HOW TO INTERACT WITH FSL EQUATION

Aug 29, 2005 2 Replies
C

Hi folks,



Sorry for this basic question, I have no experience with wireless issues, but I am learning and trying to understand the theory of antennas:



My question is what is the FSL really means?



I cam to know that FSL = 36.6 + 20 LOG F + 20 LOG D



For 5.8 GHZ and 14 miles this would yield a 133.86 db, now is this



133.86 is attenuation? Is it related to the receiver sensitivity at the destination? How can I use this number?


Let me think loud, I understand it this way



If I have EIRP of 36 dB in the transmission end, then this 36 db will be subtracted from the 133.86, hence I will get a -97.86 db, so my receiver should be sensitive enough to receive a power of 1.99 e-10 Watts.



Thanks, ConceptZone



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J

Free Space Loss. It's really a measure of inverse square law. Double the distance and you loose 4 times (-6dB) the power. I don't wanna go into the field equations, but double the frequency and you also lose 4 times the power. See the bottom half of:

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the derivation.

Yep, where FSL is in dB, F is in Mhz, and D is in miles.

Yes.

No. See below.

The purpose of doing everything in dB is to allow simple addition and subtraction to be used for calculations. Gain is added. Loss is subtracted. You start at the transmitter that put's out XX dBm of power (decibels above 1 milliwatt). The coax cable loss is subtracted. The tx antenna gain is added. The free space loss is subtracted. The receive antenna gain is added. The receive coax loss is subtracted. The difference between the receiver sensitivity and what's left of the signal is the fade margin or operating margin. See:

to FSL in dB and everything should add up.

Sorta. You're ignoring the receive antenna gain and the coax and connector losses. The -98dB (watch your significant figures) is the power that the receive *ANTENNA* gets, not what the receiver input gets.

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C

Really Thank you so much on such a valuable information :)

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