Communications MCQ Set 1
1. Which of the following is not a drawback of regenerative repeater?
d) Long haul applications
Answer: d [Reason:] 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. State whether the given statement is true or false.
Answer: 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?
Answer: a [Reason:] Optical amplifiers can be employed in three configurations. These are simplex mode, duplex mode, multi-amplifier configuration.
4. Repeaters are bidirectional. State whether the given statement is true or false.
Answer: 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.
Answer: c [Reason:] 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 [Reason:] 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
Answer: c [Reason:] 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.
b) Optical amplifiers
Answer: a [Reason:] 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.
b) Optical amplifiers
Answer: b [Reason:] 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 [Reason:] 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 [Reason:] Noise figure judges the performance factor of the devices. It is the in and out ratio of signal to noise degradation for any device.
12. How many factors govern the noise figure of the device?
Answer: a [Reason:] 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 – 5dB
c) 7 – 11 dB
d) 12 – 14 dB
Answer: c [Reason:] 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.
Communications MCQ Set 2
1. Electrical devices in optical network are basically used for _____________
a) Signal degradation
b) Node transfer
c) Signal control
Answer: c [Reason:] The optical infrastructure in networks constitutes a transparent network in which electronic devices are present. They are basically used for signal control. The other use includes providing interconnection to other networks.
2. Signals are defined as ________________ if their significant instants occur at nominally the same rate, any variation being constrained within specific limits.
Answer: a [Reason:] With any multiplexing strategies, come some setbacks. This includes the differentiation in the frequencies occurring throughout a network. This is called as plesiochronous transmission.
3. The bit stuffing in the plesiochronous digital hierarchy is complex and uneconomic. State whether the given statement is true or false.
Answer: b [Reason:] The bit stuffing is a complex process as it does not provide individual channel extraction. For individual channel extraction, the whole de-multiplexing procedure is to be performed again. This is both uneconomic and complex.
4. A ____________ digital hierarchy was required to enable the international communications network to evolve in the optical fiber era.
Answer: d [Reason:] The plesiochronous digital hierarchy was uneconomic and complex in execution. To reduce the complexity and efficient bit stuffing purpose, a synchronous digital hierarchy was required. It transformed the international communications into an optical fiber era.
5. The standardization towards a synchronous optical network termed SONET commenced in US in_______
Answer: a [Reason:] Synchronous optical network mechanism was efficient and its standardization process mainly started in 1985. Some modifications in the plesiochronous hierarchy were retained and some new features were added in the optical era to achieve efficient bit stuffing.
6. ______________ is a packetized multiplexing and switching technique which combines the benefits of circuit and packet switching.
a) Synchronous mode
b) Asynchronous transfer mode
c) Circuit packet
d) Homogeneous mode
Answer: b [Reason:] ATM transfers the information in fixed size units called cells. Each cell contains the information identifying the source of the transmission. It generally contains less data than packets.
7. The ___________ sits at the top of hierarchy of the OSI layer model.
a) Session layer
b) Transport layer
c) Application layer
d) Data link layer
Answer: c [Reason:] Application layer is the seventh layer and sits at the top of the hierarchy. It provides a means for a user to access information on or utilize the network by receiving a service.
8. The ____________ controls the dialogs between intelligent devices.
a) Physical layer
b) Transport layer
c) Application layer
d) Session layer
Answer: d [Reason:] The session layer is fifth in the OSI layer model. It controls the sessions between the devices and manages the connections between the remote and local application.
9. The network layer looks after the flow and error control mechanism. State whether the given statement is true or false.
Answer: b [Reason:] The network layer is the third level in the OSI layer model. It provides procedural and functional method for transferring data sequences from source to destination.
10. The physical layer is located at the bottom of the OSI model. State whether the given statement is true or false.
Answer:a [Reason:] The physical layer defines all the electrical, optical and media specifications for devices. Hence, it is located at the bottom of the OSI model. It establishes and terminates the connection between the media devices.
11. In order to access for end-to-end networking of optical channels to transparently convey information, the _____________ is employed in the OTN structure.
a) Presentation layer
b) Session layer
d) OCh layer
Answer: d [Reason:] The OCh stands for optical channel. It provides end-to-end access in networking of optical channels. It includes the multiplexing section to support multi-user networking.
12. An advanced type of reconfigurable OTN is referred to as an _______________
a) Automatic OTN
b) Auto-generated photon
c) Automatically switched optical network
d) Optical reimbursement
Answer: c [Reason:] Automatically switched optical network (ASON) is capable of switching the optical channels automatically when requested. It is specified in the ITU-T Recommendation G.8080 and it is basically a transport layer.
13. The __________ is a network layer that contains both addressing and control information to enable packets to be routed within a network.
b) Internet protocol (IP)
d) SONET/SDH protocol
Answer: b [Reason:] Internet protocol forms a part of the network layer. It controls the logical architecture within the network and addresses the issues accordingly. It routes the packets within a network.
14. The mapping of IP frames in SDH/SONET is accomplished in ___________ stages.
Answer: c [Reason:] Mapping requires three stages. In the first stage, point-to-point protocol is used. The second and the third stage includes synchronous mapping of data onto the SDH/SONET frame.
Communications MCQ Set 3
1. The cable must be designed such that the strain on the fiber in the cable does not exceed__________
Answer: c [Reason:] The constraints included in cable design are stability, protection, strength and jointing of the fibers. The fiber cable does not get affected if the strain exerted on it is below 0.2%. Although, it is suggested that the permanent strain on the fiber should be less than 0.1%.
2. How many categories exists in case of cable design?
Answer: b [Reason:] Cable design is separated into three categories. They are fiber buffering, cable structural and strength and cable sheath and water barrier. After successfully going through these tests, an optical cable is designed.
3. How many types of buffer jackets are used in fiber buffering?
Answer: a [Reason:] The buffer jacket is designed to protect the fiber from micro-bending losses. There are three types of buffer jackets used in fiber buffering. They are tight buffer jackets, loose tube buffer jackets and filled loose tube buffer jacket.
4. Loose tube buffer jackets exhibits a low resistance to movement of the fiber. State whether the given statement is true or false.
Answer: a [Reason:] Loose tube buffering is achieved by using a hard, smooth, flexible material in the form of extruded tube. The buffer tube is smooth from inside. Thus, it exhibits a low resistance to movement of the fiber. Also, it can be easily stripped for jointing or fiber termination.
5. An inclusion of one or more structural members in an optical fiber so as to serve as a cable core foundation around which the buffer fibers may be wrapped is called _____________
Answer: d [Reason:] Optical fiber is made structurally stronger by adding one or more strength members. The core fiber is trapped with buffered fibers or they are slotted in the core foundation. This approach is called as stranding.
6. Which of the following is not a strength member used in optical cable?
a) Steel wire
c) Aramid yarns
d) Glass elements
Answer: b [Reason:] Strength members or tensile members are added to the fiber to make it stronger and durable. These members include solid steel wire, dielectric aramid yarns (Kevlar), glass elements etc. Germanium is not a structural or strength member.
7. When the stranding approach consists of individual elements (e.g. single-fiber or multi fiber loose tube buffer) than the cable is termed as
a) Optical unit cable
b) Coaxial cable
c) Layer cable
d) Bare glass cable
Answer: c [Reason:] The stranding approach consists of a fiber core foundation around which the buffered fibers are wrapped. The cable elements are stranded in one, two or several layers around the central structural member. When the stranding is composed of individual elements, then the cable is termed as layer cable. If the cable core consists of stranding elements each of which comprises a unit of stranding elements, then it is termed as optical unit cable.
8. The primary function of the structural member is load bearing. State whether the given statement is true or false.
Answer: b [Reason:] The primary function of the structural member is not load bearing. It’s function is to provide suitable accommodation for the fiber ribbons within the cables. These fiber ribbons lie in the helical grooves or slots formed in the surface of the structural members.
9. What is the Young’s modulus of Kevlar, an aromatic polyester?
a) 9 ×1010Nm-2
b) 10 ×1010Nm-2
c) 12 ×1010Nm-2
d) 13 ×1010Nm-2
Answer: d [Reason:] Kevlar is used as a strength member in an optical fiber. The Young’s modulus of Kevlar is very high which gives it strength to weight ratio advantage four times that of steel. Kevlar is coated with extruded plastic to provide a smooth surface which in turn prevents micro-bending losses.
10. The cable is normally covered with an outer plastic sheath to reduce _______________
Answer: a [Reason:] Abrasion is the process of scraping or wearing something away. If the cable is not coated with plastic sheath, it gives rise to effects such as abrasion and crushing. The most common plastic sheath material used in covering a cable is polyethylene (PE).
Communications MCQ Set 4
1. A multimode step index fiber has a normalized frequency of 72. Estimate the number of guided modes.
Answer: b [Reason:] A step-index fiber has a constant refractive index core. The number of guided modes in a step-index fiber are given by M = (V*V)/2. Here M denotes the number of modes and V denotes normalized frequency.
2. A graded-index fiber has a core with parabolic refractive index profile of diameter of 30μm, NA=0.2, λ=1μm. Estimate the normalised frequency.
Answer: b [Reason:] Normalized frequency for a graded index fiber is given by V= 2Πa(NA)/λ. Substituting and calculating the values, we get option b. Here, V denotes normalized frequency and NA= numerical aperture.
3. A step-index fiber has core refractive index 1.46 and radius 4.5μm. Find the cutoff wavelength to exhibit single mode operation. Use relative index difference as 0.25%.
Answer: c [Reason:] The cutoff wavelength is the wavelength beyond which no single mode operation takes place. On solving λc= 2Πan12∆/ V, we get option c. Here, V=2.405, n1= refractive index of core, a=radius of core.
4. A single-mode step-index fiber or multimode step-index fiber allows propagation of only one transverse electromagnetic wave.
Answer: True [Reason:] Single mode step index fiber is also called as mono-mode step index fiber. As the name suggests, only one mode is transmitted and hence it has the distinct advantage of low intermodal dispersion.
5. One of the given statements is true for intermodal dispersion. Choose the right one.
a) Low in single mode and considerable in multimode fiber
b) Low in both single mode and multimode fiber
c) High in both single mode and multimode fiber
d) High in single mode and low in multimode fiber
Answer: a [Reason:] Single mode propagates only one wave or only one mode is transmitted. Therefore, intermodal dispersion is low in single mode. In multimode fibers, higher dispersion may occur due to varying group velocities of propagating modes.
6. For lower bandwidth applications,
a) Single mode fiber is advantageous
b) Photonic crystal fibers are advantageous
c) Coaxial cables are advantageous
d) Multimode fiber is advantageous
Answer: d [Reason:] In multimode fibers, intermodal dispersion occurs. The group velocities often differ which gradually restricts maximum bandwidth attainability in multimode fibers.
7. Most of the optical power is carried out in core region than in cladding. State true or false:
Answer: a [Reason:] In an ideal multimode fiber, there is no mode coupling. The optical power launched into a particular mode remains in that mode itself. The majority of these modes are mostly confined to fiber core only.
8. Meridional rays in graded index fibers follow
a) Straight path along the axis
b) Curved path along the axis
c) Path where rays changes angles at core-cladding interface
d) Helical path
Answer: b [Reason:] Meridional rays pass through axis of the core. Due to the varying refractive index at the core, the path of rays is in curved form.
9. What is the unit of normalized frequency?
d) It is a dimensionless quantity
Answer: d [Reason:] Normalized frequency of optical fiber is the frequency which exists at cut-off condition. There is no propagation and attenuation above cut-off. It is directly proportional to numerical aperture which is a dimensionless quantity; hence itself is a dimensionless quantity.
10. Skew rays follow a
a) Hyperbolic path along the axis
b) Parabolic path along the axis
c) Helical path
d) Path where rays changes angles at core-cladding interface
Answer: c [Reason:] The ray which does not pass through the fiber axis is termed as skew ray. Unlike Meridional rays, skew rays are more in number which makes them follow a round path called as helical path.
Communications MCQ Set 5
1. _________ heterodyne detection does not require phase matching between the incoming signal and the local oscillator.
Answer: b [Reason:] For heterodyne detection, a beat note signal produces the IF signal which is obtained using the square law optical detector. Hence, it does not require phase matching.
2. In ___________ detection, the phase of the local oscillator signal is locked to the incoming signal.
Answer: a [Reason:] Phase diversity and multiport detection is considered to be a form of heterodyne detection. In case of homodyne detection, the incoming signal is bundled with the phase of the local oscillator signal.
3. The introduction of nonlinear element within the ______ is necessary to enable efficient carrier recovery.
Answer: d [Reason:] Carrier recovery is done by slightly reducing the depth of the phase modulation. This is done by the introduction of the nonlinear within the phase locked loop(PLL).
4. What is the main attraction of the optical homodyne detection?
a) Receiver sensitivity
b) Transmission power
c) Modulation scheme
d) Line width
Answer: a [Reason:] The main attraction of optical homodyne detection is the potential 3dB improvement in the receiver sensitivity. It also eases the receiver bandwidth requirement considerably.
5. The phase difference between the local oscillator and the source must be ________ for high sensitivity reception.
d) Near zero
Answer: d [Reason:] The phase difference must be held near zero for high sensitivity reception. As it moves towards 90 degrees, the output signal current becomes zero and the detection process will cease.
6. How many strategies of homodyne demodulation have proved to be successful for coherent optical fiber reception?
Answer: d [Reason:] Two homodyne demodulation strategies have been successful for optical fiber reception. They are the use of either a pilot carrier or a decision-driven optical phase locked loop.
7. In the __________ receiver, the incoming signal is not precisely shifted to the baseband.
Answer: a [Reason:] In homodyne detection, the incoming signal is precisely shifted to the baseband. In intra-dyne receiver, the incoming signal is shifted to a frequency lower than the transmission rate.
8. The use of PLL can be avoided in the intra-dyne detection. State whether the given statement is true or false.
Answer: b [Reason:] In intra-dyne receiver, the incoming signal is lower than the transmission rate. In place of PLL, a baseband filter in the form of electrical filtering can be used.
9. Phase diversity reception is also referred to as ______________
a) Homodyne detection
b) Noisy detection
c) Homodyne detection
d) Multiport detection
Answer: d [Reason:] In phase diversity reception, phase is not locked. It also has bandwidth advantage over homodyne detection as it converts the incoming signal directly to baseband.
10. The received signal through polarization diversity reception is linearly polarized. State whether the given statement is true or false.
Answer: a [Reason:] The received optical signal through polarization diversity reception is elliptically polarized. It is also uncontrollable unlike coherent receivers.