Optical Fiber Technology
Numerical Aperture (NA) What is numerical aperture (NA)? Numerical aperture is the measure of the light gathering ability of optical fiber The higher the NA, the larger the core of light acceptance of the fiber and the easier it is to couple the light signal into the fiber At the same time, the higher the numerical aperture, the lower the bandwidth The two specifications must be balanced for optimum performance
Numerical Aperture Specifying numerical aperture 62.5/125 0.275 +- 0.015
Index of Refraction C= 3 10 8 meters per second, but it is reduced when it passes through matter. The index of refraction n: n = c υ c υ speed of light in a vacuum, 3 108 m/s speed of light in the given material λλ 0 υ 0 λ = 0 f = c ν υ λ λ 0 υ = λ f n = 0 = 1 n λ λ 0 wavelength of light in a vacuum wavelength of light in the given material
Index of refraction and speed of light for various materials. Index of Refraction Speed of Light Free space (vacuum) 1.0 3 10 8 m/s Air at sea level 1.003 2.99 10 8 m/s Ice 1.31 2.29 10 8 m/s Water 1.33 2.26 10 8 m/s Glass (minimum) 1.45 2.07 10 8 m/s Glass (maximum) 1.80 1.67 10 8 m/s Diamond 2.42 1.24 10 8 m/s
Refraction with Snell's Law n sinθ = n sin 1 1 2 θ 2 θ 1 : The incident angle (from the surface normal) θ 2 : The angle of refracted light (from the surface normal) n 1 : index of refraction in the incident medium n 2 : index of refraction in the refracting medium Light that is not absorbed or refracted will be reflected. The incident ray, the reflected ray, the refracted ray, and the normal to the surface will all lie in the same plane.
Critical Angle We want to find the critical case of total internal reflection at the corecladding boundary. Using Snell s Law with ϕ 2 = 90º, we can find the critical angle ϕ CR : n2 n2 ( ) sin ϕ CR =, or ϕ CR = arcsin n1 n1 Air n 0 Unguided ray Cladding n 2 φ 2 φ 2 = 90º if φ = φ CR Core n 1 θ ŕ φ φ φ θ r θ i θ i Incident ray Reflected ray Cladding
Numerical Aperture -- Mathematically Since we can relate θ r, CR to angle ϕ CR by simple geometry, and we can make the approximate n 0 = 1, this equation can be simplified: The negated and shifted sine function is identical to the cosine, and we can relate this cosine to the sine by the trigonometric identity: sin this sine is replaced in terms of n 1 and n 2 : sin ( ) θ n sinθ = n sin ϕ i,cr π 2 = 1 r,cr 1 CR π 2 ( ) 2 ( θ ) = n sin ϕ = n cos( ϕ ) = n 1 ( ϕ ) i, CR 1 CR 1 CR 1 sin n 2 2 2 ( ) = n = n n NA sin θ, 1 i CR 1 = n 1 2 1 2 CR
For n1 n2, we can simplify the numerical aperture calculation: ( ) ( ) ( ) ( ) ( ) Λ = = + = 2 2 2 sin 1 1 2 1 1 2 1 1 2 1 2 1,CR n n n n n n n n n n n n θ i 1 2 1 n n n = Λ For Δ <<1
Acceptance Angle θ a is the maximum angle to the axis at which light may enter the fiber in order to be propagated, and is often referred to as the acceptance angle for the fiber. NA can be specified in terms of acceptance angle as, NA = n o sin θ a = (n 12 n 22 ) 1/2
Numerical Aperture Example 2.1 A silica optical fiber with a core diameter large enough to be considered by ray theory analysis has acore refractive index of 1.50 and a cladding ref. index of 1.47. Determine: a) critical angle b) NA c) Acceptance angle
Numerical Aperture Example 2.1 Solution: a) θc = sin -1 n 2 /n 1 = sin -1 1.47/1.5 = 78.5 o b) NA = (n 12 -n 22 ) 1/2 = (1.5 2 1.47 2 ) 1/2 = 0.30 c) θ a = sin -1 NA = sin -1 0.30 = 17.4 o
Numerical Aperture -- Example For instance, if n1 = 1.5 and Λ =0.01, then the numerical aperture is 0.212 and the critical angle θ cr, is about 12.5 degrees. See also example 2.2 and 2.3
Loss and Bandwidth -- Attenuation Attenuation ranges from 0.1 db/km (single-mode silica fibers) to over 300 db/km (plastic fiber) There are two reasons for attenuation: Scattering; Absorption Attenuation (db/km) 2.5 2.0 850 nm Window Attenuation (db) = OH Absorption Peak 10 log 10 P P 2 1 1.5 1.0 0.5 1300 nm Window 1550 nm Window 0.0 800 900 1000 1100 1200 1300 1400 1500 1600 1700 Wavelength (nm)
Loss and Bandwidth Loss or attenuation is a limiting parameter in fiber optic systems Fiber optic transmission systems became competitive with electrical transmission lines only when losses were reduced to allow signal transmission over distances greater than 10 km Fiber attenuation can be described by the general relation: P out = P in α L where α is the power attenuation coefficient per unit length
Loss and Bandwidth Attenuation is conveniently expressed in terms of db/km α ( db km) 10 = log L 10 = log L 10 = L = 4.34α ( αl) log ( e) Power is often expressed in dbm (dbm is db from 1mW) 10 10 P P out in Pine P αl 10 mw P = 10 mw = 10log10 = 10 dbm 1 mw 27 10 P = 27 dbm = 1 mw 10 = 501 in 10 mw
Loss and Bandwidth Example: 10mW of power is launched into an optical fiber that has an attenuation of α=0.6 db/km. What is the received power after traveling a distance of 100 km? 50 10 P out = ( 10 ) 1 mw = 10 nw Initial power is: P in = 10 dbm Received power is: P out = P α L in =10 dbm (0.6)(100) = -50 dbm
Loss and Bandwidth Example: 8mW of power is launched into an optical fiber that has an attenuation of α=0.6 db/km. The received power needs to be - 22dBm. What is the maximum transmission distance? Initial power is: P in = 10log 10 (8) = 9 dbm Received power is: P out = 1mW 10-2.2 = 6.3 μw P out -P in = 9dBm - (-22dBm) = 31dB = 0.6 L L=51.7 km
Causes of Attenuation Attenuation, or losses, in a fiber link come from a variety of sources Bending losses Absorption Atomic Absorption Scattering Rayleigh Scattering Mie-Scattering Brillouin Scattering
Absorption The portion of attenuation resulting from the conversion of optical power into another energy form, such as heat. Every material absorbs some light energy The amount of absorption can vary greatly with wavelength It depends very strongly on the composition of a substance
Absorption is uniform The same amount of the same material always absorbs the same fraction of light at the same wavelength. Absorption is cumulative The total amount of material the light passes through Material absorbs the same fraction of the light for each unit length
Atomic Absorption The atoms of any material are capable of absorbing specific wavelengths of light. because of their electron orbital structure. As light passes along an optical fibre. more and more light is absorbed by the atoms as it continues on its path
Intrinsic Absorption is caused by basic fiber-material properties. Intrinsic absorption sets the minimal level of absorption.
Extrinsic Absorption. is caused by impurities introduced into the fiber material. Extrinsic absorption also occurs when hydroxyl ions (OH-) are introduced into the fiber. Water in silica glass forms (Si-OH) bond
Material Absorption Material absorption Intrinsic: caused by atomic resonance of the fiber material Ultra-violet Infra-red: primary intrinsic absorption for optical communications Extrinsic: caused by atomic absorptions of external particles in the fiber Primarily caused by the O-H bond in water that has absorption peaks at λ=2.8, 1.4, 0.93, 0.7 μm Interaction between O-H bond and SiO 2 glass at λ=1.24 μm The most important absorption peaks are at λ=1.4 μm and 1.24 μm
Scattering
Scattering The interaction of light with density fluctuations within a fiber The inhomogeneities of the refractive index of the media are responsible for this phenomena. Light traveling through the fiber interacts with the density areas. Light is then partially scattered in all directions.
Types of Scattering Rayleigh Scattering Mie-Scattering Brillouin Scattering Raman Scattering
Rayleigh Scattering Is the scattering of light by particles smaller than the wavelength of the light Occurs when the size of the density fluctuation (fiber defect) is less than one-tenth of the operating wavelength of light. is more effective at short wavelengths Therefore the light scattered down to the earth at a large angle with respect to the direction of the sun's light is predominantly in the blue end of the spectrum.
intensity of the scattered light is inversely proportional to the fourth power of the wavelength
Mie Scattering If the size of the defect is greater than one-tenth of the wavelength of light, the scattering mechanism is called Mie scattering. Mie scattering, caused by these large defects in the fiber core. scatters light out of the fiber core. However, in commercial fibers, the effects of Mie scattering are insignificant
Rayleigh and Mie Scattering
Brillouin scattering spontaneous Brillouin scattering simulated Brillouin scattering
Spontaneous Brillouin scattering Scattering of light through Index variations induced by the pressure differences of an acoustic wave traveling through a transparent material. spontaneous Brillouin scattering, can also be described using the quantum physics: a photon from a pump lightwave is transformed in a new Stokes photon of lower frequency and a new phonon adding to the acoustic wave.
Absorption and Scattering Loss
External Losses Bending loss Radiation loss at bends in the optical fiber Insignificant unless R<1mm Larger radius of curvature becomes more significant if there are accumulated bending losses over a long distance Coupling and splicing loss Misalignment of core centers Tilt Air gaps End face reflections Mode mismatches
BENDING LOSSES
BENDING RADIUS The bend radius that causes loss due to light leaking from the core. When you exceed the minimum bend radius, your signal strength will drop. Typical radius is three to five inches.
Microbends Small microscopic bends Microbend loss increases attenuation because low-order modes become coupled with high-order modes that are naturally lossy Loss caused by microbending can still occur even if the fiber is cabled correctly
Macrobend losses Radius of curvature is large compared to the fiber diameter. During installation, if fibers are bent too sharply, macrobend losses will occur
Loss on Standard Optical Fiber Wavelength SMF28 62.5/125 850 nm 1.8 db/km 2.72 db/km 1300 nm 0.35 db/km 0.52 db/km 1380 nm 0.50 db/km 0.92 db/km 1550 nm 0.19 db/km 0.29 db/km
Indoor/Outdoor cables
Dispersion Dispersive medium: velocity of propagation depends on frequency Dispersion causes temporal pulse spreading Pulse overlap results in indistinguishable data Inter symbol interference (ISI) Dispersion is related to the velocity of the pulse
Material Dispersion Since optical sources do not emit just a single frequency but a band of frequencies, then there may be propagation delay differences between the different spectral components of the transmitted signal. The delay differences may be caused by material dispersion and waveguide dispersion. For a source with rms spectral width σ λ and mean wavelength λ, the rms pulse broadening due to material dispersion σ m is given by σ L 2 σ λ λ m λ c dn 1 d 2
Material Dispersion The Material Dispersion for optical fibers is sometimes quoted as a value for 2 2 dn1 λ ( ) 2 dλ or simply 2 dn1 2 dλ It may be given in terms of a material dispersion parameter M defined as: 2 1 d m dn1 M τ λ = = Ld c d 2 λ λ expressed in units of ps nm -1 km -1 Where τ m is the pulse delay due to material dispersion
Example 2 2 dn1 A glass fiber exhibits material dispersion given by λ ( ) 2 dλ of 0.025. Determine the material dispersion parameter at a wavelength of 0.85 μm, and estimate the rms pulse broadening per kilometer for a good LED source with an rms spectral width of 20nm at this wavelength.
Solution The material dispersion parameter may be obtained dn1 1 2 dn1 M = λ = λ c d 2 c d 2 λ λ λ = 0.025 x x 5 2.998 10 850 snm km 1 1 = 98.1psnm km 1 1 The rms pulse broadening is given as 2 σ L dn1 σ λ m λ 2 c d λ Therefore in terms of material dispersion parameter M σ m σ λ LM Hence, the rms pulse broadning per kilometer due to material dispersion σ 12 1 m(1 km) = 20198.110 x x x = 1.96nskm
Example 2 Estimate the rms pulse broadening per kilometer for the fiber in the above example when the optical source used is an injection laser with a relative spectral width σ λ /λ of 0.0012 at a wavelength of 0.85 μm
Solution The rms spectral width may be obtained from the relative spectral width by σ λ = 0.0012 λ = 0.0012 x 0.85 x 10-6 = 1.02nm The rms pulse broadening in terms of material dispersion parameter is given by σ σ m λ LM σ m = 1.02 x 1x 98.1 x 10-12 = 0.10 ns km -1 Hence the rms pulse broadening is reduced by a factor of 20 compared with the LED source in the previous example
Polarization mode Dispersion (PMD) Polarization mode dispersion (PMD) is another complex optical effect that can occur in singlemode optical fibers. Single-mode fibers support two perpendicular polarizations of the original transmitted signal. If a were perfectly round and free from all stresses, both polarization modes would propagate at exactly the same speed, resulting in zero PMD.
Polarization mode dispersion (PMD) However, practical fibers are not perfect, thus, the two perpendicular polarizations may travel at different speeds and, consequently, arrive at the end of the fiber at different times. The fiber is said to have a fast axis, and a slow axis. The difference in arrival times, normalized with length, is known as PMD (ps/km 0.5 ).
Polarization Mode Dispersion (PMD) Polarization mode dispersion is an inherent property of all optical media. It is caused by the difference in the propagation velocities of light in the orthogonal principal polarization states of the transmission medium. The net effect is that if an optical pulse contains both polarization components, then the different polarization components will travel at different speeds and arrive at different times, smearing the received optical signal.
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Assignment The material dispersion parameter for a glass fiber is 20 ps nm -1 km -1 at a wavelength of 1.5 μm. Estimate the pulse broadening due to material dispersion within the fiber when a light is launched from an injection laser source with a peak wavelength of 1.5 μm and an rms spectral width of 2nm into a 30 km length of fiber. The material distribution in an optical fiber defined d 2 n 1 /dλ 2 is 4.0 x 10-2 μm -2. Estimate the pulse broadening per kilometer due to material dispersion within the fiber when it is illuminated with an LED source with a peak wavelength of 0.9 μm and an rms spectral width of 45 nm. Questions 2.2, 2.4, 2.5 Ramaswami