NCTI Fiber Install & Activation Exam:
Complete Questions&Answers Study
Guide with Rationales
Section 1: Fiber-Optic Fundamentals & Modulation
1. What is the basis for most digital modulation of light waves?
Answer: OOK (On-Off Keying)
Rationale: OOK is the simplest form of digital modulation, where the
presence of light represents a "1" and the absence represents a "0".
Its simplicity makes it the foundation for most digital fiber-optic
systems.
2. What is PSK (Phase Shift Keying)?
Answer: A modulation technique where the intensity of light waves
remains constant, but the phase shifts 180° when the digital signal
changes from 0 to 1 (and vice versa).
Rationale: By shifting the phase instead of turning the light on and
off, PSK can provide better performance in some optical systems.
3. What is DPSK (Differential Phase Shift Keying)?
, Answer: A form of PSK that encodes digital values as changes
(differences) in signal phase, making it more robust than standard
PSK.
Rationale: DPSK is more tolerant of noise because it looks for phase
changes rather than absolute phase values, making it suitable for
long-haul networks where signals may be weak.
4. What is a drawback of DPSK?
Answer: It requires almost twice the optical bandwidth of NRZ-OOK.
Rationale: The increased bandwidth requirement is a trade-off for
the improved noise immunity DPSK provides.
5. What is DQPSK (Differential Quadrature Phase Shift Keying)?
Answer: A modulation technique using four phase shifts, allowing
each shift to represent 2 bits (00, 01, 10, 11), effectively doubling the
data speed.
Rationale: By encoding two bits per symbol, DQPSK achieves higher
data rates within the same bandwidth.
6. Why is DQPSK less susceptible to dispersion?
Answer: Its narrower optical bandwidth makes it less susceptible to
chromatic and polarization mode dispersion.
Rationale: Narrower optical bandwidth means the light pulses are
less spread out as they travel, maintaining signal integrity.
7. What data rates can DQPSK handle in a 50 GHz channel?
Answer: Up to 40 Gbps.
Rationale: This high data rate allows for the creation of closely
spaced channels, which is essential for DWDM systems.
, 8. Why do most broadband cable networks use analog modulation in
the downstream?
Answer: Because digital modulation of the downstream would
require too much processing and high data rates.
Rationale: Analog modulation is a more efficient and cost-effective
method for broadcasting the large volume of video and data signals
in cable networks.
Section 2: Optical Transmitters, Receivers & Components
9. In which component of a fiber-optic communication system do
changes to the intensity of the optical signal occur?
Answer: The optical transmitter.
Rationale: The transmitter contains the light source (laser or LED)
that converts the electrical signal into an optical signal, modulating its
intensity.
10. What is one drawback of using Fabry-Perot (F-P) lasers in high-
speed data (HSD) networks?
Answer: They emit several discrete wavelengths, known as side
modes.
Rationale: These multiple wavelengths can cause dispersion, which
limits the laser's effectiveness in high-speed, long-distance networks.
11. Why are laser diodes most effective when coupled to single-mode
fiber (SMF)?
Answer: They have a small spectral width, allow for efficient coupling,
and offer fast modulation speeds.
Complete Questions&Answers Study
Guide with Rationales
Section 1: Fiber-Optic Fundamentals & Modulation
1. What is the basis for most digital modulation of light waves?
Answer: OOK (On-Off Keying)
Rationale: OOK is the simplest form of digital modulation, where the
presence of light represents a "1" and the absence represents a "0".
Its simplicity makes it the foundation for most digital fiber-optic
systems.
2. What is PSK (Phase Shift Keying)?
Answer: A modulation technique where the intensity of light waves
remains constant, but the phase shifts 180° when the digital signal
changes from 0 to 1 (and vice versa).
Rationale: By shifting the phase instead of turning the light on and
off, PSK can provide better performance in some optical systems.
3. What is DPSK (Differential Phase Shift Keying)?
, Answer: A form of PSK that encodes digital values as changes
(differences) in signal phase, making it more robust than standard
PSK.
Rationale: DPSK is more tolerant of noise because it looks for phase
changes rather than absolute phase values, making it suitable for
long-haul networks where signals may be weak.
4. What is a drawback of DPSK?
Answer: It requires almost twice the optical bandwidth of NRZ-OOK.
Rationale: The increased bandwidth requirement is a trade-off for
the improved noise immunity DPSK provides.
5. What is DQPSK (Differential Quadrature Phase Shift Keying)?
Answer: A modulation technique using four phase shifts, allowing
each shift to represent 2 bits (00, 01, 10, 11), effectively doubling the
data speed.
Rationale: By encoding two bits per symbol, DQPSK achieves higher
data rates within the same bandwidth.
6. Why is DQPSK less susceptible to dispersion?
Answer: Its narrower optical bandwidth makes it less susceptible to
chromatic and polarization mode dispersion.
Rationale: Narrower optical bandwidth means the light pulses are
less spread out as they travel, maintaining signal integrity.
7. What data rates can DQPSK handle in a 50 GHz channel?
Answer: Up to 40 Gbps.
Rationale: This high data rate allows for the creation of closely
spaced channels, which is essential for DWDM systems.
, 8. Why do most broadband cable networks use analog modulation in
the downstream?
Answer: Because digital modulation of the downstream would
require too much processing and high data rates.
Rationale: Analog modulation is a more efficient and cost-effective
method for broadcasting the large volume of video and data signals
in cable networks.
Section 2: Optical Transmitters, Receivers & Components
9. In which component of a fiber-optic communication system do
changes to the intensity of the optical signal occur?
Answer: The optical transmitter.
Rationale: The transmitter contains the light source (laser or LED)
that converts the electrical signal into an optical signal, modulating its
intensity.
10. What is one drawback of using Fabry-Perot (F-P) lasers in high-
speed data (HSD) networks?
Answer: They emit several discrete wavelengths, known as side
modes.
Rationale: These multiple wavelengths can cause dispersion, which
limits the laser's effectiveness in high-speed, long-distance networks.
11. Why are laser diodes most effective when coupled to single-mode
fiber (SMF)?
Answer: They have a small spectral width, allow for efficient coupling,
and offer fast modulation speeds.
