I am studying cubesat communication system. (RF noise floor is -90dBm in my receiver location) Please let me know how can I evaluate it.
if you have a 20 dBm signal radiated from a roughly omnidirectional antenna 500 km away it will be spread over about 4pi*(5e5)^2 sq metres, which is about 3e12 squ metres. The power density will be about -105 dBm per sq metre, so if your noise floor really is -90 dBm you are 20 or 30 dB short if you have an effective area of your antenna of 1 sq m. You need an effective antenna area of somewhere close to 100 sq metres. The effective area is the gain times a square wavelength divided by 4pi.
You may be able to get some improvement in SNR by integrating (effectively, narrowing the bandwidth). But I don't know if you have already taken this into account. The cubesat may also have some antenna gain, but I doubt it. I expect it is close to unity gain (the same as an omnidirectional).
You need to build a link budget to estimate the received SNR or Eb/No and compare with the min. SNR needed by the modulation of the communication signal, in this case 2-FSK. The min. SNR can be estimate from theoretical BER/PER vs SNR curves with some margin for inefficient coding or from data sheets of SATCOM radios. Here is a guide for developing your CubeSat link budget: https://www.itu.int/en/ITU-R/space/workshops/2016-small-sat/Documents/Link_budget_uvigo.pdf
There is no line of sight on Earth at this distance. You have to include frequencies as well to take into account diffraction, refraction etc. at low frequencies you will have longer distance…
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