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Communications MCQ Set 1

1. When considering source-to-fiber coupling efficiencies, the ________ is an important parameter than total output power.
a) Numerical aperture
b) Radiance of an optical source
c) Coupling efficiency
d) Angular power distribution

View Answer

Answer: b [Reason:] Radiance is the optical power radiated into a unit solid angle per unit emitting surface area. Since this optical power is dependent on radiance, radiance is much important factor than optical power.

2. It is a device that distributes light from a main fiber into one or more branch fibers.
a) Optical fiber coupler
b) Optical fiber splice
c) Optical fiber connector
d) Optical isolator

View Answer

Answer: a [Reason:] Nowadays, requirements to divide combined optical signals for applications are increasing. Optical fiber coupler is one such device that is used for dividing and combining optical signals. It is generally used in LANs, computer networks etc.

3. Optical fiber couplers are also called as ________________
a) Isolators
b) Circulators
c) Directional couplers
d) Attenuators

View Answer

Answer: c [Reason:] Optical fiber couplers are passive devices. The power transfer in couplers takes place either through the fiber core cross-section by butt jointing the fibers or by using some form of imaging optics between the fibers. It distributes light from one fiber to many fibers and hence it is also called as a directional coupler.

4. How many types of multiport optical fiber couplers are available at present?
a) Two
b) One
c) Four
d) Three

View Answer

Answer: d [Reason:] Multiport optical fiber couplers are subdivided into three types. These are three and four port couplers, star couplers and wavelength division multiplexing (WDM) couplers. These couplers distribute light among the branch fibers with no scattering loss.

5. The optical power coupled from one fiber to another is limited by
a) Numerical apertures of fibers
b) Varying refractive index of fibers
c) Angular power distribution at source
d) Number of modes propagating in each fiber

View Answer

Answer: d [Reason:] When two fibers are coupled to each other, the optical power is limited by number of modes propagating in each fiber. For example, when a fiber propagating with 500 modes is connected to a fiber that propagates only 400 modes, then at maximum, only 80% of power is coupled into the other fiber.

6. ________ couplers combine the different wavelength optical signal onto the fiber or separate the different wavelength optical signal output from the fiber.
a) 3-port
b) 2*2-star
c) WDM
d) Directional

View Answer

Answer: c [Reason:] WDM coupler is abbreviated as wavelength division multiplexing coupler. It is a category of multiport optical fiber couplers. It is designed to permit a number of different peak wavelength optical signals to be transmitted in parallel on a single fiber.

7. How many fabrication techniques are used for 3 port fiber couplers?
a) One
b) Two
c) Three
d) Four

View Answer

Answer: b [Reason:] There are two fabrication techniques available for three port couplers. First is a lateral offset method which relies on the overlapping of the fiber end faces and the other is the semi-transparent mirror method. Using these techniques, three port couplers with both multimode and single-mode fibers can be fabricated.

8. The most common method for manufacturing couplers is ____________
a) Wavelength division multiplexing
b) Lateral offset method
c) Semitransparent mirror method
d) Fused bi-conical taper (FBT) technique

View Answer

Answer: d [Reason:] The FBT technique is basic and simple. In this technique, the fibers are generally twisted together and then spot fused under tension such that the fused section is elongated to form a bi-conical taper structure. A three port coupler can be obtained by removing one of the input fibers.

9. Couplers insertion loss is same as that of excess loss. State whether the given statement is true or false.
a) True
b) False

View Answer

Answer: b [Reason:] Excess loss is defined as the ratio of input power to output power. The insertion loss is defined as the loss obtained for a particular port-to-port optical path. Thus, the insertion loss and excess loss are different in nature.

10. A four-port multimode fiber FBT coupler has 50 μWoptical power launched into port 1. The measured output power at ports 2,3 and 4 are 0.003, 23.0 and 24.5 μW respectively. Determine the excess loss.
a) 0.22 dB
b) 0.33 dB
c) 0.45 dB
d) 0.12 dB

View Answer

Answer: a [Reason:] Excess loss is a ratio of power input to power output of the fiber and it is given by Excess loss = 10log10 P1/(P3+P4) WhereP1, P3, P4 = output power at ports 1,3 and 4 resp.

11. A four-port FBT coupler has 60μW optical power launched into port one. The output powers at ports 2, 3, 4 are 0.0025, 18, and 22 μW respectively. Find the split ratio?
a) 42%
b) 46%
c) 52%
d) 45%

View Answer

Answer: d [Reason:] Split ratio indicates the percentage division of optical power between the outputs ports. It is given by Split ratio =[ P3/(P3+P4)]*100% Where P3 and P4 are output powers at ports 3 and 4 respectively.

12. How many manufacturing methods are used for producing multimode fiber star couplers?
a) Two
b) One
c) Three
d) Five

View Answer

Answer: a [Reason:] The manufacturing methods of star couplers are mixer-rod technique and FBT technique. In the mixer-rod method, a thin platelet of glass is employed, which mixes light from one fiber, dividing it among the outgoing fibers. FBT method involves twisting, heating and pulling of fiber.

13. Calculate the splitting loss if a 30×30 port multimode fiber star coupler has 1 mW of optical power launched into an input port.
a) 13 dB
b) 15 dB
c) 14.77 dB
d) 16.02 dB

View Answer

Answer: c [Reason:] The splitting loss is related to the number of output ports N of a coupler. It is given by- Splitting loss (Star coupler) = 10log10N (dB).

14. A _____________ coupler comprises a number of cascaded stages, each incorporating three or four-port FBT couplers to obtain a multiport output.
a) Star
b) Ladder
c) WDM
d) Three-port

View Answer

Answer: a [Reason:] A star coupler can be realized by constructing a ladder coupler. It consists of many cascaded stages. If a three-port coupler is used, then a ladder coupler does not form symmetrical star coupler. It is a useful device to achieve a multiport output with low insertion loss.

15. A number of three-port single-mode fiber couplers are used in the fabrication of a ladder coupler with 16 output ports. The three-port couplers each have an excess loss of 0.2 dB along with a splice loss of 0.1 dB at the interconnection of each stage. Determine the excess loss.
a) 1.9 dB
b) 1.4 dB
c) 0.9 dB
d) 1.1 dB

View Answer

Answer: d [Reason:] The number of stages M within the ladder design is given by 2M=16. Hence M=4. Thus, excess loss is given by- Excess loss = (M×loss in each 3-port coupler) + (Number of splices×Loss in each stage) Where number of splices = 3 (as the value of M is equal to 4).

Communications MCQ Set 2

1. A permanent joint formed between two different optical fibers in the field is known as a
a) Fiber splice
b) Fiber connector
c) Fiber attenuator
d) Fiber dispersion

View Answer

Answer: a [Reason:] The jointing of two individual fibers is called as fiber splicing. It is used to establish long-haul optical fiber links by joining two small length fibers.

2. How many types of fiber splices are available?
a) One
b) Two
c) Three
d) Four

View Answer

Answer: b [Reason:] Splices are divided into two types depending upon the splicing technique used. These are fusion splicing (welding) and mechanical splicing.

3. The insertion losses of the fiber splices are much less than the Fresnel reflection loss at a butted fiber joint. State whether the given statement is true or false.
a) True
b) False

View Answer

Answer: a [Reason:] The Fresnel reflection loss is usually more because there is no large step change in refractive index with the fusion splice as it forms a continuous fiber connection. Also, some method of index matching tends to be utilized with mechanical splices.

4. What is the main requirement with the fibers that are intended for splicing?
a) Smooth and oval end faces
b) Smooth and square end faces
c) Rough edge faces
d) Large core diameter

View Answer

Answer: b [Reason:] A curved mandrel is used which cleaves the fiber to achieve end preparation. The edges must be smooth and have square face at the end for splicing purpose.

5. In score and break process, which of the following is not used as a cutting tool?
a) Diamond
b) Sapphire
c) Tungsten carbide
d) Copper

View Answer

Answer: d [Reason:] The score and break process is also called as scribe and break. It involves the scribing of the fiber surface under tension with a cutting tool. Copper is not used as a cutting tool.

6. The heating of the two prepared fiber ends to their fusing point with the application of required axial pressure between the two optical fibers is called as
a) Mechanical splicing
b) Fusion splicing
c) Melting
d) Diffusion

View Answer

Answer: b [Reason:] Fusion splicing is also called as welding. It refers to the welding of two fiber ends. It is essential for fusion splicing that the fiber ends are adequately positioned and aligned in order to achieve good continuity of the transmission medium at the junction point.

7. Which of the following is not used as a flame heating source in fusion splicing?
a) Microprocessor torches
b) Ox hydric burners
c) Electric arc
d) Gas burner

View Answer

Answer: d [Reason:] Micro-plasma torches uses argon and hydrogen and alcohol vapor. The most widely used heating source is an electric arc. Thus, gas burner is not used in fusion splicing.

8. The rounding of the fiber ends with a low energy discharge before pressing the fibers together and fusing with a stronger arc is called as
a) Pre-fusion
b) Diffusion
c) Crystallization
d) Alignment

View Answer

Answer: a [Reason:] Pre-fusion involves rounding of fiber ends. It removes the requirement for fiber end preparation which has a distinct advantage in the field environment. It is utilized with multimode fibers giving average splice losses of 0.09dB.

9. _____________ is caused by surface tension effects between the two fiber ends during fusing.
a) Pre-fusion
b) Diffusion
c) Self-alignment
d) Splicing

View Answer

Answer: c [Reason:] The two fiber ends are close but not aligned before fusion. During fusion, the surface tension affects the fiber ends to get aligned. After fusion, they are aligned in such a way that a transmission medium can get a good continuity.

10. Average insertion losses as low as _________ have been obtained with multimode graded index and single-mode fibers using ceramic capillaries.
a) 0.1 dB
b) 0.5 dB
c) 0.02 dB
d) 0.3 dB

View Answer

Answer: a [Reason:] Mechanical techniques for splicing involve the use of an accurately produced rigid tube in which fiber ends are permanently bonded. It utilizes a ceramic capillary in which an epoxy resin is injected through a transverse bore to provide mechanical sealing and index matching. This technique which uses ceramic capillaries provides insertion losses as low as 0.1dB.

11. _____________ are formed by sandwiching the butted fiber ends between a V-groove glass substrate and a flat glass retainer plate.
a) Springroove splices
b) V-groove splices
c) Elastic splices
d) Fusion splices

View Answer

Answer: b [Reason:] In V-groove splices, a V-groove glass substrate is used with a flat glass plate. The name V-groove suggests that the fiber ends are spliced in a V-shape. These splices provide losses as low as 0.01dB.

12. Mean splice insertion losses of 0.05 dB are obtained using multimode graded index fibers with the Springroove splice. State whether the given statement is true or false.
a) True
b) False

View Answer

Answer: a [Reason:] Springroove utilizes a bracket containing two cylindrical pins which act as alignment guide for two fiber ends. An elastic element is used to press the fibers into a groove. The assembly is secured with a drop of epoxy resin. It provides a loss of 0.05 dB and has found a practical use in Italy.

13. Alignment accuracy of the order ___________ is obtained using the three glass rod alignment sleeve.
a) 0.23 μm
b) 0.15 μm
c) 0.05 μm
d) 0.01 μm

View Answer

Answer: c [Reason:] Alignment accuracies as high as 0.05 μmare necessary to obtain low losses. The mode-field diameter for single-mode fiber is in the range 8 to 10μm. The three glass rod alignment provides higher accuracies than rotary splice sleeve.

14. In case of multiple fusion, splice losses using an electric arc fusion device with multimode graded index fiber range from
a) 0.01 to 0.04 dB
b) 0.19 to 0.25 dB
c) 0.12 to 0.15 dB
d) 0.04 to 0.12 dB

View Answer

Answer: d [Reason:] In multiple fusions, an electric arc fusing device allows splicing of 12 fibers simultaneously. It takes a tool time of 6 minutes, which requires only 30 seconds per splice. The splice losses for single mode fiber are of 0.04 dB as maximum whereas for graded index fibers, losses are up to 0.12dB.

Communications MCQ Set 3

1. The spectral dependence on gain is always constant. State whether the given statement is true or false.
a) True
b) False

View Answer

Answer: b [Reason:] The spectral dependence on gain is mostly not constant. Thus the spectral bandwidth for erbium-doped silica fibers is restricted to around 300 GHz.

2. ESA ________ the pumping efficiency of device.
a) Increases
b) Does not affects
c) Reduces
d) Has negligible effect on

View Answer

Answer: c [Reason:] In erbium fiber amplifier photons at pump wavelength promotes the electrons in upper lasing level into a high state of excitation. These electrons decay non-radiate to intermediate levels and then back to upper lasing level thereby reducing pumping efficiency.

3. Signal amplification is obtained in ____________
a) Erbium-doped fluoro-zir-carbonate fiber multimode
b) Rare-earth-doped fiber amplifiers
c) Raman fiber systems
d) Brillouin fiber amplifier

View Answer

Answer: a [Reason:] To avoid excited state absorption (ESA). We should use different glass technology in place by using a 488 nmpump wavelength; erbium-doped multimode fluoro zir carbonate fiber provides gain at 1.525 μmwavelengths.

4. It is possible to construct a single rare-earth-doped fiber amplifier which provides amplification for all-bands. State whether the given statement is true or false.
a) True
b) False

View Answer

Answer: b [Reason:] Each material has different absorption emission properties to absorb energy either in single or multi steps. Also it possesses property to emit light in one or more narrow spectral ranges. Thus we cannot construct a single earth-doped fiber for all bands.

5. _______________ is constructed using erbium-doped glass.
a) An erbium-based micro fiber amplifier
b) Rare-earth-doped fiber amplifiers
c) Raman fiber systems
d) Brillouin fiber amplifier

View Answer

Answer: a [Reason:] As compared to other glass, erbium-based micro fiber amplifier is more advantageous. This amplifier provides high optical gain over just a few centimeters of fiber over many meters.

6. ____________ uses Er3+-doped erbium glass.
a) An erbium-based micro fiber amplifier
b) Rare-earth-doped fiber amplifiers
c) Raman fiber systems
d) Brillouin fiber amplifier

View Answer

Answer: a [Reason:] The erbium-based micro fiber amplifier uses Er3+-doped erbium glass. It supports the doping constructions of erbium ions at high levels as compared to conventional glasses.

7. The most advantageous amplification is ____________
a) An erbium-based micro fiber amplifier
b) Rare-earth-doped fiber amplifiers
c) Raman fiber systems
d) Brillouin fiber amplifier

View Answer

Answer: c [Reason:] As compared to all the amplifications, Raman amplification is more advantageous. It has self-phase matching between pump of signal together with broad gain bandwidth as compared to other nonlinear processes.

8. _________ is also known as lump Raman amplifiers.
a) An erbium-based micro fiber amplifier
b) Rare-earth-doped fiber amplifiers
c) Raman fiber systems
d) Discrete Raman amplifiers

View Answer

Answer: d [Reason:] Discrete Raman Amplifiers are lumped elements. This lumped element is to be inserted in transmission line to provide gain.

9. _______________ extends the pump power into transmission line fiber.
a) An erbium-based micro fiber amplifier
b) Rare-earth-doped fiber amplifiers
c) Raman fiber systems
d) Distributed Raman amplification

View Answer

Answer: d [Reason:] In Distributed Raman amplification, all pump power is confined to lumped element. And it is distributed when the amplification takes place among several kilometers.

10. _____________ are called hybrid Raman amplifier.
a) Lumped and distributed Raman Amplifiers
b) Rare-earth-doped fiber amplifiers
c) Raman fiber systems
d) Distributed Raman amplification

View Answer

Answer: a [Reason:] Lumped and distributed Raman Amplifiers can be combined together to be used in wideband application. This combination increases overall amplified spectral bandwidth.

11. In ___________ the ASE contributes most of noise.
a) An erbium-based micro fiber amplifier
b) Rare-earth-doped fiber amplifiers
c) Raman fiber systems
d) Distributed Raman amplification

View Answer

Answer: d [Reason:] ASE contributes most of noise in Raman Amplification. The common sources of noise include beating of signal with ASE, mixing, self-phase modulation and cross-plane modulation.

12. In _____________ Rayleigh scattering can be reduced.
a) An erbium-based micro fiber amplifier
b) Rare-earth-doped fiber amplifiers
c) Raman fiber systems
d) Distributed Raman amplification

View Answer

Answer: d [Reason:] Rayleigh scattering adverse effects can be reduced in Raman Amplification. This can be done by employing two or more stages of amplification over single stage amplification over fiber.

13. Compute the fiber nonlinear coefficient of a parametric optical amplifier having parametric peak gain of 63.6 dB, signal power of 1.6W, length 520.
a) 2.78×10-2 W-1km-1
b) 9.61×10-3 W-1km-1
c) 3.25×10-3 W-1km-1
d) 5.61×10-4 W-1km-1

View Answer

Answer: b [Reason:] The fiber nonlinear coefficient can be found by γ=Gp(dB)-log10(0.25)/P pl×L× 1/10log10(2.718)2 = 63.6+6/1.6×1.6×520×1/ 8.7 = 9.61×10-3 W-1km-1.

14. Compute signal power for parametric amplifier having length of 500, nonlinear gain coefficient 12.6×10-3 W-1km-1and parametric peak gain of 63.9 dB.
a) 0.245 W
b) 0.012 W
c) 0.19 W
d) 0.342 W

View Answer

Answer: b [Reason:] Signal power is given by Pp=Gp(dB)-log10(0.25)/γL× 1/10log10(2.718)2= 63.9+6/12.6×10-3×1/ 8.7 = 0.012 W.

15. Compute the gain of parametric amplifier having signal power of 1.6W, length of 500, non-linear coefficient of 10.19 * 10-3W-1km-1
a) 34.890
b) 19.15
c) 18.22
d) 16.11

View Answer

Answer: c [Reason:] Quadratic gain is given by- Gp(dB)=10log10(γPplL)2 Where L=length of amplifier Ppl=signal power γ=nonlinear coefficient.

Communications MCQ Set 4

1. Which of the following statements best explain the concept of material absorption?
a) A loss mechanism related to the material composition and fabrication of fiber.
b) A transmission loss for optical fibers.
c) Results in attenuation of transmitted light.
d) Causes of transfer of optical power

View Answer

Answer: a [Reason:] Material absorption is a loss mechanism which results in dissipation of transmitted optical power as heat in a waveguide. It can be caused by impurities or interaction with other components of core.

2. How many mechanisms are there which causes absorption?
a) One
b) Three
c) Two
d) Four

View Answer

Answer: b [Reason:] Absorption is a loss mechanism. It may be intrinsic, extrinsic and also caused by atomic defects.

3. Absorption losses due to atomic defects mainly include-
a) Radiation
b) Missing molecules, oxygen defects in glass
c) Impurities in fiber material
d) Interaction with other components of core

View Answer

Answer: b [Reason:] Atomic defects are imperfections in the atomic structure of fiber material. Atomic structure includes nucleus, molecules, protons etc. Atomic defects thus contribute towards loss of molecules, oxygen etc.

4. The effects of intrinsic absorption can be minimized by-
a) Ionization
b) Radiation
c) Suitable choice of core and cladding components
d) Melting

View Answer

Answer: c [Reason:] Intrinsic absorption is caused by interaction of light with one or more components of the glass i.e. core. Thus, if the compositions of core and cladding are chosen suitably, this effect can be minimized.

5. Which of the following is not a metallic impurity found in glass in extrinsic absorption?
a) Fe2+
b) Fe3+
c) Cu
d) Si

View Answer

Answer: d [Reason:] In the optical fibers, prepared by melting techniques, extrinsic absorption can be observed. It is caused from transition metal element impurities. In all these options, Si is a constituent of glass and it cannot be considered as an impurity to glass itself.

6. Optical fibers suffer radiation losses at bends or curves on their paths. State true or false
a) True
b) False

View Answer

Answer: a [Reason:] Optical fibers suffer radiation losses due to the energy in the bend or curves exceeding the velocity of light in cladding. Hence, guiding mechanism is inhibited, which in turn causes light energy to be radiated from the fiber.

7. In the given equation, state what αr suggests;
a) Radius of curvature
b) Refractive index difference
c) Radiation attenuation coefficients
d) Constant of proportionality

View Answer

Answer: c [Reason:] Above equation represents the fiber loss. This loss is seen at bends and curves as the fibers suffer radiation losses at curves. These radiation losses are represented by a radiation attenuation coefficient (αr).

8. A multimode fiber has refractive indices n1= 1.15, n2=1.11 and an operating wavelength of 0.7μm. Find the radius of curvature?
a) 8.60μm
b) 9.30μm
c) 9.1μm
d) 10.2μm

View Answer

Answer: b [Reason:] The radius of curvature of the fiber bend of a multimode fiber is given by optical-communication-questions-answers-material-absorption-fiber-bend-losses-silicon-glass-fibers-q8 Where, Rc = radius of curvature n1, n2= refractive indices λ= wavelength.

9. A single mode fiber has refractive indices n1=1.50, n2= 2.23, core diameter of 8μm, wavelength=1.5μm cutoff wavelength= 1.214μm. Find the radius of curvature?
a) 12 mm
b) 20 mm
c) 34 mm
d) 36 mm

View Answer

Answer: c [Reason:] The radius of curvature of the fiber bend of a single mode fiber is given by- optical-communication-questions-answers-material-absorption-fiber-bend-losses-silicon-glass-fibers-q9 Where R= radius of curvature, n1, n2= refractive indices, λc= cutoff wavelength, λ= operating wavelength.

10. How the potential macro bending losses can be reduced in case of multimode fiber?
a) By designing fibers with large relative refractive index differences
b) By maintaining direction of propagation
c) By reducing the bend
a) By operating at larger wavelengths

View Answer

Answer: a [Reason:] In the case of multimode fibers, radius of curvature is directly proportional to core refractive index and operating wavelength. In order to reduce the macro bending losses, the operative wavelength must be small and fibers must have large relative refractive index difference. Losses are inversely proportional to refractive index differences.

11. Sharp bends or micro bends causes significant losses in fiber. State true or false
a) True
b) False

View Answer

Answer: a [Reason:] Sharp bends usually have a radius of curvature almost near to the critical radius. The fibers with the radius near to the critical radius cause significant losses and hence they are avoided.

Communications MCQ Set 5

1. How many design considerations are considered while determining the receiver performance?
a) Three
b) Two
c) One
d) Four

View Answer

Answer: a [Reason:] Three main considerations are utilized for determining the receiver performance. Noise performance is a major design consideration providing a limitation to the sensitivity. Other two considerations are bandwidth and dynamic range.

2. FET preamplifiers provide higher sensitivity than the Si-bipolar device. State whether the given statement is true or false.
a) True
b) False

View Answer

Answer: a [Reason:] At low speeds, the FET preamplifiers provide higher sensitivity than the Si-bipolar device. It is apparent that below 10Mbits-1the Si MOSFET preamplifier provides a lower noise performance than GaAs MESFET.

3. What is the abbreviation of HBT?
a) Homo-junction unipolar transistor
b) Homo-junction bipolar transistor
c) Hetero-junction bipolar transistor
d) Hetero-Bandwidth transcendence

View Answer

Answer: c [Reason:] HBT is abbreviated as Hetero-junction bipolar transistor. It comprises a selectively doped hetero-junction FET. It is a high-speed, low-noise transistor device.

4. What type of receivers are used to provide wideband operation, low-noise operation?
a) APD optical receivers
b) Optoelectronic integrated circuits (OEICs)
c) MESFET receivers
d) Trans-impedance front-end receivers

View Answer

Answer: b [Reason:] A strategy for the provision of wideband, low-noise receivers involves the use of p-i-n photodiode detector along with the monolithic integration of the device with semiconductor alloy FETs. It has an operating wavelength of 1.1 to 1.6 μmranges.

5. ___________ circuits extends the dynamic range of the receiver.
a) Monolithic
b) Trans-impedance
c) Automatic Error Control (AEC)
d) Automatic Gain Control (AGC)

View Answer

Answer: d [Reason:] AGC circuit extends the dynamic range by diverting excess photocurrent away from the input of the receiver. The receiver dynamic range is an important performance parameter as it provides a measure of the difference between the sensitivity and its overload level.

6. The sensitivity of the low-impedance configuration is ____________
a) Good
b) Poor
c) Great
d) Same as that of high-impedance configuration

View Answer

Answer: b [Reason:] A receiver saturation level is determined by the value of the photodiode bias resistor. The photodiode bias resistor valve is indirectly proportional to the sensitivity but is directly proportional in low impedance configuration. The low resistor value provides less sensitivity in the low-impedance configuration.

7. What is generally used to determine the receiver performance characteristics?
a) Noise
b) Resistor
c) Dynamic range & sensitivity characteristics
d) Impedance

View Answer

Answer: c [Reason:] Dynamic range and sensitivity characteristics involve a graph of received power level and the value of feedback resistor. The high value of photodiode bias resistor in the high impedance front end causes high sensitivity and a narrow dynamic range. These factors prove useful for determining the performance characteristics of receiver.

8. The __________ technique eliminates the thermal noise associated with the feedback resistor in the trans-impedance front end design.
a) Compensation
b) Resonating impedance
c) Electromagnetic
d) Optical feedback

View Answer

Answer: d [Reason:] The optical feedback strategy proves most useful at low transmission rate. The use of optically coupled feedback has demonstrated dynamic ranges of around 40 dB for p-i-n receivers operating at modest bit rates. It removes thermal noise associated with the feedback resistor.

9. The removal of the feedback resistor in the optical feedback technique allows reciever sensitivity of the order of _______________
a) -54 dBm at 2Mbit/sec
b) -12 dBm at 2Mbit/sec
a) -64 dBm at 2Mbit/sec
a) -72 dBm at 2Mbit/sec

View Answer

Answer: c [Reason:] The removal of feedback resistor in the optical feedback technique allows low noise performance. Low noise performance, in turn, affects sensitivity. The receiver sensitivity gets high of the order of -64 dBm at 2Mbit/sec transmission rates.

10. The optical feedback technique is useful at low transmission rates. State whether the given statement is true or false.
a) True
b) False

View Answer

Answer: a [Reason:] The optical feedback technique is useful at low transmission rates because in this case the feedback resistors employed are smaller than the optimum value for low-noise performance. This is done to maintain the resistor at a practical size of 1MΩ. Large values of feedback resistor limits the dynamic range.

11. How many types of optical amplifier technologies are available.
a) One
b) Three
c) Four
d) Two

View Answer

Answer: d [Reason:] There are two basic optical amplifier technologies available. They are semiconductor optical amplifiers and fiber amplifiers. Both these devices are utilized in the pre-amplification role.

12. The optimum filter bandwidth is typically in the range ________________
a) 0.1 to 0.3 nm
b) 0.5 to 3 nm
c) 0.1 to 0.3 μm
d) 0.5 to 3 μm

View Answer

Answer: b [Reason:] The optimum fiber bandwidth is determined by detector noise, transmission rate and the transmitter chirp characteristics. It is typically in the range of 0.5 to 3 nmas it depends upon the filter insertion loss.

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