Category: Optical Communications Objective Questions

Optical Communications Problems on “Dispersion Management and Soliton Systems”.

1. Calculate second order dispersion coefficient for path length L₂ 20km and L₁ 160km. Dispersion coefficient for L₂ is 17.
a) -2.125ps/nm km
b) -3.25ps/nm km
c) -3.69ps/nm km
d) -1.28ps/nm km
Answer: a
Explanation: The second order dispersion coefficient for path length is given by-
β₂₁ = -β₂₂L₂/L₁. Here, β₂₂ = Dispersion coefficient forL₂, L₂ and L₁ are path lengths.

2. Calculate the path length L₂ if L₁is 160, dispersion coefficient of L₂ is 17, dispersion coefficient of L₁ is -2.25 ps/nmkm.
a) 40 km
b) 20 km
c) 30 km
d) 10 km
Answer: b
Explanation: The path length L₂ is given by-
L₂ = β₂₁L₁/-β₂₂. Here, β₂₂ = Dispersion coefficient forL₂, β₂₁ = Dispersion coefficient for L₁, L₂ and L₁ are path lengths.

3. Calculate path length L₁ if L₂ is 20, dispersion coefficient of L₂ is 17, dispersion coefficient of L₁ is -2.25 ps/nmkm.
a) 180 km
b) 30 km
c) 160 km
d) 44 km
Answer: c
Explanation: The path length L1is given by-
L₁ = β₂₁L₂/-β₂₂. Here, β₂₂ = Dispersion coefficient forL₂, β₂₁ = Dispersion coefficient for L₁, L₂ and L₁ are path lengths.

4. Calculate second order dispersion coefficient for path length L₁ 20 km and L₁ 160 km. Dispersion coefficient for L₁ is -2.125*10^-12s/nmkm.
a) 20
b) 19
c) 18
d) 17
Answer: d
Explanation: The second order dispersion coefficient for path length is given by-
β₂₂=-β₂₁L₂/L₁. Here, β21 = Dispersion coefficient forL1, L₂ and L1 are path lengths.

5. Calculate dispersion slope for second path fiber if L₁ is 150, L₂ is 10 and s₁ is 0.075.
a) 1.125
b) 2.125
c) 3.125
d) 1.9
Answer: a
Explanation: Dispersion slope for second path fiber is s₂ = -s₁(L₁/L₂). Here s1 and s2 are dispersion slopes for L₁, L₂. L₂ and L₁ are path lengths.

6. Calculate dispersion slope for first path fiber if L₁ is 160, L₂ is 20 and s₂ is 0.6ps/nm km.
a) 0.1
b) 0.432
c) 0.236
d) 0.075
Answer: d
Explanation: Dispersion slope for first path fiber is s₁ = -s₂(L₁/L₂). Here s1 and s2 are dispersion slopes for L₁, L₂. L₂ And L₁ are path lengths.

7. Calculate L₂ if dispersion slope for first path fiber is 0.075 and L₁ is 160 km and s₂ is -0.6ps/nm km.
a) 20 km
b) 30 km
c) 40 km
d) 50 km
Answer: a
Explanation: L₂ is determined by –
L2 = (-s₁/s₂)*L₁. Here s₁ and s₂ are dispersion slopes for L₁, L₂. L₂ and L₁ are path lengths.

8. Calculate L₁ if dispersion slope for first path fiber is 0.075 and L₂ is 20 km and s₂ is -0.6ps/nm km.
a) 170 km
b) 160 km
c) 180 km
d) 175 km
Answer: b
Explanation: L₁ is determined by –
L₂ = (-s₁/s₂)* L₂. Here s₁ and s₂ are dispersion slopes for L₁, L₂. L₂ and L₁ are path lengths.

9. Calculate separation of soliton pulses over a bit period length if R₂ pulse width is 6 ps for bit period of 70 ps.
a) 5.9
b) 5.7
c) 5.8
d) 5.4
Answer: c
Explanation: The separation of soliton pulses over a bit period length is calculated by –
q₀ = T₀/2ς. Here ς = pulse width and T₀ = bit period.

10. Calculate RZ pulse width if bit period is 60ps and separation of soliton pulses is 5.4.
a) 5.5ps
b) 8.1ps
c) 4.3ps
d) 2.3ps
Answer: a
Explanation: RZ pulse width can be calculated by –
ς = T₀/q₀. Here ς = pulse width and T₀ = bit period.

11. Calculate bit period if RZ pulse width is 50ps and separation of soliton pulses is 5.6.
a) 570ps
b) 540ps
c) 430ps
d) 560ps
Answer: d
Explanation: Bit period can be calculated by –
T₀ = 2T₂q₀. Here T₂=pulse width and T₀=bit period.

12. Calculate value of dimensionless parameter if bit period is 45ps and RZ pulse width is 4 ps.
a) 5.625
b) 5.0
c) 4
d) 6.543
Answer: a
Explanation: Dimensionless parameter is given by –
q₀ = T₀/2ς. Here ς=pulse width and T₀=bit period.

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Optical Communications Multiple Choice Questions on “FET Pre – Amplifiers”.

1. ____________ is the lowest noise amplifier device.
a) Silicon FET
b) Amplifier-A
c) Attenuator
d) Resonator-B
Answer: a
Explanation: FET operates by controlling the current flow with an electric field produced by an applied voltage on the gate of the device. Silicon FET is fabricated for low noise devices. It is the lowest noise amplifier device available.

2. FET device has extremely high input impedance greater than _________
a) 10⁷ Ohms and less than 10⁸
b) 10⁶ Ohms and less than 10⁷
c) 10¹⁴ Ohms
d) 10²³ Ohms
Answer: c
Explanation: FET operation involves the applied voltage on the gate of the device. The gate draws virtually no current, except for leakage, giving the device extremely high input impedance.

3. The properties of a bipolar transistor are superior to the FET.
a) True
b) False
Answer: b
Explanation: bipolar transistor operates by controlling the current flow with an electric field produced with a base current. The properties of a bipolar transistor are limited by its high trans-conductance than the FET.

4. Bipolar transistor is more useful amplifying device than FET at frequencies _____________
a) Above 1000 MHz
b) Equal to 1 MHz
c) Below 25 MHz
d) Above 25 MHz
Answer: d
Explanation: In FETs, the current gain drops to values near unity at frequencies above 25MHz. The trans-conductance is fixed with decreasing input impedance. Therefore, bipolar transistor is more useful amplifying device at frequencies above 25MHz.

5. High-performance microwave FETs are fabricated from ___________
a) Silicon
b) Germanium
c) Gallium arsenide
d) Zinc
Answer: c
Explanation: Since the mid- 1970s, the development of high-performance microwave FETs found its way. These FETs are fabricated from gallium arsenide and are called as GaAs metal Schottky field effect transistors (MESFETs).

6. Gallium arsenide MESFETs are advantageous than Silicon FETs.
a) True
b) False
Answer: a
Explanation: Gallium arsenide MESFETs are Schottky barrier devices. They operate with both low noise and high gain at microwave frequencies (GHz). Silicon FETs cannot operate with wide bands.

7. The PIN-FET hybrid receivers are a combination of ______________
a) Hybrid resistances and capacitances
b) Pin photodiode and low noise amplifier (GaAs MESFETs)
c) P-N photodiode and low noise amplifier (GaAs MESFETs)
d) Attenuator and low noise amplifier (GaAs MESFETs)
Answer: b
Explanation: The PIN-FET or p-i-n/FET receiver utilizes a p-i-n photodiode along with a low noise preamplifier (GaAs MESFETs). It is fabricated using thick-film integrated circuit technology. This hybrid integration reduces the stray capacitance to negligible levels.

8. PIN-FET hybrid receiver is designed for use at a transmission rate of _____________
a) 130 Mbits^-1
b) 110 Mbits^-1
c) 120 Mbits^-1
d) 140 Mbits^-1
Answer: d
Explanation: At 140 Mbits^-1, the performance of PIN-FET hybrid receiver is found to be comparable to germanium and alloy APD receivers. A digital equalizer is necessary as the high-impedance front end effectively integrates the signal at 140 Mbits^-1.

9. It is difficult to achieve higher transmission rates using conventional __________
a) Voltage amplifier
b) Waveguide Structures
c) PIN-FET or APD receivers
d) MESFET
Answer: c
Explanation: It is difficult to achieve higher transmission rates due to limitations in their gain bandwidth products. Also, the trade-off between the multiplication factor requirement and the bandwidth limits the performance of conventional receivers.

10. Which receiver can be fabricated using PIN-FET hybrid approach?
a) Trans-impedance front end receiver
b) Gallium arsenide receiver
c) High-impedance front-end
d) Low-impedance front-end
Answer: a
Explanation: Trans-impedance front-end receivers are fabricated using the PIN-FET hybrid approach. An example of such receivers consists of a GaAs MESFET and two complementary bipolar microwave transistors.

11. A silicon p-i-n photodiode utilized with the amplifier and the receiver is designed to accept data at a rate of ___________
a) 276Mbits^-1
b) 274 Mbits^-1
c) 278Mbits^-1
d) 302Mbits^-1
Answer: b
Explanation: A silicon p-i-n photodiode is used with the low-noise preamplifier. This preamplifier is based on a GaAs MESFET. Thus, a receiver using p-i-n photodiode accepts a data rate of 274 Mbits^-1 giving a sensitivity around -35dB_m.

12. What is usually required by FETs to optimize the figure of merit?
a) Attenuation of barrier
b) Matching with the depletion region
c) Dispersion of the gate region
d) Matching with the detector
Answer: d
Explanation: Total capacitance is given by C_t = C_d + C_a. The figure of merit is optimized when C_d=C_a. This requires FETs to be matched with the detectors. This requires FETs to be matched with the detectors. This procedure is usually not welcomed by the device and is not permitted in current optical receiver design.

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Optical Communications Quiz on “Narrow – Linewidth and Wavelength – Tunable Lasers”.

1. Which of these factors are critical in affecting the system performance in the case of coherent optical fiber transmission?
a) Laser line-width and stability
b) Refractive index and index difference
c) Core cladding diameter
d) Frequency
Answer: a
Explanation: The system employing intensity modulation does not consider line-width and stability as the factors of utmost importance. In coherent optical source transmission, laser line-width and stability are critical factors. These factors affect the system performance and are in the range of 0.5-1 Megahertz.

2. _______________ occurs as a result of the change in lasing frequency with gain.
a) Frequency multiplication
b) Dispersion
c) Attenuation
d) Line-width broadening
Answer: d
Explanation: Line-width broadening is a fundamental consequence of spontaneous emission process. It is related to the fluctuations in the phase of the optical fields. These phase fluctuations are due to the phase noises associated with the spontaneous emission process.

3. Laser cavity length can be extended by ___________
a) Increasing the refractive index
b) Reducing frequency
c) Introduction of external feedback
d) Using GRIN-rod lenses
Answer: c
Explanation: the lasers having long external cavity are referred to as LEC lasers. The extension of the laser cavity length by introduction of external feedback can be achieved by using an external cavity with a wavelength dispersive element.

4. What is the purpose of wavelength dispersive element is LEC lasers?
a) Wavelength selectivity
b) Reduction of line-width
c) Frequency multiplication
d) Avalanche multiplication
Answer: a
Explanation: A wavelength dispersive element is a part of the laser cavity. It is required because the long resonator structure has very closely spaced longitudinal modes which necessitates additional wavelength selectivity.

5. An effective method to reduce the line-width is to make the cavity longer.
a) True
b) False
Answer: a
Explanation: As the laser power increases, the device line-width decreases. The output power f laser cannot be mode arbitrarily large. Thus, the line-width is reduced by making the cavity longer. Longer cavity also enables increased wavelength selectivity.

6. Which devices are used to modulate the external cavity in order to achieve the higher switching speeds?
a) Electromagnetic
b) Acousto-optic
c) Dispersive
d) Lead
Answer: b
Explanation: The devices are tuned mechanically to extend the cavity of laser. The disadvantage of using mechanically tuned devices is low. Thus, electro-optic devices are used to modulate the external cavity in order to achieve higher switching speeds.

7. How many techniques are used to tune monolithic integrated devices (lasers)?
a) Five
b) One
c) Two
d) Three
Answer: c
Explanation: There are two techniques which can be employed to tune monolithic integrated devices. In the first method, the mode selectivity of a coupled cavity structure is used. Other method is used to a refractive index change in the device cavity provided by application of an electric field.

8. _________________ laser can be produced when a coupler section is introduced between the amplifier and phase sections of a structure.
a) SG-DBR
b) GCSR
c) Y 4-shifted
d) DSM
Answer: b
Explanation: DBR lasers are capable of wavelength tuning. Grating assisted co-directional coupler with sampled reflector (GCSR). Laser is capable of a tuning range greater than 40 nm. It consists of a co-directional coupler between the amplifier and the phase section.

9. The rare-earth-doped fiber lasers have spectral line-width in the range of _________________
a) 0.1 to 1 nm
b) 1.2 to 1.5 nm
c) 6 to 10 nm
d) 2 to 2.3 nm
Answer: a
Explanation: The rare-earth-doped fiber lasers have spectral line-width in the range of 0.1 to 1 nm. These line-widths are too long for high speed transmission is possible in this range.

10. The lasing line-width of Fox-smith resonator is ____________________
a) Less than 1 MHz
b) 1 MHz
c) 2 MHz
d) Greater than 3 MHz
Answer: a
Explanation: Fox-smith resonator employs a fused coupled fabricated from erbium-doped fiber. Narrower spectral line-width can be obtained using a resonator. It provides favorable line-widths than semiconductor laser.

11. What is the widest tuning range obtained in optical fiber laser structure?
a) 60 nm
b) 80 nm
c) More than 100 nm
d) 100 nm
Answer: c
Explanation: A tuning range greater than 100 nm by using an erbium-doped photonic crystal fiber. A wider tuning range greater than 100 nm is obtained at wavelength 1.55 nm.

12. How many techniques can be used to increase the injection cavity length?
a) One
b) Two
c) Three
d) Four
Answer: b
Explanation: Two techniques can be used to increase the injection laser cavity length. These are using laser chips and by extending a cavity with a passive medium such as air, glass etc.

13. The mechanism which results from a refractive index change in the passive waveguide layer is called as ___________
a) Absorption
b) Spontaneous emission
c) Monolithic inversion
d) Bragg wavelength control
Answer: d
Explanation: A wider wavelength tuning length is obtained by separating the Bragg region in the passive waveguide and by introducing a phase region within a waveguide control mechanism provides phase control. It takes place by some changes in a passive waveguide layer.

14. How many sections are included in a sampling grating distributed Bragg-reflector laser (SG-DBR)?
a) Four
b) Five
c) Three
d) Two
Answer: b
Explanation: In SG-DBR laser, five sections are longitudinally integrated together on a semiconductor substrate. These five sections include two diffraction Bragg grating sections, a gain, a phase and an amplifier section.

15. Fiber based lasers provide diffraction-limited power at higher levels than solid-state laser.
a) True
b) False
Answer: a
Explanation: In fiber lasers, the active gain medium is doped with rare earth elements. These lasers have active regions several kilometers long and thus provide high optical gain. Solid-state lasers, on the other hand, provide diffraction limited power at lower levels.

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Optical Communications Multiple Choice Questions on ” Optical Fiber Cables”.

1. When optical fibers are to be installed in a working environment, the most important parameter to be considered is?
a) Transmission property of the fiber
b) Mechanical property of the fiber
c) Core cladding ratio of the fiber
d) Numerical aperture of the fiber
Answer: b
Explanation: Nowadays, optical fibers are used alternatively to electric transmission lines. They are installed safely and maintained in all environments including underground areas. This requires mechanical strengthening of fibers in order to ensure proper transmission.

2. It is not important to cover these optical fibers required for transmission.
a) True
b) False
Answer: b
Explanation: Unprotected optical fibers have number of losses regarding its strength and durability. Bare glass fibers are brittle and have small cross-section area that makes them highly susceptible to damages while handling and maintenance. Thus, to improve tensile strength, optical fibers should be covered by surrounding them with number of protective layers.

3. Optical fibers for communication use are mostly fabricated from ___________
a) Plastic
b) Silica or multicomponent glass
c) Ceramics
d) Copper
Answer: b
Explanation: Silica or a compound of glass are brittle and have almost perfect elasticity until reaching their breaking point. Strength of these materials is high. Thus, optical fibers are fabricated from these materials.

4. An Si-O bond with a Young’s modulus of 9*10¹⁰Nm^-1 have an elliptical crack of depth 7nm. The surface energy is 2.29 J. Estimate fracture stress for silica fiber.
a) 4.32*10⁹Nm^-1
b) 6.32*10⁹Nm^-1
c) 5.2*10⁹Nm^-1
d) 3*10⁹Nm^-1
Answer: a
Explanation: For an elliptical crack, the fracture stress is given by-
S_f = (2Eγ_p/πC)^1/2
Where S_f = fracture stress
γ_p = surface energy
C = depth of crack.

5. Calculate percentage strain at break for a Si-O bond with a fracture strength of 3.52*10¹⁰Nm^-1 and Young’s modulus of 9 *10⁹Nm^-1.
a) 3.1 %
b) 2.8 %
c) 4.5 %
d) 3.9 %
Answer: d
Explanation: Young’s modulus is given by-
E = Stress/Strain
To calculate strain from the above formula, we have to divide stress by Young’s modulus. Therefore, Strain = Stress/E.

6. Stress corrosion must be considered while designing and testing optical fiber cables.
a) True
b) False
Answer: a
Explanation: Stress corrosion means growth of flaws due to stress and water. This occurs as a result of molecular bonds at the tip of crack being attacked by water. Hence, it is important to have a protection against water to avoid stress corrosion.

7. Which statistics are used for calculations of strengths of optical fibers?
a) Edwin statistics
b) Newton statistics
c) Wei-bull statistics
d) Gamma statistics
Answer: c
Explanation: Calculations of strengths are conducted using Wei-bull statistics in case of optical fibers. It describes the strength behavior of a system that is dependent on the weakest link of the system. The Wei-bull statistics gives the probability of failure of the optical fiber at a given strength.

8. What does n denotes in the equation given below, if v_c is the crack velocity; A is the constant for the fiber material and KI is the strength intensity factor?

vc = AKIn

a) Refractive index
b) Stress corrosion susceptibility
c) Strain
d) Young’s modulus
Answer: b
Explanation: The above equation allows estimation of the time to failure of a fiber under stress corrosion conditions. The constant n is called as stress corrosion susceptibility. It is typically in the range of 15 to 50 for a glass.

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Optical Communications Multiple Choice Questions on “Optical Switching Networks”.

1. Optical switching can be classified into ________ categories.
a) Two
b) Three
c) Four
d) One
Answer: a
Explanation: Optical switching is classified into two categories same as that of electronic switching.
The two categories are circuit switching and packet switching.

2. ___________________ are the array of switches which forms circuit switching fabrics.
a) Packet arrays
b) Optical cross connects
c) Circuit arrays
d) Optical networks
Answer: b
Explanation: Optical cross-connects incorporate switching connections or light paths. These larger arrays can switch signals from one port to another.

3. ___________ is an example of a static circuit-switched network.
a) OXC
b) Circuit regenerator
c) Packet resolver
d) SDH/SONET
Answer: d
Explanation: The circuit is said to be static when the network resources remain dedicated to the circuit connection. This should be followed during the entire transfer and the complete message follows the same path.

4. What is the main disadvantage of OCS?
a) Regenerating mechanism
b) Optical session
c) Time permit
d) Disability to handle burst traffic
Answer: d
Explanation: In traffic conditions, data is sent in the form of bursts of different lengths. Thus, the resources cannot be readily assigned. The OCS cannot efficiently handle burst traffic.

5. Optical electro-conversions takes place in _________________ networks.
a) Sessional
b) Optical packet-switched
c) Optical circuit-switched
d) Circular
Answer: c
Explanation: In an optical packet-switched network, data is transported in the optical domain. This is done without intermediate optic-electrical conversions. Optical electro-conversions takes place in circuit-switched networks.

6. How many functions are performed by an optical packet switch?
a) Four
b) Three
c) Two
d) One
Answer: a
Explanation: An optical packet switch performs four basic functions. These include routing, forwarding, switching and buffering.

7. ____________ provides data storage for packets to resolve contention problems.
a) Switching
b) Routing
c) Buffering
d) Reversing
Answer: c
Explanation: Switching involves directing the packets. Routing provides network connectivity while forwarding and reversing involves defining a packet. Buffering usually provides data storage for packets.

8. What is usually required by a packet to ensure that the data is not overwritten?
a) Header
b) Footer
c) Guard band
d) Payload
Answer: c
Explanation: A packet consists of a header and the payload. The label points to an entry in the lookup table. A guard band is usually included to ensure the data is not overwritten.

9. Routing technique is faster than the labeling technique. State whether the given statement is true or false.
a) False
b) True
Answer: a
Explanation: Labeling suggests where the packet should be directed. Routing routes the data in the given direction. Thus, labeling technique is efficient and faster than the routing technique.

10. ______________ provides efficient designation, routing, forwarding, switching of traffic through an optical packet-switched network.
a) Label correlation
b) Multiprotocol label switching
c) Optical correlation
d) Routing
Answer: b
Explanation: Multiprotocol label switching (MPLS) was first proposed by CISCO systems. Earlier, it was called as tag switching. MPLS uses labels to forward, switch, designate the traffic.

11. MPLS is independent of layer 2 and 3 in the OSI model. State whether the given statement is true or false.
a) True
b) False
Answer: a
Explanation: MPLS is flexible in the current protocol landscape. It supports Ethernet, frame relay as a data link layer but is independent of layer 2 and 3 in the OSI model.

12. Which of the following service is provided by Multiprotocol label switching (MPLS)?
a) Data forwarding
b) Routing
c) VPN’s
d) Switching
Answer: c
Explanation: One of the important services provided by MPLS is IP virtual private networks. All others are provided by packet switched networks. These VPN’s provide a secure, dedicated wide area network (WAN) in order to connect the offices all over the world.

13. Burst header cell is also known as _____________
a) Burst channel
b) Burst header circuit
c) Burst regenerator
d) Burst header packet
Answer: d
Explanation: Burst header cell consists of information regarding switching and destination address. It works with the use of transmission units called as data bursts.

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Optical Communications Multiple Choice Questions on “Application of Optical Amplifiers “.

1. Which of the following is not a drawback of regenerative repeater?
a) Cost
b) Bandwidth
c) Complexity
d) Long haul applications
Answer: d
Explanation: The regenerative repeaters are useful in long haul applications. However, such devices increase the cost and complexity of the optical communication system. It act as a bottleneck by restricting the system operational bandwidth.

2. The term flexibility, in terms of optical amplifiers means the ability of the transmitted signal to remain in the optical domain in a long haul link.
a) True
b) False
Answer: a
Explanation: Repeaters are usually used to maintain the transmitted signal in the optical domain. But, it has its own drawbacks. Thus, flexible systems which include optical amplifiers are used for such purpose.

3. How many configurations are available for employment of optical amplifiers?
a) Three
b) Four
c) Two
d) Five
Answer: a
Explanation: Optical amplifiers can be employed in three configurations. These are simplex mode, duplex mode, multi-amplifier configuration.

4. Repeaters are bidirectional.
a) True
b) False
Answer: b
Explanation: Repeaters are unidirectional. Optical amplifiers have the ability to operate simultaneously in both directions at the same carrier wavelength.

5. It is necessary to ____________ the optical carriers at different speeds to avoid signal interference.
a) Inculcate
b) Reduce
c) Intensity-modulate
d) Demodulate
Answer: c
Explanation: Optical amplifiers are bidirectional. They operate in both directions at the same carrier wavelength. In order to avoid interference, the optical carriers should be intensity modulated.

6. The _________________ increases the system reliability in the event of an individual amplifier failure.
a) Simplex configuration
b) Duplex configuration
c) Serial configuration
d) Parallel multi-amplifier configuration
Answer: d
Explanation: The optical amplifiers with spectral bandwidths in the range 50 to 100 nm allow amplifiers to be more reliable than repeaters. The parallel multi-amplifier configuration increases system reliability and relaxes the linearity.

7. Which of the following is not an application of optical amplifier?
a) Power amplifier
b) In-line repeater amplifier
c) Demodulator
d) Preamplifier
Answer: c
Explanation: Optical amplifiers have a wide variety of applications in the transmitter as well as receiver side. It is used as the power amplifier in the transmitter side and as preamplifier at the receiver side.

8. _________ reconstitutes a transmitted digital optical signal.
a) Repeaters
b) Optical amplifiers
c) Modulators
d) Circulators
Answer: a
Explanation: Optical amplifiers simply act as gain blocks on an optical fiber link. However, in contrast, the regenerative repeaters reconstitute a transmitted digital optical signal.

9. _____________ are transparent to any type of signal modulation.
a) Repeaters
b) Optical amplifiers
c) Modulators
d) Circulators
Answer: b
Explanation: The main benefit of acting as a gain block for optical amplifier is that it can be transparent to modulation bandwidth. However, both the noise and signal distortions are continuously amplified.

10. _________________ imposes serious limitations on the system performance.
a) Fiber attenuation
b) Fiber modulation
c) Fiber demodulation
d) Fiber dispersion
Answer: d
Explanation: The fiber dispersion calculation does not take into account the non-regenerative nature of the amplifier repeaters. In this, the pulse spreading and the noise is accumulated.

11. __________ is the ratio of input signal to noise ratio to the output signal to noise ratio of the device.
a) Fiber dispersion
b) Noise figure
c) Transmission rate
d) Population inversion
Answer: b
Explanation: Noise figure judges the performance factor of the devices. It is the in and out the ratio of signal to noise degradation for any device.

12. How many factors govern the noise figure of the device?
a) Four
b) Three
c) Two
d) One
Answer: a
Explanation: Noise figure is governed by factors such as the population inversion, the number of transverse modes in the amplifier cavity, the number of incident photons on the amplifier and the optical bandwidth of the amplified spontaneous emissions.

13. What is the typical range of the noise figure?
a) 1 – 2 dB
b) 3 – 5 dB
c) 7 – 11 dB
d) 12 – 14 dB
Answer: c
Explanation: Typical noise figures range from 7 to 11 dB. The SOAs are generally at the bottom end of the range and the fiber amplifiers towards the top end.

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