Radio Frequency Calculus
Last Post: January 15, 2008:

Hello,
I come here to seek help. I was reading the Chapter 2 last night of the official book and i could not understand the explanation given to measure the output power.
question 5. 50mW, 3 dB of loss, rating 6dbi.
output power equals b. 100mW
could please someone develop, step by step, how to get to that number ?
I would appreciate it,
thanks
Rafael Lucca 
Ah this RF math is driving me crazy lol.
From the above example:
A 10 mW signal after a 24 dB gain is a 2500 mW signal.
10 mW x 10 x 10 x 10 / 2 / 2 = 2500 mW = 2.5 Watts
10mW with a gain of 24dB.
How about this way:
10mw +10dB +10dB +10dB  6dB
So step by step:
100
1000
10000
Now 6dB = 3dB 3dB
10000 3dB = 5000
5000 3dB = 2500mW
I understand this part well. Now I'm having issues understanding "dBm" to mW. In the book there is an example:
"Remember that 1mW=0dBm, so a 20dBm product would be equal to 1mWx10X10, or 100mW. Likewise, a 9dBm product would be equal to 1mWx2x2x2, or 8mW. dBm is a reference to the relationship between decibels and linear watts."
I understand the first part how 20dBm is equal to 100mW, using the rule of 10. But I can't seem to grasp how 9dBm became = 8mW?
I have tried to use the formula's from the CWNA book too:
mw=log1 (dbm/10) but I don't really remember logs and inverse logs any more. Since I don't understand that formula yet but I'd like to, here is what I am doing:
1mW=0dBm
+9dBm=+3dB +3dB +3dB
Since; +3dB = Double the Power so we can make it:
+9dBm=+3dB +3dB +3dB = 1mWx2x2x2=8mW.
Similarly if it is 5dBm how would I calculate that?
because I can't really do:
+5dBm=+3dBm +3dm because it is not +6dBm
What do you guys think am I on the right track here? 
5dB is the same as 3dB + 3dB + 3dB + 3dB + 3dB  10dB
You are on the right track. 
Criss_Hyde Escribi?3:
5dB is the same as 3dB + 3dB + 3dB + 3dB + 3dB  10dB
You are on the right track.
Thanks for the answer Criss. So basically as long as I can get to the original value by using 3's and 10's it will give me the correct answer? 
Yes, for exam purposes.
The rule of 10 is exactly correct. The rule of 3 is an approximation. 
Thanks so much again for the clarification. I'm still going to learn the formulas because I rather know stuff from scratch and that "how and why" bugs me until I find out.
BTW: on a side note I see you are in Leesburg VA, cool place I have been there twice. 
Hopefully the last question:
Ok so I understand dBm to mW. Now trying to understand mW to dBm.
So if i have 8mW how can I convert it into dBm's? 
Here's a good dBmtomW (and vice versa) conversion calculator:
http://www.moonblinkwifi.com/dbm_to_watt_conversion.cfm
Typically you won't have to do mW to dBm conversions for CWNP exams but if you just remember some often occuring ones, then that will get you within the right ballpark (i.e. close enough to do a good guess). Here's some notes I made regarding this:
Joel 
Thanks Joel, will be taking my test next month will post back if I passed or not. Wish me luck and thanks again.

I need some help with this too, I think I get it but want to be sure.
A wireless device has output of 200 mW. It is connected to a cable with a 9 dBi gain and a 6 dB loss. What is the EIRP? 400 dBi
200 * (10^.9)[about 8] / (10^.6)[about 4] = 400
or is it
200 / (10^.9) * (10^.6) = 100
and the answers are in mW, correct?