0
$\begingroup$

Let's assume we have two BTS transmitters. One works with power of $10\ {\rm W}$ and the other with $200\ {\rm W}$. Both transmitters produce the same $2\ {\rm GHz}$ electromagnetic wave.

What's the difference between the two waves from a physical point of view? Just the number of photons? Why is the "wavelength range" [1] from a $200\ {\rm W}$ transmitter greater? (if the photon energy is related to the frequency, and in both cases the wave frequencies are the same)?

[1] I mean "electromagnetic wave range" (distance from a BTS station where you can still use the signal from the station (e.g. a mobile phone))

Improve this question
$\endgroup$
3
  • $\begingroup$ what do you call " wavelength range" and who says how it relates to power? $\endgroup$
    – trula
    Commented Aug 3, 2020 at 14:44
  • $\begingroup$ @trula I mean "electromagnetic wave range" (distance from a BTS station where you can still use the signal from the station (e.g. a mobile phone)). "who says how it relates to power" - so what if the parameters of the wave are the same (the only difference is in the power)? $\endgroup$
    – User
    Commented Aug 3, 2020 at 14:57
  • $\begingroup$ If the frequency is 2 GHz, then the wavelength is ~6.667 m, regardless of the power transmitted. $\endgroup$
    – The Photon
    Commented Aug 3, 2020 at 15:11

1 Answer 1

Reset to default
2
$\begingroup$

What's the difference between the two waves from a physical point of view?

The amplitude of the higher power wave is greater.

If you measured the electric field associated with the stronger wave with a field strength meter, you'd find a higher value.

In the quantum view, that means you're receiving more photons per second from the stronger transmitter, but this is almost never relevant to understanding RF phenomena.

Why is [the distance from a BTS station where you can still use the signal] from the station [with] a 200 W transmitter greater?

Because (to take a simplified scenario) the transmitter emits its power in all directions, not in a tight beam to your receiver. So the further you get from the transmitter, the smaller the fraction of its output power you are able to receive with your receiver. So the power you receive is decreasing as you get farther from the transmitter. This is the famous "r-squared power law" of radiative phenomena. (You'll also find that even if you try to transmit a tight beam, due to refraction that beam will spread and you'll still end up with received power dropping off as $r^2$)

But your receiver has a characteristic input noise, and a minimum signal-to-noise ratio at which it can operate. So when the input power gets too low, it will fail to receive the signal correctly.

Combining those two considerations, you get the reason why a higher-power transmitter can be received at a greater range.

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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