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OK, I couldn't resist:
photon energy = hf = 0.4136e-14 electron volts x f
supposing 500 MHz (about channel 19), hf = about 2e-12 ev
putting that into joules (so we can work in joules/second or watts):
photon energy = 3.3e-31 joules.
Then assuming 500 kW effective radiated power, to a receiver in the beam this looks like 5e5 joules/second or 1.5e36 photons/second into a full sphere.
Distance to nearest star, proxima centauri = 3.8e16 meters, so the sphere at that radius has an area of 1.8e34 meters.
To reconstruct the signal we need about 12 million samples per second, and each sample needs about 900 photons to get a 30:1 signal to noise ratio. (This assumes a photon-counting detector, and ignores thermal noise - a big assumption, but let's go with it.)
That means we need a total of 10.8e9 photons per second, so we need a dish of area
(1.8e34)(10.8e9/1.5e36)=130e6 square meters, or about 6.4 kilometers radius.
If we only want to detect the carrier with a 1 Hz bandwidth, we could get by with a one or two meter dish (really would use a more conventional UHF antenna).
This says that the signal does not get very far in galactic terms as a watchable program. It also says something about why space probes use digital signaling and error-correcting codes, so they do not have to reach the high signal to noise ratio needed for analog transmission.
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