# Receiver and transmitter in RF/optic satelite communciation: distance vs datarate

I am doing research in laser satellite communication and am trying to understand a difference between RF and optical/laser communication systems.

With an increase in distance between a transmitter and a receiver in an RF system, attainable data rate will decrease (all else being equal); the beam becomes wider. The signal strength decreases as the distance increases, right?

In a laser satellite system, we have a much narrower beam, but clouds, sun, and other obstacles can obstruct the signal. Does the distance affect the data rate in laser communication system?

I started writing this answer when the question was posted in Space SE, and moved it here as the question moved.

In a laser satellite system we have a much narrower beam... does the distance effect the data rate in laser communication system?

The same calculations (and physics) apply to both laser and radio link budgets. They both start out as collimated beams and over large distances they both expand in diameter linearly with distance, which means their power per unit area drops as $$1/r^2$$.

As that answer points out, the detection schemes can be different because with light we can do single photon counting/timing, whereas for radio we use normal solid state amplifiers with cooled, low noise front-end stages. That won't make a huge difference in a simple calculation but it's important to keep in mind.

...but clouds/sun/other obstacles can obstruct the signal...

Yes indeed!

Answer(s) to Will future deep space optical communications "ground stations" actually be in space, or on the ground? suggest that it's going to be a lot easier to put a few optical deep space communications satellites in Earth orbit and relay to the ground by conventional radio channels.

These stations are basically look like and will operate like small space telescopes, with apertures of a few tens of centimeters, even for very deep space missions.

There are ways to distribute ground stations that will have a high probability of clear skies: Have these optical satellite ground station locations been chosen for clear skies? but the deep space signals will be very weak so unlike earth orbiting satellite laser communications, these can only be done at night.

...distance vs data rate?

In case you need some actual equations or ballpark numbers, see the answers to :

# However...

The following is copied from my Space SE answer:

This is interesting!

At first I thought that optical communication always wins because the $$\lambda/d$$ for a 30 cm diameter telescope at 850 nm is about 350,000 whereas for a 3 meter dish on a deep space spacecraft at 8 GHz or 32 GHz Ka band is only 80 or 320. That factor of 1000 in $$\lambda/d$$ is a factor of a million in signal strength at the other end, or 60 dB.

That multiplicative factor of a million goes a long way, but the problem is that the current detection schemes for radio and optical are very different.

A radio receiver/detector couples the electric field of the incoming wave into a voltage and that squared, divided by the amplifier's impedance is a power ($$V^2/R$$).

In other words, the received radio power is also the power in the detection circuit, that we compare to the noise equivalent power (NEP) of the amplifier, which will be about $$k_B T \times \Delta f$$ where $$k_B$$ is the Boltzmann constant.

The signal to noise ratio (S/N) is just the ratio of the received power to the noise equivalent power of the receiver front end.

Let's say we are running at the very edge with a S/N = 1. If the received power drops by a factor of 10 (distance is $$\sqrt{10}$$ further) then we have to cut $$\Delta f$$ also by a factor of 10 to maintain the same S/N.

### Photon signal detection

Right now the standard method of converting an optical signal into an electrical signal is to use some kind of photodiode. Most photons that get into the photodiode are absorbed and produce an electron-hole pair. These are collected as an electrical current.

The number of pairs produced and thus the current is proportional to the indcident optical power, okay so far, but the electrical power in the amplifier is equal to the current squared divided by the impedance! ($$I^2R$$)

This means that the electrical power we must compare to the NEP is proportional to the square of the optical power!

Thus once one opens the hood on this problem, one sees that the power collected by the antenna is only half the problem; the method of conversion to electrical signals is so different for optical vs radio that at some very far distance radio may be able to win using conventional detection technology.

### But what about UN-conventional detection technology?

There are a few things to consider that can make optical communication's future at extremely large distances brighter.

Exceeding classical capacity limit in quantum optical channel (also researchgate) is reference #8 in Toyoshima et al.

The amount of information transmissible through a communications channel is determined by the noise characteristics of the channel and by the quantities of available transmission resources. In classical information theory, the amount of transmissible information can be increased twice at most when the transmission resource (e.g. the code length, the bandwidth, the signal power) is doubled for fixed noise characteristics. In quantum information theory, however, the amount of information transmitted can increase even more than twice. We present a proof-of-principle demonstration of this super-additivity of classical capacity of a quantum channel by using the ternary symmetric states of a single photon, and by event selection from a weak coherent light source. We also show how the super-additive coding gain, even in a small code length, can boost the communication performance of conventional coding technique.

Also, since detectors can count individual photons and record their exact arrival time to picosecond precision and some lasers can generate picosecond pulses at micro- and nano-second intervals, there is a lot of opportunity to use the time structure to help boost S/N in a way that is not possible with radio waves, since counting individual radio photons is far more challenging.

For more on that, see

• i have read some reseach and noticed the data rate is increased in small distance, and descreased in big distance ( Mars mission, for ex). If we have a big distance, about > 10^6, we will use RF system Sep 23 at 5:51