# Is it possible for LEO satellites to detect a usable signal from regular mobile phones on the ground?

## Background

Recently I came across a company called ASTS that claims they will be able to launch LEO satellites with the capability to talk with regular mobile phones. I initially wrote their claims off, thinking their claims to be kind of preposterous. After all, if it was possible, why hasn't anyone done it yet?

After looking into it a bit more though, the claims didn't sound as outlandish anymore. I came across this post: How to calculate data rate of Voyager 1? and at least in terms of signal strength, it seems feasible! Using the link budget equation in that post:

$$P_{RX} = P_{TX} + G_{TX} - L_{FS} + G_{RX}$$

• $$P_{TX}$$ seems to be about 200 mW. According to 5G NR UE Power Classes, Power Class 3 seems to be for Handheld User Equipment, which has a max TRP of 23 dBm, which converts to about 200 mW. I've read online that cellphones can apparently go as high as 1W, but wasn't able to verify that in any standards, so let's be pessimistic and only use 200mW. Converted, that's about -7 dBW.
• $$G_{TX}$$ I'm not exactly sure how to calculate. Optimistically this would be 0 or higher, but reading up on Planar Inverted-F Antennas, which I believe are used in most modern handsets seem to have poor gain. These are also onmi-directional antennas and I feel like they'd probably radiate in a donut-pattern. Maybe -5dBi to be on the safe side?
• $$L_{FS}$$ would depend on the frequency used, but it seems like 4G/LTE commonly uses band 1 which is 1920-1980MHz for uplink. The satellites seem to be around 500km in LEO orbit, which gives us a free space path loss of 152.1dB for 1920MHz.
• $$G_{RX}$$ is still somewhat up in the air, but the company's next satellite, BlueWalker 3, supposedly has a phased array antenna that's 330m² with a maximum gain of up to 36dBi. Supposedly the real satellites will be even bigger and have gains of 40dBi+ depending on the frequency.

Adding all this up, we get $$-7 \text{ dBW} - 5 \text{ dBi} - 152.1 \text{ dB} + 36 \text{ dBi} = -128.1 \text{ dBW}$$, which converts to $$-98.10 \text{ dBm}$$.

-98.10 dBm doesn't seem like a good signal the satellite would be getting, but it does seem usable. Wikipedia states that -100dB is the "Minimal received signal power of wireless network (802.11 variants)" and other sources rate anything above -110 dBm on LTE networks to be fair to poor.

## Question

And ok. That's about as far as I got into the calculations and ran into trouble with the next part of uhoh's answer on calculating the data rate.

1. Regarding noise, the post says:

When receiving the signal, the limit to the data rate is the ratio of received signal power to the total noise power (received plus system).

What does total noise power (received plus system) mean? Does this mean that, given how sensitive the satellite is, it may pick up two signals of the same frequency from hundreds of kilometers away on Earth's surface? Considering that those signals normally wouldn't see other, does this make it effectively impossible for the satellite to distinguish between a signal that was intended for the satellite versus dozens others that were intended to reach a cell tower? Is that received noise or system noise?

1. Then the calculation

the noise equivalent power will be about $$k_B T \times \Delta f$$ where $$k_B$$ is the Boltzmann constant.

What the heck is $$\Delta f$$? From a comment below that post it seems like $$\Delta f$$ is the bandwidth, but how did the OP decide on a arbitrary number of 1kHz? Can I pick any number I want for bandwidth and just keep plugging it into this equation:

$$10 \times \log_{10} \left( (1.38 \times 10^{-23} \text{ J/K}) \times \text{ Satellite Temp (K) } \times \text{ Bandwidth (Hz)} \right)$$

until I get something that's bigger than 10dB after subtracting it from −128.1dBW? How do I convert bandwidth into bits per second? Additionally, does anyone know what's a good value I can use for temperature for a satellite in space?

But yeah, I think the first part of my question is asking if this technology is scalable (since I feel like there'd be a lot of noise from phones that don't want to talk to satellites), and the second is determining how fast the the uplink connection is going to be.

If anyone can help that'd be great, thanks! I'd love to see this technology work - it'd be amazing to always get a signal when you're driving on the highway or when doing more remote activities like hiking.

## Update

Thanks for the current responses! From what I've gathered so far:

Q1. Will the satellite be able to deal with noise and interference from other handsets or base stations while in orbit?

It seems like yes. Beamforming and spot beams from the phased away will keep the individual cell areas away from terrestrial cell towers.

Q2. What would the uplink data rate be?

Following the answer from @Ng Ph's post, assuming no interference:

• using LTE band 3 / 1710 – 1785 MHz (changing from 1920 MHz above due to band 1 only supporting a minimum bandwidth of 5 MHz)
• using a bandwidth of 1.4 MHz
• assuming that the satellite temperature is 70ºC / 343.15ºK (The Earth's average noise temperature is 300K, but going to assume unfavorable circumstances, and that the solar panels behind the antenna will generate a lot of heat)

Link budget: $$-127.1 \text{ dBW}$$

$$-7 \text{ dBW} - 5 \text{ dBi} - 151.1 \text{ dB} + 36 \text{ dBi}$$

Signal-to-Noise (dB): $$14.81 \text{ dB}$$

$$-127.1 \text{ dBW} - 10 \times \log_{10} \left( (1.38 \times 10^{-23} \text{ J/K}) \times 343.15 \text{ K} \times 1080000 \text{ Hz} \right)$$

Subtract one from SNR as modern LTE networks follow Shannon's bound to within 1dB

Signal-to-Noise (Linear): $$19.95$$

$$\text{SNR dB} = 10 \times log_{10} (\text{SNR linear})$$

$$13.81 \text{ dB} = 10 \times log_{10} (\text{SNR linear})$$

$$\text{SNR linear} = 10^{\frac{13.81}{10}} = 19.95$$

Shannon's equation expects SNR in linear terms, so we need to convert from dB

Channel capacity: $$4739988 \text{ bits/s}$$

$$1080000 \text{ Hz} \times log_{2} \left( 1 + 19.95 \right)$$

Therefore it seems like a possible uplink transfer speed could be $$4.7 \text{ Mb/s}$$, given the assumptions above.

****Assuming I did the calculations correctly: Is it possible for LEO satellites to detect a usable signal from regular mobile phones on the ground?

• – uhoh
Commented Sep 26, 2021 at 2:25
• – uhoh
Commented Sep 26, 2021 at 2:27
• Just to mention it: There are two more concerns which makes the transmission troublesome: Correcting for the always changing Doppler shift and the latency due to the distance - it's larger than the usual protocol allows for. Commented Sep 26, 2021 at 15:29
• @asdfex Yup! I'm aware of those, thanks for mentioning! I do think those are additional technical challenges - but I didn't think it was as important as whether a standard mobile phone could actually emit a strong enough signal for a satellite to detect. Commented Sep 26, 2021 at 16:51
• Spy satellites are capable of resolving visual spectrum signal likely considerably weaker than what the cellphone can provide in the radio spectrum, so I'm pretty sure a satellite given precise location of the phone and superior quality directional antenna could establish a two-way link with that specific phone. Anything "wider" would rapidly become untenable.
– SF.
Commented Sep 27, 2021 at 10:50

What the heck is Δ𝑓?

You guessed it correctly: it is the bandwidth occupied by the signal. It is determined by the system you are dealing with. Let's take LTE (the standard used by our smartphones -aka 4G) as example. The standard stipulates that your device must support several CHANNEL bandwidths: 1.4,3,5,10,15,20 MHz. To derive the respective occupied bandwidth (Δ𝑓 in uhoh's link-budget), you have to subtract the guard-band (usually x% of the channel bandwidth). 10% is a good rule-of-thumb value for guard-bands. As a numerical example, if you have 5MHz channel, then your signal occupies a Δ𝑓 of 4.5 MHz.

For a more detailed calculation for LTE, you can read this tutorial

How do I convert bandwidth into bits per second?

You need to know the "spectrum efficiency" of your transmission link. It has as unit bits/second/Hz. So, bitrate = Δ𝑓 x (spectrum efficiency). If your system has an efficiency, say 5 bits/second/Hz, then with 4.5 MHz bandwidth you will get 4.5x5= 22.5 Mbps.

You need to know your signal-to-noise ratio (S/N), because Shannon taught us that there is a limit to the spectrum efficiency, given a S/N. State-of-the-art digital transmissions (such as LTE) follow this Shannon's bound to approximately 1 dB. Hence, suppose your link-budget gives a S/N. Then subtract 1 dB from that value and apply Shannon's bound:

to get the achievable spectrum efficiency (C/B), as a rule-of-thumb.

• Ah, thanks for clarifying Δ𝑓. || Did the calculation that Tom Spilker do in that comment give the signal-to-noise ratio? The voyager post also called it noise effective power, which also confused me... was 3db the S/N ratio there? || 10 × log_10 ((1.38×10^-23) × 20 × 10^3) = -185.6 dBW || -182.6dBW - -185.6dBW = 3db SNR? || Am I able to use those same equations for my purpose here? Commented Sep 25, 2021 at 22:04
• Yes, you can, provided your assumption of noise temperature(T) is correct. It should be higher for LEO, since basically it is the average Earth noise temperature (for Voyager it is the average space noise temperature). Also remember that thermal noise is Gaussian and flat ("white noise"). If you have interference you have to investigate how it affects your receiver. In general if noise is dominant, you can add (linearly) the interference power to the noise power. Commented Sep 26, 2021 at 15:33
• Thanks! I updated my question with those calculations. Can you check if it seems like I did everything correctly? Commented Sep 26, 2021 at 18:20
• Hmm, I am a bit confused though. If I use those calculations on the Voyager question, 1000 Hz * log_2 ( 1 + 3 ) gives me 2000 bits/s. And Voyager's transmission is only known to be 160 bits/s. Commented Sep 26, 2021 at 18:36
• Be careful, the SNR in Shannon equation is not in dB. As numerical example, with SNR=13 dB, the upper limit of the spectrum efficiency is log_2(1+20)~4.39 bits/s/Hz. You can ideally extract in excess of 4 Mbps in 1MHz. With the same signal power, if you have 2MHz, you can ideally extract 6.9Mbps. This is because your SNR(dB) drops by 3dB and your ideal spectrum efficiency is log_2(11). Bottom line: if you have bandwidth, use it! Commented Sep 26, 2021 at 19:47

What does total noise power (received plus system) mean?

Noise that's received by the antenna (everything on-frequency that isn't your desired signal, from other transmitters to lightning storms to the Sun), plus noise generated inside the receiver itself (all amplifiers are imperfect and add some noise; all resistors generate thermal noise; other electronics (especially digital) inside the receiver can create noise; and any kind of nonlinearity in the receive path can result in signals from completely different frequencies bleeding into your signal).

Does this mean that, given how sensitive the satellite is, it may pick up two signals of the same frequency from hundreds of kilometers away on Earth's surface?

Perhaps — that depends on the antenna gain. Higher gain means narrower beamwidth (antennas don't create power out of nowhere, they just concentrate power more or less tightly).

Considering that those signals normally wouldn't see other, does this make it effectively impossible for the satellite to distinguish between a signal that was intended for the satellite versus dozens others that were intended to reach a cell tower?

Perhaps. Cell phones have clever mathematical ways (CDMA) to minimize the interference between different devices that are using the same frequency, but they're not really meant for this scenario. Maybe the answer lies in getting a frequency allocation on one of the supported bands that isn't used by any terrestrial carriers in a given country. This would be ludicrously expensive, but then again, so is space.

Is that received noise or system noise?

That would be received noise.

But yeah, I think the first part of my question is asking if this technology is scalable

On the surface, it doesn't really seem like it... high gain requires narrow beamwidth, avoiding the interference problem you mention requires narrow beanwidth, but

A) LEO isn't geostationary. A satellite at 500km is zooming across the sky over the course of 10 or 20 minutes. How does the system find the users who want it, point the beam at them, and keep it there?

B) How many of these spot beams can one satellite generate? Or put another way, how many of these satellites do you need to ensure continuous coverage for all of the users, given (again) that the satellites are quickly moving across the sky, so the ones visible from a given point are constantly changing?

Seems like a stretch to me, although I'm sure their goal is to convince investors that they have good answers to those questions.

• Thanks for clarifying what noise it is! Regarding your post though, do you think the antenna in space being a giant phased array would help with the issues you highlighted in A and B? A phased array should be able to detect what direction a signal that hit it came from, and be able to lock on and transmit back to that location on the ground in theory, right? I'm not sure how much spot beams such a satellite/antenna would be able to generate, but for now, even being able to connect to one handset on the ground would be amazing progress to see in the future! Commented Sep 26, 2021 at 16:12
• As for frequency allocation, ASTS plans on working with telecoms to use normal terrestrial carrier bands. This is probably another technical challenge, but I have came across whitepapers that do describe integrating terrestrial and mobile networks, i.e.: itu.int/dms_pubrec/itu-r/rec/m/… Commented Sep 26, 2021 at 16:46
• For question (A), ASTS has to operate with a mobile operator in a given country (who holds the license for the frequency). This operator would tell ASTS to point their beams when its satellites overfly the said country. Obviously, it would be where the mobile operator's coverage is weak (probably low-density of population). For (B), they can progressively add satellites to the constellation. Commented Sep 27, 2021 at 12:08
• @0xbad1d3a5, the idea being to use satellite "cells" to complement a terrestrial coverage (of a business partner having the frequency license), the co-frequency interference you are thinking of is not possible (or of the same order than that of terrestrial interference between towers). ASTS system does not operate in stand-alone and is not intended to compete with terrestrial system in populated areas (dense BS network). Commented Sep 28, 2021 at 10:37
• @0xbad1d3a5, and this in turn means a large antenna in order to achieve "small cells". This is one of the drawbacks of ASTS proposal: large antennas => risk of break-up => debris in-orbit. Their demo sat (BlueWalker3) weighs nearly 1.5T, about 10 times a 0neWeb satellite. Another technical diificulty is the precise pointing of the antenna. Still another one is power-hungry phased array. Commented Sep 28, 2021 at 10:46

What does total noise power (received plus system) mean?

Usually it means everything that is not the signal you want. Noise can come from interference from other radio transmitters, other electronics devices and thermal noise inside the receiver system itself.

A satellite based system does have some potential advantages compared to ground towers:

Reception area

A high gain antenna limits the area from which interference is received. Looking down from above, there is a limited number of devices within that area - especially in rural areas. The antenna gain of 40 dBi means that the reception area is 10⁴ times smaller than a full sphere. With 40 dBi gain and 500 km orbit, it gives a circle with radius of 10 km:

r = √(4π R² G / π) = √(4π (500km)² 10⁻⁴ / π) = 10 km

In comparison, ground towers have a horizontal reception and transmission pattern that often extends to the horizon. For a 50 meter high tower, that would be a sector that extends 25 km at a typical angle of about 45 degrees.

With these estimates the areas turn out approximately equal. There is some further benefit in that the horizontally transmitting ground towers disturb each other much more than they disturb a satellite.

Thermal noise power grows linearly with the absolute temperature. Cooled receivers are often used in space communications to reduce the amount of noise.

I couldn't find information whether these satellites in particular will employ cooled receivers, but being in space they potentially could. With vacuum insulation, the needed cooling power is small, and suitably directed radiators could get quite cold to start with. It is not an easy task, but on the other hand satellite cooling is always a problem that needs careful design.

System noise cannot be directly added to the other interference noise because it is statistically independent. Instead the larger of the two dominates: if system noise level is 3 dB below received interference, it only adds 0.5 dB to the received noise

It is likely that received interference already dominates in ground towers, but the potentially lower received interference in a satellite could require similarly lower system noise to get the full benefit.

• Can you elaborate what you mean when you say "the reception area is a circle with radius of 7 km"? Was that a typo? 153.94km² sounds like awfully small coverage for a satellite 500km in space. Especially if a tower can cover 25km at an angle of 45º - wouldn't that effectively put its coverage at 1/4πr² = 490.875km²? Unless I'm mistaking what you mean by "reception area". For cooling, I feel like they'd be hard-pressed to do anything fancy or at all, considering to launch a phased array antenna that big it needs to unfold in space and have solar panels behind it, likely generating more heat. Commented Sep 26, 2021 at 17:01
• @0xbad1d3a5 I added the formula.. and in process noticed that I had made a mistake of 2π vs. 4π, giving 10km instead of 7km. Regarding the tower area, there are 8 sectors of 45 deg in a circle, giving 1/8πr² = 245km². But that is just a rough guess at what a rural area tower could cover - urban area towers are shorter and have more sectors.
– jpa
Commented Sep 27, 2021 at 4:13
• Oops. Don't know how I mistook 45º for 1/4 instead of 1/8. I'm still not sure how you arrived at those numbers for satellite coverage though. What formula is √(4π R² G / π)? 10km radius still sounds really small to me, especially considering one Iridium satellite can cover a circular area with a diameter of 4700km with an antenna gain of 28dB: tinyurl.com/yybznbyv (page 2, [...] create 48 spot beams [...] covering a circular area with a diameter of approximately 4,700 kilometers). If it makes any difference, the antennas will be phased array antennas and not parabolic antennas. Commented Sep 27, 2021 at 5:17
• @0xbad1d3a5 The formula follows from how dBi is defined as the gain relative to isotropic antenna, see e.g. ahsystems.com/articles/… . In modern phased array antennas, the beams are created in software - even if each one has a radius of 10 km, the satellite can have as many of them as it has computational power. The satellite covers a very wide area, but each signal chain rejects noise outside of its specific area.
– jpa
Commented Sep 27, 2021 at 5:40
• If I'm understanding you correctly, the phased array has a large area, but "listens" for signals in 10km radius at a time in a smaller spot beam / reception area (or does it listen to all spot beams at the same time)? Does this mean that a signal that's say, frequency 1800MHz in spot beam location x wouldn't interfere with another signal that's also 1800MHz in spot beam location y in the coverage area of the phased array? Commented Sep 27, 2021 at 5:53