RF power calculations
If I want to calculate the power loss in tranmission for RF of 2.4 GHz or path loss, what formula should I use...
Someone suggested Path loss= 40dB + 20 log(distance in metres)... I do not quite understand where the 40 dB term comes from?
I would greatly helped if you could give me a clearer way to calculated power at a certain distance from the transmitter...
Thanking you in advance..
Shantanu...
found a complete info here
http://www.awe-communications.com/Pr...oku/hatoku.htm
Hi,
here the simple formula to calculate the loss in a certain distance:
a(dB)=20lg(4*Π*d/λ)
a is in dB
d in meter
λ in meter (0.125m @ 2.4GHz)
You also have to know the radiated power. Like 1W transmitted power would equal to 30dBm. Add the antenna gain if there is one, for example 6dBi. So the effective radiated power is 36dBm.
there is a loss of 60dB in 10m distance on 2.4GHz. That way, you would measure
-24dBm (=34dBm-60dB).
You may add the gain of the RX antenna if there is one. Like if you have an antenna with 6dBi gain, then you could measure -18dBm of power in 10m distance.
ciao, elo
The answer from elo gives the problem clear
attenuation in dB = 20log10(4 pi d/ lambda)
f =2.45 GHz gives lambda 1/8 meter
we get
att = 20log10(4 pi/lambda) + 20 log10(d)
= 40.05 + 20 log10 (d)
regards
Another way to say the same is:
Free Space Loss(FSL) dB = 32.44 + 20 log (dist in km) + 20 log (freq in MHz)
or 32.44 + 20 log (dist in m) + 20 log (freq in GHz).
This equation comes from the Radar Friis Formula.
Pr=PtGtGr/(2 pi/ lambda) which expresed in dB is
Pr(dBm) = Pt(dBm) + Gt + Gr -FSL
Every body once in a while ask where that "loss" come from?
The loss come from the fact the antenna is not infinite directional.
This means the antenna transmite in all directions and the energy
that is not transmited in the main direction of the link is "loss"