Optical Communication MCQs – Optical Fiber Systems 1 : Intensity Modulation ( Optical Communication ) MCQs

Optical Communication MCQs – Optical Fiber Systems 1 : Intensity Modulation ( Optical Communication ) MCQs

Latest Optical Communication MCQs

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Optical Communication MCQs – Optical Fiber Systems 1 : Intensity Modulation ( Optical Communication ) MCQs

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The Optical Transmitter Circuit

1. _____________ must be operated in stimulated emission region.
a) Injection laser
b) LED’s
c) Detector
d) Receiver
Answer: a
Explanation: Injection laser is a threshold device. In stimulated emission region, continuous optical output power levels are in the range of 1 to 10mW.

2. Coherent radiation is relatively __________
a) Parabolic
b) Elliptic
c) Directional
d) Rectangular
Answer: c
Explanation: Most of the light output is coupled into optical fibre. This is because of the isotropic distribution of narrow-line width, coherent radiation is directional.

3. _____________ are capable of launching powers between 0.5 and several mW.
a) LED’s
b) Injection laser
c) Attenuator
d) Reflector
Answer: b
Explanation: Coupling efficiency up to 30% may be obtained by placing a fiber close to laser mirror. These can approach 90% with suitable lens and optical coupling arrangements. So they can launch 0.5 to several mW of optical power into fiber.

4. LED’s display good linearity.
a) True
b) False
Answer: a
Explanation: LED’s appear to be suited to analog transmission. This is because of its output which is directly proportional to the drive current.

5. Which behaviour may prove as a limitation for injection lasers and LED’s?
a) Isotropic
b) Radioactive
c) Thermal
d) Photosensitive
Answer: c
Explanation: The thermal behaviour of the injection lasers and the LED’s limits their operation within the optical transmitter. The main problem is caused by the variation of injection laser threshold current.

6. Optical output power from an LED is directly proportional to the device junction temperature.
a) False
b) True
Answer: b
Explanation: Output power is dependent on the junction temperature in case of LED’s. Most LED’s exhibit a decrease in the optical output power following an increase in junction temperature.

7. _____________ from the LED is dependent on the effective minority carrier lifetime in the semiconductor material.
a) Spontaneous emission
b) Stimulated emission
c) Absorption
d) Diffusion
Answer: a
Explanation: The speed of the response of the LED is dictated by the respective emission mechanism. Spontaneous emission is related to the carrier lifetime and hence dictating the speed of response.

8. The _________ of the LED is twice that of the effective minority carrier lifetime.
a) Dwell time
b) Reflection scatters
c) Sensitivity
d) Rise time
Answer: d
Explanation: The response of the optical fiber source is specified in terms of the rise time. This rise time is reciprocally related to the device frequency response.

9. The finite spectral width of the optical source causes ___________
a) Depletion
b) Frequency burst
c) Pulse broadening
d) Efficient reflection
Answer: c
Explanation: The finite spectral width causes pulse broadening due to material dispersion on an optical fiber communication link. This results in a limitation on the bandwidth-length product.

10. The coherent emission from an injection laser has a line width of ________
a) 2 nm
b) 3nm
c) 8 nm
d) 1nm
Answer: d
Explanation: An optical source such as an injection laser is a narrow line width device as compared to the LED. It has a narrow line width of 1 nm or less.

11. Extinction ratio is denoted by symbol __________
a) ε
b) σ
c) β
d) ρ
Answer: a
Explanation: Extinction ratio is defined as the ratio of the optical energy emitted in the 0 bit period to that emitted during the 1 bit period. It is denoted by ε.

12. The use of low impedance driving circuit may increase _____________
a) Noise
b) Width
c) Intensity
d) Switching speed
Answer: d
Explanation: Pulse shaping is usually required to increase the switching speed. However, increased switching speed may be obtained from an LED without a speed-up element by use of a low-impedance driving circuit.

The Optical Receiver Circuit

 

1. ____________ limits receiver sensitivity.
a) Noise
b) Depletion layer
c) Avalanche
d) Current
Answer: a
Explanation: 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
Answer: c
Explanation: An optical signal is always fed to a detector. A detector is an optoelectronic 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
Answer: b
Explanation: 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
Answer: a
Explanation: 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
Answer: d
Explanation: 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
Answer: b
Explanation: 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
Answer: d
Explanation: A TTL interface stage is always used with the operational amplifier. A device that requires higher accuracy often tends to depend on gain-bandwidth product.
The choice of amplifier for receiver accuracy is dependent on gain-bandwidth product.

8. The multiplication factor for the APD varies with the device temperature.
a) True
b) False
Answer: a
Explanation: 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
Answer: b
Explanation: 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
Answer: c
Explanation: Generally, eight channels are provided simultaneously in a band DGE. These are for the attenuation purpose of channels along with gain equalization.

System Design Considerations

1. __________ is the unique property of the glass fiber.
a) Transmission
b) Opaque property
c) Ductile
d) Malleable
Answer: a
Explanation: Glass fibers have a unique property as a transmission medium which enables their use in the communication. The major transmission characteristics are dispersion and attenuation.

2. __________ limits the maximum distance between the optical fiber transmitter and receiver.
a) Attenuation
b) Transmission
c) Equipment
d) Fiber length
Answer: a
Explanation: Attenuation along with dispersion and the conductor size are some of the factors that limit the maximum distance between the optical transmitter and the receiver. The associated constraints within the equipment also affect the distance.

3. The ___________ incorporates a line receiver in order to convert the optical signal into the electrical regime.
a) Attenuator
b) Transmitter
c) Repeater
d) Designator
Answer: c
Explanation: Repeaters are a mediator between transmitter and receiver. The weak signal is strengthened back by the repeaters on its path to the receiver.

4. A regenerative repeater is called as ____________
a) Repetitive repeater
b) Regenerator
c) Attenuator
d) Gyrator
Answer: b
Explanation: When digital transmission techniques are used, the repeater also regenerates the original digital signal in the electrical signal before it is retransmitted as an optical signal via a line transmitter.

 

Integrated Optics And Photonics MCQs

 

5. The wavelength range of __________ will be fruitful for the operating wavelength of the system referring to the system performance.
a) 0.8 – 0.9 μm
b) 1.1 – 2 μm
c) 5.2 – 5.7 μm
d) 3.1 – 3.2 μm
Answer: a
Explanation: It is useful if the operating wavelength of the system is established to range of 0.8-0.9 μm. This will be dictated by the overall requirements for the system performance, cost, etc.

6. How many encoding schemes are used in optical fiber communication system design requirements?
a) Three
b) One
c) Two
d) Four
Answer: c
Explanation: Encoding schemes are used for digital transmission of data. These are bi-phase and delay modulation codes. They are also called as Manchester and Miller codes respectively.

7. In ________ the optical channel bandwidth is divided into non-overlapping frequency bands.
a) Time division multiplexing
b) Frequency division multiplexing
c) Code division multiplexing
d) De-multiplexing
Answer: b
Explanation: In FDM, the non-overlapping frequency bands are divided to the individual frequencies. These individual signals can be extracted from the combined FDM signal by electrical filtering at the receiver terminal.

8. A multiplexing technique which does not involve the application of several message signals onto a single fiber is called as _________
a) Time division multiplexing
b) Frequency division multiplexing
c) Code division multiplexing
d) Space division multiplexing
Answer: d
Explanation: In SDM, each signal channel is carried on a separate fiber within a fiber bundle or multi-fiber cable form. The cross coupling between channels is negligible.

9. Which of the following is not an optical fiber component?
a) Fiber
b) Connector
c) Circulator
d) Detector
Answer: c
Explanation: Circulator is a device used in electromagnetic theory. All others are optical components.

10. ________technique involves an increase in the number of components required.
a) Time division multiplexing
b) Space division multiplexing
c) Code division multiplexing
d) Frequency division multiplexing
Answer: b
Explanation: SDM involves good optical isolation due to the negligible cross coupling between channels. It uses separate fiber and thus requires more number of components.

11. Time division multiplexing is inverse to that of frequency division multiplexing.
a) True
b) False
Answer: a
Explanation: TDM involves distribution of channels in time slots whereas FDM involves bands that are run on different frequencies. Both of these techniques improve accuracy and reduce complexity.

Digital System Planning Considerations

1. Sampling rate for each speech channel on 32-channel PCM is 8 KHz each encoded into 8 bits. Determine number of bits in a frame.
a) 64
b) 128
c) 32
d) 256
Answer: d
Explanation: Number of bits in a frame can be calculated as follows:
Bits in a frame = No. of channels * Sampling rate for each channel.

2. Sampling rate for each speech channel on 32-channel PCM is 8 KHz each encoded into 8 bits. Determine the transmission rate for system with 256 bits in a frame.
a) 2.96 Mbits/s
b) 2.048 Mbits/s
c) 3.92 Mbits/s
d) 4 Mbits/s
Answer: b
Explanation: Transmission rate can be determined by-
Transmission rate = Sampling rate * No. of bits in a frame.

3. Sampling rate for each speech channel on 32-channel PCM is 8 KHz each encoded into 8 bits. Determine the bit duration with transmission rate of 2.048 M bits/s.
a) 388 ns
b) 490 ns
c) 488 ns
d) 540 ns
Answer: c
Explanation: Bit duration is the reciprocal of the transmission rate. Thus, it is given by-
Bit duration = 1/transmission rate.

4. The bit duration is 488 ns. Sampling rate for each channel on 32-channel PCM is 8 KHz encoded into 8 bits. Determine the time slot duration.
a) 3.2 μs
b) 3.1 μs
c) 7 μs
d) 3.9 μs
Answer: d
Explanation: Time slot duration is given by –
Time slot duration = Encoded bits * bit duration.

5. Sampling rate for each speech channel on 32-channel PCM is 8 KHz each encoded into 8 bits. Determine duration of frame with time slot duration of 3.9μs.
a) 125 μs
b) 130 μs
c) 132 μs
d) 133 μs
Answer: a
Explanation: Duration of a frame is determined by –
Duration of a frame = 32 * time slot duration.

6. Sampling rate for each speech channel on 32-channel PCM is 8 KHz each encoded into 8 bits. Determine the duration of multi-frame if duration of a frame is 125μs.
a) 2ms
b) 3ms
c) 4ms
d) 10ms
Answer: a
Explanation: Multi-frame duration can be determined by –
Multi-frame duration = 16 * Duration of a single frame.

7. Determine excess avalanche noise factor F(M) if APD has multiplication factor of 100, carrier ionization rate of 0.02.
a) 3.99
b) 3.95
c) 4.3
d) 4
Answer: b
Explanation: Excess avalanche noise factor is computed by –
F (M) = k*M + (2-1/M) (1-k), where k is ionization rate and M is the multiplication factor.

8. Compute average number of photons incident at receiver in APD if quantum efficiency is 80%, F (M) = 4, SNR = 144.
a) 866
b) 865
c) 864
d) 867
Answer: c
Explanation: Average number of photons arezm=[2βςF(M)]*[S/N*η]
Here, η = quantum efficiency, S/N = signal to noise ratio.

9. Determine incident optical power if zm=864, wavelength = 1μm.
a) -85 dBm
b) -80 dBm
c) -69.7 dBm
d) -60.7 dBm
Answer: d
Explanation: Incident optical power is P0=zmhcBT/2λ. Here zm=average number of photons, hc=Planck’s constant.

10. Determine wavelength of incident optical power if zm=864, incident optical power is -60.7 dB, BT=1 * 107.
a) 1 μs
b) 2 μs
c) 3 μs
d) 4 μs
Answer: a
Explanation: Wavelength is determined by λ=zmhcBT/2P0. Here zm=average number of photons, hc=Planck’s constant, P0=incident optical power.

11. Determine total channel loss if connector loss at source and detector is 3.5 and 2.5 dB and attenuation of 5 dB/km.
a) 34 dB
b) 35 dB
c) 36 dB
d) 38 dB
Answer: a
Explanation: The total channel loss is CL=(αfcj)L + αcr. Here αcr=loss at detector and source combined, αfc = attenuation in dB/km.

12. Determine length of the fiber if attenuation is 5dB/km, splice loss is 2 dB/km, connector loss at source and detector is 3.5 and 2.5.
a) 5 km
b) 4 km
c) 3 km
d) 8 km
Answer: b
Explanation: Length of the fiber is L = CL/(αfcj) – αcr. Here αcr = loss at detector and source combined, αfc = attenuation in dB/km.

13. Determine total RMS pulse broadening over 8 km if RMS pulse broadening is 0.6ns/km.
a) 3.6 ns
b) 4 ns
c) 4.8 ns
d) 3 ns
Answer: c
Explanation: Total RMS pulse broadening is given by –
σT = σ*L Where σ = rms pulse broadening and L = length of the fiber.

14. Determine RMS pulse broadening over 8 km if total RMS pulse broadening is 5.8ns/km.
a) 0.2ns/km
b) 0.1ns/km
c) 0.4ns/km
d) 0.72ns/km
Answer: d
Explanation: RMS pulse broadening is given by –
σ = σT/L where σ = rms pulse broadening and L = length of the fiber.

Analog Systems

1. Determine dispersion equalization penalty if total RMS pulse broadening is 4.8ns, BT is 25 Mbits/s.
a) 0.03 dB
b) 0.08 dB
c) 7 dB
d) 0.01 dB
Answer: a
Explanation: Dispersion equalization penalty is denoted by DL. It is given by-
DL = 2 (2σTBT√2)4. Here σT=RMS pulse broadening.

2. Determine RMS pulse broadening with mode coupling if pulse broadening is 0.6 over 8km.
a) 1.6ns
b) 1.7ns
c) 1.5ns
d) 1.4ns
Answer: b
Explanation: Total RMS pulse broadening with mode coupling is given by-
σT = σ√L. Here σT = RMS pulse broadening, L = length of the fiber.

3. Determine dispersion equalization penalty with mode coupling of 1.7ns if BT is 25 Mbits/s.
a) 4.8 * 104dB
b) 4 * 104dB
c) 4.2 * 104dB
d) 3.8 * 104dB
Answer: c
Explanation: Dispersion equalization penalty is denoted by DL. With mode coupling, it is given by-
DL=2 (2σTBT√2)4. Here σT=RMS pulse broadening.

4. Determine dispersion equalization penalty without mode coupling if BT is 150 Mbits/s and total rms pulse broadening is 4.8ns.
a) 34 dB
b) 33 dB
c) 76.12 dB
d) 34.38 dB
Answer: d
Explanation: Dispersion equalization penalty is denoted by DL(WM). It is given by-
DL(WM) = 2 (2σTBT√2)4. Here σT = RMS pulse broadening, (WM) = without mode coupling.

5. Determine ratio of SNR of coaxial system to SNR of fiber system if peak output voltage is 5V, quantum efficiency of 70%, optical power is 1mW, wavelength of 0.85μm.
a) 1.04 * 104
b) 2.04 * 104
c) 3.04 * 104
d) 4.04 * 104
Answer: a
Explanation: Ratio of SNR of coaxial system to SNR of fiber system is given by-
Ratio = V2hc/2KTZ0ηPiλ. Here, η=quantum efficiency, Pi = 0ptical power in mW, V=optical output voltage.

6. Determine the peak output voltage if efficiency is 70%, wavelength is 0.85μm and output power is 1mW.
a) 7V
b) 8V
c) 5V
d) 6V
Answer: b
Explanation: Peak output voltage is given by-
V2 = (2KTZ0ηPiλ * Ratio)/hc. Here, η = quantum efficiency, Pi=0ptical power in mW, V=optical output voltage.

 

Optical Amplification, Wavelength Conversion And Regulation MCQs

 

7. Determine the efficiency of a coaxial cable system at 17 degree Celsius with peak output voltage 5V, 0.85 μm wavelength and SNR ratio of 1.04 * 104.
a) 80%
b) 70%
c) 40%
d) 60%
Answer: b
Explanation: The efficiency of a coaxial cable system is η=V2hc/2KTZ0ηPiλ * Ratio. Hereη=quantum efficiency, Pi = 0ptical power in mW, V=optical output voltage.

8. Determine the wavelength of a coaxial cable system operating at temperature 17 degree Celsius at output voltage of 5V, 100Ω impedance, optical power of 1mW, 70% quantum efficiency.
a) 0.39μm
b) 0.60μm
c) 0.85μm
d) 0.98μm
Answer: c
Explanation: The wavelength can be determined by –
λ = V2hc/2KTZ0ηPi * Ratio. Hereη=quantum efficiency, Pi = 0ptical power in mW, V = optical output voltage.

9. Determine the impedance of a coaxial cable system operating at temperature 17 degree Celsius at output voltage of 5V, 0.85μmwavelength, optical power of 1mW, 70% quantum efficiency and SNR ratio of 1.04 * 104.
a) 80Ω
b) 50Ω
c) 90Ω
d) 100Ω
Answer: d
Explanation: The impedance is given by-Z0=V2hc/2KTPi * Ratio. Hereη=quantum efficiency, Pi = Optical power in mW, V=optical output voltage.

10. The 10-90% rise times for components used in D-IM analog optical link is given. (LED=10ns, Intermodal=9ns/km, Chromatic=2ns/km, APD = 3ns). Link is of 5km. Determine the total rise time.
a) 62ns
b) 53ns
c) 50ns
d) 52ns
Answer: d
Explanation: Total rise time is given by-
Tsyst=1.1[Ts2+Tn2+Tc2+TD2]1/2. Here Ts = rise time, Tn = intermodal time, Tc = Chromatic time.

11. The 10-90% rise times for components used in D-IM analog optical link is given. (LED=10ns, Intermodal=9ns/km, Chromatic=2ns/km, APD = 3ns). Link is of 5km. It has an optical bandwidth of 6MHz. Determine maximum permitted system rise time.
a) 58.3ns
b) 54ns
c) 75ns
d) 43.54ns
Answer: a
Explanation: The maximum permitted system rise time is given by-
Tsyst(Max) = 0.35/Bopt. Here, Bopt=Optical Bandwidth.

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.

Application of Optical Amplifiers

1. Which of the following is not a drawback of regenerative repeater?
a) Cost
b) Bandwidth
c) Complexity
d) Long haul applications
Answer: d
Explanation: The regenerative repeaters are useful in long haul applications. However, such devices increase the cost and complexity of the optical communication system. It act as a bottleneck by restricting the system operational bandwidth.

2. The term flexibility, in terms of optical amplifiers means the ability of the transmitted signal to remain in the optical domain in a long haul link.
a) True
b) False
Answer: a
Explanation: Repeaters are usually used to maintain the transmitted signal in the optical domain. But, it has its own drawbacks. Thus, flexible systems which include optical amplifiers are used for such purpose.

3. How many configurations are available for employment of optical amplifiers?
a) Three
b) Four
c) Two
d) Five
Answer: a
Explanation: Optical amplifiers can be employed in three configurations. These are simplex mode, duplex mode, multi-amplifier configuration.

4. Repeaters are bidirectional.
a) True
b) False
Answer: b
Explanation: Repeaters are unidirectional. Optical amplifiers have the ability to operate simultaneously in both directions at the same carrier wavelength.

5. It is necessary to ____________ the optical carriers at different speeds to avoid signal interference.
a) Inculcate
b) Reduce
c) Intensity-modulate
d) Demodulate
Answer: c
Explanation: Optical amplifiers are bidirectional. They operate in both directions at the same carrier wavelength. In order to avoid interference, the optical carriers should be intensity modulated.

6. The _________________ increases the system reliability in the event of an individual amplifier failure.
a) Simplex configuration
b) Duplex configuration
c) Serial configuration
d) Parallel multi-amplifier configuration
Answer: d
Explanation: The optical amplifiers with spectral bandwidths in the range 50 to 100 nm allow amplifiers to be more reliable than repeaters. The parallel multi-amplifier configuration increases system reliability and relaxes the linearity.

7. Which of the following is not an application of optical amplifier?
a) Power amplifier
b) In-line repeater amplifier
c) Demodulator
d) Preamplifier
Answer: c
Explanation: Optical amplifiers have a wide variety of applications in the transmitter as well as receiver side. It is used as the power amplifier in the transmitter side and as preamplifier at the receiver side.

8. _________ reconstitutes a transmitted digital optical signal.
a) Repeaters
b) Optical amplifiers
c) Modulators
d) Circulators
Answer: a
Explanation: Optical amplifiers simply act as gain blocks on an optical fiber link. However, in contrast, the regenerative repeaters reconstitute a transmitted digital optical signal.

9. _____________ are transparent to any type of signal modulation.
a) Repeaters
b) Optical amplifiers
c) Modulators
d) Circulators
Answer: b
Explanation: The main benefit of acting as a gain block for optical amplifier is that it can be transparent to modulation bandwidth. However, both the noise and signal distortions are continuously amplified.

10. _________________ imposes serious limitations on the system performance.
a) Fiber attenuation
b) Fiber modulation
c) Fiber demodulation
d) Fiber dispersion
Answer: d
Explanation: The fiber dispersion calculation does not take into account the non-regenerative nature of the amplifier repeaters. In this, the pulse spreading and the noise is accumulated.

11. __________ is the ratio of input signal to noise ratio to the output signal to noise ratio of the device.
a) Fiber dispersion
b) Noise figure
c) Transmission rate
d) Population inversion
Answer: b
Explanation: Noise figure judges the performance factor of the devices. It is the in and out the ratio of signal to noise degradation for any device.

12. How many factors govern the noise figure of the device?
a) Four
b) Three
c) Two
d) One
Answer: a
Explanation: Noise figure is governed by factors such as the population inversion, the number of transverse modes in the amplifier cavity, the number of incident photons on the amplifier and the optical bandwidth of the amplified spontaneous emissions.

13. What is the typical range of the noise figure?
a) 1 – 2 dB
b) 3 – 5 dB
c) 7 – 11 dB
d) 12 – 14 dB
Answer: c
Explanation: Typical noise figures range from 7 to 11 dB. The SOAs are generally at the bottom end of the range and the fiber amplifiers towards the top end.

Dispersion Management and Soliton Systems

1. Calculate second order dispersion coefficient for path length L2 20km and L1 160km. Dispersion coefficient for L2 is 17.
a) -2.125ps/nm km
b) -3.25ps/nm km
c) -3.69ps/nm km
d) -1.28ps/nm km
Answer: a
Explanation: The second order dispersion coefficient for path length is given by-
β21 = -β22L2/L1. Here, β22 = Dispersion coefficient forL2, L2 and L1 are path lengths.

2. Calculate the path length L2 if L1is 160, dispersion coefficient of L2 is 17, dispersion coefficient of L1 is -2.25 ps/nmkm.
a) 40 km
b) 20 km
c) 30 km
d) 10 km
Answer: b
Explanation: The path length L2 is given by-
L2 = β21L1/-β22. Here, β22 = Dispersion coefficient forL2, β21 = Dispersion coefficient for L1, L2 and L1 are path lengths.

3. Calculate path length L1 if L2 is 20, dispersion coefficient of L2 is 17, dispersion coefficient of L1 is -2.25 ps/nmkm.
a) 180 km
b) 30 km
c) 160 km
d) 44 km
Answer: c
Explanation: The path length L1is given by-
L1 = β21L2/-β22. Here, β22 = Dispersion coefficient forL2, β21 = Dispersion coefficient for L1, L2 and L1 are path lengths.

4. Calculate second order dispersion coefficient for path length L1 20 km and L1 160 km. Dispersion coefficient for L1 is -2.125*10-12s/nmkm.
a) 20
b) 19
c) 18
d) 17
Answer: d
Explanation: The second order dispersion coefficient for path length is given by-
β22=-β21L2/L1. Here, β21 = Dispersion coefficient forL1, L2 and L1 are path lengths.

5. Calculate dispersion slope for second path fiber if L1 is 150, L2 is 10 and s1 is 0.075.
a) 1.125
b) 2.125
c) 3.125
d) 1.9
Answer: a
Explanation: Dispersion slope for second path fiber is s2 = -s1(L1/L2). Here s1 and s2 are dispersion slopes for L1, L2. L2 and L1 are path lengths.

6. Calculate dispersion slope for first path fiber if L1 is 160, L2 is 20 and s2 is 0.6ps/nm km.
a) 0.1
b) 0.432
c) 0.236
d) 0.075
Answer: d
Explanation: Dispersion slope for first path fiber is s1 = -s2(L1/L2). Here s1 and s2 are dispersion slopes for L1, L2. L2 And L1 are path lengths.

7. Calculate L2 if dispersion slope for first path fiber is 0.075 and L1 is 160 km and s2 is -0.6ps/nm km.
a) 20 km
b) 30 km
c) 40 km
d) 50 km
Answer: a
Explanation: L2 is determined by –
L2 = (-s1/s2)*L1. Here s1 and s2 are dispersion slopes for L1, L2. L2 and L1 are path lengths.

8. Calculate L1 if dispersion slope for first path fiber is 0.075 and L2 is 20 km and s2 is -0.6ps/nm km.
a) 170 km
b) 160 km
c) 180 km
d) 175 km
Answer: b
Explanation: L1 is determined by –
L2 = (-s1/s2)* L2. Here s1 and s2 are dispersion slopes for L1, L2. L2 and L1 are path lengths.

9. Calculate separation of soliton pulses over a bit period length if R2 pulse width is 6 ps for bit period of 70 ps.
a) 5.9
b) 5.7
c) 5.8
d) 5.4
Answer: c
Explanation: The separation of soliton pulses over a bit period length is calculated by –
q0 = T0/2ς. Here ς = pulse width and T0 = bit period.

10. Calculate RZ pulse width if bit period is 60ps and separation of soliton pulses is 5.4.
a) 5.5ps
b) 8.1ps
c) 4.3ps
d) 2.3ps
Answer: a
Explanation: RZ pulse width can be calculated by –
ς = T0/q0. Here ς = pulse width and T0 = bit period.

11. Calculate bit period if RZ pulse width is 50ps and separation of soliton pulses is 5.6.
a) 570ps
b) 540ps
c) 430ps
d) 560ps
Answer: d
Explanation: Bit period can be calculated by –
T0 = 2T2q0. Here T2=pulse width and T0=bit period.

12. Calculate value of dimensionless parameter if bit period is 45ps and RZ pulse width is 4 ps.
a) 5.625
b) 5.0
c) 4
d) 6.543
Answer: a
Explanation: Dimensionless parameter is given by –
q0 = T0/2ς. Here ς=pulse width and T0=bit period.

Optical Communication MCQs – Optical Fiber Systems 1 : Intensity Modulation ( Optical Communication ) MCQs

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