Optical Communication Set 6

Electrical Engineering

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This set of Optical Communication Multiple Choice Questions & Answers (MCQs) focuses on Optical Communication Set 6

Q1 | Determine the output current of RAPD having multiplication factor of 39 and photocurrent of 0.469μA.
  • 17.21
  • 10.32
  • 12.21
  • 18.29
Answer: 18.29
Explaination:
the multiplication factor of photodiode is given by-
Q2 | Compute the photocurrent of RAPD having multiplication factor of 36.7 and output current of 7μA.
  • 0.01 μa
  • 0.07 μa
  • 0.54 μa
  • 0.9 μa
Answer: 0.01 μa
Explaination:
the multiplication factor of photodiode is given by-
Q3 | How many circuits are present in an equivalent circuit for the digital optical fiber receiver?
  • four
  • one
  • three
  • two
Answer: four
Explaination:
a full equivalent circuit for the digital optical fiber receiver includes four circuits. these are the detector circuit, noise sources, and amplifier and equalizer circuit.
Q4 |                      compensates for distortion of the signal due to the combined transmitter, medium and receiver characteristics.
  • amplification
  • distortion
  • equalization
  • dispersion
Answer: equalization
Explaination:
equalization adjusts the balance between frequency components within an electronic signal. it compensates for distortion of the signal. the distortion may be due to the transmitter, receiver etc.
Q5 | The phase frequency response of the system should be                          in order to minimize inter-symbol interference.
  • non-linear
  • linear
  • more
  • less
Answer: linear
Explaination:
an equalizer is used as frequency shaping filter. the phase frequency response of the system should be linear to acquire the desired spectral shape for digital systems. this, in turn, minimizes the inter- symbol interference.
Q6 | Noise contributions from the sources should be minimized to maximize the receiver sensitivity.
  • true
  • false
Answer: true
Explaination:
noise sources include transmitter section, medium and the receiver section. as the noise increases, the sensitivity at the receiver section decreases. thus, noise contributions should be minimized to maximize the receiver sensitivity.
Q7 | How many amplifier configurations are frequently used in optional fiber communication receivers?
  • one
  • two
  • three
  • four
Answer: three
Explaination:
three amplifier configurations are used in optical fiber communication receivers. these are voltage amplifiers, semiconductor optical amplifier and current amplifier. voltage amplifier is the simplest and most common amplifier configuration.
Q8 | How many receiver structures are used to obtain better receiver characteristics?
  • two
  • one
  • four
  • three
Answer: three
Explaination:
the various receiver structures are low-impedance front end, high-impedance front end and trans-impedance front-end. the noise in the trans-impedance amplifier will always exceed than the front end structure.
Q9 | The high-impedance front-end amplifier provides a far greater bandwidth than the trans-impedance front-end.
  • true
  • false
Answer: true
Explaination:
the noise in the trans- impedance amplifier exceeds that incurred by the high-impedance amplifier. hence, the trans-impedance front-end provides a greater bandwidth without equalization than the high- impedance front end.
Q10 | A high-impedance amplifier has an effective input resistance of 4MΩ. Find the maximum bandwidth that may be obtained without equalization if the total capacitance is 6 pF and total effective load resistance is 2MΩ.
  • 13.3 khz
  • 14.2 khz
  • 15.8 khz
  • 13.9 khz
Answer: 13.3 khz
Explaination:
the maximum bandwidth obtained without equalization is given by – b = 1/2Πrtlct
Q11 | The major advantage of the trans- impedance configuration over the high- impedance front end is                              
  • greater bandwidth
  • less bandwidth
  • greater dynamic range
  • less dynamic range
Answer: greater dynamic range
Explaination:
greater dynamic range is a result of the different attenuation mechanism for the low-frequency components of the signal. this attenuation is obtained in the trans-impedance amplifier through the negative feedback and therefore the low frequency components are amplified by the closed loop. this increases the dynamic range.
Q12 | The trans-impedance front end configuration operates as a                      with negative feedback.
  • current mode amplifier
  • voltage amplifier
  • attenuator
  • resonator
Answer: current mode amplifier
Explaination:
the trans-impedance configuration overcomes the drawbacks of the high-impedance front end. it utilizes a low-noise, high-input-impedance amplifier with negative feedback. it operates as a current mode amplifier where high impedance is reduced by negative feedback.
Q13 |                          is also known as frequency- shaping filter.
  • resonator
  • amplifiers
  • attenuator
  • equalizer
Answer: equalizer
Explaination:
equalizer, often called as frequency-shaping filter has a frequency response inverse to that of the overall system frequency response. in wideband systems, it boosts the high frequency components to correct the overall amplitude of the frequency response.
Q14 | The major advantage of the trans- impedance configuration over the high- impedance front end is                              
  • greater bandwidth
  • less bandwidth
  • greater dynamic range
  • less dynamic range
Answer: greater dynamic range
Explaination:
greater dynamic range is a result of the different attenuation mechanism for the low-frequency components of the signal. this attenuation is obtained in the trans-impedance amplifier through the
Q15 |                          affects both the fiber attenuation and dispersion.
  • refractive index
  • micro-bending
  • connectors
  • splices
Answer: micro-bending
Explaination:
effects such as micro-bending with a resultant mode coupling affect both the fiber attenuation and dispersion. it does not provide the overall characteristics of the transmission link.
Q16 | Which of the following is not included in the optical fiber link measurement test?
  • attenuation measurement
  • dispersion measurement
  • splice loss measurement
  • receiver sensitivity
Answer: receiver sensitivity
Explaination:
it is necessary to perform some tests on the optical fiber link to enhance productivity. apart from receiver sensitivity, other measurement methods are required to test the fiber link.
Q17 | In case of field measurements, the equipment must have                        power consumption keeping in mind the battery operation.
  • low
  • high
  • negligible
  • maximum
Answer: low
Explaination:
the design criteria allows you to distinguish in parameters required for adaptation of battery operation and equipment handling. the power consumption must be low for an equipment to handle.
Q18 | Which of the following are not considered as environmental conditions required for field measurements?
  • temperature
  • humidity
  • mechanical load
  • power
Answer: power
Explaination:
the equipment must be reliable and provide accurate measurements under extreme environmental conditions such as humidity, temperature and mechanical load. power is an internal factor.
Q19 | Which of the following cannot be used in equipment for field measurements?
  • fiber
  • connector
  • external triggering
  • environmental factor
Answer: external triggering
Explaination:
the equipment cannot usually make use of external triggering and regulating circuits between the transmitter and receiver. this is because of their wide spacing on the majority of the optical links.
Q20 | Which sensors are used for alteration of spectral range in equipment?
  • wide-area photodiodes
  • circulators
  • gyrators
  • photogenic sensors
Answer: wide-area photodiodes
Explaination:
wide area photodiodes such as silicon, germanium diodes are used for alteration of spectral range. it is generally preferred to have a measurement range from
Q21 | The handheld optical power meter has a measurement accuracy of                  
  • 0.01 db
  • 0.25 db
  • 0.8 db
  • 1 db
Answer: 0.25 db
Explaination:
the optical power meter detects the fiber type and switches to optical power measurements. it provides an accuracy of about +(or -) 25 db.
Q22 |                            may be used for measurement of the absolute optical attenuation on a fiber link.
  • silicon photodiodes
  • ingaasp photodiodes
  • optical power meters
  • gyrators
Answer: optical power meters
Explaination:
optical power meter employs cut-back technique. it is used for the measurement of the optical attenuation.
Q23 | A large-area photodiode is utilized in the receiver to eliminate any effects from differing fiber and faces.
  • true
  • false
Answer: true
Explaination:
the modulating voltage maintains the equilibrium between the transmitter and the receiver side. a large area photodiode is required to eliminate differing fiber and faces to maintain the equilibrium.
Q24 | Optical time domain reflectometry is also called a backscatter measurement method.
  • false
  • true
Answer: true
Explaination:
otdr technique is used in both on-field and laboratory applications. it is also called a backscatter measurement method as it provides the measurement of attenuation on an optical link down its entire length.
Q25 | When considering source-to-fiber coupling efficiencies, the                  is an important parameter than total output power.
  • numerical aperture
  • radiance of an optical source
  • coupling efficiency
  • angular power distribution
Answer: radiance of an optical source
Explaination:
radiance is the optical power radiated into a unit solid angle per unit emitting surface area. since this optical power is dependent on radiance, radiance is much important factor than optical power.
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