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Communications MCQ Set 1

1. What is a fundamental necessity in the fabrication of fibers for light transmission?
a) Same refractive index for both core and cladding.
b) Pump source
c) Material composition of fiber
d) Variation of refractive index inside the optical fiber

View Answer

Answer: d [Reason:] For fabrication of fibers, two different transparent materials to light over a wavelength range of 0.8 to 1.7μm are required. Fiber should exhibit low attenuation, absorption and scattering losses. The variation of refractive indices in a fiber is a necessity for fiber fabrication.

2. Which materials are unsuitable for the fabrication of graded index fiber?
a) Glass-like-materials
b) Mono-crystalline structures
c) Amorphous material
d) Silica based material

View Answer

Answer: b [Reason:] In case of graded index fiber, it is essential that the refractive index of the material is varied by suitable doping with another compatible material. These two materials should have mutual solubility over a wide range of concentration. This is achieved only in glass-like-materials.

3. How many different categories are available for the methods of preparing optical glasses?
a) 1
b) 2
c) 3
d) 4

View Answer

Answer: b [Reason:] The methods of preparing optical glasses are divided into two categories. One is the conventional glass refining technique and other is vapor-phase-deposition method.

4. What is the first stage in liquid-phase-technique?
a) Preparation of ultra-pure material powders
b) Melting of materials
c) Decomposition
d) Crystallization

View Answer

Answer: a [Reason:] In liquid-phase-technique melting, the first stage includes preparation of ultra-pure material powders. These are usually oxides or carbonates which decomposes during glass melting.

5. Which processes are involved in the purification stage in liquid-phase-technique?
a) Filtration, Co-precipitation, Re-crystallization
b) Decomposition, Filtration, Drying
c) Doping, Drying, Decomposition
d) Filtration, Drying, Doping

View Answer

Answer: a [Reason:] the compounds such as oxides and carbonates are formed during the glass melting. The purification accounts for a large proportion of material cost. These compounds are commercially available. The purification involves filtration, co-precipitation, re-crystallization and drying.

6. At what temperature range, does the melting of multi components glass systems takes place?
a) 100-300 degree Celsius
b) 600-800 degree Celsius
c) 900-1300 degree Celsius
d) 1500-1800 degree Celsius

View Answer

Answer: c [Reason:] The glass materials in the powdered form and have relatively low melting point. Thus, the glass materials are melted at relatively low temperatures in the range of 900-1300 degree Celsius.

7. Fiber drawing using preform was useful for the production of graded index fibers. State whether the given statement is true or false.
a) True
b) False

View Answer

Answer: b [Reason:] A technique for producing fine optical fiber waveguides is to make a preform using the rod in the tube process. This technique was useful for the production of step-index fibers with large core diameters. In this technique, achievement of low attenuation is not critical as there is a danger of including bubbles at the core-cladding interface.

8. The minute perturbations and impurities in the fiber drawing process using preform technique can result in very high losses of
a) Between 500 and 1000 dB/km
b) Between 100 and 300 dB/km
c) Between 1200 and 1600 dB/km
d) More than 2000 dB/km

View Answer

Answer: a [Reason:] The minute perturbations and impurities in the fiber include formations of bubbles and involvement of particulate matter. The losses due to such impurities can be between 500 and 1000 dB/km.

9. The liquid-phase melting technique is used for the production of fibers
a) With a core diameter of 50μm.
b) With a core diameter less than 100μm.
c) With a core diameter more than 200μm.
d) With a core diameter of 100μm.

View Answer

Answer: c [Reason:] The multicomponent glass fibers prepared continuously by liquid-phase melting technique have losses in the range of 5 and 20 dB/km at a wavelength of 0.85μm. This method is thus used for preparation of fibers with a large core diameter. Also this technique is used for continuous production of fibers.

10. Graded index fibers produced by liquid-phase melting technique are less dispersive than step-index fibers. State whether the given statement is true or false.
a) True
b) False

View Answer

Answer: a [Reason:] Liquid-phase melting technique does not offer optimum parabolic profile fibers. This parabolic profile yields minimum pulse dispersion. Graded index fibers prepared using liquid-phase melting techniques are less dispersive but do not have the bandwidth-length products of optimum profile fibers.

Communications MCQ Set 2

1. ____________ is the lowest noise amplifier device.
a) Silicon FET
b) Amplifier-A
c) Attenuator
d) Resonator-B

View Answer

Answer: a [Reason:] 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) 107Ohms and less than 108
b) 106Ohms and less than 107
c) 1014Ohms
d) 1023Ohms

View Answer

Answer: c [Reason:] 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. State whether the given statement is true or false.
a) True
b) False

View Answer

Answer: b [Reason:] 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

View Answer

Answer: d [Reason:] 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

View Answer

Answer: c [Reason:] 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. State whether the given statement is true or false.
a) True
b) False

View Answer

Answer: a [Reason:] 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)

View Answer

Answer: b [Reason:] 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) 130Mbits-1
b) 110Mbits-1
c) 120Mbits-1
d) 140Mbits-1

View Answer

Answer: d [Reason:] At 140Mbits-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 140Mbits-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

View Answer

Answer: c [Reason:] 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

View Answer

Answer: a [Reason:] 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

View Answer

Answer: b [Reason:] 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-1giving a sensitivity around -35dBm.

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

View Answer

Answer: d [Reason:] Total capacitance is given by Ct=Cd+Ca. the figure of merit is optimized when Cd=Ca. 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.

Communications MCQ Set 3

1. A measure of amount of optical fiber emitted from source that can be coupled into a fiber is termed as
a) Radiance
b) Angular power distribution
c) Coupling efficiency
d) Power-launching

View Answer

Answer: c [Reason:] Coupling efficiency depends upon the type of fiber attached to the source which should consider the parameters such as numerical aperture, core size, R.I. profile, radiance, core-cladding index difference. All these parameters relate to the performance of the fibers determined by power coupled into the fiber to power emitted by the source. This is called coupling efficiency ηwhich is given by η= PF/Ps Where PF= power coupled into the fiber Ps= power emitted by the source.

2. The ratio r = (n1 – n)/ (n1 – n) indicates
a) Fresnel reflection
b) Reflection coefficient
c) Refraction coefficient
d) Angular power distribution coefficient

View Answer

Answer: b [Reason:] The ratio, r = (n1-n)/ (n1-n) is known as Reflection coefficient. It relates the amplitude of the reflected ray to the amplitude of the incident wave.

3. A GaAs optical source having a refractive index of 3.2 is coupled to a silica fiber having a refractive index of 1.42. Determine Fresnel reflection at interface in terms of percentage.
a) 13.4%
b) 17.4%
c) 17.6%
d) 14.8%

View Answer

Answer: d [Reason:] If the fiber end and the source are in close physical contact, the reflection is given by r = ((n1-n)/ (n1-n))2 Multiplying r by 100, we get the value of r in terms of percentage.

4. A particular GaAs fiber has a Fresnel reflection magnitude of 17.6% i.e. 0.176. Find the power loss between the source and the fiber?
a) 0.86 dB
b) 0.78 dB
c) 0.84 dB
d) 0.83 dB

View Answer

Answer: c [Reason:] The optical losses in decibels at the joint is given by Loss = -10log10(1-r) Where L = loss due to Fresnel reflection R = magnitude of Fresnel reflection.

5. Two joined step index fibers are perfectly aligned. What is the coupling loss of numerical aperture are NAR= 0.26 for emitting fiber?
a) -0.828 dB
b) -0.010 dB
c) -0.32 dB
d) 0.32 dB

View Answer

Answer: b [Reason:] Coupling loss for two joined step index fibers is given by LF(NA)= -10 log (NAR/NAE) 2 Where LF= coupling loss NAR= Numerical aperture of receiving fiber NAE= Numerical aperture of emitting fiber.

6. Two joined graded index fibers that are perfectly aligned have refractive indices αR=1.93for receiving fiber αE= 2.15 for emitting fiber. Calculate the coupling loss
a) 0.23 dB
b) 0.16 dB
c) 0.82 dB
d) 0.76 dB

View Answer

Answer: a [Reason:] Coupling loss for two joined and perfectly aligned graded index fiber is given by LF(α)= -10log10αRE+2) / αER+2) Where LF(α)= Coupling loss αR= refractive index of receiving fiber αE= refractive index of emitting fiber.

7. How many types of misalignments occur when joining compatible fiber?
a) One
b) Two
c) Five
d) Three

View Answer

Answer: d [Reason:] There are three layers of fiber misalignments and they are: Longitudinal, lateral and angular misalignments.

8. Losses caused by factors such as core-cladding diameter, numerical aperture, relative refractive index differences, different refractive index profiles, fiber faults are known as
a) Intrinsic joint losses
b) Extrinsic losses
c) Insertion losses
d) Coupling losses

View Answer

Answer: a [Reason:] There are inherent connection problems while joining fibers. These connection problem cause different losses in the fibers and are called as Intrinsic joint losses.

9. A step index fiber has a coupling efficiency of 0.906 with uniform illumination of all propagation modes. Find the insertion loss due to lateral misalignment?
a) 0.95 dB
b) 0.40 dB
c) 0.42 dB
d) 0.62 dB

View Answer

Answer: c [Reason:] The insertion loss due to lateral misalignment is given by Loss10t=-10log10t η10t Where, Loss10t= insertion loss due to lateral misalignment η10t= Coupling efficiency.

10. A graded index fiber has a parabolic refractive index profile (α=2) and core diameter of 42μm. Estimate an insertion loss due to a 2 μm lateral misalignment when there is index matching and assuming there is uniform illumination of all guided modes only.
a) 0.180
b) 0.106
c) 0.280
d) 0.080

View Answer

Answer: d [Reason:] The misalignment loss (assuming there is uniform illumination of all guided modes) is given by Lt= 0.85(y/a) Where y=lateral misalignment a=core radius.

11. Determine coupling efficiency if the misalignment loss in a graded index fiber is 0.102
a) 0.136
b) 0.898
c) 0.982
d) 0.684

View Answer

Answer: b [Reason:] If the misalignment loss is known, the coupling efficiency is defined by η= 1-Lt Where η= coupling efficiency Lt= misalignment loss.

12. In a single mode fiber, the losses due to lateral offset and angular misalignment are given by 0.20 dB and 0.46 dB respectively. Find the total insertion loss
a) 0.66 dB
b) 0.26 dB
c) 0.38 dB
d) 0.40 dB

View Answer

Answer: a [Reason:] The total insertion loss in a single mode fiber is given by TT = TL+ Ta Where, TT = total insertion loss TL = lateral offset loss Ta = Angular misalignment loss.

13. The intrinsic loss through a multimode fiber joint is independent of direction of propagation. State whether the given statement is true or false
a) True
b) False

View Answer

Answer: b [Reason:] Intrinsic loss is defined as the summation of lateral offset loss and angular misalignment loss. In case of multimode fibers, the intrinsic loss is dependent on the refractive index gradient. The intrinsic loss through a single mode fiber joint is independent of direction of propagation.

Communications MCQ Set 4

1. A technique used for determining the total fiber attenuation per unit length is ________ method.
a) Frank
b) Cut-off
c) cut-back
d) Erlangen

View Answer

Answer: c [Reason:] Fiber attenuation techniques have been developed in order to determine the total fiber attenuation. This fiber attenuation is from both scattering and absorption losses. Cut-back method is used as per unit length medium.

2. The system designer finds greatest interest in the ______________
a) Overall fiber attenuation
b) Fiber dispersion
c) Latitude of the fiber
d) Durability

View Answer

Answer: a [Reason:] Fiber attenuation results from various aspects such as signal degradation and physical factors such as environment. The system designer has to look upon the overall fiber attenuation while the relative magnitude is important for the development team.

3. How many parameters are usually worked upon by the measurement techniques in attenuation?
a) Three
b) Two
c) One
d) Five

View Answer

Answer: b [Reason:] The measurement techniques used to obtain fiber attenuation usually gives two parameters. One is spectral loss characteristic and the other is the spot measurement. The spot measurement is usually referred to as the loss at a single wavelength.

4. What type of a light source is usually present in the cut-back method?
a) Tungsten or xenon
b) LED
c) Laser
d) Photo-sensor

View Answer

Answer: a [Reason:] A cut-back method usually calculates the spectral loss which eventually is used as a parameter to obtain total fiber attenuation. It consists of a white light source which is a tungsten halogen or xenon lamp.

5. The device used to remove any scattered optical power from the core is __________
a) Mode setup terminator
b) Nodal spectrum
c) Mode stripper
d) Attenuator

View Answer

Answer: c [Reason:] A mode stripper is usually incorporated at the output end of the fiber. It removes the optical power that is scattered from the core to cladding down the fiber length.

6. What is the hierarchy of the process at the receiving end of the cut-back technique?
a) Photodiode – photo-detector – lock-in amplifier
b) Photodiode – lock-in amplifier
c) Photodiode – photo-detector – Attenuator
d) Photo-detector – lock-in amplifier – receiver

View Answer

Answer: a [Reason:] The optical power at the receiving end is detected using the p-i-n or avalanche photodiode. The photo-detector is then index matched to the fiber output end. It is then fed to the lock-in amplifier.

7. What is the unit of measurement of the optical attenuation per unit length?
a) dB-km
b) dB/km
c) km/dB
d) V

View Answer

Answer: b [Reason:] The optical fiber attenuation per unit length is measured by unit dB/km. dB refers to the electrical parameter used to calculate the attenuation in the form of losses and spot measurements.

8. Determine the attenuation per kilometer for a fiber whose length is 2 km, output voltage is 2.1 V at a wavelength of 0.85μm. The output voltage increases to 10.7 V when the fiber is cut-back to leave 2 metres. Determine the attenuation per km for the fiber at wavelength 0.85μm.
a) 2.8dB/km
b) 3.1dB/km
c) 3.5dB/km
d) 8dB/km

View Answer

Answer: c [Reason:] The attenuation per kilometer can be obtained by – αdB= 10 log10(V2/V1)/(L1-L2) dB/km where L1-L2=1.998 and V2,V1are output voltages.

9. ___________ are used to allow measurements at a selection of different wavelengths.
a) Diaphragms
b) Spot attenuators
c) Belts
d) Interference filters

View Answer

Answer: d [Reason:] The interference filters are located on a wheel at a length. These allow measurement at different wavelengths simultaneously and are accurate to a known level.

10. Cut-back technique is destructive. State whether the given statement is true or false.
a) True
b) False

View Answer

Answer: a [Reason:] The cut-back method has some drawback. It is suitable for laboratory measurements but is far from ideal for attenuation measurements in the field. Hence it is termed as destructive.

11. Which technology is used by the backscatter measurement method?
a) Refraction
b) Francis flat recovery
c) Optical time domain reflectometry
d) Optical frequency

View Answer

Answer: c [Reason:] Backscatter method for attenuation measurement is the most popular non-destructive method. It uses optical time domain reflectometry and provides best results in laboratory as well as field measurements.

12. ________________ measurements checks the impurity level in the manufacturing process.
a) Material reflectometry
b) Material absorption loss
c) Material attenuation loss
d) Calorimetric loss

View Answer

Answer: b [Reason:] The material absorption loss measurements check the absorption losses. It checks the impurity level in the fiber at the manufacturing stage. Thus, it is efficient than the attenuation measurement methods.

13. _____________ may be achieved by replacing the optical fibers with thin resistance wires.
a) Diffraction
b) Segmentation
c) Calorimetric calibration
d) Electrical calibration

View Answer

Answer: d [Reason:] Electrical measurement measurements are also efficient. The calibration is done by many methods. Electrical calibration involves the use of resistance wires in place of optical fibers.

14. A scattering cell consists of ______ square solar cells called as Tynes cell.
a) Five
b) Four
c) Six
d) Three

View Answer

Answer: c [Reason:] Fiber scattering measurements use scattering cell to detect the light which is scattered. This cell consists of six square solar cells which are called as Tynes cell. It contains index-matching fluid.

15. ___________ removes the light propagating in the cladding.
a) Cladding mode strippers
b) Core strippers
c) Mode enhancers
d) Attenuators

View Answer

Answer: a [Reason:] The inaccuracies in the measurements resulting from the scattered light are removed by cladding mode strippers. These strippers are placed before and after the scattering cell.

Communications MCQ Set 5

1. Demountable fiber connectors are more difficult to achieve than optical fiber splices. State whether the given statement is true or false.
a) True
b) False

View Answer

Answer: a [Reason:] 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

View Answer

Answer: b [Reason:] 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

View Answer

Answer: c [Reason:] 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

View Answer

Answer: d [Reason:] 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

View Answer

Answer: a [Reason:] 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

View Answer

Answer: b [Reason:] 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

View Answer

Answer: b [Reason:] 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. State whether the given statement is true or false.
a) False
b) True

View Answer

Answer: a [Reason:] 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

View Answer

Answer: b [Reason:] 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.