reasonable assumptions.
Answer:
Let y-axis correspond to longitude of Earth Station, therefore the
co-ordinates are (0, rCos(theta, rSin(theta))
let Phi be the longitude of the satellite (w.r.t Earth Station)
Here, Phi = (30-5.17) = 25 (roughly)
Co-ordinates are [(r+h)sin(Phi), (r+h)Cos(Phi), 0]
Now r = 6378km and r+h = 42612 (depending on G values, mass of earth etc
taken)
therefore h = 35786km
ES^2 = (X1-X2)^2 + (Y1-Y2)^2 + (Z1-Z2)^2
Therefore the range to satellite is:
ES^2 = [0 - 42164Sin(25)]^2 + [65378*0.64279 - 42164*0.9063]^2 +
[6378*0.76604]^2
which gives ES = 38.796km. Problem solved
But what of the rest of the question??
Power received by the Satellite = Prx * Area of Antenna/Area of beam (W)
= [500*(pi/4)*1^2] /[pi/4(3.88*10^5 * 2.6*10^3)] = 4.8*10^-8
remember that pi*r^2 = pi*(D/2)^2 = pi/4 * D^2 and D = 1;
SO, the noise power at the satellite = k(Tae+Tamp)B where B = bandwidth
Assume the Noise Figure for the amp is 1.5dB
Therefore the noise power = 1.38*10^-23*(290+120)*5*10^8 = 3.5pW.
So the CNR = 41dB
Hope this helps
Phil
"Richard Henry" <rphenry@home.com> wrote in message
news:45Vjc.2058$fE.724@fed1read02...
Trigonometric.
BTW, your problem statment doesn't say so, but I assume the "Atlantic
Ocean
Satellite" is in geosynchronous orbit.
"Philip Newman" <nojunkmail@ntlworld.com> wrote in message
news:0uRjc.80$RI.53@newsfe1-win...
This is to do with telecommunications, and hopefully someone should know
the
answer to this!
Assume a ground-station has a longitude of 5.17degW and a latitude of
50.05degN
There is an Atlantic Ocean Satellite with longitude 30.00degW.
How can I calculate the range of the satellite from the earth station?
I have had a look at some literature, but nothing seems to be what I
need.
What equations do I need?
Phil