COMM.ENG OPTICAL FIBER

Transcription

1 COMM.ENG OPTICAL FIBER

2 Objectives To Explain the basic concept of light transmission and reflection along an optical fiber line as one of the guided media including Total internal reflection dispersion Transmission properties

3 Is a dielectric channel used to guide light by total internal reflection Is a thin, flexible, transparent fiber that acts as a waveguide, or "light pipe", to transmit light between the two ends of the fiber

4 1-History 1960 : first laser for OFCS : development of many laser systems for a OFCS based on line of sight through the atmosphere (requires clear atmosphere & light LOS) 1970 : low loss glass fiber waveguide for OFCS Early communications (revolution invention compared to coaxial cables) by eyes Advantages of optical fibers: -Smaller and lighter than other types of cables -Very high tensile strength -Higher information capacity often Mbps and ready to achieve 2 Gbps -Lower losses -Greater repeater spacing Disadvantages of optical fibers: -High cost -Needs more expensive TX & RX -More difficult and expensive to splice than wire

5 TX Optical carrier 2-Optical communication system Tx coupler Channel Rx coupler Modulator baseband Input sensor passband Transmission media Optical fiber RECEIVER Output sensor (photo-detector) baseband Destination information message Input sensor: Converts message message into electric signal (baseband signal ), microphone Source information Modulator: Converts baseband signal into format appropriate for channel transmission ; impresses the signal onto the optical carrier Optical carrier: Generates the light-wave onto which the information is carried (laser diode (LD) or light emitting diode (LED)). The carrier source is intensity modulated which means that as the input current changes the output optical power changes in the same way Coupler : Couples light from the source to the fiber channel Channel: Glass or plastic fibers that are the transmission medium Receiver coupler : Transfers the optical power from the fiber to the photo-detector Photo-detector (o/p sensor): Converts optical power to electrical current, ideally the current should be a replica of the current used to modulate the light source Message destination : devices such as speakers, video monitors and computers

6 3-Construction of optical fiber line An optical fiber (American spelling) or fibre (British spelling): is cylindrical dielectric waveguide that transmits light along its axis, by the process of total internal reflection -The fiber consists of a denser core surrounded by a cladding layer -For total internal reflection to confine the optical signal in the core, the refractive index of the core must be greater than that of the cladding n 2 Low index cladding n 1 High index core n 1 > n 2 -The boundary between the core and cladding may either be abrupt, in step-index fiber, or gradual, in graded index fiber -The fiber is encased in an insulating jacket which protects it from moisture and provide some mechanical strength

7 4-Total internal reflection (TIR) The light which is transmitted usually changes direction when it enters the second material. This bending of light is called refraction and it depends upon the fact that light travels at one speed in one material and at a different speed in a different material n 1/ o o c / v 1/ r o r o n r r r where n is the refractive index c is the speed of light in a vacuum v is the speed of light in the material As a result each material has its own refractive index which we use to help us calculate the amount of bending which takes place. Refractive index is defined as: n r

8 Critical angle Form Snell s low n 1 sinθ 1 = n 2 sinθ 2 Since n 1 >n 2 then sinθ 2 = (n 1 /n 2 )sinθ 1 θ 2 > θ 1 when θ 1 = θ c ; the critical angle at which TIR occurs then: n 1 sinθ c = n 2 sin90 o =n 2 n 2 n 1 θ 1 θ 2 hence sinθ c = n 2 /n 1 if θ 1 > θ c then the ray is completely reflected in medium 1 so ; Hence conditions for light ray propagating through an optical fiber core are : 1-the core index is dense and greater then the - cladding index 2-incident angle of ray with normal to the core-- cladding interface θ 1 > θ c n 2 n 1 θ c θ 2 = 90 o

9 Acceptance angle Is the maximum angle θ imax which gives just total internal reflection at the core gladding interface -When θ i = θ imax θ= θ c B -When θ i < θ imax θ> θ c The ray propagate through TIR n 2 n o n 1 θ< θ c Cladding θ> θ c A Propagating θ i > θ imax B θ i < θ imax -When θ i >θ imax θ< θ c The ray penetrates the cladding and lost A Core i.e., the maximum angle at which it may enter the guide and travel by total internal reflection (TIR) is termed acceptance angle

10 n a sinθ i = n 1 sinθ t sinθ i = n 1 sinθ t The minimum value of θ gives reflection is θ c which gives maximum value of θ i sinθ imax = n 1 cosθ c =n 1 1-sin 2 θ c =n 1 1-(n 2 /n 1 ) 2 =n 1 (n 1 -n 2 )(n 1 +n 2 )/n 1 2 =n 1 (n 1 -n 2 )2n 1 /n 1 2 =n 1 2(n 1 -n 2 ) /n 1 n a =1 θ θ θ t θ i n 2 n 1 sinθ imax = n 1 2, = (n 1 -n 2 ) /n 1 is the refractive index fractional difference Numerical aperture NA= n a sinθ imax NA =sinθ imax = n 1 2 = (n 12 -n 22 ) It measure the light collecting ability of the fiber

11 Example Derive an expression for the maximum value of n 2 as a function of n 1 that permits all light incident on the end face of a fiber to be propagated. Assume n a = 1 and that the end face is perpendicular to the fiber axis. Calculate this limiting value of n 2 when n 1 = 1.46 Solution NA= n a sinθ imax = (n 12 -n 22 ) Example Under this condition θ imax = 90 o (n 12 -n 22 )=1 n 2 = (n 12-1) If n 1 = 1.46 then n 2 = A silica optical fiber with a core diameter large enough to be considered by ray theory analysis has a core refractive index of 1.5 and cladding refractive index of 1.47.Determine the critical angle at the core cladding interface Solution sinθ c = n 2 /n 1 = 1.47/1.5=0.98 θc = o

12 a r n a r n r n 2 1 ) ( a r n a r n r n 2 1 ) ( a r n a r a r n r n p 2 1 ) ) / ( 2 (1 ) ( α is the profile parameter which gives the characteristic refractive index profile of the core. It is a general expression for any refractive index profile inside the core for α= is step index α=2 is parabolic profile α=1 is triangular profile α 5-Types of cylindrical optical fiber

13 Monomode step index -The refractive index of the fiber steps up as we move from the cladding to the core of the fiber -The diameter of the core is about 5-10 µm -This narrow core limits the propagation of waveguide modes to a single one as the angle required for the higher modes would not be achievable in this configuration Multimode step index -The refractive index of the fiber steps up as we move from the cladding to the core of the fiber -The diameter of the core is about µm and this is wide enough to allow several different waveguide modes to propagate down the fiber Multimode graded index -The refraction index is -higher at the centre and decreases outwards in a specified nonlinear fashion toward the lower refractive index of the cladding -The best results for multimode optical propagation are obtained when we have a near parabolic refractive index profile (α=2)

14 COMMUNICATION ENG. Monomode step index Multimode step index Multimode graded index Advantages Minimum dispersion because all rays propagating down the fiber take approximately the same path, consequently, 1-a pulse of light entering the cable can be reproduced at the receiving end very accurately 2- larger bandwidths 3- higher bit rates 1-Inexpensive and simple to manufacture 2- Easy to couple light into and out because of larger numerical aperture and core diameter 3-Allow the use of incoherent optical sources (most LEDs) which are difficult to couple to single mode fibers 4-Tolerance requirement on connectors between fibers is much less stringent than the case with single mode fibers Disadvantages 1-Difficult to couple the light source 2- Needs leaser source 3-Difficult to manufacturing 1-Different passes for the rays which distort the pulse 2-Lower bandwidth and bit rates have far less intermodal dispersion than multimode SI and also have transmission bandwidths capacity much more than equivalent SI multimode fibres. Why? Rays travelling near the axis have a shorter path than rays travelling into the outer region of the core. However, near axial rays are transmitted through a region with higher refractive index and so travel with a slower speed than the outer rays, this compensate for the path difference and reduces dispersion in the fibre

15 More about GI fiber The gradual decrease in refractive index can be imagined as many layers with slightly different refractive indices n3 n2 n1 Beam path n The main disadvantages of GI index fibre is that its numerical aperture will be a function of the radial distance from the fibre axis and will be less than its corresponding size in SI fibres

16 Mode: 6-Number of modes propagation in optical fiber the allowed rays that can incident on the core-cladding interface that can propagate constructively. There are discrete incident angles on that interface which satisfy these conditions ( these modes are derived using Maxwell equations like any waveguide)

17 Condition of single mode SI fiber: 0 V < 2.405, V= 2π a NA / λ is the normalized frequency GI fiber: 0 V < 2.405(1+2/α) 1/2 Number of modes : The number of modes that can be propagated in SI fiber GI fiber M SI = V 2 /2 M GI = M SI (α/(α+2)) α is the profile parameter Therefore for a parabolic index profile number of modes supported are half that of a step index with the same V The fraction of power that can travel in the cladding : P clad / P tot = 4/(3 M)

18 Example A step index fiber of core radius and index 5um and 1.46 respectively. If the refractive index fractional difference is 0.01, then find the numerical aperture and the maximum operating frequency for single mode propagation Example A step index fiber of core diameter 50um and numerical aperture 0.25 carries light of wavelength 850 nm. Determine the number of modes and the fractional power travels to the cladding. Then determine the number of modes if its GI fiber with parabolic profile

19 7-Single modes fiber -The advantage of the propagation of single mode in optical fiber is the avoiding the dispersion due to the delay difference between different modes in multimode fiber -Single mode optical fiber has many parameters such as: (1) Cut off normalized frequency V c V c =2.405 (2) Cut off wavelength λ c Is the wavelength above which the fiber becomes single mode λ c = 2π a NA / V c Hence, and for a SI fiber λ c /λ=v/v c λ c =λv/2.405

20 Example Determine the maximum core diameter of a step index fiber of relative refractive index difference 1.5% and core refractive index 1.48 carries light of wavelength 0.85 um to be suitable for single mode operation. Determine the new maximum core diameter if the relative refractive index difference is reduced by a factor of 10

21 (3)- Modal field diameter (MFD) and spot size -In SI & GI fibers near cut off wavelength λ c, the field is well approximated by Gaussian distribution E(r) E(0) MFD : is the distance between 1/e=0.37 field amplitude points( 1/e 2 =0.135 for the power) of the corresponding values at the fiber axis ω o E(0)/e Spot size ω o : is the mode field radius, the nominal half width of the input excitation field i.e., MFD =2 ω o MFD = 2ω Radius r [um] o The optimum values of the spot size related to the core radius a and the normalized frequency V as: ω o /a = V -3/ V -6 ω o 2 /a = (1/n 1 k)(2/δ) 1/2 for SI fiber for GI near parabolic profile

22 Example Estimate the fiber core diameter for a single mode SI fiber which has a MFD of 11.6um when the normalized frequency is 2.2 Solution a = ω o /( V -3/ V -6 ) =5.8x10-6 /( (2.2) -3/ (2.2) -6 = 4.95um

23 8-Mode coupling It is coupling (transfer) of the energy traveling in one mode to another due a specific perturbation This perturbation might be due to deviation of the fiber axis, variation of the core diameter, straightness, irregularities of the core cladding interface, bending, refractive index variation, Irregularity Bending

24 9-Optical fiber cable installations 11/22/2014 LECTURES 24

25 10-Transmission characteristics of fiber optic 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 [w] P o =P i -αl [db]

26 Example A fiber has a coupled power -8dBm and attenuation of 6 dbm/km, and a length 2 km. Calculate the output power dbm/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

27 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

28 Attenuation (db/km) Glass Absorption in UV 1 Glass Absorption in IR Intrinsic absorption 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

29 -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 2.73mm (resonant wavelength for absorption).other wavelength causes large absorptions at 1.37, 1.23 and 0.95mm. Therefore for efficient propagation those wavelength must be avoided 11/22/2014 LECTURES 29

30 (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 -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 11/22/2014 LECTURES 30

31 - 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 11/22/2014 LECTURES 31

32 (3) Bend losses -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 -The loss can be represented by a radiation attenuation coefficient which is given by : α b =c 1 exp(-c 2 R) θ c θ c R where R is the radius of curvature of the fiber bend and c 1, c 2 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 : For single mode fibers the critical radius of curvature R CS approximately by : can be 11/22/2014 LECTURES 32 where λ c is the cutoff wavelength of the single mode fiber

33 Based on these relations the criteria to reduce macro bending losses is by : -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 db/km The overall attenuation λ [nm] λ Type Size,μm db/km 11/22/2014 LECTURES SI SI GI GI GI GI SM SM 62.5/ / /125 50/ /125 50/125 X/125 X/

34 Example A62.5/125GI parabolic fiber of core and cladding index and respectively operating at 1.3um. If the radius of curvature of the turn is 2cm, find the power lost in that turn 11/22/2014 LECTURES 34

35 (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% 11/22/2014 LECTURES 35

36 (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

37 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 = = 0.36 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

38 (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

39 (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)

40 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 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 α R < α T otherwise =0 db dt dr 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

41 -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 / /4.2 ) -2 ] =-10log = 0.35 db

42 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 =

43 = = = =

44 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 if 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 -This phenomena is clear in the following example:

45 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 Time Distinguishable pulses Composite pattern output at d 1 Amplitude Amplitude 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

46 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

47 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

48 Types of dispersion Dispersion Material Intramodal (Chromatic) waveguide Intermodal (mode) 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). 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 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

49 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

50 Example Intermodal dispersion

51 -Material dispersion -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) -Exists in all fibers and is a function of the source line width Spectrum of I/P pulse I/P pulse O/P pulse

52 -Waveguide dispersion -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 and is a function of the source line width

53 -Modal dispersion Due to propagation delay differences between modes within the fiber. It has a greatest effect of pulse broadening in step index multimode fibers High order mode Low order mode Intensity Light pulse Cla dding Core Broadene d light pulse Intensity Axial Spread, 0 t t 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 S.O. Kasap, Optoelectronics (Prentice Hall)