0
\$\begingroup\$

I've been thinking absent-mindedly about how mobile phone communication works when I wondered how something as power efficient as a mobile phone can have such a powerful radio. I am guessing that most celltowers have a transmit power of, say, 100w (I have no evidence, but the point is that it's a lot), and they have a range of >6km. What I want to know is how can such a power constrained radio (I'm guessing max 2w) transmit a receivable signal to the phone tower?

I have tried googling it but of course it just comes up with a bunch of very basic and irrelevant descriptions of how a celltower works.

Of course I have absolutely no knowledge of radio stuff besides the highschool physics stuff, so I am not able to do any calculations to see whether my question is unjustified.

\$\endgroup\$
2
\$\begingroup\$

Cell towers only transmit around 10 watts usually. Sometimes up to 50 or so in urban areas. Your phone can transmit up to 2 watts.

Transmit power does obviously have a big effect on the range of a signal, but it is not nearly as much as you would think due to the 1/r2 relationship of radio waves propagating out from the source. If your transmitter puts out 4 times as much power, you only get twice the range. 16 times the power, only 4 times the range. 100 times the power... 10 times the range.

The main advantage a cell tower has over your phone is in the size, power, complexity, and quality of the low-noise amplifier in its receiver. Since the tower is not limited so much by size and cost, they can put a really high quality amplifier in the tower (probably a noise figure of 1-2 dB...) and very effective low-loss filters in front of the amp (probably a cavity filter which is 10x the size and weight of a cell phone all by itself). This has a huge effect on the quality of the received signal compared to the cheap, small stuff that has to fit into the cell phone.

You might think that the tower's big antenna means it can "hear" small signals better compared to the phone's tiny antenna, but this is not true. The path from the cell tower to the phone is 100% symmetrical to the reverse path. In other words, when the tower transmits to the phone it gets a big benefit from its large antenna because the power is transmitted mostly in the direction of the phone, but this same benefit is in effect when the tower is receiving signals from the phone. The phone's tiny antenna has equally lousy performance when receiving signals from the tower as it does when it is transmitting back out to the tower.

\$\endgroup\$
5
  • \$\begingroup\$ So am I getting the TL;DR right here? The cell tower can pick out the phone's crappy signal really well (because of the good amplifiers and filters) from the background noise where as a phone could not do the same \$\endgroup\$
    – Jachdich
    Jan 7 '20 at 20:00
  • \$\begingroup\$ Exactly bro, one love \$\endgroup\$
    – A Azam
    Jan 7 '20 at 20:03
  • \$\begingroup\$ Cool, I'm not 100% sure whether to accept this answer or the other one. What do you think? \$\endgroup\$
    – Jachdich
    Jan 7 '20 at 20:05
  • \$\begingroup\$ Trust your heart, believe in yourself. \$\endgroup\$
    – A Azam
    Jan 7 '20 at 20:10
  • \$\begingroup\$ Problem is, they both answer my question but in different ways. Might have to look on meta to see what I should do in that case. \$\endgroup\$
    – Jachdich
    Jan 7 '20 at 20:10
1
\$\begingroup\$

Here what we call propagation equations : https://en.wikipedia.org/wiki/Friis_transmission_equation.

To explain it with hands, the Base Station (BTS) (your celltower) is emitting a signal with a power of 100 W (+50 dBm) using an anisotropic antenna ( all directions). But this is attenuated ( power decreased) with the square of the distance to your cell phone . Suppose the attenuation is 75 dBm. This meens your cellphone will receive +50 dBm - 75 dBm = -25 dBm.

Suppose your cellphone has a sensitivity of -100 dBm (can detect any signal above this level). So your phone can easily receive the -25 dbm.

If you think about it in the other side, from Cellphone to BTS : Suppose a emitting power of 1 mW ( 0 dBm) , the BTS will receive 0 - 75 dBm = -75 dBm ( attenuation is the same as the fist case) and suppose the sensitivity of this BTS is -120 dBm . The -75 dBm received by the BTS is fare above it limit and the signal will be detected easily.

Sorry for my English.

    -
\$\endgroup\$
5
  • \$\begingroup\$ Your English is perfect :) \$\endgroup\$
    – Jachdich
    Jan 7 '20 at 20:01
  • \$\begingroup\$ Thanks for the answer, the maths there is interesting: I'm not familiar with what recieving and transmitting power is feasible in terms of dBm, but do I understand correctly that as the base station has a higher gain than the phone that it's possible to receive much weaker signals? \$\endgroup\$
    – Jachdich
    Jan 7 '20 at 20:04
  • \$\begingroup\$ I think that even if cell phone and BTS have the same sensitivity, you have the effect your describe in your question. The challenge is not really the emitting power of the cell but its sensitivity . Devices are designed in such way the noise factor ( noise added by the cell) is at the minimum and this allow detect signal not so far from the noise floor . But this challenge, according to me, does't need much hardware . An other think that increase dramacally the efficiency of the system is all this digital signal processing and filtering and high perfomed modulation technics. \$\endgroup\$
    – Sisyphus
    Jan 7 '20 at 20:20
  • \$\begingroup\$ Ah yes I understand, thanks for the explanation \$\endgroup\$
    – Jachdich
    Jan 7 '20 at 20:22
  • \$\begingroup\$ For question of health, devices need to emit a minimum of power. This put the challenge on the receive sensitivity. \$\endgroup\$
    – Sisyphus
    Jan 7 '20 at 20:24

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.