Category: Optical Communications Objective Questions

Optical Communications Multiple Choice Questions on “Modulation Formats”.

1. _____________ is essentially a crude form of Amplitude shift keying.
a) Analog modulation
b) Digital intensity modulation
c) Photodetector
d) Receiver structure
Answer: b
Explanation: Many techniques have been developed to amplitude modulate an optical signal. Digital intensity modulation used in direct detection systems is essentially a crude form of ASK in which the received signal is detected using square law detector.

2. Almost _________ of the transmitter power is wasted in the use of external modulators.
a) Half
b) Quarter
c) One-third
d) Twice
Answer: a
Explanation: All external modulators suffer the drawback that around half of the transmitted power is wasted. To avoid this, non-synchronous detection can be employed.

3. The line width in the range ________ of bit rate is specified for ASK heterodyne detection.
a) 8%
b) 2 to 8%
c) 10 t0 50%
d) 70%
Answer: c
Explanation: The ASK modulation scheme can be used with laser sources exhibiting the line widths comparable with the bit transmission rate. For ASK heterodyne detection, line width range of 10 to 50% is usually specified.

4. ______________ is also referred to as on-off keying (OOK).
a) FSK
b) DSK
c) PSK
d) ASK
Answer: d
Explanation: Amplitude shift keying (ASK) involves the locking and assembling of the amplitude of the wave. It involves the carrier wave along with the amplitude wave or transmitted wave and hence referred to as on-off keying.

5. ________ does not require an external modulator.
a) FSK
b) DSK
c) PSK
d) ASK
Answer: a
Explanation: FSK involves the frequency deviation property of the directly modulated semiconductor laser used in wideband systems. Unlike ASK, it does not require an external modulator, which in turn, avoids the wastage of transmitted power.

6. The frequency deviation at frequencies above 1 MHz is typically ____________
a) 10 to 20 mA^-1
b) 100 to 500 mA^-1
c) 1000 to 2000 mA^-1
d) 30 to 40 mA^-1
Answer: b
Explanation: The carrier modulation effect occurs at the frequencies above 1 MHz. At the phase of carrier modulation, the frequency deviation is about 100 to 500 mA^-1.

7. ___________ offers the potential for improving the coherent optical receiver sensitivity by increasing the choice of signalling frequencies.
a) MFSK
b) MDSK
c) MPSK
d) MASK
Answer: a
Explanation: Multilevel FSK includes 4-level or 8-level FSK. It improves the receiver sensitivity by reducing the deviation and increasing the usage of signalling frequencies.

8. Eight level FSK and binary PSK yields an equivalent sensitivity.
a) False
b) True
Answer: b
Explanation: Binary PSK and 8-level FSK provides an equivalent sensitivity. The main drawback of 8-level FSK is that it yields an equivalent sensitivity to binary PSK at the expense of a greater receiver bandwidth requirement.

9. External modulation for ________ modulation format allows the most sensitive coherent detection mechanism.
a) FSK
b) DSK
c) PSK
d) ASK
Answer: c
Explanation: External modulation for PSK is usually straightforward. It is therefore utilized to provide the modulation format which allows the most sensitive coherent detection mechanism.

10. _________ can potentially provide spectral conservation through the use of multilevel signalling.
a) M-ary PSK
b) MFSK
c) ASK
d) DFSK
Answer: a
Explanation: In M-ary schemes, the spectral efficiency is increased by the factor log₂ M.this is purely for M-level schemes which can provide multilevel signalling patterns.

11. The digital transmission on implementation of polarization modulation which involves polarization characteristics of the transmitted optical signal is known as _____________
a) Frequency shift keying
b) Amplitude shift keying
c) Phase shift keying
d) Polarization shift keying
Answer: d
Explanation: Polarization shift keying is abbreviated as PolSK. PolSK requires additional receiver complexity than other modulation formats.

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Optical Communications Multiple Choice Questions on “Optical Amplifiers – Semiconductor Optical Amplifiers”.

1. For linear as well as in nonlinear mode _______________ are most important network elements.
a) Optical amplifier
b) Optical detector
c) A/D converter
d) D/A converters
Answer: a
Explanation: In single-mode fiber system, signal dispersion is very small, hence there is attenuation. These systems don’t require signal regeneration as optical amplification is sufficient so optical amplifier are most important.

2. The more advantages optical amplifier is ____________
a) Fiber amplifier
b) Semiconductor amplifier
c) Repeaters
d) Mode hooping amplifier
Answer: b
Explanation: Semiconductor optical amplifiers are having smaller size. They can be integrated to produce subsystems. Thus are more profitable than other optical amplifier.

3. ________________ cannot be used for wideband amplification.
a) Semiconductor optical amplifier
b) Erbium-doped fiber amplifier
c) Raman fiber amplifier
d) Brillouin fiber amplifier
Answer: d
Explanation: Brillouin fiber amplifiers provide a very narrow spectral bandwidth. These bandwidth can be around 50 MHz, hence cannot be employed for wideband amplification.

4. ____________ is used preferably for channel selection in a WDM system.
a) Semiconductor optical amplifier
b) Erbium-doped fiber amplifier
c) Raman fiber amplifier
d) Brillouin fiber amplifier
Answer: d
Explanation: Brillouin fiber provides amplification of a particular channel. This amplification can be done without boosting other channels besides that particular channel.

5. For used in single-mode fiber __________ are used preferably.
a) Semiconductor optical amplifier
b) Erbium-doped fiber amplifier
c) Raman fiber amplifier
d) Brillouin fiber amplifier
Answer: a
Explanation: Semiconductor optical amplifiers have low power consumption. There single mode structure makes them appropriate and suitable for used in single mode fiber.

6. Mostly ____________ are used in nonlinear applications.
a) Semiconductor optical amplifier
b) Erbium-doped fiber amplifier
c) Raman fiber amplifier
d) FPAs
Answer: d
Explanation: FPAs have a resonant nature. This can be combined with their high internal fields. They provide pulse shaping and bi-stable elements. Thus, are used widely in nonlinear application.

7. _______________ is superior as compared to _________________
a) TWA, FPA
b) FPA, TWA
c) EDFA, FPA
d) FPA, EDFA
Answer: a
Explanation: In TWA operating in single-pass amplification mode, the Fabry-Perot resonance is suppressed by facet reflectivity reduction. This affects in increasing of amplifier spectral bandwidth. This makes them less dependence of transmission characteristics on fluctuations in biased current, input signal polarization. Thus FPA are superior to TWA.

8. ______________ are operated at current beyond normal lasing threshold current, practically.
a) Semiconductor optical amplifier
b) Erbium-doped fiber amplifier
c) Raman fiber amplifier
d) Brillouin fiber amplifier
Answer: a
Explanation: The anti-reflection facet coatings affects in the form of increasing lasing current threshold. This causes SOAs to be operated at current beyond normal lasing threshold current.

9. An uncoated FPA has peak gain wavelength 1.8μm, mode spacing of 0.8nm, and long active region of 300 v. Determine RI of active medium.
a) 4.25×10⁶
b) 3.75×10⁷
c) 3.95×10⁷
d) 4.25×10⁹
Answer: b
Explanation: n=λ²/2δλL=1.8×10^-6/2×0.8×10^-9×300×10^-6=3.75×10⁷.

10. Determine the peak gain wavelength of uncoated FPA having mode spacing of 2nm,and 250μmlong active region and R.I of 3.78.
a)2.25×10^-4
b)4.53×10^-8
c)1.94×10^-6
d)4.25×10⁹
Answer: c
Explanation: The peak gain wavelength is given by
λ²=n2δλL=3.78×2×2×10^-9×250×10^-6=1.94×10^-6m.

11. An SOA has net gain coefficient of 300, at a gain of 30dB. Determine length of SOA.
a) 0.32 m
b) 0.023 m
c) 0.245 m
d) 0.563 m
Answer: b
Explanation: The length of SOA is determined by
L = G_s(dB)/10×g×log_e = 30/10×300×0.434`= 0.023 m.

12. An SOA has length of 35.43×10^-3m, at 30 dB gain. Determine net gain coefficient.
a) 5.124×10^-3
b) 1.12×10^-4
c) 5.125×10^-3
d) 2.15×10^-5
Answer: c
Explanation: The net gain coefficient of SOA is given by
g = L×10×log_e/G_s(dB) = 35.43×10^-3×10×0.434/30
=5.125×10^-3.

13. An SOA has mode number of 2.6, spontaneous emission factor of 4, optical bandwidth of 1 THz. Determine noise power spectral density.
a) 1.33×10^-3
b) 5.13×10¹²
c) 3.29×10^-6
d) 0.33×10^-9
Answer: a
Explanation: The noise power spectral density P_ast is
P_ast = mn_sp(G_s-1) hfb
= 2.6×4(1000-1)×6.63×10^-34×1.94×10¹⁴×1×10¹²
= 1.33×10^-3W.

14. An SOA has noise power spectral density of 1.18mW, spontaneous emission factor of 4, optical bandwidth of 1.5 THz. Determine mode number.
a) 1.53 × 10²⁸
b) 6.14 × 10¹²
c) 1.78 × 10¹⁶
d) 4.12 × 10^-3
Answer: a
Explanation: The mode number is determined by
m = P_ast/n_sp(Gs-1) hfB
= 1.18×10^-3/4(1000-1)×6.63×10^-34×1.94×10¹⁴×1.3×10¹²
= 1.53 × 10^-34.

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Optical Communications Multiple Choice Questions on “LED Power and Efficiency”.

1. The absence of _______________ in LEDs limits the internal quantum efficiency.
a) Proper semiconductor
b) Adequate power supply
c) Optical amplification through stimulated emission
d) Optical amplification through spontaneous emission
Answer: c
Explanation: The ratio of generated electrons to the electrons injected is quantum efficiency. It is greatly affected if there is no optical amplification through stimulated emission. Spontaneous emission allows ron-radiative recombination in the structure due to crystalline imperfections and impurities.

2. The excess density of electrons Δnand holes Δpin an LED is ____________
a) Equal
b) Δpmore than Δn
c) Δn more than Δp
d) Does not affects the LED
Answer: a
Explanation: The excess density of electrons ΔnandΔp (holes) is equal. The charge neutrality is maintained within the structure due to injected carriers that are created and recombined in pairs. The power generated internally by an LED is determined by taking into considering the excess electrons and holes in p- and n-type material respectively.

3. The hole concentration in extrinsic materials is _________ electron concentration.
a) much greater than
b) lesser than
c) equal to
d) negligible difference with
Answer: a
Explanation: In extrinsic materials, one carrier type will be highly concentrated than the other type. Hence in p-type region, hole concentration is greater than electron concentration in context of extrinsic material. This excess minority carrier density decays with time.

4. The carrier recombination lifetime becomes majority or injected carrier lifetime.
a) True
b) False
Answer: b
Explanation: The initial injected excess electron density and τrepresents the total carrier recombination time. In most cases, Δnis a small fraction of majority carriers and contains all minority carriers. So in these cases, carrier recombination lifetime becomes minority injected carrier lifetime τ_i.

5. In a junction diode, an equilibrium condition occurs when ____________
a) Δngreater than Δp
b) Δnsmaller than Δp
c) Constant current flow
d) Optical amplification through stimulated emission
Answer: c
Explanation: The total rate at which carriers are generated in sum of externally supplied and thermal generation rates. When there is a constant current flow in this case, an equilibrium occurs in junction diode.

6. Determine the total carrier recombination lifetime of a double heterojunction LED where the radioactive and nonradioactive recombination lifetime of minority carriers in active region are 70 ns and 100 ns respectively.
a) 41.17 ns
b) 35 ns
c) 40 ns
d) 37.5 ns
Answer: a
Explanation: The total carrier recombination lifetime is given by
τ = τ_rτ_nr/τ_r+τ_nr = 70× 100/70 + 100 ns = 41.17 ns
Where
τ_r = radiative recombination lifetime of minority carriers
τ_nr = nonradioactive recombination lifetime of minority carriers.

7. Determine the internal quantum efficiency generated within a device when it has a radiative recombination lifetime of 80 ns and total carrier recombination lifetime of 40 ns.
a) 20 %
b) 80 %
c) 30 %
d) 40 %
Answer: b
Explanation: The internal quantum efficiency of device is given by
η_int = τ/τ_r = 40/80 ×100 = 80%
Where
τ = total carrier recombination lifetime
τ_r = radiative recombination lifetime.

8. Compute power internally generated within a double-heterojunction LED if it has internal quantum efficiency of 64.5 % and drive current of 40 mA with a peak emission wavelength of 0.82 μm.
a) 0.09
b) 0.039
c) 0.04
d) 0.06
Answer: b
Explanation: The power internally generated within device i.e. double-heterojunction LED can be computed by
P_int = η_int hci/eλ = 0.645×6.626×10^-34×3×10⁸×40×10^-3/ 1.602×10^-19 × 0.82 × 10^-6
= 0.039 W
Where
η_int = internal quantum efficiency
h = Planck’s constant
c = velocity of light
i = drive current
e = electron charge
λ = wavelength.

9. The Lambertian intensity distribution __________ the external power efficiency by some percent.
a) Reduces
b) Does not affects
c) Increases
d) Have a negligible effect
Answer: a
Explanation: In Lambertian intensity distribution, the maximum intensity I₀is perpendicular to the planar surface but is reduced on the sides in proportion to the cosine of θ i.e. viewing angle as apparent area varies with this angle. This reduces the external power efficiency. This is because most of the light is tapped by total internal refraction when radiated at greater than the critical angle for crystal air interface.

10. A planar LED fabricated from GaAs has a refractive index of 2.5. Compute the optical power emitted when transmission factor is 0.68.
a) 3.4 %
b) 1.23 %
c) 2.72 %
d) 3.62 %
Answer: c
Explanation: The optical power emitted is given by
P_e = P_intF_n²/4n_x² = P_int (0.680×1/4×(2.5)²) = 0.0272 P_int.
Hence power emitted is only 2.72 % of optional power emitted internally.
Where,
F_n² = transmission factor
nx = refractive index.

11. A planar LED is fabricated from GaAs is having a optical power emitted is 0.018% of optical power generated internally which is 0.018% of optical power generated internally which is 0.6 P. Determine external power efficiency.
a) 0.18%
b) 0.32%
c) 0.65%
d) 0.9%
Answer: d
Explanation: Optical power generated externally is given by
η_cp = (0.018P_int/2P_int)*100
Where,
P_int = power emitted
η_cp = external power efficiency.

12. For a GaAs LED, the coupling efficiency is 0.05. Compute the optical loss in decibels.
a) 12.3 dB
b) 14 dB
c) 13.01 dB
d) 14.6 dB
Answer: c
Explanation: The optical loss in decibels is given by-
Loss = -10log₁₀ η_c
Where,
η_c = coupling efficiency.

13. In a GaAs LED, compute the loss relative to internally generated optical power in the fiber when there is small air gap between LED and fiber core. (Fiber coupled = 5.5 * 10^-4P_int)
a) 34 dB
b) 32.59 dB
c) 42 dB
d) 33.1 dB
Answer: b
Explanation: The loss in decibels relative to P_int is given by-
Loss = -10log₁₀P_c/P_int
Where,
P_c = 5.5 * 10^-4P_int.

14. Determine coupling efficiency into the fiber when GaAs LED is in close proximity to fiber core having numerical aperture of 0.3.
a) 0.9
b) 0.3
c) 0.6
d) 0.12
Answer: a
Explanation: The coupling efficiency is given by
η_c = (NA)² = (0.3)² = 0.9.

15. If a particular optical power is coupled from an incoherent LED into a low-NA fiber, the device must exhibit very high radiance.
a) True
b) False
Answer: a
Explanation: Device must have very high radiance specially in graded index fiber where Lambertian coupling efficiency with same NA is about half that of step-index fibers. This high radiance is obtained when direct bandgap semiconductors are fabricated with DH structure driven at high current densities.

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Optical Communications Multiple Choice Questions on “Cable Design”.

1. The cable must be designed such that the strain on the fiber in the cable does not exceed __________
a) 0.002%
b) 0.01%
c) 0.2%
d) 0.160%
Answer: c
Explanation: The constraints included in cable design are stability, protection, strength and jointing of the fibers. The fiber cable does not get affected if the strain exerted on it is below 0.2%. Although, it is suggested that the permanent strain on the fiber should be less than 0.1%.

2. How many categories exists in case of cable design?
a) Two
b) Three
c) One
d) Four
Answer: b
Explanation: Cable design is separated into three categories. They are fiber buffering, cable structural and strength and cable sheath and water barrier. After successfully going through these tests, an optical cable is designed.

3. How many types of buffer jackets are used in fiber buffering?
a) Three
b) One
c) Two
d) Four
Answer: a
Explanation: The buffer jacket is designed to protect the fiber from micro-bending losses. There are three types of buffer jackets used in fiber buffering. They are tight buffer jackets, loose tube buffer jackets and filled loose tube buffer jacket.

4. Loose tube buffer jackets exhibits a low resistance to movement of the fiber.
a) True
b) False
Answer: a
Explanation: Loose tube buffering is achieved by using a hard, smooth, flexible material in the form of extruded tube. The buffer tube is smooth from inside. Thus, it exhibits a low resistance to movement of the fiber. Also, it can be easily stripped for jointing or fiber termination.

5. An inclusion of one or more structural members in an optical fiber so as to serve as a cable core foundation around which the buffer fibers may be wrapped is called _____________
a) Attenuation
b) Splicing
c) Buffering
d) Stranding
Answer: d
Explanation: Optical fiber is made structurally stronger by adding one or more strength members. The core fiber is trapped with buffered fibers or they are slotted in the core foundation. This approach is called as stranding.

6. Which of the following is not a strength member used in optical cable?
a) Steel wire
b) Germanium
c) Aramid yarns
d) Glass elements
Answer: b
Explanation: Strength members or tensile members are added to the fiber to make it stronger and durable. These members include solid steel wire, dielectric aramid yarns (Kevlar), glass elements etc. Germanium is not a structural or strength member.

7. When the stranding approach consists of individual elements (e.g. single-fiber or multi fiber loose tube buffer) than the cable is termed as _____________
a) Optical unit cable
b) Coaxial cable
c) Layer cable
d) Bare glass cable
Answer: c
Explanation: The stranding approach consists of a fiber core foundation around which the buffered fibers are wrapped. The cable elements are stranded in one, two or several layers around the central structural member. When the stranding is composed of individual elements, then the cable is termed as layer cable. If the cable core consists of stranding elements each of which comprises a unit of stranding elements, then it is termed as optical unit cable.

8. The primary function of the structural member is load bearing.
a) True
b) False
Answer: b
Explanation: The primary function of the structural member is not load bearing. It’s function is to provide suitable accommodation for the fiber ribbons within the cables. These fiber ribbons lie in the helical grooves or slots formed in the surface of the structural members.

9. What is the Young’s modulus of Kevlar, an aromatic polyester?
a) 9 ×10¹⁰Nm^-2
b) 10 ×10¹⁰Nm^-2
c) 12 ×10¹⁰Nm^-2
d) 13 ×10¹⁰Nm^-2
Answer: d
Explanation: Kevlar is used as a strength member in an optical fiber. The Young’s modulus of Kevlar is very high which gives it strength to weight ratio advantage four times that of steel. Kevlar is coated with extruded plastic to provide a smooth surface which in turn prevents micro-bending losses.

10. The cable is normally covered with an outer plastic sheath to reduce _______________
a) Abrasion
b) Armor
c) Friction
d) Dispersion
Answer: a
Explanation: Abrasion is the process of scraping or wearing something away. If the cable is not coated with plastic sheath, it gives rise to effects such as abrasion and crushing. The most common plastic sheath material used in covering a cable is polyethylene (PE).

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Optical Communications Interview Questions and Answers on “Electromagnetic Mode Theory for Optical Propagation”.

1. Which equations are best suited for the study of electromagnetic wave propagation?
a) Maxwell’s equations
b) Allen-Cahn equations
c) Avrami equations
d) Boltzmann’s equations
Answer: a
Explanation: Electromagnetic mode theory finds its basis in electromagnetic waves. Electromagnetic waves are always represented in terms of electric field E, magnetic field H, electric flux density D and magnetic flux density B. These set of equations are provided by Maxwell’s equations.

2. When λ is the optical wavelength in vacuum, k is given by k=2Π/λ. What does k stand for in the above equation?
a) Phase propagation constant
b) Dielectric constant
c) Boltzmann’s constant
d) Free-space constant
Answer: a
Explanation: In the above equation, k = 2Π/λ, also termed as wave equation, k gives us the direction of propagation and also the rate of change of phase with distance. Hence it is termed as phase propagation constant.

3. Constructive interference occur when total phase change after two successive reflections at upper and lower interfaces is equal to? (Where m is integer)
a) 2Πm
b) Πm
c) Πm/4
d) Πm/6
Answer: a
Explanation: The component of phase waves which is in x direction is reflected at the interference between the higher and lower refractive index media. It is assumed that such an interference forms a lowest order standing wave, where electric field is maximum at the center of the guide, decaying towards zero.

4. When light is described as an electromagnetic wave, it consists of a periodically varying electric E and magnetic field H which are oriented at an angle?
a) 90 degree to each other
b) Less than 90 degree
c) Greater than 90 degree
d) 180 degree apart
Answer: a
Explanation: In case of electromagnetic wave which occur only in presence of both electric and magnetic field, a particular change in magnetic field will result in a proportional change in electric field and vice versa. These changes result in formation of electromagnetic waves and for electromagnetic waves to occur both fields should be perpendicular to each other in direction of wave travelling.

5. A monochromatic wave propagates along a waveguide in z direction. These points of constant phase travel in constant phase travel at a phase velocity V_p is given by?
a) V_p=ω/β
b) V_p=ω/c
c) V_p=C/N
d) V_p=mass/acceleration
Answer: a
Explanation: Velocity is a function of displacement. Phase velocity V_p is a measure of angular velocity.

6. Which is the most important velocity in the study of transmission characteristics of optical fiber?
a) Phase velocity
b) Group velocity
c) Normalized velocity
d) Average velocity
Answer: b
Explanation: Group velocity is much important in relation to transmission characteristics of optical fiber. This is because the optical wave propagates in groups or form of packets of light.

7. What is refraction?
a) Bending of light waves
b) Reflection of light waves
c) Diffusion of light waves
d) Refraction of light waves
Answer: a
Explanation: Unlike reflection, refraction involves penetration of a light wave from one medium to another. While penetrating, as it passes through another medium it gets deviated at some angle.

8. The phenomenon which occurs when an incident wave strikes an interface at an angle greater than the critical angle with respect to the normal to the surface is called as ____________
a) Refraction
b) Partial internal reflection
c) Total internal reflection
d) Limiting case of refraction
Answer: c
Explanation: Total internal reflection takes place when the light wave is in the more dense medium and approaching towards the less dense medium. Also, the angle of incidence is greater than the critical angle. Critical angle is an angle beyond which no propagation takes place in an optical fiber.

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Optical Communications Assessment Questions and Answers on “Optical Network Deployment”.

1. A _________________ is a network connecting several regional or national networks together.
a) Long-haul network
b) Domain network
c) Short-haul network
d) Erbium network
Answer: a
Explanation: Long-haul networks are also called as core or backbone networks. These networks connect regional or national networks together on a large scale. This can be extended to extended long haul networks.

2. What is the range of transmission of extended long haul network?
a) 200-400 km
b) 600-1000 km
c) 1000-2000 km
d) 2000-4000 km
Answer: c
Explanation: Extended long haul networks comprise of DWDM links. The transmission ranges may vary depending upon the complexity of the network. The extended long haul network’s transmission range varies from 1000 to 2000 km.

3. What is the range of transmission of ultra-long haul network?
a) 200-400 km
b) 600-1000 km
c) 1000-2000 km
d) 2000-4000 km
Answer: d
Explanation: Ultra haul networks comprise of DWDM links which provides them maximum range. The transmission ranges may vary depending upon the complexity of the network. The ultra-long haul network’s transmission range varies from 2000 to 4000 km.

4. Which feature plays an important role in making the longer haul networks feasible?
a) Channeling
b) Forward error control
c) Backward error control
d) Interconnection
Answer: b
Explanation: The longer haul networks can be made feasible by improvements in the DWDM systems and using forward error control mechanism. Such networks operate at channel rates in G-bits.

5. Which of the following is not an element of a submerged cable system?
a) Repeater
b) Branching unit
c) Gain equalizer
d) Attenuator
Answer: d
Explanation: The submerged cable system consists of a dry and wet plant. The elements associated with it include a repeater, branching unit, gain equalizer and a line amplifier. Attenuator is not present in cable system.

6. ___________ provides interconnection between the United States and European countries.
a) TAT
b) WTE
c) PFE
d) POP
Answer: a
Explanation: TAT is abbreviated as transatlantic optical fiber cables. TAT-14 is the newest version which is used as a medium of interconnection between the countries.

7. TAT-14 employs a DWDM bidirectional ring configuration.
a) False
b) True
Answer: b
Explanation: TAT-14 was first used in the year 2000. Its transmission capacity is more that the previous TAT versions. It’s DWDM configuration enables it to connect the various countries of Europe with the United States.

8. A single fiber in TAT-14 can carry _________ wavelength channels.
a) One
b) Twelve
c) Sixteen
d) Ten
Answer: c
Explanation: TAT-14 ‘ s single fiber carries a total of 16 wavelength channels. Each channel can allow a transmission rate of 10 Gigabits. It possesses a high operational capacity.

9. Optical MAN’S are usually structured in _______ topologies.
a) Ring
b) Bus
c) Mesh
d) Star
Answer: a
Explanation: MAN’s are characterized by changing traffic patterns requiring the networks to be fast. Also, MAN must be cost effective in terms of both operation and maintenance. Hence, they are structured in ring topologies.

10. The ________ network is an element of public telecommunication network that connects access nodes to individual users or MAN’s.
a) Ring
b) Access
c) Mesh
d) Nodal
Answer: b
Explanation: Access network is usually a last link in the network. It provides the strategies to connect to end-point users on both sides of connection.

11. _____________ is a technique that combines two or more network resources for redundancy or higher throughput.
a) Signal bonding
b) Attenuation
c) Re-signaling
d) Channel bonding
Answer: d
Explanation: Channel bonding combines two interfaces. It increases the overall bandwidth by the number of channels bonded. The data-rates are similar in this technique.

12. The upstream traffic in EPON is managed by employing a TDM approach.
a) True
b) False
Answer: a
Explanation: EPON upstream traffic is divided into time slots. The time slots are dedicated to each ONU(Optical network units) in order not to interfere with the data.

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