For electromagnetic waves, what is the relation between the the carrier frequency, attenuation (absorption loss) and propagation distance? I know that modulating high frequency carriers are required to reduce antenna size, and to transmit several signals simultaneously.

My question is : does the attenuation increase with these high frequecies when transmitted through air? If so, how is this problem solved?


60GHz is associated with absorption by Oxygen.

Rain and Snow reduce range.

Radio waves can penetrate fresh water. The Great Lakes was a place where submarine crews trained during WW2 and could send and receive underwater. Salt water is another situation entirely. Frequencies need to be very low. There is often a good natural wave guide just above the ocean surface.

Foliage can attenuate.

The situation in the optical and IR regions are complicated. There is a program called MODTRAN that models absorption in detail.

For 60GHz, attenuation can actually be a virtue, not a problem. You can pack more receivers and transmitters closer together, at a given complexity.

  • $\begingroup$ +1 Nice practical examples! What's the deal with 60 GHz and Oxygen? Is this wavelength compared to molecule size or something? $\endgroup$ – VMMF Apr 11 '18 at 0:37
  • $\begingroup$ rfglobalnet.com/doc/…. Probably little millimeter sized Pac-Men chomping away. $\endgroup$ – Stanley Pawlukiewicz Apr 11 '18 at 1:03

I should add to the previous answer, that even though there are several causes that produce a greater attenuation as we increase the frequency, in vacuum we will also have a limited propagation distance for a given frequency.

Not due to the propagation phenomenon itself, but due to the physical dimensions of the receiver antenna and thus the effective area of the antenna, which limits the received power. It is called free-space loss and is independent on the composition of the propagation media.


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