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If cellphones have such a low power transmitter (around 200mW for LTE) and also doesn't have directionality then how can they talk back to the tower even though they are so powerful and built specifically with aim for better direction. And how sensitive are the tower receivers?

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closed as too broad by winny, Dmitry Grigoryev, mkeith, pipe, Maple Sep 20 '18 at 16:45

Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer. Avoid asking multiple distinct questions at once. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

  • \$\begingroup\$ They are not “so powerful”. Average power is 10W in a tower \$\endgroup\$ – PDuarte Sep 16 '18 at 15:21
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    \$\begingroup\$ @ObsessionWithElectricity Do not confuse power with energy. Small batteries provide low energy capacity but high power can be easily achieved. \$\endgroup\$ – PDuarte Sep 16 '18 at 18:11
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    \$\begingroup\$ That is why it only transmit 6 W when it absolutely has to, to reach the tower. \$\endgroup\$ – The Photon Sep 16 '18 at 18:16
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    \$\begingroup\$ Think about communicating with a space probe (like Voyager). The ground station can transmit a very strong signal, but the remote probe only has limited power available and transmits back a very weak signal. We use very large antennas, careful receiver designs, maybe even cryogenic electronics, to receive those signals. But yet we can receive them. The cell phone situation is similar but without so much distance or so much disparity between the base and remote transmitter power. \$\endgroup\$ – The Photon Sep 16 '18 at 18:18
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    \$\begingroup\$ It isn't just the antenna. It also depends on the receiver circuitry, and the processing that goes on after the signal is received. \$\endgroup\$ – JRE Sep 16 '18 at 18:42
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I'm out of date on how modern mobile telephone networks handle multiple simultaneous voice calls but I believe each phone is allocated a time slot and there may be up to 100 time slots but it could be forty (something nagging in the back of my mind!).

However it doesn't matter for answering this question; we can assume 100 time slots and that means the phone has to compress its outgoing voice message and transmit it in one hundredth of the time that the old fashioned analogue system had. That old system allocated a frequency for each handset and is now out-of-date.

Let's assume that the voice data could be compressed to 10 kbps and given that it has to be transmitted in one-hundredth the time it will need to store data and transmit it at about 1 Mbps. Again, don't hang me on this if I'm not up to the latest 4G methods because that isn't important for this question I feel.

So, the tower receiver is handling a data rate of around 1 Mbps continuously to service up to 100 simultaneous voice calls. How much received power does the tower antenna need to adequately recover the data you would consider next.

The thermal noise per hertz for 50 ohms is -174 dBm and we wish to exceed this with sufficient margin to make data reception viable so, a well trodden formula says we need 100 times this power or -154 dBm. But that is "per Hz" and we need to have a bandwidth of about 1 MHz to adequately convert an RF signal carrying 1 Mbps data back to usable data. So the formula (at an ambient of 25 degC) becomes: -

Received power in dBm = -154 dBm + 10 log\$_{10}\$(data rate)

For the example of 1 MHz BW, we need a received signal in the order of -94 dBm. Given that RF engineers normally load this with a thing called fade margin (20 dB to 30 dB more) we would probably be OK statistically with -74 dBm.

Then there is link loss and the formula for free space is this:-

Link loss (dBm) = 32.45 + 20 loglog\$_{10}\$(Fc) + 20 log log\$_{10}\$(distance in km)

Fc is in MHz and is the carrier frequency. I'm going to assume that the telephone carrier frequency is about 2 GHz but don't hang me if I'm too far out.

If the distance is 3 km then the link loss is 32.45 dB + 66 dB +9.54 dBm. Overall that is about 108 dB lost between transmitter and receiver but we will have antenna gain and this could drop to maybe 102 dB or better.

We need -74 dBm therefore we need to transmit a level that is 102 dB higher or 28 dBm. 28 dBm is a power of 631 mW. Not a million miles from the 200 mW but the important thing is that you know have the fundamentals to put the real numbers into.

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  • \$\begingroup\$ How did you calculated link loss? \$\endgroup\$ – ObsessionWithElectricity Sep 17 '18 at 9:19
  • \$\begingroup\$ @ObsessionWithElectricity it's the standard Friis transmission link loss equation expressed in decibel form. Try this calculator to save you a little time. \$\endgroup\$ – Andy aka Sep 17 '18 at 9:21
  • \$\begingroup\$ There's also this one that directly gives link loss. \$\endgroup\$ – Andy aka Sep 17 '18 at 9:26
  • \$\begingroup\$ BTW so the frequency does matter in terms of range of signal?. I read someone posted on SE that frequency doesn't matter. I always doubted him. \$\endgroup\$ – ObsessionWithElectricity Sep 17 '18 at 9:40
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    \$\begingroup\$ Frequency does not matter for power density (to a first order), power in W/square meter (or mW/cm^2, or whatever) is independent of frequency (at least below the high microwave bands), but aerials get smaller for a given gain as frequency rises, so the capture area as a fraction of the area of the sphere falls with rising frequency (Antenna design being otherwise unchanged). \$\endgroup\$ – Dan Mills Sep 17 '18 at 11:12

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