dBi Calculation Question

Feb 08, 2005 2 Replies

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



It is the calculation for a circuit as below.



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



Yep, it is simple. Of course you have to be confident enough to know when you are looking at a typo! They left off one "x2", because 12 dB is one 2x for each 3 dB, so that is a total of *4* x2's that should be in that string.

Looks like your understanding of it is perfect, and what you lack is understanding that you do understand it! :-)

Thanks guys - still a little worried about this but I will face it down come final week before exam.

- I will probably put another post up here later...lol!

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