Received power for free space optics

[Nur Ziadah]

Hello everyone,

I have calculated the received power for free space optic (FSO) using the equation:

Lsystem (system loss) is set to 8dB. PTotal can be calculated as:

where Ntx (number of receiver) = 1 and PTx (transmitted power) =7.78 dBm. LGeo can be calculated as:

where d^2R (receiver diameter) = 0.07, l=1, θ (divergence angle) =0.05 and Nr (number of receiver) =1.

The problem is, I got received power=10.72 dBm which is illogical value. As I know, the received power must be lower than transmitted power.
I hope that anyone may help me to understand this situation.
For your information, I refer this paper for the calculation: https://ieeexplore.ieee.org/document/6015903
Thank you.

 

[trurle]

You have above the θl<d^2R, which is outside of the formula applicability range.
Your beam is not illuminating the entire receiver area, therefore the error.

 

[Nur Ziadah]

You have above the θl<d^2R, which is outside of the formula applicability range.
Your beam is not illuminating the entire receiver area, therefore the error.

What is the applicable range?

 

[trurle]

What is the applicable range?

Range (all values) of input variables producing a valid output.

 

[Nur Ziadah]

You mean that θl must greater than d^2R in order to produce a valid output?

 

[tech99]

You have above the θl<d^2R, which is outside of the formula applicability range.
Your beam is not illuminating the entire receiver area, therefore the error.

I am find that optical engineers seem to re-invent the wheel all the time; we have the simple Friis formula for microwave. This uses Ptx, Prx, antenna gains and path loss. The method used above seems very unclear to me. But I am impressed that you noticed that we do not have far field conditions.