I would've thought that as 5GHz is a higher frequency, and it carries more energy, it would be able to pass through walls much more easily compared to a 2.4GHz frequency- similar to how short frequencies cannot pass through the atmosphere (and be received by satellites) but high frequencies are able to "penetrate" through due to their higher energy.

  • 2
    $\begingroup$ Do you have a source saying that 5GHz are shorter-ranged than 2.4GHz? If yes, please add that to the post. $\endgroup$
    – ACuriousMind
    Commented Sep 11, 2014 at 15:01
  • $\begingroup$ There do seem to be a couple of sources agreeing with the notion that 5 GHz has less range than 2.4 GHz: Netgear support page and speed guide forum. $\endgroup$ Commented Sep 11, 2014 at 15:16
  • $\begingroup$ Pedant alert: frequencies are quick (high) or slow (low) while wavelengths are short or long. Also, sources saying short (low?) frequencies pass through atmosphere less than high frequencies would be good especially as the radio band is pretty low frequency and doesn't seem to have trouble with the atmosphere or houses. $\endgroup$
    – user121330
    Commented Sep 11, 2014 at 17:26
  • 1
    $\begingroup$ Would Electrical Engineering be a better home for this question? $\endgroup$
    – Qmechanic
    Commented Sep 11, 2014 at 18:04
  • $\begingroup$ I think so @Qmechanic $\endgroup$ Commented Sep 29, 2015 at 17:11

2 Answers 2


Section 2.9 (bottom half on page 11) of this document describes why a 5GHz router suffers greater attenuation than a 2.4 GHz router, particularly so in residential and office settings. Attenuation is greater at 5 GHz than at 2.4 GHz for signals that need to pass through walls, doors, and glass.

From the above reference:

enter image description here

More energy per photon does not necessarily mean longer range. Taking this to an even greater extreme, attenuation in the visible range is significantly higher than in the microwave frequencies used by routers. We don't make visible frequency routers because a visible frequency signal doesn't pass through walls and doors, period. The signal isn't just attenuated. It's completely gone after passing through first 1/10 mm or of material.

  • $\begingroup$ +1 - The greater attenuation is actually considered a benefit for short range communication: you have less interference from your neighbors' 5GHz WiFi than from their 2.4 GHz because it's attenuated more. $\endgroup$
    – Floris
    Commented Sep 11, 2014 at 15:41
  • $\begingroup$ Note that section 2.8 of the document you link provides support for the argument put forward by @user31748. >"Antenna aperture characteristics at 5.25 GHz (vs. 2.54 GHz) are likely to slow the migration of generic wireless appliances to the 5 GHz UNII band. An ideal isotropic antenna at 5.8 GHz captures less energy than at 2.4 GHz, therefore, a device moving to 5.8 GHz will have less range (all other factors remaining constant)." $\endgroup$
    – Floris
    Commented Sep 11, 2014 at 15:47
  • $\begingroup$ By "visible frequency" do you mean light in the human-visible portion of the spectrum? Also what is meant by "We don't visible frequency routers..."? $\endgroup$
    – JYelton
    Commented Jan 19, 2017 at 17:05
  • $\begingroup$ @JYelton -- There was a word missing in "We don't visible frequency routers ..." That should have been "We don't make visible frequency routers ..." (Fixed) $\endgroup$ Commented Jan 19, 2017 at 17:51
  • 2
    $\begingroup$ Well also visible-frequency routers would be annoying. $\endgroup$
    – Devsman
    Commented Jan 19, 2017 at 18:04

Use Friis' formula http://en.wikipedia.org/wiki/Friis_transmission_equation to estimate the received power. Notice that it has the transmit and receive antenna gains and these are nearly ~1 for simple dipole (or monopole) antennas that are omni-directional in a plane perpendicular to the antenna current flow. Given the antennas the received power is proportional to the square of the wavelength, hence the higher received power at 2.4GHz than at 5.8GHzwith the same transmit powers.

  • $\begingroup$ This makes sense - a quarter wave monopole will be larger (and thus "grab more energy") for lower wavelengths. But the attenuation argument (David Hammen's answer) is possibly more valid for residential settings. $\endgroup$
    – Floris
    Commented Sep 11, 2014 at 15:49

Not the answer you're looking for? Browse other questions tagged or ask your own question.