FACULTY OF ENGINEERING

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1 Reg. No:. KIGALI INSTITUTE OF SCIENCE AND TECHNOLOGY INSTITUT DES SCIENCES ET TECHNOLOGIE Avenue de l'armée, B.P Kigali, Rwanda INSTITUTE EXAMINATIONS ACADEMIC YEAR SEMESTER II MAIN EXAMINATION FACULTY OF ENGINEERING ELECTRICAL AND ELECTRONICS ENGINEERING FOURTH YEAR ETE SEMESTER II (PART-TIME) EEE 3422 FIBER OPTIC COMMUNICATION DATE:.. TIME: 2 HOURS MAXIMUM MARKS = 60 INSTRUCTIONS 1. This paper contains ONE (1) question in SECTION A and THREE (3) questions in SECTION B 2. Answer - ALL questions in SECTION A and - any TWO (2) questions from SECTION B 3. No written materials allowed. 4. Write all your answers in the answer booklet provided. 5. Do not forget to write your Registration Number. 6. Do not write any answers on this question paper. 1

2 SECTION A (Compulsory) Question 1 (20 Marks) a) State three disadvantages of Optical Fiber Communications. (3 marks) b) Give 2 examples of dopants which: (i) Increase the refractive index of the core (1 example). (1 mark) (ii) Decrease the refractive index of the cladding (1 example). (1 mark) c) Compare a single-mode step index fibers and multimode step index fibers while used as an optical channel. (5 marks) d) Compare the light emitted from LASERs and LEDS. (2 marks) e) With the aid of a block diagram, discuss the function of the major elements of an optical fiber transmitter. (4 marks) f) Design a bus optical network topology and explain its working principle. (4 marks) SECTION B (Attempt any two questions) Question 2 (20 Marks) a) Using simple ray theory, design a mechanism for the transmission of light within an optical fiber. Briefly discuss with the aid of a suitable diagram what is meant by the acceptance angle for an optical fiber. Show how this is related to the fiber numerical aperture and the refractive indices for the fiber core and cladding. (5 marks) b) Briefly describe the two processes by which light can be emitted from an atom. Discuss the requirement for population inversion in order that stimulated emission may dominate over spontaneous emission. Illustrate your answer with an energy level diagram. (7 marks) c) Derive the expression for the output photocurrent of a basic coherent detector in both heterodyne and homodyne detection. (8 marks) 2

3 Question 3 (20 Marks) a) Discuss the mechanism of optical feedback to provide oscillation and hence amplification within the laser. (4 marks) b) A D-IM analog optical fiber link of length 2 Km employs an LED which launches mean optical power of 10 dbm into a multimode optical fiber. The fiber cable exhibits a loss of 3.5 with splice losses calculated at 0.7 In addition, there is a connector loss at the receiver of 1.6 db. The p-i-n photodiode receiver has a sensitivity of 25 dbm for an SNR of 50 db and with a modulation index of 0.5. It is estimated that a safety margin of 4 db is required. Assuming there is no dispersion-equalization penalty: (i) Perform an optical power budget for the system operating under the above conditions and ascertain its viability. (ii) Estimate any possible increase in link length which may be achieved using an injection laser source which launches mean optical power of 0 dbm into the fiber cable. In this case, the safety margin must be increased to 7 db. (8 marks) c) Compare orthogonal frequency division multiplexing (OFDM) with conventional frequency division multiplexing (FDM). (8 marks) Question 4 (20 Marks) a) Derive an expression for the coupling efficiency of a surface-emitting LED into a step index fiber, assuming the device to have a Lambertian output. Determine the optical loss in decibels when coupling the optical power emitted from the device into a step index fiber with an acceptance angle of 14. It may be assumed that the LED is smaller than the fiber core and that the two are in close proximity. (10 marks) b) Explain the Four-channel OTDM fiber system. (10 marks) 3

4 MARKING SCHEME Question 1 a) State three disadvantages of Optical Fiber Communications. (3 marks) b) Give 2 examples of dopants which: (i) Increase the refractive index of the core (1 example). (1 mark) (ii) Decrease the refractive index of the cladding (1 example). (1 mark) c) Compare a single-mode step index fibers and multimode step index fibers while used as an optical channel. (5 marks) d) Compare the light emitted from LASERs and LEDS. (2 marks) e) With the aid of a block diagram, discuss the function of the major elements of an optical fiber transmitter. (4 marks) f) Design a bus optical network topology and explain its working principle. (4 marks) Solution 1 a) Disadvantages of Optical Fiber Communications (3 marks) b) (i) Dopants such as GeO2 and P2O5 increase the refractive index of silica and are suitable for the core. (1 mark) (ii) Dopants such as B2O3 and fluorine decrease the refractive index of silica and are suitable for the cladding. (1 mark) 4

5 c) The single-mode step index fiber has the distinct advantage of low intermodal dispersion (broadening of transmitted light pulses), as only one mode is transmitted, whereas with multimode step index fiber considerable dispersion may occur due to the differing group velocities of the propagating modes. This in turn restricts the maximum bandwidth attainable with multimode step index fibers, especially when compared with single-mode fibers. (2 marks) However, for lower bandwidth applications multimode fibers have several advantages over single-mode fibers. These are: (i) The use of spatially incoherent optical sources (e.g. most light-emitting diodes) which cannot be efficiently coupled to single-mode fibers; (ii) Larger numerical apertures, as well as core diameters, facilitating easier coupling to optical sources; (iii) Lower tolerance requirements on fiber connectors. (3 marks) d) A laser emits a relatively narrow angular spread of Light while a LED transmits light within a relatively wide cone (2 marks) e) The principal components of a general optical fiber transmitter for either digital or analog transmission are shown in the system block schematic of Figure 1. Figure 1: The principal components of an optical fiber transmitter The transmit terminal equipment consists of an information encoder or signal shaping circuit preceding a modulation or electronic driver stage which operates the optical source. Light emitted from the source is launched into an optical fiber incorporated within a cable which constitutes the transmission medium. (4 marks) 5

6 f) The representation of a bus structure is shown in Figure 2. Figure 2: Example of a bus structure with optical amplifiers and one OADM A number (n) of WDM channels emitted from the M-Tx enters the OADM. A subset (n*) of WDM channels is dropped and added by the OADM. The number n* of dropped and added channels may range between 0 and n. When n* = n, all WDM channels are dropped and added. If n* = 0, then no channel is added or dropped, i.e. the OADM is just a through-way network element. This scheme can be generalized by incorporating a sequence of optical amplifiers and optical add/drop multiplexers (OADMs). (4 marks) Question 2 a) Using simple ray theory, describe the mechanism for the transmission of light within an optical fiber. Briefly discuss with the aid of a suitable diagram what is meant by the acceptance angle for an optical fiber. Show how this is related to the fiber numerical aperture and the refractive indices for the fiber core and cladding. (5 marks) b) Briefly describe the two processes by which light can be emitted from an atom. Discuss the requirement for population inversion in order that stimulated emission may dominate over spontaneous emission. Illustrate your answer with an energy level diagram. (7 marks) c) Derive the expression for the output photocurrent of a basic coherent detector in both heterodyne and homodyne detection. (8 marks) Solution 2 a) An optical fiber consists of a very thin fiber at its centre known as Core of refractive index n 1 surrounded by a coaxial middle region which is made of material less dense than the core material known as Cladding of refractive index n 2 (n 1 > n 2 ). If the light enters in such a fiber at an angle θ greater than the critical angle at the interface corecladding, it will undergo total internal reflection and is reflected at the same angle to the normal. The light will then be contained within the fiber and will propagate to the far end by a series of reflections. 6

7 Acceptance angle and Numerical Aperture Consider a meriditional ray (ray that passes through the axis of the fiber core). The light rays contained within the core having a full angle of along the fiber. are accepted and transmitted The Acceptance angle is expressed as: The numerical aperture NA of the fiber is defined as the sine of the acceptance angle. Let ( )( ) As Δ Normalized difference between the refractive indices of core and cladding If (5 marks) 7

8 b) This emission process can occur in two ways: (i) By spontaneous emission in which the atom returns to the lower energy state in an entirely random manner; (ii) By stimulated emission when a photon having an energy equal to the energy difference between the two states (E 2 E 1 ) interacts with the atom in the upper energy state causing it to return to the lower state with the creation of a second photon. Spontaneous emission: An atom stays in the excited state about seconds and reverts to the lower state emitting a photon of energy E 2 E 1 = hf. The emission of a photon by an atom without an external agent is called spontaneous emission. (A = Atom, = Atom in excited state) Stimulated emission: When an atom is at the excited level E 2, an interaction with a photon of energy E 2 E 1 = hf induces its deexciatation to a lower energy level E 1 with emission of an additional photon of the same frequency f. Thus 2 photons instead of one move on. This phenomenon of forced photon emission by an excited atom due to the action of external agent is called stimulated (induced) emission. (A = Atom, = Atom in excited state) Figure 3: Energy state diagram showing: (b) spontaneous emission; (c) stimulated emission. The black dot indicates the state of the atom before and after a transition takes place Population inversion However, to achieve optical amplification it is necessary to create a non-equilibrium distribution of atoms such that the population of the upper energy level is greater than that of the lower energy level (i.e. N 2 > N 1 ). This condition, which is known as population inversion, is illustrated in Figure 4. 8

9 Figure 4: Populations in a two-energy-level system: (a) Boltzmann distribution for a system in thermal equilibrium; (b) a non-equilibrium distribution showing population inversion (7 marks) c) Coherent detection principle A simple coherent receiver model for ASK is displayed in Figure 5. Figure 5: Basic coherent receiver model The low-level incoming signal field e S is combined with a second much larger signal field e L derived from the local oscillator laser. It is assumed that the electromagnetic fields obtained from the two lasers (i.e. the incoming signal and local oscillator devices) can be represented by cosine functions and that the angle ϕ = ϕ S ϕ L represents the phase relationship between the incoming signal phase ϕ S and the local oscillator signal phase ϕ L defined at some arbitrary point in time. Hence, as depicted in Figure 5, the two fields may be written as: (1) and 9

10 (2) where E S is the peak incoming signal field and ω S is its angular frequency, and E L is the peak local oscillator field and ω L is its angular frequency. The angle ϕ (t) representing the phase relationship between the two fields contains the transmitted information in the case of FSK or PSK. However, with ASK ϕ (t) is constant and hence it is simply written as ϕ in Eq. (4.6). For heterodyne detection, the local oscillator frequency ω L is offset from the incoming signal frequency ω S by an intermediate frequency such that: (3) where ω IF is the angular frequency of the IF. The IF is usually in the radio-frequency region and may be a few tens or hundreds of megahertz. By contrast, within homodyne detection there is no offset between ω S and ω L and hence ω IF = 0. In this case the combined signal is therefore recovered in the baseband. The two wavefronts from the incoming signal and the local oscillator laser must be perfectly matched at the surface of the photodetector for ideal coherent detection. In the case of both heterodyne and homodyne detection, the optical detector produces a signal photocurrent I p which is proportional to the optical intensity (i.e. the square of the total field for the square-law photodetection process) so that: (4) Substitution in the expression (4.11) from Eqs (4.6) and (4.7) gives: Assuming perfect optical mixing expansion of the right hand side of the expression shown in Eq. (4.10) gives: (5) Removing the higher frequency terms oscillating near the frequencies of 2ω S and 2ω L which are beyond the response of the detector and therefore do not appear in its output, we have: 10 (6)

11 Then recalling that the optical power contained within a signal is proportional to the square of its electrical field strength, expression (4.11) may be written as: (7) where P S and P L are the optical powers in the incoming signal and local oscillator signal respectively. Furthermore, a relationship was obtained between the output photocurrent from an optical detector and the incident optical power P o of the form: Hence the expression in (4.12) becomes: quantum efficiency of the photodetector, e is the charge on an electron, h is Planck s constant and f is optical frequency. (8) where η is the When the local oscillator signal is much larger than the incoming signal, then the third a.c. term in Eq. (4.13) may be distinguished from the first two d.c. terms and Ip can be replaced by the approximation I S where: (9) Equation (4.14) allows the two coherent detection strategies to be considered. For heterodyne detection ω S ω L and substituting from Eq. (4.8) gives: (10) indicating that the output from the photodetector is centered on an IF. This IF is stabilized by incorporating the local oscillator laser in a frequency control loop. Temperature stability for the signal and local oscillator lasers is also a factor which must be considered. The stabilized IF current is usually separated from the direct current by filtering prior to electrical amplification and demodulation. For the special case of homodyne detection, however, ω S = ω L and therefore Eq. (4.14) reduces to: (11) or (12) 11

12 where R is the responsivity of the optical detector. In this case the output from the photodiode is in the baseband and the local oscillator laser needs to be phase locked to the incoming optical signal. Question 3 (8 marks) a) Discuss the mechanism of optical feedback to provide oscillation and hence amplification within the laser. (4 marks) b) A D-IM analog optical fiber link of length 2 Km employs an LED which launches mean optical power of 10 dbm into a multimode optical fiber. The fiber cable exhibits a loss of 3.5 with splice losses calculated at 0.7 In addition, there is a connector loss at the receiver of 1.6 db. The p-i-n photodiode receiver has a sensitivity of 25 dbm for an SNR of 50 db and with a modulation index of 0.5. It is estimated that a safety margin of 4 db and with a modulation index of 0.5. It is estimated that a safety margin of 4 db is required. Assuming there is no dispersion-equalization penalty: (i) Perform an optical power budget for the system operating under the above conditions and ascertain its viability. (ii) Estimate any possible increase in link length which may be achieved using an injection laser source which launches mean optical power of 0 dbm into the fiber cable. In this case, the safety margin must be increased to 7 db. (8 marks) c) Compare orthogonal frequency division multiplexing (OFDM) with conventional frequency division multiplexing (FDM). (8 marks) Solution 3 a) Optical feedback To achieve this laser action, it is necessary to contain photons within the laser medium and maintain the conditions for coherence. This is accomplished by placing or forming mirrors (plane or curved) at either end of the amplifying medium, as illustrated in Figure 6. Figure 6: The basic laser structure incorporating plane mirrors 12

13 The optical cavity formed is more analogous to an oscillator than an amplifier as it provides positive feedback of the photons by reflection at the mirrors at either end of the cavity. Hence the optical signal is fed back many times while receiving amplification as it passes through the medium. Very high radiation density should be present in the medium. The density is made larger by enclosing the emitted radiation in an optical resonant cavity. An optical resonant cavity is a pair of optically plane mirrors; one of them is fully reflecting (100%) while the other is partially reflecting (90%) and a small fraction of it is transmitted through it as a laser. b) (4 marks) 13 (8 marks)

14 c) Orthogonal frequency division multiplexing (OFDM) is a multicarrier transmission technique which is based on frequency division multiplexing (FDM). In conventional FDM, multiplefrequency signals are transmitted simultaneously in parallel where the data contained in each signal is modulated onto subcarriers and therefore the subcarrier multiplexed signal typically contains a wide range of frequencies. Each subcarrier is separated by a guard band to avoid signal overlapping. The subcarriers are then demodulated at the receiver by using filters to separate the frequency bands. By contrast, OFDM employs several subcarrier frequencies orthogonal to each other (i.e. perpendicular) and therefore they do not overlap. Hence this technique can squeeze multiple modulated carriers tightly together at a reduced bandwidth without the requirement for guard bands while at the same time keeping the modulated signals orthogonal so that they do not interfere with each other, as illustrated in Figure 7. In the upper spectral diagram, 10 non-overlapping subcarrier frequency signals arranged in parallel depicting conventional FDM are shown, each being separated by a finite guard band. OFDM is displayed in the bottom spectral diagram where the peak of one signal coincides with the trough of another signal. Each subcarrier, however, must maintain the Nyquist criterion separation with the minimum time period of T (i.e. a frequency spread of 1/T) for each subcarrier. OFDM uses the inverse fast Fourier transform (IFFT) for the purpose of modulation and the fast Fourier transform (FFT) for demodulation. Moreover, this is a consequence of the FFT operation by which subcarriers are positioned perpendicularly and hence the reason why the technique is referred to as orthogonal FDM. It may be observed that a large bandwidth saving in comparison with conventional FDM is identified in Figure 7 resulting from the orthogonal placement of the subcarriers. Figure 7: Orthogonal frequency division multiplexing (OFDM) compared with conventional frequency division multiplexing (FDM) (8 marks) 14

15 Question 4 a) Derive an expression for the coupling efficiency of a surface-emitting LED into a step index fiber, assuming the device to have a Lambertian output. Determine the optical loss in decibels when coupling the optical power emitted from the device into a step index fiber with an acceptance angle of 14. It may be assumed that the LED is smaller than the fiber core and that the two are in close proximity. (10 marks) b) Explain the Four-channel OTDM fiber system. (10 marks) Solution 4 a) Although the possible internal quantum efficiency can be relatively high, the radiation geometry for an LED which emits through a planar surface is essentially Lambertian in that the surface radiance is constant in all directions. The Lambertian intensity distribution is illustrated in Figure 8 where the maximum intensity I 0 is perpendicular to the planar surface but is reduced on the sides in proportion to the cosine of the viewing angle θ as the apparent area varies with this angle. Figure 8: The Lambertian intensity distribution typical of a planar LED A further loss is encountered when coupling the light output into a fiber. Considerations of this coupling efficiency are very complex; however, it is possible to use an approximate simplified approach. If it is assumed for step index fibers that all the light incident on the exposed end of the core within the acceptance angle θa is coupled, then for a fiber in air: (2.5) 15

16 Also, incident light at angles greater than θa will not be coupled. For a Lambertian source, the radiant intensity at an angle θ, I(θ), is given by (see Figure 2.7): (2.6) where I 0 is the radiant intensity along the line θ = 0. Considering a source which is smaller than, and in close proximity to, the fiber core, and assuming cylindrical symmetry, the coupling efficiency η c is given by: Hence substituting from Eq. (2.6): (2.7) Furthermore, from Eq. (2.5): (2.8) Numerical Aperture NA = Sin (ϴ A ) = Sin (14 0 ) = Coupling loss ( ) ( ) (10 marks) 16

17 b) A block schematic of an OTDM system which has demonstrated 160 Gbit s 1 transmission over 100 km is shown in Figure 9. The principle of this technique is to extend ETDM (Electrical TDM) by optically combining a number of lower speed electronic baseband digital channels. In the case illustrated in Figure 9, the optical multiplexing and demultiplexing ratio is 1: 4, with a baseband channel rate of 40 Gbit s 1. Hence the system can be referred to as a four-channel OTDM system. Figure 9: Four-channel OTDM fiber system The four optical transmitters in Figure 5.1 were driven by a common 40 GHz clock using quarter bit period time delays. Mode-locked semiconductor laser sources which produced short optical pulses (around 2 ps long) were utilized at the transmitters to provide low duty cycle pulse streams for subsequent time multiplexing. Data was encoded onto these pulse streams using integrated optical intensity modulators which gave return-to-zero transmitter outputs at 40 Gbit s 1. These I/O devices were employed to eliminate the laser chirp which would result in dispersion of the transmitted pulses as they propagated within the single-mode fiber, thus limiting the achievable transmission distance. The four 40 Gbit s 1 data signals were combined using an OTDM multiplexer. Although four optical sources were employed, they all emitted at the same optical wavelength and the 40 Gbit s 1 data streams were bit interleaved to produce the 160 Gbit s 1 signal. At the receive terminal, the incoming signal was decomposed into the 40 Gbit s 1 baseband components in a demultiplexer. Hence single-wavelength 160 Gbit s 1 optical transmission was obtained with electronics which only required a maximum bandwidth of about 40 GHz, as return-to-zero pulses were employed. The transmitter and receiver sections shown in Figure 5.1 employed electro-absorption modulators to provide for operation at the high transmission rate and furthermore negative dispersion fibers were also incorporated to compensate for the positive dispersion of the standard single-mode fiber (SSMF). Moreover, a field trial employing such transmitters and receivers at a transmission rate of 160 Gbit s 1 over deployed SSMF has been successfully carried out. (10 marks) 17