Microsoft word - ee555_exam_jan2008_solutions.doc

DUBLIN CITY UNIVERSITY
Optical Communications System Design
Exam Solutions

Question 1

Write a brief note on each of the following topics (each part carries 5 marks):
a) State the main differences between a LED to a LASER [2.7.4] • A LED emits spontaneous radiation which is broad in its spectral nature • A laser emits stimulated radiation, which has a narrower linewidth and is coherent (same phase, polarisation, wavelength, direction) • Laser’s are more efficient due to stimulated emission dominating over • A laser is a threshold device, a LED isn’t • A laser has a strong dependence of threshold current and hence output power, on temperature, whilst an LED has a weak temperature dependence • Although the two devices can generate similar amounts of optical power, the better directed laser beam means that more light from a laser can be coupled into fibre. [5 marks]
b) Molecular Beam Epitaxy (MBE) [3.2.2] • Evaporation technique that is used in the manufacturing process for • System consists of deposition chamber that is maintained at very low pressure and one or more effusion cells that contain very pure base materials. • Electron beam directed into effusion cell heats material to liquid state, resulting in atoms evaporating out of materials, exiting through shutter and depositing on the wafer. Shutter is computer controlled allowing precise control over the thickness of each layer • Evaporated atoms interact with each other only when they reach the • Adv: ability to form multiple layers on wafter during one process step, • Disadv: low film growth resulting in low productivity and high expense [5 marks]
c) Operation of a Mach-Zehnder Interferometer as used for External Modulation [4.1.1] • Uses electro-optic effect (refractive index of material is changed by the application of an applied electric field) in lithium niobate to alter the optical path length and cause either constructive or destructive interference at the output node of the device • Structure of device consists of a block of lithium niobate from which two waveguides are cut with electrodes placed on either side of the waveguide (interaction length) • Light enters the device, it is split equally among the two branches and propagates over the same distance until recombining at the output node. • Electric field, which is proportional to the electrical data being transmitted applied to one arm of the modulator, altering the optical path length, resulting in either constructive or destructive interference at output node. • Adv: signal nearly chirp free, low temporal jitter, large bandwidth • Disadv: increased cost and insertion loss [5 marks]
i. Simplest and involve direct fibre connection between nodes iii. Might need to use amplifiers with distance exceeds 20km iv. Station processes electrical data, with opto-electronic i. Information distributed to a group of subscribers using either ii. Star: Nodes connected by point-to-point links to a central node called the hub (star formation). Stars connected by point-to-point links. Central node can be active or passive: active, signal converted to electrical domain and the electrically distributed to individual nodes; passive, distribution takes place in optical domain iii. Bus: single fibre carries all data transmitted. Access to bus achieved using optical couplers, with can either active or passive as with the star. Problem can be the power loss for a large number of users, might need optical amplifiers i. Large number of users within a local area ii. Allows users to access network randomly and transmit to any iii. Transmission distances are less than 10km, so fibre loss is not iv. Bus, Star or Ring can be used. Ring is were consecutive nodes are connected by point-to-point links to form a closed ring. Node is an active device that monitors information transmitted around ring for data, listening for its address [5 marks]
• Low insertion loss: takes into account coupling loss, waveguide loss and excess loss. Will be important in a large network as it can affect overall power budget • Switching speed: amount of time taken from when command is given to where insertion loss reaches 90% of value. Required switching time depends on the particular application: protection millisecond; packet nanosecond; OTDM picosecond • Crosstalk: measure of amount of interference between channels, with low level of crosstalk and extinction ratio representing high signal quality. Typical values around 40dB • Wavelength response: wavelength dependent response would allow switch to carry out wavelength selection and detection without the need for external filtering. Reduce cost and overall loss of switch • Bit rate and protocol transparency: ensures that switch can operate at future data rates and protocols without the need for upgrades [5 marks]
Question 2

a) Laser structure question….poor carrier and optical confinement… comment on heterojunction lasers, stripe geometry, buried heterostructure [2.8, 3.1] • Disadvantage of simple PN junction are their inherenet efficiency due to the very poor optical and carrier confinement, resulting in the need for large threshold currents which increases cost and excessive heating • Improved using heterojunction devices, which has a lightly doped or intrinsic active region sandwiched between two different bandgap semiconductors which have a wider band gap and lower refractive index. • Bandgap difference forms a potential barriers in both the CB and VB which prevents injected carriers from diffusing away from the active region. This allows active region to be made very narrow. • The change of refractive index provides a more efficient waveguide structure compared to homojunctions, confining the optical carriers to the active region. This also helps reduce the amount of optical carrier absorption that can occur outside the active region. This helps to reduce the threshold current level required for device operation. • Narrow stripe that extends the entire longitudinal direction of the laser • Removes gain in any of the lateral modes, and confines population inversion to narrow region reducing operating current, reducing heat, improving reliability • Example is a MESA stripe: involves etching away the undesired regions of the semiconductor and coating with an oxide insulator either side of the active region. Confines injected carriers to a narrow channel running from the contact, through top material and into active region • Provides better lateral confinement than MESA, where active region surrounded on all four sides by materials of lower refractive index. • Regrown areas act as dielectric waveguide due to their lower refractive index, which has the affect of confining the optical power just as the lateral heterojunction confines the carriers. [8 marks]
b) Describe how a DFB lasers operates, and state the major difference between a • Employs a wavelength selective component close to the active region • Feedback provided throughout the devices length, and not just by the • A grating is etched into the top layer surrounding the active region, running the entire longitudinal direction of the device. The grating feedback is wavelength specific, allowing DFB output to be single-moded. • Grating is a periodic variation of the refractive index of the guiding layer, with the feedback provided by backward Bragg scattering. The reflected wave then couples with other propagating waves. • The wavelength that is reflected is governed by the Bragg condition, with the desired wavelength experiencing constructive interference. The number of periods used in the grating determines the wavelength. • The grating is produced during the formation of the DH structure by chemically etching away one of the epitaxial layers and using holographic techniques to form the corrugation structure • During DFB operation, some of the light leaks out form the active region and enters the surrouding cladding regions (evanescent light). This light is then filtered by the Bragg grating, with the feedback only providing optical gain in one of the modes propagating within the active region. • The tuning range can be slightly tuned using temperature control. The spectral width can be as narrow as 1MHz. • Gratings used in DBR are placed outside the active region • Has 3/4 separate different sections (gain, mirror, phase) • Wider tuning range achieved by injecting current into the phase and mirror sections, chanign the carrier density which changes the refractive index • Allow wavelength tuning of 40nm over the entire C-band • Suffer increased losses and the optical signal generated within the active region has to propagate through the unpumped phase and mirror sections • Can be overcome by incorporating SOA into device design to boost the [8 marks]
c) A 10 channel wavelength division multiplexed system is designed to operate over a distance of 200km using single mode fibre with a loss coefficient of 0.2dB/km. The link contains three optical amplifiers, each with a gain of 15dB, and the transmitted power per channel is 0.032mW. Given that the loss of each optical filter before the receiver is 2dB, and the loss encountered by each channel on passing through a 1 x 10 or a 10 x 1 coupler is 6dB, calculate the required receiver sensitivity for each wavelength in mW’s allowing for a power margin of 4dB. • Power out = Power In – Loss + Gain = -15dBm - (fibre loss + filter loss + coupler x 2 + power margin) + [9 marks]
Question 3
WDM
a) Write a brief note on the following topics with relation to wavelength division multiplexed optical communications systems; [8.1] i. Maximum number of channels that can be transmitted • Amplification range of the EDFA limited to the c-band. Thus in order to increase the amount of data that is transmitted the wavelength channels have to be positioned as close as possible without interfering • Thus one of the most important parameters will be the spectral width and wavelength stability, with external modulation of single-moded lasers preferred [4 marks]
ii. Optical Amplification of the WDM signal • Used to overcome fibre loss
• Boosters are used immediately after the transmitter or multiplexer
to increase the power of the signal entering the fibre. Inline amplifiers are then used to amplify the signal periodically between the transmitter and the receiver. Pre-amplifiers are used before the receiver/Demultiplexer to increase received power and transmission distance. • EDFA are used as they amplify over the entire C-band, with high • Problems associated with introduction of nonlinear effects, noise added and the non-uniformity of the gain spectrum [4 marks]
iii. Demultiplexing of the WDM signal at the receiver • Two ways to carry it out: optical filters or use a AWG
• For filtering, a passive fibre coupler splits the received signal into
N different copies where N corresponds to the number of data channels transmitted. Each copy of the WDM signal then enters a filter which isolates one wavelength channel. The isolated signal is then detected using a photodetector. Important parameters are insertion loss, filter bandwidth. Problems are need for large number of filters if transmitting large number of channels, and insertion loss • AWG works by splitting the input WDM signal by diffraction via a lens and waveguide into N different copies. Each copy than propagates through an arrayed waveguide, where adjacent waveguides have a specific length difference. This introduces a phase shift between adjacent channels. When signals exit arrayed waveguide, they are spread by diffraction again at a specified wavelength. As different copies have experienced different phase shifts, destructive interference in the output waveguide allows certain copies of the same wavelength to be eliminated, thereby extracting individual wavelengths.It has a lower insertion loss for a larger number of wavelengths when compared to optical filters [4 marks]
b) The dispersion paramters of a single mode optical fibre is D=15ps/(km.nm) at a wavelength of 1550nm. If this fibre is used to transmit an NRZ signal at a bit rate of 10Gb/s, what is the required bandwidth of the receiver when the transmission distance is 20km? The laser transmitter used is at a wavelength of 1550nm, has a spectral linewidth of 0.1nm, and a 3dB bandwidth of 12GHz. Assume that overall risetime must be kept below 70% of the bit period for a NRZ communication system. • Tsys=(Tt/x^2+Tfibre^2+Tr/x^2)^1/2 • Tsys= bit period x 0.7=70ps • Tt/x=0.35/12GHz=29ps • Tfibre=D x L x sigma_lambda=30ps • 70ps=29ps^2+30ps^2+Tr/x^2 • Tr/x=56ps • BW of Tr/x=0.35/62ps=6.25GHz [13 marks]
Question 4
OTDM
• OTDM allows high aggregate data rates to be transmitted over a single wavelength by using ultrashort optical pulse to represent data and multiplexing these optical data pulses in the time domain, instead of the wavelength domain as in WDM [5 marks]
• Multiplexes in the time domain by allocating each channel specific bit slots in the overall multiplexed signal. • The main component is a RZ pulse source. The optical pulse train generated at a repetition rate R is split into N copies of itself by a passive optical coupler, where N is the number of channels being transmitted. Each copy of the pulse train is then individually modulated with electrical data ata a repetition rate R. As the modulators are operating at the individual channel data rate, they are readily accessible using current electronic components. • The modulated optical signal then passes through a fixed fibre delay length which delays each channel by 1/RN relative to adjacent channels in the system. This ensures that the optical data channels arrices the output at a time corresponding to its allocated bit slot in the overall OTDM signal. The N optical data channels are then recombined using a second fibre coupler resulting in the OTDM data signal [8 marks]
c) What would be the required pulse width for a bit-interleaved OTDM pulse source operating at 10GHz, transmitting 16channels be? Remember to avoid ISI, the bit slot is about 3 times the pulse width. Bit Slot per channel=100ps/16channels=6.25ps Using the same pulse width found, how many channels could be accommodated if the repetition rate was increased to 40GHz? [5 marks]
d) Why is an optical pulse source of the most important elements of an high- speed OTDM network? List the important criteria that has to be met [9.3.1] • Important element as the overall data rate of the system is essentially determined by the temporal separation between data channels o Pulse duration: determines the upper limit of the bit rate and must be short enough to support the desired overall transmission rate. For Tb/s sub-picosecond pulses are required. o Spectral width: pulses should as a spectrally pure as possible in order to minimise fibre dispersion and maximise transmission distance. Ideally the pulse source is required to be transform limited, that is, the spectral width of the generated optical pulses are as small as possible for the associated pulse width o Timing jitter: it is the random fluctuation in the pulse repetition period and can be responsible for degradation of the temporal resolution, thereby limiting the number of channels in an OTDM network. Should be less than 7% of the width of the temporal bit slot for BER of 10e-12 o Stability: temperature fluctuations can effect the optical path length which can result in the bit-interleaving process suffereing from significant crosstalk between adjacent pulses. o Other parameters: tunable wavelength and repetition rate, and [7 marks]
Question 5
BER: - gives quantitative analysis of system performance - Essentially the number of bits received in error divided by total number - Usually around 10e-9 for optical communications system - BW of detector and amplifier also affect the BER Eye Diagram: - Superposition of all possible bit sequences overlaid - Opening of the eye gives a good indication of the overall system - Eye closes vertically due to receiver noise - Eye closes horizontally due to limited BW in the system [5 marks]
Calculate the maximum transmission distance when T1=40ps, T0=2ps, B2=-20ps2/km [6 marks]
c) A hybrid radio/fibre system is to be used to transmit an RF data signal (at a frequency of 28 GHz) from a central station to base stations. From the base station, the microwave signal is sent over air to the network users. The system uses a laser diode followed by an external modulator, to which the RF data is applied. Given the following information, calculate the minimum transmission power out of the laser in the central station such that the system operates correctly: • The RF power level applied to the modulator is such that the modulation index of the optical signal is 1 (100 %) • The insertion loss of the modulator is 3dB • The attenuation coefficent of the fibre is 0.2dB/km • Length of fibre=20km • The responsivity of the detector is 0.8 A/W • The detected RF signal passes through a 50 Ohm electrical amplifier with a gain of 16 dB before being transmitted over air using an antenna • Transmitted RF power must be greater than –20 dBm for correct operation • Fibre dispersion and detector noise can be neglected • Electrical power in a sinewave with a peak-to-peak voltage VP is (VP)2/2R Power in Base Station - from antenna back, we have -20dBm, RF amplifier 16dB gain, need -36dB - convert -36dB electrical to current, and then use responsitivity of detector o 10^-36/10=0.251uW o Using equation given, Vp=sqrt(2*R*P)=5.01mV o I=Vp/50=100uA (100e-6A) - Optical Power from responsitivty (0.8A/W) Power Budget Part - Power out = Power In – Loss o Power out=-9dBm, Loss = 0.2 x 20km= 4dB o Power in = Power Out + Loss o Power in = -9dBm + 4dB = -5dBm - Power into fibre = Power out of laser – modulator insertion loss - Power out of laser = Power into fibre + modulator insertion loss - Power out of laser =-5dBm + 3dB = -2dBm [12 marks]

Source: http://www.rince.ie:8888/ee454/1351-EE/version/default/part/AttachmentData/data/EE555_Exam_Jan2008_Solutions.pdf