Category: Optical Communications Objective Questions

Optical Communications Multiple Choice Questions on “Multicarrier Systems”.

1. A major attribute of coherent optical transmission was its ability to provide _______________ for future multicarrier systems and networks.
a) Attenuation
b) Dispersion
c) Frequency selectivity
d) Noisy carriers
Answer: c
Explanation: A coherent optical transmission involves the wavelength, frequency and the distance as its main factors. It provides frequency and wavelength selectivity with narrow channel spacing’s for future multicarrier systems and networks.

2. The channel width is narrow for the coherent systems than the WDM systems.
a) True
b) False
Answer: a
Explanation: The coherent optical systems involve wavelength selectivity with narrow channel spacing. The conventional WDM systems use a far more relaxed channel spacing than the coherent systems.

3. The technique within the coherent multicarrier systems used to broadcast the optical signals over the network is the use of passive _____________
a) Star coupler
b) Optical resonator
c) Optical regenerator
d) Local oscillator
Answer: a
Explanation: In multicarrier systems, the channels are separated via various techniques. The use of passive star coupler ensures that the optical signals are broadcasted over the network within the coherent system.

4. Estimate the minimum transmitter power if number of photons per bit are 150, wavelength 1.3 micrometer with an optical bandwidth of 20 THz.
a) 0.2 mW
b) 0.5mW
c) 1 mW
d) 2.3mW
Answer: b
Explanation: The minimum transmitter power is given by –
P_tx = N_p hf B
Here, P_tx = transmitter power, f = frequency, h = Planck’s constant and B = bandwidth.

5. The performance degradation due to nonlinear phase noise is referred to as _____________
a) Munich effect
b) Linear bipolar effect
c) Gordon Mollenauer effect
d) Delta effect
Answer: c
Explanation: The phase noise is a variant due to the RZ signal. The nonlinear phase noise causes severe performance degradation in terms of bandwidth and frequency.

6. ______________ facilitates the doubling of the feasible spectral efficiency through the transmission of independent information in each of the two orthogonal polarizations.
a) WDM
b) Gordon Mollenauer effect
c) EDFA control
d) POLMUX
Answer: d
Explanation: POLMUX is abbreviated as Polarization multiplexing. It provides a different approach to the multilevel modulation. It requires polarization control at the receiver side.

7. ____________ is a transparent multiplexing technique.
a) POLMUX
b) PDM
c) WDM
d) Munich
Answer: a
Explanation: It is a transparent technique as it is not dominated by polarization mode dispersion or polarization-dependant loss. It provides many advantages to the multilevel modulation format in comparison to the non-POLMUX signals at the same data rate.

8. Which of the following is not a drawback of POLMUX?
a) Polarization-sensitive detection at the receiver
b) Receiver complexity
c) Cross-polarization nonlinearities
d) Multilevel modulation
Answer: d
Explanation: The drawbacks of the POLMUX include receiver complexity, polarization sensitivity at the receiver side and the cross polarization nonlinearities. It is advantageous to the multilevel modulation scheme.

9. A multicarrier modulation format in which there has been growing interest to compensate for impairments in optical fiber transmission systems is _______________
a) OFDM
b) EDM
c) WDM
d) ADM
Answer: a
Explanation: OFDM is abbreviated as orthogonal frequency division multiplexing. It combats both fiber chromatic dispersion and polarization mode dispersion.

10. It is suggested that the technique with high white noise is an attractive option for use in long haul systems.
a) False
b) True
Answer: a
Explanation: Long haul systems require a technique which boosts the distance covered by the signals with less use of carrier signal. The technique which exhibits high spectral density is an attractive option for the long haul systems.

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Optical Communications Multiple Choice Questions on “Integrated Optics and Photonics Technologies”.

1. Integrated technology for optical devices are developed within optical fiber communication.
a) True
b) False
Answer: a
Explanation: Integration of optical devices enable fabrication of the whole system onto a single chip. Integration of such devices has become a confluence of several optical terms.

2. When both active and passive devices are integrated on a single chip, in multilayered form, then these devices are known as _____________
a) IP devices
b) IO devices
c) Wavelength converters
d) Optical parametric amplifiers
Answer: a
Explanation: IP technology enables fabrication of subsystems and systems. This is all realized on a single substrate. The integration on a single chip is done in IP technology.

3. _________ is a further enhancement of ________
a) IP, IO
b) IO, IP
c) IO, wavelength converters
d) IP, wavelength converters
Answer: a
Explanation: IP seems to be a miniaturization process and integration of optical systems on a single chip. IO devices are formed when both active and passive elements are interconnected. Thus, IP is a developed version of IO.

4. Thin transparent dielectric layers on planar substrates are used in _________ and ______ devices.
a) Wavelength converters and amplification devices
b) IP and IO
c) IP and wavelength converters
d) IO and amplification devices
Answer: b
Explanation: IP and IO provide an alternative to conversion of optical signal back to electrical signal. Thin transparent dielectric layers act as optical waveguides to produce small-scale and miniature circuits.

5. __________ did not make significant contribution to earlier optical fiber systems.
a) IO
b) IP
c) Wavelength amplifiers
d) Couplers
Answer: a
Explanation: IO is based on single mode optical waveguides. Thus it is incompatible with multimode fiber systems. Thus, IO has less importance than IP.

6. Side or edge-emitting or conducting optical devices cannot be integrated on same substrate.
a) True
b) False
Answer: b
Explanation: In serial integration of device, different elements of optical chip can be interconnected in a consecutive manner. Thus, integration of side or edge emitting optical devices can be done on a single substrate.

7. Hybrid ________ integration demands _________ IP circuits to be produced on a single substrate.
a) IP, single-layered
b) IO, multilayered
c) IP, multilayered
d) IO, multilayered
Answer: c
Explanation: To gain control of optical signals, elements can be directly attached to IP circuit. Both active and passive devices should be on the same substrate. To make devices compatible with 3d structures of other IP/IO devices, hybrid IP integration demands multilayered IP circuits.

8. Using SOI integration technique __________ components can be coupled to IP devices.
a) Passive
b) Layered
c) Demounted
d) Active
Answer: d
Explanation: SOI is used to produce micro-waveguide bends and couplers thereby maintaining compatibility with silicon fabrication techniques. Thus, active components like optical sources, detectors can be coupled to other IP devices using SOI technique.

9. Who invented the IO technology?
a) Albert Einstein
b) Anderson
c) M.S Clarke
d) Robert
Answer: b
Explanation: The birth of IO can be traced back to the basic ideas outlined by Anderson in 1966. He suggested the micro-fabrication technology which in turn led to the term integrated optics in 1969.

10. Electronic circuits have a practical limitation on speed of operation at a frequency of around _________
a) 10¹⁰Hz
b) 10¹²Hz
c) 10¹⁴Hz
d) 10¹¹Hz
Answer: a
Explanation: The speed of operation of electronic devices or circuits results from their use of metallic conductors to transport electronic charges and build up signals. It has a limitation to speed of operation of frequency around 10¹⁰Hz.

11. The use of light as an electromagnetic wave of high frequency provides high speed operation around ____________ times the conceivable employing electronic circuits.
a) 10⁸Hz
b) 10⁵Hz
c) 10⁶Hz
d) 10⁴Hz
Answer: d
Explanation: The use of light with its property as an electromagnetic wave offers the possibility of high speed operation. For this, the frequency should be high as 10¹⁴to 10¹⁵Hz.

12. How many layers are possessed by waveguide structures of silica-on-silicon(SOS)?
a) Two
b) Three
c) Four
d) One
Answer: b
Explanation: The SOS is a part of IP technology. The waveguide structures provided by it comprises of three layers. They are buffer, the core and the cladding.

13. The ________________ is a versatile solution-based technique for making ceramic and glass materials.
a) SOL gel process
b) SSL gel process
c) SDL gel process
d) SAML gel process
Answer: a
Explanation: The SOL gel process involves the transition of system from a liquid to a gel. The SOL gel process along with SOS technique is used for the fabrication of ceramic fibers, film coatings and waveguide based optical amplifiers.

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Optical Communications Multiple Choice Questions on “Device Types”.

1. ____________ converts the received optical signal into an electrical signal.
a) Detector
b) Attenuator
c) Laser
d) LED
Answer: a
Explanation: A detector is an essential component of an optical fiber communication system. It dictates the overall system performance. Its function is to convert optical signal into an electrical signal. This electrical signal is then amplified before further processing.

2. The first generation systems of optical fiber communication have wavelengths between ___________
a) 0.2 and 0.3 μm
b) 0.4 and 0.6 μm
c) 0.8 and 0.9 μm
d) 0.1 and 0.2 μm
Answer: c
Explanation: The first generation systems operated at a bit-rate of 45 Mbps with repeater spacing of 10 km. It operates at wavelengths between 0.8 and 0.9μm. These wavelengths are compatible with AlGaAs laser and LEDs.

3. The quantum efficiency of an optical detector should be high.
a) True
b) False
Answer: a
Explanation: The detector must satisfy stringent requirements for performance and compatibility. The photo detector thus produces a maximum electrical signal for a given amount of optical power; i.e. the quantum efficiency should be high.

4. Which of the following does not explain the requirements of an optical detector?
a) High quantum efficiency
b) Low bias voltages
c) Small size
d) Low fidelity
Answer: d
Explanation: The size of the detector must be small for efficient coupling to the fiber. Also, ideally, the detector should not require excessive bias voltages and currents. The fidelity and quantum efficiency should be high.

5. How many device types are available for optical detection and radiation?
a) One
b) Two
c) Three
d) Four
Answer: b
Explanation: Two types of devices are used for optical detection and radiation. These are external photoemission and internal photoemission devices. External photoemission devices are too bulky and require high voltages for operation. Internal devices provide good performance and compatibility.

6. The ___________ process takes place in both extrinsic and intrinsic semiconductors.
a) Avalanche multiplication
b) External photoemission
c) Internal photoemission
d) Dispersion
Answer: c
Explanation: During intrinsic absorption, the received photons excite electrons from the valence band and towards the conduction band in the semiconductor. Extrinsic absorption involves impurity centers created with the material. Generally, intrinsic absorption is preferred for internal photoemission.

7. ____________ are widely used in first generation systems of optical fiber communication.
a) p-n diodes
b) 4-alloys
c) 3-alloys
d) Silicon photodiodes
Answer: d
Explanation: The first generation systems operates at wavelengths 0.8 and 0.9 μm. Silicon photodiodes have high sensitivity over the 0.8-0.9 μm wavelength band with adequate speed, long term stability. Hence, silicon photodiodes are widely used in first generation systems.

8. Silicon has indirect band gap energy of __________________
a) 1.2 eV
b) 2 eV
c) 1.14 eV
d) 1.9 eV
Answer: c
Explanation: Silicon’s indirect band gap energy of 1.14 eV gives a loss in response above 1.09μm. To avoid this, narrower bandgap materials are used. Hence, silicon’s usefulness is limited to first generation systems and not for second and third generation systems.

9. Which of the following detector is fabricated from semiconductor alloys?
a) Photoconductive detector
b) p-i-n detector
c) Photodiodes
d) Photoemission detectors
Answer: a
Explanation: The detectors fabricated from semiconductor alloys can be used for longer wavelengths. Photoconductive detector and hetero-junction transistor have found favor as a potential detector over a wavelength range of 1.1 to 1.6μm.

10. Silicon photodiodes provide high shunt conductance.
a) True
b) False
Answer: b
Explanation: Semiconductor photodiodes provide best solution for detection in optical fiber communications. Silicon photodiodes have high sensitivity, negligible shunt conductance and low dark current.

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Optical Communications Multiple Choice Questions on “Fiber Connectors”.

1. Demountable fiber connectors are more difficult to achieve than optical fiber splices.
a) True
b) False
Answer: a
Explanation: Fiber connectors must maintain tolerance requirements similar to splices in order to couple light efficiently between the fibers. Also, fiber connectors must accomplish this in a removable fashion. The connector design must allow repeated connection and disconnection without any problems of fiber alignment.

2. What is the use of an index-matching material in the connector between the two jointed fibers?
a) To decrease the light transmission through the connection
b) To increase the light transmission through the connection
c) To induce losses in the fiber
d) To make a fiber dispersive
Answer: b
Explanation: The index-matching material used might be epoxy resin. It increases the light transmission through the connection while keeping dust and dirt from between the fibers. It also provides optimum optical coupling.

3. How many categories of fiber connectors exist?
a) One
b) Three
c) Two
d) Four
Answer: c
Explanation: Fiber connectors are separated into two broad categories. They are butt-jointed connectors and expanded beam connectors. Butt-jointed connectors rely upon alignment of the two fiber ends butted to each other whereas expanded beam connectors uses interposed optics at the joint.

4. The basic ferrule connector is also called as _____________
a) Groove connector
b) Beam connector
c) Multimode connector
d) Concentric sleeve connector
Answer: d
Explanation: The basic ferrule connector is the simplest connector. The ferrules are placed in an alignment sleeve within the connector. The alignment sleeve is concentric which allows the fiber ends to be butt-jointed.

5. What is the use of watch jewel in cylindrical ferrule connector?
a) To obtain the diameter and tolerance requirements of the ferrule
b) For polishing purposes
c) Cleaving the fiber
d) To disperse a fiber
Answer: a
Explanation: Ferrule connectors have a watch jewel in the ferrule end face. It is used instead of drilling of the metallic ferrule end face which takes time. It is used to obtain close diameter and tolerance requirements of the ferrule end face whole easily.

6. The concentricity errors between the fiber core and the outside diameter of the jeweled ferrule are in the range of ___________ with multimode step-index fibers.
a) 1 to 3μm
b) 2 to 6μm
c) 7 to 10μm
d) 12 to 20μm
Answer: b
Explanation: The fiber alignment accuracy of the basic ferrule connector is dependent on the ferrule hole into which the fiber is inserted. The concentricity errors in the range of 2 to 6μm gives insertion losses in the range 1 to 2dB with multimode step index fibers.

7. The typical average losses for multimode graded index fiber and single mode fiber with the precision ceramic ferrule connector are _____________ respectively.
a) 0.3 and 0.5 dB
b) 0.2 and 0.3 dB
c) 0.1 and 0.2 dB
d) 0.4 and 0.7 dB
Answer: b
Explanation: Unlike metal and plastic components, the ceramic ferrule material is harder than the optical fiber. Thus, it is unaffected by grinding and polishing process. This factor enables to provide the low-loss connectors which have low losses as low as 0.2 and 0.3 dB in case of optical fibers.

8. Bi-conical ferrule connectors are less advantageous than cylindrical ferrule connectors.
a) FalseStat
b) True
Answer: a
Explanation: Cylindrical and bi-conical ferrule connectors are assembled in housings to form a multi-fiber configuration. The force needed to insert multiple cylindrical ferrules can be large when multiple ferrules are involved. The multiple bi-conical ferrule connectors are more advantageous as they require less insertion force.

9. In connectors, the fiber ends are separated by some gap. This gap ranges from ____________
a) 0.040 to 0.045 mm
b) 0.025 to 0.10 mm
c) 0.12 to 0.16 mm
d) 0.030 to 0.2mm
Answer: b
Explanation: In connectors, gaps are introduced to prevent them from rubbing against each other and becoming damaged during connector fixing/engagement. The gap ranges from 0.025 to 0.10 mm so as to reduce the losses below 8dB for a particular diameter fiber say 50μm.

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Optical Communications Multiple Choice Questions on “Photonic Crystal Fibers & Attenuation”.

1. Photonic crystal fibers also called as ___________
a) Conventional fibers
b) Dotted fibers
c) Stripped fibers
d) Holey fibers
Answer: d
Explanation: Photonic crystal fibers contain a fine array of air holes running longitudinally down the fiber cladding. The microstructure within the fiber is highly periodic.

2. Conventional optical fibers has more transmission losses than photonic crystal fibers.
a) True
b) False
Answer: a
Explanation: Conventional optical fibers have several hundreds of losses in transmission. Photonic crystal fibers have resulted in reduction in overall transmission losses.

3. Losses in photonic crystal fibers are reduced to a level of ___________
a) 0.1dB/km
b) 0.2dB/km
c) 0.3dB/km
d) 0.4dB/km
Answer: c
Explanation: Conventional fibers have losses of several hundred decibels per km. The invention of photonic crystal tubes has reduced the losses by hundreds of decibels.

4. The high index contrast enables the PCF core to be reduced from around 8 μmin conventional fiber to ___________
a) Less than 1μm
b) More than 5μm
c) More than 3μm
d) More than 2μm
Answer: a
Explanation: PCF’s have a wider range of optical properties in comparison with standard fibers. The lesser the core, more is the intensity of light in the core and enhances the non-linear effects.

5. The periodic arrangement of cladding air holes in photonic band gap fibers provides for the formation of a photonic band gap in the ___________
a) H-plane of fiber
b) E-plane of fiber
c) E-H-plane of fiber
d) Transverse plane of fiber
Answer: d
Explanation: Photonic band gap fibers are a class of micro structured fiber in which periodic arrangement of air holes is required. As a PBG fiber exhibits a 2-dimensional band gap, than the wavelengths within this band gap cannot propagate perpendicular to the fiber axis.

6. In index-guided photonic crystal fiber structure, the dark areas are air holes. What does white areas suggests?
a) Air
b) Silica
c) Water
d) Plasma
Answer: d
Explanation: Index-guided photonic crystal fibers have greater index contrast because the cladding contains air-holes having refractive index 1. Both index guided and conventional fibers arises from the manner in which guided mode interacts with the cladding region.

7. Which is the unit of measurement of attenuation in optical fibers?
a) km
b) dB
c) dB/km
d) Coulomb’s
Answer: c
Explanation: Attenuation is also referred to as transmission loss. Channel attenuation helped to determine the maximum transmission distance prior to signal restoration. Attenuation is usually expressed in logarithmic unit of decibel. It is given by
α_dBL = 10 log₁₀P_i / P_o
Where α_dB = signal attenuation per unit length
P_i & P_o = Input and output power.

8. The optical fiber incurs a loss in signal power as light travels down the fiber which is called as ___________
a) Scattering
b) Attenuation
c) Absorption
d) Refraction
Answer: b
Explanation: When the light is passed through the fiber, it travels a large amount of distance before it starts fading. It needs restoration in the path. This loss or fading is called as Attenuation.

9. If the input power 100μW is launched into 6 km of fiber, the mean optical power at the fiber output is 2μW. What is the overall signal attenuation through the fiber assuming there are no connectors or splices?
a) 15.23dB
b) 16.98dB
c) 17.12dB
d) 16.62dB
Answer: b
Explanation: Signal attenuation is usually expressed in decibels. It is given by
Signal attenuation=10 log₁₀P_i / P_o
Where, P_i & P_o = Input and output power.

10. A device that reduces the intensity of light in optical fiber communications is ___________
a) compressor
b) Optical attenuator
c) Barometer
d) Reducer
Answer: b
Explanation: A compressor compresses the signal before transmission. It does not affect the intensity of light. Optical attenuator is a device that affects the intensity of light and incurs a loss in transmission.

11. A decibel may be defined as the ratio of input and output optical power for a particular optical wavelength.
a) True
b) False
Answer: a
Explanation: Signal attenuation refers to the loss in transmission and it needs a logarithmic unit to express. Decibel is mainly used for comparing two power levels. It has the advantage that the operations of multiplication and division reduce to addition and subtraction.

12. When the input and output power in an optical fiber is 120μW & 3μW respectively and the length of the fiber is 8 km. What is the signal attenuation per km for the fiber?
a) 3dB/km
b) 2dB/km
c) 1dB/km
d) 4dB/km
Answer: b
Explanation: Signal attenuation per unit length is given by
α_dBL = 10 log₁₀P_i / P_o
α_dBL = 16 dB
α_dB = 16 dB/L = 2dB/km.

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Optical Communications Multiple Choice Questions on “Receiver Sensitivities”.

1. ASK with heterodyne detection can achieve same SNR limit as ASK with asynchronous detection.
a) True
b) False
Answer: a
Explanation: The coherent optical systems’ receiver sensitivities depend on the SNR limit and the 3dB improvement in the receiver bandwidth. The ASK with heterodyne and asynchronous detection allows for the same receiver SNR limit.

2. In a ______________ receiver, the analyses of signal and noise phenomena are more complicated than in the IM-DD case.
a) Homodyne
b) Heterodyne
c) Circular
d) Heterogeneous
Answer: b
Explanation: The optical output appearance shows the analysis of signal and noise phenomenon. In a heterodyne receiver, the analyses of signal and noise phenomena are more complicated than in the IM-DD case because the optical detector output appears as an IF signal and not as a baseband signal.

3. FSK modulation is attributed to the use of __________ frequencies unlike ASK modulation.
a) One
b) Three
c) Two
d) Four
Answer: c
Explanation: FSK heterodyne detection has receiver sensitivity of 3dB more than tha ASK. Hence, it is attributed to the use of two frequencies unlike ASK where only one frequency is used.

4. BER of FSK modulation scheme is ________ as/to the ASK modulation scheme.
a) Twice
b) Thrice
c) Unequal
d) Same
Answer: d
Explanation: FSK modulation uses two frequencies. The similar BER can be obtained with the two modulation schemes when the average power is transmitted.

5. ___________ can also be used in place of multilayered filters in the dual filter direct detection FSK receiver.
a) Bragg gratings
b) Ceramic gratings
c) Aluminum arsenide
d) Bragg diodes
Answer: a
Explanation: A dual optical filter uses two multilayered dielectric filters. Bragg gratings are an excellent alternative to the dual dielectric filters as they perform the same function as dual dielectric multilayer filters.

6. FSK synchronous detection is _________ more sensitive than asynchronous heterodyne detection.
a) 0.24 dB
b) 0.45 dB
c) 0.9 dB
d) 0.12 dB
Answer: b
Explanation: During FSK detection, the probability of error is increased with respect to direct detection as a result of amplified noise from the orthogonal polarization. Thus, the synchronous detection is more sensitive than the asynchronous heterodyne detection.

7. The asynchronous heterodyne detection is _________ more sensitive than the dual filter direct detection FSK receiver.
a) 0.9 dB
b) 0.23 dB
c) 0.43 dB
d) 0.40 dB
Answer: d
Explanation: The orthogonal polarization leads to the amplified noise structures and behavior in the FSK receiver. This leads to the loss in the sensitivity and accuracy of the FSK receiver.

8. The use of __________ was undertaken to separate the polarization in an analog to digital conversion.
a) FSK
b) DSP
c) ASK
d) DP-FSK
Answer: b
Explanation: DSP is abbreviated as digital signal processing. Digital signal processing involves the cutting, trimming and manipulating of signals in the process of analog to digital conversion. The signals are converted in the form of numerous pulses.

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