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This question already has an answer here:

The Friis transmission equation can be given as:

Pr = Pt + Gt + Gr + 20 log10 (λ / 4πR)

Where the variables represent:

Pr: Power at the receiver
Pt: Power at the transmitter (in e.g. dBm)
Gt: gain of the transmission system (in decibels)
Gr: gain of the receiver
R: "radius" of signal, i.e. simply the distance between transmitter and receiver
λ: the wavelength of the signal

The λ term is very surprising, isn't it? It very conveniently makes the units work out, but why should the frequency of a signal affect how "rapidly" it dissipates?

I know that ionspheric and other terrestrial effects tend to make HF signals propagate farther than VHF and higher, but what accounts for this "increased range" of low frequency signals even in theoretical free space?!

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marked as duplicate by Kevin Reid AG6YO Dec 16 '16 at 3:43

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ Ah, so it is. Feel free to moderate as you feel appropriate. $\endgroup$ – natevw - AF7TB Dec 16 '16 at 3:19
  • $\begingroup$ But if I just merge, your answer will be under the other question and it won't make sense because your answer is assuming the (different) symbols defined in the question. Would you be willing to edit the answer to make sense in the new context afterward? (Or, separately, do you think that yours isn't needed given the other one?) $\endgroup$ – Kevin Reid AG6YO Dec 16 '16 at 3:36
  • $\begingroup$ I don't know all the options here, but I'm fine with just closing this as a duplicate. Then if people do find this one by search term coincidence, the content is still here, but points to the original one. $\endgroup$ – natevw - AF7TB Dec 16 '16 at 3:38
  • $\begingroup$ Yes, that's the “default” choice. But I could also move your answer to the other question, and I'd like to do that if it has a useful different perspective — but it doesn't sound like you think it does, and what I see in it is mainly about the particular expression of the formula. I'll just close as duplicate for now. $\endgroup$ – Kevin Reid AG6YO Dec 16 '16 at 3:42
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This is actually a side effect from the receiver/transmitter gain terms in the overall equation! In short, as communication frequency gets lower, the antennas must grow bigger to keep the same gain. A bigger antenna can capture more energy overall. Therefore, lower frequency signals will be "louder" at the receiver — iff its antenna is adjusted to compensate.

The Wikipedia article on Free-space path loss explains this frequency dependency) in more detail.

So in a sense, the equation has been simplified in a misleading way. More intuitively, instead of λ ending up part of the R term, the Gr term would be replaced with variables corresponding to the antenna aperture and that signal wavelength λ. In fact, one form of the Friis equation makes this clearer:

Pr = Pt * (At * Ar) / (r^2 * λ^2)

The "adding up gains" equation in the question basically re-groups this ratio (which is much more intuitively understood) into terms of decibels (which are much more practical to use).

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  • $\begingroup$ This is a very important insight. May be it helps to consider the following case: 1. Transmitter antenna is omni. 2. Receiving antenna is large relative to wavelengths. Then, energy received only depends on receiving antenna area, as long as 2 still holds. $\endgroup$ – Jon Kåre Hellan LA4RT Dec 16 '16 at 10:27

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