Download INTRODUCTION TO FIBER OPTIC SYSTEM

INTRODUCTION TO FIBER OPTIC SYSTEM
CONTENTS
1.0
1.1
1.2
1.3
1.4
1.5
1.6
1.7
PAGE
Introduction to Fiber Optic……………………………………………. …..3
Light………………………………………………………………………..4
Applications of Fiber-Optic………………………………………………..5
Fiber-Optic System Block Diagram……………………………………......6
Fiber-Optic Cable Construction……………………………………………7
Basic Optical Laws…………………………………………………………13
Light Ray Propagation in a Fiber-Optic Cable……………………………..19
Classification of Fiber………………………………………………………20
1
INTRODUCTION TO FIBER-OPTIC SYSTEM
OBJECTIVES
General Objective
:
To understand the concept of the Fiber-Optic System.
Specific Objectives
At the end of the unit you will be able to :

define the terms optical and light.

name six typical communications applications for fiber-optic
cable.

draw a basic block diagram of a fiber-optic communications
system and tell what each part of it does.

explain how light is propagated through the fiber-optic cable.

name the three basic types of fiber-optic cables, and state the
two materials from which they are made.
2
1.0
Introduction to Fiber-Optic
Light wave communication was first considered more than 100 years ago. The
implementation of optical communication using light waveguides was restricted to very
short distance prior to 1970. Corning Glass Company achieved a breakthrough in 1970 by
producing a fused silica (SiO2) fiber with a loss approximately 20 dB/km. The
development of semiconductor light source also started to mature at about that time,
allowing the feasibility of transmission over a few kilometers to be demonstrated. Since
1970, the rate of technological progress has been phenomenal, and optical fibers are now
used in transoceanic service. Besides the long-distance routes, fibers are used in inter
exchange routes, and the subscriber loop in the final link in what will eventually be the
global interconnection chain. Optical fibers are associated with high-capacity
communications. A lot of attention is presently being given to optical fibers to provide a
very extensive broadband ISDN.
Fiber optics is defined as that branch of optics that deals with the
transmission of light through ultrapure fibers of glass, plastic, or some
other form of transparent media. From a decorative standpoint, most of
us are familiar with the fiber optic lamp, which uses bundles of thin
optical fibers illuminated from the base end of the lamp by a light
source.
The light source is made to vary in color, which can be seen at the opposite ends
of the fiber as a tree of illuminating points radiating various colors of the
transmitted light. Although the lamp is used for decorative purposes only, it
serves as an excellent model of how light can be transmitted through the fiber.
3
1.1
Light
Light is a kind of electromagnetic radiation. The
basic characteristic of electromagnetic radiation
is its frequency or wavelength. Light frequencies fall between microwaves and xrays, as shown in Figure1.1.
Figure1.1 Electromagnetic Frequency Spectrum
Light frequency spectrum can be divided into three general bands:
1. Infrared : Band of light wavelengths that are too long to be seen by the
human eye.
2. Visible :
Band of light wavelengths to which the human eye will respond.
3. Ultraviolet : Band of light wavelengths that are too short to be seen by the
human eye.
Light waves are commonly specified in terms of wavelength instead of frequency.
Units typically used are the nanometer or micrometer ( 1 micrometer = 1micron).
4
1.2 Applications of Fiber-Optic
Fiber-optic communications systems are being used more and more each day.
Their primary use is in long-distance telephone systems and cable TV system.
Fiber-optic cables are no more expensive or complex to install than standard
electrical cables, yet their information-carrying capacity is many times greater.
In all cases, the fiber-optic cables replace conventional coaxial or twisted-pair cables.
Below are some applications which use fiber-optic cables :1.
TV studio to transmitter interconnection eliminating a microwave radio
link.
2.
Closed-circuit TV systems used in buildings for security.
3.
Secure communications systems at military bases.
4.
Computer networks, wide area and local area.
5.
Shipboard communications.
6.
Aircraft communications.
7.
Aircraft controls.
8.
Interconnection of measuring and monitoring instruments in plants and
laboratories.
9.
Data acquisition and control signal communications in industrial process
control systems.
10.
Nuclear plant instrumentation.
11.
College campus communication.
12.
Utilities ( electrical , gas, and so on) station communications.
13.
Cable TV systems replacing coaxial cable.
5
1.3
Fiber-Optic System Block Diagram
Figure1.2 shows a simplified block diagram of an optical fiber communications
link. The three primary building blocks of the link are the transmitter, the
receiver, and the fiber guide.
Fiber
Light
Source
Coder
Repeater
(long distance)
Light pulses
Fiber
Light
Detector
Decoder
Figure1.2 Simplified fiber optic communications block diagram
The transmitter consists of an analog to digital converter (coder), and a light
source. The A/D converter is used to convert continuous analog signals such as
voice or video (TV) signals into a series of digital pulses.
6
The digital pulses are then used to flash a powerful light source off and on very
rapidly. The light source is either a light-emitting diode (LED) or an injection
laser diode (ILD).The light-beam pulses are then fed into a fiber-optic cable
where they are transmitted over long distances.
The optical fiber consists of a glass or plastic fiber core, a cladding, and a
protective jacket.
Repeaters are used to ensure the signals can be transmitted efficiently when the
two stations are separated far enough from each other.
The receiver includes a light detector or photocell and a decoder. The light
detector is very often either a PIN (p-type-intrinsic-n-type) diode or an APD
(avalanche photodiode). The light detector, acting as the receiving element,
converts the received light pulses back to pulses of electrical current. The
electrical pluses are amplified and reshaped back into digital form that is fed to a
decoder such as a D/A converter, where the original voice or video is recovered.
1.4
Fiber-optic cable construction
An optical fiber is a long, cylindrical, transparent material that confines and
propagates light waves shown in Figure 2.3 and Figure 2.4 . It is comprised of
three layers; the centre core that carries the light, the cladding layer covering the
core that confines the light to the core, and the coating that provides protection
for the cladding. The fiber itself is generally regarded as the core and its cladding.
The material composition of these two layers can be any of the following :

Glass cladding and glass core

Plastic cladding and glass core

Plastic cladding and plastic core
Coating materials include lacquer, silicone, and acrylates.
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Figure1.3
Typical Fiber Optic Cabling
The index of refraction in the silica core is about 1.5 and the cladding is slightly
less, at about 1.48. The index of refraction of air is 1.003. The fiber coating is
normally colored using manufacturer’s standard color codes to facilitate the
identification of fiber. Optical fibers can also be made completely from plastic or
other materials. They are usually less expensive but have higher attenuation (loss)
and limited application.
Figure1.4 Basic construction of a fiber-optic cable
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Example1.1
The optical spectrum is made up of three parts. Name them.
Solution to Example1.1
In the optical spectrum, the three parts are infrared, visible and ultraviolet.
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QUIZ 1A
Answer the following questions.
1.1
The major use of fiber-optic cables is ______________________.
1.2
True or False. Fiber-optic cable has more loss than electric cable over long
distance.
1.3
True or False. Fiber-optic cable is smaller, lighter, and stronger than
electric cable.
1.4
Voice and video signals are converted into ____________ before being
transmitted by a light beam.
Choose the letter which best answers each question.
1.5
1.6
The core of a fiber-optic cable is made of ….
a.
Air
b.
Glass
c.
Diamond
Which of the following is not a common application of fiber-optic
cable?
a.
Computer networks
b.
Telephone systems
c.
Consumer TV
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1.7
1.8
1.9
The core of a fiber-optic cable is surrounded by …
a.
Cladding
b.
Wire braid shield
c.
Paper
Which of the following is not part of the optical spectrum?
a.
Infrared
b.
Gamma-rays
c.
Ultraviolet
The speed of light in plastic compared to the speed of light in air
is….
a.
Less
b.
More
c.
The same
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Feedback To QUIZ 1A
1.1
telephone systems
1.2
false
1.3
true
1.4
binary or digital pulses
1.5
c
1.6
a
1.7
b
1.8
a
1.9
a
I have done all the
questions in Quiz1A
successfully.
Well done! And now you
can proceed to the next
lesson
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1.5
Basic Optical Laws
When light traveling in a transparent material meets the surface of another
transparent material two things happen:a)
some of the light is reflected
b)
some of the light is transmitted into the second transparent material
The light which is transmitted usually changes direction when it enters the second
material. This bending of light is called refraction and it depends upon the fact
that light travels at one speed in one material and at a different speed in a different
material. As a result each material has its own Refractive Index which we use to
help us calculate the amount of bending which takes place. Refractive index is
defined as:
n =C
Equation 1.1

where
n is the refractive index
C is the speed of light in a vacuum
 is the speed of light in the material
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The indexes of refraction of several common materials are given in Table1.1.
Medium
Index of Refraction
Vacuum
1.0
Air
1.0003 ( 1.0 )
Water
1.33
Ethyl alcohol
1.36
Fused quartz
1.46
Glass fiber
1.5-1.9
Diamond
2.0-2.42
Silicon
3.4
Gallium-arsenide
3.6
Index of refraction is based on a wavelength of light emitted from a
sodium flame (5890 Å)
Table 1.1
Typical Indexes of Refraction
How a light ray reacts when it meets the interface of two transmissive materials
that have different indexes of refraction can be explained with Snell’s law.
Snell’s law simply states
n1 sin 1 = n2 sin 2
Equation 1.2
where n1 = refractive index of material 1 (unit less)
n2 = refractive index of material 2 (unit less)
1 = angle of incidence (degrees)
2 = angle of refraction (degrees)
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A refractive index for Snell’s law is shown in figure 1.5. At the interface, the
incident ray may be refracted toward the normal or away from it, depending on
whether n1 is less than or greater than n2 .
( n1  n2 )
Normal
( n1 = n2 )
Unrefracted Ray
2
( n1  n2 )
Refracted Ray
2
Medium n2
Medium n1
1
(angle of incidence)
Incident ray
Figure 1.5 Refractive model for Snell’s law
1.5.1 Critical Angle
The critical angle is defined as the minimum angle of incidence at which a light
ray may strike the interface of two media and result in an angle of refraction of
90 or greater, as shown in Figure 1.6. This definition pertains only when the
light ray is traveling from a more dense medium into a less dense medium. The
critical angle can be derived from Snell’s law as follows:
n1 sin 1 = n2 sin 2
sin 1 = n2 sin 2
15
n1
When 2 = 90 which result to sin 2 = 1, then 1 = C , Therefore,
n 
Critical Angle : C = sin 1  2 
 n1 
Equation
1.3
Normal
2
n2 less dense
Refracted ray
(more to less dense)
n1 more dense
1=C
(minimum)
Incident ray
Figure 1.6 Critical Angle Refraction
1.5.2 Total Internal Reflection
The transmitted ray now tries to travel in both materials simultaneously for
various reasons this is physically impossible so there is no transmitted ray and all
the light energy is reflected. This is true for any value of 1, the angle of
incidence is equal to or greater than c Figure1.7 shows the Total Internal
Reflection (TIR).
16
We can define the two conditions necessary for TIR to occur:
1. The refractive index of the first medium is greater than the refractive index of
the second one.
2. The angle of incidence, 1, is greater than or equal to the critical angle, c
The phenomenon of TIR causes 100% reflection. In no other situation in nature,
where light is reflected, does 100% reflection occur. So TIR is unique and very
useful.
Normal
n2 less dense
n1 more dense
1C
Total Internal
Reflection of
Incident Ray
Incident ray
Figure 1.7 The Total Internal Reflection
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1.5.3 Numerical Aperture
One of the properties of a fiber, which we need to know, is called the Numerical
Aperture. The numerical aperture is defined as:
Equation 1.4
and
NA 
Also,
sin 1 NA   A
Where
n12  n22
Equation 1.5
NA
=
numerical aperture (unitless)
n1
=
refractive index of the glass core (unitless)
n2
=
refractive index of the cladding (unitless)
A
=
acceptance angle (degrees)
It is a measure of the light gathering power of the fiber. It lies between 0 and 1. A
numerical aperture of 0 means that the fiber gathers no light (corresponding to A
= 0o). A numerical aperture of 1 means that the fiber gathers all the light that falls
onto it (corresponding to A = 90o).
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1.6 Light Ray Propagation in a Fiber-Optic Cable
The angle A in the Figure1.8 is called the Acceptance Angle. Any light entering
the fiber at less than this angle will meet the cladding at an angle greater than c.
If light meets the inner surface of the cladding (the core-cladding interface) at
greater than or equal to c then TIR occurs. So all the energy in the ray of light is
reflected back into the core and none escapes into the cladding. The ray then
crosses to the other side of the core and because the fiber is more or less straight,
the ray will meet the cladding on the other side at an angle which again causes
TIR. The ray is then reflected back across the core again and the same thing
happens. In this way the light zigzags its way along the fiber. This means that the
light will be transmitted to the end of the fiber.
Cladding
n=1.47
TIR
1
A
Core
n=1.5
Cladding
Figure1.8 Propagation of light in a fiber-optic
1.6.1 Mode of Propagation and Index profile
In fiber-optic terminology, the word mode can be defined as path. If there is only
one path for light to take down the cable, it is called single mode. If there are
more paths that can be used in a fiber-optic cable, it is called multimode.
Where by the index profile of an optical-fiber is a graphical representation of the
value of the refractive index across the fiber. There are two basic types of index
profile: Step and graded. A step-index fiber has a central core with a uniform
refractive index less than that of the central core. In a graded-index fiber there is
19
no cladding, and the refractive index of the core is non-uniform; it is highest at the
center and decreases gradually with distance toward the outer edge.
1.7 Classification of Fiber
Essentially, fiber-optic can be classified into three types of configurations: singlemode step index, multi-mode step index and multimode graded index, as shown
in figure 1.9.





 For long distance
 Difficult to work
with.
 Phone companies
and CATV
companies
For short distance
Easy to work with.
LANs
Provides more
bandwidth than (c)
Most common and
widely used type
 For short
distance
 Easy to work
with.
 LANs
 For very high
Figure
1.9 Core index profiles: (a) single-mode step index; (b) multi-mode step index; (c) multipulse rates
mode graded index
20
HINTS
& HELP
The simplest way of manipulating light is to reflect it. The
direction of reflected light can be predicted by applying the
law of reflection : the angle of incidence is equal to the
angle of reflection.
Example1.2
In figure1.5 , let medium 1 be glass and medium 2 be ethyl alcohol. For an angle
of incidence of 30 , determine the angle of refraction .
Solution to Example 1.2
From Table1.1,
n1 (glass) =
1.5
n2 (ethyl alcohol) =
1.36
Rearranging Equation 2.2 and substituting for n1, n2 and 1 gives us
n1
sin 1  sin  2
n2
1.5
sin 30  0.5514  sin  2
1.36
 2  sin 1 0.5514  33.47
21
Example1.3
Determine the critical angle for the fiber describe in Example1.2.
Solution to Example1.3
Using the Equation 1.2,
n 
 C = sin 1  2 
 n1 
 C = sin 1 
1.36 

 1 .5 
 65.05
Example1.4
Determine the numerical aperture for the fiber describe in Example1.2.
Solution to Example 1.4
Using the Equation 1.4,
NA 

n12  n22
1.52
 1.36 
2
 0.633
22
QUIZ 1B
TEST OUR UNDERSTANDING BEFORE YOU CONTINUE WITH THE
NEXT LESSON…!
Answer the following questions.
1.10
The device that converts the light pulses into an electrical signal is a
_________________.
1.11
Light is a type of _________________ radiation.
Choose the letter which best answers each question.
1.12
Total internal reflection takes place if the incident ray strikes the
interface at an angle with what relationship to the critical angle?
1.13
a.
Less than
b.
Greater than
c.
Zero
The operation of a fiber-optic cable is based on the principle of …
a.
Reflection
b.
Dispersion
c.
Absorption
23
1.14
1.15
Which of the following is not a common type of fiber-optic cable?
a.
Single-mode step index
b.
Multi-mode graded index
c.
Single-mode graded index
The ratio of the speed of light in air to the speeds of light in another
substance is called the
1.16
1.17
1.18
a.
Speed factor
b.
Index of reflection
c.
Index of refraction
Refraction is the
a.
Reflection of light waves
b.
Distortion of light waves
c.
Bending of light waves
Which type of fiber-optic cable is the most widely used?
a.
Multimode step-index
b.
Single-mode step-index
c
Multimode graded-index
Which type of fiber-optic cable is best for very high-speed data?
a.
Multimode step-index
b.
Single-mode step-index
c.
Multimode graded-index
24
Feedback To Quiz1B
1.10
light detector or photocell
1.11
electromagnetic
1.12
c
1.13
b
1.14
a
1.15
b
1.16
c
1.17
c
1.18
a
1.19
c
CONGRATULATIONS !!!
Now you can proceed to the
next activity
25
KEY FACTS
1. Fiber-optics is a communications technology using transmission of light over
glass or plastic fibers.
2. Total Internal Reflection is a used in fiber-optics, governing how light rays
propagate through a transparent medium by reflecting off its boundaries.
3. The multimode step index fiber cable is widely used at short to medium
distances at relatively low pulse frequencies. This cable is also the easiest to make
and the least expensive of the fiber-optic cable.
26
SELF-ASSESSMENT
You are approaching success. Try all the questions in this self-assessment
section and check your answers with those given in the Feedback on SelfAssessment given on the next page.
Question 1-1
a.
List three general bands in the light frequency spectrum.
Question 1-2
a.
Fiber-optic communications system is primary used in
_________________________.
Question 1-3
a.
Outline the primary building blocks of a fiber-optic system.
b.
Briefly describe the construction of a fiber-optic cable.
Question 1-4
a.
State Snell’s law for refraction and outline its significant in fiberoptic cables.
b.
Define the following terms : refractive index, critical angle and
total internal reflection.
c.
For a glass (n = 1.5) / quartz (n=1.41) interface and an angle
incidence of 38, determine the angle of refraction, the critical
angle and the numerical aperture for the cable.
d.
A glass fiber has an index of refraction of 1.55 is surrounded by
water whose index of refraction is 1.33. Compute the critical
angle,c , above which total internal reflection occurs in the glass
slab.
27
e.
A glass fiber has an index of refraction of 1.62. It surrounded by
cladding material having an index of refraction of 1.604. Compute
the critical angle, c.
Feedback To Self-Assessment
Have you tried the questions????? If “YES”, check our answers now.
Answer 1-1
a.
Infrared, visible and ultraviolet.
Answer 1-2
a.
Long-distance telephone systems and cable TV system.
Answer 1-3
a.
Coder
Light
source
Fiber Optic
Cable
Decoder
b.
Light
Detector
It is comprised of three layers ; the centre core that carries the light, the
cladding layer covering the core that confines the light to the core, and the
coating that provides protection for the cladding.
28
Answer 1-4
a.
Snell’s law simply states
n1 sin 1 = n2 sin 2
where n1 = refractive index of material 1 (unit less)
n2 = refractive index of material 2 (unit less)
1 = angle of incidence (degrees)
2 = angle of refraction (degrees)
Snell’s law is applied in TIR concept when :
i.
The refractive index of the first medium is greater than the refractive
index of the second one.
ii.
The angle of incidence, 1, is greater than or equal to the critical angle, c
b.
refractive index
:
Determines the amount of bending that light
undergoes
when entering a different medium.
critical angle :
The minimum angle of incidence at which a
light ray may strike the interface of two media and result in an angle of
refraction of 90 or greater.
total internal reflaction.
:
The principle of how light rays
propagate through a transparent medium by reflecting off its boundaries.
c.
2 = 41
d.
C = 59
C = 70
NA = 0.512
29
e.
C = 82
30