CWNA question
Last Post: February 6, 2006:
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You have a 20dBm transmitter, a cable with loss of 8dBm and antenna gain with 12 dBi gain which amplifier would provide maximum EIRP tha doesnot exceed 36dBm?
- 8 dBm
-10 dBm
-14 dBm
-20 dBm
Please let me know result with calculation
Thanks
-Saj -
[quote="amar"]You have a 20dBm transmitter, a cable with loss of 8dBm and antenna gain with 12 dBi gain which amplifier would provide maximum EIRP tha doesnot exceed 36dBm?
- 8 dBm
-10 dBm
-14 dBm
-20 dBm
Please let me know result with calculation
Thanks
-Saj
I guessed it has 14dBm But just wanted the how to calculate
Thanks
-Saj -
amar Escribi?3:
You have a 20dBm transmitter, a cable with loss of 8dBm and antenna gain with 12 dBi gain which amplifier would provide maximum EIRP tha doesnot exceed 36dBm?
- 8 dBm
-10 dBm
-14 dBm
-20 dBm
Please let me know result with calculation
Thanks
-Saj
I guessed it has 14dBm But just wanted the how to calculate
Thanks
-Saj
20dBm transmitter - 8dB cable loss + 12dB antenna gain = 24 dBm EIRP, so inserting a 14 dB amp will push you over the limit by 2 dBm (24 dBm + 14 dB= 38 dBm EIRP)- better use the 10 dB amplifier... -
Stan Brooks,
Thanks for the explanation, I appreciate your advise
-Saj -
One thing that might help is to keep in mind calculating RF math is that all the numbers are additive.
So, you're really adding a -8dBm cable loss and then adding a +12dBi antenna gain to your 20dBm transmitter.
WLANstan - I know adding -8dBm vs. subtracting 8dBm is kind of just semantics, but it will help in understanding how to calculate these scenarios.
Good luck! -
CCollins,
Thanks for the update.
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
-Saj -
Amar,
you can look at this 2 ways - the +3's & 10' approximation is the simplest:
100mW -3dB = 100mW/2 = 50mW
50mW +10dB = 50mW*10 = 500mW
This rule is simply: +/- 3dB ~= *// 2
+/-10dB ~= *//10
The key to RF math (I think) is remembering that mW are absloute units of measurement, while dB are units of relative difference - dB are essentially meaningless without a reference point to give them meaning.
The other way to calculate this is to conver mW to dBm (dBm = dB relative to 1mW):
100mW = 20dBm
now you can just add:
20dBm -3dB + 10dB = 27dBm
the tricky part is converting back from dBm to mW, you need to represent the dB value in terms of 10's and 3's:
27dBm = 1mW + 10dB + 10dB + 10dB - 3dB
= 10 * 10 * 10 / 2 mW
= 1000/2 mW
= 500 mW
for 500 mW this is a reasonably straightforward conversion, but some can be a bit more tricky, hence I recommend the first method over the second.
The table on pg 55 of ed.3 CWNA study guide can useful if you want to try this method, however.
Hope this has clarified RF math for you.
:-)
Erin.
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