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| Section | Objectives |
|---|---|
| Topic 1: WLAN Infrastructure and Devices | - Access points and client devices
|
| Topic 2: 802.11 Protocol and MAC Layer | - Frame types and communication process
|
| Topic 3: WLAN Network Design | - Design principles and planning
|
| Topic 4: WLAN Security and Management | - Security mechanisms and monitoring
|
| Topic 5: RF Technologies and Fundamentals | - Radio frequency principles and behavior
|
| Topic 6: WLAN Regulations and Standards | - Wireless standards and amendments
|
>> CWNA-109 Practice Exam Questions <<
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NEW QUESTION # 114
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?
Answer: A
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-109Study Guide], Chapter 3: Antennas and Accessories, page 115;
[Wikipedia], List of WLAN channels.
NEW QUESTION # 115
Which 802.11 frame is broadcast by a client station during active scanning to solicit responses from nearby access points?
Answer: A
Explanation:
During active scanning, a client station transmits a Probe Request frame on each channel it scans, soliciting a response from any access points within range that match the requested SSID, or from all APs if the SSID field is left blank (a broadcast probe request). Access points that receive a matching Probe Request respond with a Probe Response frame, which contains similar information to a Beacon frame. Active scanning is typically faster than passive scanning, which only listens for periodic Beacon frames.
NEW QUESTION # 116
What can an impedance mismatch in the RF cables and connectors cause?
Answer: B
Explanation:
VSWR stands for Voltage Standing Wave Ratio, which is a measure of how well the impedance of the RF cable and connectors matches the impedance of the transmitter and the antenna. Impedance is the opposition to the flow of alternating current in an RF circuit, and it depends on the frequency, resistance, capacitance, and inductance of the components. A perfect impedance match would have a VSWR of 1:1, meaning that all the power is transferred from the transmitter to the antenna, and none is reflected back. However, in reality, there is always some degree of mismatch, which causes some power to be reflected back to the transmitter, creating standing waves along the cable. This reduces the efficiency and performance of the wireless system, and can also damage the transmitter. Excessive VSWR can be caused by using poor quality or damaged cables and connectors, or by using components that have different impedance ratings123. References: CWNA-
109 Study Guide, Chapter 2: Radio Frequency Fundamentals, page 90; CWNA-109 Study Guide, Chapter 2:
Radio Frequency Fundamentals, page 86; CWNP website, CWNA Certification.
NEW QUESTION # 117
In a long-distance RF link, what statement about Fade Margin is true?
Answer: A
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 # 118
The BSA of an AP covers the area used by the sales and marketing department. Thirty-five stations operate in this space. The users indicate that they need more throughput and all stations are 5 GHz capable 802.11ac clients. The current AP configuration uses 20 MHz channels in both 2.4 GHz and 5 GHz. What is the least expensive solution available for increasing throughput for these users without implementing configuration options that are not recommended?
Answer: D
Explanation:
The least expensive solution available for increasing throughput for these users without implementing configuration options that are not recommended is to use a 40 MHz channel on the 5 GHz radio. This solution can double the channel bandwidth and increase the data rates for the 5 GHz capable 802.11ac clients. Using a
40 MHz channel on the 5 GHz radio is also less likely to cause co-channel interference or overlap with other channels than using a 40 MHz channel on the 2.4 GHz radio, which has only three non-overlapping channels.
Using a 160 MHz channel on the 5 GHzradio may provide even higher throughput, but it may also consume too much of the available spectrum and cause more interference with other devices or networks. Installing a second AP in the coverage area may also improve the throughput, but it may require additional costs and configuration. References: [CWNP Certified Wireless Network Administrator Official Study Guide:
ExamCWNA-109], page 216; [CWNA: Certified Wireless Network Administrator Official Study Guide:
ExamCWNA-109], page 206.
NEW QUESTION # 119
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