OFCS TRANSMISSION CHARACTERISTICS OF OPTICAL FIBERS

OFCS TRANSMISSION CHARACTERISTICS OF OPTICAL FIBERS

Like any communication system there are some important factors affecting performance of optical fibers as a transmission medium. The most interest are those attenuation and bandwidth 1. Attenuation Is the ratio of the input (transmitted) optical power into the fiber to output (received) optical power from the fiber P o =P i 10 -αl/10 P o =P i -αl [W] [db] 9/9/2014 LECTURES 2

Example A fiber has a coupled power -8dBm and attenuation of 6 dbm/km, and a length 2 km. Calculate the output power 6dBm/km Example A fiber coupled power -8dBm and attenuation of 6 dbm/km. Find the fiber length if the output power is -30dBm. 6dBm/km 9/9/2014 LECTURES 3

A number of mechanisms are responsible for the signal attenuation within optical fibers: 1-Material absorption 2-Scattering 3-Fiber bend losses 4-losses due coupling the source to the fibers 5-losses due to mismatching between the fibers (Fresnel) 6-losses due to splices and connectors 7-Losses due to deviation of geometrical and optical parameters 8-losses due to misalignments 9-Modal coupling radiation losses 10- Leaky mode losses (1) Material absorption: Due to photon absorption in interaction with atoms or molecules of the material. It happens due to material composition and fabrication process impurities which causes attenuation in the transmitted optical power in the form of heat due to absorption It is divided into two types: -Intrinsic due to interaction with the main components of the glass -Extrinsic due to interaction with the impurities in the glass 9/9/2014 LECTURES 4

Attenuation (db/km) 100 10 Glass Absorption in UV 1 Glass Absorption in IR 0.1 0.01 Intrinsic absorption 0.5 0.6 0.7 1 1.2 1.5 2 3 5 10 Wavelength ( m) -Intrinsic absorption occurs in the UV region and have peaks also in the IR in the 7 to 12 m region. This type of absorption is insignificant because it is out of operation band of optical fiber 9/9/2014 LECTURES 5

-Extrinsic absorption is a major source of loss in practical fiber. There are two types are the main sources of impurity absorption : -Transition metal ions -OH ions Transition metals (e.g : copper, iron, etc ) absorbs strongly in the region of interest and so must not exceed a few parts per billion to ensure losses are kept below 20dB/km OH absorption occurs because of excess water content and peak absorption occurs at 2.73um (resonant wavelength for absorption).other wavelength causes large absorptions at 1.37, 1.23 and 0.95um. Therefore for efficient propagation those wavelength must be avoided 9/9/2014 LECTURES 6

(2) Scattering: -Linear scattering due to random refractive index through the material causes some optical power transfer from one propagating mode to another, this tends to attenuation of the transmitted light as the transfer may be to a leaky or radiated mode which does not continue to propagate within the fiber core, but is radiated from the fiber - Raleigh scattering: Is the scattering of the photons at the random boundaries due to inhomogenities or the randomness of the refractive index fluctuations -The glass was formed using heat, which caused a random movement of the molecules and when solidified the molecules were frozen in their random locations. This yields a random refractive index through the material 9/9/2014 LECTURES 7

-This applies when a wave travel through a medium having scattering objects much smaller than the wavelength (molecular level) -Thus, Raleigh scattering increases with the decrease in wavelength and was found to be proportional to λ -4 and can be approximated by the following expression : L=1.7(0.85/λ) 4 db/km - Mie scattering: Is the scattering of the photons at the inhomogenities due to nonperfect cylindrical structure such as irregularities in the core cladding interface, core cladding refractive index difference, diameter fluctuations, strain and bubbles -This applies when a wave travel through a medium having a wavelength comparable to the size of inhomogenities -It depends on the fiber material, design and manufacture -It can be reduced by: 1-Removing imperfections due to glass manufacturing process; 2-Carefully control the fiber coating 3-Increasing the fiber guidance by increasing the relative refractive index difference 9/9/2014 LECTURES 8

(3) Bend losses -Macro bend -Optical fibers suffer radiation losses at bends or curves. At that curvature most of the lower and higher order modes will incident at an angle less the critical, which let them out in the cladding, providing more power loss in the fiber -Macro bend are bends that are large enough to be seen by the human eye -Generally more at the cable level or for fibers, the bends necessary to fit fibers inside splice closures or patch panels -Macro bending testing is done by wrapping the fiber or cable around a mandrel of a specified diameter -The loss can be represented by a radiation attenuation coefficient which is given by : α b =c 1 exp(-c 2 R) where R is the radius of curvature of the fiber bend and c1, c2 are constants which are independent of R

-There is also a critical radius of curvature R c where large bending losses can occur given by : where λ c is the cutoff wavelength of the single mode fiber For single mode fibers the critical radius of curvature R CS can be approximately by : -Based on these relations the criteria to reduce macro bending losses is : Designing fibers with large relative refractive index difference Operating at the possible shortest wavelength -The losses due to bending can be determined as : P 2 / P 1 = 1- ((α+2)/2α )[2a/R +(3λ/4πn 2 R) 2/3 ] where P 1 and P 2 are the power before and after bending respectively It is possible that the cables could be run around corners better and fibers would not have as much loss when stuffed into small spaces

-Micro bend -These are identical in effect to the macro bend but differ in size and cause Their radius is equal to, or less than, the diameter of the bare fiber which is very small indeed -Micro bend loss refers to small scale "bends" in the fiber, often from pressure exerted on the fiber itself as when it is cabled and the other elements in the cable press on it -i.e., these are generally a manufacturing problem. For example if the fiber gets too cold, the outer layers will shrink and get shorter. If the core/cladding shrinks at a slower rate, it is likely to kink and cause a micro bend -There is no real test for micro bending Micro bending loss needs careful attention to the fiber coatings, if applied in several layers) which protects the fiber from being bent 9/9/2014 LECTURES 11

Example A66.5/125GI parabolic fiber of core and cladding index 1.448 and 1.443 respectively operating at 1.3um. If the radius of curvature of the turn is 2cm, find the power lost in that turn Example Assume a multimode graded index fiber has a refractive index at the core axis of 1.46 with a cladding refractive index of 1.45. The critical radius of curvature which allows large bending losses to occur is 84 µm when the fiber is transmitting light of a particular wavelength. Determine the wavelength of the transmitted light λ =0.82um 9/9/2014 LECTURES 12

db/km 6 4 2 The overall attenuation 800 1000 1200 1400 1600 λ [nm] 1330 1550 λ Type Size,μm db/km 800 850 1300 1550 SI SI GI GI GI GI SM SM 62.5/125 62.5/125 62.5/125 50/125 62.5/125 50/125 X/125 X/125 5.0 4.0 3.3 2.7 0.9 0.7 0.5 0.2 9/9/2014 LECTURES 13

(4) Losses due coupling the source to the fibers The power coupled to a fiber depends on many factors: the source diameter, variations of refractive index due to fabrications, variation in core diameter irregularities at interface, variation of the index profile and variation of the numerical aperture Due to spatial distribution of the source The power coupled to the fiber P c is related to the power emitted from the source P s by what is called the coupling efficiency and is defined as : P c / P s = (NA) 2 min[1,(a /r s ) 2 ]= η SI source r s Source radiation pattern lost power r s is the LED radius a =2n 1 2 [1-(2/(α+2))(r s /a) α ]= η GI In SM fiber a is small, and since NA is also small, hence η is very small, so its very important to use LASER DIODE in SI and also in MM because its spread is narrow and most of the power is inside the acceptable angle of the fiber η laser = 30-50% 9/9/2014 LECTURES 14

(5) Fresnel reflection at fiber to fiber joint When the two jointed fiber ends are smooth and perpendicular to the fiber axes, and the two fiber axes are perfectly aligned, there is a reflection causes loss or attenuation due to mismatching of the refractive index of the medium between the two jointed fibers. The fraction of the light power reflected at a single interface is: R=[(n 1 -n)/(n 1 +n)] 2 n is the refractive index of the medium between the two fibers and the transmission ratio is defined as : η t = P c / P emitted = 1-R or in db is η t =10log(1-R) or the optical loss due to Fresnel reflection at a single interface is Loss Fres = -10log(1-R) This loss should be taken into consideration at both fiber interface 9/9/2014 LECTURES 15

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Example Two identical optical fibers of core index 1.5. If the end faces of two fibers are butted together and their axes are perfectly aligned, then calculate the optical loss due Fresnel reflection when there is an air gap between the fiber end faces Solution R= [(1.5-1)/(1.5+1)] 2 = 0.04 Loss Fres = -10log(1-R) = 0.18 db Then the total losses due to Fresnel reflection at the two faces = 0.18+0.18 = 0.36 Matching transformer When two junctions( fiber and fiber, fiber and source, ) are different in index of refraction we insert a matching material between them to achieve minimum reflection as: Example n m ' n 1 n 1

(6) Losses due to splices and connectors Any communication systems have requirements for joining and terminations of the transmission medium. The number of intermediate connections or joints is dependent upon the link length between repeaters. In optical fiber the joints are: -Fiber splices (like the soldered joints in other systems) -Fiber demountable connectors (like plugs and sockets in other systems) These types are used to couple the light from one fiber to the adjoint one -Fiber coupler: splits all the light ( or proportion) from the main fiber into two or more fibers. Also it combines the light from branch fibers into the main fiber The joint losses is critically dependent on the alignment of the two fibers 9/9/2014 LECTURES 18

(7)Losses due to deviation of geometrical and optical parameters There are inherent connection problems when jointing fiber with : -different core and /or cladding diameter; -different NA and/or relative refractive index difference Δ; -different refractive index profile α; -different spot size ω o ; -fiber faults like core ellipticity, core concentricity, The losses caused by these factors and Fresnel reflection loss are usually referred to what is called INTRINSIC joint losses The best results are achieved with compatible (same) fibers which are manufactured to the lowest tolerance, but still the problem of the quality of alignment provided by jointing mechanisms (EXTRINSIC joint loss) 9/9/2014 LECTURES 19

Intrinsic losses In addition to the reflection loss the losses due to different geometry and properties of the fiber can be formulated as: -In multimode fiber joints Considering all the modes are equally excited in multimode (SI or GI) fibers, the loss from core diameter mismatch can be represented as: Loss cd = -10log(d R /d T ) 2 for d R < d T otherwise =0dB dt dr and the loss from refractive index profile mismatch can be represented as: Loss α = -10log (α R (α T +2))/ (α T (α R +2))] for d R < d T otherwise =0 db and the loss from NA mismatch can be represented as: NA T NA R Loss NA = -10log(NA R /NA T ) 2 for NA R < NA T otherwise =0 db 9/9/2014 LECTURES 20

-In single mode fiber joints The intrinsic coupling loss due to spot size mismatch can be represented as: Loss sz = -10log[4(ω or /ω ot + ω ot /ω or ) -2 ] db where ω or and ω ot are the spot sizes of the receiving and transmitting fibers respectively Example Two single mode fibers with mode field diameter of 11.2um and 8.4um are to be connected together. Assuming no extrinsic losses, determine the intrinsic loss due to the modal field diameter mismatch. Solution Loss sz = -10log[4(4.2 /5.6+ 5.6 /4.2 ) -2 ] =-10log 0.922 = 0.35 db 9/9/2014 LECTURES 21

Example When the mean optical power launched into an 8km length of fiber is 120uW, the mean optical power at the fiber output is 3 uw. Determine: -the overall signal attenuation or loss in db through the fiber without connectors or splices; -the signal attenuation per kilometer of the fiber; -the overall signal attenuation for 10 km optical link using the same fiber with splices at 1km intervals, each giving an attenuation of 1dB; -the numerical input/output power ratio in the previous link Solution = 9/9/2014 LECTURES 22

= = = = 9/9/2014 LECTURES 23

(8)Misalignment losses (extrinsic loss) Longitudinal: the separation between the fiber Lateral (radial, axial): offset perpendicular to the core axis Angular : the angle between the core axes 0.1dB for z=10um z 1dB for y=10um y 1dB for θ=4 o -5 o Longitudinal Lateral Angular θ GI fiber MMSI fiber 9/9/2014 LECTURES 24

-In multimode fiber joints -MMSI: Lateral misalignment coupling efficiency Lateral misalignment loss Loss lat = -10logη lat Angular misalignment coupling efficiency Angular misalignment loss Loss ang = -10logη ang 9/9/2014 LECTURES 25

Example SI fiber has a refractive index of 1.5 with a core diameter of 50um. The fiber is jointed with a lateral misalignment between the core axes of 5um. Determine the insertion loss at the joint due to lateral misalignment: a-there is a small air gab at the joint; b-the gab is considered index matched Solution a- - b- 9/9/2014 LECTURES 26

Example SI fiber has a refractive index of 1.48 with numerical aperture of 0.2. The fiber is jointed with 5 o air angular misalignment of the core axes. Determine the insertion loss at the joint due to an angular misalignment Solution 9/9/2014 LECTURES 27 = =

-MMGI: Lateral misalignment coupling efficiency L t = 0.75y/a η lat = 1- L t where for uniform optical power distribution for all guided modes for uniform optical power distribution for all guided modes and leaky modes Lateral misalignment loss Loss lat = -10logη lat -In single fiber joints Considering Gaussian or near Gaussian shape of the modes propagating in single mode fibers (SI or GI) and the spot size of the two coupled fibers are the same, the total lateral and angular loss can be represented as: ω o is the normalized spot size for single mode fiber, for the mode LP 10 is different and is given by 9/9/2014 LECTURES 28

Example SM fiber has a refractive index of 1.46 with numerical aperture of 0.1. The normalized frequency is 2.4 and the core diameter is 8um. Determine the insertion loss at the joint due lateral misalignment of 1 um and to an angular misalignment of 1 o consider LP 10 mode Solution 9/9/2014 LECTURES 29

2. Bandwidth of the fiber The other characteristic of primary interest is the bandwidth of the fiber which is limited by the signal dispersion within the fiber. Once the attenuation is reduced to acceptable levels, attention is directed towards the dispersive properties of the fiber Dispersion Dispersion with the fiber cause broadening of the transmitted light pulses as they travel along the channel -During an optical transmission of a digitally modulated signal, dispersion with the fiber cause broadening of the transmitted light pulses as they travel along the channel. As a result if we have a stream of digital pulses, each pulse broadens and overlapped with its neighbors and becomes indistinguishable at the receiver input -Since the broadening increases with the distance traveled along the fiber, we define the parameter BW x length of the fiber: Typical values: 20 MHz km (MMSI), 1GHz km (MMGI), 100 MHz km (SMSI) -This phenomena is clear in the following example: 9/9/2014 LECTURES 30

Amplitude Consider the input digital pattern to the fiber shown and notice the output at a distance d 1 and further distance d 2 Input digital bit pattern Amplitude Amplitude Time Distinguishable pulses Composite pattern output at d 1 Time Indistinguishable pulses no zero level output at d 2 Intersymbol interference Time Thus, pulse broadening causes overlapping between pulses and eventually the pulses can become indistinguishable 9/9/2014 LECTURES 31

This limits the maximum bit rate β T to be carried by the optical fiber A conservative estimate of which assumes a pulse duration of τ and that pulse spreading can be up to τ (broadening) is given by (no overlap at all): 1 T 2 A more accurate estimation of the maximum bit rate for an optical channel with dispersion can be obtained by taking into consideration that the light pulses at the output are Gaussian in shape with rms width σ. This allows a slight overlap while still avoiding any penalties and errors due to inter-symbol interference and low SNR The maximum bit rate in this case is 0.2 T (max) bits / s It is very important to point out that this formula gives a reasonable good approximation for other pulse shapes which may occur on the channel resulting from various dispersive mechanisms within the fiber and may be assumed to represent the rms impulse response for the channel 9/9/2014 LECTURES 33

Example A multimode graded index fiber exhibits total pulse broadening of 0.1us over a distance of 15km. Estimate: a) The max. possible BW assuming no inter symbol interference; b) The pulse dispersion per unit length; c) The BW-length product of the fiber. Solution a) T 1 2 = 1/(0.2x10-6 )=5 MHz b) Dispersion = dispersion /total length = 0.1 x 10-6 /15= 6.67 ns km -1 c) BW x length = 5 MHz x 15km=75 MHz km 9/9/2014 LECTURES 34

Types of dispersion Dispersion Intramodal (Chromatic) Intramodal: due to finite spectral line width of the optical source. The optical source emits a band of frequencies (LD emits fraction percent of the central frequency while LED emits significant percentage) We know that light of different wavelengths is refracted by differing amounts There will be a propagation delay differences between the different spectral components of the transmitted signal which in turn causes broadening of each transmitted mode and hence intramodal dispersion 9/9/2014 LECTURES 35

Example Intramodal dispersion Consider the source have a certain spectral width, the pulse generated will consist of a sum of identical pulses which are only different in their wavelengths. For simplicity consider we have only three wavelength components coming out of the source which constitute the pulse Components of input pulse Components of input pulse λ 1 Input pulse λ 1 output pulse λ 2 λ 2 λ 3 Source signal λ 3

Intermodal dispersion Dispersion Intramodal (Chromatic) Intermodal (mode) Intermodal: due to the propagation delay differences between the modes within a multimode fiber. (MMSI fiber exhibits a large amount of mode dispersion which gives the greatest pulse broadening) The single mode operation does not give intermodal dispersion and therefore pulse broadening is solely due to the intramodal dispersion 9/9/2014 LECTURES 37

Example Intermodal dispersion

-Material dispersion Dispersion Material Intramodal (Chromatic) Intermodal (mode) -Is the broadening due to different group velocities of the various spectral components launched from the source into the fiber -It occurs when the phase velocity of the plane wave propagating in the dielectric medium varies nonlinearly with the wavelength (d 2 n/dλ 2 0), i.e., due to atomic structure -Exists in all fibers and is a function of the source line width 9/9/2014 Spectrum of I/P pulse I/P pulse LECTURES O/P pulse

-The group delay (per unit length) is the inverse of the group velocity; τ g =1/v g -The pulse delay due to material dispersion in a fiber of length L is given by -For a source with rms spectral width σ λ and a mean λ, the rms pulse broadening due to material dispersion σ m is obtained as follows: -We define material dispersion factor M as M=(λ/c) d 2 n 1 /dλ 2 and the pulse broadening can be written as σ m = σ c =σ λ LM 9/9/2014 LECTURES 40

We can observe from the graph of material dispersion of Silica that, to minimize the material dispersion: -Longer wave length around 1.3um is used because it gives low material dispersion; tends to zero -LD with narrow spectral is used rather than LED Material dispersion parameter M (Silica) Region of negligible material dispersion λ in um 9/9/2014 LECTURES 41

Example 11- A glass fiber material dispersion expressed by λ 2 (d 2 n 1 /dλ 2 ) of 0.025. It is used with a good LED source of rms spectral width 20 nm at a wavelength of 0.85um. Estimate -The material dispersion parameter (factor) -The rms pulse broadening per kilometer for the used source and also when the optical source is LD with relative spectral width of 0.0012 at the same wavelength Solution -Concerning the material dispersion parameter or factor M M=(λ/c) d 2 n 1 /dλ 2 = (1/λc)λ 2 d 2 n 1 /dλ 2 = 0.025/(3x10 5 x0.85x10-6 )= 9.8x10-2 s m -1 km -1 -Concerning the material dispersion for LED and LD: for LED σ m /1km =σ λ M = 20 x 10-9 x M = 19.6x 10-10 s km -1 for LD with σ λ =0.0012λ =0.0012 x 0.85 x 10-6 =1.02 x 10-9 m σ m /1km=σ λ M = 1.02 x 10-9 x M = 0.9996x 10-10 s km -1 i.e., 20 times less than LED

-Waveguide dispersion Dispersion Material Intramodal (Chromatic) Waveguide Intermodal (mode) -Is the broadening due to variation in the group velocity with wavelength for a particular mode -It occurs when (d 2 β/dλ 2 0) -Exists in multimode fibers but can be significant in single mode fibers 9/9/2014 LECTURES 43

-Modal dispersion Dispersion Material Intramodal (Chromatic) Waveguide Intermodal (mode) Modal High order mode Low order mode Due to propagation delay differences between modes within the fiber. It has a greatest effect of pulse broadening in step index multimode fibers 9/9/2014 Intensity 0 Light pulse t Cla dding Core Axial Broadene d light pulse Intensity Spread, Schematic illustration of light propagation in a slab dielectric waveguide. Light pulse entering the waveguide breaks up into various modes which then propagate at different group velocities down the guide. At the end of the guide, the modes combine to constitute the output light pulse which is broader than the input light pulse. 1999 S.O. Kasap, Optoelectronics (Prentice Hall) t

Assume the length of the fiber is L; the pulse spread due to intermodal dispersion is due to the difference in time between the two extreme rays shown in the previous figure : θ θ c θ i Guide n 1 >n 2 Cladding n 2 but θ c but in terms of NA The rms broadening for MMSI fiber can be derived as σ nsi = 2 3c In terms of NA it can be written as NA 2 σ nsi = 4 3n1 c 9/9/2014 LECTURES 45

Using mode theory analysis it can be shown that the intermodal dispersion in GI fibers with parabolic index profile is given by : This correspond to an increase in transmission time for the slowest mode of Δ 2 /8 relative to the fastest mode σ ngi = For optimum index profile, rms intermodal broadening is σ ngi = 20 3 c The overall fiber dispersion in multimode is given by where σ c is the chromatic dispersion (which is all causes related to waveguide and material), σ n is from intermodal dispersion In most cases waveguide dispersion is ignored and the chromatic dispersion is mainly due to material dispersion

Example An 11 km optical fiber link consisting of the optimum near parabolic profile graded index fiber exhibits rms intermodal pulse broadening of 346 ps over its length. If the fiber has a relative refractive index of 1.5%, estimate the core axis refractive index. Hence determine the numerical aperture for the fiber. Solution 9/9/2014 LECTURES 47

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