Category: Optical Communications Objective Questions

Optical Communications Multiple Choice Questions on “Receiver Structures”.

1. How many circuits are present in an equivalent circuit for the digital optical fiber receiver?
a) Four
b) One
c) Three
d) Two
Answer: a
Explanation: A full equivalent circuit for the digital optical fiber receiver includes four circuits. These are the detector circuit, noise sources, and amplifier and equalizer circuit.

2. __________ compensates for distortion of the signal due to the combined transmitter, medium and receiver characteristics.
a) Amplification
b) Distortion
c) Equalization
d) Dispersion
Answer: c
Explanation: Equalization adjusts the balance between frequency components within an electronic signal. It compensates for distortion of the signal. The distortion may be due to the transmitter, receiver etc.

3. ____________ is also known as frequency-shaping filter.
a) Resonator
b) Amplifiers
c) Attenuator
d) Equalizer
Answer: d
Explanation: Equalizer, often called as frequency-shaping filter has a frequency response inverse to that of the overall system frequency response. In wideband systems, it boosts the high frequency components to correct the overall amplitude of the frequency response.

4. The phase frequency response of the system should be ____________ in order to minimize inter-symbol interference.
a) Non-Linear
b) Linear
c) More
d) Less
Answer: b
Explanation: An equalizer is used as frequency shaping filter. The phase frequency response of the system should be linear to acquire the desired spectral shape for digital systems. This, in turn, minimizes the inter-symbol interference.

5. Noise contributions from the sources should be minimized to maximize the receiver sensitivity.
a) True
b) False
Answer: a
Explanation: Noise sources include transmitter section, medium and the receiver section. As the noise increases, the sensitivity at the receiver section decreases. Thus, noise contributions should be minimized to maximize the receiver sensitivity.

6. How many amplifier configurations are frequently used in optional fiber communication receivers?
a) One
b) Two
c) Three
d) Four
Answer: c
Explanation: Three amplifier configurations are used in optical fiber communication receivers. These are voltage amplifiers, semiconductor optical amplifier and current amplifier. Voltage amplifier is the simplest and most common amplifier configuration.

7. How many receiver structures are used to obtain better receiver characteristics?
a) Two
b) One
c) Four
d) Three
Answer: d
Explanation: The various receiver structures are low-impedance front end, high-impedance front end and trans-impedance front-end. The noise in the trans-impedance amplifier will always exceed than the front end structure.

8. The high-impedance front-end amplifier provides a far greater bandwidth than the trans-impedance front-end.
a) True
b) False
Answer: a
Explanation: The noise in the trans-impedance amplifier exceeds that incurred by the high-impedance amplifier. Hence, the trans-impedance front-end provides a greater bandwidth without equalization than the high-impedance front end.

9. A high-impedance amplifier has an effective input resistance of 4MΩ. Find the maximum bandwidth that may be obtained without equalization if the total capacitance is 6 pF and total effective load resistance is 2MΩ.
a) 13.3 kHz
b) 14.2 kHz
c) 15.8 kHz
d) 13.9 kHz
Answer: a
Explanation: The maximum bandwidth obtained without equalization is given by –
B = 1/2ΠR_TLC_T
Where,
R_TL = Total load resistance
C_T = Total capacitance.

10. A high-input-impedance amplifier has following parameters (Total effective load resistance = 2MΩ, Temperature = 300 K). Find the mean square thermal noise current per unit bandwidth for the high-impedance configuration.
a) 8.9×10^-27A²/Hz
b) 8.12×10^-27A²/Hz
c) 8.29×10^-27A²/Hz
d) 8.4×10^-27A²/Hz
Answer: c
Explanation: the mean square thermal noise current per unit bandwidth for the high-impedance configuration is given by –
i_T²= 4KT/R_TL
Where, K = constant
T = Temperature (Kelvin)
R_TL = total effective load resistance.

11. The mean square thermal noise current in the trans-impedance configuration is _________ greater than that obtained with the high-input-impedance configuration.
a) 30
b) 20
c) 15
d) 10
Answer: b
Explanation: 13 dB noise penalties are incurred with the trans-impedance amplifier over that of the high-input-impedance configuration. It is the logarithmic function of the noise current value. However, the trans-impedance amplifiers can be optimized for noise performance.

12. The major advantage of the trans-impedance configuration over the high-impedance front end is ______________
a) Greater bandwidth
b) Less bandwidth
c) Greater dynamic range
d) Less dynamic range
Answer: c
Explanation: Greater dynamic range is a result of the different attenuation mechanism for the low-frequency components of the signal. This attenuation is obtained in the trans-impedance amplifier through the negative feedback and therefore the low frequency components are amplified by the closed loop. This increases the dynamic range.

13. The trans-impedance front end configuration operates as a __________ with negative feedback.
a) Current mode amplifier
b) Voltage amplifier
c) Attenuator
d) Resonator
Answer: a
Explanation: The trans-impedance configuration overcomes the drawbacks of the high-impedance front end. It utilizes a low-noise, high-input-impedance amplifier with negative feedback. It operates as a current mode amplifier where high impedance is reduced by negative feedback.

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Optical Communications Multiple Choice Questions on “Non – Semiconductor Lasers”.

1. ____________________ lasers are presently the major laser source for optical fiber communications.
a) Semiconductor
b) Non-Semiconductor
c) Injection
d) Solid-state
Answer: c
Explanation: Injection laser coupling using discrete lasers have proved to fruitful. Such lenses provide for relaxation for an alignment tolerances normally required for fiber coupling. Certain non-semiconductor sources are making its lace in the optical fiber communication. At slowly present, injection lasers are mostly used as laser sources.

2. In Nd: YAG lasers, the maximum doping levels of neodymium is ____________
a) 0.5 %
b) 1.5 %
c) 1.8 %
d) 2 %
Answer: b
Explanation: The Nd: YAG laser structure is formed by doping of yttrium- aluminum -garnet (YAG) with neodymium. The energy levels for lasing transition and pumping are provided by neodymium ions. The maximum doping level of neodymium in YAG is around 1.5 %.

3. Which of the following is not a property of Nd: YAG laser that enables its use as an optical fiber communication source?
a) Single mode operation
b) Narrow line-width
c) Long lifetime
d) Semiconductors and integrated circuits
Answer: d
Explanation: Nd: YAG laser is a non-semiconductor laser. It does not include the use of semiconductors and thus cannot take advantage of well-developed technology associated with integrated circuits. Single mode operation, narrow line-width, lifetime are the properties that are useful for optical communication.

4. The Nd: YAG laser has a narrow line-width which is ________________
a) < 0.01 nm
b) > 0.01 nm
c) > 1 mm
d) > 1.6 mm
Answer: a
Explanation: The Nd: YAG laser has several properties which make it an active optical source. One of such properties is its narrow line-width. It is less than 0.01 nm which is useful for reducing dispersion of optical links.

5. The strongest pumping bands is a four level system of Nd: YAG laser at wavelength of range_________________
a) 0.25 and 0.56 nm
b) 0.75 and 0.81 nm
c) 0.12 and 0.23 nm
d) 1 and 2 nm
Answer: b
Explanation: The Nd: YAG laser is a four level system. It consists of number of pumping bands and fluorescent transitions. The strongest pumping bands are the wavelengths of 0.75μm and 0.81μm. and gives lasing transition at 1.064μm and 1.32μm. Single mode emission is usually obtained at these wavelengths.

6. The Nd: YAG laser is costlier than earth-doped glass fiber laser.
a) True
b) False
Answer: a
Explanation: The most important requirement of the Nd: YAG laser is pumping and modulation. These two requirements tend to give a cost disadvantage in comparison with earth-doped glass fiber laser. Also it is easier and less expensive to fabricate glass fiber in earth-doped laser.

7. It is a resonant cavity formed by two parallel reflecting mirrors separated by a mirror separated by a medium such as air or gas is?
a) Optical cavity
b) Wheatstone’s bridge
c) Oscillator
d) Fabry-perot resonator
Answer: d
Explanation: Resonant cavity is formed between two mirrors where fiber core doped with earth ions is placed. This cavity is 250-500 μm long and 5 to 15 μm wide. A Fabry-perot resonator oscillates at resonant frequency for which there is high gain.

8. In a three level system, the threshold power decreases inversely with the length of the fiber gain medium.
a) True
b) False
Answer: b
Explanation: If the imperfection losses are low then in a four level system the threshold power decreases inversely with the length of the fiber gain medium. A three level consists of an optimum length. This optimum length gives the minimum threshold power which is independent of the value of imperfection losses.

9. Which of the following co-dopant is not employed by neodymium and erbium doped silica fiber lasers?
a) Phosphorus pent oxide
b) Germania
c) Nitrogen
d) Alumina
Answer: c
Explanation: Silica based glass fibers are proved to be the best host material till date. These silica fibers are doped with neodymium and erbium. These dopants include co-dopants such as phosphorus pent-oxide, germanium and alumina.

10. Dopants levels in glass fiber lasers are generally ___________
a) Low
b) High
c) Same as that of GRIN rod lens laser
d) Same as that of semiconductor laser
Answer: a
Explanation: Dopant levels are low in glass fibers (nearly 400 parts per million). This is because of increasing in concentration quenching which increases with the doping level. It may cause the reduction in the population of the upper lasing level as well as crystallization within the glass matrix.

11. _______________ fibers include addition of lead fluoride to the core glass in order to raise the relative refractive index.
a) Solid-state
b) GaAs
c) Semiconductor
d) ZBLANP
Answer: d
Explanation: Up-conversion pumping of laser material is used to convert an infrared laser output to a visible laser output. ZBLANP is host material on which laser action at all wavelengths can be obtained by pumping. The relative refractive index is increased by addition of lead fluoride which makes it a very interesting host material.

12. The lasing output of the basic Fabry-perot cavity fiber is restricted to between ____________
a) 1 and 2 nm
b) 5 and 10 nm
c) 3 and 6 nm
d) 15 and 30 nm
Answer: b
Explanation: the gain spectrum of rare earth ions may be seen over a wavelength range of 50 nm. The lasing output will thus be narrow unless the dielectric on the mirror is arranged. Such a narrow line-width is not used for a broadband optical source.

13. In Fabry-perot laser, the lower threshold is obtained by ___________
a) Increasing the refractive index
b) Decreasing the refractive index
c) Reducing the slope efficiency
d) Increasing the slope efficiency
Answer: c
Explanation: The finesse of Fabry-perot cavity provides a measure of its filtering properties. When the finesse is high the splitting ratio is low thus lowering the laser threshold in an optical cavity without mirror. In Fabry-perot laser, mirrors are present and thus lower threshold is obtained by reducing the slope efficiency.

14. When did the non-semiconductor laser developed?
a) 1892
b) 1946
c) 1985
d) 1993
Answer: c
Explanation: Non-semiconductor sources are crystalline and glass wave-guiding structures. They are doped with rare earth ions and are good optical sources. The development of these sources started in the year 1985. Example: Nd: YAG laser.

15. Y₃A_l5 O₁₂ is a molecular formula for _____________
a) Ytterbium aluminate
b) Yttrium oxide
c) Ytterbium oxy-aluminate
d) Yttrium-aluminum garnet
Answer: d
Explanation: The atomic number of Yttrium is 39. It is the base element of Yttrium-aluminum garnet. Y₃A_l5 O₁₂, doped with rare earth ion neodymium to form Nd: YAG laser structure.

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Optical Communications Multiple Choice Questions on “Optical Fibers”.

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
Answer: a
Explanation: 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^-1 at wavelength 0.69μm
b) 3.2 to 30 dB km^-1 at wavelength 0.59μm
c) 2.6 to 50 dB km^-1 at wavelength 0.85μm
d) 1.6 to 60 dB km^-1 at wavelength 0.90μm
Answer: c
Explanation: 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. Multimode step index fiber has a large core diameter of range is ___________
a) 100 to 300 μm
b) 100 to 300 nm
c) 200 to 500 μm
d) 200 to 500 nm
Answer: a
Explanation: 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
Answer: b
Explanation: 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.
Answer: b
Explanation: 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
Answer: a
Explanation: 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
Answer: b
Explanation: 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) 100 μm/140 μm
c) 85 μm/125 μm
d) 50 μm/125μm
Answer: b
Explanation: 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.
a) True
b) False
Answer: b
Explanation: 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, which is the most beneficial index profile?
a) Step index
b) Graded index
c) Step and graded index
d) Coaxial cable
Answer: b
Explanation: 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
Answer: a
Explanation: 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
Answer: d
Explanation: 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
Answer: b
Explanation: 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
Answer: c
Explanation: SSMFs are utilized for operation in O-band only. It shows high dispersion in the 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
Answer: b
Explanation: Standard single mode fibers with a step index profile are 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.

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Optical Communications Multiple Choice Questions on “Wavelength Routing Networks”.

1. Which of the following is used to provide wavelength signal service among the nodes?
a) Regularization
b) Optical enhancing
c) Hopping
d) Pulse breakdown
Answer: c
Explanation: The optical layer is dependent on wavelength. The entire physical interconnected network provides wavelength signal service among the nodes using hopping technique.

2. How many types of hopping are present?
a) Two
b) One
c) Three
d) Four
Answer: a
Explanation: There are two types of hopping. They are single hop and multihop. These techniques provide wavelength dependent service for interconnected physical network among the nodes.

3. How many switching layers are possessed by MG-OXC?
a) Two
b) Three
c) One
d) Six
Answer: b
Explanation: An MG-OXC has three switching layers. They are wavelength cross-connect (WXC), waveband cross-connects (BXC), and fiber cross-connects (FXC). These layers help to terminate the wavebands and individual wavelength channels.

4. _____________ supports a great number of wavelength channels and reduces the number of switches within the optical network.
a) Waveband switching
b) Optical remuneration
c) Optical genesis
d) Wavelength multiplexing
Answer: a
Explanation: Waveband switching reduces the number of ports within the optical network. It reduces the complexity of numerous wavelength-driven channels and makes it efficient.

5. Individual wavelength channels and wavebands are terminated through ________________ layers.
a) WXC and PXC
b) WXC and FXC
c) BXC and FXC
d) WXC and BXC
Answer: d
Explanation: The individual wavelength channels are terminated and the terminated waveband is then de-multiplexed. The de-multiplexing is in the form of individual channels which are sent to WXC layer as inputs.

6. The routing and wavelength assignment problem addresses the core issue of _____________
a) Traffic patterns in a network
b) Wavelength adjustment
c) Wavelength continuity constraint
d) Design problem
Answer: c
Explanation: The routing and wavelength assignment problem includes selecting a suitable path and allocating an available wavelength. These problems fall into two categories of sequential or combinational selections.

7. How many techniques of implementation are there for routing wavelength assignment (RWA)?
a) Two
b) Six
c) Three
d) Four
Answer: a
Explanation: The implementation of RWA can be static and dynamic. This depends on the traffic patterns in the network. Static RWA techniques are semi-permanent and dynamic RWA techniques are random in nature.

8. ____________ deals with establishing the light path in frequently varying traffic patterns.
a) Wavelength routing
b) Wavelength multiplexing
c) Static RWA
d) Dynamic RWA
Answer: d
Explanation: In dynamic RWA, the traffic patterns are not known. Thus, the connection requests are initiated in random fashion. Its random nature depends on the network state at the time of request.

9. Static RWA problem is also known as _____________
a) Routing problem
b) Virtual topology problem
c) Static wavelength problem
d) Light path problem
Answer: b
Explanation: Static RWA problem refers to the connection problems which remain connected for a smaller duration of time. Thus, network resources are assigned to each connection. It is also called as virtual topology design problem.

10. The ___________ provides information about the physical path and wavelength assignment for all active light paths.
a) Network state
b) RWA
c) LAN topology
d) Secluded communication protocol
Answer: a
Explanation: The physical path i.e. route is associated with the routing problem. Each connection is provided with network resources to reduce complexity in functioning. The network state is basically required to provide information related to routing and assignment problems.

11. ________________ plays an important role in determining the blocking probability of a network.
a) CGA algorithm
b) Semi-pristine environment
c) RWA algorithm
d) Pass key protocol
Answer: c
Explanation: RWA algorithm’s efficiency is calculated on the basis of no blocking or lowest blocking probability. It also provides the information about the availability of the path between the source and destination.

12. Wavelength assignment in RWA is independent on the network topology.
a) True
b) False
Answer: b
Explanation: RWA algorithm deals with the wavelength assignment, physical path and blocking probability. Network topology plays a crucial role in the wavelength assignment. The network state and topology enables the RWA algorithm to function smoothly.

13. Static RWA technique is semi-permanent.
a) True
b) False
Answer: a
Explanation: The connections employs in static RWA are semi-permanent but remain active for a relatively longer period of time. The traffic patterns are known in advance and thus the optimization can be done by assigning network resources to each connection.

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Optical Communications Multiple Choice Questions on “Multiplexing Strategies”.

1. What is the full form of ETDM?
a) Electronic tube di-cyclic mechanism
b) Electrical time division multiplexing
c) Emphasis tier division mechanism
d) Electrical tube dielectric medium
Answer: b
Explanation: ETDM is the major baseband digital strategy. It allows for greater exploitation of available fiber bandwidth.

2. The practical limitations of the speed of electronic circuits have been pushed towards operational frequencies around ___________
a) 100 MHz
b) 120 MHz
c) 100GHz
d) 80 Hz
Answer: c
Explanation: The speed of the circuitry in the fiber optic communication plays an important role in its performance. It is pushed around 100 GHz frequency allowing for 100 Gbit/s feasibility.

3. A strategy used for increasing the bitrate of digital optical fiber systems beyond the bandwidth capabilities of the drive electronics is known as ___________
a) Optical time division multiplexing
b) Electrical time division multiplexing
c) Frequency division multiplexing
d) Code division multiplexing
Answer: a
Explanation: OTDM is favourable for long distance transmission of signal. It is designed to push the bitrate of the fiber systems beyond the bandwidth limits to gain performance.

4. ____________ semiconductor laser sources provide low duty cycle pulse streams for subsequent time multiplexing.
a) Diameter preferred
b) Mode locked
c) Divine
d) Depletion
Answer: b
Explanation: Mode locked semiconductor laser sources were used at the transmitter side. They provide effective distribution of time multiplexing providing low duty cycle pulse streams.

5. ______________ are the devices which are employed to eliminate the laser chirp.
a) Optical intensity modulators
b) Demodulators
c) Circulators
d) Optical Isolators
Answer: a
Explanation: Optical intensity modulators eliminate the laser chirp. This laser chirp may result in dispersion of the transmitted pulses as they propagate within the single mode fiber, thus limiting the achievable transmission distance.

6. _____________ provides operation at high transmission rate.
a) Optical intensity modulators
b) Demodulators
c) Circulators
d) Electro-absorption modulators
Answer: d
Explanation: Electro-absorption modulators are employed at the transmitter and receiver sections. They provide operation at high transmission rate and for field trial.

7. In __________ the microwave frequency are modulated with an optical carrier and transmitted using a single wavelength channel.
a) Subcarrier multiplexing
b) TDM
c) FDM
d) Code division multiplexing
Answer: a
Explanation: Optical Subcarrier multiplexing (SCM) is transmitted using a single wavelength channel. It enables multiple broadband signals to be transmitted over single-mode fiber.

8. Which of the following techniques is easy to implement?
a) Amplitude shift keying
b) Phase shift keying
c) Frequency shift keying
d) SCM
Answer: c
Explanation: Frequency shift keying has an advantage of being simple to implement at the modulator as well as demodulator side. It is formed by up converting to a narrowband channel at high frequency employing frequency.

9. Which of the following is the disadvantage of SCM?
a) Source nonlinearity
b) Linearity
c) Distortion
d) Narrow bandwidth
Answer: a
Explanation: The problem associated with SCM is source nonlinearity. The distortion caused by this becomes noticeable when several subcarriers are transmitted from a single optical source.

10. In CATV, the signal must be received with a carrier to noise ratio of between __________
a) 90 and 100 dB
b) 10 and 30 dB
c) 60 and 70 dB
d) 45 and 55 dB
Answer: d
Explanation: The CATV multichannel spectrum tends to minimize the required bandwidth. The carrier to noise ratio must be between to avoid degradation of picture quality.

11. The IF signal can be input to a demodulator to recover the baseband signal.
a) True
b) False
Answer: a
Explanation: The IF signal is obtained through SCM at the receive terminals. The baseband video signal in a CATV is obtained through IF signal by using it with a demodulator input.

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Optical Communications Multiple Choice Questions on “Receiver Noise”.

1. Which are the two main sources of noise in photodiodes without internal gain?
a) Gaussian noise and dark current noise
b) Internal noise and external noise
c) Dark current noise & Quantum noise
d) Gaussian noise and Quantum noise
Answer: c
Explanation: The two main sources of noise in photodiodes without internal gain are dark current noise and quantum noise. They are regarded as shot noise on the photocurrent. These noise are together called as analog quantum noise.

2. The dominating effect of thermal noise over the shot noise in photodiodes without internal gain can be observed in wideband systems operating in the range of ________
a) 0.4 to 0.5 μm
b) 0.8 to 0.9 μm
c) 0.3 to 0.4 μm
d) 0.7 to 0.79 μm
Answer: b
Explanation: When the photodiode is without internal avalanche gain, the detector load resistor and active elements’ thermal noise in the amplifier tends to dominate. It is seen in wideband systems operating in the 0.8 to 0.9 μmwavelength band. This is because the dark currents in the silicon diodes can be made very small.

3. A silicon p-i-n photodiode incorporated in an optical receiver has following parameters:

Quantum efficiency = 70%
Wavelength = 0.8 μm
Dark current = 3nA
Load resistance = 4 kΩ
Incident optical power = 150nW.
Bandwidth = 5 MHz

Compute the photocurrent in the device.
a) 67.7nA
b) 81.2nA
c) 68.35nA
d) 46.1nA
Answer: a
Explanation: The photocurrent is given by
I_p = ηP₀eλ/hc
Where η = Quantum efficiency
P₀ = Incident optical power
e = electron charge
λ = Wavelength
h = Planck’s constant
c = Velocity of light.

4. In a silicon p-i-n photodiode, if load resistance is 4 kΩ, temperature is 293 K, bandwidth is 4MHz, find the thermal noise in the load resistor.
a) 1.8 × 10^-16A²
b) 1.23 × 10^-17A²
c) 1.65 × 10^-16A²
d) 1.61 × 10^-17A²
Answer: d
Explanation: The thermal noise in the load resistor is given by –
i_t² = 4KTB/R_L
Where T = Temperature
B = Bandwidth
R_L = Load resistance.

5. ________________ is a combination of shunt capacitances and resistances.
a) Attenuation
b) Shunt impedance
c) Shunt admittance
d) Thermal capacitance
Answer: c
Explanation: Admittance is a measure of how easily a circuit will allow a current to flow. It is the inverse of impedance and is measured in Siemens. It is a combination of shunt capacitances and resistances.

6. ______________ is used in the specification of optical detectors.
a) Noise equivalent power
b) Polarization
c) Sensitivity
d) Electron movement
Answer: a
Explanation: Noise equivalent power is defined as the amount of incident optical power per unit bandwidth required to produce an output power equal to detector output noise power.
Noise equivalent power is the value of incident power which gives an output SNR of unity.

7. A photodiode has a capacitance of 6 pF. Calculate the maximum load resistance which allows an 8MHz post detection bandwidth.
a) 3.9 kΩ
b) 3.46 kΩ
c) 3.12 kΩ
d) 3.32 kΩ
Answer: d
Explanation: The load resistance is given by-
R_L = 1/2πC_dB
Where
B = Post detection bandwidth
C_d = Input capacitance
R_L = Load resistance.

8. The internal gain mechanism in an APD is directly related to SNR. State whether the given statement is true or false.
a) True
b) False
Answer: a
Explanation: The internal gain mechanism in an APD increases the signal current into the amplifier. This improves the SNR because the load resistance and amplifier noise remains unaffected.

9. ____________ is dependent upon the detector material, the shape of the electric field profile within the device.
a) SNR
b) Excess avalanche noise factor
c) Noise gradient
d) Noise power
Answer: b
Explanation: Excess avalanche noise factor is represented as F (M). Its value depends upon the detector material, shape of electric field profile and holes and electrons inclusion. It is a function of multiplication factor.

10. For silicon APDs, the value of excess noise factor is between _________
a) 0.001 and 0.002
b) 0.5 and 0.7
c) 0.02 and 0.10
d) 1 and 2
Answer: c
Explanation: The excess noise factor (K) is same as that of the multiplication factor. In case of holes, the smaller values of K produce high performance and therefore the performance is achieved when k is small. For silicon APDs, k = 0.02 to 0.10.

11. __________ determines a higher transmission rate related to the gain of the APD device.
a) Attenuation
b) Gain-bandwidth product
c) Dispersion mechanism
d) Ionization coefficient
Answer: b
Explanation: Gain-bandwidth product is defined as Gain multiplied by the bandwidth. Gain is a dimensionless quantity but the gain-bandwidth product is therefore measured in the units of frequency.

12. _________________ APDs are recognized for their high gain-bandwidth products.
a) GaAs
b) Alloy-made
c) Germanium
d) Silicon
Answer: d
Explanation: Silicon APDs possess a large asymmetry of electron and hole ionization coefficient. Thus, they possess high gain-bandwidth products. These APDs do not operate at high transmission rates.

13. APDs do not operate at signal wavelengths between 1.3 and 1.6μm.
a) True
b) False
Answer: a
Explanation: APDs having high gain-bandwidth products do not operate at signal wavelengths between 1.3 and 1.6 μm.Hence, these APDs are not prefered for use in receivers operating at high transmission rates.

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