GANADIPATHY TULSI’S JAIN ENGINEERING COLLEGE Form No: Chittoor – Cuddalore Road, Kaniyambadi (Post), Vellore – 632102. GTEC/ACADEMIC/ 2014 Academic Year 2013 – 14: Odd Semester Question Bank Rev No: 0 Department of Electronics & Communication Engineering Year/ Sem: IV/VII Sub Code/Name: EC 2402/ Optical Communication and Networking UNIT – I Part A -Two Mark Questions: 1. What are the advantages and disadvantages of the ray optics theory? [Nov-Dec 2008] Ans: The large scale optical effects such as reflection and refraction can be analyzed by simple geometrical process of ray tracing. The rays show the direction of energy flow in the light beam. The wave fronts are separated by one Wavelength. 2. A typical relative refractive index difference for an optical fiber designed for long distance transmission is 1%. Estimate the numerical aperture for the fiber when the core index is 1.47. [Nov-Dec 2008] n1=1.47,n2=1.46 NA=(n12-n22)1/2 NA=0.1711. 3. Define acceptance angle and critical angle of the fiber. [ April/May 2009] Acceptance Angle: The maximum angle with which a ray of light can enter through the entrance end of the fiber and still be totally internally reflected is called acceptance angle of the fiber. Critical Angle: If the angle of incidence increase a point will eventually be reached where the light ray in air is parallel to the giro surface. This is known as critical angle. 4. What is fiber birefringence and fiber beat length? .[ April/May 2009] Ans: Fiber Birefringence: Imperfections in the fiber are cOmmor such as asymmetrical lateral stress, non-circular imperfect variations of refractive index profile. These :imperfections break the circulat symmetcry of ideal fiber and mode propagate with different phase velocity and the different between their Refractive index is called fiber GANADIPATHY TULSI’S JAIN ENGINEERING COLLEGE Form No: Chittoor – Cuddalore Road, Kaniyambadi (Post), Vellore – 632102. GTEC/ACADEMIC/ 2014 Academic Year 2013 – 14: Odd Semester Question Bank Rev No: 0 birefringence 5. What is Tunnel effect? [Nov-Dec 2009] The leaky modes are continuously radiating their power out of the core as they propagate along the fiber.This power radiation out of the waveguide in quantum mechanical phenomenon is referred as tunnel effect. 6. What is skew ray? [Nov-Dec 2009] Skew rays are not transmitted through the fiber axis.The skew rays follow a helical path in the optical fiber.It is very difficult to track the skew rays as they do not lie in a single plane. 7. A multimode step index fiber with a core diameter of 80µm and a relative index difference of 1.5% is operating at a wavelength of 0.85µm. If the core index difference is 1.48.Determine [April/May 2010 ] a.Normalized frequency of fiber b.the number of guided modes. a. Normalized frequency V = (2π/λ an1(2Δ)1/2) =75.8 b.Ms=V2/2 =2873 8. Define acceptance angle and numerical aperture of a fiber.[ April/May 2010] Acceptance angle It the maximum angle to the axis at which light may enter the fiber inorder to be Propagated. It is greater than the critical angle. Numerical aperture It is the relationship between the acceptance angle and the refractive indices of the three media involved namely core,cladding and air. The numerical aperture is a dimensionless quantity which is less than unity,which values ranging from 0.14 to 1.50 GANADIPATHY TULSI’S JAIN ENGINEERING COLLEGE Form No: Chittoor – Cuddalore Road, Kaniyambadi (Post), Vellore – 632102. GTEC/ACADEMIC/ 2014 Academic Year 2013 – 14: Odd Semester Question Bank Rev No: 0 9. What is the energy of the single photon of the light whose ƛ=1550nm,in ev? [Nov-Dec 2011] The peak emission wavelength ƛ in micrometers can be expressed as a function of the band-gap energy Eg in electron volts ƛ(µm)=1.240/Eg(ev) Eg(ev)=1.240/(1550*10-3) = 0.8 ev 10. Assume that there is a glass rod of R.I 1.5,surrounded by air.find the critical incident angle? [Nov-Dec 2011] n1 = 1.5 and n2 = 1.0(air) ɸc= sin-1 (n2/n1) = sin-1 1.5/1.0 ɸc= 0.66o 11. A typical refractive index difference for an optical fiber designed for long distance transmission is 1%.Determine the NA when the core index is 1.46.Calaculate the critical angle at the core-cladding interface with in the fiber. [May - June 2012]. (i) NA = n1(2*Δ)1/2 = 1.46*(2*0.01)1/2 (ii)Critical angle: NA=(n12-n22)1/2 = 1.4448 ɸc=sin-1(n2/n1) = sin-1(1.4448/1.46) = 81.9o 12. A Step –index fiber has a normalized frequency V=26.6 at a 1300nm wavelength.If the core radius is 25µm,Find out numerical aperture? [May - June 2012]. GANADIPATHY TULSI’S JAIN ENGINEERING COLLEGE Form No: Chittoor – Cuddalore Road, Kaniyambadi (Post), Vellore – 632102. GTEC/ACADEMIC/ 2014 Academic Year 2013 – 14: Odd Semester Question Bank Rev No: 0 Normalized frequency V = (2π/λ an1(2Δ)1/2) = 2π/λ a NA NA = Vλ/2πa =0.22 13. Calculate the cutoff wavelength of a single mode fiber with core radius of a 4µm and del=0.003.[Nov-Dec 2012] Cut off wavelength=(2πan1(2 )1/2)/Vc) For SM, normalised frequency V c =2.405 and consider n1=1 =(2*3.14*4*1(2*0.003)1/2)/2.405 =1.9468/2.405 =0.809µm 14. For a fiber with core refractive index of 1.54 and fractional refractive index difference of 0.01,calculate its NA. [Nov-Dec 2012] NA=n1(2*Δ)1/2 =1.54*(2*0.01) ½ =0.217 15. For n1=1.55 and n2=1.52, calculate the critical angle and numerical aperture.[MayJune2013] Solution: (a)the critical angle ɸc at the core cladding interface ɸc=sin-1(n1/n2) = 78.5o (b)the NA is NA=(n12-n22)1/2 NA=0.30 16. What is linearly polarized mode? [May-June2013] GANADIPATHY TULSI’S JAIN ENGINEERING COLLEGE Form No: Chittoor – Cuddalore Road, Kaniyambadi (Post), Vellore – 632102. GTEC/ACADEMIC/ 2014 Academic Year 2013 – 14: Odd Semester Question Bank Rev No: 0 in a step – index fiber the difference in the core and cladding indices of refraction is very small (Δ<<1) this the basis of the weakly guiding fiber approximation. Usually Δ is less then 0.33 (3%) for optical communication. With this weakly guiding assumption, only four field components (HE, EH, TE and TM) need to be considered and their expressions become significantly similar. The field components are called linearly polarized (LP) components Part B – Four/Eight/ Sixteen Mark Questions: 1. (a) (i) Explain with a neat block diagram the fundamentals of optical fiber communication. (8) [Nov-Dec 2008] An optical fiber transmission link comprises the elements as shown The key assumptions are transmitter consisting of a light source and its associated drive circuitry. Additional components include optical amplifiers, connectors, splices, couplers and regenerators. Analogous to copper cables, optical fibre cables can be installed either aerially, in ducts, undersea or buried directly in the ground. For undersea installations. The slicing and repeater-installation functions are carried out on board a specially designed cable-laying ship. Early technology mode exclusive use of the 800 to 900 tun wavelength band, since, in this region, the fibers made at that time exhibited a local minimum in the attenuation curve, and optical sources and photodetectors operating at these wavelengths were available. This region is referred to as the first window. By reducing the concentration of hydroxyl ions and metallic impurities GANADIPATHY TULSI’S JAIN ENGINEERING COLLEGE Form No: Chittoor – Cuddalore Road, Kaniyambadi (Post), Vellore – 632102. GTEC/ACADEMIC/ 2014 Academic Year 2013 – 14: Odd Semester Question Bank Rev No: 0 in the fibre material, they were able to fabricate optical fibers with very loss in the 11004o1600nm region. Two windows are defined here: the second window, centred around 1310nm, and the third window, centered around I 550w-n. Once the cbale is installed, a light source that is dimensionally compatible with the fibre core is used to launch optical power into the fiber. Semiconductor LED's and La diodes are suitable for this purpose, since their light output can he modulat..xl rapidly by simply varying the bias current at the desired transmission rate. thereby producing an optical signal. The electric input signals to the transmitte circuitry for the optical source can be of either analog or digital form. For high rate systems, direct modulation of the source can lead to unacceptable signal distortion. In this case, an external modulator is used to vary the amp 'tilde of a continuous light output from a Laser diode source. After an optical signal is launched into a fiber, it will become progressively attenuated and distorted with increasing distance because of scattering, absorption, dispersion mechanism in the glass material. 1. (a) (ii) Discuss the mode theory of circular waveguides. (8) [Nov-Dec 2008] Ans: In this, it is necessary to solve the Maxwell's equation subject to the cylindrical boundary conditions at the interface between the core and the cladding of the fiber. When solving Maxwell's equation for hollow metallic waveguides, only transverse electric (TE) modes and transverse magnetic (TM) modes are found. This gh..:s rise to hybrid modes, which makes optical waveguide analysis more complex than metallic waveguide analysis. • Although the theory of light propagation in optical fibers is well understood, a complete description of the guided and radiation modes is rather complex since it involves six component hybrid electromagnetic fields that have very involved mathematical expressions. A simplification these expressions can be carried out in practice, since ftt:derms are eusually constructed so that ni m n2 < < I. The field components are called Linearly Polarized modes are given as LP where j and integers designating mode solutions. 1. (b) i) Discuss berfily about linearly polarized modes. [Nov-Dec 2008] Arts: In practice, since fibers are usually constructed so that n,- n2 « I. The field components are called Linearly Polarized modes (LP) and are labelled as LPPI where j and GANADIPATHY TULSI’S JAIN ENGINEERING COLLEGE Form No: Chittoor – Cuddalore Road, Kaniyambadi (Post), Vellore – 632102. GTEC/ACADEMIC/ 2014 Academic Year 2013 – 14: Odd Semester Question Bank Rev No: 0 in are integers designating mode lOis scheme for the lowest-order-modes, integers of each stderivedIn from an HE,s, mode and each LP mode comes from TE E). and TMO„, and HEurn modes. Thus, the fundamental LP si mode corresponds to VIE„ mode. An important parameter connected with the cut-off condition is the V number defined by (b) ( ii) Dr aw t he st ruct ur e s o f single and mU lt imo de st ep ind e x a nd gr ad ed inde x f ibe r t he ir t yp ic a l dimensions. (6) f iber s w it h 1. (b)(iii) Mention the advantages of opt ical fiber communication systems. (4) [Nov-Dec 2008] Ans: 1) Attenuation makes wide range of distance possible 2) Smaller size and lighter weight 3) Electromagnetic isolation No physical electrical connection is required 5) More reliable 6) No cross-talk Greater bandwidth Very good dielectrics GANADIPATHY TULSI’S JAIN ENGINEERING COLLEGE Form No: Chittoor – Cuddalore Road, Kaniyambadi (Post), Vellore – 632102. GTEC/ACADEMIC/ 2014 Academic Year 2013 – 14: Odd Semester Question Bank Rev No: 0 Data rate is much higher 9) 10)The cost per channel is lower Systems 2. a i ) D e r i v e a n e x p r e s s i o n t o d e t e r m i n e t h e m o d e s propagating in step index fiber.. (11) [ April/May 2009] Ans: A standard mathematical procedure for solving equation is to use the separation-of variable's method, which assume's the solution of the form Ez = A f .,(t) Fa) F 3 (z) F4(1) GANADIPATHY TULSI’S JAIN ENGINEERING COLLEGE Form No: Chittoor – Cuddalore Road, Kaniyambadi (Post), Vellore – 632102. GTEC/ACADEMIC/ 2014 Academic Year 2013 – 14: Odd Semester Question Bank Rev No: 0 2.a) ii)Calculate the numerical aperture, cut-off parameter and number of modes supported by a fiber having core = 1.54, Cladding = 1.5, core radius 25 and operating wavelength 1300nm.(5). [ April/May 2009]. NA=(n12-n22)1/2 =0.348 Cut-off parameter V = (2πa)/(λ) NA = 42.0276 Number of modes M=2.092*1017 2.b.What are the various features of graded index fiber?Explain the refractive index profile and ray transmission in a multimode graded index fiber. [ April/May 2009]. Ray theory transmission Total internal reflection GANADIPATHY TULSI’S JAIN ENGINEERING COLLEGE Form No: Chittoor – Cuddalore Road, Kaniyambadi (Post), Vellore – 632102. GTEC/ACADEMIC/ 2014 Academic Year 2013 – 14: Odd Semester Question Bank Rev No: 0 GANADIPATHY TULSI’S JAIN ENGINEERING COLLEGE Form No: Chittoor – Cuddalore Road, Kaniyambadi (Post), Vellore – 632102. GTEC/ACADEMIC/ 2014 Academic Year 2013 – 14: Odd Semester Question Bank Rev No: 0 GANADIPATHY TULSI’S JAIN ENGINEERING COLLEGE Form No: Chittoor – Cuddalore Road, Kaniyambadi (Post), Vellore – 632102. GTEC/ACADEMIC/ 2014 Academic Year 2013 – 14: Odd Semester Question Bank Acceptance Angle Rev No: 0 GANADIPATHY TULSI’S JAIN ENGINEERING COLLEGE Form No: Chittoor – Cuddalore Road, Kaniyambadi (Post), Vellore – 632102. GTEC/ACADEMIC/ 2014 Academic Year 2013 – 14: Odd Semester Question Bank Numerical Aperture Rev No: 0 GANADIPATHY TULSI’S JAIN ENGINEERING COLLEGE Form No: Chittoor – Cuddalore Road, Kaniyambadi (Post), Vellore – 632102. GTEC/ACADEMIC/ 2014 Academic Year 2013 – 14: Odd Semester Question Bank Rev No: 0 GANADIPATHY TULSI’S JAIN ENGINEERING COLLEGE Form No: Chittoor – Cuddalore Road, Kaniyambadi (Post), Vellore – 632102. GTEC/ACADEMIC/ 2014 Academic Year 2013 – 14: Odd Semester Question Bank Rev No: 0 GANADIPATHY TULSI’S JAIN ENGINEERING COLLEGE Form No: Chittoor – Cuddalore Road, Kaniyambadi (Post), Vellore – 632102. GTEC/ACADEMIC/ 2014 Academic Year 2013 – 14: Odd Semester Question Bank Rev No: 0 GANADIPATHY TULSI’S JAIN ENGINEERING COLLEGE Form No: Chittoor – Cuddalore Road, Kaniyambadi (Post), Vellore – 632102. GTEC/ACADEMIC/ 2014 Academic Year 2013 – 14: Odd Semester Question Bank 3..Explain mode propagation in circular waveguides.Obtain its wave equation and modal equations for step index fibers.[April/May 2010] Rev No: 0 GANADIPATHY TULSI’S JAIN ENGINEERING COLLEGE Form No: Chittoor – Cuddalore Road, Kaniyambadi (Post), Vellore – 632102. GTEC/ACADEMIC/ 2014 Academic Year 2013 – 14: Odd Semester Question Bank Rev No: 0 GANADIPATHY TULSI’S JAIN ENGINEERING COLLEGE Form No: Chittoor – Cuddalore Road, Kaniyambadi (Post), Vellore – 632102. GTEC/ACADEMIC/ 2014 Academic Year 2013 – 14: Odd Semester Question Bank Rev No: 0 GANADIPATHY TULSI’S JAIN ENGINEERING COLLEGE Form No: Chittoor – Cuddalore Road, Kaniyambadi (Post), Vellore – 632102. GTEC/ACADEMIC/ 2014 Academic Year 2013 – 14: Odd Semester Question Bank Rev No: 0 4.(a)(i) what is numerical aperture of an optical fiber? Deduce an expression for the same. [NovDec 2011]    Numerical aperture is used to describe the light-gathering or light collecting ability of an optical fiber. Referred as figure of merit,used to measure the magnitude of an acceptance angle. The numerical aperture for light entering the glass fiber from an air medium is described as NA=sin Ɵa Where, Ɵa=acceptance angle(degrees) And also NA=(n12-n22)1/2 Acceptance angle interms of NA is Ɵa=sin-1(NA) GANADIPATHY TULSI’S JAIN ENGINEERING COLLEGE Form No: Chittoor – Cuddalore Road, Kaniyambadi (Post), Vellore – 632102. GTEC/ACADEMIC/ 2014 Academic Year 2013 – 14: Odd Semester Question Bank Rev No: 0 NA=sinƟa=(n12-n22)1/2 = ((n1+n2)(n1-n2))1/2 NA=n1(2Δ)1/2 Δ is called index difference Where ii) calculate NA of silica fiber with its core R.I (n1) of 1.48 and cladding R.I of 1.46.what should be the new value of n1 in order to change the NA to 0.23 solution: NA=(n12-n22)1/2 a) = (1.482-1.462)1/2 NA=0.24 (NA)2=n12-n22 b) n12=n22+(NA)2 n1=1.48 4.(b) i) explain the phenomenon of total internal refelction using snell‘s law with figures and calculations. [Nov-Dec 2011] When the light ray strikes the interface at an angle(incident angle) greater than the critical angle, the light ray does not pass through the interface into the glass, a mirror effect is existed at the interface. When this occurs,the angle of reflection ɸ2 is equal to tha angle of incidence ɸ1 as if a real mirror were used.this action is known as total internal reflection. Total internal reflection occurs only in materials in which the velocity of light is slower than in air. The ray has an angle of incidence at the interface which is greater than the critical angle (ɸ1>ɸc) and is totally reflected back into the air at the same time angle to the normal.this action is known as total internal reflection. Two necessary conditions for TIR to occur are:  The R.I of first medium (n1)>R.I of second medium (n2) GANADIPATHY TULSI’S JAIN ENGINEERING COLLEGE Form No: Chittoor – Cuddalore Road, Kaniyambadi (Post), Vellore – 632102. GTEC/ACADEMIC/ 2014 Rev No: 0 Academic Year 2013 – 14: Odd Semester Question Bank  ɸ1>ɸc (ii)distinguish step index from graded index fibers s.no parameter SI fiber GI fiber 1 Data rate Slow Higher 2 Coupling efficiency Higher Lower 3 Ray path By total internal reflection Light ray travels in oscillatory fashion 4 NA Remains same Changes continuously with distance from fiber axis 5 Pulse spreading Pulse spreading by fiber length is more Pulse spreading is less 6 Typical light source LED LED,laser 7 Attenuation of light Less typically 0.34 dB/km at 1.3µm More 0.6 to 1 dB/km at 1.3µm 8 Bandwidth efficiency 10-20 MHz/km 1 GHz/km 9 Material used Normally plastic or glass is preferred Only glass is preferred 10 Index variation Δ=(n1-n2)/n1 Δ=(n12-n22)/2n12 11 Applications Subscriber local network communication Local and wide area network 5..a)(i) Draw and explain the acceptance angle and numerical aperture of an optical fiber and drive expressions for both. [May - June 2012] GANADIPATHY TULSI’S JAIN ENGINEERING COLLEGE Form No: Chittoor – Cuddalore Road, Kaniyambadi (Post), Vellore – 632102. GTEC/ACADEMIC/ 2014 Academic Year 2013 – 14: Odd Semester Question Bank Rev No: 0 (ii)A fiber has a core radius of 25µm,core refractive index of 1.48 and relative refractive index difference is 0.01.if the operating wavwelength is 0.84µm,find the value of normalised frequency and the number of guided modes.Determine the number of guided modes if Δ is reduced to 0.03. [May - June 2012] Solution: (a) The normalised frequency V=(2πan1(2Δ )1/2)/λ =(2π*25*10-6*1.48*(2*0.01)1/2)/0.84*10-6 =(6.28*37*10-6*0.1414)/0.84*10-6 =32.85/0.84 =39. The total number of guided mode Ms=V2/2 =1521/2 =760.5 Hence this fiber has a V number of approximately 39,giving nearly 761 guided modes. (b) The normalised frequency V=(2πan1(2Δ )1/2)/λ =(2π*25*10-6*1.48*(2*0.003)1/2)/0.84*10-6 =(6.28*37*10-6*0.0774)/0.84*10-6 =17.98/0.84 =21.4 =22 The total number of guided mode Ms=V2/2 =484/2=242 Hence this fiber has a V number of approximately 22,giving nearly 242 guided modes. 5.b.(i) Draw and explain the refractive index profile and ray transmission in ingle mode and multimode step index fibers and graded index fibers.write the expresions for the numerical aperture and number of guided modes for a graded indexfiber. [May - June 2012] (ii) A step index fiber a core diameter 7µm,and core refractive index of 1.49 Estimate the shortest wavelength of light which allows single mode operation when the relative index difference for the fiber is 1%.[May - June 2012] GANADIPATHY TULSI’S JAIN ENGINEERING COLLEGE Form No: Chittoor – Cuddalore Road, Kaniyambadi (Post), Vellore – 632102. GTEC/ACADEMIC/ 2014 Academic Year 2013 – 14: Odd Semester Question Bank Rev No: 0 Solution: Cut off wavelength λ=(2πan1(2 )1/2)/Vc) For single mode,Normalized frequency V c=2.40 ={2*3.14*1.49*3.5*{2*0.0.}1/2}/2.405 ={6.28*5.21*{0.02} 1/2 }/2.405 ={32.71*0.1414}/2.405 =4.625/2.405 λ =1.923µm 6.a. Describe the ray theory the optical fiber communication by total internal reflection.State the application of snells law in it. [Nov-Dec 2012] (or) Ray theory transmission Total internal reflection GANADIPATHY TULSI’S JAIN ENGINEERING COLLEGE Form No: Chittoor – Cuddalore Road, Kaniyambadi (Post), Vellore – 632102. GTEC/ACADEMIC/ 2014 Academic Year 2013 – 14: Odd Semester Question Bank Rev No: 0 GANADIPATHY TULSI’S JAIN ENGINEERING COLLEGE Form No: Chittoor – Cuddalore Road, Kaniyambadi (Post), Vellore – 632102. GTEC/ACADEMIC/ 2014 Academic Year 2013 – 14: Odd Semester Question Bank Rev No: 0 GANADIPATHY TULSI’S JAIN ENGINEERING COLLEGE Form No: Chittoor – Cuddalore Road, Kaniyambadi (Post), Vellore – 632102. GTEC/ACADEMIC/ 2014 Academic Year 2013 – 14: Odd Semester Question Bank Rev No: 0 6.b.(i)ASI fiber with silica-core refractive index of 1.458,V=75 and NA=0.3 is to be oprate at 820nm.What should be its core size and cladding refractive index Calculate the total number of modes entering this fiber. [Nov-Dec 2012] Given: n1=1.458, V=75, NA=0.3 &λ=820 nm (a)Core Size: The normalized frequency for the fiber is V=(2π/ λ)a(NA) Radius of core a=V λ /2π(NA) =75*820*10-9/2*3.14*0.3 =61.5*10-6/1.884 =32.64µm Hence the maximum core diameter for step-index is now approximately 65µm (b)Cladding refractive index (n2) NA=n11-n22 N2=n12-(NA)2 =(1.458)2-(0.3)2 =2.126-0.09 =2.036=1.427 (c)The total number of guided mode: Ms=v2/2 GANADIPATHY TULSI’S JAIN ENGINEERING COLLEGE Form No: Chittoor – Cuddalore Road, Kaniyambadi (Post), Vellore – 632102. GTEC/ACADEMIC/ 2014 Academic Year 2013 – 14: Odd Semester Question Bank Rev No: 0 =5625/2=2812.5 Hence this fiber has a V number of approximately 75,giving nearly 2813 guided modes. 6.(b)(ii) Drive the expression for the linearly polarised modes in optical fibers and obtain the equation for V number. [Nov-Dec 2012] GANADIPATHY TULSI’S JAIN ENGINEERING COLLEGE Form No: Chittoor – Cuddalore Road, Kaniyambadi (Post), Vellore – 632102. GTEC/ACADEMIC/ 2014 Academic Year 2013 – 14: Odd Semester Question Bank Rev No: 0 7. a calculate the NA of fiber having n1=1.46 and n2=1.49 and another fiber having n1=1.458 and n2=1.405 which fiber has greater acceptance angle? [May-June2013] (a) NA=(n12-n22)1/2 =0.58 Acceptance angle Ɵa=sin-1(NA) = 35.4 (b) NA for another fiber NA=(n12-n22)1/2 =0.51 And acceptance angleƟa=sin-1(NA) GANADIPATHY TULSI’S JAIN ENGINEERING COLLEGE Form No: Chittoor – Cuddalore Road, Kaniyambadi (Post), Vellore – 632102. GTEC/ACADEMIC/ 2014 Academic Year 2013 – 14: Odd Semester Question Bank Rev No: 0 =30.7 7.b)i)explain the ray theory of a fiber with a special mention about TIR,acceptance angle and NA. [May-June2013] Total internal reflection GANADIPATHY TULSI’S JAIN ENGINEERING COLLEGE Form No: Chittoor – Cuddalore Road, Kaniyambadi (Post), Vellore – 632102. GTEC/ACADEMIC/ 2014 Academic Year 2013 – 14: Odd Semester Question Bank Rev No: 0 GANADIPATHY TULSI’S JAIN ENGINEERING COLLEGE Form No: Chittoor – Cuddalore Road, Kaniyambadi (Post), Vellore – 632102. GTEC/ACADEMIC/ 2014 Academic Year 2013 – 14: Odd Semester Question Bank Rev No: 0 GANADIPATHY TULSI’S JAIN ENGINEERING COLLEGE Form No: Chittoor – Cuddalore Road, Kaniyambadi (Post), Vellore – 632102. GTEC/ACADEMIC/ 2014 Academic Year 2013 – 14: Odd Semester Question Bank Acceptance Angle Rev No: 0 GANADIPATHY TULSI’S JAIN ENGINEERING COLLEGE Form No: Chittoor – Cuddalore Road, Kaniyambadi (Post), Vellore – 632102. GTEC/ACADEMIC/ 2014 Academic Year 2013 – 14: Odd Semester Question Bank Numerical Aperture Rev No: 0 GANADIPATHY TULSI’S JAIN ENGINEERING COLLEGE Form No: Chittoor – Cuddalore Road, Kaniyambadi (Post), Vellore – 632102. GTEC/ACADEMIC/ 2014 Academic Year 2013 – 14: Odd Semester Question Bank Rev No: 0 GANADIPATHY TULSI’S JAIN ENGINEERING COLLEGE Form No: Chittoor – Cuddalore Road, Kaniyambadi (Post), Vellore – 632102. GTEC/ACADEMIC/ 2014 Academic Year 2013 – 14: Odd Semester Question Bank Rev No: 0 GANADIPATHY TULSI’S JAIN ENGINEERING COLLEGE Form No: Chittoor – Cuddalore Road, Kaniyambadi (Post), Vellore – 632102. GTEC/ACADEMIC/ 2014 Academic Year 2013 – 14: Odd Semester Question Bank Rev No: 0 7.b.ii)Describe single mode fibers and their mode field diameter Single Mode Fibers The core size of single mode fibers is small. The core size (diameter) is typically around 8 to 10 micrometers (&mu;m). A fiber core of this size allows only the fundamental or lowest order mode to propagate around a 1300 nanometer (nm) wavelength. GANADIPATHY TULSI’S JAIN ENGINEERING COLLEGE Form No: Chittoor – Cuddalore Road, Kaniyambadi (Post), Vellore – 632102. GTEC/ACADEMIC/ 2014 Academic Year 2013 – 14: Odd Semester Question Bank Rev No: 0 Single mode fibers propagate only one mode, because the core size approaches the operational wavelength (&lambda;). The value of the normalized frequency parameter (V) relates core size with mode propagation. In single mode fibers, V is less than or equal to 2.405. When V &le; 2.405, single mode fibers propagate the fundamental mode down the fiber core, while high-order modes are lost in the cladding. For low V values (&le;1.0), most of the power is propagated in the cladding material. Power transmitted by the cladding is easily lost at fiber bends. The value of V should remain near the 2.405 level. Single mode fibers have a lower signal loss and a higher information capacity (bandwidth) than multimode fibers. Single mode fibers are capable of transferring higher amounts of data due to low fiber dispersion. Basically, dispersion is the spreading of light as light propagates along a fiber. Dispersion mechanisms in single mode fibers are discussed in more detail later in this chapter. Signal loss depends on the operational wavelength (&lambda;). In single mode fibers, the wavelength can increase or decrease the losses caused by fiber bending. Single mode fibers operating at wavelengths larger than the cutoff wavelength lose more power at fiber bends. They lose power because light radiates into the cladding, which is lost at fiber bends. In general, single mode fibers are considered to be lowloss fibers, which increase system bandwidth and length. Mode Field Diameter (MFD) The typical core diameter of communication single mode fibers is from 8~10um for operating wavelength 1.31um to 1.5um. Fiber with a core diameter less than about ten times the wavelength of the propagating light cannot be modeled using geometric optics as we did in the explanation of step-index multimode fiber. Instead, it must be analyzed as an electromagnetic structure, by solution of Maxwell's equations as reduced to the electromagnetic wave equation. So even though the fiber cladding confines the light within the fiber core, some light does penetrate into the cladding, despite the fact that it nominally undergoes total internal reflection. This occurs both in single mode and multimode fibers, but this phenomenon is more significant in single mode fibers. GANADIPATHY TULSI’S JAIN ENGINEERING COLLEGE Form No: Chittoor – Cuddalore Road, Kaniyambadi (Post), Vellore – 632102. GTEC/ACADEMIC/ 2014 Academic Year 2013 – 14: Odd Semester Question Bank Rev No: 0 UNIT – II Part A -Two Mark Questions: 1. Define group Delay ? [Nov-Dec 2008] Ans : As the signal propagated along Inc fiber each spectral component can be assumed to travel i ndependently and to undergo a time delay or goup delay per unit length in the direction of propagation given by 2. A multi mode graded index fiber exhibits total pulse broadening of 0.1 ps over a distance of 15 km. Estimate the maximum possible bandwidth on the link assuming RZ Coding without intersymbol interference. [Nov-Dec 2008] 3. A 100 km fiber is used in a communicat ion syst em. The fiber has 3.0 dB/km loss. What will be the output power, when the input power fed at the input fiber is 500 i.tW? [ April/May 2009] GANADIPATHY TULSI’S JAIN ENGINEERING COLLEGE Form No: Chittoor – Cuddalore Road, Kaniyambadi (Post), Vellore – 632102. GTEC/ACADEMIC/ 2014 Academic Year 2013 – 14: Odd Semester Question Bank Rev No: 0 4. What is the need for mode coupling in optical fiber? [ April/May 2009] Ans: The mode coupling will increase the distortion less rapidly after a certain inital length of fibre due to this mode coupling and differential mode losses. 5. What are microbends?How they are formed? [Nov-Dec 2009] Microbends are repetitive small scale fluctuations in the radius of curvature of the fiber axis. They are formed either by non uniformities in the manufacture of the fiber or by non uniform lateral pressures created during cabling of fibers. 6. What is intramodal dispersion?How the effect can be minimized? [April-May 2010] Intramodal dispersion is due to group velocity being a function of wavelength. The increasing spectral width of optical source will increase the intramodal dispersion. 7. Define the attenuation coefficient of the fiber. [Nov-Dec 2011] The standard formula for expressing the total power loss in an optical fiber cable is α(dB)=10log10(pi/po) Where, α(dB) – toal reduction in power level(attenuation) in decibels po – cable output power(watts) and pi – cable input power (watts) 8. Calculate the cut off wavelength of an optical signal through a fiber with its core refractive index of 1.50 and that of cladding = 1.46. the core radius of 25µm.the normalized frequency is 2.405. [Nov-Dec 2011] Cut off wavelength ƛc= 2πan1(2Δ)1/2/vc Δ=(n1-n2)/n1 = (1.50-1.46)/1.50 = 0.0267 GANADIPATHY TULSI’S JAIN ENGINEERING COLLEGE Form No: Chittoor – Cuddalore Road, Kaniyambadi (Post), Vellore – 632102. GTEC/ACADEMIC/ 2014 Academic Year 2013 – 14: Odd Semester Question Bank Rev No: 0 For single mode,normalized frequency vc=2.405 = (2*3.14*1.50*25*(2*0.0267)1/2)/2.405 = 62.5959/2.405 ƛc= 26 µm 9. What factors cause Rayleigh scattering in optical fibers?[May - June 2012] Rayleigh scattering is the dominant intrinsic loss mechanism in the ultra violet region.Its tail extends up to infrared region. Rayleigh scattering is a fundamental loss mechanism arising from local microscopic fluctuations in density.Density fluctuations lead to random fluctuations of the refractive index on a scale smaller than the optical wavelength (lemda) .Light scattering in such a medium is known as Rayleigh scattering. 10. Consider a 20 km long optical fiber that has an attenuation of 0.8 dB/km at 1300 nm.Find the optical power pout if 200µw of optical power is launched into the fiber? [May - June 2012] The input power in dBm units: Pin=10 log[Pin(w)/1mW]=10 log[200*10-6/1*10-3] = -7 dBm The output power level in dBm at z=30km is Pout(dBm)= 10 log[Pout(w)/1mW] = 10 log[Pin(w)/1mW]-αz =-7.0dBm-(0.8 dB/km)(30 km) = -7.0dBm-24 dBm Pout(dBm)=-31.0dBm GANADIPATHY TULSI’S JAIN ENGINEERING COLLEGE Form No: Chittoor – Cuddalore Road, Kaniyambadi (Post), Vellore – 632102. GTEC/ACADEMIC/ 2014 Academic Year 2013 – 14: Odd Semester Question Bank   Rev No: 0 11. What is reason for chromatic dispersion? [Nov-Dec 2012] This dispersion arises due to the variation of refractive index of the core material as a function of wavelength or frequency of light. This causes a wavelength dependence of the group velocity of any given mode,i.e.pulse spreading occurs even when different wavelengths follow the sama path. 12. What are the most important non-linear effects of optical fiber communication? [Nov-Dec 2012] This non-linear scattering causes the optical power from one mode to be transferred in either the forward or backward direction to the same.or other modes,at a different frequency. 13. what is Rayleigh scattering? [May-June2013] Rayleigh scattering is a fundamental loss mechanism arising from local microscope fluctuations in density. Density fluctuations lead to random fluctuations of the refractive index on a scale smaller then the optical wavelength λ. Light scattering in such a medium is known as Rayleigh scattering. 14. what is mean by mechanical splice? [May-June2013] This mechanical splice fiber alignment may be achieved by various methods including the use of tubes around the fiber end (tube splices) or V-grooves into which the butted fibers are placed (groove splices) . All these techniques seek to optimize the splice performance (i.e reduce the insertion loss at the joint) through both fiber end prepration and alignment of the two joint fibers. Part B – Four/Eight/ Sixteen Mark Questions: 1. (a) (i) Explain t he scat t er ing and bending tosses t hat occur in an opt ical fiber wit h relevant diagrams and .expressio ns. (8) [Nov-Dec 2008] Ans: Rayleigh S cat t ering Loss:T he glass, which is used in t he fabr icat ion o f fibres, has man y microscopic inho mogeneit ies and mat er ial Rayleigh influct uat io ns o f t he silica mat er ial cont ent s. As a result , a port io n gt liiir socuat t let r liiig t he glassfibr e get s scat t ered. This pheno menon is called PIt produces — att enuat io n dependence. The scat t eri ng lo ss due t o dens it y GANADIPATHY TULSI’S JAIN ENGINEERING COLLEGE Form No: Chittoor – Cuddalore Road, Kaniyambadi (Post), Vellore – 632102. GTEC/ACADEMIC/ 2014 Academic Year 2013 – 14: Odd Semester Question Bank Rev No: 0 fluct uat ion only can be expressed as: Bending Losses: Microbending and macr obending are t wo t ypes o f bending lo sses: Microb ending Losses: Microbends have s mall rando m deviat io ns about a no mina l st raight line posit io n. These are obser ved in cabled fibre. T his cross -coupling leads t o t he lo ss t hrough cladding. On so me occasio ns t hese bend ings may be quit e sharp and also will he int roducing cert ain lo ss of light energy. in t he mibre. GANADIPATHY TULSI’S JAIN ENGINEERING COLLEGE Form No: Chittoor – Cuddalore Road, Kaniyambadi multi (Post), Vellore – 632102. Where u is defined and a is the fiber radius. For mode fibers o GTEC/ACADEMIC/ the waveguide dispersion is generally very small c mpared 2014 Academic Year 2013 – 14: Odd Semester Question Bank Rev No: 0 1 a) ii ) D iscu s s po la r iz at io n mo de d is p er s io n a nd it s l i m it at io ns. Ans: Birefringence can result from intrinsic factors such as geometric irregularities of the fibre core or internal stress on it. In addition, external , twisting or pinching of the fiber can also lead Where u is defined and a is the fiber radius. For multimode fibers the waveguide dispersion is generally very small compared w ith material dispersion and can therefore be neglected 1. (b) Write a brief note on pulse 2008] br oadening in g raded index fibers. (15) [Nov-Dec Ans: The analysis of pulse broadening in graded index waveguides is more involved owing to the radial variation in core refractive index. The feature of this grading of the refractive index profile is that it offers multimode propagation in a relatively large core together with the possibility of a very low intermodal delay distortion. This combination allows the transmission of high data rates over long distances while still maintaining a resonable degree of light launching and coupling case. Since the index of refraction is lower at the outer edges of the core, light rays will travel foster to this region than in the centre of the core where the refractive index is higher. This can be seen from the fundamental relationship V = c/n, where V is the speed of light in a medium of refractive index n. Thus, the ray congruence categorizing the higher- order mode will tend to travel further than the fundamental ray congruence, but at a faster rate. 2. What are the losses on signal attenuation mechanisms in a fiber?Explain in detail.[April-May 2009].[April-May 2010] Attenuation GANADIPATHY TULSI’S JAIN ENGINEERING COLLEGE Form No: Chittoor – Cuddalore Road, Kaniyambadi (Post), Vellore – 632102. GTEC/ACADEMIC/ 2014 Academic Year 2013 – 14: Odd Semester Question Bank Material Absorption losses Intrinsic absorption Rev No: 0 GANADIPATHY TULSI’S JAIN ENGINEERING COLLEGE Form No: Chittoor – Cuddalore Road, Kaniyambadi (Post), Vellore – 632102. GTEC/ACADEMIC/ 2014 Academic Year 2013 – 14: Odd Semester Question Bank Rev No: 0 GANADIPATHY TULSI’S JAIN ENGINEERING COLLEGE Form No: Chittoor – Cuddalore Road, Kaniyambadi (Post), Vellore – 632102. GTEC/ACADEMIC/ 2014 Academic Year 2013 – 14: Odd Semester Question Bank Rev No: 0 GANADIPATHY TULSI’S JAIN ENGINEERING COLLEGE Form No: Chittoor – Cuddalore Road, Kaniyambadi (Post), Vellore – 632102. GTEC/ACADEMIC/ 2014 Academic Year 2013 – 14: Odd Semester Question Bank Extrinsic Absorption Linear scattering losses Mie Scattering Rev No: 0 GANADIPATHY TULSI’S JAIN ENGINEERING COLLEGE Form No: Chittoor – Cuddalore Road, Kaniyambadi (Post), Vellore – 632102. GTEC/ACADEMIC/ 2014 Academic Year 2013 – 14: Odd Semester Question Bank Non-linear scattering losses Rev No: 0 GANADIPATHY TULSI’S JAIN ENGINEERING COLLEGE Form No: Chittoor – Cuddalore Road, Kaniyambadi (Post), Vellore – 632102. GTEC/ACADEMIC/ 2014 Academic Year 2013 – 14: Odd Semester Question Bank Stimulated brillouin scattering Stimulated Raman scattering Rev No: 0 GANADIPATHY TULSI’S JAIN ENGINEERING COLLEGE Form No: Chittoor – Cuddalore Road, Kaniyambadi (Post), Vellore – 632102. GTEC/ACADEMIC/ 2014 Academic Year 2013 – 14: Odd Semester Question Bank Rev No: 0 3. Explain the effects of signal distortion in optical waveguide. .[April-May 2009]. .[April-May 2012]. Intramodal or chromatic dispersion GANADIPATHY TULSI’S JAIN ENGINEERING COLLEGE Form No: Chittoor – Cuddalore Road, Kaniyambadi (Post), Vellore – 632102. GTEC/ACADEMIC/ 2014 Academic Year 2013 – 14: Odd Semester Question Bank Rev No: 0 Intermodal dispersion 4. Describe the linear and non linear scattering losses in optical fiber.[Nov-Dec 2009] Linear scattering losses GANADIPATHY TULSI’S JAIN ENGINEERING COLLEGE Form No: Chittoor – Cuddalore Road, Kaniyambadi (Post), Vellore – 632102. GTEC/ACADEMIC/ 2014 Academic Year 2013 – 14: Odd Semester Question Bank Mie Scattering Non-linear scattering losses Rev No: 0 GANADIPATHY TULSI’S JAIN ENGINEERING COLLEGE Form No: Chittoor – Cuddalore Road, Kaniyambadi (Post), Vellore – 632102. GTEC/ACADEMIC/ 2014 Academic Year 2013 – 14: Odd Semester Question Bank Stimulated brillouin scattering Rev No: 0 GANADIPATHY TULSI’S JAIN ENGINEERING COLLEGE Form No: Chittoor – Cuddalore Road, Kaniyambadi (Post), Vellore – 632102. GTEC/ACADEMIC/ 2014 Academic Year 2013 – 14: Odd Semester Question Bank Stimulated Raman scattering Rev No: 0 GANADIPATHY TULSI’S JAIN ENGINEERING COLLEGE Form No: Chittoor – Cuddalore Road, Kaniyambadi (Post), Vellore – 632102. GTEC/ACADEMIC/ 2014 Academic Year 2013 – 14: Odd Semester Question Bank Rev No: 0 5. (a) (i) What do you mean by pulse broadening? Explain its effect on information carrying capacity of the fiber. [Nov-Dec 2011] A light pulse will broaden as it travels along the fiber.this pulse broadening will cause overlap with neighbouring pulses. At certain distance the pulses are not individually distinguished at the receiver and errors will occur.The information capacity of an optical waveguide is usually specified by the bandwidth-distance product(BDP) or bandwidth-length product(BLP) in MHz-km. As the length of optical fiber increases , the bandwidth decrease in proportion. For S.I BDP is 20MHz.km and for G.I it is 2.5GHz.km The information carrying capacity can be determined by the short light pulses propagating along the fiber. (ii)An LED operating at 850 nm has a spectral width of 45nm.what is the pulse spreading in ns/km due to material dispersion? What is the pulse spreading when a laser diode having a 2 nm spectral width is used? The material dispersion is 90 ps/nm km. GANADIPATHY TULSI’S JAIN ENGINEERING COLLEGE Form No: Chittoor – Cuddalore Road, Kaniyambadi (Post), Vellore – 632102. GTEC/ACADEMIC/ 2014 Academic Year 2013 – 14: Odd Semester Question Bank Rev No: 0 Solution: (i)rms pulse broadening per kilometer due to material dispersion σ m=σl LM σm(1km)= 45*1*90*10-12 = 4.05 nskm-1 (ii)rms pulse broadening per kilometer due to material dispersion σm=σƛLM σm(1km)= 2*1*90*10-12 = 1.8 nskm-1 5. (b)(i)what is meant by fiber splicing? Explain fusion splicing of optical fibers. [Nov-Dec 2011] A fiber splice is a permanent or semi-permanent joint between two fibers and the process of joinig two fiber is calledmas splicing It is used to create a long optical links or in situations where frequent connection and disconnections are not needed.splicces offer lower attenuation and lower back reflections than connectors and are less expensive. The factors to be considered in splicing are  geometrical difference between two fibers  fiber misalignment at the joint and  mechanical strength of the splice. Fusion splicing: Fusion splicing involves butting two cleaned fiber end faces and heating them until they melt together or fuse.  There are four steps o Initial setting o Arrangement of smooth surface by pre fusion o Pressed together o Accomplishment of splice Fiber ends are firstprealigned and butted together under a microscope with micromanipulators GANADIPATHY TULSI’S JAIN ENGINEERING COLLEGE Form No: Chittoor – Cuddalore Road, Kaniyambadi (Post), Vellore – 632102. GTEC/ACADEMIC/ 2014 Academic Year 2013 – 14: Odd Semester Question Bank Rev No: 0 The fusion splicing of single fiber involves the heating of the two prepared fiber ends to their fusing point with the application of sufficient axial pressure between the two optical fibers The butt joint is then heated with an electric arc or a laser pulse so that the ends are momentarily melted and bonded together.this technique can produce very low splice losses. (ii)explain expanded beam fiber connectors with a neat schematic. The expanded beam connector consisting of two lens for collimating and refocusing the light from one fiber into the other The use of this interposed optic makes the achievement of leteral alignment much less critical than with a butt jointed fiber connector. The collimating lens converts the light from the fiber into a parallel beam light and focusing lens converts the parallel beam of light into a focused beam of light on to the core of the receiving fiber. The fiber to lens distance equal to the focal length of the lens. Expanded beam connectorsare useful for multimode fiber connection and edge connection for printed circuit boarde where lateral and longtitudinal 6. (b)(i) Explain mechanical splices with neat diagrams. [May - June 2012] (ii)Write a brief note on fiber alignment and joint loss[Nov-Dec 2012] Fiber alignment and joint loss GANADIPATHY TULSI’S JAIN ENGINEERING COLLEGE Form No: Chittoor – Cuddalore Road, Kaniyambadi (Post), Vellore – 632102. GTEC/ACADEMIC/ 2014 Academic Year 2013 – 14: Odd Semester Question Bank Rev No: 0 7. (a)(i) Describe the linear and non-linear scattering losses in optical fibers. [Nov-Dec 2012] (ii).An LED operating at 850nm as a spectral width of 45nm.What is the pulse spreading in ns/Km due to material dispersion?What is the pulse spreading when a laser diode having a 2nm spectral width is used? The material dispersion is 90 ps/nmKm. Solution: (i)Rms pulse broadening per kilometre due to material dispersion Σm=σλLM Σm(1 km)=45*1*90*10-12 =4.05 nskm-1 GANADIPATHY TULSI’S JAIN ENGINEERING COLLEGE Form No: Chittoor – Cuddalore Road, Kaniyambadi (Post), Vellore – 632102. GTEC/ACADEMIC/ 2014 Academic Year 2013 – 14: Odd Semester Question Bank Rev No: 0 (ii)Rms pulse broadening per kilometre due to material dispersion Σm=σλ LM Σm(1 km)=2*1*90*10-12 =1.8 nskm-1 8.a)i)Derive an expression for material dispersion and waveguide dispersion and explain them[May-June2013] Material Dispersion: Material dispersion occurs because the index of refraction varies as a function of the optical wavelength. Or alternately, we can say that β varies with n(λ). Material dispersion is an intramodal effect and of particular importance for single-mode and LED systems. Let the variation in the index of refraction be n(λ) ii) various types of fiber connectors and couplers GANADIPATHY TULSI’S JAIN ENGINEERING COLLEGE Form No: Chittoor – Cuddalore Road, Kaniyambadi (Post), Vellore – 632102. GTEC/ACADEMIC/ 2014 Academic Year 2013 – 14: Odd Semester Question Bank Rev No: 0 GANADIPATHY TULSI’S JAIN ENGINEERING COLLEGE Form No: Chittoor – Cuddalore Road, Kaniyambadi (Post), Vellore – 632102. GTEC/ACADEMIC/ 2014 Academic Year 2013 – 14: Odd Semester Question Bank 8.(b)i)draw and explain the various fiber alignment and joint losses[May-June2013] ii)write notes on fiber splicing technique Fiber splices Rev No: 0 GANADIPATHY TULSI’S JAIN ENGINEERING COLLEGE Form No: Chittoor – Cuddalore Road, Kaniyambadi (Post), Vellore – 632102. GTEC/ACADEMIC/ 2014 Academic Year 2013 – 14: Odd Semester Question Bank Rev No: 0 GANADIPATHY TULSI’S JAIN ENGINEERING COLLEGE Form No: Chittoor – Cuddalore Road, Kaniyambadi (Post), Vellore – 632102. GTEC/ACADEMIC/ 2014 Academic Year 2013 – 14: Odd Semester Question Bank Rev No: 0 GANADIPATHY TULSI’S JAIN ENGINEERING COLLEGE Form No: Chittoor – Cuddalore Road, Kaniyambadi (Post), Vellore – 632102. GTEC/ACADEMIC/ 2014 Academic Year 2013 – 14: Odd Semester Question Bank Rev No: 0 UNIT – III Part A -Two Mark Questions: 1. Distinguish direct and indirect bandgrap materials. [Nov-Dec 2008] Ans: In direct bandgap materials, the electron and hole have the same momentum value. In indirect bandgap material, the conduction band minimum and the valence band maximum energy level occur at different values of momentum. 2. An LED has radiative and non-radiative recombination times of 30 and 100 ns respectively. Determine the internal quantum efficiency. [Nov-Dec 2008] 3. Define responsivity of a photo detector: [Nov-Dec 2008] Ans : It is defined as the ratio of output photo current to the incident optical power 4. . Compare the performance of APD and PIN diode. [Nov-Dec 2008] When an incident photon has 1) The carrier multiplication yields newly created carriers which d Ans: are alsoA accelerate PD by hi gh an energy greater than or equal to the band gap P I N Di o d e energy, it give up its energy el ectri c ugh t hus gai ning and excite an electron to enough energy to cause impact conduction band. This ionization which is known as process generates the avalanche effect. photocarriers. GANADIPATHY TULSI’S JAIN ENGINEERING COLLEGE Form No: Chittoor – Cuddalore Road, Kaniyambadi (Post), Vellore – 632102. GTEC/ACADEMIC/ 2014 Rev No: 0 Academic Year 2013 – 14: Odd Semester Question Bank Beat Fed Arkee Parl Winn Fir R2ortil Analogous to the pio performance of an API) is characterized by its responsivity. Riot),. photodiode the 5. What is the need for mode coupling in optical fiber?[April-May2009] Ans: The mode coupling will increase the distortion less rapidly after a certain inital length of fibre due to this mode coupling and differential mode losses. 6. When a LED has 2 V applied to its terminals, It draws 100 mA and produces 2 mW of optical power. Determine the LED conversion efficiency from eletrical to optical power? [April-May2009] Ans: Given: Input to LED = 2Y, I= 100 rnA, Output Power = 2 mW LED Conversion efficiency 7. . What is the principle of ciperation of LASER? [April-May2009] Ans: The principle of operation of LASER is population inversion the most photons incident on the system. GANADIPATHY TULSI’S JAIN ENGINEERING COLLEGE Form No: Chittoor – Cuddalore Road, Kaniyambadi (Post), Vellore – 632102. GTEC/ACADEMIC/ 2014 Academic Year 2013 – 14: Odd Semester Question Bank Rev No: 0 8. Define quantum effiieney and responsitivity of photodetector. [April-May2009] ANS: • The internal quantum efficiency is the fraction of the electron- hole pairs that recombine radiatively. If the radiative recombination rate is R and the non-radiative recombination ratios R,, then the internal quantum efficiency is the ratio of the radiative recombination rate to the recombination rate.  The performance of a avalanche_photodiode is characterized by its responsivity, R apd = nqm/ hv = RdM. 9. . A silicon photodiode has a quantum efficiency of 65% at a wavelength. of 900 nm. If 0.5 uN optical power produces a multiplied photocurrent of 10 iA. Determine its primary photon current and multiplication factor. [April-May2009] Given: q = 65%, X = 900 nm,1„,=10M, p=0.5 FtN, 1,,=2 M =? 10. What are the characteristics that a semiconductor laser diode should possess? [Nov-Dec 2011] For optical power systems, a semiconductor laser diodes are used as an optical source.because its output radiation is highly monochromatic and the light beam is very directional. 11. Photons of energy 1.53*10 -49 J are incident on a photodiode which has a responsivity of 0.75 A/W. if the optical power leval is 10µm,then find out the photo current generated? [Nov-Dec 2011] Solution: Ro=0.75 A/W Po=10µw Ip=RO PO = (0.75 A/W)(10µW) IP=7.5µA GANADIPATHY TULSI’S JAIN ENGINEERING COLLEGE Form No: Chittoor – Cuddalore Road, Kaniyambadi (Post), Vellore – 632102. GTEC/ACADEMIC/ 2014 Academic Year 2013 – 14: Odd Semester Question Bank Rev No: 0 12. Define a ‗quantum efficiency‘ of a photo detector and write its expression. [Nov-Dec 2011] The quantum efficiency is defined as ―the number of electron hole carrier pairs generated per incident photon of energy hv‖ and is given by Ƞ= number of electron hole pairs generated/number of incident photons = (IP/q)/(PO/hv) Where, IP =photocurrent q = charge of an electron PO= opicalpowert incident on the photo detector hv = energy of the iincident photon 13. What are the advantages of LED? [May - June 2012] 1. Simpler fabrication 2. COST:The simpler construction of the LED leads to much reduced cost. 3. Reliability 4. Generally less temperature dependence. 5. SIMPLER DRIVE CIRCUITRY:This is due to the generally lower device currents and reduced temperature dependance which makes temperature compensation circuits unnecessary. 6. LINEARITY:Ideally the LED has a linear light output against current characteristic. 14. Photons of energy 1.53*10 -19J are incident on a photodiode which has a responsivity of 0.65A/W.If the optical power level is 10µw,then find out the photo current generated? [May - June 2012] Solution: Given: R0=0.65A/W P0=10µw Ip=R0P0 [R=Ip/P0] =(0.65A/W)(10µw) Ip =6.5µA 15. Compare and contrast between surface and edge emitting LEDs. [Nov-Dec 2012] GANADIPATHY TULSI’S JAIN ENGINEERING COLLEGE Form No: Chittoor – Cuddalore Road, Kaniyambadi (Post), Vellore – 632102. GTEC/ACADEMIC/ 2014 Academic Year 2013 – 14: Odd Semester Question Bank Rev No: 0  In surface emitting LEDs,the plane of active light-emitting region is perpendicular to the fiber axis.  Edge emitting LEDs emit a more directional light pattern than the surface emitting LEDs.  In order to reduce the losses caused by absorption in the active layer and to make the beam more directional,the light is collected from the edge of the LED.Such a device is known as edge emitting LED.  Edge emitters have a substantially better modulation bandwidth of the order of hundres of mege hertz than comparable surface-emitting structures with the same drive level.  The coupling losses with surface emitters are greater,and they have narrow bandwidths.  Edge emitter couple 7.5 times more power into low NA fiber than a comparable surface emitter. 16. What is the significations of intrinsic layer in PIN diodes? [Nov-Dec 2012] The PIN photo detector structure consists of p and n regions separator by a very lightly ndopped intrinsic (i)region.When the photo diode is reverse based,the intrinsic region od diode is fully depleted of carriers. The depletion region must be sufficiently thick to allow a large fraction of incident light(photon) to be absorbed in order to achieve maximum carrier-pair generation. 17. Calculate the band gapenergy for an led to emit 850 nm. [May-June2013] We know that, λ (µm) = 1.240/Eg(ev) Eg(ev) =1.240/0.850=1.459 18. Define detector response time. [May-June2013] The response time of a photo diode aredepends on  Transit time of photo carriers within the depletion region  Diffusion time of photo carriers outside the depletion region  RC time constant of the photodiode and its associated circuit Part B – Four/Eight/ Sixteen Mark Questions: 1. (a) (i) Draw and explain the different structures used to achieve carrier andopticalconfinement in laser diodes. (8) [Nov-Dec 2008] Ans: For optical fiber communication systems requiring bandwidths greater than approximately 200MHz, the semiconductor injection laser diode is preferred over the LED. Laser diodes GANADIPATHY TULSI’S JAIN ENGINEERING COLLEGE Form No: Chittoor – Cuddalore Road, Kaniyambadi (Post), Vellore – 632102. GTEC/ACADEMIC/ 2014 Academic Year 2013 – 14: Odd Semester Question Bank Rev No: 0 typically have response time less than Ins, have optical bandwidths of 2nm or less, and,in general, are capable of coupling several ten of milliwatts of useful luminescent power into optical fibers with small core and small mode-field diameters. Virtually all laser diodes in use are multilayered heterojunction diodes. The doluble heterojunction LED configuration evolved from the successful demonstration of both carrier and optical confinement in heterojunction injection laser diodes. The more rapid evolvement and utilization of LEDs as compared with laser diodes lies in the inherently simpler construction, the smaller temperature dependence of the emitted optical power, and the absence of catastrophic degradation in LEDs. In other laser diode type, commonly referred to as the distributed- feedback laser the cleaved facit are not required for optical feedback. The stimulated emission rate into a given mode is proportional to the intensity of the radiation in that mode. The radiation intensity at a photon energy hy varies exponentially with the distance z that it traverses along the lasing cavity according to the relationship. where a- is the effective absorption coefficient of the material in the optical path and r is the optical-confinement factor - that is, the fraction of optical power in the active layer. 1. (a) (ii) Discuss the effects of temperature on the performance of a laser diode. (4) [Nov-Dec 2008] Ans: An important factor to consider in the application of laser diodes is the temperature dependence of the threshold current. This parameter increases with temperature in all types of semiconductor lasers because of various complex temperature-dependent factors. However, the Where To is a measure of the relative temperature insensitivity and I is a GANADIPATHY TULSI’S JAIN ENGINEERING COLLEGE Form No: Chittoor – Cuddalore Road, Kaniyambadi (Post), Vellore – 632102. GTEC/ACADEMIC/ 2014 Academic Year 2013 – 14: Odd Semester Question Bank Rev No: 0 constant. Fora conventional stripe-geometry GaAlAs Laser diode. To is typically 120 165°C in the vicinity of room temperature. An example of a laser diode with T„ = 135°C and lz -= 52A. The smaller dependences oflth on temperature have been demonstrated for GaAlAs quantum-well heterostructure lasers. For these lasers, To can be as high as 437°C. Consequently, if a constant optical output power level is to be maintained as the temperature of the laser changes or as the laser ages, it is necessary to adjust the dc bias current level. 1 (b) (i) Derive expressio ns for t he power coupled frort i a sur fa ( surface emitt ing LED int o st ep index and graded index fibers. (10) [Nov-Dec 2008] Ans: "Ihe coupled power can be found using the GANADIPATHY TULSI’S JAIN ENGINEERING COLLEGE Form No: Chittoor – Cuddalore Road, Kaniyambadi (Post), Vellore – 632102. GTEC/ACADEMIC/ 2014 Academic Year 2013 – 14: Odd Semester Question Bank Rev No: 0 relationship. Where the are and solid acceptance angle of the fiber define the limits of the integrals. The total coupled power is then determined by summing up the contributions from each individual emitting-point source of incremental area dOz r dr; that is, integrating over the emitting area. 1. (b) (ii) Explain the mechanical misalignments that can occur between two joined fibers with necessary diagrams. (6) [Nov-Dec 2008] Mechanical alignment is a major poblem when joining two fibers, owing to their microscopic size. A standard multimode graded-index fiber core is 50-100 p in in diameter. Radiation losses result from mechanical misalignments because the radiation ocne of the emitting fiber does not match the acceptance once of the receiving fiber. The magnitude of the radiation loss depends on the degree of misalignments: Longitudinal separation occurs when the fibers hav& the same axis but have a gap S between their end faces. Angular misalignments results when the two axes form ana ngle so that the fiber end faces are no longer parallel. Axial displacement results when the axes of the two fibers are separated by a distance d. The most common misalignment occuring in practice; which also causes the .greatest power loss, is axial displacement. 2. (a) Discuss with necessary expressions the different types of noise that affect the performance of a photodetector. (10) [Nov-Dec 2008] A ns: In fiber optic communication systems, the photodiode is generally required to detect very weak optical signals. The power Signal-to-noise ratio S/N at the output of an optical receiver is defined by S/N = Signal Power from Photocurrent/ Photo detector Noise Power + Amplifier Noise Power GANADIPATHY TULSI’S JAIN ENGINEERING COLLEGE Form No: Chittoor – Cuddalore Road, Kaniyambadi (Post), Vellore – 632102. GTEC/ACADEMIC/ 2014 Academic Year 2013 – 14: Odd Semester Question Bank Rev No: 0 To achieve a high ratio the following conditions to be met: I. The photo detector must have a high quantum efficiency to generate a large signal power. 2. The photodetector and amplifier noises should be kept as low as possible. The sensitivity of a photodetector in an optical fiber conununication system is describable in terms of the minimum detectable optirtal power. This is the optical power necessary to produce a photocurrent. 3.Explain the basic LED configurations used as optical source.Derive the expression for quantum efficiency and optical power generated in LED‘s.[April-May 2009] [April-May 2010] [May-June2013] GANADIPATHY TULSI’S JAIN ENGINEERING COLLEGE Form No: Chittoor – Cuddalore Road, Kaniyambadi (Post), Vellore – 632102. GTEC/ACADEMIC/ 2014 LED structures Surface emitter Academic Year 2013 – 14: Odd Semester Question Bank Rev No: 0 GANADIPATHY TULSI’S JAIN ENGINEERING COLLEGE Form No: Chittoor – Cuddalore Road, Kaniyambadi (Post), Vellore – 632102. GTEC/ACADEMIC/ 2014 Edge emitter Academic Year 2013 – 14: Odd Semester Question Bank Rev No: 0 GANADIPATHY TULSI’S JAIN ENGINEERING COLLEGE Form No: Chittoor – Cuddalore Road, Kaniyambadi (Post), Vellore – 632102. GTEC/ACADEMIC/ 2014 Academic Year 2013 – 14: Odd Semester Question Bank Rev No: 0 GANADIPATHY TULSI’S JAIN ENGINEERING COLLEGE Form No: Chittoor – Cuddalore Road, Kaniyambadi (Post), Vellore – 632102. GTEC/ACADEMIC/ 2014 Academic Year 2013 – 14: Odd Semester Question Bank Rev No: 0 4.With a neat sketch explain the operation of PIN photodiode and Avalanche diode.[April-May 2009] [May-June2013] PIN photodiode GANADIPATHY TULSI’S JAIN ENGINEERING COLLEGE Form No: Chittoor – Cuddalore Road, Kaniyambadi (Post), Vellore – 632102. GTEC/ACADEMIC/ 2014 Academic Year 2013 – 14: Odd Semester Question Bank Rev No: 0 GANADIPATHY TULSI’S JAIN ENGINEERING COLLEGE Form No: Chittoor – Cuddalore Road, Kaniyambadi (Post), Vellore – 632102. GTEC/ACADEMIC/ 2014 Academic Year 2013 – 14: Odd Semester Question Bank Rev No: 0 GANADIPATHY TULSI’S JAIN ENGINEERING COLLEGE Form No: Chittoor – Cuddalore Road, Kaniyambadi (Post), Vellore – 632102. GTEC/ACADEMIC/ 2014 Academic Year 2013 – 14: Odd Semester Question Bank Rev No: 0 GANADIPATHY TULSI’S JAIN ENGINEERING COLLEGE Form No: Chittoor – Cuddalore Road, Kaniyambadi (Post), Vellore – 632102. GTEC/ACADEMIC/ 2014 Academic Year 2013 – 14: Odd Semester Question Bank Avalanche photodiode Rev No: 0 GANADIPATHY TULSI’S JAIN ENGINEERING COLLEGE Form No: Chittoor – Cuddalore Road, Kaniyambadi (Post), Vellore – 632102. GTEC/ACADEMIC/ 2014 Academic Year 2013 – 14: Odd Semester Question Bank Rev No: 0 GANADIPATHY TULSI’S JAIN ENGINEERING COLLEGE Form No: Chittoor – Cuddalore Road, Kaniyambadi (Post), Vellore – 632102. GTEC/ACADEMIC/ 2014 Rev No: 0 Academic Year 2013 – 14: Odd Semester Question Bank 5. (a) (i)compare LED with a laser diode. [Nov-Dec 2011] s.no Parameter LED LASER 1 Output beam Coherent In-coherent 2 Couping efficiency High Very low 3 Priniciple of operation Stimulated emission Spontaneous emission GANADIPATHY TULSI’S JAIN ENGINEERING COLLEGE Form No: Chittoor – Cuddalore Road, Kaniyambadi (Post), Vellore – 632102. GTEC/ACADEMIC/ 2014 Rev No: 0 Academic Year 2013 – 14: Odd Semester Question Bank 4 Output power High Low 5 Data rate High Low 6 Cost Expensive Less 7 Life time 104hours 105 hours 8 Circuit complexity Complex Simple 9 Temperature dependant More Less 10 Apllications Low data rate high (ii)with the help of a neat diagram explain the construction and working of a surface emitting LED. [Nov-Dec 2011][May-June 2012] surface-emitting LED is also known as the Burrus LED in honor of C. A. Burrus, its developer. In SLEDs, the size of the primary active region is limited to a small circular area of 20 &mu;m to 50 &mu;m in diameter. The active region is the portion of the LED where photons are emitted. The primary active region is below the surface of the semiconductor substrate perpendicular to the axis of the fiber. A well is etched into the substrate to allow direct coupling of the emitted light to the optical fiber. The etched well allows the optical fiber to come into close contact with the emitting surface. In addition, the epoxy resin that binds the optical fiber to the SLED reduces the refractive index mismatch, increasing coupling efficiency. Example of the SLED structure. GANADIPATHY TULSI’S JAIN ENGINEERING COLLEGE Form No: Chittoor – Cuddalore Road, Kaniyambadi (Post), Vellore – 632102. GTEC/ACADEMIC/ 2014 Academic Year 2013 – 14: Odd Semester Question Bank Rev No: 0 5.(b)(i)explain the structure and working of a silicon APD. [Nov-Dec 2011] [Nov-Dec 2012] APD internally multiplies the primary signal photo current before it enters the input circuitry of the following amplifier.it has interenal gain and responsivity is better than PIN.the photo current is multiplied to increase the receiver sensitivity. Impact ionization: The photo generatd carriers transverse a region where a very high electric field is present.a photo generated electron or hole can gain enough energy in this high electric field and excite new electron hole pairs.this process is called impact ionization and this phenomenon leads to avalanche break dowm in ordinary RB diodes. Avalanche effect: Due to ionization effect new carriers are generated.the newly created carriers also accelerated by the high electric field.this effect is called avalanche effect. Reach through avalanche photo diode(RAPD) A commonly used structure for achieving carrier multiplication with very little excess noise is the reach through construction Operation: 1. When a low bias voltage is applied,most of the potential drop is across Pn+junction.the depletion layer increase with increasing reverse bias voltage. 2. At particular voltage avalanche break down occurs ,the depletion layer just reach through to nearly intrinsic region 3. Rapd is operated at fully depleted mode.when light enters the device it is absorbed in π region.hence it is called collection region 4. this absorbed photon gives up energy and electron hole pair are created and accelerated by high electric field. 5. The carrier multiplication takes place GANADIPATHY TULSI’S JAIN ENGINEERING COLLEGE Form No: Chittoor – Cuddalore Road, Kaniyambadi (Post), Vellore – 632102. GTEC/ACADEMIC/ 2014 Academic Year 2013 – 14: Odd Semester Question Bank Rev No: 0 Ionization rate The average number of electron hole pairs created by a carrier per unit distance travelled K=β/α Avalanche multiplication(M) Carrier multiplication takes place by impact ionization or avalanche effect. M=Im/Ip Responsivity(R) RAPD=(ƞq/hv)M Benefits: o Excellent linearity o Less sensitive Demerits: o If power received at the receiver is more than 1 microwatts,then APD is not used o Gain decreases as temperature rises o Increased cost o High noise contribution (ii)define S/N ratio of a photodetector.what conditions should be met to achieve a high SNR? The performance of the receiver is evaluated by power signal to noise ratio at the output of an optical receiver (S/N)=(signal power from photo current)/(photodector noise power+amplifier noise power) To get high signal to noise ratio (i)the photo detector must have a high quantum efficiency to generate a large signal power and (ii)photo detector and amplifier noise should be very low. To get high sensitivity of a photo detector(sensitivity is inversely proportional to the minimum detectable optical power) S/N ratio should be more. 6.A silicon p-i-n photodiode incorporated into an optical receiver has a quantum efficiency of 60% when operating at a wavelength of 0.9µm.The dark current in the device is 3nA and the load resistance is 4kΩ.The incident optical power is 200nW and the post detection bandwidth of the receiver is 5MHz Calculate the root mean square (rms) shotnoise and thermal noise currents generated.[May-June 2012] Solution: GANADIPATHY TULSI’S JAIN ENGINEERING COLLEGE Form No: Chittoor – Cuddalore Road, Kaniyambadi (Post), Vellore – 632102. GTEC/ACADEMIC/ 2014 Academic Year 2013 – 14: Odd Semester Question Bank (i) Primary photocurrent (Ip): Ip=RPo =(ὴq/hv)*po =(ὴqlamda/hc)*po =((0.60)(1.6*10-19c)(0.9*10-6 m)/6.625*10-34 j.s)(3*108m/s))*200*10-9w =172.8*10-34/19.875*10-26 =8.7*10-6 A Ip =8.7µA (ii)Mean – square quantum (or) shot noise current for a pin photodiode is (I2Q)=2qIpB =2(1.6*10-19c)(8.7*10-6A)(5*106Hz)=139.2*10-19A2 (I2Q)=13.92*10-18A2 (or) (I2Q)2=3.73nA (iii) Mean-square thermal noise current (I2T)=(4KBT/RL)*B =(4(1.38*10-23J/K)(293K)/4000)*5*106Hz =8086.8*10-17/4000 =2*10-17 (iT2)=0.2*10-18A2 (or) (i2T)2=0.447nA Rev No: 0 GANADIPATHY TULSI’S JAIN ENGINEERING COLLEGE Form No: Chittoor – Cuddalore Road, Kaniyambadi (Post), Vellore – 632102. GTEC/ACADEMIC/ 2014 Academic Year 2013 – 14: Odd Semester Question Bank Rev No: 0 UNIT – IV Part A -Two Mark Questions: 1. Define threshold level. [Nov-Dec 2011] A decision circuit compares the signal in each time slot with a certain reference voltage known as the threshold level. (i) Received signal > threshold=1 (ii) Received signal<threshold=0 2. Define quantum limit. [May - June 2012] To find the minimum received optical power required for a specific bit error rate performance in a digital system. This minimum received power level is known as the quantum limit. 3. 1. 2. 3. What are the methods used to measure fiber refractive index profile? [May - June 2012] Interferometric method Near field scanning method and Refracted near field method 4. What is dark current? [Nov-Dec 2012] It is the current that continues to flow through the base circuit of the device when no light is incident on the photo diode.This is a combination of bulk and surface current. The bulk dark current arises from electrons and/or holes which are thermally generated in the PN junction of the photodiode. 5. List out the various error sources. [Nov-Dec 2012] [May-June2013] o Quantum noise o Bulk dark current o Surface leakage current o Thermal noise o Amplifier noise 6. what is known as quantum limit? [May-June2013] To find the minimum received optical power required for a specified bit error rate performance in a digital system.this minimum received power level is known as the quantum limit. GANADIPATHY TULSI’S JAIN ENGINEERING COLLEGE Form No: Chittoor – Cuddalore Road, Kaniyambadi (Post), Vellore – 632102. GTEC/ACADEMIC/ 2014 Academic Year 2013 – 14: Odd Semester Question Bank Rev No: 0 Part B – Four/Eight/ Sixteen Mark Questions: 1. a)i) explain the fiber optic receiver operation using a simple model and its equivalent circuit[Nov-Dec 2011] [Nov-Dec 2012] Receiver structures Low impedance front end GANADIPATHY TULSI’S JAIN ENGINEERING COLLEGE Form No: Chittoor – Cuddalore Road, Kaniyambadi (Post), Vellore – 632102. GTEC/ACADEMIC/ 2014 Academic Year 2013 – 14: Odd Semester Question Bank Rev No: 0 GANADIPATHY TULSI’S JAIN ENGINEERING COLLEGE Form No: Chittoor – Cuddalore Road, Kaniyambadi (Post), Vellore – 632102. GTEC/ACADEMIC/ 2014 Academic Year 2013 – 14: Odd Semester Question Bank 1)b)explain the measurement technique used in the case of (i)numerical aperture[Nov-Dec 2011] [Nov-Dec 2012] [May - June 2012] Fiber numerical aperture measurement Rev No: 0 GANADIPATHY TULSI’S JAIN ENGINEERING COLLEGE Form No: Chittoor – Cuddalore Road, Kaniyambadi (Post), Vellore – 632102. GTEC/ACADEMIC/ 2014 Academic Year 2013 – 14: Odd Semester Question Bank Rev No: 0 GANADIPATHY TULSI’S JAIN ENGINEERING COLLEGE Form No: Chittoor – Cuddalore Road, Kaniyambadi (Post), Vellore – 632102. GTEC/ACADEMIC/ 2014 Academic Year 2013 – 14: Odd Semester Question Bank Rev No: 0 (ii)refractive index profile[May –June 2013] INTERFEROMETRIC METHODS: Interference microscopes (mach-zehnder) are widely used to determine the refractive index profiles of optical fibers. A thin slice fiber (slab) with both ends accurately polished to obtain optically flat surfaces. The slab with often immersed in an index matching fluid and an assembly is examined with an inter face microscope. Using either a transmitted light interferometer or a reflected light form interferometer. In both cases light form the microscope travels normal to the prepared fiber faces (parallel to the fiber axis) When the phase of the incident light is compared with the phase of the emerging light, a field of parallel interference fringes is observed. The fringe displacements for the points within the fiber core are then measured using a as reference the parallel fringes out side the fiber core. (In the fiber cladding). The refractive index difference between a point in the fiber core ( e.g. the core axis ) and the cladding can be optained from the fringeshift q, which corresponds to a number of fringe displacement. The different in refractive index δn is given by Δn= (qλ/x) Where, X is the thickness of the fiber slab and λ is the incident optical wavelength. (iii)fiber cut off wavelength (iv)fiber diameter Fiber diameter measurement GANADIPATHY TULSI’S JAIN ENGINEERING COLLEGE Form No: Chittoor – Cuddalore Road, Kaniyambadi (Post), Vellore – 632102. GTEC/ACADEMIC/ 2014 Academic Year 2013 – 14: Odd Semester Question Bank Rev No: 0 GANADIPATHY TULSI’S JAIN ENGINEERING COLLEGE Form No: Chittoor – Cuddalore Road, Kaniyambadi (Post), Vellore – 632102. GTEC/ACADEMIC/ 2014 Academic Year 2013 – 14: Odd Semester Question Bank Rev No: 0 GANADIPATHY TULSI’S JAIN ENGINEERING COLLEGE Form No: Chittoor – Cuddalore Road, Kaniyambadi (Post), Vellore – 632102. GTEC/ACADEMIC/ 2014 Academic Year 2013 – 14: Odd Semester Question Bank Rev No: 0 GANADIPATHY TULSI’S JAIN ENGINEERING COLLEGE Form No: Chittoor – Cuddalore Road, Kaniyambadi (Post), Vellore – 632102. GTEC/ACADEMIC/ 2014 Academic Year 2013 – 14: Odd Semester Question Bank Rev No: 0 UNIT –V Part A -Two Mark Questions: 1. Define modal noise and mode partition noise. [May-June 2009] [May-June 2010] Ans: Modal Noise: It arises when the light from a coherent laser is oupled into a multimodc fiber operating at 400 MbPs and higher. It ainly occurs due to mechanical vibrations and fluctuations in the equency of optical source. The Partition Noise: The mode partition noise is associate d with intensity fluctuation's in the longitudinal modes of a laser diode. It becomes more pronounced for the higher bit rates. 2. . What are the advantages, of using soliton signals through fiber? [May-June 2009] Anis: Group velocity dispersion causes most pulses to broader in time as they propogate through an optical fiber. However a particular puls shape-known as a soliton makes advantages of non-linear effect's in silica, particular self-phase modulation. 3. What are the three common topologies used for the fiber optical network? Give the schematic of any one network? [Nov-Dec 2011] Three common topologies used for the fiber optic networks.these are (i) (ii) (iii) Linear bus topology Ring topology Star topology Bus topology A bus topology provides multipoint communication.in this case, long cable called bus forms the backbone to all nodes. 4. Define optical CDMA[Nov-Dec 2011] Optical code-divison multiple access (CDMA) scheme can provide multiple access to a network,without using wavelength-sensitive components as in WDM,and without employing very high-speed electronic data processing devices as are needed in TDM networks. 5. Write the functions of path overhead. [May - June 2012] GANADIPATHY TULSI’S JAIN ENGINEERING COLLEGE Form No: Chittoor – Cuddalore Road, Kaniyambadi (Post), Vellore – 632102. GTEC/ACADEMIC/ 2014 Academic Year 2013 – 14: Odd Semester Question Bank Rev No: 0 The nine path-overhead(POH) bytes are always in a column ans can be located anywhere in the SPE in SONET/SDH. The POH supports performance monitoring by the end equipment,syatus,signallabelling,a tracing function,and a user channel. 6. What are the drawbacks of broadcast and select networks for wide area network applications? [May - June 2012] 1. There is an electro optical conversion takes place between transmitter and reeiver. 2. There is no direct connection between nodes. 7. What is WDM? [Nov-Dec 2012] Wavelength division multiplexing(WDM) involves the transmission of a number of different peak wavelength optical in parallel on a signal optical fiber. 8. Enumerate the various SONET/SDH layers? [Nov-Dec 2012] SONET/SDH has four optical interface layers. They are o Path layer o Line layer o Section layer o Photonic layer 9. What is a broadcast and selected network?[May-June 2013] In broadcast and selected networks,a node sends its transmission to the star on the available wavelength, using a laser which produces an optical information stream.the information streams from multiple sources are equally combined by the star and the signal power of each stream is equally split and forwarded to all of the nodes on their receive nodes Ex:ethernets,tokenrings,FDDI networks 10. what is solitons? [May-June 2013] Solitons are the p[ulses that travel along the fiber without change in shape or amplitude or velocity.The term soliton refers to special kinds of waves that can propagate undistorted over long distances and remain unaffected after collision with each other. Part B – Four/Eight/ Sixteen Mark Questions: 1. (a) Explain the salient features of solitons using relevant expressions and diagrams. (16) [Nov-Dec 2008] [April-May 2010] Ans: The term soliton" refers to special kinds of waves that can propagate undistorted over long GANADIPATHY TULSI’S JAIN ENGINEERING COLLEGE Form No: Chittoor – Cuddalore Road, Kaniyambadi (Post), Vellore – 632102. GTEC/ACADEMIC/ 2014 Academic Year 2013 – 14: Odd Semester Question Bank Rev No: 0 distances and remain unaffected after collisions with each other. The family of pulses that do not change in shape are called fundamental solitons, and those that undergo periodic shape changes are called higher-order solutions. In either case, attenuation in the fiber will eventually decrease the soliton energy. Soliton Pulses: No optical pulse is monochromatic, since it excites a spectrum of frequencies. If an optical source emits power in a wavelength band Ax its spectral speed is A V. When such a pulse traverses a medium with a positive GVD parameter 132 for the constituent frequencies, the leading part of the pulse is shifted toward a longer wavelength, so the speed in that portio increases. Conversely, in the trailing half, tne frequency rises so the speed decreases. This causes the trailing edge to be further delayed. These effects will severely limit high-speed long-distance transmission lithe system is operated in this condition. To. derive the evolution of the pulse shape required for soliton transmission, one needs to consider the NLS equation. For the 3 right-hand terms 1) The first term represents GVD effects of the fiber. 2) The second non-linear term denotes the fact that the refractive index of the fiber depends on the light intensity. 3) The third term represents the effects of energy loss or gain. The solution for the fundamental solition is given by u(z, t) = sech (t) exp 1jz12). here sech(t) is the hyperbolic secant function. This is a bell-shaped pulse. The time scale is given in units normalized to the 1/e width of the pulse. d = Iti(012 dz = sech2(t) dz. W For the non-linear process, For the dispersion effect. Since such a phase shift changes neither the temporal nor the spectral shape of a pulse, the soliton remains completely non-dispersive in both the temporal and frequency domains. Soliton Parameters: The FWHM of a pulse is defined as the full width of the pulse at its half-maximum power level. Thus the FWI-IM Ts of the fundamental soliton pulse in normalized time is found from the relationship sec112( t )= 1/2 with T =T1(2T0), where To is the basic normalized time unit. This yields GANADIPATHY TULSI’S JAIN ENGINEERING COLLEGE Form No: Chittoor – Cuddalore Road, Kaniyambadi (Post), Vellore – 632102. GTEC/ACADEMIC/ 2014 Academic Year 2013 – 14: Odd Semester Question Bank Rev No: 0 Soliton width and spacing: The return-to-zero (RZ) format is used. This condition thus constraints the achievable bit rate. Since there is a limit on how narrow. a soliton pulse can be generated 1) (b) (i) Discuss the concept of WDM with a neat diagram. (6) [Nov-Dec 2008] [April-May 2010] Ans: In standard point-to-point links a single fiber line hasone optical soutte at its transmitting end and one photodetector at the receiving end. To find the optical bandwidth corresponding to a particular spectral width in these regions, we use the relationship C=2,..y, which relates the wavelength x to the carrier frequency y, where C is the speed of light. Where the derivation in frequency A y corresponds the wavelength deviation A x around x. Since the spectral width of a high-quality source occupies only a narrow optical bandwidth, the two low-loss windows provide many additional operating regions. The integrities of the independent messages fro meach source are maintained for subsequent conversion to electric signals at the receiving end. Since WDM is essentially frequency-division multiplexing at optical carrier frequencies. The literature often uses the term dense WIDIv1, in contrast to conventional or regular WDM. A key feature of WDM is that the discrete wavelengths from an orthogonal set of carriers that can be separated GANADIPATHY TULSI’S JAIN ENGINEERING COLLEGE Form No: Chittoor – Cuddalore Road, Kaniyambadi (Post), Vellore – 632102. GTEC/ACADEMIC/ 2014 Academic Year 2013 – 14: Odd Semester Question Bank Rev No: 0 routed, and switched without interfering with each other. The performance of active devices can be control led electronically, thereby providing a large degree of networ flexibility. Active WDM components include tunable optical filters, tunable sources, and optical amplifiers. A multiplexer is needed to combine these optical outputs into a serial spectrum of closely spaced wavelength signals and couple them onto a single fiber. At the receiving end, a demo Itiplexer is required to separate the optical signals into appropriate detection channels for signal processing. Since the optical signals that are combined generally do not emit any signififant amount of optical power outside of the designated channels spectral width, interchannel crosstalk factors are relatively unimportant at the transmitting end. 1. (b) (ii) Give a brief account of the different types of losses to be considered in the design of an optical link. (6) [Nov-Dec 2008] Arts: An optical power loss model is analysed. The capital poWer received m the photodetector depends on the amount of light coupled into the fiber and the losses oecuring, in the fiber at the connectors and splices. Each of these loss elements is expressed in dB as Loss = 10 log Pin/Pout The link loss budget simply considers the Pr that is allowed between the light source and the photodetector, and allocates this loss to cable attenuation, connector loss, splice loss and system margin. Thus if Ps is the optical power emerging from the end of a fiber flylead attached to the light source, P r = Ps - PR = 2tc + a ft + system margin. The splice loss is incorporated into the cable loss for simplicity. 1. (b) (iii) Draw and explain the basic format of an STS-N SONET frame. (4) [Nov-Dec 2008] [April-May 2009] GANADIPATHY TULSI’S JAIN ENGINEERING COLLEGE Form No: Chittoor – Cuddalore Road, Kaniyambadi (Post), Vellore – 632102. GTEC/ACADEMIC/ 2014 Academic Year 2013 – 14: Odd Semester Question Bank Rev No: 0 Aim: With the advent of fibre optic transmission lines, the next step in the evolution of the digital time-division-multiplexing scheme was a standard signal format called synchronous optical network (SONET). Commonly used SONET and SDH transmission rates: SONET Electrical Level Line Rate (M Ws) SIM Equivalent OC - I STS -- 3I 51.84 STS STM - 1 OC - 3 155.52 STS - 3 OC - 12 STS- 12 622.08 STM - 4 OC - 24 STS - 24 1244.16 STM - 8 OC - 48 STS - 48 2488.32 STM - 16 OC -96 STS -96 4976.64 STM -32 OC - 192 STS - 192 9953.28 STM -64 2. Explain the operation of WDM Components ?[April-May 2009] Operational Principle of WDM What is WDM: The technology of combining a number of wavelengths onto the same fiber is known as wavelength-division-multiplexing or WDM DWDM: dense wavelength-division-multiplexing (channel spacing indense) GANADIPATHY TULSI’S JAIN ENGINEERING COLLEGE Form No: Chittoor – Cuddalore Road, Kaniyambadi (Post), Vellore – 632102. GTEC/ACADEMIC/ 2014 Academic Year 2013 – 14: Odd Semester Question Bank Rev No: 0 ฀ ITU (International Telecommunication Union) specify channel spacing in terms of frequency - C band (conventional band) : 1524 – 1560 nm - L band (longer wavelength band) : 1570 – 1610 nm - frequency reference 193.10 THz (1552.524 nm) - Channel spacing: 200 GHz, 100 GHz, 50 GHz, 25 GHz 3.Explain the architecture of SONET and discuss nonlinear effects on network performance. [Nov-Dec 2011][April-May 2013] The basic foundation of SONET consists of groups of DS-0 signals (64Kbits/sec) that are multiplexed to create a 51.84Mbit/sec signal, which is also known as STS-1 (Synchronous Transport Signal). STS-1 is an electrical signal rate that corresponds to the Optical Carrier line rate of OC-1, SONET's building block. Subsequent SONET rates are created by interleaving (at the byte level) STS-1 signals to create a concatenated, or linked, signal. For example, three STS1 frames can form an STS-3 frame (155Mbits/sec). Rates above STS-3 can be created by either GANADIPATHY TULSI’S JAIN ENGINEERING COLLEGE Form No: Chittoor – Cuddalore Road, Kaniyambadi (Post), Vellore – 632102. GTEC/ACADEMIC/ 2014 Rev No: 0 Academic Year 2013 – 14: Odd Semester Question Bank directly multiplexing STS-1 signals or by byte-interleaving STS-3 signals. The following table lists the hierarchy of the most common SONET/SDH data rates: Optical Level Electrical Level Line Rate (Mbps) Payload Rate (Mbps) Overhead Rate (Mbps) SDH Equivalent OC-1 STS-1 51.840 50.112 1.728 - OC-3 STS-3 155.520 150.336 5.184 STM-1 OC-12 STS-12 622.080 601.344 20.736 STM-4 OC-48 STS-48 2488.320 2405.376 82.944 STM-16 OC-192 STS-192 9953.280 9621.504 331.776 STM-64 OC-768 STS-768 39813.120 38486.016 1327.104 STM-256 The "line rate" refers to the raw bit rate carried over the optical fiber. A portion of the bits transferred over the line are designated as "overhead". The overhead carries information that provides OAM&P (Operations, Administration, Maintenance, and Provisioning) capabilities such as framing, multiplexing, status, trace, and performance monitoring. The "line rate" minus the "overhead rate" yields the "payload rate" which is the bandwidth available for transferring user data such as packets or ATM cells. The SONET/SDH level designations sometimes include a "c" suffix (such as "OC-48c"). The "c" suffix indicates a "concatenated" or "clear" channel. This implies that the entire payload rate is available as a single channel of communications (i.e. the entire payload rate may be used by a single flow of cells or packets). The opposite of concatenated or clear channel is "channelized". In a channelized link the payload rate is subdivided into multiple fixed rate channels. For example, the payload of an OC-48 link may be subdivided into four OC-12 channels. In this case the data rate of a single cell or packet flow is limited by the bandwidth of an individual channel. Protocol Layers GANADIPATHY TULSI’S JAIN ENGINEERING COLLEGE Form No: Chittoor – Cuddalore Road, Kaniyambadi (Post), Vellore – 632102. GTEC/ACADEMIC/ 2014 Academic Year 2013 – 14: Odd Semester Question Bank Rev No: 0 The SONET/SDH standard includes a definition of a transmission protocol stack which solves the operation and maintenance problems often found when dealing with networks that have component streams lacking a common clock. The photonic layer is the electrical and optical interface for transporting information over fiber optic cabling. It converts STS electrical signals into optical light pulses (and vice versa, at the receiving end). The section layer transports STS frames over optical cabling. This layer is commonly compared with the Data-Link layer of the OSI model, which also handles framing and physical transfer. The line layer takes care of a number of functions, including synchronization and multiplexing for the path layer above it. It also provides automatic protection switching, which uses provisioned spare capacity in the event of a failure on the primary circuit. The highest level, the path layer, takes services such as DS-3, T1, or ISDN and maps them into the SONET/SDH format. This layer, which can be accessed only by equipment like an add/drop multiplexer (a device that breaks down a SONET/SDH line into its component parts), takes care of all end-to-end communications, maintenance, and control. Topology SONET/SDH supports several topologies, including point to point, a hub and spoke star configuration, and the ring topology. The ring topology, which is by far the most popular, has been used for years by such network technologies as FDDI and Token Ring and has proven quite robust and fault-tolerant. A SONET/SDH ring can contain two pairs of transmit and receive fibers. One pair can be designated as active with the other one functioning as a secondary in case of failure. SONET/SDH rings have a "self-healing" feature that makes them even more appealing for long distance connections from one end of the country to another. One of SONET/SDH's most interesting characteristics is its support for a ring topology. Normally, one piece of fiber -- the working ring -- handles all data traffic, but a second piece of fiber -- the protection ring remains on standby. Should the working ring fail, SONET/SDH includes the capability to automatically detect the failure and transfer control to the protection ring in a very short period of time, often in a fraction of a second. For this reason, SONET/SDH can be described as a self-healing network technology. Rings normally will help SONET/SDH service to reach the "five nines" availability level. However, the usefulness of rings also depends on their physical location. If the rings are located next to each other, then if a back-hoe from a construction company takes out one of your fibers, GANADIPATHY TULSI’S JAIN ENGINEERING COLLEGE Form No: Chittoor – Cuddalore Road, Kaniyambadi (Post), Vellore – 632102. GTEC/ACADEMIC/ 2014 Academic Year 2013 – 14: Odd Semester Question Bank Rev No: 0 it is quite likely that the second one will go as well. Thus, your rings should be physically separated from each other as much as possible in order to achieve high uptime. 4. Explain the principle of solitons and discuss the soliton parameters with necessary expressions and diagrams. [May - June 2012] Solitons A single, consensus definition of a soliton is difficult to find. Drazin & Johnson (1989, p. 15) ascribe three properties to solitons: 1. They are of permanent form; 2. They are localised within a region; 3. They can interact with other solitons, and emerge from the collision unchanged, except for a phase shift. More formal definitions exist, but they require substantial mathematics. Moreover, some scientists use the term soliton for phenomena that do not quite have these three properties (for instance, the 'light bullets' of nonlinear optics are often called solitons despite losing energy during interaction). Dispersion and non-linearity can interact to produce permanent and localized wave forms. Consider a pulse of light traveling in glass. This pulse can be thought of as consisting of light of several different frequencies. Since glass shows dispersion, these different frequencies will travel at different speeds and the shape of the pulse will therefore change over time. However, there is also the non-linear Kerr effect: the refractive index of a material at a given frequency depends on the light's amplitude or strength. If the pulse has just the right shape, the Kerr effect will exactly cancel the dispersion effect, and the pulse's shape won't change over time: a soliton. See soliton (optics) for a more detailed description. Many exactly solvable models have soliton solutions, including the Korteweg–de Vries equation, the nonlinear Schrödinger equation, the coupled nonlinear Schrödinger equation, and the sineGordon equation. The soliton solutions are typically obtained by means of theinverse scattering GANADIPATHY TULSI’S JAIN ENGINEERING COLLEGE Form No: Chittoor – Cuddalore Road, Kaniyambadi (Post), Vellore – 632102. GTEC/ACADEMIC/ 2014 Rev No: 0 Academic Year 2013 – 14: Odd Semester Question Bank transform and owe their stability to the integrability of the field equations. The mathematical theory of these equations is a broad and very active field of mathematical research. Some types of tidal bore, a wave phenomenon of a few rivers including the River Severn, are 'undular': a wavefront followed by a train of solitons. Other solitons occur as the undersea internal waves, initiated by seabed topography, that propagate on the oceanic pycnocline. Atmospheric solitons also exist, such as the Morning Glory Cloud of the Gulf of Carpentaria, where pressure solitons travelling in atemperature inversion layer produce vast linear roll clouds. The recent and not widely accepted soliton model in neuroscience proposes to explain the signal conduction within neurons as pressure solitons. A topological soliton, also called a topological defect, is any solution of a set of partial differential equations that is stable against decay to the "trivial solution." Soliton stability is due to topological constraints, rather than integrability of the field equations. The constraints arise almost always because the differential equations must obey a set of boundary conditions, and the boundary has a non-trivialhomotopy group, preserved by the differential equations. Thus, the differential equation solutions can be classified into homotopy classes. There is no continuous transformation that will map a solution in one homotopy class to another. The solutions are truly distinct, and maintain their integrity, even in the face of extremely powerful forces. Examples of topological solitons include the screw dislocationin a crystalline lattice, the Dirac string and the magnetic monopole in electromagnetism, the Skyrmion and the Wess–Zumino–Witten model in quantum field theory, and cosmic strings and domain walls in cosmology. Prepared By: [D.Santhosh Kumari] AP/ECE HOD /ECE GANADIPATHY TULSI’S JAIN ENGINEERING COLLEGE Form No: Chittoor – Cuddalore Road, Kaniyambadi (Post), Vellore – 632102. GTEC/ACADEMIC/ 2014 Academic Year 2013 – 14: Odd Semester Question Bank Rev No: 0