1. Optical Fiber Communications
Objectives
To discuss the key advantages of optical fiber communication
To introduce optical fiber communication system
To discuss the light source in optical fiber communication systems
To describe the principle of LED
To describe the principle of laser
To illustrate light propagation in optical fibers
To explain total internal reflection
To introduce the concept of numerical aperture
To study the transmission properties of optical fibers
To explain the working principle of photodetector
To study optical fiber communication system design

2. Communication Channels
The major demand placed on telecommunication systems is for more information-carrying capacity because the volume of information produced (and required) increases rapidly.
Especially for digital communication systems, they require more channel capacity than analog systems  the need for higher information-carrying capacity.
Information-carrying capacity is proportional to channel (transmission) bandwidth  the channel bandwidth needs to be increased.

3. Communication Channels
However, the channel bandwidth is limited by
----- the frequency of the signal carrier.
The higher the carrier’s frequency, the greater the channel bandwidth and the higher the information-carrying capacity of the system.
The rule-of-thumb for estimating the transmission bandwidth is that its maximum value is approximately 10% of the carrier frequency.
Hence, if an electrical signal uses a 1 GHz carrier, then its maximum bandwidth is about 100 MHz.

4. Optical Fiber Communications
The information bandwidth is limited to be equal to
the carrier frequency or some fraction of the carrier
frequency.
The carrier wave with high frequency needs to be
selected.

5. Communication Channels
A copper wire can carry a signal up to several hundred kHz over several tens of kms of distance.
A coaxial cable can propagate a signal up to several hundreds of MHz.
Radio transmission is in the range of 500 kHz to 100 MHz. Microwaves, including satellite channels, operate up to 100 GHz.
Optical communications uses light as the carrier, light frequency is between 100 and 1000 THz (T = 1012).
Therefore, optical systems have the largest capacity for information transmission.

6. Optical Fiber Communications
Optical Fibres
They are normally made of hair-thin high purity silica glass, covered with plastic.
What are the key advantages of optical fiber communication system?
1. Wide bandwidth
The amount of information transmitted is directly related to the
bandwidth of the modulated carrier. The increasing of the carrier
frequency increases the available transmission bandwidth. The
optical frequency range has a usable bandwidth of 105 times that
of a carrier in RF range.
2. Low loss
Optical fibers have lower transmission losses than copper cables.
In a copper cable, the attenuation increases with modulation
frequency: the higher the frequency, the greater the loss.

7. Optical Fiber Communications
Bandwidth is an indication of the transmission rate at
which information can be sent.
Loss indicates how far the information can be sent. 
The combination of high bandwidth and low loss of
optical fiber communication system means more data
can be sent over longer distances, thereby decreasing the
number of wires and the repeaters required, and hence
decreasing the system cost and complexity.

8. Communication Channels
Other advantages of optical fibers:
(1) are not susceptible to electromagnetic interference (because they are insulators) and therefore have small crosstalk;
(2) high security (cannot be tapped, no sparks);
(3) are cheaper (abundant raw material);
(4) have lower weight, smaller size and are more flexible (thus are easier to install); and
(5) are corrosion resistant (thus have longer operating lifetimes).

9. Communication Channels
The disadvantage of optical fibre:
Coupling (for signal distribution) and
connecting (to other fibres) cost
is higher than coaxial copper cable.

10. Optical Fiber Communications
The typical optical communications system essentially
consists of a transmitter with a diode laser, a receiver
with a photo-diode and an optical fibre serving as the
transmitting medium.
Block diagram of the optical fiber communication system

11. Optical Fiber Communications
Light sources
The fundamental function of optical source
in optical fiber communications is to
convert electrical energy in the form of
current into optical energy.
A simple and low cost light source used in
short haul optical communication is light
emitting diode (LED)

12. Optical Fiber Communications
Light emitting diode (LED)
· An LED is a semiconductor p-n junction under forward bias.
· In forward bias condition, electrons cross the pn junction from the n-type material and recombine with holes in the p-type material. The recombining electrons release energy in the form of heat and light.

13. Optical Fiber Communications
Light emitting diode (LED)
Energy diagram:

14. Optical Fiber Communications
Light emitting diode (LED)
· In LED, the dominant photon generation is spontaneous emission in which the electron drops to the lower energy level in an entirely random way. The output spectrum of an LED is relatively wide.

15. Optical Fiber Communications
Spontaneous emission
An atom in an excited level can make a downward transition spontaneously (i.e., on its own) by emitting a photon corresponding to the energy difference between the two levels.

16. Optical Fiber Communications
During spontaneous emission, the amount of energy emitted is equal to the difference between the two energy levels, E2 and E1, which is also the photon energy.
Photon energy
h = E2 – E1
where h = 6.626×10-34 J•s is Planck’s constant

17. Optical Fiber Communications
Example
Suppose you use an LED whose energy gap equals 2.5 eV. What is its radiating wavelength?
Solution
1 eV = ×10-19 J
Since the energy gap is the photon energy Ep, and
Ep = h = hc /λ,
then as
hc = 6.626×10-34 J•s ×3×108 m/s ≈20×10-26 m•J
Ep = 2.5 eV = 2.5 ×1.602×10-19 ≈ 4×10-19 J
We have
λ = hc / Ep = 20×10-26 m•J / 4×10-19 J = 5×10-7 m = 500 nm

18. Optical Fiber Communications
The most frequently used light source in optical communication systems is laser.
Why we use a laser as the light source?
monochromatic: suitable for elimination of white noise
coherent: suitable for coherent detection
(c) high power: improves signal to noise ratio
(d) small divergence: improves efficiency of transmission
(e) small source size: suitable for use with optical fibres

19. Optical Fiber Communications
Laser
An acronym for light amplification by stimulated emission of radiation.
Stimulated emission
An atom in an excited level can make a downward transition in the presence of external stimulation by emitting a photon corresponding to the energy difference between the two levels. The emitted photon is in phase with the incident photon.

20. Optical Fiber Communications
Stimulated emission

21. Optical Fiber Communications
Absorption
An atom in a lower energy state can absorb photons and make an upward transition to the higher energy level.

22. Optical Fiber Communications
Population inversion
In laser operation, light amplification should be achieved, this needs that the population of the upper energy level is greater than that of the lower energy level, this condition is known as population inversion.

23. Optical Fiber Communications
How to realize population inversion?
--- By pumping techniques.
Pumping is to excite atoms into the upper energy
level and hence obtain a nonequilibrium distribution
by using the external source, such as a current source,
a light source etc.

24. Optical Fiber Communications
Furthermore, a resonant cavity is needed to build up stimulated emission by use of feedback.
Optical feedback and laser oscillation
Light amplification occurs when a photon colliding with an atom in the excited energy state causes the stimulated emission of a second photon, the two photons are in phase, and then both these photons release two more.
A positive feedback mechanism has to be used to increase the net gain and achieve a laser light output.

25. Optical Fiber Communications

26. Optical Fiber Communications
Light propagation in optical fibers
The simplest way to view light in fiber optics is by ray theory. In
this theory, the light is treated as a simple ray, shown by a line.
An arrow on the line shows the direction of propagation.
The speed of light in vacuum is: c = 300,000 km/s
However, the speed of light in medium is more slowly, v = c / n.
The ratio of the velocity of light, c, in vacuum, to the velocity of light in the medium, v, is the refractive index, n.
n = c / v

27. Optical Fiber Communications
Light traveling from one material to another causes
the change of speed, which results in the change of
light traveling direction.
This deflection of light is called refraction.

28. Optical Fiber Communications
The relation between incident ray and reflection ray:
r = i (Law of reflection)
The relation between incident ray and refraction ray:
n1sini = n2sint (Snell’s law)
where n1 and n2 are refractive indices of the incident and transmission regions, respectively.
An interesting phenomenon can be found in the light refraction.

29. Optical Fiber Communications

30. Optical Fiber Communications
Total internal reflection
From Snell’s law, n1sini = n2sint
if n1 > n2, then sini = (n2/n1) sint < sint,
which leads to i < t, i.e.
the angle of refraction is always greater than the
angle of incidence.
Thus, when the angle of refraction t = 90, as
sini = (n2/n1) sint = (n2/n1) sin90 = n2/n1 < 1
the angle of incidence, i, must be less than 90.

31. Optical Fiber Communications
Critical angle
The angle of incidence that yields an angle of refraction t = 90 is called the critical angle, C.
sinC = n2/n1
When the angle of incidence is greater than the
critical angle, the light will be reflected back into
the originating dielectric medium.
This is known as total internal reflection.

32. Optical Fiber Communications
Example
A Beam of light is incident on a plane boundary between two dielectrics. The incident-ray angle is at 10º to the boundary normal and the transmitted beam is at 12º. Which of the two media has the higher refractive index?
Solution
From Snell’s law
sint / sini = n1 / n2
since t > i, then sint > sini
and then we must have n1 > n2.

33. Optical Fiber Communications
Optical fibers:
An optical fiber consists of a core surrounded by a cladding.
There are two types of fibers: step-index fibers and graded-index fibers.

34. Optical Fiber Communications
Light guiding
In order to propagate a long distance in the optical fiber, the light beam must satisfy the conditions for total internal reflection
Conditions for total internal reflection in optical fiber
Refractive index of fiber core, n1, is greater than refractive index of fiber cladding n2, i.e. n1 > n2
The incident angle is larger than the critical angle. i > C

35. Optical Fiber Communications
How the light enter the optical fiber?
Acceptance angle
Acceptance angle, a, is the maximum angle over which light rays entering the fiber will be guided along its core.

36. Optical Fiber Communications
The acceptance angle is usually measured as the numerical aperture (NA).
Numerical aperture
At the air-core interface,
n0sina = n1sin2 = n1sin(90 - C)
= n1cosC = n1(1 – sin2C)1/2
= n1[1 – (n2/n1)2]1/2 = (n12 - n22)1/2 = NA
The value of NA represents the light collecting ability.

37. Optical Fiber Communications
Example
What is the numerical aperture of silica fiber with n1 = 1.48 and n2 = 1.46 ?
What is the numerical aperture of plastic fiber where n1 = and n2 = ?
Solution
(a) NA = (n12 - n22)1/2 = (1.482 – 1.462)1/2 =
NA = (n12 - n22)1/2 = ( – )1/2 =
* Notice the difference in the values of NA for these two fibers

38. Optical Fiber Communications
Light propagation in graded-index fiber
It guides light by refraction.
Its refractive index decreases gradually away from its center, dropping to the same as the cladding at the edge of the core.
The change in refractive index causes refraction, bending light rays back toward the axis as they pass through layers with lower refractive index.

39. Optical Fiber Communications
The light beam with larger i experience total internal reflection earlier, light beam with smaller i travels longer distance until it experience total internal reflection.

40. Optical Fiber Communications
Recall that
In order to propagate a long distance in the optical
fiber, the light beam must satisfy the conditions for
total internal reflection
Question
Can all the light beams propagate in the optical fiber for a long distance even if they satisfy the conditions for total internal reflection?

41. Optical Fiber Communications
Answer
Only a set of separate beams at distinct propagation angles can propagate in the optical fiber for a long distance.
These different beams are called modes.

42. Optical Fiber Communications
Modes in waveguides
From the theoretical standpoint, an optical fiber is a waveguide, which confines light waves so they travel along the fiber.
Modes are stable patterns in which a wave can travel along a waveguide.

43. Optical Fiber Communications
Two light beams in superposition

44. Optical Fiber Communications
If two rays have the same initial phase, along with AC and BDEF respectively. The optical path difference (OPD) between the two rays is x = 2ndcosi.
Electromagnetic theory shows that there is a phase shift, , on reflection, so the total phase difference between the two rays is
t = kx = 2kndcosi - 2
where k = 2/ is the propagation constant.
When t = 2kndcosi - 2 = 2m
the two light beams produce a constructive interference. However, this happens only for certain angle of incidence as
Rays with i satisfying above equation can travel along the waveguide and we say that they correspond to a waveguide mode.

45. Optical Fiber Communications
Optical Fibres
The fibre that is dominantly used for long-distance
transmission is a step-index single mode optical fibre .
The “single-mode” fibre transmits only one light ray.
The fiber core diameter of the single mode fiber is
8 – 12 m.

46. Optical Fiber Communications

47. Optical Fiber Communications
Transmission properties of optical fibers
The most important transmission properties of optical
fiber are attenuation and dispersion.
Attenuation limits how far a signal can travel through a
fiber before it becomes too weak to be detected.
How to measure the fiber attenuation?
By the amount of light lost between input and output.

48. Optical Fiber Communications
Fibre Attenuation
Fibre attenuation is a function of wavelength and it
gives a measure of the loss suffered by the light in
the fibre per km of length travelled. The
attenuation constant, , is given by
where L is the length of the fibre, Pin is the input light power and Pout is the output light power.

49. Optical Fiber Communications
Main types of fiber attenuation:
Absorption, scattering and light coupling loss
Absorption
Absorption is related to the material composition and the fabrication process for the fiber, which results in the dissipation of some of the transmitted optical power as heat in the waveguide.
Optical power → heat
 optical power loss

50. Optical Fiber Communications
Scattering
Scattering refers to the process by which the light wave encounters a particle smaller than its wavelength, with the results that energy is sent to a new direction.

51. Optical Fiber Communications
Bend loss
Bend loss is the loss resulting from bend. Bend can cause the change of incident angle at which the light hits the core-cladding boundary.

52. Optical Fiber Communications
Dispersion
Dispersion is the
spreading of a light pulse.
Dispersion limits digital transmission speed by causing pulses to overlap, so they cannot be distinguished.
The bit rate must be low enough to ensure that pulses do not overlap.

53. Optical Fiber Communications
Three main types of dispersion
Material dispersion
Material dispersion occurs because the refractive index of the material changes with the optical wavelength.
As n = n(), and n = c / v,
then v = c / n()
Different wavelength elements travel at different velocities through a fiber, even in the same mode.

54. Optical Fiber Communications
Waveguide dispersion
It is equivalent to the angle between the ray and the fiber axis varying with wavelength.
From
For a given mode (e.g. m = 1)
Different  lead to different values of 1 and hence results in different transmission time (in the same mode).

55. Optical Fiber Communications
Mode dispersion
Mode dispersion arises
because rays
follow different paths
through the fiber and
consequently arrive at the
other end of the fiber at
different times.
Different modes travel
with different speeds.

56. Optical Fiber Communications
Optical Detectors
The optical detector performs the opposite function from the source:
It converts optical energy to electrical energy.
The commonly used detector is the photodiode, which produces current in response to incident light.
P-N photodiode
· The simplest photodiode is the p-n photodiode. It is essentially a reverse biased p-n junction.

57. Reverse bias the diode
When the diode is reverse biased, the negative terminal of the battery attracts holes away from the pn junction, while the positive terminal attracts electrons away from the junction.  the depletion region widens
more positive ions are created in the n region and more negative ions in the p region.

58. Optical Fiber Communications
A photon with an energy h  Eg incident in or near the depletion region will excite an electron from valence band into the conduction band, and hence leave an empty hole in the valence band (photogeneration of an electron-hole pair).
Carrier pairs so generated near the junction are separated and swept (drift) under the influence of the electrical field to produce a current in the external circuit.

59. Optical Fiber Communications

60. Optical Fiber Communications
Fibre-optic System Design
The key system requirements which must be specified
in system design are:
(a) The desired (or possible) transmission distance;
(b) The data rate (channel bandwidth); and
The bit error rate.
Thus, link power budget and bandwidth budget analyses need to be carried out to ensure that the desired system performance can be achieved.

61. Optical Fiber Communications
The Power Budget
The power budget calculation determines whether the
power produced by the transmitter will be adequate to
overcome the fibre loss and the other losses so that an
adequate amount of light reaches the detector
(in order to achieve a minimum bit error rate).

62. Optical Fiber Communications
The Bandwidth Budget
The other consideration is to determine whether the
system bandwidth will be adequate.
Since there is an inverse relationship between the
bandwidth and the rise time, the system rise time limit
can be expressed as
where ts, tt, tf, and tr are respectively the rise times of
the system, the transmitter, the fibre and the receiver.

63. Optical Fiber Communications
Rise time
The rise time, tr, is the time it takes for the output to
change from 10% to 90% of its final value when the
input is a step function

64. The specifications of the components used in a 80-km fibre-optic
_______________________________________________________________________________________________________________
Example:
The specifications of the components used in a 80-km fibre-optic
transmission link are given as follows:
Transmitter : Emission wavelength = nm
Spectral width = 0.1 nm
Output power delivered to fibre
= dBm (1 mW = 0 dBm)
Rise-time = 0.1 ns = 100 ps
Receiver : Rise-time = 0.1 ns = 100ps
Minimum power at 10 Gbps (BER = 10-9)
=  30 dBm
Fibre: Total length: 80 km, Attenuation = 0.3 dB/km,
Dispersion = 2 ps/nm/km at 1550 nm
Connector loss per km of fibre: dB
Determine whether the system can successfully transmit a bit rate of
10 Gbps by performing power and rise-time budget calculations.
Assume a maximum bit rate of 0.7 / (total system rise-time).

65. The input power to receiver is  34 dBm (10  24  20 dB)
Solution:
The input power to receiver is  34 dBm (10  24  20 dB)
which is less than the required power ( 30 dBm).
Therefore, there is a shortfall in the power budget.
The system rise-time is: (1 ps = s)

66. Therefore, the maximum bit rate is 0.7 / 142  10-12 = 4.9  109 or
Solution:
Therefore, the maximum bit rate is 0.7 / 142  = 4.9  109 or
4.9 Gbps. This is not within the required bandwidth budget (i.e. it is
less than 10 Gbps, which means that the system would also not
conform to requirement as far as bandwidth is concerned.) The
bandwidth and the power budget requirements are not met, therefore,
this system is not viable unless some changes to loss or power and
also the rise-time are made.

67. Questions (Optical fiber Communications)
What are the key advantages of optical fiber communication system?
Describe an optical fiber. What is its function in a telecommunication system?
What is refractive index? What is the refractive index of a vacuum?
Describe the functions of core and the cladding in an optical fiber. Why are their refractive indexes different? Which one has to be greater and why?
Why is it necessary to meet the total internal reflection requirement inside an optical fiber?
What is the acceptance angle? Why do we need to know what this angle is?

68. Questions (Optical fiber Communications)
7. What is attenuation in an optical fiber? List of three major causes of attenuation in an optical fiber and explain its mechanisms.
8. What is dispersion? What are main types of dispersion in an optical fiber?
9. What is spontaneous emission?
10. Explain the working principle of LED.
11. What is stimulated emission?
12. What is population inversion? How to realize it?
13. What is the function of optical detector?
14. Explain the operating principle of a P-N photodiode.
15. How to carry out power budget and bandwidth budget analysis in the optical fiber communication system design?

69. Problems (Optical Fiber Communications)
1. Suppose a laser diode radiates red light with wavelength of 650nm. What is the energy of a single photon?
2. Assume you have a glass rod surrounded by air as shown. Find the critical incident angle.
3. What is the acceptance angle for the fiber when n1 = 1.48 and n2 = 1.46?
4. For a specific fiber NA = and n1 = Find the critical propagation angle which is equal to 90º - θC .

70. Solutions (Optical Fiber Communications)
Ep = hf = hc/
= (6.6   3  108)/650  10-9 = 3.04  J
2. From Snell’s law, n1 sin1 = n2 sin2
Since n1 = 1.6, n2 = 1.0, while 2 = 90, we have
1c = arcsin (1/1.6) = 38.68

71. Solutions (Optical Fiber Communications)
3. From Snell’s law, n0 sina = n1 sin2.
For air, n0 = The critical angle sinc = n2/n1.
Since 2 = 90 - c, sinc = cos2;
hence cos2 = n2/n1.
Thus sin2 = [1- (n2/n1)2]1/2
sina = n1 [1- (n2/n1)2]1/2 = 1.48 [1 – (1.46/1.48)2]1/2 =
a = 
The whole acceptance angle is: 2a = 28.07

72. Problems (Optical Fiber Communications)
4. n2 = [n12 - NA2]1/2 = [1.492 – ]1/2 = ,
c = arcsin (n2 / n1) = arcsin ( ) = 79.4
90º - c = 10.6