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RE: Sun noise

> How much sun noise should i expect on 2.4 GHz with the
> following set-up:
> - 90 cm offset dish
> - +/- 0,2 dB estimated loss in adapters
> - 0,7 dB NF preamp
> Merry Christmas and 73's
> Eric

Eric -- let's calculate it from first principles:

Your dish is 90 cm and the wavelength is 13 cm. Therefore your dish has a
beamwidth of a bit more than 13/90 radians = 8.3 degrees. For a circular
aperture, Airy's criterion would predict the diffraction limited beam to be
1.22 times this large, or about 10.1 degrees in diameter. To be simple, I'll
call this 10 degrees.

The radio sun at 13 cm, depending on sunspot count, is about 40,000 Kelvin
and is about 1 degree in diameter. Therefore the sun fills about (1/10)^2 or
~1% of the beam. Therefore, the sun should contribute about (1/100)*40000 =
400 degrees Kelvin if you dish has perfect (100%) aperture efficiency.

Since I know nothing about the quality of the feed and dish, I will guess
that you actually have about 50% aperture efficiency (meaning that half of
the power is in the main beam and half is scattered into sidelobes).
Therefore the sun probably contributes ~400/2 or about 200 Kelvin.

You say the preamp has 0.7 db NF and you have 0.2 dB connector loss. You do
not say how long the short piece of coax between the feed and the preamp is,
so I will guess that it adds another 0.1 dB of loss. Therefore your real
Noise Figure is 0.7+0.2+0.1 = 1 dB.

Many graphical and slide rule devices make the next calculation be easier,
but we go back to first principles to find that the correspondence between
Noise Figure and Temperature is
      T = 290 * [10^(Noise Figure/10)-1] Kelvin
Which, for NF=1, is 75 Kelvin as an estimate of your system temperature.

So this says that the sun should have S/N = 200/75 = 2.66 times the noise in
the receiver. Therefore if you have a true power detector on the output of
the receiver and set it to 1 volt = receiver noise off the sun, it should go
up from 1 volt to (1+2.66) =3.66 volts when pointing at the sun, or a
Yfactor of 3.66:1.

You can check this by making the dish look at large trees with many leaves.
These trees will contribute about 200-250 K of noise, but since they fill
the entire ~10 degree beam, they will also contribute about the same noise
as we predict for the sun.

Another idea is to have a 90 cm wide person stand in front of the dish
completely covering the dish. The person will contribute about 300-320
Kelvin of noise. If you happen to transmit, this makes the person bring a
new meaning to the words "dummy load" ;<}

Hope this helped -- seasons greetings de Tom W3IWI

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