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VK2ZRG > SETI 06.12.04 14:45l 98 Lines 3778 Bytes #999 (0) @ WW
BID : 980_VK2ZRG
Read: GUEST DO1EH
Subj: Receiving TV signals on Mars
Path: DB0FHN<DB0RGB<OK0PPL<DB0RES<ON0AR<7M3TJZ<ZL2TZE<VK5UJ<VK5BRC<VK5ATB<
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Sent: 041206/1201Z @:VK2WI.#SYD.NSW.AUS.OC #:55859 [SYDNEY] FBB7 $:980_VK2ZRG
From: VK2ZRG@VK2WI.#SYD.NSW.AUS.OC
To : SETI@WW
VK2ZRG/TPK 1.83d Msg #:980 Date:06-12-04 Time:10:37Z
VK2ZRG/TPK 1.83d Msg #:978 Date:06-12-04 Time:10:25Z
R:041203/1511Z @:WT3V.#CNJ.NJ.USA.NA #:31850 [Lakehurst] $:31850_WT3V
Hello Warren and readers,
You have it right Warren, the key word IS "broadband".
I used 5 MHz in the calculation, so the numbers are not too far off.
MASERS are almost old hat nowadays, the in things are liquid helium cooled
LNA's. You can get noise temperatures well under 10 Kelvins this way.
What I was trying to do in my bulletin, was to show the silliness of the
`eavesdropping on ET' idea, when applied to broadband, high power, signals
such as TV and radar transmissions on Earth. (And, yes, I am aware of the
narrow bandwidths and consequent slow data rates used for space probes.)
My original calculation was a few dB out, I admit, but not enough to make
much difference.
Here are the numbers for calculating the antenna size required for a 10 dB
signal to noise ratio, for a 5 MHz wide TV signal received on Mars, when
Mars is near its closest approach to Earth...say 60 million kilometres.
Path loss at 600 MHz for 60 million kilometres with isotropic antennas.
Loss in dB = 20 * log(F) + 20 * log(D km) + 32.4
20 * log(60) = 55.6
20 * log(60E6) = 155.6
32.4
Path loss = 243.6 dB
System temperature = 50 Kelvins.
Assume no liquid helium available on Mars, so use uncooled LNA with
25 Kelvin noise temperature. (about 0.35 dB NF)
Use 25 Kelvins for effective antenna temperature. This includes
all losses plus any QRM from Earth, plus 3 Kelvins background noise.
Noise in 5 MHz BW = -114.6 dBm
For a temperature of 50 Kelvins, noise in 1 Hz bandwidth = -181.6 dBm
For 5 MHz bandwidth, add 67 dB, giving noise power of -114.6 dBm
Gain required = 139 dB
System gain required for 10 dB signal to noise = 243.6 - 114.6 + 10
This equals 139 dB.
1 megawatt ERP = + 90 dBm. (Say 50 Kw into 13 dBi antenna)
RX antenna gain required = 139 - 90 = 49 dBi
Dish diameter for 49 dBi at 600 MHz = 60 metres (for 10 dB S/N in 5MHz BW)
My error before, was that I said 100 metres.
Gain of a parabolic dish antenna may be easily calculated from the capture
area of an isotropic antenna. This is 1 / (4*Pi) or 0.07958 square Lambda.
So to calculate a 60 metre dish at 600 MHz, you get these numbers.
(Wavelength for 600 MHz = 0.5 metres.)
Radius of dish in wavelengths = 30 metres / 0.5 metres = 60
Area in square wavelenghts = 60 * 60 * Pi = 11310
Effective area = 11310 * 0.55 = 6220.5 square wavelengths
So gain over isotropic antenna = 6220.5 * 4 * Pi = 78169
gain in decibels = 10 * log(78169) = 48.93 dB
The only way to improve on these numbers is to use more TX power or
a lower RX system temperature. If you reduce the system temperature to
12.5 Kelvins (a big ask), then you gain just 6 dB. So the dish diameter
could be reduced to 30 metres for 12.5 Kelvins system temperature.
Re the doppler shift; shouldn't be a problem with a 5 MHz bandwidth.
Re another missive from Warren.
Gain of the Arecibo antenna at 440 MHz may be calculated using the above
formulas. Lambda = 300/440 = 0.6818 metres. Diameter = 305 metres.
So (152.5 / 0.61818)^2 * Pi * 0.55 = 86444.7 square wavelengths.
Gain over isotropic = 86444.7 * 4 * Pi = 1086296
Gain in dB = 10 * log(1086296) = 60.4 dB.
If I've got the numbers wrong, I apologise. I just pulled the formulas
out of the grey matter between my ears. I didn't spend a year researching
these formulas on the internet. Hi!
Merry Cricket, Ho Ho Ho
73s from Ralph VK2ZRG@VK2WI.#SYD.NSW.AUS.OC
/ack
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