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.
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:
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.
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.
protected by Qmechanic♦ Sep 29 '15 at 17:48
Thank you for your interest in this question.
Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).
Would you like to answer one of these unanswered questions instead?