Category: Optical Communications Objective Questions

Optical Communications Multiple Choice Questions on “Analog Systems”.

1. Determine dispersion equalization penalty if total RMS pulse broadening is 4.8ns, B_T 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 D_L. It is given by-
D_L = 2 (2σ_TB_T√2)⁴. Here σ_T=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-
σ_T = σ√L. Here σ_T = RMS pulse broadening, L = length of the fiber.

3. Determine dispersion equalization penalty with mode coupling of 1.7ns if B_T 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 D_L. With mode coupling, it is given by-
D_L=2 (2σ_TB_T√2)⁴. Here σ_T=RMS pulse broadening.

4. Determine dispersion equalization penalty without mode coupling if B_T 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 D_L(WM). It is given by-
D_L(WM) = 2 (2σ_TB_T√2)⁴. Here σ_T = 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/2KTZ₀ηP_iλ. Here, η=quantum efficiency, P_i = 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 = (2KTZ₀ηP_iλ * Ratio)/hc. Here, η = quantum efficiency, P_i=0ptical power in mW, V=optical output voltage.

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/2KTZ₀ηP_iλ * Ratio. Hereη=quantum efficiency, P_i = 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/2KTZ₀ηP_i * Ratio. Hereη=quantum efficiency, P_i = 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/2KTP_i * Ratio. Hereη=quantum efficiency, P_i = 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-
T_syst=1.1[T_s²+T_n²+T_c²+T_D²]^1/2. Here T_s = rise time, T_n = intermodal time, T_c = 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-
T_syst(Max) = 0.35/B_opt. Here, B_opt=Optical Bandwidth.

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Optical Communications Multiple Choice Questions on ” Single Frequency Injection Lasers”.

1. __________________ may be improved through the use of frequency-selective feedback so that the cavity loss is different for various longitudinal modes.
a) Frequency selectivity
b) Longitudinal mode selectivity
c) Electrical feedback
d) Dissipated power
Answer: b
Explanation: Improved longitudinal mode selectivity can be achieved using structures which gives loss discrimination between the desired and all the unwanted modes. Thus, mode discrimination can be seen. To allow for stable mode operation, the use of frequency-selective feedback approach is undertaken.

2. Device which apply the frequency-selective feedback technique to provide single longitudinal operation are referred to as ________________
a) DSM lasers
b) Nd: YAG lasers
c) Glass fiber lasers
d) QD lasers
Answer: a
Explanation: DSM lasers are also known as single frequency lasers. Such devices provide single longitudinal mode operation hence called as dynamic single mode lasers. These lasers reduce fiber intra-modal dispersion within high speed systems.

3. Which of the following does not provide single frequency operation?
a) Short cavity resonator
b) DSM lasers
c) Coupled cavity resonator
d) Fabry-Perot resonator
Answer: d
Explanation: DSM lasers, short cavity resonators, coupled cavity resonators employ frequency selective feedback approach and provide single mode operation. However, the Fabry-Perot resonator allows several longitudinal modes to exist within the gain spectrum of the device.

4. A method for increasing the longitudinal mode discrimination of an injection laser which is commonly used?
a) Decreasing refractive index
b) Increasing the refractive index
c) Increasing cavity length
d) Shortening of cavity length
Answer: d
Explanation: The longitudinal mode discrimination of an injection laser is indirectly proportional to the cavity length. Thus, as the cavity length is shortened, the mode discrimination will get increase. If the cavity length is reduced from 250 to 25 units, the mode spacing is increased from 1 to 10 nm.

5. Conventional cleaved mirror structures are difficult to fabricate with the cavity lengths below __________
a) 200 μm and greater than 150 μm
b) 100 μm and greater than 50 μm
c) 50 μm
d) 150 μm
Answer: c
Explanation: cleaved laser mirrors are used in Fabry-Perot resonator which does not give result for shorter cavity lengths. These lengths may vary from 20 μm to 50μm. Hence micro-cleaved or etched resonator is used for shorter cavity lens.

6. In the given equation, corrugation period is given by lλ_b/2N_e. If λb is the Bragg wavelength, then what does ‘l’ stand for?
a) Length of cavity
b) Limitation index
c) Integer order of grating
d) Refractive index
Answer: c
Explanation: The period of corrugation is given by
Period of corrugation = lλ_b/2N_e
Where, λ_b = Bragg wavelength
L = integer order of grating.

7. The first order grating (l=1) provide the strongest coupling within the device.
a) True
b) False
Answer: a
Explanation: The period of corrugation is given by lλ_b/2N_e includes order of grating. The second grating provide larger spatial period and thus helps in fabrication. If the order of grating is 1, then the device is coupled at high level.

8. The semiconductor lasers employing the distributed feedback mechanism are classified in _________________ categories.
a) One
b) Two
c) Three
d) Four
Answer: b
Explanation: Considering the device operation, semiconductor lasers are classified into two broad categories referred to as distributed feedback laser and distributed Bragg reflector laser. In the DFB laser, optical grating is applied over the entire active region, whereas in DBR lasers, the grating is etched only near the cavity ends.

9. DBF-BH lasers exhibit low threshold currents in the range of ________________
a) 40 to 50 mA
b) 21 to 30 mA
c) 2 to 5 mA
d) 10 to 20 mA
Answer: d
Explanation: DFB lasers are used to provide single frequency semiconductor optical sources. DFB-BH lasers, developed in laboratories exhibit high modulation speeds, output power but low threshold currents in the range of 10 to 20 mA.

10. Fabry-Perot devices with BH geometries high modulation speeds than DFB-BH lasers.
a) True
b) False
Answer: b
Explanation: DFB-BH lasers exhibit low threshold currents but high output power and modulation speeds. DFB-BH laser is fabricated by etching or grating. Fabry-perot devices provide modulation speeds of M-bits per seconds whereas, DFB-BH lasers provides modulation speeds of G-bits/sec.

11. The InGaAsP/InP double channel planar DFB-BH laser with a quarter wavelength shifted first order grating provides a single frequency operation and incorporates a phase shift of ______________
a) π/2 Radians
b) 2π Radians
c) π Radians
d) 3π/2 radians
Answer: a
Explanation: The performance of DFB laser is improved by modifying a grating, which in turn introduces an optical phase shift. The phase shift depends on the wavelength used. A quarter wavelength shifted first order grating incorporates the phase shift of π/2 in the corrugation at the center of laser cavity.

12. The narrow line-width obtained under the CW operation for quarter wavelength shifted DFB laser is ________________
a) 2 MHz
b) 10 MHz
c) 3 MHz
d) 1 MHz
Answer: c
Explanation: A quarter wavelength shifted DFB laser provides a large gain difference between the central mode and side modes. It provides improved dynamic single mode stability. Narrow line-width of around 3 MHz can be obtained under CW operation.

13. Line-width narrowing is achieved in DFB lasers by a strategy referred as _______________
a) Noise partition
b) Grating
c) Tuning
d) Bragg wavelength detuning
Answer: d
Explanation: Line-width narrowing is achieved in DFB lasers by detuning the lasing wavelength towards the shorter wavelength side of gain peak. It increases the differential gain between the central mode and nearest side mode. This strategy is called as Bragg wavelength detuning.

14. _________________ is a technique used to render the non-conducting material around the active cavity by producing permanent defects in the implanted area.
a) Dispersion
b) Ion de-plantation
c) Ion implantation
d) Attenuation
Answer: c
Explanation: Ion implantation approach concentrates the injection current in active region. Current confinement is realized by ion implantation. Ions are implanted into a selective area of a semiconducting material to make it non-conducting.

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Optical Communications Questions and Answers for Experienced people on “Preparation of Optical Fibers – Liquid Phase Techniques”.

1. What is a fundamental necessity in the fabrication of fibers for light transmission?
a) Same refractive index for both core and cladding
b) Pump source
c) Material composition of fiber
d) Variation of refractive index inside the optical fiber
Answer: d
Explanation: For fabrication of fibers, two different transparent materials to light over a wavelength range of 0.8 to 1.7μm are required. Fiber should exhibit low attenuation, absorption and scattering losses. The variation of refractive indices in a fiber is a necessity for fiber fabrication.

2. Which materials are unsuitable for the fabrication of graded index fiber?
a) Glass-like-materials
b) Mono-crystalline structures
c) Amorphous material
d) Silica based material
Answer: b
Explanation: In case of graded index fiber, it is essential that the refractive index of the material is varied by suitable doping with another compatible material. These two materials should have mutual solubility over a wide range of concentration. This is achieved only in glass-like-materials.

3. How many different categories are available for the methods of preparing optical glasses?
a) 1
b) 2
c) 3
d) 4
Answer: b
Explanation: The methods of preparing optical glasses are divided into two categories. One is the conventional glass refining technique and other is vapor-phase-deposition method.

4. What is the first stage in liquid-phase-technique?
a) Preparation of ultra-pure material powders
b) Melting of materials
c) Decomposition
d) Crystallization
Answer: a
Explanation: In liquid-phase-technique melting, the first stage includes the preparation of ultra-pure material powders. These are usually oxides or carbonates which decomposes during glass melting.

5. Which processes are involved in the purification stage in liquid-phase-technique?
a) Filtration, Co-precipitation, Re-crystallization
b) Decomposition, Filtration, Drying
c) Doping, Drying, Decomposition
d) Filtration, Drying, Doping
Answer: a
Explanation: The compounds such as oxides and carbonates are formed during the glass melting. The purification accounts for a large proportion of material cost. These compounds are commercially available. The purification involves filtration, co-precipitation, re-crystallization and drying.

6. At what temperature range, does the melting of multi components glass systems takes place?
a) 100-300 degree Celsius
b) 600-800 degree Celsius
c) 900-1300 degree Celsius
d) 1500-1800 degree Celsius
Answer: c
Explanation: The glass materials in the powdered form and have relatively low melting point. Thus, the glass materials are melted at relatively low temperatures in the range of 900-1300 degrees Celsius.

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

8. The minute perturbations and impurities in the fiber drawing process using preform technique can result in very high losses of _____________
a) Between 500 and 1000 dB/km
b) Between 100 and 300 dB/km
c) Between 1200 and 1600 dB/km
d) More than 2000 dB/km
Answer: a
Explanation: The minute perturbations and impurities in the fiber include formations of bubbles and involvement of particulate matter. The losses due to such impurities can be between 500 and 1000 dB/km.

9. The liquid-phase melting technique is used for the production of fibers ___________
a) With a core diameter of 50μm
b) With a core diameter less than 100μm
c) With a core diameter more than 200μm
d) With a core diameter of 100μm
Answer: c
Explanation: The multicomponent glass fibers prepared continuously by liquid-phase melting technique have losses in the range of 5 and 20 dB/km at a wavelength of 0.85μm. This method is thus used for preparation of fibers with a large core diameter. Also this technique is used for the continuous production of fibers.

10. Graded index fibers produced by liquid-phase melting techniques are less dispersive than step-index fibers.
a) True
b) False
Answer: a
Explanation: Liquid-phase melting technique does not offer optimum parabolic profile fibers. This parabolic profile yields minimum pulse dispersion. Graded index fibers prepared using liquid-phase melting techniques are less dispersive but do not have the bandwidth-length products of optimum profile fibers.

To practice all areas of Optical Communications for Experienced people, .

Optical Communications Multiple Choice Questions on “Optical Network Concepts”.

1. Each stage of information transfer is required to follow the fundamentals of ____________
a) Optical interconnection
b) Optical hibernation
c) Optical networking
d) Optical regeneration
Answer: c
Explanation: Optical networking uses optical fiber as a transmission medium. It provides a connection between users to enable them to communicate with each other by transporting information from a source to a destination.

2. ____________ is a multi-functional element of optical network.
a) Hop
b) Optical node
c) Wavelength
d) Optical attenuation
Answer: b
Explanation: An optical node is a multi-functional element which acts as a transceiver unit capable of receiving, transmitting and processing the optical signal. The optical nodes are interconnected with optical fiber links.

3. A signal carried on a dedicated wavelength from source to destination node is known as a ___________
a) Light path
b) Light wave
c) Light node
d) Light source
Answer: a
Explanation: A light path is a dedicated path from a source to a destination. The data can be sent over the light paths as soon as connections are set up. A controlling mechanism is present to control the data flow.

4. The fundamentals of optical networking are divided into _______ areas.
a) Two
b) One
c) Four
d) Three
Answer: d
Explanation: The fundamentals divided into three areas contain mainly optical network terminology. The other two areas include functions and types of optical network node and switching elements and the wavelength division multiplexed optical networks.

5. The optical networking fundamentals are _____________ of the transmission techniques.
a) Dependent
b) Independent
c) Similar
d) Dissimilar
Answer: b
Explanation: The optical networking fundamentals include transfer of data. Irrespective of the difference in the transmission techniques, the fiber networking fundamentals remain the same.

6. The network structure formed due to the interconnectivity patterns is known as a ____________
a) Network
b) Struck
c) Topology
d) D-pattern
Answer: c
Explanation: A topology is a combination of patterns interconnected to each other. It provides connection patterns to users at different places. It embarks on the principle of multi-usability.

7. In the __________ topology, the data generally circulates bi-directionally.
a) Mesh
b) Bus
c) Star
d) Ring
Answer: b
Explanation: In a bus topology, data is input via four port couplers. The couplers couples and stations the data bi-directionally and are removed from the same ports.

8. The ring and star topologies are combined in a ________ configuration.
a) Mesh
b) Fringe
c) Data
d) Singular
Answer: a
Explanation: The mesh configuration is a combination of ring and star topologies. It is referred to as full-mesh when each network node is interconnected with all nodes in the network.

9. The full-mesh configuration is complex.
a) False
b) True
Answer: b
Explanation: The full-mesh topology is a combination of two or more topologies. It is often preferred for the provision of either a logical or virtual topology due to its high flexibility and interconnectivity features.

10. How many networking modes are available to establish a transmission path?
a) Three
b) One
c) Two
d) Four
Answer: c
Explanation: There are two networking modes often referred to the networking. These are connection-oriented and connectionless networking modes. These include an end-to-end and bidirectional communication environment between source and destination.

11. Packet switching is also called as ___________
a) Frame switching
b) Cell switching
c) Trans-switching
d) Buffer switching
Answer: b
Explanation: In packet or cell switching, messages are sent in small packets called cells. Cells from different sources are statistically multiplexed and are sent to the destinations.

12. ___________ mode is temporary, selective and continuous.
a) Cell switching
b) Buffer switching
c) Cache
d) Circuit switching
Answer: d
Explanation: An end-to-end connection is required for a circuit switching to take place. The transmissions are continuous and are in real time. Once the transmission is complete, the connection is ended.

13. A _______________ is a series of logical connections between the source and destination nodes.
a) Cell circuit
b) Attenuation circuit
c) Virtual circuit
d) Switched network
Answer: c
Explanation: A virtual circuit consists of different routes which provide connections between sending and receiving devices. These routes can change at any time and the incoming return route does not have to mirror the outgoing route.

14. ____________ refers to the process whereby a node finds one or more paths to possible destinations in a network.
a) Routing
b) Framing
c) Lightning
d) Cloning
Answer: a
Explanation: Routing refers to the path finding process in a network. In this, the control and data functions are performed to identify the route and to handle the data during the journey from source to destination.

15. How many stages are possessed by the control plane?
a) Two
b) Three
c) Four
d) Five
Answer: b
Explanation: The routing process called as control plane has three stages. These are neighbor discovery, topology discovery and path selection. These stages enable the network in routing mechanisms efficiently.

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Optical Communications Multiple Choice Questions on “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 arez_m=[2β_ςF(M)]*[S/N*η]
Here, η = quantum efficiency, S/N = signal to noise ratio.

9. Determine incident optical power if z_m=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=z_mh_cB_T/2λ. Here z_m=average number of photons, h_c=Planck’s constant.

10. Determine wavelength of incident optical power if z_m=864, incident optical power is -60.7 dB, B_T=1 * 10⁷.
a) 1 μs
b) 2 μs
c) 3 μs
d) 4 μs
Answer: a
Explanation: Wavelength is determined by λ=z_mh_cB_T/2P₀. Here z_m=average number of photons, h_c=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 C_L=(α_fc+α_j)L + α_cr. Here α_cr=loss at detector and source combined, α_fc = 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 = C_L/(α_fc+α_j) – α_cr. Here α_cr = loss at detector and source combined, α_fc = 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 –
σ_T = 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 = σ_T/L where σ = rms pulse broadening and L = length of the fiber.

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