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My apologies in advance if this is the wrong place to ask this question. I was told that there were some RF communications experts here, so i figured there might be some FSO experience here as well.

Anyway, I am trying to put together a link budget for a 1550nm laser communication system in low earth orbit (~600km), and I have been seeing some contradictory information regarding the contribution of atmospheric attenuation (absorption, scattering, scintillation). I am not looking for an exact figure at this point, but I do need to have some idea in order to spec out the rest of the system. I have read from multiple sources that the minimum (clear air) attenuation for light in this region is about 0.2 dB/km.

In "Free-Space Laser Communications: Principles and Advances", the following example link budget is proposed:

enter image description here

Unfortunately, no justification is provided for the "Clear air transmission loss" term, which is about 11km at 0.2dB/km, rather than 483km.

In another system, a comparable attenuation term is used enter image description here

Is the atmospheric attenuation this low because it decreases with altitude (and 0.2 dB/km is the attenuation coefficient near sea level)? How do you go about making that computation. If the altitude dependence is significant, it is not discussed in the book. If the effect of atmospheric attenuation becomes negligible beyond a certain altitude, where would that be?

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    \$\begingroup\$ What height of column of air at 1 atmosphere pressure would weigh enough to generate that pressure? Handy envelope and some scribbles suggests it's about 8.33km. \$\endgroup\$
    – user16324
    Commented Sep 26, 2016 at 21:53
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    \$\begingroup\$ Rain and fog will make a big difference. \$\endgroup\$
    – Andy aka
    Commented Sep 26, 2016 at 22:12
  • \$\begingroup\$ @BrianDrummond I hadn't looked at it that way but that makes a lot of sense. Definitely gets me in the ball-park of what I am expecting. I would upvote you if I was higher level haha. Expanding on this, if i was to take the difference between the pressure at, say, 100km and sea level I could determine the equivalent horizontal distance in the same method. \$\endgroup\$
    – user2387855
    Commented Sep 26, 2016 at 22:21
  • \$\begingroup\$ Search on term "scale height" to get equivalent liner atmosphere height. No guarantee that this is 100% good for comms due to possible characteristic change with pressure. \$\endgroup\$
    – Russell McMahon ♦
    Commented Sep 27, 2016 at 2:54
  • \$\begingroup\$ @user2387855: Here is some information for you: informit.com/articles/article.aspx?p=26141&ranMID=24808 \$\endgroup\$
    – Uwe
    Commented Sep 27, 2016 at 7:35

2 Answers 2

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Unfortunately, no justification is provided for the "Clear air transmission loss" term, which is about 11km at 0.2dB/km, rather than 483km.

Clear AIR transmission loss happens only in the lower layers of the atmosphere, which are the ones where the air is. That's where the 11 km figure comes from.

The troposphere contains 80% of the total mass of air of the whole atmosphere. The average height of the troposphere is 12 km, and ranges from 8 to 18 km depending on the latitude and the season of the year. I've seen 11 km being used in a lot of satcom link budgets, it's a pretty conventional value.

Troposphere height

If the effect of atmospheric attenuation becomes negligible beyond a certain altitude, where would that be?

At the tropopause (the layer between the troposphere and the stratosphere).

Earth atmosphere profile

Image source: wikipedia article on Atmosphere of Earth.

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Check out this reference: http://www.phy.davidson.edu/FacHome/jny/Optics/Burle%20Electro_Optics.pdf. This is the Burle Electro-Optics Handbook. The chapter on atmospheric transmittance, pp 81-108, and particularly the graph on pg 83 address your question. (Looks like rather low attenuation at 1550 nm up through a clear atmosphere, but if you used a bit shorter wavelength, the attenuation would increase rapidly.)

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