Category: Optical Communications Objective Questions

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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Optical Communications Multiple Choice Questions on “Receiver Structures”.

1. How many circuits are present in an equivalent circuit for the digital optical fiber receiver?
a) Four
b) One
c) Three
d) Two
Answer: a
Explanation: A full equivalent circuit for the digital optical fiber receiver includes four circuits. These are the detector circuit, noise sources, and amplifier and equalizer circuit.

2. __________ compensates for distortion of the signal due to the combined transmitter, medium and receiver characteristics.
a) Amplification
b) Distortion
c) Equalization
d) Dispersion
Answer: c
Explanation: Equalization adjusts the balance between frequency components within an electronic signal. It compensates for distortion of the signal. The distortion may be due to the transmitter, receiver etc.

3. ____________ is also known as frequency-shaping filter.
a) Resonator
b) Amplifiers
c) Attenuator
d) Equalizer
Answer: d
Explanation: Equalizer, often called as frequency-shaping filter has a frequency response inverse to that of the overall system frequency response. In wideband systems, it boosts the high frequency components to correct the overall amplitude of the frequency response.

4. The phase frequency response of the system should be ____________ in order to minimize inter-symbol interference.
a) Non-Linear
b) Linear
c) More
d) Less
Answer: b
Explanation: An equalizer is used as frequency shaping filter. The phase frequency response of the system should be linear to acquire the desired spectral shape for digital systems. This, in turn, minimizes the inter-symbol interference.

5. Noise contributions from the sources should be minimized to maximize the receiver sensitivity.
a) True
b) False
Answer: a
Explanation: Noise sources include transmitter section, medium and the receiver section. As the noise increases, the sensitivity at the receiver section decreases. Thus, noise contributions should be minimized to maximize the receiver sensitivity.

6. How many amplifier configurations are frequently used in optional fiber communication receivers?
a) One
b) Two
c) Three
d) Four
Answer: c
Explanation: Three amplifier configurations are used in optical fiber communication receivers. These are voltage amplifiers, semiconductor optical amplifier and current amplifier. Voltage amplifier is the simplest and most common amplifier configuration.

7. How many receiver structures are used to obtain better receiver characteristics?
a) Two
b) One
c) Four
d) Three
Answer: d
Explanation: The various receiver structures are low-impedance front end, high-impedance front end and trans-impedance front-end. The noise in the trans-impedance amplifier will always exceed than the front end structure.

8. The high-impedance front-end amplifier provides a far greater bandwidth than the trans-impedance front-end.
a) True
b) False
Answer: a
Explanation: The noise in the trans-impedance amplifier exceeds that incurred by the high-impedance amplifier. Hence, the trans-impedance front-end provides a greater bandwidth without equalization than the high-impedance front end.

9. A high-impedance amplifier has an effective input resistance of 4MΩ. Find the maximum bandwidth that may be obtained without equalization if the total capacitance is 6 pF and total effective load resistance is 2MΩ.
a) 13.3 kHz
b) 14.2 kHz
c) 15.8 kHz
d) 13.9 kHz
Answer: a
Explanation: The maximum bandwidth obtained without equalization is given by –
B = 1/2ΠR_TLC_T
Where,
R_TL = Total load resistance
C_T = Total capacitance.

10. A high-input-impedance amplifier has following parameters (Total effective load resistance = 2MΩ, Temperature = 300 K). Find the mean square thermal noise current per unit bandwidth for the high-impedance configuration.
a) 8.9×10^-27A²/Hz
b) 8.12×10^-27A²/Hz
c) 8.29×10^-27A²/Hz
d) 8.4×10^-27A²/Hz
Answer: c
Explanation: the mean square thermal noise current per unit bandwidth for the high-impedance configuration is given by –
i_T²= 4KT/R_TL
Where, K = constant
T = Temperature (Kelvin)
R_TL = total effective load resistance.

11. The mean square thermal noise current in the trans-impedance configuration is _________ greater than that obtained with the high-input-impedance configuration.
a) 30
b) 20
c) 15
d) 10
Answer: b
Explanation: 13 dB noise penalties are incurred with the trans-impedance amplifier over that of the high-input-impedance configuration. It is the logarithmic function of the noise current value. However, the trans-impedance amplifiers can be optimized for noise performance.

12. The major advantage of the trans-impedance configuration over the high-impedance front end is ______________
a) Greater bandwidth
b) Less bandwidth
c) Greater dynamic range
d) Less dynamic range
Answer: c
Explanation: Greater dynamic range is a result of the different attenuation mechanism for the low-frequency components of the signal. This attenuation is obtained in the trans-impedance amplifier through the negative feedback and therefore the low frequency components are amplified by the closed loop. This increases the dynamic range.

13. The trans-impedance front end configuration operates as a __________ with negative feedback.
a) Current mode amplifier
b) Voltage amplifier
c) Attenuator
d) Resonator
Answer: a
Explanation: The trans-impedance configuration overcomes the drawbacks of the high-impedance front end. It utilizes a low-noise, high-input-impedance amplifier with negative feedback. It operates as a current mode amplifier where high impedance is reduced by negative feedback.

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Optical Communications Multiple Choice Questions on “Non – Semiconductor Lasers”.

1. ____________________ lasers are presently the major laser source for optical fiber communications.
a) Semiconductor
b) Non-Semiconductor
c) Injection
d) Solid-state
Answer: c
Explanation: Injection laser coupling using discrete lasers have proved to fruitful. Such lenses provide for relaxation for an alignment tolerances normally required for fiber coupling. Certain non-semiconductor sources are making its lace in the optical fiber communication. At slowly present, injection lasers are mostly used as laser sources.

2. In Nd: YAG lasers, the maximum doping levels of neodymium is ____________
a) 0.5 %
b) 1.5 %
c) 1.8 %
d) 2 %
Answer: b
Explanation: The Nd: YAG laser structure is formed by doping of yttrium- aluminum -garnet (YAG) with neodymium. The energy levels for lasing transition and pumping are provided by neodymium ions. The maximum doping level of neodymium in YAG is around 1.5 %.

3. Which of the following is not a property of Nd: YAG laser that enables its use as an optical fiber communication source?
a) Single mode operation
b) Narrow line-width
c) Long lifetime
d) Semiconductors and integrated circuits
Answer: d
Explanation: Nd: YAG laser is a non-semiconductor laser. It does not include the use of semiconductors and thus cannot take advantage of well-developed technology associated with integrated circuits. Single mode operation, narrow line-width, lifetime are the properties that are useful for optical communication.

4. The Nd: YAG laser has a narrow line-width which is ________________
a) < 0.01 nm
b) > 0.01 nm
c) > 1 mm
d) > 1.6 mm
Answer: a
Explanation: The Nd: YAG laser has several properties which make it an active optical source. One of such properties is its narrow line-width. It is less than 0.01 nm which is useful for reducing dispersion of optical links.

5. The strongest pumping bands is a four level system of Nd: YAG laser at wavelength of range_________________
a) 0.25 and 0.56 nm
b) 0.75 and 0.81 nm
c) 0.12 and 0.23 nm
d) 1 and 2 nm
Answer: b
Explanation: The Nd: YAG laser is a four level system. It consists of number of pumping bands and fluorescent transitions. The strongest pumping bands are the wavelengths of 0.75μm and 0.81μm. and gives lasing transition at 1.064μm and 1.32μm. Single mode emission is usually obtained at these wavelengths.

6. The Nd: YAG laser is costlier than earth-doped glass fiber laser.
a) True
b) False
Answer: a
Explanation: The most important requirement of the Nd: YAG laser is pumping and modulation. These two requirements tend to give a cost disadvantage in comparison with earth-doped glass fiber laser. Also it is easier and less expensive to fabricate glass fiber in earth-doped laser.

7. It is a resonant cavity formed by two parallel reflecting mirrors separated by a mirror separated by a medium such as air or gas is?
a) Optical cavity
b) Wheatstone’s bridge
c) Oscillator
d) Fabry-perot resonator
Answer: d
Explanation: Resonant cavity is formed between two mirrors where fiber core doped with earth ions is placed. This cavity is 250-500 μm long and 5 to 15 μm wide. A Fabry-perot resonator oscillates at resonant frequency for which there is high gain.

8. In a three level system, the threshold power decreases inversely with the length of the fiber gain medium.
a) True
b) False
Answer: b
Explanation: If the imperfection losses are low then in a four level system the threshold power decreases inversely with the length of the fiber gain medium. A three level consists of an optimum length. This optimum length gives the minimum threshold power which is independent of the value of imperfection losses.

9. Which of the following co-dopant is not employed by neodymium and erbium doped silica fiber lasers?
a) Phosphorus pent oxide
b) Germania
c) Nitrogen
d) Alumina
Answer: c
Explanation: Silica based glass fibers are proved to be the best host material till date. These silica fibers are doped with neodymium and erbium. These dopants include co-dopants such as phosphorus pent-oxide, germanium and alumina.

10. Dopants levels in glass fiber lasers are generally ___________
a) Low
b) High
c) Same as that of GRIN rod lens laser
d) Same as that of semiconductor laser
Answer: a
Explanation: Dopant levels are low in glass fibers (nearly 400 parts per million). This is because of increasing in concentration quenching which increases with the doping level. It may cause the reduction in the population of the upper lasing level as well as crystallization within the glass matrix.

11. _______________ fibers include addition of lead fluoride to the core glass in order to raise the relative refractive index.
a) Solid-state
b) GaAs
c) Semiconductor
d) ZBLANP
Answer: d
Explanation: Up-conversion pumping of laser material is used to convert an infrared laser output to a visible laser output. ZBLANP is host material on which laser action at all wavelengths can be obtained by pumping. The relative refractive index is increased by addition of lead fluoride which makes it a very interesting host material.

12. The lasing output of the basic Fabry-perot cavity fiber is restricted to between ____________
a) 1 and 2 nm
b) 5 and 10 nm
c) 3 and 6 nm
d) 15 and 30 nm
Answer: b
Explanation: the gain spectrum of rare earth ions may be seen over a wavelength range of 50 nm. The lasing output will thus be narrow unless the dielectric on the mirror is arranged. Such a narrow line-width is not used for a broadband optical source.

13. In Fabry-perot laser, the lower threshold is obtained by ___________
a) Increasing the refractive index
b) Decreasing the refractive index
c) Reducing the slope efficiency
d) Increasing the slope efficiency
Answer: c
Explanation: The finesse of Fabry-perot cavity provides a measure of its filtering properties. When the finesse is high the splitting ratio is low thus lowering the laser threshold in an optical cavity without mirror. In Fabry-perot laser, mirrors are present and thus lower threshold is obtained by reducing the slope efficiency.

14. When did the non-semiconductor laser developed?
a) 1892
b) 1946
c) 1985
d) 1993
Answer: c
Explanation: Non-semiconductor sources are crystalline and glass wave-guiding structures. They are doped with rare earth ions and are good optical sources. The development of these sources started in the year 1985. Example: Nd: YAG laser.

15. Y₃A_l5 O₁₂ is a molecular formula for _____________
a) Ytterbium aluminate
b) Yttrium oxide
c) Ytterbium oxy-aluminate
d) Yttrium-aluminum garnet
Answer: d
Explanation: The atomic number of Yttrium is 39. It is the base element of Yttrium-aluminum garnet. Y₃A_l5 O₁₂, doped with rare earth ion neodymium to form Nd: YAG laser structure.

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Optical Communications Multiple Choice Questions on “Optical Fibers”.

1. Multimode step index fiber has ___________
a) Large core diameter & large numerical aperture
b) Large core diameter and small numerical aperture
c) Small core diameter and large numerical aperture
d) Small core diameter & small numerical aperture
Answer: a
Explanation: Multimode step-index fiber has large core diameter and large numerical aperture. These parameters provides efficient coupling to inherent light sources such as LED’s.

2. A typically structured glass multimode step index fiber shows as variation of attenuation in range of ___________
a) 1.2 to 90 dB km^-1 at wavelength 0.69μm
b) 3.2 to 30 dB km^-1 at wavelength 0.59μm
c) 2.6 to 50 dB km^-1 at wavelength 0.85μm
d) 1.6 to 60 dB km^-1 at wavelength 0.90μm
Answer: c
Explanation: A multimode step index fibers show an attenuation variation in range of 2.6 to 50dBkm^-1. The wide variation in attenuation is due to the large differences both within and between the two overall preparation methods i.e. melting and deposition.

3. Multimode step index fiber has a large core diameter of range is ___________
a) 100 to 300 μm
b) 100 to 300 nm
c) 200 to 500 μm
d) 200 to 500 nm
Answer: a
Explanation: A multimode step index fiber has a core diameter range of 100 to 300μm. This is to facilitate efficient coupling to inherent light sources.

4. Multimode step index fibers have a bandwidth of ___________
a) 2 to 30 MHz km
b) 6 to 50 MHz km
c) 10 to 40 MHz km
d) 8 to 40 MHz km
Answer: b
Explanation: Multimode step index fibers have a bandwidth of 6 to 50 MHz km. These fibers with this bandwidth are best suited for short -haul, limited bandwidth and relatively low-cost application.

5. Multimode graded index fibers are manufactured from materials with ___________
a) Lower purity
b) Higher purity than multimode step index fibers.
c) No impurity
d) Impurity as same as multimode step index fibers.
Answer: b
Explanation: Multimode graded index fibers have higher purity than multimode step index fiber. To reduce fiber losses, these fibers have more impurity.

6. The performance characteristics of multimode graded index fibers are ___________
a) Better than multimode step index fibers
b) Same as multimode step index fibers
c) Lesser than multimode step index fibers
d) Negligible
Answer: a
Explanation: Multimode graded index fibers use a constant grading factor. Performance characteristics of multimode graded index fibers are better than those of multimode step index fibers due to index graded and lower attenuation.

7. Multimode graded index fibers have overall buffer jackets same as multimode step index fibers but have core diameters ___________
a) Larger than multimode step index fibers
b) Smaller than multimode step index fibers
c) Same as that of multimode step index fibers
d) Smaller than single mode step index fibers
Answer: b
Explanation: Multimode graded index fibers have smaller core diameter than multimode step index fibers. A small core diameter helps the fiber gain greater rigidity to resist bending.

8. Multimode graded index fibers with wavelength of 0.85μm have numerical aperture of 0.29 have core/cladding diameter of ___________
a) 62.5 μm/125 μm
b) 100 μm/140 μm
c) 85 μm/125 μm
d) 50 μm/125μm
Answer: b
Explanation: Multimode graded index fibers with numerical aperture 0.29 having a core/cladding diameter of 100μm/140μm. They provide high coupling frequency LED’s at a wavelength of 0.85 μm and have low cost. They are also used for short distance application.

9. Multimode graded index fibers use incoherent source only.
a) True
b) False
Answer: b
Explanation: Multimode graded index fibers are used for short haul and medium to high bandwidth applications. Small haul applications require LEDs and low accuracy lasers. Thus either incoherent or incoherent sources like LED’s or injection laser diode are used.

10. In single mode fibers, which is the most beneficial index profile?
a) Step index
b) Graded index
c) Step and graded index
d) Coaxial cable
Answer: b
Explanation: In single mode fibers, graded index profile is more beneficial as compared to step index. This is because graded index profile provides dispersion-modified-single mode fibers.

11. The fibers mostly not used nowadays for optical fiber communication system are ___________
a) Single mode fibers
b) Multimode step fibers
c) Coaxial cables
d) Multimode graded index fibers
Answer: a
Explanation: Single mode fibers are used to produce polarization maintaining fibers which make them expensive. Also the alternative to them are multimode fibers which are complex but accurate. So, single-mode fibers are not generally utilized in optical fiber communication.

12. Single mode fibers allow single mode propagation; the cladding diameter must be at least ___________
a) Twice the core diameter
b) Thrice the core diameter
c) Five times the core diameter
d) Ten times the core diameter
Answer: d
Explanation: The cladding diameter in single mode fiber must be ten times the core diameter. Larger ratios contribute to accurate propagation of light. These dimension ratios must be there so as to avoid losses from the vanishing fields.

13. A fiber which is referred as non-dispersive shifted fiber is?
a) Coaxial cables
b) Standard single mode fibers
c) Standard multimode fibers
d) Non zero dispersion shifted fibers
Answer: b
Explanation: A standard single mode fiber having step index profile is known as non-dispersion shifted fiber. As these fibers have a zero dispersion wavelength of 1.31μm and so are preferred for single-wavelength transmission in O-band.

14. Standard single mode fibers (SSMF) are utilized mainly for operation in ___________
a) C-band
b) L-band
c) O-band
d) C-band and L-band
Answer: c
Explanation: SSMFs are utilized for operation in O-band only. It shows high dispersion in the range of 16 to 20ps/nm/km in C-band and L-band. So SSMFs are used in O-band.

15. Fiber mostly suited in single-wavelength transmission in O-band is?
a) Low-water-peak non dispersion-shifted fibers
b) Standard single mode fibers
c) Low minimized fibers
d) Non-zero-dispersion-shifted fibers
Answer: b
Explanation: Standard single mode fibers with a step index profile are called non dispersion shifted fiber and it is particularly used for single wavelength transmission in O-band and as if has a zero-dispersion wavelength at 1.31μm.

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Optical Communications Multiple Choice Questions on “Wavelength Routing Networks”.

1. Which of the following is used to provide wavelength signal service among the nodes?
a) Regularization
b) Optical enhancing
c) Hopping
d) Pulse breakdown
Answer: c
Explanation: The optical layer is dependent on wavelength. The entire physical interconnected network provides wavelength signal service among the nodes using hopping technique.

2. How many types of hopping are present?
a) Two
b) One
c) Three
d) Four
Answer: a
Explanation: There are two types of hopping. They are single hop and multihop. These techniques provide wavelength dependent service for interconnected physical network among the nodes.

3. How many switching layers are possessed by MG-OXC?
a) Two
b) Three
c) One
d) Six
Answer: b
Explanation: An MG-OXC has three switching layers. They are wavelength cross-connect (WXC), waveband cross-connects (BXC), and fiber cross-connects (FXC). These layers help to terminate the wavebands and individual wavelength channels.

4. _____________ supports a great number of wavelength channels and reduces the number of switches within the optical network.
a) Waveband switching
b) Optical remuneration
c) Optical genesis
d) Wavelength multiplexing
Answer: a
Explanation: Waveband switching reduces the number of ports within the optical network. It reduces the complexity of numerous wavelength-driven channels and makes it efficient.

5. Individual wavelength channels and wavebands are terminated through ________________ layers.
a) WXC and PXC
b) WXC and FXC
c) BXC and FXC
d) WXC and BXC
Answer: d
Explanation: The individual wavelength channels are terminated and the terminated waveband is then de-multiplexed. The de-multiplexing is in the form of individual channels which are sent to WXC layer as inputs.

6. The routing and wavelength assignment problem addresses the core issue of _____________
a) Traffic patterns in a network
b) Wavelength adjustment
c) Wavelength continuity constraint
d) Design problem
Answer: c
Explanation: The routing and wavelength assignment problem includes selecting a suitable path and allocating an available wavelength. These problems fall into two categories of sequential or combinational selections.

7. How many techniques of implementation are there for routing wavelength assignment (RWA)?
a) Two
b) Six
c) Three
d) Four
Answer: a
Explanation: The implementation of RWA can be static and dynamic. This depends on the traffic patterns in the network. Static RWA techniques are semi-permanent and dynamic RWA techniques are random in nature.

8. ____________ deals with establishing the light path in frequently varying traffic patterns.
a) Wavelength routing
b) Wavelength multiplexing
c) Static RWA
d) Dynamic RWA
Answer: d
Explanation: In dynamic RWA, the traffic patterns are not known. Thus, the connection requests are initiated in random fashion. Its random nature depends on the network state at the time of request.

9. Static RWA problem is also known as _____________
a) Routing problem
b) Virtual topology problem
c) Static wavelength problem
d) Light path problem
Answer: b
Explanation: Static RWA problem refers to the connection problems which remain connected for a smaller duration of time. Thus, network resources are assigned to each connection. It is also called as virtual topology design problem.

10. The ___________ provides information about the physical path and wavelength assignment for all active light paths.
a) Network state
b) RWA
c) LAN topology
d) Secluded communication protocol
Answer: a
Explanation: The physical path i.e. route is associated with the routing problem. Each connection is provided with network resources to reduce complexity in functioning. The network state is basically required to provide information related to routing and assignment problems.

11. ________________ plays an important role in determining the blocking probability of a network.
a) CGA algorithm
b) Semi-pristine environment
c) RWA algorithm
d) Pass key protocol
Answer: c
Explanation: RWA algorithm’s efficiency is calculated on the basis of no blocking or lowest blocking probability. It also provides the information about the availability of the path between the source and destination.

12. Wavelength assignment in RWA is independent on the network topology.
a) True
b) False
Answer: b
Explanation: RWA algorithm deals with the wavelength assignment, physical path and blocking probability. Network topology plays a crucial role in the wavelength assignment. The network state and topology enables the RWA algorithm to function smoothly.

13. Static RWA technique is semi-permanent.
a) True
b) False
Answer: a
Explanation: The connections employs in static RWA are semi-permanent but remain active for a relatively longer period of time. The traffic patterns are known in advance and thus the optimization can be done by assigning network resources to each connection.

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