Communications MCQ Set 1
1. ____________ converts the received optical signal into an electrical signal.
Answer: a [Reason:] A detector is an essential component of an optical fiber communication system. It dictates the overall system performance. Its function is to convert optical signal into an electrical signal. This electrical signal is then amplified before further processing.
2. The first generation systems of optical fiber communication have wavelengths between ___________
a) 0.2 and 0.3 μm
b) 0.4 and 0.6 μm
c) 0.8 and 0.9 μm
d) 0.1 and 0.2 μm
Answer: c [Reason:] The first generation systems operated at a bit-rate of 45 Mbps with repeater spacing of 10 km. It operates at wavelengths between 0.8 and 0.9μm. These wavelengths are compatible with AlGaAs laser and LEDs.
3. The quantum efficiency of an optical detector should be high. State whether the given statement is true or false.
Answer: a [Reason:] The detector must satisfy stringent requirements for performance and compatibility. The photo detector thus produces a maximum electrical signal for a given amount of optical power; i.e. the quantum efficiency should be high.
4. Which of the following does not explain the requirements of an optical detector?
a) High quantum efficiency
b) Low bias voltages
c) Small size
d) Low fidelity
Answer: d [Reason:] The size of the detector must be small for efficient coupling to the fiber. Also, ideally, the detector should not require excessive bias voltages and currents. The fidelity and quantum efficiency should be high.
5. How many device types are available for optical detection and radiation?
Answer: b [Reason:] Two types of devices are used for optical detection and radiation. These are external photoemission and internal photoemission devices. External photoemission devices are too bulky and require high voltages for operation. Internal devices provide good performance and compatibility.
6. The ___________ process takes place in both extrinsic and intrinsic semiconductors.
a) Avalanche multiplication
b) External photoemission
c) Internal photoemission
Answer: c [Reason:] During intrinsic absorption, the received photons excite electrons from the valence band and towards the conduction band in the semiconductor. Extrinsic absorption involves impurity centers created with the material. Generally, intrinsic absorption is preferred for internal photoemission.
7. ____________ are widely used in first generation systems of optical fiber communication.
a) p-n diodes
d) Silicon photodiodes
Answer: d [Reason:] The first generation systems operates at wavelengths 0.8 and 0.9 μm. Silicon photodiodes have high sensitivity over the 0.8-0.9 μm wavelength band with adequate speed, long term stability. Hence, silicon photodiodes are widely used in first generation systems.
8. Silicon has indirect band gap energy of __________________
a) 1.2 eV
b) 2 eV
c) 1.14 eV
d) 1.9 eV
Answer: c [Reason:] Silicon’s indirect band gap energy of 1.14 eV gives a loss in response above 1.09μm. To avoid this, narrower bandgap materials are used. Hence, silicon’s usefulness is limited to first generation systems and not for second and third generation systems.
9. Which of the following detector is fabricated from semiconductor alloys?
a) Photoconductive detector
b) p-i-n detector
d) Photoemission detectors
Answer: a [Reason:] The detectors fabricated from semiconductor alloys can be used for longer wavelengths. Photoconductive detector and hetero-junction transistor have found favor as a potential detector over a wavelength range of 1.1 to 1.6μm.
10. Silicon photodiodes provides high shunt conductance. State whether the given statement is true or false.
Answer: b [Reason:] Semiconductor photodiodes provide best solution for detection in optical fiber communications. Silicon photodiodes have high sensitivity, negligible shunt conductance and low dark current.
Communications MCQ Set 2
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.
Answer: d [Reason:] 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.
Answer: b [Reason:] 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 [Reason:] 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 [Reason:] 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 [Reason:] 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.
Answer: a [Reason:] 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.
Answer: b [Reason:] 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.
Answer: c [Reason:] Average number of photons arezm=[2βςF(M)]*[S/N*η]
Hereη=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 [Reason:] 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 [Reason:] 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 [Reason:] The total channel loss isCL=(αfc+αj)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 [Reason:] Length of the fiber is L=CL/(αfc+αj) – 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 [Reason:] 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.
Answer: d [Reason:] RMS pulse broadening is given by –
σ=σT/L where σ = rms pulse broadening and L= length of the fiber.
Communications MCQ Set 3
1. What is dispersion in optical fiber communication?
a) Compression of light pulses
b) Broadening of transmitted light pulses along the channel
c) Overlapping of light pulses on compression
d) Absorption of light pulses
Answer: b [Reason:] Dispersion of transmitted optical signal causes distortion of analog as well as digital transmission. When the optical signal travels along the channel, the dispersion mechanism causes broadening of light pulses and thus in turn overlaps with their neighboring pulses.
2. What does ISI stands for in optical fiber communication?
a) Invisible size interference
b) Infrared size interference
c) Inter-symbol interference
d) Inter-shape interference
Answer: c [Reason:] Dispersion causes the light pulses to broaden and overlap with other light pulses. This overlapping creates an interference which is termed as inter-symbol interference.
3. For no overlapping of light pulses down on an optical fiber link, the digital bit rate BT must be:
a) Less than the reciprocal of broadened pulse duration
b) More than the reciprocal of broadened pulse duration
c) Same as that of than the reciprocal of broadened pulse duration
Answer: a [Reason:] The digital bit rate and pulse duration are always inversely proportional to each other.
B T < = ½ Γ
Where B T = bit rate
2Γ= duration of pulse.
4. The maximum bit rate that may be obtained on an optical fiber link is 1/3Γ. State whether the given statement is true or false.
Answer: b [Reason:] The digital bit rate is function of signal attenuation on a link and signal to noise ratio. For the restriction of interference, the bit rate should be always equal to or less than 1/2Γ.
5. 3dB optical bandwidth is always ___________ the 3dB electrical bandwidth.
a) Smaller than
b) Larger than
c) Negligible than
d) Equal to
Answer: b [Reason:] Optical bandwidth is half of the maximum data rate. For non-return:0 (NRZ), bandwidth is same as bit rate. The bandwidth B for metallic conductors is defined by electrical 3dB points. Optical communication uses electrical circuitry where signal power has dropped to half its value due to modulated portion of modulated signal.
6. A multimode graded index fiber exhibits a total pulse broadening of 0.15μsover a distance of 16 km. Estimate the maximum possible bandwidth, assuming no intersymbol interference.
a) 4.6 MHz
b) 3.9 MHz
c) 3.3 MHz
d) 4.2 MHz
Answer: c [Reason:] The maximum possible bandwidth is equivalent to the maximum possible bitrate. The maximum bit rate assuming no inter-symbol interference is given by
B T = ½ Γ
Where B T = bandwidth.
7. What is pulse dispersion per unit length if for a graded index fiber, 0.1μs pulse broadening is seen over a distance of 13 km?
Answer: b [Reason:] The dispersion mechanism causes broadening of light pulses. The pulse dispersion per unit length is obtained by dividing total dispersion of total length of fiber.
Dispersion = 0.1*10-6/13= 7.69 ns/km.
8. Chromatic dispersion is also called as intermodal dispersion. State whether the given statement true or false.
Answer: b [Reason:] Intermodal delay is a result of each mode having a different group velocity at a single frequency. The intermodal delay helps us to know about the information carrying capacity of the fiber.
9. Chromatic dispersion is also called as intermodal dispersion. State true or false
Answer: b [Reason:] Intermodal delay, the name only suggests, includes many modes. On the other hand chromatic dispersion is pulse spreading that takes place within a single mode. Chromatic dispersion is also called as intermodal dispersion.
10. The optical source used in a fiber is an injection laser with a relative spectral width σλ/λ of 0.0011 at a wavelength of 0.70μm. Estimate the RMS spectral width.
a) 1.2 nm
b) 1.3 nm
c) 0.77 nm
d) 0.98 nm
Answer: c [Reason:] The relative spectral width σλ/λ= 0.01 is given. The rms spectral width can be calculated as follows:
σλ /λ = 0.0011
σλ = 0.0011λ
= 0.0011*0.70*10-6 = 0.77 nm.
11. In waveguide dispersion, refractive index is independent of
a) Bit rate
b) Index difference
c) Velocity of medium
Answer: d [Reason:] In material dispersion, refractive index is a function of optical wavelength. It varies as a function of wavelength. In wavelength dispersion, group delay is expressed in terms of normalized propagation constant instead of wavelength.
Communications MCQ Set 4
1. Which technology development has helped the field of optical fiber communication?
a) Glass technology
b) Component technology
Answer: b [Reason:] Substantial developments in the component technology have allowed the initial difficulties in the optical fiber communication to go away. The coherent factor experienced most of the difficulties.
2. __________ dictates the performance characteristics required from components and devices which are to be utilized in coherent optical fiber systems.
a) System considerations
b) Bluetooth technology
d) Practical constraints
Answer: d [Reason:] Practical constraints inhibit the development of coherent optical fiber communications. These constraints are derived from factors associated with the elements of the coherent optical fiber communication.
3. Coherent optical transmission is degraded by the ________ associated with the transmitter and local oscillator lasers.
a) Phase noise
b) White noise
Answer: a [Reason:] Phase noise is determined by the laser line width. The phase noise associated with both the transmitter and the mid-tier section severely degrades the coherent optical transmission as well as reception.
4. ___________ improves the spectral purity of the device output and noise current.
a) Power dissipation
b) Laser line width reduction
c) Laser line width injection
d) Phase noise
Answer: b [Reason:] Laser line width determines the level of phase noise and long term phase stability. The reduced phase noise is obtained using narrow-line width devices. This improves the spectral purity as well as reduces the noise current.
5. ____________ is the principal cause of line width broadening in the coherent devices.
a) Electromagnetic field
b) Power dissipation
c) Injection laser phase noise
d) Gaussian noise
Answer: c [Reason:] Injection laser phase noise affects the system performance. The system performance considerations include receiver noise, power loss and line width broadening.
6. Which technique was started for narrowing of injection laser line widths?
a) External resonator cavity
b) Long-hauled oscillator
Answer: a [Reason:] Many approaches evolved in time for laser line width problem. The one which sustained and showed effects was the use of external resonator cavity in the lasers.
7. The line width tolerance is wider for heterodyne receivers. State whether the given statement is true or false.
Answer: b [Reason:] The laser line width requirements depend on the modulation format, coherent detection mechanism which includes the use of heterodyne and homodyne receivers. The line width tolerance is wider for heterodyne receivers when employing FSK modulation.
8. ___________ is an alternative to reduce phase noise and line width requirements.
a) Homodyne detection
b) Heterodyne detection
c) FSK modulation
d) Phase diversity reception
Answer: d [Reason:] The more sensitive coherent transmission techniques are most affected by phase noise problem. A specially configured reception technique called as phase diversity reception technique is used to overcome phase noise problem.
9. ______________ is the progressive spatial separation between the two polarization modes as they propagate along the fiber.
a) Fiber birefringence
b) Fiber dispersion
c) Fiber separation
d) Fiber coupling
Answer: a [Reason:] In a perfectly formed fiber, both modes would travel together. But, in practice, the fiber contains random manufacturing irregularities. This result in a progressive spatial separation called as fiber birefringence.
10. How many compensator devices are required to provide full polarization-state control?
Answer: d [Reason:] At least two compensator devices are required to provide full polarization-state control. They can be placed in either the incoming signal path or the local oscillator output path.
11. Which technique was found to be providing an infinite range of polarization control?
a) Homodyne detection
b) Fiber squeezers
c) Heterodyne detection
d) Power dissipation
Answer: b [Reason:] Four fiber squeezers provide an infinite range of adjustment or endless polarization control. Stress was applied to the fiber in the local oscillator path using the squeezers which are angled at 45 degrees to each other.
12. What is the main drawback of the squeezer?
a) Damages the fiber
d) Signal degradation
Answer: a [Reason:] The squeezers are simple to configure. The main drawback of squeezer is that they tend to damage the fiber and could not be engineered into reliable transducers for practical systems.
13. The use of balanced receiver compensates the losses due to coupling optics. State whether the given statement is true or false.
Answer: a [Reason:] The losses due to coupling optics and the suppression of the excess noise in the local oscillator signal are eliminated by the use of balanced receiver. It is also called as balanced-mixer receiver.
14. ___________ is the phenomenon which occurs in the single carrier systems due to small refractive index changes induced by the optical power fluctuations.
a) SBS gain
b) Self-phase modulation
c) FSK modulation
Answer: b [Reason:] It occurs only in the single-carrier systems. It affects the phase of the transmitted signal.
Communications MCQ Set 5
1. What is the use of interposed optics in expanded beam connectors?
a) To achieve lateral alignment less critical than a butt-joined fiber connector
b) To make a fiber loss free
c) To make a fiber dispersive
d) For index-matching
Answer: Expanded beam connector utilize interposed optics at the joint in order to expand the beam from transmitting fiber end before reducing it to a size compatible with the receiving fiber end. It helps to achieve lateral alignment less critical than a butt-jointed connector. Also, the longitudinal separation is critical in expanded beam connectors.
2. The expanded beam connectors use ____________ for beam expansion and reduction.
a) Square micro-lens
b) Oval micro-lens
c) Spherical micro-lens
d) Rectangular micro-lens
Answer: c [Reason:] Expanded beam connectors use the principle of transmission of digital data to the receiver. It uses spherical micro-lens to first expand the beam from the transmitting end and reduces the beam at the receiving end.
3. Lens-coupled expanded beam connectors exhibit average losses of _________ in case of single mode and graded index fibers.
a) 0.3 dB
b) 0.7 dB
c) 0.2 dB
d) 1.5 dB
Answer: b [Reason:] Lens-coupled expanded beam connectors use spherical micro-lenses. The average losses are in the range of 1dB. With the antireflection coating on the lenses, the losses are reduced to 0.7 dB in case of single mode fibers.
4. Sapphire ball lens expanded beam design is successful than spherical lens coupled design. State whether the given statement is true or false.
Answer: a [Reason:] Spherical lens coupled design exhibits losses in the range 0.7 dB to 1dB. Sapphire ball lens expanded beam design achieved successful single mode fiber connection with losses as low as 0.4dB.
5. The fiber is positioned at the ________ of the lens in order to obtain a collimated beam and to minimize lens-to-lens longitudinal misalignment effects.
b) Focal length
d) Exterior circumference
Answer: b [Reason:] The expanded beam connector also uses a molded spherical lens. A lens alignment sleeve is used to minimize the effects of angular misalignment. The fiber is positioned at the focal length of the lens to achieve losses as low as 0.7dB.
6. ___________ exhibits a parabolic refractive index profile with a maximum at the axis similar to graded index fiber.
a) Lens coupled design
b) Sapphire ball lens
c) Spherical micro-lens
d) GRIN-rod lens
Answer: d [Reason:] GRIN-rod lens geometry has a parabolic refractive index profile. It facilitates efficient beam expansion and collimation within expanded beam connectors. It finds its applications in fiber couplers and source-to-fiber coupling.
7. The GRIN-rod lens can produce a collimated output beam with a divergent angle αof between _____________ from a light source situated on, or near to, the opposite lens face.
a) 1 to 5 degrees
b) 9 to 16 degrees
c) 4 to 8 degrees
d) 25 to 50 degrees
Answer: a [Reason:] GRIN-rod lens comprises of a cylindrical glass rod typically 0.2 to 2 mm in diameter. It exhibits a parabolic refractive index profile. It facilitates efficient beam expansion and collimation with an angle in the range 1 to 5 degrees.
8. In the given equation, if r is the radial distance, n is the refractive index; what does z stands for?
dr2/dz2= (1/n) (d n/dr)
a) Focal length
b) Distance along the optical axis
c) Axial angle
Answer: b [Reason:] The above equation is known as paraxial ray equation which governs the ray propagation through the GRIN-rod lens. GRIN-rod lens geometry is parabolic in nature. Thus z is the distance along the optical axis of a parabolic profile.
9. The majority of the GRIN-rod lenses have diameters in the range of ____________
a) 2 to 2.5 mm
b) 3 to 4 mm
c) 0.1 to 0.4 mm
d) 0.5 to 2 mm
Answer: d [Reason:] The GRIN-rod lenses performance directly depends on the radial distance. The diameters in the range of 0.5 to 2 mm may be employed with either single mode or multimode fibers. They are available with numerical apertures of 0.37, 0.46 and 0.6.
10. Which of the following factors does not cause divergence of the collimated beam from a GRIN-rod lens?
a) Lens cut length
b) Size of fiber core
c) Refractive index profile
d) Chromatic aberration
Answer: c [Reason:] Various factors contribute to the divergence of the collimated beam from a GRIN-rod lens. Error in lens cut length, finite size of the fiber core and chromatic aberration are the factors that cause divergence.
11. GRIN-rod lens connectors have loss characteristics which are independent of the modal power distribution in the fiber. State whether the given statement is true or false.
Answer: a [Reason:] GRIN-rod lens geometry is analogous to butt-jointed multimode fiber connectors. The loss characteristics of butt-jointed connectors are dependent on modal power distribution in the fiber.