Earth-Moon-Earth Signal Attenuation

259

02:31:55

This is a comparison of what the signal attenuation of a moon bounce looks like. Since the moon is a very poor reflector and the path loss between the earth and moon and back again is huge, an attenuation rate of -252dB@144MHz is expectant. This is what such an attenuation looks like. You'd probably have more power induced in your body from standing across the room from a small transformer or something. Anyway NASA has successfully bounced and received a signal from the moon at this attenuation ratio with a transmission power of only 3mW. Hah, I've been studying telecommunications for years and it never ceases to amaze me how far we've come since the damn of the radio a mere 100-150 years ago. Anyway not sure if the comparison is correct, because the source didn't say if the attenuation was in the power level or voltage level, and what was the reference level to begin with. I took a reference level of 7V/50ohm in this example. 

published 7 years ago

kiani

7 years ago

Hoe about the signals from the hubble telescope and mars and pluto ¡ don"t really know what NASA is receiving. Radio waves going through gama rays, to reach earth from the moon. Really?! Such madive electromagnetic fields frim the sun. and a 3mW NASA signal?! Hmm

hurz

7 years ago

EveryCircuit reference level 0dB is 1Vp your 7.07V are 17dB above that so your resistor network does cause 269dB attenuation.

eekee

7 years ago

Odd... in the late 80s I got the impression amateurs were experimenting with moon bounce. Anyway, there's a crucial piece of info missing from your description, @thebugger: What waveband was NASA's bounced signal? That will make a huge difference to losses in the atmosphere and ionosphere.

thebugger

7 years ago

Kiani, what? :D I didn't get the question. Hurz, you're right, but as I said, the source article didn't mention a power level, so I just assumed a standard value. Should've gone with NASA's power of 3mW. And eekee, the article wasn't very specific about the waveband. Must be VHF or UHF, because lower frequencies than that get reflected back to earth, and higher frequencies have significant atmospheric attenuation to begin with.

hurz

7 years ago

500mW is a standard value? Standard is 1mW, or for hugh power 1Watt and called 0dBm In your case of 3mW you might need 1mW at 50 Ohm. Which is 223.4mV at 50 Ohm. If you want 1Watt as reference then you need 10Vp at 50 Ohm which is 7.07Vrms at 50 Ohm. So what “standard“ you are talking?

thebugger

7 years ago

1W. Fixed it. I made it with the rms value by mistake

Addramyrz

6 years ago

Also depends on the time of day. The atmosphere changes a lot depending on if it's being ionized.

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