This document outlines the objectives and outcomes of the course EC8751-Optical Communication. The key objectives are to study optical fiber modes, materials, fabrication, transmission characteristics, optical sources and detectors, receiver systems, and measurements. The outcomes are to understand basic fiber elements, analyze dispersion and polarization techniques, design optical components, construct receiver systems, and design communication systems and networks. It provides textbook references and outlines topics like fiber structure, types, applications, generation of optical fiber communication systems, and fiber materials.

2. EC8751-OPTICAL COMMUNICATION
OBJECTIVES:
 To study about the various optical fiber modes, configuration, Fiber materails
and Fiber Fabrication
 To learn about transmission characteristics of optical fibers
 To learn about the various optical sources and detectors
 To learn about optical receiver and to explore various idea about optical fiber
measurements and also various coupling techniques
 To enrich the knowledge about optical communication systems and networks

3. OUTCOMES:
At the end of the course, the you should be able to:
 Realize basic elements in optical fibers, different modes and configurations
 Analyze the transmission characteristics associated with dispersion and
polarization techniques
 Design optical sources and detectors with their use in optical
communication system
 Construct fiber optic receiver systems, measurements and coupling
techniques
 Design optical communication systems and its networks

4. Sources
TEXT BOOKS:
1. P Chakrabarti, "Optical Fiber Communication”, McGraw Hill Education (India)Private
Limited, 2016. (UNIT I, II, III)
2. Gred Keiser,"Optical Fiber Communication”, McGraw Hill Education (India) Private
Limited. Fifth Edition, Reprint 2013. (UNIT I, IV, V)
REFERENCES:
1. John M.Senior, ―Optical fiber communication”, Pearson Education, Second Edition.2007.
2. Rajiv Ramaswami, ―Optical Networks ― , Second Edition, Elsevier , 2004.
3. J.Gower, ―Optical Communication System”, Prentice Hall of India, 2001.
4. Govind P. Agrawal, ―Fiber-Optic Communication Systems”, Third Edition, John Wiley & Sons,
2004.

5. OUTLINE:
*OpticalFiberCommunication??
*WhyOpticalFiber??
*OpticalFiberStructure
*ProsandCons(+and-)
*TypesofFibers
*Applications

6. General and Optical Communication systems

7. Spectrum

8. Need for Optical Fiber Communication

9. Evolution of Optical Fiber

10. Generation of Optical Fiber
 First generation system - Bit rate of 45 Mb/s with repeater spacing of up to 10 km
 Second generation - Developed for commercial use in the early 1980s, operated at 1.3 μm and
used InGaAsP semiconductor lasers.
Multi mode fiber dispersion, and in 1981 the single-mode fiber was revealed to
greatly improve system performance
By 1987, these systems were operating at bit rates of up to 1.7 Gb/s with
repeater spacing up to 50 km (31 mi).
 Third generation - Fiber optic systems operated at 1.55 μm and had losses of about 0.2 dB/km
Indium Gallium Arsenide photodiode
Overcame earlier difficulties with pulse-spreading at that wavelength using
conventional InGaAsP semiconductor lasers.
Dispersion shifted fibers designed to have minimal dispersion at 1.55 μm
Third generation systems to operate commercially at 2.5 Gb/s with repeater
spacing in excess of 100 km (62 mi)

11. Generation of Optical Fiber Contd…
 Fourth generation of fiber optic communication systems used optical amplification
to reduce the need for repeaters and wavelength-division multiplexing to increase
data capacity
• Doubling of system capacity every six months starting in 1992 until a bit rate
of 10 Tb/s was reached by 2001
• In 2006 a bit-rate of 14 Tb/s was reached over a single 160 km (99 mi) line
using optical amplifiers
 Fifth generation of fiber-optic communications is on extending the wavelength range
over which a WDM system can operate
• The conventional wavelength window, known as the C band, covers the
wavelength range 1.53 –1.57 μm, and dry fiber has a low-loss window
promising an extension of that range to 1.30–1.65 μm

12. OPTICAL FIBER
COMMUNICATION
It is the method of
communication in
which signal is
transmitted in the form
of light .
Optical Fiber is used
as a medium of
transmitting the light
signal from one place
to another.
TRANSMITTER RECEIVEROPTICAL FIBER

13. Elements of Optical Communication System
Optical
Receiver
Electronics
Optical
Transmitter
Regenerator
Drive
Circuit
Light
Source
Fiber
flylead
Transmitter
Electrical
input signal
Connector
Optical coupler
or beam splitter
Optical
Splicer
Optical
fiber
To other
equipment
Photo
Detector
Signal
Restorer
Amplifier
Electrical
signal out
Fiber
flylead
Receiver
Optical
Amplifier
Electrical signal
Optical signal

14. An OPTICAL FIBER is a flexible and transparent fiber made
up of glass ( silica) or plastic which has diameter
slightly thicker than that of human hair
Optical Fibre uses light pulses instead of electrical
pulses to transmit information, thus delivers
hundreds of times higher bandwidth than
traditional electrical systems
It is a cylindrical dielectric waveguide (nonconducting
waveguide) that transmits light along its axis, by the
process of total internal reflection

15. Fiber Structure
The Fiber consists of a core surrounded by a cladding layer,
both of which are made of dielectric materials

16. Fiber Structure

17. PROS & CONS Of
OPTICAL FIBER
COMMUNICATIO
N

18. 1.Large Transmission
Capacity
2.Low Attenuation
3.Easy Amplification
Pros
4.Low Cost

19. 5.Light Weight
9.No Interference
8.Greater BW
6.High Speed
7.Better Reliability
10.Longer Distance
Pros

20. Advantages of Optical Fiber Communication
1. Information bandwidth is more.
2. Optical fibers are small in size and light weighted.
3. Optical fibers are more immune to ambient electrical noise, electromagnetic
interference.
4. Cross talk and internal noise are eliminated in optical fibers.
5. There is no risk of short circuit in optical fibers.
6. Optical fibers can be used for wide range of temperature.
7. A single fiber can be used to send many signals of different wavelengths using
Wavelength Division Multiplexing (WDM).
8. Optical fibers are generally glass which is made up of sand and hence they are
cheaper than copper cables.
9. Optical fibers are having less transmission loss and hence less number of
repeaters are used.
10. Optical fibers are more reliable and easy to maintain.

21. PLASTICGLASS
Graded Index
Fiber
Single Mode
Fiber
Multi Mode
Fiber
Step Index Fiber
Types of Optical Fibers

22. • It has a smaller core diameter of 8-12μm
Single Mode Fiber
• It has a larger core diameter than that of single
mode fiber optic cable
• It allows multiple pathways and several
wavelengths of light to be transmitted
Multimode Fiber
• It allows a single wavelength and pathway for
light to travel, which greatly decreases light
reflections and lowers attenuation.

23. Signal Transmission in
Single and Multimode Fibers

24. Types of Fibers with its dimensions and Index Profile

25. Applications of Optical Fiber
Telephone Signals
Transmission
Internet
Communication
Cable Television
Signal Transmission
Medical Diagnosis
Detect
Nuclear Radiations
Military
Applications
Railway Monitoring

26. Latest Applications of Optical Fiber
Night time effect Night lampShiny optical
fiber frock

27. Disadvantages !!
3.Special Testing Equipment is required
1. High Initial Cost
2.Difficulty in Joining and Testing of fiber

28. Silica and Plastic as Fiber Optic Materials
Silica Fibers
 Both Core And Cladding are of Glass
 Very Pure Sio2 or Fused Quartz
 Ge or Phosphorus to Increase the Index of Refraction
 Boron or Fluorine to Decrease the Index of Refraction
 Silica Fibers Mainly Used due to their Low Intrinsic Absorption at Wavelengths of
Operation
 Any Other Remaining Impurities cause Attenuation and Scattering
Attenuation
Spectrum
of Silica Fibers

29. Fiber Optic Materials Contd..
Plastic Fibers
 Plastic core and plastic cladding
 Polymethyl Methacrylate (most commonly used)
 Flexible and Light
 Widely used in short distance applications
Plastic-Clad Fibers
Glass as core and plastic as cladding
Which is better? (Plastic or Silica)
 Plastic less expensive, flexible, lighter
 Plastic is larger in diameter, so easy to connect across joints
 Plastic is less efficient than Silica
 Plastic has more attenuation, and less bandwidth making it more suitable for shorter distances.
Attenuation Spectrum of Plastic
Fibers

30. Basic Definition of Frequency and Wavelength
 The radio waves and light are electromagnetic waves. The rate at which they alternate in polarity is called
their frequency(f) measured in Hz.
 The speed of electromagnetic wave in free space is approximately 3*108 m/sec.
 The distance travelled during each cycle is called as Wavelength(λ).
• In fiber optics, it is more convenient to use the
wavelength of light instead of the frequency with
light frequencies, wavelength if often stated in
micron or nanometres

31. Ray Theory(Laws of Optics) Transmission
Reflection
Refraction
Refractive Index(n)
Snell’s Law
 Critical Angle(φC)
 Total Internal Reflection(TIR)
 Acceptance Angle (Ɵa or φ0)
 Numerical Aperture(NA)

32. REFLECTION
Incident Angle:
The angle between the incident wave
and the normal is called the angle of
Incidence (φ1)
Reflected Angle:
The angle between the reflected wave
and normal is called the angle of
reflection (φ2)
Law of Reflection:
The angle of incidence is equal to the
angle of reflection
Law of Reflection

33. Refraction
 When wave passes through less dense
medium to more dense medium, the
wave is refracted (bent) towards the
normal.
 The refraction (bending) takes place
because light travels at different speed
in different mediums.
 The speed of light in free space is
higher than in water or glass.
Refraction

34. Refractive Index (n)
 The amount of refraction or bending that occurs at the interface of two
materials of different densities is usually expressed as refractive index of two
materials
 Refractive index is also known as index of refraction and is denoted by n
 Based on material density, the refractive index is expressed as the ratio of the
velocity of light in free space to the velocity of light of the dielectric material
(substance). i.e
The refractive index (n)
for
Vacuum and air n= 1.003
Water n=1.33
Glass n=1.52
Diamond n=2.42

35. Snell’s Law
 If φ1 and φ2 be the angles of incidence and angle of refraction respectively. Then according to
Snell’s law, a relationship exists between the refractive index of both materials given by,
n1 sin φ1 = n2 sin φ2
 The refracted wave will be towards the normal when n1 < n2 and will away from it when n1 > n2
Refractive Model for Snell’s Law
 Snell’s law states how light ray reacts when it
meets the interface of two media having
different indexes of refraction
 Let the two medias have refractive indexes n1
and n2, where n1 > n2

36. Critical Angle(φC)
The critical angle is defined as the minimum angle
of incidence (φ1) at which the ray strikes the
interface of two media and causes an angle of
refraction (φ2) equal to 90o . It is denoted as φC.
Hence at critical angle, φ1 = φC and φ2 = 90
°
Using Snell’s law : n1 sin φ1 = n2 sin φ2
Critical Angle

37. Total Internal Reflection(TIR)
Ray Propagation by TIR

38. Condition for TIR
Total Internal Reflection at the fiber wall can occur only if two conditions are
Condition 1
The index of refraction of glass fiber must be slightly greater than the index
refraction of material surrounding the fiber (cladding).
If refractive index of glass fiber = n1
and refractive index of cladding = n2
then n1 > n2
Condition 2
The angle of incidence (φ1)of light ray must be greater than Critical Angle (φ

39. Acceptance Angle (Ɵa or φ0 )
Formation of Acceptance Cone of a fiber Acceptance Cone
 The Cone of acceptance is the angle within which
the light is accepted into the core and is able to
travel along the fiber. The launching of light wave
becomes easier for large acceptance cone.
 Then, Acceptance angle is the maximum angle from
the fiber axis at which light may enter the fiber so
that it will propagate, or travel, in the core of the fiber
Acceptance Angle

40. Derivation of Acceptance Angle (φ0 )
Applying Snell’s law to external incidence angle,
n0 sin φ0 = n1 sin φ1
But φ1 = (90 – φC )
sin φ1 = sin (90 – φC ) = cos Φc
Substituting sin φ1 in above equation.
n0 sin φ0 = n1 cos φC
Applying Pythagorean theorem to ΔPQR

41. Derivation of Acceptance Angle (φ0 ) Contd…
The maximum value of external incidence angle for which light will propagate in the fiber is
When the light rays enters the fibers from an air medium n0 = 1. Then above equation reduces
to,
The angle φ0 is called as Acceptance Angle and φ0max defines the maximum angle in which the
light ray may incident on fiber to propagate down the fiber.

42. Numerical Aperture(NA)
 The numerical aperture (NA) of a fiber is a figure of merit which
represents
its light gathering capability.
 Larger the numerical aperture, the greater the amount of light
accepted by fiber.
 The acceptance angle also determines how much light is able to be
enter the fiber and hence there is relation between the numerical
aperture and the cone of acceptance.
Numerical aperture (NA) = sin φ0max

43. Numerical Aperture(NA) Contd…
By the formula of NA, note that the numerical aperture is effectively
dependent only on refractive indices of core and cladding material. NA is
not a function of fiber dimension
The index difference (Δ) and the numerical aperture (NA) are related
to the core and cladding indices:
also
since

44. Problem no.1
A light ray is incident from medium-1 to medium-2. If the refractive
indices of medium-1 and medium-2 are 1.5 and 1.36 respectively then
determine the angle of refraction for an angle of incidence of 30o

45. Problem no.2
Calculate the NA, acceptance angle and critical angle of the fiber having
n1 (Core refractive index) = 1.50 and refractive index of cladding n2 =
1.45.

46. Problem no.3
A light ray is incident from glass to air. Calculate the critical angle (φC).
Solution:
Refractive index of glass n1 = 1.50 and
Refractive index of air n2 = 1.00
At critical angle, φ1 = φC and φ2 = 90 °
Using Snell’s law : n1 sin φ1 = n2 sin φ2
Critical angle(φC) = sin-1(1/1.5)
= 41.81°
o

47. Problem no 4
Calculate the numerical aperture and acceptance angle for a fiber cable of which ncore = 1.5 and ncladding
= 1.48. The launching takes place from air.
Solution:

48. Practice questions with Final Solution
1.The relative refractive index difference for an optical fiber is 0.05.If the
entrance end
of fiber is facing the air medium and refractive index of core is 1.46,
estimate its
Numerical Aperture.(NA=0.46)
2. A silica optical fiber has a core refractive index of 1.50 and a cladding
refractive index of
1.47. Determine,(a)the critical angle at the core cladding interface,(b)the NA
for the fiber and
(c) the acceptance angle in air for the fiber.(φc =78.5°, NA=0.3, Ɵa=17.4°)
3. Calculate NA,∆ and Ɵa of a fiber having refractive index of core is 1.44,

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