CWNP CWNA-109 Exam Dumps - Actual Questions Answers

OFDMA is a new modulation scheme introduced in 802.11ax that allows multiple users to share the same channel by dividing it into smaller subchannels called resource units (RUs). This improves the efficiency and capacity of the WLAN by reducing contention and overhead. However, to use OFDMA, both the AP and the client must support 802.11ax and negotiate the parameters of the subchannel allocation. Therefore, the customer needs to upgrade the clients that require OFDMA to 802.11ax devices12.

The other options are not correct because they do not reflect the reality of OFDMA. Option B is incorrect because OFDMA is a mandatory feature of 802.11ax for both downlink and uplink transmissions, and all 802.11ax APs must support it1. Option C is incorrect because OFDM and OFDMA are different modulation schemes, and OFDM does not allow multiple users to share the same channel. Option D is incorrect because 802.11ac devices cannot support OFDMA through driver upgrades, as they lack the hardware and firmware capabilities to do so2.

References: 1:CWNA-109Official Study Guide, page 144 2: OFDMA

Cellular/LTE/5G is the best method for implementing a network backhaul for communications across the Internet in a remote area that is affected by a natural disaster. This is because cellular/LTE/5G networks are wireless and do not depend on physical infrastructure that may be damaged or unavailable in such scenarios. Cellular/LTE/5G networks also offer high-speed data transmission and wide coverage area, which are essential for emergency response operations. 802.11 bridging to the response team headquarters is not feasible because it requires line-of-sight and has limited range. Turning up the output power of the WLAN at the response team headquarters is not effective because it may cause interference and does not guarantee reliable connectivity. Temporary wired DSL is not practical because it requires installing cables and equipment that may not be available or accessible in a remote area. References: CWNA-109 Study Guide, Chapter 7: Wireless LAN Topologies, page 2031

The power level at the Intentional Radiator (IR) is 250 mW. The IR is the point where the RF signal leaves the transmitter and enters the antenna system. To calculate the power level at the IR, we need to consider the output power level of the transmitter, the loss of the cable, and the gain of the antenna. The formula is:

Power level at IR (dBm) = Output power level (dBm) - Cable loss (dB) + Antenna gain (dBi)

We can convert the output power level of 50 mW to dBm by using the formula:

Power level (dBm) = 10 * log10(Power level (mW))

Therefore, 50 mW = 10 * log10(50) = 16.99 dBm

We can plug in the values into the formula:

Power level at IR (dBm) = 16.99 - 3 + 16 = 29.99 dBm

We can convert the power level at IR from dBm to mW by using the inverse formula:

Power level (mW) = 10^(Power level (dBm) / 10)

Therefore, 29.99 dBm = 10^(29.99 / 10) = 999.96 mW

However, since we need to round off the answer to the nearest integer value, we get:

Power level at IR (mW) = 1000 mW

References: [CWNP Certified Wireless Network Administrator Official Study Guide: ExamCWNA-109], page 67; [CWNA: Certified Wireless Network Administrator Official Study Guide: ExamCWNA-109], page 57.