# Why does path loss decrease as frequency decreases? [duplicate]

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)
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?!

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

• Ah, so it is. Feel free to moderate as you feel appropriate. – natevw - AF7TB Dec 16 '16 at 3:19
• 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?) – Kevin Reid AG6YO Dec 16 '16 at 3:36
• 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. – natevw - AF7TB Dec 16 '16 at 3:38
• 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. – Kevin Reid AG6YO Dec 16 '16 at 3:42

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)