Forum

what happens in case Given wireless LAN transmitter emits 100mW signal connected to with cable loss 3dB and if a cable connected to an antenna with 10dBi gain
What is the resultant EIRP of antenna element?

I am guessing it has 500mW but need to know how this calculation works

-Amar

You have the right answer, so here is one way to get it.

I always follow down the path of the signal and factor for each cable/connector on the way. So...

The signal is now 100mw. When it reaches the end of the cable, it has now lost 3dBs. With the rules of 10's and 3's, I know that a 3dB loss is 1/2 the power, so now it is at 50mw. The next thing the signal encounters is a 10dBi antenna, which is always gain. So again with the rules of 10's and 3's, I know that 10dB gain is 10 times the power, so now I take my 50mw times 10 and I get 500mw.

So, if you understand this, here is another question for you.

You have an AP with a 50mw transmitter. You have a cable with 2.3 dB's of loss and a lightning arrestor with .7 dB's of loss. You now have a 7dBi antenna. What is your EIRP in mw?

:) If you get this one, you understand RF Math just fine.

What is the rule of 10's and 3's ?
could you give more details about this rule?

The rule of 10's and 3's is for simple calculations of RF math. Note that the "3's" rule is an approximation, but is accepted for general use.

RF math is used to calculate gains and losses. For example, cables have resistance, so there is a loss assoicated with cable. An antenna provides for gain, so that also needs to be calculated. One of the uses of RF math is to determine whether your system is within FCC (or your country's equivalent) guidelines.

Here is a quick chart.

+3 dB = 2x the power
+10 dB = 10x the power
-3 dB = 1/2 power
-10 dB = 1/10th the power

So, if you have 100mw transmitter and you have 3dB of cable loss, then at the end of the cable you have 1/2 the power, or 50mw. If that cable is attached to an antenna with 10dBi (don't worry about the i part of dBi yet) of gain, then you take the 50mw coming out of the cable and multiply by 10, just as the rules of 10's and 3's states.

If you plan on studying wireless networking and topics such as these, I recommend the CWNA study guide. Even if you don't plan on getting certified, it is a great book on wireless networking. Note, that if you get the book to check the errata on this website. There is an error on pages 56-57, which is in the RF math section.

Good luck and let us know if this works out for you.

Thanks GTHill
-Saj

Always welcome!

GTHill,
You have an AP with a 50mw transmitter. You have a cable with 2.3 dB's of loss and a lightning arrestor with .7 dB's of loss. You now have a 7dBi antenna. What is your EIRP in mw?


Is this right:(2.3+.7=3.0)

50/2=25mw

EIRP: 25*7= 175mw

-Saj

You got the first half correct! Remember that you using the rules of 10's and 3's you will only times by 10 or 2 or divide by 10 or 2, never anything else. So, you are correct that you now have 25mW going into your antenna. What does the 7dBi antenna do to your 25mW? The first thing is to figure out how you use 10's and 3's to get to the number 7. Remember, you can add and subtract dB's to get to your goal.

Amar, you have two different directions that will yield the same result expressed in two differenct measurement values.

For each 3 db of loss the power is decreased by half. When there is a 10 db change ( in your example it is an increase ) and the other measurement is in milliwatts , the 10 db change is equal to multiplying the value in milliwatts which would be 500 mw.

The second method would have been to convert the 100mw to db which is equal to 20db. The solutions is straight algebra. A+B+C= EIRP 20 + (-3) + 10 = 27db which is the same as 500 mw.

RF math is straight forward if you remember the rules of 3s and 10s.

Regards,

George