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CWNP CWNA-109 Exam Syllabus Topics:

Section	Objectives
WLAN Network Design	- Design principles and planning 1. Site survey fundamentals 2. Coverage vs capacity planning
WLAN Infrastructure and Devices	- Access points and client devices 1. AP architecture and deployment modes 2. Antennas and MIMO concepts
RF Technologies and Fundamentals	- Radio frequency principles and behavior 1. RF signal propagation characteristics 2. Wavelength, frequency, and power concepts
802.11 Protocol and MAC Layer	- Frame types and communication process 1. CSMA/CA and channel access methods 2. Management, control, and data frames
WLAN Security and Management	- Security mechanisms and monitoring 1. Authentication and encryption methods (WPA2/WPA3 overview) 2. Troubleshooting and performance monitoring
WLAN Regulations and Standards	- Wireless standards and amendments 1. Regulatory domains (FCC, ETSI, etc.) 2. IEEE 802.11 standards overview

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CWNP Wireless Network Administrator (CWNA) Sample Questions (Q36-Q41):

NEW QUESTION # 36
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?

A. 13 dBm
B. 26 dBm
C. 23 dBm
D. 10 dBm

Answer: C

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 # 37
What modulation and multiplexing technique, used by 802.11a/g/n/ac/ax, divides a wireless channel into multiple narrow, orthogonal subcarriers to transmit data in parallel?

A. DSSS
B. FHSS
C. OFDM
D. CCK

Answer: C

Explanation:
Orthogonal Frequency Division Multiplexing (OFDM) is a modulation technique that divides a wireless channel into many smaller, overlapping but mathematically orthogonal subcarriers. Data is transmitted in parallel across these subcarriers simultaneously, which improves spectral efficiency and provides resistance to multipath interference. OFDM was first used in WLANs by the 802.11a amendment and has remained the foundational modulation technique for 802.11g,
802.11n, 802.11ac, and 802.11ax (with the addition of OFDMA).

NEW QUESTION # 38
You are implementing a VHT-capable AP. Which one of the following channels is available in the 802.11-
2016 standard that was not available before the ratification of 802.11 ac?

A. 0
B. 1
C. 2
D. 3

Answer: D

Explanation:
Channel 144 is a new channel that was added to the 5 GHz band by the 802.11ac amendment, which defines the VHT (Very High Throughput) PHY for WLANs. Channel 144 has a center frequency of 5720 MHz and a bandwidth of 20 MHz. It can also be combined with adjacent channels to form wider channels of 40 MHz, 80 MHz, or 160 MHz. Channel 144 is available in some regions, such as North America and Europe, but not in others, such as Japan and China . References: [CWNA-109 Study Guide], Chapter 3: Antennas and Accessories, page 121; [CWNA-109 Study Guide], Chapter 3: Antennas and Accessories, page 115;
[Wikipedia], List of WLAN channels.

NEW QUESTION # 39
What cause of hidden nodes within a BSS would be more likely lead to an increase in collisions and retries?

A. Data frames too large for the physical environment
B. Client stations broadcasting with too much power
C. Access points broadcasting with too little power
D. Large 802.11 cells with physically distributed stations

Answer: C

NEW QUESTION # 40
In a long-distance RF link, what statement about Fade Margin is true?

A. A Fade Margin is unnecessary on a long-distance RF link if more than 80% of the first Fresnel zone is clear of obstructions.
B. The Fade Margin is a measurement of signal loss through free space and is a function of frequency and distance.
C. The Fade Margin of a long-distance radio link should be equivalent to the receiver's low noise filter gain.
D. Fade Margin is an additional pad of signal strength designed into the RF system to compensate for unpredictable signal fading.

Answer: D

Explanation:
Fade Margin is an additional pad of signal strength designed into the RF system to compensate for unpredictable signal fading. It is the difference between the receiver's sensitivity and the actual received signal level. A higher Fade Margin indicates a more robust link that can withstand interference, attenuation, or other factors that may reduce the signal strength. A lower Fade Margin means that the link is more susceptible to failure or performance degradation. Fade Margin is usually expressed in decibels (dB) and can be calculated by subtracting the receiver sensitivity from the received signal level. References: 1, Chapter 2, page 51; 2, Section 2.1

NEW QUESTION # 41
......

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