Free CWNA-109 Exam Braindumps (page: 16)

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As an RF wave propagates through space, the wave front experiences natural expansion that reduces its signal strength in an are

  1. What describes the rate at which this expansion happens?
  2. Fresnel zone thinning
  3. Ohm's law
  4. Inverse square law
  5. MU-MIMO

Answer(s): C

Explanation:

The inverse square law states that the signal strength of an RF wave is inversely proportional to the square of the distance from the source. This means that as the distance from the transmitter increases, the signal strength decreases rapidly.


Reference:

Wireless Network Administrator Official Study Guide, Chapter 3, page 64.



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?

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

Answer(s): C

Explanation:

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.



Which unit of measurement, as formally defined, is an absolute unit that is used to quantify received signal power levels on a logarithmic scale?

  1. SNI
  2. VSWR
  3. dBm
  4. dBi

Answer(s): C

Explanation:

The unit of measurement that is an absolute unit and is used to quantify received signal power levels on a logarithmic scale is dBm. dBm stands for decibel-milliwatt and represents the power level relative to 1 milliwatt (mW). dBm is an absolute unit because it has a fixed reference point and does not depend on the input power level. dBm is used to measure the received signal power levels on a logarithmic scale because it can express large variations in power levels with small numbers and make calculations easier. For example, a 10 dB increase in power level means a 10-fold increase in power, and a 20 dB increase means a 100-fold increase in power.


Reference:

[CWNP Certified Wireless Network Administrator Official Study Guide: Exam CWNA-109], page 66; [CWNA: Certified

Wireless Network Administrator Official Study Guide: Exam CWNA-109], page 56.



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?

  1. 25 mW
  2. 250 mW
  3. 500 mW
  4. 1000 mW

Answer(s): B

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 antenn a. 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


Reference:

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



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Quentin commented on November 14, 2024
I noticed that some comments were related to answers not being 100% correct. But for me as long as questions are real and same as the actual exam I was okay.
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