Atmospheric attenuation. Use a value of 300 dB/km for the attenuation of a beam
ID: 1699909 • Letter: A
Question
Atmospheric attenuation. Use a value of 300 dB/km for the attenuation of a beam at ~ I THz (compare slide 19 from the THz lecture - this is roughly a value for mid- latitude summer at 1.05 THz. If 10 mW of power in a collimated beam is transmitted through the atmosphere, how much power is received at a target 100 m away? (Ignore 1/r2 losses for a collimated beam and assume that all the beam hits the target). If the target reflectivity is 10%, how much power returns to the emitter? (Again, ignore 1/r2 losses for a collimated beam, and assume all the returned beam is collected).Explanation / Answer
The reason we like to use dB is that you can divide it easily. So if it is 300 dB/km it will be 30 dB in 1/10km
each 10 dB is 10x
so 10 mW and 10 dB loss is 1mW, 20 dB loss is 0.1 mW, 30 dB loss is 0.01 mW.
Target receives 0.01 mW
The target reflectivity is 10% which is the same as 10 dB, so we have 0.001 mW reflected. Now it needs to get back to the emitter. This will be another 30 dB, so another 1,000x smaller or 0.000001 mW or 0.001 micro Watt, or 1 nano Watt
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.