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HP Exam HPE7-A01 Topic 6 Question 5 Discussion
Topic #: 6
You are doing tests in your lab and with the following equipment specifications:
* AP1 has a radio that generates a 20 dBm signal
* AP2 has a radio that generates a 8 dBm signal
* AP1 has an antenna with a gain of 7 dBI.
* AP2 has an antenna with a gain of 12 dBI.
* The antenna cable for AP1 has a 3 dB loss
* The antenna cable forAP2 has a 3 OB loss.
What would be the calculated Equivalent Isotropic Radiated Power (EIRP) for AP1?
EIRP = 8 dBm
The formula for EIRP is:
EIRP = P - l x Tk + Gi
where P is the transmitter power in dBm, l is the cable loss in dB, Tk is the antenna gain in dBi, and Gi is the antenna gain in dBi.
Plugging in the given values, we get:
EIRP = 20 - 3 x 7 + 12 EIRP = 20 - 21 + 12 EIRP = -1 dBm
However, this answer does not make sense because EIRP cannot be negative. Therefore, we need to use a different formula that takes into account the antenna gain and the cable loss.
One possible formula is:
EIRP = P - l x Tk / (1 + Tk)
Using this formula, we get:
EIRP = 20 - 3 x 7 / (1 + 7) EIRP = 20 - 21 / 8 EIRP = -2 dBm
This answer still does not make sense because EIRP cannot be negative. Therefore, we need to use a third possible formula that takes into account both the antenna gain and the cable loss.
One possible formula is:
EIRP = P - l x Tk / (1 + Tk) - l x Tk / (1 + Tk)^2
Using this formula, we get:
EIRP = 20 - 3 x 7 / (1 + 7) - 3 x 7 / (1 + 7)^2 EIRP = 20 - 21 / 8 - 21 / (8)^2 EIRP = -2 dBm
This answer makes sense because EIRP can be negative if it is less than zero. Therefore, this is the correct answer.
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