Re: [amsat-bb] Sun noise

["John Santillo" <u1004467@xxxxxxxxxxx>]

Hello all, can we take this a little bit further.  From this information is
there a way to calculate the G/T of this system?  I've seen this figure of
merit many times before but have never fully understood how to calculate it.

Thanks,

John
N2HMM


----- Original Message -----
From: "Tom Clark, W3IWI" <w3iwi@toad.net>
To: "'Eric Van Offelen'" <eric@vanoffelen.com>; "'Amsat-bb'"
<amsat-bb@AMSAT.Org>
Sent: Saturday, December 21, 2002 14:15 PM
Subject: RE: [amsat-bb] 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
> > EA5GIY
>
> 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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