dBi Calculation Question

Feb 08, 2005 3 Replies

Guys I am working on my CWNA paper and have hit a little point that I cannot get beyond (although you will probably think this is pretty simple).



It is the calculation for a circuit as below.



RF Circuit



AP--connector (A) -----Cable----Connector (B)----cable---- Connector (C)---Antenna (D)



AP is 100mW and the paper is explaining the calculation as below.



AP Point A Ponit B Point C Point D



100mW -3dB -3dB -3dB +12dBi =100mW /2 /2 /2 (x2x2x2) =100mW /2 /2 /2 x16 =50mW /2 /2 x16 =25mW /2 x16 =200mW

It was my understanding that all dB units (including dBi) are relative units and can be added and subtracted from other dB units - therefore the overall dB should be +3db = doubling the power - hey presto the right answer!



What I cant fathom is the calc shown above - why have they suggested a multiple of x2x2x2 ?



Can someone please explain why......only a little point but bloody annoying when your are learning this stuff for the first time!



Thanks



What they are showing is that 100mw/2 = 50mw, 50mw/2=25mw,

25mw/2=12.5mw which is = to -9db

the x2x2x2 means +12dbi would take the 12.5mw x 2= 25mw, 25mw x

2=50mw, 50mw x 2 = 100mw = up 9db, just where it started. plus 3 extra dbi = 200mw

Make since.... you are right in your calculation, there is a +3db gain, started with 100mw,,, now x2 = 200mw

relative

therefore

suggested

This should probably say x2x2x2x2. Are you getting this from the CWNA study book? Is so what version?

pretty

below.

bloody

I just checked my CWNA second version and it says x2x2x2x2

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