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Relation of attenuation and swamping

  1. Mar 28, 2017 #1
    1. The problem statement, all variables and given/known data
    Question: If the attenuation between Earth and satellite is 190dB, with reference to attenuation explain why there should be a difference between uplink and downlink?
    (I am a senior high school student so I haven't learned in too much depth)

    3. The attempt at a solution
    My question is what role does attenuation play in the difference between uplink and downlink. All I have been taught up till now is that if the frequencies are close in the frequency spectrum, they might interfere. But what relation does attenuation have with frequency? (If somebody could provide a little intuitive approach towards "swamping" before using it in a post", that would be very helpful)
  2. jcsd
  3. Mar 28, 2017 #2
    I think the main point here is that the RX in the satellite is pointing at a warm Earth (at about 300K) which is noisy. We need enough received power to overcome the noise. That is why the Earth station has a powerful TX and a large dish having high gain. The receiving terminal on the Earth, however, is looking up into cold space, hence low noise, and a small dish can be used.
    Noise power = Bolzmann's Constant x Temp in Kelvin x Bandwidth in Hertz.
    Regarding interference between uplink and downlink, they are normally in different frequency bands, so isolation is not a big problem.
  4. Mar 29, 2017 #3
    Thank you very much. But I think you haven't really told me about what relation is there between attenuation and the difference in frequencies. I am aware of the reason why high-powered signals are required but not really sure if there's a connection between attenuation and difference in frequencies.
  5. Mar 29, 2017 #4
    The path loss in free space is caused by the spreading of the signal, and can be found in decibels from from,
    PL = 10 log (4 pi L / lambda)^2
    where L is the distance and lambda is the wavelength, both in metres.
    This gives the attenuation between isotropic antennas. An isotropic antenna radiates equally in all directions.
    So you can see it increases 6dB every time the wavelength is halved (or the frequency doubled).
    The reason for this is that, in a similar way to a dipole, an isotropic antenna gets smaller as the frequency is raised.
    But the gain of a dish antenna increases with frequency, in the following way.
    G = 10 log E x (pi D / lambda)^2
    where E is aperture efficiency (about 0.6) and D is diameter in metres.
    So the gain increases by 6dB every time the wavelength is halved or the frequency doubled.
    As the path loss between TX and RX is the free space loss minus two antenna gains, you can see it will decrease by 6dB every time the frequency is doubled.
    The result is sometimes summarised in the Friis Formula,
    Loss = 10 log (L x lambda) ^2 / (At x Ar)
    where At is the effective area of the transmitting antenna and Ar is that of the receiving antenna, L is distance in metres and lambda is wavelength in metres.
    From this you can again see that if lambda is halved then PL decreases by 6dB.
  6. Mar 30, 2017 #5
    Okay I understand. The power loss is inversely proportional to frequency. Can you also tell me, if the attenuation is constant and very high, and if I send a very high frequency signal, will it observe a decrease in frequency?
    Last edited: Mar 30, 2017
  7. Mar 30, 2017 #6
    The frequency will always remain that of the transmitter.
    I think you are getting confused by thinking about the energy of photons.
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