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Title: New CWNA-109 Test Price & Learning CWNA-109 Materials
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CWNP CWNA-109 Exam Syllabus Topics:
TopicDetails
Topic 1
  • WLAN Regulations and Standards: The topic discusses the roles of WLAN and networking industry organizations. It also addresses the concepts of various Physical Layer (PHY) solutions, spread spectrum technologies, and 802.11 WLAN functional concepts.
Topic 2
  • WLAN Network Security: It addresses the concepts of weak security options, security mechanisms for enterprise WLANs, and security options and tools used in wireless networks.
Topic 3
  • WLAN Protocols and Devices: It focuses on terminology related to the 802.11 MAC and PHY, the purpose of the three main 802.11 frame types, MAC frame format, and 802.11 channel access methods.
Topic 4
  • WLAN Network Architecture and Design Concepts: This topic deals with describing and implementing Power over Ethernet (PoE). Furthermore, the topic covers different wireless LAN architectures, coverage requirements, roaming considerations, and common proprietary features in wireless networks.
Topic 5
  • RF Validation and WLAN remediation: This topic covers RF interference, WLAN performance, the basic features of validation tools, and common wireless issues.

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CWNP Wireless Network Administrator (CWNA) Sample Questions (Q101-Q106):NEW QUESTION # 101
A WLAN transmitter that emits a 50 mW signal is connected to a cable with 3 dB loss. If the cable is connected to an antenna with 9dBi gain, what is the EIRP at the antenna element?
Answer: B
Explanation:
To calculate the EIRP at the antenna element, we need to add the transmitter output power, subtract the cable loss, and add the antenna gain. All these values need to be converted to dBm first, if they are not already given in that unit. In this case, we have:
Transmitter output power = 50 mW = 10 log (50) dBm = 16.99 dBm Cable loss = 3 dB Antenna gain = 9 dBi EIRP = Transmitter output power - Cable loss + Antenna gain EIRP = 16.99 - 3 + 9 EIRP = 22.99 dBm Rounding up to the nearest integer, we get 23 dBm as the EIRP at the antenna element12. References: CWNA-
109 Study Guide, Chapter 2: Radio Frequency Fundamentals, page 92; CWNA-109 Study Guide, Chapter 2:
Radio Frequency Fundamentals, page 88.

NEW QUESTION # 102
In a mesh BSS (MBSS), according to the 802.11 standard, what device connect the mesh to an Ethernet network?
Answer: D
Explanation:
a mesh portal is a device that connects a mesh BSS (MBSS) to an Ethernet network, such as the Internet. A mesh portal acts as a bridge between the wired and wireless domains, and allows the mesh stations to communicate with external networks. A mesh portal is also a mesh point, which means it can forward traffic within the MBSS.
The other options are not correct. Option A. Mesh Gate is a device that connects a mesh BSS (MBSS) to another mesh BSS or another wireless network, such as an infrastructure BSS or an ad hoc network2. A mesh gate acts as a gateway between different wireless domains, and allows the mesh stations to communicate with other wireless networks. A mesh gate is also a mesh point, which means it can forward traffic within the MBSS. Option B. Mesh Switch is not a valid term in the 802.11 standard. Option C. Mesh Router is also not a valid term in the 802.11 standard.

NEW QUESTION # 103
What statement about 802.3, Clause 33 Power over Ethernet is true?
Answer: C
Explanation:
https://www.cablinginstall.com/a ... -vs-cat-5e-poe.html The statement that the lowest voltage drop is achieved when using CAT6 cable instead of Cat5 or CAT5e is true about 802.3, Clause 33 Power over Ethernet. Power over Ethernet (PoE) is a technology that allows electrical power to be delivered over Ethernet cables along with data signals. PoE is defined by IEEE 802.3, Clause 33 and has several variants, such as PoE (802.3af), PoE+ (802.3at), and PoE++ (802.3bt). PoE works by using a device called PSE (Power Sourcing Equipment) that injects power into the Ethernet cable and a device called PD (Powered Device) that receives power from the Ethernet cable. The PSE can be either an endpoint device, such as a switch or a router, or a midspan device, such as an injector or a splitter, that is inserted between two Ethernet devices. The PD can be any device that requires power, such as an access point, a camera, or a phone.
One of the factors that affects PoE performance is voltage drop, which is the reduction of voltage that occurs as current flows through a cable due to its resistance. Voltage drop can cause power loss and inefficiency in PoE systems, as well as damage to PDs if the voltage falls below their minimum requirement. To minimize voltage drop, it is recommended to use high-quality cables with low resistance and short length. Among the common types of Ethernet cables, CAT6 has the lowest resistance and therefore the lowest voltage drop compared to Cat5 or CAT5e. CAT6 also has higher bandwidth and data rate than Cat5 or CAT5e, making it more suitable for PoE applications. References: 1, Chapter 7, page 263; 2, Section 4.4

NEW QUESTION # 104
A dual-band 802.11ac AP must be powered by PoE. As a class 4 device, what power level should be received at the AP?
Answer: C
Explanation:
PoE has different standards that define different power levels for PSEs and PDs. The original standard, IEEE
802.3af, defines two classes of PSEs: Class 3 (15.4 W) and Class 4 (30 W). The newer standard, IEEE 802.3 at, also known as PoE+, defines four classes of PSEs: Class 0 (15.4 W), Class 1 (4 W), Class 2 (7 W), and Class 3 (12.95 W). The power level received at the PD is always lower than the power level provided by the PSE, due to cable resistance and power dissipation. The IEEE standards specify the minimum power level that must be received at the PD for each class of PSE. For a Class 4 PSE, the minimum power level received at the PD is 25.5 W910. References: CWNA-109 Study Guide, Chapter 7: Power over Ethernet (PoE), page
295; CWNA-109 Study Guide, Chapter 7: Power over Ethernet (PoE), page 289.

NEW QUESTION # 105
An 802.11 WLAN transmitter that emits a 50 mW signal is connected to a cable with 3 dB of loss. The cable is connected to an antenna with 16 dBi of gain. What is the power level at the Intentional Radiator?
Answer: D
Explanation:
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
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