RCDD Chapter 1 Principles Of Transmission
(96).webp "RCDD Chapter 1 Principles Of Transmission - Quiz")
Questions and Answers
- 1.
WDM uses a series of lenses to refract and direct light pulses into a single optical fiber that carries the combined wavelengths
- A.
True
- B.
False
Correct Answer
A. TrueExplanation
Wavelength Division Multiplexing (WDM) is a technology that allows multiple wavelengths of light to be combined and transmitted over a single optical fiber. It achieves this by using a series of lenses to refract and direct light pulses into the fiber. This process enables multiple signals to be transmitted simultaneously, increasing the capacity and efficiency of the fiber. Therefore, the statement is true.Rate this question:
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- 2.
What are the two types of attenuators
- A.
Fixed
- B.
Random
- C.
Variable
- D.
Adjustable
Correct Answer(s)
A. Fixed
C. VariableExplanation
The question asks for the two types of attenuators. The correct answer is "Fixed" and "Variable". Fixed attenuators have a fixed level of attenuation and cannot be adjusted, while variable attenuators allow for adjustable levels of attenuation.Rate this question:
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- 3.
Loss variations due to temperature changes can sometimes be as high as
- A.
1 dB/km
- B.
2 dB/km
- C.
3 dB/km
- D.
4 dB/km
Correct Answer
B. 2 dB/kmExplanation
Loss variations due to temperature changes can sometimes be as high as 2 dB/km. This means that for every kilometer of distance, the loss can vary by up to 2 decibels due to changes in temperature. This variation is significant and can impact the performance and reliability of the system.Rate this question:
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- 4.
Wavelength areas that are most suitable for optical communications are called:
- A.
Spectrums
- B.
Windows
- C.
Layers
- D.
Stacks
Correct Answer
B. WindowsExplanation
In optical communications, "windows" refer to specific wavelength ranges in the electromagnetic spectrum that are most suitable for transmitting data. These windows are chosen because they have low signal attenuation and minimal interference from other sources. By using these windows, optical signals can travel long distances without significant loss of signal quality. Therefore, "windows" is the correct term to describe the wavelength areas that are most suitable for optical communications.Rate this question:
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- 5.
The outer most part of a fiber is called the cladding
- A.
True
- B.
False
Correct Answer
B. FalseExplanation
The outer most part of a fiber is not called the cladding. The cladding is actually the layer surrounding the core of the fiber, which helps to confine the light within the core and prevent signal loss. The outermost part of a fiber is typically referred to as the coating, which provides protection to the fiber and may have additional functions such as enhancing flexibility or resistance to environmental factors.Rate this question:
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- 6.
Visible light wavelength is:
- A.
200-400 nm
- B.
300-700 nm
- C.
400-700 nm
- D.
500-800 nm
Correct Answer
C. 400-700 nmExplanation
Visible light is a form of electromagnetic radiation that can be detected by the human eye. The wavelength of visible light determines its color. The range of wavelengths that humans can perceive as visible light is approximately 400-700 nm. This range includes colors from violet (shorter wavelength) to red (longer wavelength).Rate this question:
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- 7.
The bandwidth of a multimode system is a function of:
- A.
Chloric dispersion
- B.
Chromatic dispersion
- C.
Medial dispersion
- D.
Modal dispersion
Correct Answer(s)
B. Chromatic dispersion
D. Modal dispersionExplanation
The bandwidth of a multimode system is determined by chromatic dispersion and modal dispersion. Chromatic dispersion occurs when different wavelengths of light travel at different speeds, causing the signal to spread out and degrade over long distances. Modal dispersion, on the other hand, happens when light rays take different paths within a multimode fiber, leading to different arrival times at the receiver and causing signal distortion. Both chromatic dispersion and modal dispersion limit the maximum data rate that can be transmitted over a multimode system.Rate this question:
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- 8.
Bandwidth is the information carrying capacity of a system.
- A.
True
- B.
False
Correct Answer
A. TrueExplanation
Bandwidth refers to the maximum amount of data that can be transmitted over a network or communication channel in a given amount of time. It represents the information carrying capacity of a system, indicating how much data can be transferred at any given moment. Therefore, the statement "Bandwidth is the information carrying capacity of a system" is true.Rate this question:
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- 9.
What is the numerical aperture of 62.5/125 fiber
- A.
0.20
- B.
0.25
- C.
0.275
- D.
.030
Correct Answer
C. 0.275Explanation
The numerical aperture of a fiber optic cable determines its ability to gather and transmit light. In this case, the numerical aperture of the 62.5/125 fiber is 0.275. This means that the fiber optic cable has a relatively high ability to gather and transmit light, making it suitable for applications that require high-speed data transmission over short distances.Rate this question:
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- 10.
What is the center wavelength for a CD laser
- A.
850 nm
- B.
780 nm
- C.
1300 nm
- D.
1310 nm
- E.
1550 nm
Correct Answer
B. 780 nmExplanation
The center wavelength for a CD laser is 780 nm. This is the specific wavelength at which the laser emits light.Rate this question:
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- 11.
What is the numerical aperture of 50/125 fiber:
- A.
0.20
- B.
0.250
- C.
0.275
- D.
0.30
Correct Answer
A. 0.20Explanation
The numerical aperture of a fiber optic cable is a measure of the light-gathering capability of the cable. It is determined by the refractive index of the core and the cladding of the fiber. In this case, the given answer of 0.20 suggests that the numerical aperture of the 50/125 fiber is 0.20.Rate this question:
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- 12.
Mutual conductance G(line) is expressed in
- A.
Siemens (S)
- B.
Ohms
- C.
Henrys (H)
- D.
Farads (F)
Correct Answer
A. Siemens (S)Explanation
Mutual conductance, also known as transconductance, is a measure of how much the output current of a device changes in response to a change in input voltage. It is expressed in Siemens (S), which is the unit of conductance. Ohms is the unit of resistance, Henrys is the unit of inductance, and Farads is the unit of capacitance, none of which are directly related to mutual conductance. Therefore, the correct answer is Siemens (S).Rate this question:
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- 13.
A higher attenuation number is better
- A.
True
- B.
False
Correct Answer
B. FalseExplanation
A higher attenuation number indicates that more signal is being lost or weakened as it travels through a medium. Therefore, a lower attenuation number would be better as it means less signal loss and better transmission of the signal.Rate this question:
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- 14.
Attenuation to Crosstalk ratio (ACR) is calculated how:
- A.
Received noise - received signal
- B.
Minimum NEXT loss - maximum attenuation
- C.
Maximum attenuation - minimum NEXT loss
Correct Answer
B. Minimum NEXT loss - maximum attenuationExplanation
The correct answer is "Minimum NEXT loss - maximum attenuation." This calculation is used to determine the Attenuation to Crosstalk Ratio (ACR). ACR is a measure of the difference between the minimum NEXT (Near End Crosstalk) loss and the maximum attenuation. NEXT loss refers to the amount of crosstalk between adjacent cables, while attenuation measures the loss of signal strength. By subtracting the maximum attenuation from the minimum NEXT loss, we can determine the ACR, which helps assess the quality and performance of a communication channel.Rate this question:
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- 15.
NEXT loss is dominated by components in the near zone, what is defined as the near zone:
- A.
- B.
- C.
- D.
Correct Answer
B.Explanation
The near zone is defined as the region close to the source where the components of the NEXT loss are dominant. In this region, the effects of coupling, crosstalk, and other factors are more significant compared to the far zone. The near zone is characterized by a higher level of electromagnetic field interactions and shorter distances between the components, leading to increased signal degradation. Therefore, understanding and mitigating the effects in the near zone are crucial for minimizing NEXT loss and maintaining signal integrity.Rate this question:
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- 16.
The mutual capacitance C(line) is expresssed in
- A.
Henrys (H)
- B.
Farads (F)
- C.
Ohms
- D.
Siemans (S)
Correct Answer
B. Farads (F)Explanation
The mutual capacitance C(line) is expressed in Farads (F) because capacitance is a measure of the ability of a capacitor to store electric charge. Farads is the unit of capacitance in the International System of Units (SI). It represents the amount of charge that can be stored per volt of potential difference across the capacitor. Therefore, expressing mutual capacitance in Farads is the appropriate unit for measuring the capacitance of a line.Rate this question:
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- 17.
The series resistance R(line) is expressed in:
- A.
Watts
- B.
Ohms
- C.
Amps
- D.
Volts
Correct Answer
B. OhmsExplanation
The series resistance R(line) is expressed in ohms. Ohms is the unit of measurement for resistance in an electrical circuit. It represents the opposition to the flow of electric current. In a series circuit, the total resistance is the sum of the individual resistances, and it is measured in ohms. Watts, amps, and volts are units used to measure power, current, and voltage respectively, but they are not applicable for expressing resistance.Rate this question:
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- 18.
The series inductance L(line) is expressed in:
- A.
Ohms
- B.
Henrys (H)
- C.
Farads (F)
- D.
Siemens (S)
Correct Answer
B. Henrys (H)Explanation
The series inductance, denoted by L(line), is a measure of the ability of a circuit to store energy in the form of a magnetic field. It is expressed in Henrys (H), which is the SI unit for inductance. The unit is named after Joseph Henry, an American scientist who made significant contributions to the study of electromagnetism. Therefore, the correct answer is Henrys (H).Rate this question:
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- 19.
What is the rate for ATM (STS-3)
- A.
12.96 Mb/s
- B.
25.6 Mb/s
- C.
51.8 Mb/s
- D.
155 Mb/s
Correct Answer
D. 155 Mb/sExplanation
The rate for ATM (STS-3) is 155 Mb/s.Rate this question:
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- 20.
DS-1C systems are no longer being deployed.
- A.
True
- B.
False
Correct Answer
A. TrueExplanation
The statement indicates that DS-1C systems are not being used anymore. Therefore, the correct answer is True.Rate this question:
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- 21.
Good internet performance requires a down-to-upstream ratio of at least:
- A.
2:1
- B.
5:1
- C.
10:1
- D.
15:1
Correct Answer
C. 10:1Explanation
A down-to-upstream ratio of at least 10:1 is necessary for good internet performance. This means that the download speed should be at least 10 times faster than the upload speed. This is important because most internet activities, such as browsing websites, streaming videos, and downloading files, require faster download speeds. A higher down-to-upstream ratio ensures that users can access and download content quickly and efficiently.Rate this question:
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- 22.
What is the rate for ATM (STS-1)
- A.
12.96 Mb/s
- B.
25.6 Mb/s
- C.
51.8 Mb/s
- D.
155 Mb/s
Correct Answer
C. 51.8 Mb/sExplanation
The rate for ATM (STS-1) is 51.8 Mb/s.Rate this question:
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- 23.
The rate at which the digital signal state changes and the bit transmission rate are the same
- A.
True
- B.
False
Correct Answer
B. FalseExplanation
The statement is false because the rate at which the digital signal state changes, known as the baud rate, is not necessarily the same as the bit transmission rate. The baud rate refers to the number of signal changes per second, while the bit transmission rate refers to the number of bits transmitted per second. In some cases, multiple bits can be transmitted per signal change, resulting in a higher bit transmission rate than the baud rate. Therefore, the two rates can be different.Rate this question:
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- 24.
When converting digital data to digital signals, what are the two common methods of encoding:
- A.
Manchester
- B.
Winchester
- C.
Alternate mark inversion (AMI)
- D.
Alternate bit inversion (ABI)
Correct Answer(s)
A. Manchester
C. Alternate mark inversion (AMI)Explanation
The two common methods of encoding when converting digital data to digital signals are Manchester and Alternate mark inversion (AMI). In the Manchester encoding method, the signal transitions in the middle of each bit, with a high-to-low transition representing a 0 and a low-to-high transition representing a 1. This ensures that there is always a transition in the middle of each bit, providing synchronization and reducing the chance of errors. In the AMI encoding method, a zero is represented by no line signal, while a one is represented by alternating positive and negative line signals. This method also helps in synchronization and reduces the chance of errors.Rate this question:
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- 25.
The bit is the basic unit of digital data
- A.
True
- B.
False
Correct Answer
A. TrueExplanation
A bit is indeed the basic unit of digital data. It represents a binary value, either 0 or 1, and is used to store and transmit information in a digital format. All digital data, such as text, images, and videos, are ultimately represented and processed as a series of bits. Therefore, the statement "The bit is the basic unit of digital data" is correct.Rate this question:
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- 26.
The process of reconstituting the individual channels from the composite signal is called:
- A.
Duplexing
- B.
Multiplexing
- C.
Demultiplexing
- D.
Companding
Correct Answer
C. DemultiplexingExplanation
Demultiplexing is the process of separating a composite signal into its individual channels. In other words, it is the reverse process of multiplexing, where multiple signals are combined into a single composite signal. Demultiplexing is commonly used in telecommunications and data transmission systems to transmit multiple signals over a single transmission medium. By demultiplexing the composite signal at the receiving end, the individual channels can be extracted and processed separately.Rate this question:
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- 27.
When converting analog signals to digital, if you have a frequency of 4 kHz, what would your sampling rate be:
- A.
200
- B.
2000
- C.
4000
- D.
800
- E.
8000
Correct Answer
E. 8000Explanation
The sampling rate for converting analog signals to digital is typically twice the frequency of the analog signal, according to the Nyquist-Shannon sampling theorem. In this case, the analog signal has a frequency of 4 kHz, so the sampling rate would be 8 kHz, which is equivalent to 8000 samples per second.Rate this question:
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- 28.
When converting analog signals to a digital signal the sampling rate must be at least three times the highest frequency.
- A.
True
- B.
False
Correct Answer
B. FalseExplanation
When converting analog signals to a digital signal, the sampling rate must be at least twice the highest frequency according to the Nyquist-Shannon sampling theorem. This theorem states that in order to accurately reconstruct a continuous signal from its samples, the sampling rate must be at least twice the highest frequency component of the signal. Therefore, the statement that the sampling rate must be at least three times the highest frequency is incorrect.Rate this question:
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- 29.
For power over ethernet, what pins are used to deliver power, select all that apply:
- A.
1
- B.
4
- C.
5
- D.
7
- E.
8
Correct Answer(s)
B. 4
C. 5
D. 7
E. 8 -
- 30.
What is the maximum source power output level for power over ethernet
- A.
10.5 watts at nominal 24 volts
- B.
11.4 watts at nominal 24 volts
- C.
12.8 watts at nominal 48 volts
- D.
15.4 watts at nominal 48 volts
Correct Answer
D. 15.4 watts at nominal 48 voltsExplanation
The maximum source power output level for power over ethernet is 15.4 watts at nominal 48 volts. This means that the power over ethernet system can provide a maximum of 15.4 watts of power when the voltage is at 48 volts.Rate this question:
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- 31.
The common distance between H loading coils is:
- A.
1.37 km (4500 ft)
- B.
1.5 km (4900 ft)
- C.
1.7 km (5500 ft)
- D.
1.8 km (5900 ft)
Correct Answer
D. 1.8 km (5900 ft)Explanation
The common distance between H loading coils is 1.8 km (5900 ft). This means that the distance between each H loading coil is 1.8 km or 5900 ft.Rate this question:
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- 32.
The common distance between D loading points is:
- A.
.75 km (2400 ft)
- B.
1.1 km (3500 ft)
- C.
1.37 km (4500 ft)
- D.
1.8 km (5900 ft)
Correct Answer
C. 1.37 km (4500 ft)Explanation
The common distance between D loading points is 1.37 km (4500 ft). This means that each loading point is located at a distance of 1.37 km from the other loading points.Rate this question:
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- 33.
What are the two types of loading coils
- A.
D loading
- B.
B loading
- C.
F loading
- D.
H loading
Correct Answer(s)
A. D loading
D. H loadingExplanation
Loading coils are used in telecommunication systems to compensate for the attenuation of signals over long distances. They are placed at regular intervals along the transmission line to boost the signal strength. The two types of loading coils mentioned in the answer, D loading and H loading, refer to different configurations of the coils. Each type is designed to provide specific impedance matching and signal enhancement properties based on the characteristics of the transmission line and the desired signal quality.Rate this question:
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- 34.
In telephony, delays greater than _____ are perceptible if they are of sufficient strenth.
- A.
25 milliseconds
- B.
40 milliseconds
- C.
50 milliseconds
- D.
75 milliseconds
Correct Answer
C. 50 millisecondsExplanation
Delays greater than 50 milliseconds are perceptible in telephony if they are of sufficient strength. This means that if there is a delay of more than 50 milliseconds in a telephone conversation, it will be noticeable to the person speaking or listening. The strength of the delay refers to how noticeable or significant it is.Rate this question:
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- 35.
What speed do signals travel in a cable
- A.
.2c to .5c
- B.
.4c to .6c
- C.
.6c to .8c
- D.
.7c to .9c
Correct Answer
C. .6c to .8cExplanation
Signals travel at speeds ranging from .6c to .8c in a cable. This means that the speed of the signals is between 60% and 80% of the speed of light.Rate this question:
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- 36.
If the signal power has a power ratio of 100, how many dB increase will you have.
- A.
+20 dB
- B.
+40 dB
- C.
+70 dB
- D.
+100 dB
Correct Answer
A. +20 dBExplanation
A power ratio of 100 corresponds to a 20 dB increase. This is because the decibel (dB) scale is logarithmic, and a power ratio of 100 can be expressed as 10 times the logarithm (base 10) of 100, which equals 20 dB.Rate this question:
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- 37.
A doubling of a signals power will result in an increase of how many dB.
- A.
+2 dB
- B.
+3 dB
- C.
+4 dB
- D.
+5 dB
Correct Answer
B. +3 dBExplanation
When the power of a signal is doubled, it results in an increase of 3 dB. Decibels (dB) is a logarithmic unit used to measure the ratio of two power levels. In this case, a doubling of power corresponds to a ratio of 2:1, which is equivalent to a 3 dB increase. This is because the logarithm of 2 is approximately 0.301, and when multiplied by 10, it gives a value of approximately 3 dB. Therefore, the correct answer is +3 dB.Rate this question:
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- 38.
What is the frequency range for UHF:
- A.
3-30 MHz
- B.
30-300 MHz
- C.
300-3000 MHz
- D.
30-300 kHz
Correct Answer
C. 300-3000 MHzExplanation
UHF stands for Ultra High Frequency, and it refers to the frequency range between 300 and 3000 MHz. This range is commonly used for various communication purposes, including television broadcasting, mobile phones, Wi-Fi, and satellite communication. Frequencies within this range are capable of transmitting signals over longer distances and have better penetration through obstacles compared to higher frequencies.Rate this question:
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- 39.
What is the frequency range for LF audio:
- A.
30-300 kHz
- B.
3-30 MHz
- C.
3-30 kHz
- D.
30-300 MHz
Correct Answer
A. 30-300 kHzExplanation
LF audio refers to low frequency audio. The frequency range for LF audio is typically between 30-300 kHz. This range is considered to be in the lower end of the audio spectrum. Frequencies within this range are often used for various purposes such as audio transmission, communication systems, and scientific research.Rate this question:
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- 40.
Phase is a description of the reference:
- A.
Time
- B.
Voltage
- C.
Current
Correct Answer
A. TimeExplanation
The given options, "Voltage" and "Current," are both measurements of a quantity, whereas "Time" is a description of the reference. In the context of phase, time refers to the position of a waveform in relation to a reference point. It indicates the point in time at which a waveform starts or reaches a certain point. Therefore, "Time" is the correct answer as it aligns with the concept of phase being a description of the reference in terms of time.Rate this question:
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- 41.
What is the frequency range for VLF audio:
- A.
30-300 kHz
- B.
30-300 MHz
- C.
3-30 kHz
- D.
300-3000 MHz
Correct Answer
C. 3-30 kHzExplanation
VLF stands for Very Low Frequency, and it refers to the range of electromagnetic waves with frequencies between 3 and 30 kHz. This frequency range is commonly used for communication with submarines and for various scientific research purposes. It is important to note that VLF audio refers specifically to the audio signals within this frequency range that can be heard by humans.Rate this question:
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- 42.
The effectiveness of single-layer and multiple-layer braids against magnetic fields is:
- A.
Poor
- B.
Fair
- C.
Good
- D.
Excellent
Correct Answer
A. PoorExplanation
Single-layer and multiple-layer braids are not effective against magnetic fields. They provide poor protection against magnetic interference.Rate this question:
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- 43.
Multiple-layer braid shield effectiveness for radio frequency is:
- A.
Poor
- B.
Fair
- C.
Good
- D.
Excellent
Correct Answer
D. ExcellentExplanation
The multiple-layer braid shield is highly effective for radio frequency because it provides multiple layers of protection against electromagnetic interference. This shield is designed to minimize the amount of external radio frequency signals that can penetrate the cable or wire, ensuring excellent signal quality and reducing the risk of interference. The multiple layers of the braid shield create a strong barrier, making it highly effective in blocking out unwanted radio frequency signals.Rate this question:
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- 44.
Foils shield effectiveness for audio frequencies is:
- A.
Poor
- B.
Fair
- C.
Good
- D.
Excellent
Correct Answer
D. ExcellentExplanation
The given answer "Excellent" suggests that foils are highly effective in shielding audio frequencies. This means that foils are able to block or reduce the interference or electromagnetic radiation that can affect audio signals. Foils are likely to provide a high level of protection against external electromagnetic interference, resulting in a clear and undistorted audio signal.Rate this question:
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- 45.
The maximum cable attenuation coefficient singlemode inside plant fiber is:
- A.
3.5 dB/km @ 1550 nm
- B.
0.5 dB/km @ 1300 nm
- C.
1.0 dB/km @ 1310 nm
- D.
1.0 dB/km @ 1550 nm
- E.
1.5 dB/km @1300 nm
Correct Answer(s)
C. 1.0 dB/km @ 1310 nm
D. 1.0 dB/km @ 1550 nmExplanation
The maximum cable attenuation coefficient for singlemode inside plant fiber is 1.0 dB/km at both 1310 nm and 1550 nm. This means that for every kilometer of fiber, the signal strength decreases by 1.0 dB at these wavelengths. Attenuation is an important factor to consider in fiber optic communication as it affects the quality and distance of signal transmission.Rate this question:
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- 46.
The maximum cable attenuation coefficient singlemode outside plant fiber is:
- A.
3.5 dB/km @ 1550 nm
- B.
0.5 dB/km @ 1310 nm
- C.
0.5 dB/km @1300 nm
- D.
1.0 dB/km @ 1550 nm
- E.
0.5 dB/km @ 1550 nm
Correct Answer(s)
B. 0.5 dB/km @ 1310 nm
E. 0.5 dB/km @ 1550 nmExplanation
The correct answer is 0.5 dB/km @ 1310 nm, 0.5 dB/km @ 1550 nm. This means that the maximum cable attenuation coefficient for singlemode outside plant fiber is 0.5 dB/km at both 1310 nm and 1550 nm wavelengths. Attenuation coefficient refers to the amount of signal loss that occurs as light travels through the fiber optic cable. In this case, the cable has a very low attenuation coefficient, indicating that it is capable of transmitting signals over long distances without significant loss of signal strength.Rate this question:
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- 47.
Two key parameters in optical fiber cabling performance that must be verified for compatibility with the proposed electronics are:
- A.
Spectral width
- B.
Bandwidth
- C.
Numerical apperature
- D.
Attenuation
Correct Answer(s)
B. Bandwidth
D. AttenuationExplanation
Bandwidth and attenuation are two key parameters in optical fiber cabling performance that need to be verified for compatibility with the proposed electronics. Bandwidth refers to the capacity of the fiber to carry data, and it is important to ensure that it meets the requirements of the proposed electronics. Attenuation, on the other hand, refers to the loss of signal strength as it travels through the fiber. It is crucial to verify that the attenuation is within acceptable limits to ensure proper transmission of data.Rate this question:
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- 48.
EM fields are usually expressed in:
- A.
Ohm/meters
- B.
Capacitive/meters
- C.
Volt/meters
- D.
Ohm/foot
Correct Answer
C. Volt/metersExplanation
EM fields are usually expressed in Volt/meters because the electric field strength (E) is measured in volts per meter. This unit represents the amount of electric potential difference (voltage) per unit distance. It is commonly used to quantify the strength of electric fields in various applications, such as telecommunications, electromagnetic radiation, and electrical engineering.Rate this question:
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- 49.
When selecting cable shields, the overall cable size limitations also may affect the decision.
- A.
True
- B.
False
Correct Answer
A. TrueExplanation
The statement suggests that when choosing cable shields, the decision may be influenced by the size limitations of the overall cable. This implies that the size of the cable could impact the selection of suitable cable shields. Therefore, the statement is true.Rate this question:
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- 50.
Conductive nonmetallic materials are sometimes used at power and some low audio frequencies. These semiconductive shields are not normally used for applications at frequencies above:
- A.
250 kHz
- B.
300 kHz
- C.
400 kHz
- D.
500 kHz
Correct Answer
D. 500 kHzExplanation
Conductive nonmetallic materials are sometimes used as semiconductive shields at power and low audio frequencies. However, they are not typically used for applications at frequencies above 500 kHz. This suggests that conductive nonmetallic materials may not be effective in providing shielding at higher frequencies, potentially due to their limited conductivity or other properties.Rate this question:
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