CWNP CWNA-109 Online Practice Questions and Exam Preparation

CWNP CWNA-109 Online Questions & Answers

• Question 21:

  You manage a WLAN with 100 802.11ac access points. All access points are configured to use 80 MHz channels. In a particular BSS, only 40 MHz communications are seen. What is the likely cause of this behavior?

  A. All clients implement single spatial stream radios
  B. The clients are all 802.11n STAs or lower
  C. The AP is improperly configured to use only 40 MHz of the 80 MHz allocated bandwidth
  D. The short guard interval is also enabled
  B. The clients are all 802.11n STAs or lower

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  Explanation/Reference:

  https://7signal.com/802-11ac-migration-part-2-whats-nobodys-telling-you- about-80mhz-and-160mhz-channel-bonding The clients are all 802.11n STAs or lower is the likely cause of this behavior. If a WLAN with 100 802.11ac access points is configured to use 80 MHz channels, butonly 40 MHz communications are seen in a particular BSS, it means that the clients in that BSS do not support 80 MHz channels. This could be because they are using older standards, such as 802.11n or lower, that do not support 80 MHz channels. Alternatively, they could be using newer standards, such as 802.11ac or ax, but have their channel width settings limited to 40 MHz or lower due to device capabilities or configuration options. In either case, the AP will adapt to the client's channel width and use only 40 MHz of the 80 MHz allocated bandwidth to communicate with them. This will reduce the potential throughput and efficiency of the WLAN. References: , Chapter 3, page 111; , Section 3.2

• Question 22:

  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?

  A. 25 mW
  B. 250 mW
  C. 500 mW
  D. 1000 mW
  B. 250 mW

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  Explanation/Reference:

  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.

• Question 23:

  What ID is typically mapped to an AP's MAC address if a single BSS is implemented?

  A. SSID
  B. Device ID
  C. VLAN ID
  D. BSSID
  A. SSID
  B. Device ID
  C. VLAN ID
  D. BSSID

• Question 24:

  You are reconfiguring an AP to use the short guard interval. How long will the new guard interval duration be after the change?

  A. 800 ns
  B. 400 ns
  C. 104 ms
  D. 10 ms
  B. 400 ns

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  Explanation/Reference:

  The short guard interval is an optional feature of 802.11n and 802.11ac that reduces the time between OFDM symbols from 800 ns to 400 ns. This can increase the data rate by about 11%, but also requires more precise timing and synchronization between the transmitter and the receiver. The short guard interval is only used when both the AP and the client support it and agree to use it . References: [CWNA-109 Study Guide], Chapter 4: Radio Frequency Signal and Antenna Concepts, page 163; [CWNA-109Study Guide], Chapter 4: Radio Frequency Signal and Antenna Concepts, page 157.

• Question 25:

  Which one of the following channels can be used for VHT transmissions according to the 802.11 specification?

  A. 6
  B. 144
  C. 1
  D. 11
  B. 144

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  Explanation/Reference:

  The channel that can be used for VHT transmissions according to the 802.11 specification is channel 144. VHT stands for Very High Throughput and is the PHY layer specification for 802.11ac devices. VHT transmissions can use channel bandwidths of 20 MHz, 40 MHz, 80 MHz, or 160 MHz in the 5 GHz band. Channel 144 is one of the channels in the 5 GHz band that can support VHT transmissions with any of these bandwidths. Channel 6, channel 1, and channel 11 are channels in the 2.4 GHz band that cannot support VHT transmissions, as they are only compatible with legacy (802.11b/g/n), HT (802.11n), or ERP (802.11g) transmissions with up to 20 MHz bandwidth. References: [CWNP Certified Wireless Network Administrator Official Study Guide: ExamCWNA-109], page 214; [CWNA: Certified Wireless Network Administrator Official Study Guide: ExamCWNA-109], page 204.

• Question 26:

  What 802.11 PHY uses available space in very low frequency ranges that is not in use at the time by broadcast video signals?

  A. DMG
  B. SIG
  C. DSSS
  D. TVHT
  D. TVHT

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  Explanation/Reference:

  TVHT stands for Television Very High Throughput and it is a PHY defined by the 802.11af amendment. It uses the TV white space (TVWS) spectrum in the VHF and UHF bands between 54 and 790 MHz, which are not in use by broadcast video signals at the time. It can provide long-range and low-power connectivity for WLAN devices.

• Question 27:

  When antenna gain is reported in dBi, the gain of the antenna is compared to what theoretical radiator?

  A. End-fire radiator
  B. Dipole radiator
  C. Isotropic radiator D. Anthropomorphic radiator
  C. Isotropic radiator D. Anthropomorphic radiator

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  Explanation/Reference:

  An isotropic radiator is a theoretical point source of electromagnetic radiation that radiates equally in all directions. It has no physical dimensions and no preferred direction of radiation. It is used as a reference for antenna gain because it represents the ideal case of a perfect omnidirectional antenna12 Antenna gain is a measure of how well an antenna concentrates its radiated power in a certain direction. It is expressed in decibels (dB) relative to a reference antenna. Whenthe reference antenna is an isotropic radiator, the antenna gain is denoted by dBi, which stands for decibels relative to isotropic12 For example, an antenna with a gain of 3 dBi means that it radiates 3 dB more power in its main direction than an isotropic radiator would. Conversely, an antenna with a gain of -3 dBi means that it radiates 3 dB less power in its main direction than an isotropic radiator would12

• Question 28:

  A natural disaster has occurred in a remote area that is approximately 57 miles from the response team headquarters. The response team must implement a local wireless network using 802.11 WLAN access points. What is the best method, of those listed, for implementation of a network back-haul for communications across the Internet in this scenario?

  A. 802.11 bridging to the response team headquarters
  B. Cellular/LTE/5G
  C. Turn up the output power of the WLAN at the response team headquarters
  D. Temporary wired DSL
  B. Cellular/LTE/5G

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  Explanation/Reference:

  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

• Question 29:

  You were previously onsite at XYZ's facility to conduct a pre-deployment RF site survey. The WLAN has been deployed according to your recommendations and you are onsite again to perform a post-deployment validation survey.

  When performing this type of post-deployment RF site survey voice over Wi-Fi, what is an action that must be performed?

  A. Spectrum analysis to locate and identify RF interference sources.
  B. Frequency-band hopping analysis to detect improper RF channel implementations.
  C. Application analysis with an active phone call on an VoWiFi handset.
  D. Protocol analysis to discover channel use on neighboring APs.
  C. Application analysis with an active phone call on an VoWiFi handset.

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  Explanation/Reference:

  When performing a post-deployment validation survey for voice over Wi-Fi (VoWiFi), an action that must be performed is Application analysis with an active phone call on a VoWiFi handset. Application analysis is a method of testing the performance of a specific application over the WLAN by measuring parameters such as throughput, latency, jitter, packet loss, MOS score, and R-value. Application analysis with an active phone call on a VoWiFi handset can help to evaluate the quality of service (QoS) and user experience of VoWiFi calls over the WLAN. It can also help to identify any issues or bottlenecks that may affect VoWiFi calls such as interference, roaming delays, or insufficient coverage. References: [CWNP Certified Wireless Network Administrator Official Study Guide: ExamCWNA-109], page 549; [CWNA: Certified Wireless Network Administrator Official Study Guide: ExamCWNA-109], page 519.

• Question 30:

  Return Loss is the decrease of forward energy in a system when some of the power is being reflected back toward the transmitter. What will cause high return loss in an RF transmission system, including the radio, cables, connectors and antenna?

  A. The use of 50 ohm cables longer than one meter in the RF system
  B. High output power at the transmitter and use of a low-gain antenna
  C. A significant impedance mismatch between components in the RF system
  D. A Voltage Standing Wave Ratio (VSWR) of 1:1
  C. A significant impedance mismatch between components in the RF system

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  Explanation/Reference:

  Return loss is a measure of how well the components of an RF system are matched in terms of their impedance. Impedance is the opposition to the flow of alternating current in a circuit, and it depends on the frequency, resistance,

  capacitance, and inductance of the components. When the impedance of the source, the transmission line, and the load are not equal, some of the power is reflected back to the source, causing a loss of forward power. This loss is expressed

  in decibels (dB) as return loss. The higher the return loss, the lower the reflection and the better the impedance matching. Conversely, the lower the return loss, the higher the reflection and the worse the impedance matching. VSWR (Voltage

  Standing Wave Ratio) is another way of expressing the same concept. It is the ratio of the maximum voltage to the minimum voltage along a transmission line due to the interference of the incident and reflected waves. A VSWR of 1:1 means

  that there is no reflection and perfect impedance matching. A VSWR higher than 1:1 means that there is some reflection and impedance mismatch. The higher the VSWR, the higher the reflection and the lower the return loss.

  Therefore, a significant impedance mismatch between components in an RF system will cause high reflection, high VSWR, and low return loss.