Exam 350-401 topic 1 question 407 discussion

3 dB of gain = +3 dB = doubles signal strength (Let’s say, the base is P. So 10*log10(P/P) = 10*log101 = 0 dB and 10*log10(2P/P) = 10*log10(2) = 3dB -> double signal)

Remember the rule of 3s and 10s

It´s A To double the transmit power, we need to increase the power level by 3 dB. 10 dBm + 3 dB = 13 dBm

In terms of dBm, an increase of 3 dBm represents a doubling of the power. This is due to the logarithmic nature of the decibel scale, where: An increase of 10 dB means a tenfold increase in power. An increase of 3 dB means approximately a doubling of power. Given the current power level is 10 dBm: Doubling the power (an increase of 3 dBm) would result in 10 dBm + 3 dBm = 13 dBm 10 dBm+3 dBm=13 dBm.

3db - 2x

Agree with A, 3 dB doubles the signal stregth

Law of 3s

If the maximum power level assignment for global Transmit Power Control (TPC) for 802.11a/n/ac is configured to 10 dBm, then doubling the transmit power would require an increase of 3 dBm. A 3 dBm increase represents a doubling of the power level, so a power level of 13 dBm would effectively double the transmit power. However, it's important to note that TPC settings can vary depending on the specific implementation and regulatory requirements, so the actual power levels available may be different in different situations.

aggree with jj970us

Definitely A.

Definitivamente es la A

it's widely known fact +3dB means doubled

The answer is A, it is just math simply.

Page 493 of the Cert guide goes over this. Answer is A