7
$\begingroup$

I have been told that Wi-Fi, LTE etc signal strength fall of as $$\propto \frac1{\log(r)}$$ where $r$ is the distance. I am wondering why this is. I better explain what I mean with this question.

When I took my very first course in physics I had a similar question about why the intensity of acoustic vibrations (or sound) falls of $\propto r^{-2}$. I got the, very satisfying, answer that it is because the wavecrests are spreading over the surface of a sphere as the sound propagates through a homogeneous medium: $$\frac{I_0}{S}=\frac{I_0}{4\pi r^2}.$$ A very intuitive explanation that made perfect sense to me at the time.

What is the nature of these signals that make their intensity fall of in a different way from acoustic vibrations?

$\endgroup$
5
  • 8
    $\begingroup$ The intensity does fall of as $r^{-2}$. The logarithm only appears when the power is expressed in decibels (which is a logarithmic scale). $\endgroup$
    – lemon
    Jul 2, 2015 at 11:27
  • $\begingroup$ @lemon maybe make it an answer! $\endgroup$
    – innisfree
    Jul 2, 2015 at 12:04
  • $\begingroup$ @innisfree Naaaaaaaaaaaaaaaah. $\endgroup$
    – lemon
    Jul 2, 2015 at 12:15
  • 2
    $\begingroup$ The $ \frac{1}{r^2}$ relationship only holds for a point source. Granted, this is close enough when you're reasonably far from a simple antenna, but antenna arrays will have rather different power vs. distance properties $\endgroup$ Jul 2, 2015 at 13:17
  • $\begingroup$ @CarlWitthoft I agree. Wifi and LTE signals are highly directional. i.e they are not a point source, they signal is directed in one specific direction to cover a certain area, it does not spread out evenly in a spherical way like a point source wave $\endgroup$
    – Mitchell D
    Dec 21, 2015 at 10:20

1 Answer 1

2
$\begingroup$

For most sources of radiation, there are two factors that diminish the power with distance. The first of these is the inverse square law, which tells us that every time we double the distance to the source, the power per unit area drops by a factor of 4. This makes sense from the perspective of conservation of energy - the sum (integral) of energy over all directions must be constant.

But there is a second factor that becomes more important at large distances, and that is absorption. While the inverse square law assumes no losses along the path of the radiation, that is not the case in the real world. Attenuation follows an exponential law, and at sufficiently large distances from the source, attenuation will dominate the calculation and lead to an exponential drop off of source intensity with distance.

For the case of WiFi signals, the attenuation in air is small (but the attenuation in walls etc. is substantial). The equation you quote is for the dB power level ignoring attenuation. If I have

$$P \propto \frac{1}{r^2}$$

then on a dB scale,

$$\log(p/p_0) = -2 \log(r/r_0) $$

Where $p_0$ and $r_0$ are the reference power at reference distance.

Note that is NOT a $\frac{1}{\log(r)}$ relationship ... that makes no sense at all (and is demonstrably wrong, even for a dB scale).

UPDATE:

Reading this answer again a few years later, I now see that it's possible that the equation the OP was shown (and possibly misremembered) included

$$\log{P} \propto P_{dB} \propto \log{\frac{1}{r}}$$

Power in decibels does indeed scale with the log of the inverse of distance (but not the inverse of the log of the distance...). Note that whether it's distance or distance squared just changes the constant of proportionality in log space...

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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