Chap7 Fiber Communication

8/7/2019 Chap7 Fiber Communication

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Optical Fiber Communications

Objectives To discuss the importance of optical fiber communication To introduce optical fiber communication system 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 introduce the concept of modes in waveguides To study the transmission properties of optical fibers To explain the working principle of photodetector To study optical fiber communication system design

The objective of any communications system is the transfer of information from one point to another. Thisinformation transfer is accomplished most often by superimposing (modulating) the information onto an

electromagnetic wave (carrier). The modulated carrier is then transmitted (propagated) to the destination,where the electromagnetic wave is received and the information recovered (demodulated). Such systems

are often designated by the location of the carrier frequency in the electromagnetic spectrum. In radio

systems, the electromagnetic carrier wave is selected with a frequency from the radio frequency (RF)portion of the spectrum. In an optical fiber communication system, the carrier is selected from the optical

region, which includes the infrared and visible frequencies.

What are the advantages of optical fiber communication system?

1. Wide bandwidthIn any communication system, the amount of information transmitted is directly related to the

bandwidth of the modulated carrier. Thus increasing the carrier frequency increases the availabletransmission bandwidth, and therefore the information capacity of the overall system. The frequencies

in the optical range will have a usable bandwidth approximately 105 times that of a carrier in RF range.

2. Low lossOptical 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.

Bandwidth is an indication of the 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. This reduction in equipment and components decreases the system cost and complexity.

The typical optical communications system essentially consists of a transmitter with a diode laser, areceiverwith aphoto-diode and an optical fibre between the two serving as the transmitting medium.


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Block diagram of the optical fiber communication system

An optical fiber communication system uses a digital communication scheme.

Optical sources

The fundamental function of optical source in optical fiber communications is to convert electrical energy

in the form of current into optical energy.The main optical sources currently used in optical fiber communications are lasers and light emitting

diodes (LEDs).

Light emitting diode (LED)

AnLED is essentially a semiconductor p-n junction under forward bias.In this condition, electrons cross the pn junction from the n-type material and recombine with holes in the

p-type material. When recombine takes place, the recombining electrons release energy in the form of heatand light. A large exposed surface area on one layer of the semiconductive material permits thephoton to

be emitted as visible light. This process is called electroluminescence. Various impurities are added during

the doping process to establish the wavelength of the emitted light. The wavelength determines the color of

the light and if it is visible or invisible (infrared).


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In the energy diagram it can be seen that, the major carriers from both sides of the p-n junction are injected

to the other side where they become minority. The relatively large local minority carrier population close to

the junction leads to a minority carrier concentration gradient. As a consequence, the excess minority

carriers will diffuse away from the junction recombine with carriers and emit photons.

LEDs are made of gallium arsennide (GaAs), gallium arsenide phosphide (GaAsP), orgallium phosphide

(GaP). Silicon and germanium are not used because they are essentially heat-producing materials and arevery poor at producing light. GaAs LEDs emit infrared (IR) radiation, which is nonvisible, GaAsP

produces either red or yellow visible light, and GaP emits red or green visible light. LEDs that emit blue

light are also available. Red is the most common.

In LED, the dominant photon generation is spontaneous emission in which the electron drops to the lowerenergy level in an entirely random way. The output spectrum of an LED is relatively wide.

Laser

The most frequently used light source in optical communication systems is laser.

Why a laser as the light source?(a) monochromatic: suitable for elimination of white noise

(b) coherent: suitable for synchronous detection;

(c) high power: improves signal to noise ratio;

(d) small divergence: improves efficiency of transmission; and(e) small source size: suitable for use with optical fibres

Laseris an acronym forlight amplification by stimulated emission of radiation.In general, an atomic system is characterized by discrete energy level, and the constituent atoms can exist

in one of the allowed energy levels or states. The atoms can make upward or downward transitions between

any two allowed states by absorbing or releasing, respectively, an amount of energy equal to the difference


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between the two energy levels. Thus, if we consider two levels of an atomic system that participate in aninteraction with optical radiation of photon energy

h = E2 E1where h is Plancks constant, then there could be three types of interactions:

Absorption

An atom in a lower energy state can absorb photons and make an upward transition to the higher energylevel.

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.

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 inphase with the incident photon.

ExampleSuppose you use an LED whose energy gap equals 2.5 eV. What is its radiating wavelength?Solution 1 eV = 1.602 10-19 J

Since the energy gap is the photon energy Ep, and Ep = h = hc /, then ashc = 6.62610-34 Js 3108 m/s 2010-26 mJEp = 2.5 eV = 2.5 1.60210

-19 410-19 J

We have = hc /Ep = 2010-26 mJ / 4H10-19 J = 5H10-7 m = 500 nm

What is the condition for light amplification by stimulated emission?

Population inversion

To achieve light amplification, it is necessary that the population of the upper energy level is greater thanthat of the lower energy level, this condition is known as population inversion.


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How to realize population inversion?

Population inversion is achieved by pumping techniques.Pumpingis to excite atoms into the upper energylevel and hence obtain a nonequilibrium distribution by the use of external source.

Optical feedback and laser oscillationIf the population inversion is achieved, light amplification occurs when a photon colliding with an atom in

the excited energy state causes the stimulated emission of a second photon and then both these photons

release two more. When the electromagnetic wave associated with these photons are in phase, amplified

emission is obtained. A positive feedback mechanism has to be provided in the amplifying medium toincrease the net gain and achieve a laser light output.

Totalreflector

Partialreflector

Some light escapes

Some light escapesThe rest is fed back

Eventually, equilibriumis established

Growth of stimulated emission in a resonant laser cavity


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Oscillations occur in the laser cavity over a small range of frequencies where the cavity gain is sufficient toovercome the losses incurred in the amplifying medium. Hence the laser source is not monochromatic. The

spectral emission from the laser source lies within the frequency range dictated by the gain curve.

The type of laser most frequently used in optical communication systems is the semiconductor diode laser,or the fibre laser.

Why semiconductor diode lasers?

(a) They are excited conveniently: by injecting current;

(b) Their operating wavelengths suffer low attenuation in fibres: small loss;(c) Their size is compatible with fibre dimensions: improved coupling;

(d) They work continuously at room temperature;

(e) They do not have moving parts: easier to use, no maintenance;

(f) They are highly efficient;(g) They have a very long life; and

(h) They are relatively cheap.

The laser diode commonly used in optical fiber communication systems is single mode, emits light with a

wavelength of around 1.55 m, within a bandwidth of 0.2 nm. The light emitting area is of the order of

several m2. Peak power is typically around 2mW.Modulation of the light produced by the laser can be achieved by modulating the current injected into the

laser, since the output of the laser follows closely the waveform of the injected current.

Light propagation in optical fibers

The simplest way to view light in fiber optics is by ray theory. The light is treated as a simple ray, shownby a line. An arrow on the line shows the direction of propagation.


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What is commonly calledspeed of lightis actually velocity of electromagnetic energy in a vacuum such asspace. Light travels at slower velocities in other materials such as glass. Light traveling from one material

to another changes speed, which, because of wave motion, results in light changing its traveling direction.This deflection of light is called refraction.

The refractive index, n, of a dielectrical medium is defined as the ratio of the velocity of light, c, invacuum, to the velocity of light in the medium, v.

n = c / vWhen a ray of light is incident on the interface between two dielectrics of different refractive indices with

an angle of incidence i, the reflected ray will propagate along the angle of reflection r, and the transmitted

ray will travel according to the angle of refraction t, and

r= I (Law of reflection)

n1sini = n2sint (Snells law)

where n1 and n2 are refractive indices of the incident and transmission regions, respectively.

Total internal reflection

From Snells law, n1sini = n2sint

if n1 > n2, then sini = (n2/n1) sint < sint,

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

sini = (n2/n1) sint = (n2/n1) sin90 = n2/n1 < 1

the refracted ray emerges parallel to the interface between the dielectrics, the angle of incidence, i, must be

less than 90.

The angle of incidence that yields an angle of refraction t = 90 is called the critical angle,C.

sinC = n2/n1

When the angle of incidence is greater than the critical angle, the light will be reflected back into theoriginating dielectric medium. This is known as total internal reflection.


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Numerical apertureThe acceptance angle is normally measured as numerical aperture (NA).

At the air-core interface,

n0sina = n1sin2 = n1sin(90 - C) = n1cosC = n1(1 sin2C)

1/2

= n1[1 (n2/n1)2]1/2 = (n1

2 - n22)1/2 = NA

The value of NA represents the light collecting ability.

Example

(a) What is the numerical aperture of silica fiber with n1 = 1.48 and n2 = 1.46 ?(b) What is the numerical aperture of plastic fiber where n1 = 1.495 and n2 = 1.402 ?

Solution

(a) NA = (n12 - n2

2)1/2 = (1.482 1.462)1/2 = 0.2425(b) NA = (n1

2 - n22)1/2 = (1.4952 1.4022)1/2 = 0.5192

*Notice the difference in the values of NA for these two fibers

Light propagation in graded-index fiber

The graded-index fibers guide light by refraction. The fibers refractive index decreases gradually awayfrom its center, finally dropping to the same value 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 withlower refractive index.

The light beam with largeri experience total internal reflection earlier, light beam with smalleri travels

longer distance until it experience total internal reflection.

Recall that in order to propagate a long distance in the optical fiber, the light beam must satisfy theconditions for total internal reflection,but

Can all the light beams propagate in the optical fiber for a long distance if they satisfy the conditions fortotal internal reflection?

The answer is: only a set of separate beams at distinct propagation angles can propagate in the optical fiberfor a long distance. These different beams are called modes.


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*Modes in waveguidesFrom 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.

If two rays have the same initial phase at two points A and B respectively, one ray goes along the path AC

and the other travels along BDEF. The optical path difference between the two rays is 2ndcosi.

Electromagnetic theory shows that there is a phase shift, , on reflection, so the total phase differencebetween the two rays is

t = 2kndcosi - 2

where k = 2/ is the propagation constant.

When t = 2kndcosi - 2 = 2m

the two light beams produce a constructive interference. However, this happens only for certain angle of

incidence as

cos 0, 1, 2, ,im

mnkd

+ = =

Rays with angle of incidence i satisfying above equation can travel along the waveguide and we say thatthey correspond to a waveguide mode.

The number of modes supported by a fiber waveguide depends on its size and properties. In particular, if

only one mode can be supported when the fiber core diameter is very small (8 12 m), the fiber is called

singlemode fiber.*A strict analysis using Maxwells electromagnetic wave theory tells us that a mode is a specific solution of

the wave equation that satisfies the appropriate boundary conditions and has the property that its spatial

distribution does not change with propagation.

Optical FibresThe fibre that is dominantly used for long-distance transmission is a step-index single mode optical fibre .


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This is called a single-mode fibre because it transmits only one light ray because of the small size of itscore.

Transmission properties of optical fibers

The most important transmission properties of optical fiber are attenuation and dispersion.

Fiber attenuation

The attenuation of an optical fiber measures the amount of light lost between input and output. Total

attenuation is the sum of all losses. It usually is dominated by imperfect light coupling into the fiber and

absorption and scattering within the fiber.Attenuation limits how far a signal can travel through a fiber before it becomes too weak to be detected.

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.

( ) LPP inout //log10 10=


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*Three main types of dispersion

1. Material dispersionMaterial 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.

2. Waveguide dispersionIt is equivalent to the angle between the ray and the fiber axis varying with wavelength which subsequentlyleads to a variation in the transmission times for the rays, and hence dispersion.

For a given mode, t = 2kndcosi - 2 = 2m = constant

i.e. 2kndcosi = constant

or 2(2/)ndcosi = constantcosi / = constant

so that is related to i.

e.g.1

( )cos

2nkd nd

+ + = =

Different elements lead to different values of1 and hence results in different transmission time.

3. Mode dispersionMode dispersion occurs only in multimode fibers. It arises because rays follow different paths through the

fiber and consequently arrive at the other end of the fiber at different times.


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Fibre DispersionBecause the bandwidth of the optical fibre is not infinitely large, the digital pulses broaden as they travel

along the fibre. This is called chromatic dispersion.Chromatic dispersion parameter D() expresses how much a light pulse will broaden per km of length

travelled if it has a spectral width of .

Optical detectors

The opticaldetectorperforms the opposite function from the source: It converts optical energy to electricalenergy. The commonly used detector is the photodiode, which produces current in response to incident

light.

P-N photodiodeThe simplest photodiode is the p-n photodiode. It is essentially a reverse biased p-n junction.

When revere biased, the electrical field sweeps mobile carriers to their respective majority sides. A

depletion region is therefore created. This barrier has the effect of stopping the majority carriers crossingthe junction. However, the field accelerates minority carriers to the opposite side of the junction, forming

the reverse leakage current to the diode. Thus intrinsic conditions are created in the depletion region.

A photon with an energy h Eg incident in or near the depletion region will excite an electron fromvalence 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.

Responsivity of the photodiodeThe efficiency of conversion from optical energy to electrical energy is measured by the use of responsivity

of the photodiode, which is the ratio of the output current of the detector to its optic input power.

Typical values of peak responsivity for the best photodiodes working at 1.5 m are of the order of 10 A/W.


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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

(c) The bit error rate.

The design requires an analysis of the limitations imposed by fibre loss and fibre dispersion. Both vary

with the wavelength, therefore the choice of the wavelength is a major design issue.

Then the receiver, transmitter and fibre are selected without overlooking the essential requirement of their

compatibility. Cost versus performance and system reliability is also considered.Finally, link power budget and bandwidth budget analyses are carried out to ensure that the desired system

performance can be achieved.

The Power Budget

The power budget calculation determines whether the power produced by the transmitter will be adequateto 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.

The power budget calculation considers the losses between two of the nearest repeaters / regenerators, and

hence determines the spacing of the repeaters.

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 trare respectively the rise times of the system, the transmitter, the fibre and the receiver

Rise timeThe rise time 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.

222

rfts tttt ++


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tr tf

Amplitude

tw

Pulse width

90%

10%

Example:The specifications of the components used in a 80-km fibre-optic transmission link are given as follows:

Transmitter: Emission wavelength = 1550 nm

Spectral width = 0.1 nm

Output power delivered to fibre = + 10 dBm (1 mW 0 dBm)

Rise-time = 0.1 nsReceiver: Rise-time = 0.1 ns

Minimum power at 10 Gbps (BER = 10-9

) = 30 dBmFibre: Total length: 80 km,

Attenuation = 0.3 dB/kmDispersion = 2 ps/nm/km at 1550 nm

Connector loss per km of fibre: 0.25 dB

Determine whether the system can successfully transmit a bit rate of 10 Gbps by performing power andrise-time budget calculations. Assume a maximum bit rate of 0.7 / (total system rise-time).

Solution:

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

Therefore, the maximum bit rate is 0.7 / 142 10-12 = 4.9 109 or 4.9 Gbps. This is not within therequired 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.

4 4

10 256 10 142ps+ + =


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Questions (Optical Fiber Communications)1. What are the advantages of optical fiber communication system?2. Describe an optical fiber. What is its function in a telecommunication system?3. What is refractive index? What is the refractive index of a vacuum?4. 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?

5. Why is it necessary to meet the total internal reflection requirement inside an optical fiber?6. What is an acceptance angle? Why do we need to know what this angle is?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 are modes in an optical fiber?9. What is dispersion? What are main types of dispersion in an optical fiber?10. What is spontaneous emission?11. Explain the working principle of LED.12. What is stimulated emission?13.

What is population inversion? How to realize it?14. What is the function of optical detector?

15. Explain the operating principle of a P-N photodiode.

Exercise 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 = 0.275 and n1 = 1.490. Find the critical propagation angle which equals 90-C .

glass rod: n =1.61

air: n =12

Chap7 Fiber Communication

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