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Communications MCQ Set 1

1. When optical fibers are to be installed in a working environment, the most important parameter to be considered is
a) Transmission property of the fiber
b) Mechanical property of the fiber
c) Core cladding ratio of the fiber
d) Numerical aperture of the fiber

View Answer

Answer: b [Reason:] Nowadays, optical fibers are used alternatively to electric transmission lines. They are installed safely and maintained in all environments including underground areas. This requires mechanical strengthening of fibers in order to ensure proper transmission.

2. It is not important to cover these optical fibers required for transmission. State whether the given statement is true or false.
a) True
b) False

View Answer

Answer: b [Reason:] Unprotected optical fibers have number of losses regarding its strength and durability. Bare glass fibers are brittle and have small cross-section area that makes them highly susceptible to damages while handling and maintenance. Thus, to improve tensile strength, optical fibers should be covered by surrounding them with number of protective layers.

3. Optical fibers for communication use are mostly fabricated from
a) Plastic
b) Silica or multicomponent glass
c) Ceramics
d) Copper

View Answer

Answer: b [Reason:] Silica or a compound of glass are brittle and have almost perfect elasticity until reaching their breaking point. Strength of these materials is high. Thus, optical fibers are fabricated from these materials.

4. An Si-O bond with a Young’s modulus of 9*1010Nm-1 have an elliptical crack of depth 7nm. The surface energy is 2.29 J. Estimate fracture stress for silica fiber.
a) 4.32*109Nm-1
b) 6.32*109Nm-1
c) 5.2*109Nm-1
d) 3*109Nm-1

View Answer

Answer: a [Reason:] For an elliptical crack, the fracture stress is given by- Sf= (2Eγp /πC)1/2 Where Sf = fracture stress γp= surface energy C= depth of crack.

5. Calculate percentage strain at break for a Si-O bond with a fracture strength of 3.52*1010Nm-1 and Young’s modulus of 9 *109Nm-1.
a) 3.1 %
b) 2.8 %
c) 4.5 %
d) 3.9 %

View Answer

Answer: d [Reason:] Young’s modulus is given by- E = Stress/Strain To calculate strain from the above formula, we have to divide stress by Young’s modulus. Therefore, Strain = Stress/ E.

6. Stress corrosion must be considered while designing and testing optical fiber cables. State whether the given statement is true or false.
a) True
b) False

View Answer

Answer: a [Reason:] Stress corrosion means growth of flaws due to stress and water. This occurs as a result of molecular bonds at the tip of crack being attacked by water. Hence, it is important to have a protection against water to avoid stress corrosion.

7. Which statistics are used for calculations of strengths of optical fibers?
a) Edwin statistics
b) Newton statistics
c) Wei-bull statistics
d) Gamma statistics

View Answer

Answer: c [Reason:] Calculations of strengths are conducted using Wei-bull statistics in case of optical fibers. It describes the strength behavior of a system that is dependent on the weakest link of the system. The Wei-bull statistics gives the probability of failure of the optical fiber at a given strength.

8. What does n denotes in the equation given below, if vc is the crack velocity; A is the constant for the fiber material and KI is the strength intensity factor?
vc= AKIn
a) Refractive index
b) Stress corrosion susceptibility
c) Strain
d) Young’s modulus

View Answer

Answer: b [Reason:] The above equation allows estimation of the time to failure of a fiber under stress corrosion conditions. The constant n is called as stress corrosion susceptibility. It is typically in the range of 15 to 50 for a glass.

Communications MCQ Set 2

1. Multimode step index fiber has
a) Large core diameter & large numerical aperture
b) Large core diameter and small numerical aperture
c) Small core diameter and large numerical aperture
d) Small core diameter & small numerical aperture

View Answer

Answer: a [Reason:] Multimode step-index fiber has large core diameter and large numerical aperture. These parameters provides efficient coupling to inherent light sources such as LED’s.

2. A typically structured glass multimode step index fiber shows as variation of attenuation in range of
a) 1.2 to 90 dB km-1at wavelength 0.69μm
b) 3.2 to 30 dB km-1at wavelength 0.59μm
c) 2.6 to 50 dB km-1at wavelength 0.85μm
d) 1.6 to 60 dB km-1at wavelength 0.90μm

View Answer

Answer: c [Reason:] A multimode step index fibers show an attenuation variation in range of 2.6 to 50dBkm-1. The wide variation in attenuation is due to the large differences both within and between the two overall preparation methods i.e. melting and deposition.

3. A multimode step index fiber has a large core diameter of range
a) 100 to 300 μm
b) 100 to 300 nm
c) 200 to 500 μm
d) 200 to 500 nm

View Answer

Answer: a [Reason:] A multimode step index fiber has a core diameter range of 100 to 300μm. This is to facilitate efficient coupling to inherent light sources.

4. Multimode step index fibers have a bandwidth of
a) 2 to 30 MHz km
b) 6 to 50 MHz km
c) 10 to 40 MHz km
d) 8 to 40 MHz km

View Answer

Answer: b [Reason:] Multimode step index fibers have a bandwidth of 6 to 50 MHz km. These fibers with this bandwidth are best suited for short -haul, limited bandwidth and relatively low-cost application.

5. Multimode graded index fibers are manufactured from materials with
a) Lower purity
b) Higher purity than multimode step index fibers.
c) No impurity
d) Impurity as same as multimode step index fibers.

View Answer

Answer: b [Reason:] Multimode graded index fibers have higher purity than multimode step index fiber. To reduce fiber losses, these fibers have more impurity.

6. The performance characteristics of multimode graded index fibers are
a) Better than multimode step index fibers.
b) Same as multimode step index fibers.
c) Lesser than multimode step index fibers
d) Negligible

View Answer

Answer: a [Reason:] Multimode graded index fibers use a constant grading factor. Performance characteristics of multimode graded index fibers are better than those of multimode step index fibers due to index graded and lower attenuation.

7. Multimode graded index fibers have overall buffer jackets same as multimode step index fibers but have core diameters
a) Larger than multimode step index fibers.
b) Smaller than multimode step index fibers.
c) Same as that of multimode step index fibers.
d) Smaller than single mode step index fibers.

View Answer

Answer: b [Reason:] Multimode graded index fibers have smaller core diameter than multimode step index fibers. A small core diameter helps the fiber gain greater rigidity to resist bending.

8. Multimode graded index fibers with wavelength of 0.85μm have numerical aperture of 0.29 have core/cladding diameter of
a) 62.5 μm/125 μm
b) 100μm/140 μm
c) 85 μm/ 125 μm
d) 50 μm/ 125μm

View Answer

Answer: b [Reason:] Multimode graded index fibers with numerical aperture 0.29 having a core / cladding diameter of 100μm/ 140μm. They provide high coupling frequency LED’s at a wavelength of 0.85 μm and have low cost. They are also used for short distance application.

9. Multimode graded index fibers use incoherent source only. State whether the following statement is true or false.
a) True
b) False

View Answer

Answer: b [Reason:] Multimode graded index fibers are used for short haul and medium to high bandwidth applications. Small haul applications require LEDs and low accuracy lasers. Thus either incoherent or incoherent sources like LED’s or injection laser diode are used.

10. In single mode fibers, the most beneficial index profile is
a) Step index
b) Graded index
c) Step and graded index
d) Coaxial cable

View Answer

Answer: b [Reason:] In single mode fibers, graded index profile is more beneficial as compared to step index. This is because graded index profile provides dispersion-modified-single mode fibers.

11. The fibers mostly not used nowadays for optical fiber communication system are
a) Single mode fibers
b) Multimode step fibers
c) Coaxial cables
d) Multimode graded index fibers

View Answer

Answer: a [Reason:] Single mode fibers are used to produce polarization maintaining fibers which make them expensive. Also the alternative to them are multimode fibers which are complex but accurate. So, single-mode fibers are not generally utilized in optical fiber communication.

12. Single mode fibers allow single mode propagation; the cladding diameter must be at least
a) Twice the core diameter
b) Thrice the core diameter
c) Five times the core diameter
d) Ten times the core diameter

View Answer

Answer: d [Reason:] The cladding diameter in single mode fiber must be ten times the core diameter. Larger ratios contribute to accurate propagation of light. These dimension ratios must be there so as to avoid losses from the vanishing fields.

13. A fiber which is referred as non-dispersive shifted fiber is
a) Coaxial cables
b) Standard single mode fibers
c) Standard multimode fibers
d) Non zero dispersion shifted fibers

View Answer

Answer: b [Reason:] A standard single mode fiber having step index profile is known as non-dispersion shifted fiber. As these fibers have a zero dispersion wavelength of 1.31μm and so are preferred for single-wavelength transmission in O-band.

14. Standard single mode fibers (SSMF) are utilized mainly for operation in
a) C-band
b) L-band
c) O-band
d) C-band and L-band

View Answer

Answer: c [Reason:] SSMFs are utilized for operation in O-band only. It shows high dispersion in range of 16 to 20ps/nm/km in C-band and L-band. So SSMFs are used in O-band.

15. Fiber mostly suited in single-wavelength transmission in O-band is
a) Low-water-peak non dispersion-shifted fibers
b) Standard single mode fibers
c) Low minimized fibers
d) Non-zero-dispersion-shifted fibers

View Answer

Answer: b [Reason:] Standard single mode fibers with a step index profile is called non dispersion shifted fiber and it is particularly used for single wavelength transmission in O-band and as if has a zero-dispersion wavelength at 1.31μm.

Communications MCQ Set 3

1. An FBG is developed within a fiber core having a refractive index of 1.30. Find the grating period for it to reflect an optical signal with a wavelength of 1.33μm.
a) 0.51 μm
b) 0.58 μm
c) 0.61 μm
d) 0.49 μm

View Answer

Answer: a [Reason:] The grating period is denoted by Λ.It is given by- Λ=λB/ 2n Where λB= wavelength n= refractive index.

2. It is a passive device which allows the flow of optical signal power in only one direction and preventing reflections in the backward direction.
a) Fiber slice
b) Optical fiber connector
c) Optical isolator
d) Optical coupler

View Answer

Answer: c [Reason:] Ideally, an optical isolator transmits the signal power in the desired forward direction. Material imperfections in the isolator medium generate backward reflections. Optical isolators can be implemented by using FBG.

3. Which feature of an optical isolator makes it attractive to use with optical amplifier?
a) Low loss
b) Wavelength blocking
c) Low refractive index
d) Attenuation

View Answer

Answer: b [Reason:] Optical isolators are made using FBGs. Since FBGs are wavelength dependent, the optical isolators can be designed to allow or block the optical signal at particular wavelength. The wavelength blocking feature makes the optical isolator a very attractive device for use with optical amplifier in order to protect them from backward reflections.

4. Magneto-optic devices can be used to function as isolators. State whether the given statement is true or false.
a) True
b) False

View Answer

Answer: a [Reason:] Magneto-optic devices use the principle of Faraday rotation. It relates the TM mode characteristics and polarization state of an optical signal with its direction of propagation. The rotation of polarization plane is proportional to the intensity of component of magnetic field in the direction of optical signal. Therefore, it is possible to block and divert an optical signal using magnetic properties which is a function of an isolator.

5. How many implementation methods are available for optical isolators?
a) One
b) Four
c) Two
d) Three

View Answer

Answer: d [Reason:] Optical isolators can be implemented using three techniques. These are as follows: -By using FBGs -By using magnetic oxide materials -By using semiconductor optical amplifiers (SOAs).

6. A device which is made of isolators and follows a closed loop path is called as a ____________
a) Circulator
b) Gyrator
c) Attenuator
d) Connector

View Answer

Answer: a [Reason:] Isolator can be connected together to form multiport devices. A circulator is formed from isolators connected together to form a closed circular path. In circulator, the signal continues to travel in closed loop and does not get discarded unlike isolator.

7. The commercially available circulators exhibit insertion losses around ________________
a) 2 dB
b) 0.7 dB
c) 0.2 dB
d) 1 dB

View Answer

Answer: d [Reason:] A number of isolators can be used to implement a circulator. However, as the number of ports increases, the device complexity increases. Hence, three-or four-port circulators are used for optical interconnection with insertion losses around 1 dB and high isolation in the range of 40-50dB.

8. A combination of a FBG and optical isolators can be used to produce non-blocking optical wavelength division add/draw multiplexers. State whether the given statement is true or false.
a) True
b) False

View Answer

Answer: b [Reason:] Optical wavelength divisions add/draw multiplexers can be produced by a combination of a FBG and a circulator. Non-blocking NXM optical wavelengths divisions add/draw multiplexer is produced where N and M denotes the number of wavelength channels and add/drop channels.

Communications MCQ Set 4

1. ____________ limits receiver sensitivity.
a) Noise
b) Depletion layer
c) Avalanche
d) Current

View Answer

Answer: a [Reason:] Receiver noise affects receiver sensitivity. It can dictate the overall system design. The noise can be temperature, environmental factor or due to components.

2. A ____________ performs the linear conversion of the received optical signal into an electric current.
a) Receiver
b) Converter
c) Detector
d) Reflector

View Answer

Answer: c [Reason:] An optical signal is always fed to a detector. A detector is an opto – electronic converter which linearly converts the received optical signal into an electric current.

3. __________ are provided to reduce distortion and to provide a suitable signal shape for the filter.
a) Detector
b) Equalizer
c) Filters
d) Amplifier

View Answer

Answer: b [Reason:] Optical detectors are linear devices. They do not introduce distortion themselves but other components may exhibit nonlinear behaviour. To compensate for distortion, an equalizer is provided in the receiver circuit.

4. A _________ maximizes the received signal-to-noise ratio in the receiver circuitry.
a) Filter
b) Equalizer
c) Detector
d) Reflector

View Answer

Answer: a [Reason:] A filter reduces the noise bandwidth as well as inbounds noise levels. A filter maximizes the received signal-to-noise ratio while preserving the essential features of the signal. It also reduces ISI.

5. ________ can be operated in three connections.
a) Reflectors
b) Diodes
c) LED’s
d) FET’s

View Answer

Answer: d [Reason:] FET’s or bipolar transistors are operated in three useful connections. These are the common emitter, the common base or gate, and the emitter or source follower.

6. How many structures of pre-amplifiers exist?
a) Two
b) Three
c) Four
d) One

View Answer

Answer: b [Reason:] The basic structures of pre-amplifiers are observed in three forms. These are low-impedance, high-impedance and trans-impedance front end preamplifier structures.

7. What is the main factor contributing to the choice of the operational amplifier?
a) Gain
b) Impedance
c) Conductance
d) Gain-Bandwidth product

View Answer

Answer: d [Reason:] A TTL interface stage is always used with the operational amplifier. A device which requires higher accuracy often tend to depend on gain-bandwidth product. The choice of amplifier for receiver accuracy is dependant on gain-bandwidth product.

8. The multiplication factor for the APD varies with the device temperature. State whether the given statement is true or false.
a) True
b) False

View Answer

Answer: a [Reason:] Optimum multiplication factor is required for smooth voltage variance. The multiplication factor for APD varies with the device temperature thus making provision of fine control for bias voltage.

9. How many categories of dynamic gain equalizers are available?
a) One
b) Two
c) Three
d) Four

View Answer

Answer: b [Reason:] Dynamic gain equalizers are categorized into two types. These are single-channel and multichannel equalizers, thus providing operation using single or multiple wavelengths.

10. How many simultaneous channels can be provided in a band DGE(Dynamic gain equalizer)?
a) Six
b) Two
c) Eight
d) Ten

View Answer

Answer: c [Reason:] Generally, eight channels are provided simultaneously in a band DGE. These are for the attenuation purpose of channels along with gain equalization.

Communications MCQ Set 5

1. A device which converts electrical energy in the form of a current into optical energy is called as
a) Optical source
b) Optical coupler
c) Optical isolator
d) Circulator

View Answer

Answer: a [Reason:] An Optical source is an active component in an optical fiber communication system. It converts electrical energy into optical energy and allows the light output to be efficiently coupled into the Optical fiber.

2. How many types of sources of optical light are available?
a) One
b) Two
c) Three
d) Four

View Answer

Answer: c [Reason:] Three main types of optical light sources are available. These are wideband sources, monochromatic incoherent sources. Ideally the optical source should be linear.

3. The frequency of the absorbed or emitted radiation is related to difference in energy E between the higher energy state E2 and the lower energy state E1. State what h stands for in the given equation?
E =E2 – E1 = hf
a) Gravitation constant
b) Planck’s constant
c) Permittivity
d) Attenuation constant

View Answer

Answer: b [Reason:] In the given equation, difference in the energy E is directly proportional to the absorbed frequency (f) where h is used as a constant and is called as Planck’s constant. The value of h is measured in Joules/sec & is given by- h = 6.626×10-34Js.

4. The radiation emission process (emission of a proton at frequency) can occur in __________ ways.
a) Two
b) Three
c) Four
d) One

View Answer

Answer: a [Reason:] The emission process can occur in two ways. First is by spontaneous emission in which the atom returns to the lower energy state in a random manner. Second is by stimulated emission where the energy of a photon is equal to the energy difference and it interacts with the atom in the upper state causing it to return to the lower state along with the creation of a new photon.

5. Which process gives the laser its special properties as an optical source?
a) Dispersion
b) Stimulated absorption
c) Spontaneous emission
d) Stimulated emission

View Answer

Answer: d [Reason:] In Stimulated emission, the photon produced is of the same energy to the one which cause it. Hence, the light associated with stimulated photon is in phase and has same polarization. Therefore, in contrast to spontaneous emission, coherent radiation is obtained. The coherent radiation phenomenon in laser provides amplification thereby making laser a better optical source than LED.

6. An incandescent lamp is operating at a temperature of 1000K at an operating frequency of 5.2×1014 Hz. Calculate the ratio of stimulated emission rate to spontaneous emission rate.
a) 3×10-13
b) 1.47×10-11
c) 2×10-12
d) 1.5×10-13

View Answer

Answer: b [Reason:] The ratio of the stimulated emission rate to the spontaneous emission rate is given by- Stimulated emission rate/ Spontaneous emission rate = 1/exp (hf/KT)-1.

7. The lower energy level contains more atoms than upper level under the conditions of ________________
a) Isothermal packaging
b) Population inversion
c) Thermal equilibrium
d) Pumping

View Answer

Answer: c [Reason:] Under the conditions of thermal equilibrium, the lower energy level contains more atoms than the upper level. To achieve optical amplification, it is required to create a non-equilibrium distribution such that the population of upper energy level is more than the lower energy level. This process of excitation of atoms into the upper level is achieved by using an external energy source and is called as pumping.

8. __________________ in the laser occurs when photon colliding with an excited atom causes the stimulated emission of a second photon.
a) Light amplification
b) Attenuation
c) Dispersion
d) Population inversion

View Answer

Answer: a [Reason:] Laser emits coherent radiation of one or more discrete wavelength. Lasers produce coherent light through a process called stimulated emission. Light amplification is obtained through stimulated emission. Continuation of this process creates avalanche multiplication.

9. A ruby laser has a crystal of length 3 cm with a refractive index of 1.60, wavelength 0.43 μm. Determine the number of longitudinal modes.
a) 1×102
b) 3×106
c) 2.9×105
d) 2.2×105

View Answer

Answer: d [Reason:] The number of longitudinal modes is given by- q = 2nL/λ Where q = Number of longitudinal modes n = Refractive index L = Length of the crystal λ = Peak emission wavelength.

10. A semiconductor laser crystal of length 5 cm, refractive index 1.8 is used as an optical source. Determine the frequency separation of the modes.
a) 2.8 GHz
b) 1.2 GHz
c) 1.6 GHz
d) 2 GHz

View Answer

Answer: c [Reason:] The modes of laser are separated by a frequency internal δf and this separation is given by- δf = c/2nL Where c = velocity of light n = Refractive index L = Length of the crystal.

11. Doppler broadening is a homogeneous broadening mechanism. State whether the given statement is true or false.
a) True
b) False

View Answer

Answer: b [Reason:] Doppler broadening is a inhomogeneous broadening mechanism. In this broadening, the individual groups of atoms have different apparent resonance frequencies. Atomic collisions usually provide homogeneous broadening as each atom in collection has same resonant frequency and spectral spread.

12. An injection laser has active cavity losses of 25 cm-1and the reflectivity of each laser facet is 30%. Determine the laser gain coefficient for the cavity it has a length of 500μm.
a) 46 cm-1
b) 51 cm-1
c) 50 cm-1
d) 49.07 cm-1

View Answer

Answer: d [Reason:] The laser gain coefficient is equivalent to the threshold gain per unit length and is given by – gth= α + 1/L ln (1/r) Where α= active cavity loss L = Length of the cavity r = reflectivity.

13. Longitudinal modes contribute only a single spot of light to the laser output. State whether the given statement is true or false.
a) True
b) False

View Answer

Answer: a [Reason:] Laser emission includes the longitudinal modes and transverse modes. Transverse modes give rise to a pattern of spots at the output. Longitudinal modes give only a spot of light to the output.

14. Considering the values given below, calculate the mode separation in terms of free space wavelength for a laser.
Frequency separation = 2GHz
Wavelength = 0.5 μm.
a) 1.4×10-11
b)1.6×10-12
c) 1×10-12
d) 6×10-11

View Answer

Answer: b [Reason:] The mode separation in terms of free space wavelength is given by- δλ= λ 2/c δf Where δf= frequency separation λ= wavelength c = velocity of light.