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Communications MCQ Set 1

1. The _____________ is photosensitive to act as light gathering element.
a) Base-emitter junction
b) Base-collector junction
c) Collector-emitter junction
d) Base-collector junction and Base-emitter junction

View Answer

Answer: a [Reason:] Base-collector junction is photosensitive in n-p-n phototransistor and act as light gathering element. This light absorbed affects the base current and gives multiplication of primary photocurrent in device.

2. A large secondary current _________________ in n-p-n InGaAs phototransistor is achieved.
a) Between base and collector
b) Between emitter and collector
c) Between base and emitter
d) Plasma

View Answer

Answer: b [Reason:] The photo-generated holes are swept to the base. This increases the forward bias device. This generates secondary current between emitter and collector.

3. _______ emitter-base and collector-base junction capacitances is achieved by use of hetero-structure along with _________ base resistance.
a) Low, high
b) High, low
c) Low, low
d) High, negligible

View Answer

Answer: c [Reason:] In hetero-structure, there is low doping level in emitter and collector which is coupled with heavy doping base. This is due low emitter-base and collector-base junction capacitance and low base resistance. This allows large current gain.

4. A ________ is created by hetero-junction at collector-base junction.
a) Potential barrier
b) Depletion region
c) Parasitic capacitance
d) Inductance

View Answer

Answer: a [Reason:] Potential barrier is created at emitter-base junction by hetero-junction. This eliminates hole junction from base. This is achieved when junction is forward-biased and provides good emitter-base efficiency.

5. Phototransistors based on hetero-junction using _________ material are known as waveguide phototransistors.
a) InGaP
b) InGaAs
c) InGaAsP/ InAlAs
d) ErGaAs

View Answer

Answer: c [Reason:] Phototransistor using InGaAsP/ InAlAs are known as waveguide phototransistors. They function as waveguide phototransistors. They function as high performance photo-detectors at 1.3 micro-meter wavelength. They utilize a passive waveguide layer under active transistor region.

6. A phototransistor has collector current of 18 mA, incident optical power of 128 μW with a wavelength of 1.24 μm. Determine an optical gain.
a) 1.407 *102
b) 19.407 *102
c) 2.407 *102
d) 3.407 *102

View Answer

Answer: a [Reason:] The optical gain is given by- G0=hcIc/λeP0, where h=Planck’s constant, Ic=collector current, λ=wavelength, P0=incident optical power.

7. For a phototransistor having gain of 116.5, wavelength of 1.28 μm, optical power 123μW. Determine collector current.
a) 0.123 mA
b) 0.0149 mA
c) 1.23 mA
d) 0.54 mA

View Answer

Answer: b [Reason:] The collector current is given by- Ic= G0λeP0/ hc, where h=Planck’s constant, Ic=collector current, λ=wavelength, P0=incident optical power.

8. The detection mechanism in the ____________ photo-detector includes inter sub-band transitions.
a) Dwell
b) Set
c) Avalanche
d) Futile

View Answer

Answer: a [Reason:] The inter sub-band transitions are also known as type-2 transitions. It comprises of mini-bands within a single energy band, The detection mechanism in DWELL photo-detector includes inter sub-band transitions.

9. Which of the following is the difference between the n-p-n and conventional bipolar transistor?
a) Electric property
b) Magnetic property
c) Unconnected base
d) Emitter base efficiency

View Answer

Answer: c [Reason:] The n-p-n bipolar transistor differs in the following ways: base is unconnected, base-collector junction is photosensitive as a light gathering element.

10. The n-p-n hetero-junction phototransistor is grown using ______________
a) Liquid-phase tranquilizers
b) Liquid-phase epistaxis
c) Solid substrate
d) Hetero poleax

View Answer

Answer: b [Reason:] The technique LPE consists of a thin layer of n-type collector based on a p-type base layer. Liquid phase epistaxis is used in hetero-junction technology.

11. The _____________ at emitter-base junction gives good emitter base injection efficiency.
a) Homo-junction
b) Depletion layer
c) Holes
d) Hetero-junction

View Answer

Answer: d [Reason:] The hetero-junction at the emitter-base junction effectively eliminates hole injection from the base when the junction is forward biased. This gives good emitter-base injection efficiency.

12. Waveguide phototransistors utilize a ___________ waveguide layer under the _________ transistor region.
a) Active, passive
b) Passive, active
c) Homo, hetero
d) Hetero, homo

View Answer

Answer: b [Reason:] Waveguide phototransistors are based on hetero-junction structure. They function as high-performance photo-detectors and thus utilize a passive waveguide layer under the active transistor region.

13. What is the main benefit of the waveguide structure over conventional hetero-junction phototransistor?
a) High depletion region
b) Depletion width
c) Increased photocurrent, responsivity
d) Low gain

View Answer

Answer: c [Reason:] Waveguide structure offers increased photocurrent. Photocurrent is directly proportional to the responsivity; thus in turn increases responsivity.

14. Waveguide structure provides high quantum efficiency. State whether the given statement is true or false.
a) True
b) False

View Answer

Answer: b [Reason:] Responsivity and quantum efficiency follow a different path. They are indirectly proportional to each other. Thus, in waveguide structure, as the responsivity increases, quantum efficiency remains low.

15. Metal-semiconductor-metal (MSM) photo-detectors are photoconductive detectors. State whether the given statement is true or false.
a) True
b) False

View Answer

Answer: a [Reason:] MSM photo-detectors are the simplest of photo-detectors. It provides the simplest form of photo-detection within optical fiber communications and are photoconductive.

Communications MCQ Set 2

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

View Answer

Answer: a [Reason:] 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

View Answer

Answer: b [Reason:] 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

View Answer

Answer: d [Reason:] 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

View Answer

Answer: d [Reason:] 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

View Answer

Answer: a [Reason:] 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

View Answer

Answer: d [Reason:] 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

View Answer

Answer: a [Reason:] 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

View Answer

Answer: a [Reason:] 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

View Answer

Answer: b [Reason:] 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

View Answer

Answer: c [Reason:] 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

View Answer

Answer: b [Reason:] The length of SOA is determined by L = Gs(dB)/10×g×loge=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

View Answer

Answer: c [Reason:] The net gain coefficient of SOA is given by g=L×10×loge/Gs(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

View Answer

Answer: a [Reason:] The noise power spectral density Past is Past = mnsp(Gs-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

View Answer

Answer: a [Reason:] The mode number is determined by m = Past/ nsp(Gs-1) hfB = 1.18×10-3/ 4(1000-1)×6.63×10-34×1.94×1014×1.3×1012 = 1.53 ×10-34.

Communications MCQ Set 3

1. ___________ provides a series of optical processing functions.
a) Wavelength convertors
b) Wavelength amplifiers
c) Detectors
d) Bi-stable optical devices

View Answer

Answer: d [Reason:] Optical bi-stable devices include optical logic and memory elements, A-D convertors. Their response to light is nonlinear giving the basis of optical communication.

2. ___________ comprise of Fabry-Perot cavity.
a) Wavelength convertors
b) Wavelength amplifiers
c) Bi-stable optical devices
d) Detectors

View Answer

Answer: c [Reason:] Fabry-Perot cavity consists of a material in which there are variations in refractive index with optical intensity. These variations are nonlinear giving rise to bistability.

3. The optical path length in nonlinear medium is integer number of ______ wavelength.
a) Half
b) Double
c) Three-fourth
d) Single

View Answer

Answer: a [Reason:] Fabry-Perot cavity exhibits a sharp resonance to optical power passing into and through it. This is achieved when optical path length is integer number of half wavelength in nonlinear medium.

4. As compared to laser, the value of _________ in the cavity controls the optical transmission.
a) Amplification
b) Refractive index
c) Rectification
d) Reflection

View Answer

Answer: b [Reason:] The refractive index value in the Fabry-Perot cavity controls the optical transmission. This provides high optical output on resonance and low optical output off resonance.

5. ___________ are able to latch between two distinct optical states.
a) a) Wavelength convertors
b) Wavelength amplifiers
c) Detectors
d) Bistable optical devices

View Answer

Answer: d [Reason:] The transfer characteristic for Bistable optical devices’ exhibit two state hysteresis resulting from turning in and out of resonance. So they can be latched between two states responding to external signal acting as flip-flop.

6. __________ can act as AND, OR, NOT gate.
a) Wavelength convertors
b) Wavelength amplifiers
c) Detectors
d) Bistable optical devices

View Answer

Answer: d [Reason:] BOD’s exhibit 2-state hysteresis. Thus they are able to latch between two operating states (0 and 1) thereby providing logic functions.

7. _______ proves superior to _______
a) BOD’s, electronic devices
b) Electronic devices, BOD’s
c) BOD’s, convertors
d) Convertors, BOD’s

View Answer

Answer: a [Reason:] There is also a thing of picosecond switching using only Pico-joules of energy. A BOD comprises of these switching properties. Thus, it proves superior to electronic devices.

8. ________ BOD’s provides optical feedback.
a) Extrinsic
b) Intrinsic
c) Detector
d) Bistable

View Answer

Answer: b [Reason:] All optical or intrinsic devices which utilize a nonlinear optical medium between a pair of partially reflecting mirrors forming a nonlinear etalon in which feedback is provided optical.

9. ___________ devices employ artificial nonlinearity.
a) Extrinsic
b) Intrinsic
c) Hybrid
d) Bistable

View Answer

Answer: c [Reason:] Hybrid devices have artificial nonlinearity in an electro-optic medium in the cavity. This produces variations in refractive index through electro-optic effect.

10. Hybrid devices have limited ________ speed.
a) Switching
b) Planar
c) Curvature
d) Electrical

View Answer

Answer: a [Reason:] Hybrid BOD’s provides flexibility. But at the same time their switching speeds are limited by use of electrical feedback. These devices are interconnected to provide a more complex logic circuit.

11. _______ exhibit optical bistability.
a) Extrinsic lasers
b) Intrinsic lasers
c) Detectors
d) Semiconductor lasers

View Answer

Answer: d [Reason:] Semiconductor lasers have optical bistability. This is due to nonlinearities in absorption, gain, dispersion, wave guiding and the selection of output polarization.

12. ___________ is fabricated with tandem electrode.
a) Full convertor
b) Semiconductor
c) Detector diode
d) Bistable laser diode

View Answer

Answer: d [Reason:] Bistable laser diode is fabricated with tandem electrode. The tandem electrode provides two gain sections. Also it has a loss region between them.

13. Optical pulsing can be obtained using _________
a) BODs
b) WDM
c) Detector
d) Semiconductor

View Answer

Answer: a [Reason:] BODs with a very narrow bi-stable loop can provide optical pulsing. This type of device can be used to shape, clean up and amplify a noisy input pulse.

14. A weak second beam is introduced in _________
a) BOD differential amplifier
b) WDM
c) Detector
d) Semiconductor laser

View Answer

Answer: a [Reason:] A weak second beam in BOD differential amplifier is introduced into the nonlinear optical cavity. This is used to control the resonance and transmission of main beam through effects of its own stored energy.

Communications MCQ Set 4

1. P-n photodiode is forward biased. State whether the given statement is true or false.
a) True
b) False

View Answer

Answer: b [Reason:] p-n photodiode includes p and n regions. The electric field developed across the p-n junction sweeps holes and electrons to p and n regions respectively. P-n photodiode is thus reverse biased due to reverse leakage current.

2. The depletion region must be ____________ to allow a large fraction of the incident light to be absorbed in the device(photodiode).
a) Thick
b) Thin
c) Long
d) Inactive

View Answer

Answer: a [Reason:] In p-n photodiode, intrinsic conditions are created in the depletion region. The depletion region must be thick in order to achieve maximum carrier pair generation. Also, its width must be limited to enhance the speed of operation of the p-n photodiode.

3. The process of excitation of an electron from valence band to conduction band leaves an empty hole in the valence band and is called as
a) Detection
b) Absorption
c) Degeneration of an electron-hole pair
d) Regeneration of an electron-hole pair

View Answer

Answer: d [Reason:] A photon is incident in the depletion region of a device has an energy greater than or equal to the band gap energy of the fabricating material. This will cause excitation of an electron from valence to the conduction band. This creates an empty hole in valence band which is referred to as photo-generation of an electron-hole pair.

4. __________________ always leads to the generation of a hole and an electron.
a) Repulsion
b) Dispersion
c) Absorption
d) Attenuation

View Answer

Answer: c [Reason:] Absorption affects the electron and excites it to some other level say conduction band. This is called as photo-generation as absorption always leads to the generation of hole and electron. This does not mean that both contribute to the electronic transport.

5. The electron hole pairs generated in a photodiode are separated by the ____________
a) Magnetic field
b) Electric field
c) Static field
d) Depletion region

View Answer

Answer: b [Reason:] Electric field separates the electron-hole pairs in a photodiode. The electric field distribution is determined by an internal and an external field component. A reverse bias voltage is usually applied to the p-n photodiode.

6. Electric field in the depletion region should be high. State whether the given statement is true or false.
a) True
b) False

View Answer

Answer: a [Reason:] The electric field in the depletion region is always kept high in order to extract all photogenerated carriers. Only the extracted electron hole pairs contribute to the overall photocurrent.

7. The photocurrent of an optical detector should be __________
a) Less
b) More
c) Linear
d) Non-linear

View Answer

Answer: c [Reason:] A linear relationship must exist between the intensity of the incident light and the photocurrent. This makes the photodiode free of noise. It increases the system performance.

8. How many types of optical detectors are available?
a) One
b) Four
c) Two
d) Three

View Answer

Answer: d [Reason:] Three types of optical detectors are available. These are diodes, photoconductors and photo-transistors. Diodes include p-n photodiodes, p-i-n diodes, avalanche photodiodes and schottky diodes.

Communications MCQ Set 5

1. What is the exception in the similarities between the optical Ethernet and the Ethernet LAN?
a) Physical layer
b) Data-link layer
c) Refractive index
d) Attenuation mechanism

View Answer

Answer: a [Reason:] Optical Ethernet is similar to the conventional Ethernet LAN with the exception of the physical layer. Physical layer includes the flow of data in the form of binary digits. This transmission takes place on the bit level.

2. Which technology is used by optical Ethernet?
a) GP-technology
b) HJ-technology
c) IP-technology
d) GB-technology

View Answer

Answer: c [Reason:] Optical Ethernet is the fourth generation of the Ethernet family. The earlier generations include X.25, Frame Relay and ATM. Unlike these technologies, optical Ethernet uses IP-based technology.

3. When was the Gigabit Ethernet network developed?
a) 1977
b) 1988
c) 1990
d) 2002

View Answer

Answer: b [Reason:] The Gigabit Ethernet (Gbe) network was developed in 1988. It was developed by merging two technologies namely 802.3 Ethernet and ANSI X3T11 fiber channel.

4. Optical Ethernet can operate at the transmission rates as low as ______________
a) 10 M bits per second
b) 40 M bits per second
c) 100 M bits per second
d) 1000 M bits per second

View Answer

Answer: a [Reason:] Usually, high transmission rates define optical Ethernet. The ITU-T Recommendation specifies the physical layer for optical Ethernet. It can operate at transmission rates as low as 10 M bits per second.

5. How many types of optical Ethernet connections are developed?
a) Two
b) One
c) Four
d) Three

View Answer

Answer: d [Reason:] There are three different types of optical Ethernet connections. They are point-to-point, point-to-multipoint, and multipoint-to-multipoint. Multipoint refers to more number of connections on either side.

6. Which type of connection can be used as an Ethernet switch?
a) Point-to-point
b) Multipoint-to-multipoint
c) Multipoint-to-point
d) Point-to-multipoint

View Answer

Answer: b [Reason:] The multipoint-to-multipoint configuration refers to the bus, tree or mesh topology. Such a mesh can be made to work as a switching hub with non-blocking switching features. It facilitates switching between different optical Ethernet users.

7. How many aspects are included in the standard Ethernet protocol?
a) One
b) Two
c) Four
d) Three

View Answer

Answer: c [Reason:] Optical Ethernet follows standard Ethernet protocol. This protocol includes four different aspects: Frame, MAC, signaling components and the physical medium.

8. Which of the following is not included in the Ethernet frame format?
a) MAC
b) Preamble
c) Destination address
d) Source address

View Answer

Answer: a [Reason:] The Ethernet frame format includes preamble, destination and source addresses, length, data and the check sequence. MAC is the protocol which is used for sharing the network nodes.

9. The _______________ provides point-to-point access to a bidirectional single-mode optical fiber.
a) Optical regenerator
b) Optical session
c) Optical distribution node
d) Optical buffer

View Answer

Answer: c [Reason:] The ODN is abbreviated as Optical distribution node. It can access an optical fiber on a point-to-point basis. Thus, a single mode bidirectional optical fiber can be accessed by an Optical distribution node.

10. _______________ is the de-multiplexing technique used to split SONET bandwidth into logical groups.
a) SDH
b) Virtual concatenation
c) STS-1
d) Optical breakdown

View Answer

Answer: b [Reason:] Virtual concatenation (VC) is basically a splitting technique. It can split the SONET bandwidth into groups. These groups may be transported or routed independently.

11. Ethernet switches support multiprotocol label switching. State whether the given statement is true or false.
a) True
b) False

View Answer

Answer: a [Reason:] Ethernet switches support multiprotocol label switching. This feature is desired mainly in MAN. The use of such switches in LAN exceeds the network capabilities.

12. Length field in MAC frame ensures that the frame signals stay on the network in order to detect the frame within the correct time limit. State whether the given statement is true or false.
a) True
b) False

View Answer

Answer: b [Reason:] MAC frame includes Length field to identify the type or length of the network protocols. The data field is used to ensure that the frame signals stay on the network long enough to detect the frame within the desired limit.

13. The ___________ protocol is not used when the Ethernet connections are configured for a full duplex operation.
a) TCP/IP
b) MAC
c) CSMA/CD
d) DTH

View Answer

Answer: c [Reason:] In Ethernet connections, the full-duplex operation situation may lead to an increased frame dropping rate. The dropped frame cannot be detected without collision. Thus, CSMA protocol is not used in full duplex mode.

14. Optical Ethernet provides switching capabilities in layers ________
a) 1 and 2
b) 2 and 3
c) 3 and 4
d) 1 and 4

View Answer

Answer: b [Reason:] Layer 2 and 3 are data link and network layers. IP routing is usually considered to be ⅔ switched network. Thus, unlike conventional LAN, optical Ethernet provides switching capabilities between layers 2 and 3.

15. The ______________ approach can provide interconnection among multiple site locations within 40 km range.
a) 3 Gbe
b) 5 Gbe
c) 1 Gbe
d) 10 Gbe

View Answer

Answer: d [Reason:] The 10Gbe approach uses a 40 km window range. It can be used in LAN’s and MAN. It provides switch-to-switch network within data centers.