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Optical Fiber Communications

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

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Optical Fiber Communications

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  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 hundredkHz over several tens of kms of distance. A coaxial cable can propagate a signal up to several hundreds ofMHz. Radio transmission is in the range of 500 kHz to 100 MHz. Microwaves, including satellite channels, operate up to100 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 atransmitter 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•sis 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 = 1.602 ×10-19 J Since theenergy gap is the photon energy Ep, and Ep =h=hc /λ, then as hc= 6.626×10-34J•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 aspopulation 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 = 1.495 and n2 = 1.402 ? • Solution • (a) NA = (n12 - n22)1/2 = (1.482 – 1.462)1/2 = 0.2425 • NA = (n12 - n22)1/2 = (1.4952 – 1.4022)1/2 = 0.5192 • * 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 calledmodes.

  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.

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