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Aktu Optical Communication KEC-058 Btech Short Question, Notes

Discover Short Question Notes on Optical Communication from the B.Tech. AKTU Quantum Book. For effective data transfer, investigate the fundamentals of high-speed data transmission, fibre optics, and developments in communication technology.

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Unit-I: Introduction to Optical Communication

Q1. Write down the wavelength region corresponding to first, second and third windows.

Ans. 

WindowWindow rangeOperating wavelength
First window800 nm – 900 nm850 nm
Second Window1260 nm – 1360 nm1310 nm
Third Window1500 nm – 1600 nm1550 nm

Q2. Write down the advantages of optical fiber communication.

Ans. Advantages of optical fiber communication:

  • i. Enormous potential bandwidth.
  • ii. Small size and weight.
  • iii. Electrical isolation.
  • iv. Immunity to interference and crosstalk.
  • v. Signal security.

Q3. What are acceptance angel? Discuss its importance.

Ans. Acceptance angle: The acceptance angle is the maximum angle to the axis at which light may enter the fiber and begin to propagate.

Importance: The acceptance angle provides an indication of the potential size of the incidence angle needed to launch light effectively.


Q4. What is the principle used in the working of fibers as light guides?

Ans. The optical fiber is a device that uses the total internal reflection (TIR) concept to transfer light signals from one end to the other with a very small loss.


Q5. Define numerical aperture of a step index fiber.

                                                  OR

Define numerical aperture for an optical fiber mathematically and what does it signify.

Ans. A. Numerical aperture:

The sine of the maximum angle an incident ray can have before undergoing total internal reflection in the core is used to define a fiber’s numerical aperture (NA).

Thus the NA is defined as,

Define numerical aperture of a step index fiber. Optical Communication

B. Significance: NA has significance that it is very useful measure of the light-collecting ability of a fiber.


Q6. State Goos-Hanchen effect.

Ans.

  • 1. Through physical observation, it is possible to understand the phase change that results from the total internal reflection of a light beam on a planar dielectric contact.
  • 2. Close inspection reveals that the reflected beam is laterally moved from the trajectory anticipated by straightforward ray theory analysis, as shown in Fig.
  • 3. This lateral displacement is known as the Goos-Hanchen shift, after its first observer.
State Goos-Hanchen effect.

Q7. Define skew rays and meridional rays.

Ans. Skew rays: Skew rays are the rays that significantly outnumber meridional rays and go through the fiber in a helical pattern. The fiber axis blocks the passage of skew rays.

Meridional rays: A ray that is constrained to the plane comprising the optical axis of the system and the object point from which it emanated is called a meridional or tangential ray.


Q8. What is difference meridional rays and skew rays?

Ans. 

S. No. Skew raysMeridional rays
1.This particular rays communicate without going through the fiber axis.The axis of the core is traversed by these rays as they enter.
2.These types of rays are not confined to single plane.These rays are confined to meridian planes of the fiber.
3.These rays are difficult to locate as they are not confined to single plane.These rays lies in a single plane hence it is easy to track its path.

Q9. Name the fiber materials and its fabrication techniques.

Ans.

  • A. Mainly two fiber materials are used for fabrication, and these are:
    • i. Glasses.
    • ii. Plastics.
  • B. Fabrication techniques are as follows:
    • i. Outside vapour phase oxidation (OVPO).
    • ii. Vapour axial deposition (VAD).
    • iii. Modified chemical vapour deposition(MCVO).
    • iv. Plasma-activated chemical vapour deposition (PCVD).

Q10. What are step index and graded index fibers?

Ans. Step index fibers: When the refractive index of the core n1 is constant and greater than the refractive index of the cladding n2, the fiber is said to have a step index.

Graded index fibers: A fiber with a graded index is one whose core’s reflective index n1 decreases with radial distance from a maximum value at the axis to a constant value n2 outside the core radius, or an in the cladding.


Q11. What is the need of cladding?

Ans. In order to generate reflection inside the core and allow light waves to go through the fibre, the cladding’s role is to provide a lower refractive index at the core interface.


Q12. What are the advantages and disadvantages of SM fiber and MM fiber?

Ans. Advantages of SM (Single Mode) fiber:

  • 1. Expand available bandwidth.
  • 2. Restrictions on Data Dispersion and External Interference.
  • 3. Quick Transfer Speed.

Disadvantages of SM fiber:

  • 1. It is difficult to launch the light through the fiber.
  • 2. It is more expensive.

Advantages of MM (Multimode) fiber:

  • 1. High bandwidth and transfer rate.
  • 2. It is less expensive.

Advantages of MM fiber:

  • 1. More limited in both speed and distance.
  • 2. As, the number of modes increases the effect of modal dispersion increases.

Q13. What is the cut-off wavelength for single mode fiber?

Ans. The cut-off wavelength is the wavelength above which a specific fiber transforms into a single mode. It is denoted by 𝞴c.

What is the cut-off wavelength for single mode fiber?

Q14. Define Mode-Field Diameter (MFD).

Ans. 1. The mode-field diameter (MFD), which accounts for the wavelength dependent field penetration into the fiber cladding, is a crucial statistic for describing single mode fiber qualities.

2. For step index and graded single mode fibers the MFD is generally taken as the distance between the opposite 1/e = 0.37 field amplitude points and the power 1/e2 = 0.135 point.


Q15. A silica optical fiber with a core refractive index of .50 and a cladding refractive index of 1.47. Determine the numerical aperture for the fiber.

Ans. Given : n1 = 1.5, n2 = 1.47

To Find = NA

A silica optical fiber with a core refractive index of .50 and a cladding refractive index of 1.47.

Q16. Determine the acceptance angle in air for fiber, if NA is 0.30.

Ans. Given: NA = 0.30

To Find: Acceptance angle (𝞱a)

                                 𝞱a = sin-1 (NA)

                                     = sin-1(0.30) = 17.4 º


Unit-II: Signal Loss in Optical Fibers

Q1. List the properties of optical fiber that results in signal degradation.

Ans. The types of losses that result in signal degradation in optical fiber are:

1. Attenuation loss                              2. Absorption

3. Scattering                                       4. Bending loss.

5. Dispersion loss                               6. Coupling loss.


Q2. Describe scattering losses.

Ans. Scattering losses in fiber are caused by minute differences in the density of the material, compositional changes, and manufacturing defects.


Q3. Define Rayleigh scattering and Mie scattering.

An. A. Rayleigh scattering:

  • 1. In the low absorption window between the ultraviolet and infrared absorption tails, Rayleigh scattering predominates as a loss mechanism.
  • 2. It is caused by random homogeneities that occur on a scale that is smaller than the wavelength of light.

B. Mie scattering:

  • 1. Inhomogeneities that are similar in size to the directed wavelength may also exhibit linear scattering. The scattered intensity with an angular dependency can be made quite large when the scattering inhomogeneity size is more than 𝞴/10.
  • 2. Because of this, the scattering caused by these inhomogeneities is primarily d is the forward direction and is known as mie scattering.

Q4. What is bending loss?

Ans. Anytime an optical fiber bends with a finite radius of curvature, bending losses occur.


Q5. What do you mean by kerr effect?

Ans. The refractive index’s nonlinearity, or kerr nonlinearity, results in the carrier-induced phase modulation of the propagation signal, or kerr effect.


Q6. What is pulse broadening?

Ans. The spreading of the light pulses as they move down the fiber is referred to as pulse broadening.


Q7. What is the reason for pulse broadening in the case of material dispersion?

Ans. The varied group velocities of the various spectral components fired into the fiber from the optical source cause pulse broadening in the material dispersion.


Q8. Define polarization.

Ans. Transverse waves have a feature called polarization that describes the geometric orientation of the oscillations.


Q9. What do you mean by the term waveguide dispersion?

Ans. The difference in the index of refraction between the core and cladding, which results in a “drag” effect between the core and cladding sections of the power, is the source of waveguide dispersion.


Q10. Draw the diagram to show the effect of waveguide dispersion in single mode fiber.

Ans.

Draw the diagram to show the effect of waveguide dispersion in single mode fiber.

Q11. What is Intramodal dispersion?

Ans. Intramodal dispersion, also known as chromatic dispersion, is the spreading of a pulse within a single mode. This dispersion results from the guided mode’s group velocity, which depends on wavelength.


Q12. How the information capacity of an optical fiber specified? Give examples.

Ans. The bandwidth distance product in MHz-km is typically used to measure the information capacity of an optical fiber.

Example:- For a step index fiber, the various distortion effects tend to limit the bandwidth distance product to about 20 MHz- km.


Q13. How does the source spectral width affect the information carrying capacity of a fiber?

Ans. 1. The amount of information a multimode optical fiber can carry is expressed as the product of bandwidth and length (Mhz. km). The length is measured in kilometers, while the bandwidth is measured in megahertz (MHz) (km).

2. The bandwidth that the fiber can carry per kilometer of its length is indicated by the MHz.km number. The identification of the fiber must always be larger than or equal to the fiber’s bandwidth times its length.


Unit-III: Optical Fiber

Q1. What is the mechanism by which light is emitted from LED?

Ans. In an LED, electrons are injected into the typically empty conduction band of the semiconductor by the forward current flowing across the junction, and light is produced when these electrons recombine with holes in the valence band to emit a photon.


Q2. Discuss the advantage and disadvantage pf LED.

Ans. Advantages:

  • i. Simple fabrication.
  • ii. Reliability.
  • iii. Generally less temperature dependent.
  • iv. Simpler drive circuitry.

Disadvantages:

  • i. Usually lower modulation bandwidth.
  • ii. Harmonic distortion.

Q3. Name the major types of LED structure.

Ans.

  • i. Planar LEDs.
  • ii. Dome LEDs.
  • iii. Surface emitter LEDs.
  • iv. Edge emitter LEDs.
  • v. Super-luminescent LEDs.

Q4. Write the name of materials used for fabrication of LED.

Ans. i. GaAS/AlGaAs: Operate in shorter wavelength region.

ii. INGaAS/InP: Operate in longer wavelength region.


Q5. What is threshold current density of laser?

Ans. The amount of current needed to exceed the lasing threshold is known as the threshold current density.


Q6. What is threshold condition for laser oscillation?

Ans.

  • i. The overall losses are precisely balanced by the gain in the amplifying medium.
  • ii. The population inversion between the energies that produce the laser transition is required for the establishment of oscillation.
  • iii. In order to start and maintain laser oscillations, a minimum or threshold gain must be reached within the amplifying medium.

Q7. Compare the spectrum of a laser source and an LED source.

Ans. The spectral width of LED is wider i.e., 25 to 100 nm (10 to 50 THz) and the spectral width of laser is narrower i.e., less than 10-5 to 5 nm (<1 MHz to 2 MHz).


Q8. Comment on the reliability of laser (ILD).

Ans. Device reliability has been a major problem with injection laser. The degradation behaviour may be separated into two major processes:

  • i. Mirror facets that have been mechanically damaged undergo catastrophic deterioration.
  • ii. An increase in threshold current often characterizes gradual degradation.

Q9. Define mode hopping. 

Ans. All single-mode injection lasers experience the phenomenon known as mode hopping. Such a laser may run for a while in a single resonator mode under certain environmental circumstances before abruptly switching to another mode.


Q10. Define the term population inversion.

Ans. To achieve optical amplification, it is necessary to create a non-equilibrium distributions of atoms such that population of upper energy level is greater than that of lower energy level (N2 > N1).


Q11. Define external quantum efficiency.

Ans. The external quantum efficiency 𝜼ext is defined as the number of photon emitted per radiative electron hole pair recombination above threshold.

Define external quantum efficiency. Optical Communication

Q12. State chirping effect for laser diode.

Ans. When the light output level is modulated directly, the spectral width of a laser can expand dramatically. The chirping effect is the name given to this line widening.


Unit-IV: Power Launching in Fiber

Q1. What is power launching?

Ans. The sole factor affecting the optical power launched into the fiber is the source’s brightness. The source’s wavelength has no bearing on it.


Q2. List the techniques used for coupling the optical source (LED/LASERs) to fiber.

Ans.

  • i. Round-end fiber coupling.
  • ii. Cylindrical lens.
  • iii. Spherical-surfaced LED and spherical-ended fiber coupling
  • iv. Taper-ended fiber coupling.

Q3. What are the two basic requirement of optical detector?

Ans. 1. The operational wavelength’s sensitivity needs to be extremely high.

2. The photodetector should have a high quantum efficiency so that it can create the highest electrical signal for a given amount of optical power.


Q4. Give the name of any two optical communication detectors.

Ans. i. p-i-n photodiode.

ii. Avalanche photodiode.


Q5. Name the materials suitable for making photodiodes for short distance links and long haul links.

Ans. i. GaAs/AlGaAs: Operate in shorter wavelength region.

ii. InGaAsP/InP: Operate in longer wavelength region.


Q6. What are three main factors that limit the speed of response of a photodiode?

Ans.

  • i. The duration of carriers’ drift through the region of depletion.
  • ii. The time at which carriers produced outside the depletion region diffuse.
  • iii. The photodiode’s capacitance with its load causes a time constant.

Q7. Discuss the comparison between PIN and APD photodetectors.

Ans. 

S. No. PINAPD
1.Less sensitive (0-12 dB).More sensitive (5-15 dB).
2.Less reverse biased voltage (5 to 10 V).High reverse biased voltage (20 to 400 V).
3.Conversion efficiency is about 0.5 to 1.0 Amps/watt.Conversion efficiency is about 0.5 to 100 Amps/watt.

Q8. Define Avalanche effect.

Ans. A considerable amount of electrical force is given to a non-conducting (insulator) or semi-conducting (semiconductor), causing a sudden, quick increase in the current to occur.


Q9. Describe the term noise.

Ans. Unwanted electrical signal components that tend to interfere with signal processing and transmission are referred to as noise.


Q10. Mention the noise present in optical receiver.

                                               OR

What type of noise is present in optical receiver?

Ans. i. Thermal noise.

ii. Dark current noise.

iii. Quantum noise.


Q11. How does quantum noise arise?

Ans. The source of the shot noise known as quantum noise is the inability to forecast the number of electron-hole pairs that will be produced by an optical power impinge on the detector.


Q12. Define dark current noise.

Ans. 1. Thermal energy within the silicon lattice of the CCD causes dark current to form (Charge-coupled device). During time, electrons are generated that are separate from the light that strikes the detector.

2. The potential wells of the CCD capture electrons, which are then tallied as signals. Moreover, this rise in signal carries dark current noise, a statistical variation.


Q13. What is the effect of temperature on avalanche gain?

Ans. The ionization rate of electrons and holes will rise, increasing the avalanche gain, if the operating temperature drops while the applied bias voltage remains constant.


Q14. When 3 x 1011 photons are incident on a photo diode, on  average 1.2 x 1011 electrons are collected at the terminals of the device. Determine quantum efficiency.

Ans. 

Determine quantum efficiency.

Unit-V: Digital Receiver Performance

1. Define bit error rate (BER).

Ans. The ratio of errors occurring over a time period to the number of pulses delivered during that period is known as the bit error rate (BER).


Q2. What do you mean by receiver sensitivity?

Ans. The minimum power level that must reach the photodetector at a specific data rate in order to obtain the specified BER is known as receiver sensitivity.


Q3. What is meant by quantum limit? Express it mathematically.

Ans. 1. For an ideal photodetector quantum efficiency 𝜼 = 1 and has zero dark current then the minimum received. Power for a specific bit-error rate is known as quantum limit.

2. Then the probability of emitting zero electrons during the interval is 𝜏

What is meant by quantum limit? Express it mathematically.

where, N is average number of electron hole pair.


Q4. What is point-to-point link?

Ans. A point-to-point link with a transmitter on one end and a receiver on the other is the most basic type of transmission link. The least amount of optical fiber technology is required for this kind of connection.


Q5. What are the key system requirements needed to analyze the link?

Ans. i. The desired transmission distance.

ii. The data rate and channel bandwidth.

iii. The bit-error rate.


Q6. What are the components required for point-to-point link?

Ans. i. Multimode or single mode optical fiber.

ii. LED or laser diode optical source.

iii. PIN and avalanche photodiode.


Q7. Write a short note on power penalties.

Ans. The signal-to-noise ratio (SNR) of the system deviates from the ideal situation when any impairment impact is present in a link; this deviation is known as a power penalty.


Q8. List the name of power penalties in link design.

Ans. i. Chromatic dispersion penalty.

ii. Polarization-mode dispersion penalty.

iii. Extinction ratio penalty.


Q9. What are the methods used for error detection and correction in an optical link design?

Ans. A. Error correction methods:

i. Automatic repeat request                       ii. Forward error correction

B. Error detection methods:

i. Linear-error detection code.                   ii. Polynomial codes


Q10. Define isotype and an isotype heterojunctions.

Ans. 1. Isotype heterojunctions are the one formed with materials of same conductivity (p – P or N – n).

2. An isotype heterojunctions are formed with materials of different conductivities (N-p) etc.


Q11. Write any two eye pattern features.

Ans. The eye pattern measurements are made in the time domain, allowing the effects of waveform distortion to be seen right away on the display screen of BER test equipment that is typically used.


Q12. Define multiplexing.

Ans. Several analogue signals can be transmitted across one higher capacity fibre connection using the multiplexing technique. Time-division multiplexing (TDM) and frequency division multiplexing are two methods for channel or signal multiplexing in the time or frequency domain (FDM).


Q13. Enlist the passive components of WDM.

Ans. Commonly required passive components are: 

1. N x N couplers                                      2. Power splitters

3. Power taps                                            4. Star coupler


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