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VK2ZRG > BALL 30.05.05 12:53l 79 Lines 3735 Bytes #999 (0) @ WW
BID : 1335_VK2ZRG
Read: GUEST OE7FMI
Subj: Re: Moon & earth ? M5WJF...SETI
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Sent: 050528/1113Z @:VK2WI.#SYD.NSW.AUS.OC #:3230 [SYDNEY] FBB7 $:1335_VK2ZRG
From: VK2ZRG@VK2WI.#SYD.NSW.AUS.OC
To : BALL@WW
VK2ZRG/TPK 1.83d Msg #:1335 Date:28-05-05 Time:12:12Z
>R:050525/0008Z @:GB7MAX.#28.GBR.EU #:8651 [Bloxwich] $:C30641M5WJF
Wayne M5WJF wrote :-
>Some mention was made within the television programme 'Making Contact'
>(Open University), regarding the sensitivity of receiving equipment able
>to detect such signals, and I seem to remember a figure of 60 light years
>being mentioned at one point, perhaps those on packet with a better grasp
>of current radio receiver technology might wish to comment on the
>possibility of receiving the alien equivalent of at least the audio
>content of an Australian Soap from 60 light years away?
Hello Wanye and SETI fans,
Receiver sensitivity is determined by bandwidth
and noise figure. What really matters is the system noise, which includes
the antenna noise and background galactic noise. Given the frequency and the
distance, free space path loss is easily calculated. Knowing these parameters,
you can then determine how much TX power you need, and how much antenna gain.
The path loss formula with isotropic antennas. Using base 10 logs.
Loss dB = 20 * Log(F MHz) + 20 * Log(Kilometres) + 32.44
one light year is 9.46E12 Km so 69 ly = 5.68E13 km
For 60 light years at 1421 MHz you get
20 * 3.15 + 20 * 13.75 + 32.44 = 370.6 dB That's the easy part.
For the receiving side you must estimate what system temperature is possible.
For a liquid helium cooled LNA operating at a physical temperature of a few
Kelvins above absolute zero, you would hope to get around 5 Kelvins noise
temperature, or a little below 0.1 dB noise figure. A good deep space dish
antenna might be around 10 Kelvins effective temperature. And the background
sky temperature maybe 5 Kelvins. This adds up to 20 Kelvins system temperature.
Now you can calculate the input noise power to your receiver for a given
bandwidth. At 290 K the noise power in a one Hz bandwidth is -174 dBm. Assume
a 50 Hz bandwidth and the noise is 17 dB higher at -157 dBm. 20 Kelvins is
14.5 times less than 290 K so the noise will be 10 * Log(14.5) dB less
or 11.6 dB.
So the RX input noise = 174 - 17 + 11.6 = 168.6 dBm.
So for a 10 dB signal to noise ratio, the input signal will be -158.6 dBm.
For a 370.6 dB path loss, you then need 370.6 - 158.6, or 212 dB to be made
up with transmit power and antenna gains. Let's be optimistic and say that we
will use a fully steerable 100 metre dish at each end of the path. The dish
will have a perfect parabolic surface, thereby having possible maximum gain.
The calculated gain at 1421 MHz is 61 dBi and the beamwidth is 0.15 degrees.
The power required will be 212 - 122 dBm = 90 dBm or 1,000,000 watts of real
RF power. The effective radiated power will be 1,260,000,000,000 watts.
A high power UHF TV transmitter has at best 1,000,000 watts ERP.
So there is your answer Wayne. If ET is transmitting and knows exactly where
to point his antenna, and we know exactly where ET is so that we can point our
antenna, it may just be possible if ET knows morse code. Of course, knowing
what the exact frequency was would be helpful too, as well as just when to
listen. This is the reason why is possible to still receive signals from the
Voyager space craft. We know where to point the antennas and what the
frequency is! If ET is just listening too, then we won't hear anything.
As for eavesdropping on stray TV signal from Earth, forget it! It's pie in
sky dreaming. I reckon we have Buckleys chance of hearing ET.
73s from Ralph VK2ZRG@VK2WI.#SYD.NSW.AUS.OC
/ack
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