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An AP emmits 1000mW signal is connected to connector of 10dB loss, with 3dB gian anteena

EIRP ?

My answer is : 1000/10=100*3 =300

But how will it be 200mW

AM I missing some thing or my mehtod of calculation is right, since in method of 10's and 3's 10dB loss is 1/10 of power

-Saj

If you are using Gain 10s multiply the power by 10
( 100mw AP with a 10dBi gain Ant. = 1000mw or 1W ).
If you are introducing loss 10s divide by 10 ( 100mw AP with 10dBm loss in cable = 10mw ). With the 3s gain doubles and loss cuts in half ( 100mw AP with 3dBi gain ant. = 200mw and 100mw AP with 3dBm loss in cable = 50mw )

Remember decibels are measurements of change not absolute power.
10s...............................3s
1mW = 0dBm ....... 1mW = 0dBm
10mW = 10dBm ....... 2mW = 3dBm
100mW = 20dBm ....... 4mW = 6dBm
1000mW = 30dBm ....... 8mW = 9dBm

combined 10s and 3s
20mW = 13dBm
200mW = 23dBm

Here is a simpler way to solve the problem:
Keep in mind that Watts should converted to dBm in order to determine the power ratio relative to 0 dBm (1mW), then convert back to Watts if that is what you need to know

1000mW = 1W = 30dBm
30dBm ?¡é?€?¡° 10dB(cable loss) = 20 dBm

We know that 20dBm = 100mW
Add 3dBi gain by doubling (x2) 100mW x 2 = 200 mW.

OR to just work with dB terms 20dBm + 3dBi = 23 dBm EIRP (200mW)

Best Regards

Amar,
The only mistake you made was one that I see often with my students. When adding or subtracting 3dB, you multiplied by three or divided by three. Remember, the rule for three's is that if you add 3 dB, you times by TWO (double) ,if you lose 3dB you divide by TWO (one half).

Gene - good catch. I missed that the first time too...they should have made it the rule of 2's and 10's

:--)

Knowing the math behind conversions dB to Watts and vise versa helps to validate your work and knowledge. It also takes away the guess work and provides a more accurate and professional evaluation. I guess you can say it's more empirical and supportive when dealing with CTO's.

To convert watts to dB: 10 Log (watts/.001mW)

And to convert dB back to watts use this equation

((10x (dB/10))X .001mW

The 10x is the inverse Log function on your calculator. Try this to validate your 10 and 3's assumptions.

Regards
Steve