How Efficient Are Radio Signals Over Distance?

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SUMMARY

This discussion centers on the efficiency of radio signals over distance, specifically the reception of a 1 Watt signal at 10 meters using ordinary antennas. Participants highlight the importance of antenna design, particularly the need for antennas to be 1/4 of the wavelength for optimal performance. The inverse square law (1/D^2) is identified as a key principle for calculating signal strength over distance. The conversation also emphasizes the significance of polarization alignment between transmitting and receiving antennas to ensure effective signal reception.

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  • Understanding of radio wave propagation and the inverse square law
  • Knowledge of antenna types, specifically dipole and Yagi antennas
  • Familiarity with antenna polarization concepts
  • Basic principles of signal coherence and transmission efficiency
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Thomas1980
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Hey there

I've been wondering just how efficient radiosignals are... If I were to place a transmitter broadcasting 1 Watt then what would I be receiving at say 10 meters away? Of course the antennas and frequencies are adjusted to near optimal with regards to dimensions, size and frequency in both reception and transmition, so I need a best case guess... Any radioamateurs or other wise people around? :-D

Best regards

Thomas Hansen
 
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Not much gets absorbed by the atmosphere, but the problem is keeping the signal coherent. It is possible to keep some radio signals highly coherent by shaping the transmitter in a parabola.
 
With my limited knowledge on radiowaves and radiosignals I can only help a little bit: I know that it is important that the antenna is 1/4 of the wavelength of the signal you're sending/ receiving because you will then have the optimal dimensions of the antennas and therefore have the most powerful signal reception. My question is, just how much of a 1 Watt signal will you receive 10 metres away, if all of the equipment is properly adjusted? ( that's around 30 feet IIRC)
I need it to be with ordinary antennas, Parabolic is not and option.
The question is so simple, and the answer straightforward, it is probably even posible to calculate it... I just don't know how! :-(
Hope this narrows it down a little bit.

Best regards

Thomas Hansen
 
Some HAM operators sent a radio signal from the US to New Zealand,
Try 1/D^2
the inverse square law; for all your needs (if it fits of-cource)
Nice Coder
 
Try 1/D^2 the inverse square law; for all your needs (if it fits of-cource)

Indeed, but if he's using an "ordinary antenna" (dipole?) orented say north-south, there won't be much signal going in the north-south direction.
 
"Indeed, but if he's using an "ordinary antenna" (dipole?) orented say north-south, there won't be much signal going in the north-south direction." ??
you may mean that if he is using a vertically polerised dipole, that if the transmitting antenna was horizontally polerised that he wouldn't get much signal?
The minimum signal coming fro mthe transmitting antenna (at 90 degrees) is only -3DB, so its not that much (how many op-amps only have a 3db gain?)
:smile:
De Nice Codre
 
you may mean that if he is using a vertically polerised dipole, that if the transmitting antenna was horizontally polerised that he wouldn't get much signal?
I was thinking of a horizontal fullwave dipole. But the point was that his question depends on what "ordinary antennas" means.
 
on long wavelengths horizontal dipoles are predominantly used, for short wavelengths yagi's, and vertically polerised dipoles are used.

It does not matter what his definition of an 'ordinary antenna' is, as long as they are using the same antenna, with the same polerisation and are not using directional antenna's pointing away from each other!
 

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