Continue to Site

Welcome to EDAboard.com

Welcome to our site! EDAboard.com is an international Electronics Discussion Forum focused on EDA software, circuits, schematics, books, theory, papers, asic, pld, 8051, DSP, Network, RF, Analog Design, PCB, Service Manuals... and a whole lot more! To participate you need to register. Registration is free. Click here to register now.

Question about 40dB in RF power loss calculations

Status
Not open for further replies.

sbhalerao

Member level 1
Joined
May 2, 2005
Messages
37
Helped
1
Reputation
2
Reaction score
1
Trophy points
1,288
Activity points
1,787
RF power calculations

Hi to all,
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...
 

RF power calculations

found a complete info here
Code:
[I]**broken link removed**[/I]

and or search on google w/ keyword propagation loss
 

    sbhalerao

    Points: 2
    Helpful Answer Positive Rating
Re: RF power calculations

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
 

    sbhalerao

    Points: 2
    Helpful Answer Positive Rating
Re: RF power calculations

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
 

    sbhalerao

    Points: 2
    Helpful Answer Positive Rating
Re: RF power calculations

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"
 

    sbhalerao

    Points: 2
    Helpful Answer Positive Rating
Status
Not open for further replies.

Part and Inventory Search

Welcome to EDABoard.com

Sponsor

Back
Top