exercises2014-fiber by Yvan Ngassa

Preface This booklet is intended for use with the course textbook Fiber-Optic Communication Systems 4th ed. by Govind P. Agrawal, in conjunction with the lecture notes. The collection is divided in three parts; problems, answers and exams. Each problem section begins with a few (usually simpler) problems and/or derivations that are often taken from the bookâ&#x20AC;&#x2122;s list of problems. At the end of some problem sections a few old exam problems are found. The answers to these problems are given in the second part, and since the mere answer of a problem might be too brief to facilitate self-studies, complete solutions to the problems can be obtained from the teachers on request. Finally, in the last part we provide two exams including solutions. Any findings of misprints or other errors (along with other suggestions for improvement) are gratefully acknowledged.

Contents I

Problems

1 Introduction, optical fibers

2 Signal generation, transmitters

3 Signal propagation in optical fibers

4 Nonlinear impairments and solitons

5 Optical transmitters and receivers

6 Lightwave systems (Error probability, power penalties)

7 Multichannel systems

8 Optical amplifiers

9 Dispersion management

10 Coherent systems

Answers

1 Introduction, optical fibers

2 Signal generation, transmitters

3 Signal propagation in optical fibers

4 Nonlinear impairments and solitons

5 Optical tansmitters and receivers

6 Lightwave systems (Error probability, power penalties)

7 Multichannel systems

8 Optical amplifiers

9 Dispersion management

10 Coherent systems

III

Exams

Part I

Problems 1

Introduction, optical fibers

1.1 Calculate the transmission distance over which the optical power will attenuate by a factor of 10 for three fibers with losses of 0.2, 20, and 2000 dB/km. Assuming that the optical power decreases as exp(-αL), calculate α (in cm−1 ) for the three fibers. 1.2 Assume that a digital communication system can be operated at a bit rate of up to 1% of the carrier frequency. How many audio channels at 64 kbit/s can be transmitted over a microwave carrier at 5 GHz and an optical carrier at 1.55 µm? 1.3 An one-hour lecture script is stored on the computer hard disk in the ASCII-format. Estimate the total number of bits assuming a delivery rate of 200 words per minute and on average 5 letters per word. How long will it take to transmit the script at a bit rate of 1 Gbit/s? 1.4 A 1.55-µm digital communication system operating at 10 Gbit/s receives an average power of -40 dBm at the detector. Assuming that ’1’ and ’0’ bits are equally likely to occur, calculate the number of photons received within each ’1’ bit. 1.5 An analog voice signal that can vary over the range 0 - 50 mA is digitized by sampling it at 8 kHz. The first four sample values are 10, 21, 36, and 16 mA. Write the corresponding digital signal (a string of ’1’ and ’0’ bits) by using a 4-bit representation for each sample. 1.6 A 1.55 µm digital communication system is transmitting digital signals over 100 km at 20 Gbit/s. The transmitter launches 2 mW of average power into the fiber cable, having a net loss of 0.3 dB/km. How many photons are incident on the receiver during a single ’1’ bit? Assume that ’0’ bits carry no power, while ’1’ bits are in the form of a rectangular pulse occupying the entire bit slot (NRZ format). 1.7 A 0.8 µm optical receiver needs at least 1000 photons to detect the ’1’ bits accurately. What is the maximum possible length of the fiber link for a 100 Mb/s communication system designed to transmit −10 dBm of average power? The fiber loss is 2 dB/km at 0.8 µm. Assume the NRZ format and a rectangular pulse shape. 1.8 A high performance silica-fiber has a minimum attenuation of 0.2 dB/km at 1.55 µm. (a) If 1 mW of optical power is launched into the fiber, how much power is left after 45 km? (b) How much power is left if the same amount of power is launched into the fiber at 0.85 µm wavelength, where the minimum attenuation is 2 dB/km? 1.9 In a multi-mode step-index fiber, intermodal dispersion is the dominant mechanism for pulse spreading. Intermodal dispersion arises because light rays incident with different angles at the core-cladding interface travel different distances along the fiber. n2 NA=sin(θi) θi

θt

Problem 1.9: A multi-mode step-index fiber geometry.

(a) Derive an expression, using geometrical optics, for the pulse spreading, ∆T, due to intermodal dispersion as a function of the numerical aperture NA. (b) How large is the numerical aperture? (c) What is the maximum pulse spreading in a 2 km long fiber with core index 1.51 and cladding index 1.50? 1.10 A multimode fiber with a 50 µm core diameter is designed to limit the intermodal dispersion to 10 ns/km. What is the numerical aperture of this fiber? What is the limiting bit rate for transmission over 10 km at 0.88 µm? Use 1.45 for the refractive index of the cladding. 1.11 A 1.3 µm lightwave system uses a 50 km fiber link and requires at least 0.3 µW at the receiver. The fiber loss is 0.5 dB/km. The fiber is spliced every 5 km and has two connectors of 1 dB loss at both ends. The splice loss is only 0.2 dB. Determine the minimum power that must be launched into the fiber. 1.12 If 16 channels, each operating at 2.5 Gbit/s, need to be multiplexed using time-division multiplexing, how short should each optical pulse be? 1.13 If 20 channels, each operating at 10 Gbit/s, are multiplexed using wavelength-division multiplexing and the spectral efficiency is 0.4 bit/s/Hz, what is the total bandwidth of the signal? 1.14 You want to transmit 100 Mbit/s data over 100 m step-index, multimode fiber with an LED as light source providing 1 mW of optical output power. The minimum power required at the fiber output is 50 µW and the loss from source to receiver is dominated by the input coupling loss which is approximately: η = NA2 . Which of two fibers available, with ∆ = 0.5% and 1.5% respectively, and otherwise identical characteristics will satisfy the requirements? (Assume n1 = 1.5) (Exam. 970825) 1.15 At your disposal is a step-index multimode fiber with core index n1 = 1.5. You are free to choose a cladding index such that the numerical aperture is in the range 0.1 to 0.5. What will be the maximum and minimum capacity (in Mbit/(s km)) of the fiber with this restriction assuming a ray optics approximation? (Exam. 980817)

Signal generation, transmitters

2.1 Sketch how the electrical field of a carrier would change with time for the PSK and DPSK format, respectively during five bits with the pattern 01010. 2.2 Sketch the variation of optical power with time for a digital NRZ bit stream 010111101110. Assuming a bit rate of 10 Gbit/s, what is the duration of the shortest and widest pulse? 2.3 A 1.3 µm optical transmitter is used to generate a digital bit stream at a bit rate of 2 Gbit/s. Calculate the number of photons contained in a single ’1’ bit when the average power emitted by the transmitter is 4 mW. Assume that the ’0’ bits carry no energy. 2.4 Sketch the design of optical receivers used to recover an optical bit stream transmitted in the RZDPSK format. Explain how the receiver can detect phase information despite that the photodiodes generate a photocurrent proportional to the optical power only (direct detection). 2.5 A fiber optic system transmits 16-QAM at 100 Gbit/s. (a) How long will the symbol slots be? Answer in ps. (b) You use the same bit rate and modulation format, but also make use of polarization multiplexing. How long will the symbol slots be now? (c) Estimate how much longer the dispersion limited distance will be compared to a OOK-system at the same bit rate for the two cases above. 2.6 You have at your disposal a Mach-Zehnder modulator (MZM) in which the refractive index can be changed in only one of the two arms. You plan to use it to generate differential phase-shift keying. Will this be possible? Motivate your answer. 2.7 The amplitude of the electromagnetic field at the output of a MZM is a sinusoidal function of the applied voltage. How much smaller will the power in the middle of the symbol slot be for a DPSK-signal if you instead of using a full voltage swing (the peak-to-peak voltage that gives highest possible output power) use only 75% of this voltage? 2.8 You transmit 112 Gbit/s of data for a single channel if you use polarization multiplexed QPSK as modulation format and a symbol rate of 28 Gsymbol/s. What should the symbol rate be if you use polarization multiplexed 64-QAM instead and want the same bit rate? 2.9 How much higher average power do you need for QPSK and 16-QAM compared to PSK in order to have the same smallest distance between symbols in the constellations after coherent detection? 2.10 Draw the constellation diagrams of the following modulations formats, QPSK, 8-PSK, 16-QAM. Assign an arbitrary bit-to-symbol mapping. Sketch how the electrical field of a channel would change with time during the following bit sequence ”110110100010011110000110” for the three modulation formats and the chosen bit-to-symbol mapping.

Signal propagation in optical fibers

3.1 A fiber optical communication system employs a 1.3 µm InGaAsP semiconductor laser as the transmitter. To minimize dispersion, a single-mode fiber is used. (a) Determine the largest possible core diameter for the fiber with a core refractive index = 1.500 and a cladding refractive index = 1.495. (b) How many modes can propagate in the fiber if the transmitter is replaced with an AlGaAs semiconductor laser with an emission wavelength of 0.85 µm or a HeNe-laser with an emission wavelength of 0.63 µm? For waveguide dispersion, make the (realistic) assumption that the material dispersion can be neglected (dn/dω = 0). 3.2 A step-index fiber with a pure silica core has a core refractive index of 1.5000, a cladding refractive index of 1.4967, and is used to transmit light at 1.55 µm. The core radius is 5 µm. (a) How many modes can propagate in the fiber? (b) Which are the two main causes of dispersion in this fiber? (c) Determine the dispersion due to these two mechanisms. 3.3 A single-mode fiber is measured to have λ2 (d2 n/dλ2 ) = 0.02 at 0.8 µm. Calculate the dispersion parameters β2 and D. 3.4 Show that a chirped Gaussian pulse is compressed initially inside a single-mode fiber when β2 C < 0. Derive expressions for the minimum width and the fiber length at which the minimum occurs. 3.5 An optical communication system is operating with chirped Gaussian input pulses. Assume that β3 = 0 and Vω 1 in Eq. 2.4.23 in Fiber-Optic Communication Systems 4th ed., and obtain a condition on the bit rate in terms of the parameters C, β2 and L. 3.6 A single-mode fiber has an index step of 0.005. Calculate the core radius if the fiber has a cutoff wavelength of 1 µm. Estimate the spot size (FWHM) of the fiber mode and the fraction of the mode power inside the core. n1 = 1.45. 3.7 A 0.88 µm communication system transmits data over a 10 km single-mode fiber by using 10 ns (FWHM) Gaussian pulses. Determine the maximum bit rate if the light-emitting diode (LED) has a spectral FWHM of 30 nm with Gaussian shape. Use D = −80 ps/(km nm). 3.8 Estimate the limiting bit rate for a 60 km single-mode fiber link at 1.3 and 1.55 µm wavelength assuming transform-limited 50 ps (FWHM) Gaussian input pulses. Assume β2 = 0 and −20 ps2 /km and β3 = 0.1 and 0 ps3 /km at 1.3 and 1.55 µm wavelengths, respectively. Also assume that Vω 1. 3.9 Use Eq. 2.4.23 in Fiber-Optic Communication Systems 4th ed. to prove that the bit rate of an optical √ communications system operating at the zero-dispersion wavelength is limited by BL |S| σλ 2 < 1/ 8 where S = dD/dλ and σλ is the rms spectral width of the Gaussian source spectrum. Assume that C = 0 and Vω 1 in the general expression of the pulse width. 3.10 Repeat the preceding problem for the case of a single-mode semiconductor laser for which Vω 1 and show that the bit rate is limited by B(|β3 | L)1/3 < 0.324. What is the limiting bit rate for L = 100 km if β3 = 0.1 ps3 /km? 3.11 A 1.55 µm optical communication system operating at 5 Gbit/s is using Gaussian pulses of width 100 ps (FWHM) chirped such that C = −6. What is the dispersion-limited maximum fiber length? How much will it change if the pulses are unchirped? Neglect laser linewidth and assume β2 = −20 ps2 /km.

3.12 A standard single-mode optical fiber has minimum dispersion at λ ∼ 1.3 µm and minimum attenuation at λ ∼ 1.55 µm. To be able to make use of the low attenuation we would like to design a dispersion shifted optical fiber with zero dispersion at 1.55 µm. To facilitate coupling of light into the fiber, the core radius should be as large as possible. By minimizing the core-cladding index difference, the core radius will be sufficiently large. Determine the core radius and the core-cladding index difference. The fiber material is SiO2 and the refractive index is approximately 1.5. Neglect the material dispersion when calculating the contribution of waveguide dispersion. 3.13 Show that the intensity FWHM time-bandwidth-product of a chirped Gaussian pulse produced by a semiconductor laser is given by p ∆t∆f = 0.44 1 + C 2 where C is the chirp factor.

3.14 A 1.55 µm unchirped Gaussian pulse of 50 ps width (FWHM) is launched into a single-mode fiber. Calculate the FWHM after 50 km if the fiber has a dispersion of D=16 ps/(nm km). 3.15 Which of the following two systems minimizes the pulse width at the output of the system? The systems consist of a transmitter, an optical fiber, and a receiver. Both systems generate transform limited Gaussian pulses. In both cases the transmitter wavelength is 1.5 µm. Consider two different lengths of the optical fiber: L = 100 and 500 km.

Transmitted pulse rms width (ps) Dispersion (ps/(nm km))

System 1 50 20

System 2 2 1

3.16 A 1 km long polarization-maintaining single-mode fiber exhibits a birefringence of ∆n = 6 · 10−4 . Calculate the differential group delay (DGD) for this fiber at 1.55 µm, assuming that the average mode index n = 1.45 and dn/dλ = −0.01 µm−1 at this wavelength. 3.17 We have two transmitters and two types of fiber: Transmitter 1: λ = 1.55 µm, Gaussian spectral shape, spectral width (rms) 0.67 nm, no chirp Transmitter 2: λ = 1.55 µm, negligible spectral width, no chirp Fiber 1: α = 0.21 dB/km, D=25 ps/(km nm) Fiber 2: α = 0.23 dB/km, D=0 ps/(km nm), S=0.06 ps/(km nm2 ) We are transmitting Gaussian pulses, T0 = 80 ps, at a bit rate of 750 Mbit/s. Calculate the dispersion limited transmission length for all combinations of fibers and transmitters. If the signal is allowed to drop 20 dB before it is amplified, calculate the attenuation limited transmission length. Which combination (of the 4 possible combinations) gives the maximum transmission length, without amplifiers (considering both attenuation and dispersion)? (Exam. 980116) 3.18 At the input of a 5 km long optical dispersion-shifted fiber, you launch pulses from two DFB-lasers at different wavelengths, 1535 nm and 1557 nm. The zero-dispersion wavelength of the fiber is between the two laser wavelengths. Both lasers give the same pulse shape, shown in the left figure, both lasers have identical linear chirp, and a spectral width of ≈ 0.8 nm (FWHM).

At the fiber output, the pulses have experienced different dispersion, hence different broadening, which can be observed in the figure below. The pulse at the shorter wavelength has even experienced compression. Both pulses in the figure are normalized to their peak intensity. Due to dispersion, the pulses at the different wavelengths will arrive at different times at the receiver (the displacements shown in the figure are not to scale). Calculate the time difference between the pulses at the fiber output. Make reasonable assumptions. (Exam. 970530) 5

output pulse 2

Input pulse

output pulse 1

Problem 3.18: The input pulse (dashed, center) and the two output pulses (solid). The pulses are 20 ps separated just to be distinguishable in the plot, and this does not reflect any real measured separation.

3.19 How large is the difference (with sign) of the propagation phase delay time in a 1000 km long singlemode fiber with the characteristics below when comparing operation at the cut-off wavelength with operation at 1700 nm? Use n1 =1.500, n2 = 1.495, and a = 2 Âľm. (Exam. 990111)

Nonlinear impairments and solitons

4.1 Solve the NLS equation: α ∂A iβ2 ∂ 2 A 2 = iγ |A| A − A + ∂z 2 ∂t2 2 in the limit of zero dispersion and derive an expression for the SPM-induced nonlinear phase shift for pulses of arbitrary shape. How is this shift affected by fiber losses? √ 4.2 Apply the result of the preceding problem to input pulses with A(0, t) = P0 sech(t/T0 ) and plot the frequency chirp as a function of time at the output of a 25 km long fiber. Assume α = 0.2 dB/km, γ = 2 (W km)−1 and 5 ps pulses (FWHM) with 20 mW peak power. 4.3 A 1.55 µm continuous-wave signal with 6 dBm power is launched into a fiber with 50 µm2 effective mode area. After what fiber length would the nonlinear phase shift induced by SPM become 2π? Assume n2 = 2.6 · 10−20 m2 /W and neglect fiber losses. 4.4 Calculate the power launched into a 40 km long single-mode fiber for which the SPM-induced nonlinear phase shift becomes π. Assume λ = 1.55 µm, Aeff = 40 µm2 , α = 0.2 dB/km, and n2 = 2.6 · 10−20 m2 /W. 4.5 Find the maximum frequency shift occurring because of the SPM-induced chirp imposed on a Gaussian pulse of 20 ps width (FWHM) and 5 mW peak power after it has propagated 100 km. Use the fiber parameters of the preceding problem but assume no loss. 4.6 Verify by direct substitution that the soliton solution given in the lecture notes (Lecture 5, slide 23) satisfies the Nonlinear Schrodinger Equation (Eq. 9.1.2 in Fiber-Optic Communication Systems 4th ed.). 4.7 Explain why a SPM phase shift is more deteriorating for 16-QAM than for QPSK or PSK.

Optical transmitters and receivers

5.1 The active region of a 1.3 µm InGaAsP laser is 250 µm long. Find the active region gain required for the laser to reach threshold. Assume that the internal loss is 30 cm−1 , the mode index is 3.3, and the confinement factor is 0.4. (Problem 3.2 in Fiber-Optic Communication Systems 4th ed.) 5.2 Solve the rate equations in the steady state and obtain the analytic expression for P and N as a function of the injection current I. Neglect spontaneous emission for simplicity. (Problem 3.5 in Fiber-Optic Communication Systems 4th ed.) 5.3 A 250 µm long laser has an internal loss of 40 cm−1 . It operates at 1.55 µm in a single-mode, with modal index 3.3 and the group index 3.4. Calculate the photon lifetime. What is the threshold value of the electron population? Assume that the gain varies as G = GN (N − N0 ) with GN = 6 × 103 s−1 and N0 = 1 × 108 . (Problem 3.7 in Fiber-Optic Communication Systems 4th ed.) 5.4 Calculate the frequency (in GHz) and the damping time of the relaxation oscillations for the laser of Problem 5.3 operating twice above the threshold. Assume that GP = −4 × 104 s− 1 where the GP is the derivative of G with respect to P. Also assume that RSP = 2/τp . (Problem 3.10 in Fiber-Optic Communication Systems 4th ed.) 5.5 Determine the 3-dB bandwidth for the laser of Problem 5.3 biased to operate twice above threshold. What is the corresponding 3-dB electrical bandwidth? (Problem 3.11 in Fiber-Optic Communication Systems 4th ed.) 5.6 Calculate the responsivity of a p-i-n photodiode at 1.3 and 1.55 µm if the quantum efficiency is 80%. Why is the photodiode more responsive at 1.55 µm? 5.7 Photons at a rate of 1010 s−1 are incident on an APD with responsivity of 6 A/W. Calculate the quantum efficiency and the photocurrent at the operating wavelength of 1.5 µm for an APD gain of 10. 5.8 Consider a 0.8 µm receiver with a silicon p-i-n photodiode. Assume 20 MHz bandwidth, 65% quantum efficiency, 1 nA dark current, 8 pF junction capacitance, and 3 dB amplifier noise figure. The receiver is illuminated with 5 µW of optical power. Calculate the rms noise currents due to shot noise, thermal noise, and amplifier noise. Also determine the SNR. 5.9 The receiver of Problem 5.8 is used in a digital communication system that requires a SNR of at least 20 dB for satisfactory performance. What is the minimum received power when the detection is limited by (1) shot noise and (2) thermal noise? 5.10 Derive an expression for the optimum value of M for which the SNR becomes maximum by using FA (M ) = M x in Eq. (4.4.19) in Fiber-Optic Communication Systems 4th ed.. 5.11 Derive an expression for the optimum gain Mopt of an APD receiver that would maximize the receiver sensitivity by taking the excess-noise factor as M x . Plot Mopt as a function of x for σT = 0.2 µA and ∆f = 1 GHz and estimate its value for InGaAsP APDs with x ∼ 0.7. 5.12 Derive an expression for the intensity-noise-induced power penalty of a p-i-n receiver by taking into account a finite extinction ratio. Both shot noise and intensity noise contributions can be neglected compared with the thermal noise in the off state but not in the on state.

Lightwave systems (Error probability, power penalties)

6.1 Make the power budget and determine the maximum transmission distance for a 1.3 µm lightwave system operating at 100 Mb/s by using an InGaAsP LED capable of coupling 0.1 mW of average power into a single-mode fiber. Assume 1 dB/km attenuation, 0.2 dB splice loss at 2 km intervals. 1 dB connector loss at each end of fiber link, and 100 nW sensitivity for the p-i-n receiver. Allow a 6 dB system margin. 6.2 A 1.3 µm long-haul lightwave system is designed to operate at 1.5 Gbit/s. It uses semiconductor lasers capable of coupling 1 mW of average power into the single-mode fiber. The fiber-cable loss is specified as 0.5 dB/km and includes splice losses. The connectors at each end have 1 dB loss. The InGaAs p-i-n receiver has a sensitivity of 250 nW. Make the power budget with a 6 dB system margin and estimate the repeater spacing. 6.3 Use the results of Problem 5.12 to obtain an expression of the reflection-induced power penalty in the case of a finite extinction ratio rex . Reproduce the penalty curves shown in Fig. 5.9 in Fiber-Optic Communication Systems 4th ed. for the case rex = 0.1. 6.4 Make a power budget for the following 2.4 Gbit/s NRZ fiber optical communication system with a transmission distance of 40 km. The maximum BER is 10−9 and the minimum system margin is 5 dB. Data on the components used in the system are listed below: Transmitter: DFB laser, λ = 1.55 µm, output power = 10 mW (10 dBm), 7 dB coupling loss to fiber Fiber: Single-mode, available in lengths of 7 km that is spliced together, attenuation = 0.4 dB/km, dispersion = 20 ps/(km nm) Connectors: Loss = 1 dB/connector (one at the transmitter and one at the receiver) Splices: Loss = 0.2 dB/splice Detector: InGaAs-p-i-n, 1 dB coupling loss from fiber Receiver: Transimpedance, GaAs MESFET, sensitivity = -25 dBm for BER =10−9 6.5 Design a fiber optic communication system with the components listed below. The BER should be less than 10−9 at a data rate of 565 Mbit/s. RZ code (50% duty factor) with equal probability for ’1’ and ’0’ should be used. The system margin should be 5 dB. Assume for simplicity that the noise is Gaussian and that the threshold voltage is such that the error probabilities for a ’1’ and a ’0’ are equal. Neglect the power penalty caused by the non-zero extinction ratio. Laser: Peak power at a transmitted 1: 10 mW, ratio between emitted energy during bit period at a transmitted 1 and 0: 5.5, rise time: 100 ps, spectral FWHM-width: 0.7 Å, wavelength:1.55 µm Fiber: Coupling loss between laser and fiber: 3 dB, coupling loss between fiber and detector: 1 dB. The single-mode fiber is spooled on rolls with a unit length of 10 km, fractional lengths are allowed. Attenuation: 0.2 dB/km, dispersion: 20 ps/(km nm), optical connector loss: 0.5 dB/connector. Amplifiers: Gain: 40 dB, noise factor: 5 dB, RC-limited bandwidth: 1 GHz Detector: Responsivity: 0.6 mA/mW, load resistance: 50 Ω, RC-limited bandwidth:10 GHz. Assume ideal unipolar RZ-pulses superimposed on a constant background. Also, assume that the noise equivalent bandwidth of the receiver is equal to the RC-limited bandwidth. (a) Is the system attenuation or dispersion limited? (b) What is the maximum transmission distance? (c) What is the threshold voltage in the receiver decision circuit? 9

6.6 A DFB-laser with negligible spectral width together with an electro-absorption modulator is used as a transmitter in a fiber optic communication system. The width of the transmitted Gaussian pulses can be varied. The pulses are chirp free, the ’1’ and ’0’ bits are equally likely to occur and the average transmitted power is always 1 mW. The fiber is a dispersion shifted fiber (DSF) with dispersion D = 1.3 ps/nm/km, length L and attenuation α = 0.23 dB/km. The average number of photons per bit required by the receiver is N = 500. On condition that the input pulse width is optimized, for what bit rates is the system loss-limited and dispersion-limited, respectively? (Exam. 980817) 6.7 Use the result of Problem 5.11 to plot the power penalty as a function of the intensity-noise parameter r1 [see Eq. (4.7.6) in Fiber-Optic Communication Systems 4th ed. for its definition] for several values of the extinction ratio. When does the power penalty become infinite? Explain the meaning of an infinite power penalty. 6.8 A LED operating at 1300 nm injects 25 µW of optical power into a fiber. The attenuation between the LED and the photodetector is 40 dB and the photodetector quantum efficiency is 0.65. (a) What is the probability that 5 electron-hole pairs will be generated at the detector in a 1 ns interval? (b) What is the probability that no electron-hole pairs are generated at the detector during the same time interval? (c) If this time interval defines one bit of information, what is the quantum limit (minimum received optical power with a detector quantum efficiency of 100% to achieve a maximum BER of 10−9 )? What is the corresponding minimum launched optical power? Assume equal number of ’0’ and ’1’ pulses. 6.9 Derive an expression for the timing-jitter-induced power penalty by assuming a parabolic pulse shape I(t) = Ip (1 − B 2 t2 ) and a Gaussian jitter distribution with a standard deviation τ (rms value). You can assume that the receiver performance is dominated by thermal noise. Calculate the tolerable value of Bτ that would keep the power penalty below 1 dB. 6.10 Explain how the FEC technique is implemented in practice for improving the performance of a lightwave system. Define the FEC overhead and the redundancy and calculate their values if the effective bit rate of a 40 Gbit/s signal is 45 Gbit/s. 6.11 Explain the meaning of coding gain associated with an FEC technique. How much coding gain is realized if the FEC decoder improves the BER from 10−5 to 10−12 ? 6.12 We have an attenuation-limited digital fiber optic link with the following data: Data rate: Code: Output power when a 1 is transmitted: Extinction ratio: Wavelength: Coupling loss between laser and fiber: Fiber attenuation: Fiber length: Coupling loss between fiber and detector: p-i-n-photodetector quantum efficiency: Receiver noise equivalent bandwidth: Temperature: Equivalent load resistance: Amplifier noise factor:

2 Gbit/s NRZ, equal prob. for ’1’ and ’0’ 10 mW 0.1 1.3 µm 3 dB 1 dB/km 34 km 0 dB 80% 1 GHz 290 K 1 kΩ 3 dB

(a) Determine the bit error rate (BER) with optimum threshold voltage.

(b) We now replace the p-i-n-photodetector by an APD with FA (M ) = M 0.7 and apply a reverse bias voltage so that M = 300. We also adjust the threshold voltage so that the error probability is equal for a received ’1’ and ’0’. Assume that the quantum efficiency, the coupling losses, the receiver bandwidth, and the equivalent load resistance remain the same. What is now the BER? Assume all noise sources to be gaussian and that all noise sources are filtered through the noise equivalent bandwidth of the receiver. Neglect dark currents and leakage currents. 6.13 To be able to estimate the performance of a fiber optical communication system we want to know how close the system is to being ideal. In an ideal system the thermal noise is absent and the only source for errors is the statistic nature of light. Minimum error probability is obtained when there are no photons in a transmitted ’0’. The probability of receiving m photons during the time interval T is given by the Poisson distribution: P0 (m) =

Npm exp(−Np ) m!

where Np is the average number of photons registered during the time interval T. (a) How many photons per ’1’ bit are required on average in this ideal system for a BER < 10−10 and with equal probabilities for ’1’ and ’0’ ? (b) What is the required peak power from the laser transmitter under such ideal conditions for the fiber optic link described below if the BER is not allowed to exceed 10−10 ? System data: Fiber attenuation: Fiber length: Wavelength: Coupling loss between laser and fiber: Coupling loss between fiber and detector: Data rate: Noise equivalent bandwidth: Code: Neglect the dispersion in the optical fiber.

0.2 dB/km 30 km 1.5 µm 1 dB 0 dB 2 Gbit/s 2 GHz RZ (pulsewidth = 0.5 bit period)

6.14 We would like to design a binary digital decision circuit for an optical receiver. A ’1’ corresponds to 2.5 V and a ’0’ corresponds to 0 V. Assume that the noise in the receiver is dominated by thermal noise with equal rms-value for a ’1’ and ’0’ and that the probability for a ’1’ and a ’0’ is the same. (a) What is the optimum threshold voltage? (b) What is the rms noise voltage if BER = 10−9 ? (c) We now increase the optical signal level so that a ’1’ corresponds to 5 V in the decision circuit. We found that the noise at a received ’1’ was dominated by shot noise and that we could approximate the shot noise with Gaussian noise with a standard deviation twice that for the thermal noise. Determine the optimum threshold voltage. 6.15 Consider a 4 Gbit/s, NRZ, fiber optic communication system which consists of a transmitter which sends data into a 180 km fiber link with a fiber loss of 0.2 dB/km at the operating wavelength 1550 nm. The receiver consists of a 10 GHz pin-diode (Rd = 0.6 A/W) connected to an electrical amplifier with a gain of 40 dB, a noise figure of 2 dB, and a bandwidth of 3 GHz (50 Ω load) which is connected to a BER measurement equipment. The extinction ratio is -10 dB. (a) Calculate the bit-error rate (i.e. the probability of a detection error) for the system described above if the average input power to the fiber link is +18 dBm. (b) As a proficient engineer you know that the system above is not very clever since the very high input power will give rise to nonlinear phenomena in the fiber, which will preclude transmission through the fiber. Instead you decrease the input power to -2 dBm and insert an erbium-doped 11

fiber amplifier with a gain of 30 dB and a signal-spontaneous beat noise-limited noise figure of 5 dB in front of the receiver. Again, calculate the BER. Assume the amplifier to have a uniform spontaneous emission between 1.53 and 1.56 µm. (c) Maybe you can improve the system even more... Yes, an optical band-pass filter after the EDFA will be perfect! Calculate the BER if you have a 1 nm band-pass filter centered at the signal directly after the EDFA. Assume P(0/1)=P(1/0) for the BER calculation and neglect splice/connector losses. (Exam. 970530) 6.16 The average received optical power in a 3 Gbit/s optical communication system is -5 dBm. The signal wavelength is 1550 nm and the transmitted data is NRZ with equal probabilities of ’1’ and ’0’ occurrences. The received optical pulses can be assumed to be free from intensity noise but they suffer an extinction ratio of about 30%. The receiver consists of a p-i-n photo-detector and an electrical amplifier. The detector has a quantum efficiency of 95%, a dark current of 50 µA and a load resistance of 50 Ω. The electrical amplifier has a gain of 30 dB and can be assumed to be noise-free. Your task is to investigate whether a lowpass filter is needed at the receiver in order to obtain a BER better than 10−9 . If a filter is needed, determine what bandwidth it should have. The threshold in the receiver decision circuit is chosen to give P(0/1)=P(1/0), where P(0/1) is the probability that a ’0’ is decided when a ’1’ is received and P(1/0) is the vice versa. The system is operating at room temperature. Planck’s constant is 6.63 × 10−34 Js and Boltzmann’s constant is 1.38 × 10−23 J/K. (Exam. 980116) 6.17 Calculate the minimum received average power required for BER < 10−9 for an optical communication system operating at 1 Gbit/s. The carrier wavelength is 1550 nm. The data being sent is RZ with equal probabilities of ’1’ and ’0’ occurrences and negligible extinction ratio. The receiver consists of a pin-photodetector with a quantum efficiency of 80% and a load resistance of 50 Ω. The detector dark currents may be neglected. The receiver bandwidth is 1 GHz and the temperature is 300 K. Plancks constant is 6.63 × 10−34 Js, Boltzmann’s constant is 1.38 × 10−23 J/K, and the electronic charge is 1.602 × 10−19 As. (Exam. 970117)

Multichannel systems

7.1 So-called dry fibers eliminate the loss peak at 1.4 µm otherwise found in silica fibers. The region of acceptably low losses thus extend from 1.3 to 1.6 µm. Estimate the capacity of a WDM system covering this entire region using 40 Gbit/s channels spaced apart by 50 GHz. 7.2 The C and L spectral bands cover a wavelength range from 1530 to 1610 nm. How many channels can be transmitted through WDM when the channel spacing is 25 GHz? What is the effective bit rate-distance product when a WDM signal covering the two bands using 10 Gbit/s channels is transmitted over 2000 km? 7.3 The low-loss region of common silica fiber extends from 1.5 to 1.6 µm. How many channels can be transmitted by using optical FDM when the channel spacing is 10 GHz? If each channel is operated at 2 Gbit/s with a power budget of 30 dB allocated to fiber loss, calculate the effective bit rate-distance product BL of the multichannel system by assuming a loss of 0.2 dB/km. 7.4 An optical WDM communication system operates at λ = 1.5 µm and uses direct detection. It transmits Gaussian pulses with the following power envelope: 2 t P = P0 exp − 2 T0 where T0 = 50 ps. The available channel bandwidth in order to avoid crosstalk is 10 GHz (FWHM). The system is 100 km long, the fiber has a loss of 0.23 dB/km and a nonlinear coefficient γ = 1 (W km)−1 . Calculate the maximum possible value of P0 that keeps the signal within the channel bandwidth, considering only the self-phase modulation (SPM). Use the following formulas (if needed): q ∆νtot =

∆ν12 + ∆ν22 + ∆ν32 + ...

which states that the different contributions to the spectral width should be summed quadratically, and 1 dΦSP M δν = 2π dt which gives the chirp due to SPM, and ∆νSP M ≈ 2δνmax which gives the spectral broadening due to SPM. (Exam. 970530)

7.5 A high speed, 3-channel WDM system is used in a point-to-point link between two major cities. The frequencies used are equispaced 100 GHz apart. Due to Four-Wave Mixing (FWM) in the dispersion shifted fiber, new frequencies are generated according to ωnew = ωi ± ωj ± ωk where i 6= j 6= k or i = j 6= k (degenerated case). (a) How many of the signal frequencies are suffering from FWM induced crosstalk? (b) Give an example of how the crosstalk can be reduced by reallocation of the signal frequencies. (c) Give other examples of how this crosstalk can be reduced. Support your discussion with pictures! (Exam. 990111)

Optical amplifiers

8.1 The Lorentzian gain profile of an optical amplifier has a FWHM of 1 THz. Calculate the amplifier bandwidths when it is operated to provide 20 and 30 dB gain. Neglect gain saturation. 8.2 An optical amplifier can amplify a 1 µW signal to the 1 mW level. What is the output power when a 1 mW signal is incident on the same amplifier? Assume that the saturation power of the small-signal gain is 10 mW. 8.3 Explain the gain mechanism in EDFAs. Use the three-level rate equations to derive an expression for the small-signal gain. You can assume a rapid transfer of the pumped population to the excited state. 8.4 Start from Eq. (7.6.1) in Fiber-Optic Communication Systems 4th ed., and derive Eq. (7.6.4) for the sensitivity of a direct-detection receiver when an EDFA is used as a preamplifier. 8.5 Calculate the receiver sensitivity at a BER of 10−9 and 10−12 by using Eq. (7.6.4) in Fiber-Optic Communication Systems 4th ed.. Assume that the receiver operates at 1.55 µm with a 3 GHz bandwidth. The preamplifier has a noise figure of 4 dB, and a 1 nm optical filter is installed between the preamplifier and the detector. 8.6 When the optical output power from an optical amplifier is detected by a photodetector with unity quantum efficiency, the mean square noise current generated in the detector is given by 2 2 σ 2 = σT2 + σs2 + σsp−sp + σsig−sp

Show that, if the signal-spontaneous beat noise dominates for the output signal and signal-shot noise dominates for the input signal, the noise figure can be expressed as Fo ≈

2nsp ηin

when the input coupling efficiency η in is taken into account. S 8.7 Derive an expression for the output saturation power, Pout , of an optical amplifier in terms of the unsaturated amplifier gain G0 and the saturation power Ps .

Hint: The incremental increase in power dP in an incremental distance dz of the amplifier in the saturation regime is given by: dP (z) g0 = P (z) dz 1 + P (z)/Psat where P (z) is the position dependent optical power, Ps is the saturation power, and g0 is the unsaturated gain coefficient of the gain medium. 8.8 Starting with the differential equation ∂A 1 = g0 A + fn (z, t) ∂z 2 and Eq. (7.1.5) in Fiber-Optic Communication Systems 4th ed., prove that the spectral density of ASE noise added by a lumped amplifier of length la is given by SASE = nsp hν0 [exp(g0 la ) − 1] . 8.9 Derive an expression for the optical SNR at the end of a fiber link containing NA amplifiers spaced apart by a distance LA . Assume that an optical filter of bandwidth ∆νo is used to control the ASE noise. 8.10 Calculate the optical SNR at the output end of a 4000 km lightwave system designed using 50 EDFAs with 4.5 dB noise figure. Assume a fiber-cable loss of 0.25 dB/km at 1.55 µm, an input power of 1 mW, and a 2 nm-bandwidth for the optical filter. 14

8.11 Explain the concept of noise figure for an optical amplifier. Use Eq. 7.5.9 in Fiber-Optic Communication Systems 4th ed. for the total variance of current fluctuations and prove that the minimum noise figure is 3 dB for an ideal amplifier with high gain (G 1) and complete inversion (nsp = 1). 8.12 Which of the following EDFA configurations will provide the best receiver sensitivity when acting as a preamplifier setup? Assume the amplifiers operate at the signal-spontaneous beat noise limit. • One EDFA with G = 30 dB, Fn = 6 dB

• Two cascaded EDFAs: the first with G1 = 10 dB, Fn1 = 4 dB, the second with G2 = 23 dB, Fn2 = 12 dB. In between the two EDFAs there is a 3 dB connector loss. (Exam. 980817) 8.13 An optical communication system consists of 5 fiber links (below) each with a loss of L=25 dB, with erbium-doped fiber amplifiers in between to exactly compensate for the loss in each fiber section. For some reason, the second amplifier becomes defective, i.e. the pump laser in the amplifier degrades and gives only 20 mW pump power rather than the usual 60 mW. Calculate the difference in output power and effective noise figure at the end of the system, using the specifications of the amplifiers in the below plots. (Exam. 980116) L

L G

∆Pout= ? ∆Fneff= ?

Problem 8.13: A cascaded amplifier system (top) with amplifier characteristics (bottom).

8.14 An optical communication link consists (see figure, top) of two Erbium-Doped Fiber Amplifiers (EDFA) with a fiber link in between with a loss of -30 dB. The input signal average power is -30 dBm, the input signal-to-noise ratio is 25 dB, and the input signal is believed to be shot noise limited. The output spectrum directly after the first EDFA is also shown (see the figure, bottom). After the first EDFA a narrow optical band-pass filter, with no loss, is placed, leaving the noise into the second amplifier limited by signal-spontaneous beat noise. The noise figures of the two amplifiers are the same, the wavelength is 1550 nm and the resolution of the spectrum analyzer is 1 Å. What is the signal-to-noise ratio after the last EDFA? (Exam. 980116) 8.15 Your job is to optimize the gain and position of two optical amplifiers for maximum SNR at the output of a fiber-optic link. The link consists of 215 km of fiber with a loss of 0.21 dB/km. It is possible to place amplifiers at the distances 50, 90, 140 and 170 km from the transmitter (this is the case because the fiber is buried in the ground). You have two amplifiers; the gain of these can be varied from 0 dB to 25 dB, the noise figure is independent of the gain (constant) and the absolute maximum amplifier output power is 10 dBm (the amplifiers operates in the signal-spontaneous beat 15

Loss=-30 dB SNRin =25 dB NF Pin =-30 dBm

~ ~ ~

Optical bandpass filter

~ ~ ~

SNRout= ?

Optical bandpass filter

Problem 8.14: A cascaded amplifier system (top) with the output spectrum directly after the first amplifier (bottom).

noise limit). When an amplifier is inserted there will be 1.5 dB loss directly before and 1.5 dB loss directly after the amplifier. The transmitter delivers 1 mW average power and the required received power is specified to 40 µW in the ISO9000 specification you got from your supervisor. Explain with a few words how you are thinking when you maximize the SNR (less than 30 words and no equations will do just fine). (Exam. 980529) 8.16 A Raman amplifier is pumped in the backward direction using 1 W of power. Find the output power when a 1 µW signal is injected into the 5 km-long amplifier. Assume losses of 0.2 and 0.25 dB/km at the signal and pump wavelengths, respectively, Aeff = 50µm2 , and gR = 6 × 10−14 m/W. Neglect gain saturation. (Problem 7.5 in Fiber-Optic Communication Systems 4th ed.) 8.17 Discuss the origin of gain saturation in fiber Raman amplifiers. Solve Eqs. (7.3.2) and (7.3.3) in Fiber-Optic Communication Systems 4th ed. with αs = αp and derive Eq. (7.3.7) Fiber-Optic Communication Systems 4th ed. in for the saturated gain. (Problem 7.4 in Fiber-Optic Communication Systems 4th ed.)

Dispersion management

9.1 What is the dispersion limited distance for a 1.55 µm lightwave system making use of direct modulation at 10 Gbit/s? Assume that frequency chirping broadens the gaussian-shape pulse spectrum by a factor of 6 from its transform-limited width. Use D = 17 ps/(nm km) for fiber dispersion. 9.2 How much improvement in the dispersion-limited transmission distance is expected if an external modulator is used in place of direct modulation for the lightwave system of problem 9.1. 9.3 Solve Eq. 8.1.2 in Fiber-Optic Communication Systems 4th ed. by using the Fourier transform method. Use the solution to find an analytic expression for the pulse shape after a Gaussian input pulse has propagated a distance L in a fiber with β2 = 0. 9.4 Prove by using Eq. 8.1.5 that a DCF can provide dispersion compensation over the entire C-band (1530-1570 nm) when the ratio S/D for the DCF is matched to that of the fiber used to construct the transmission link. 9.5 The pre-chirp technique is used for dispersion compensation in a 10-Gbit/s lightwave system operating at 1.55 µm in an SMF, and transmitting the ’1’ bits as chirped gaussian pulses of 40 ps width (FWHM). Pulse broadening by up to 50% can be tolerated. What is the optimum value of the chirp parameter C, and how far can the signal be transmitted for this optimum value? Use D = 17 ps/(nm km). 9.6 We are transmitting chirp-free Gaussian pulses with 20 ps pulse width (FWHM) through three different fibers. The spectral width is dominated by the Fourier spectrum of the pulses, the operating wavelength is 1550 nm, and β3 = 0 for all three fibers. The fiber properties are: DSF (Dispersion-shifted fiber): D=0.37 ps/(nm km), L = 25 km DCF (Dispersion compensating fiber): D = -52 ps/(nm km), L = 2 km SMF (Standard single-mode fiber): D = 17 ps/(nm km), L = 16 km (a) Calculate the output pulse width after transmission through each of the fibers. (b) The three fibers are spliced together (forming one fiber). Calculate the pulse width after transmission through this combined fiber. Would a different ordering of the fibers give a different pulse width? Why/why not? (c) Fiber 3 is used in a local area network. Fiber of type 2 is used to compensate the dispersion of fiber 3 (a suitable length of fiber 2 is spliced to fiber 3). Optimize the length of fiber 2, so that the pulse width out from the combined fiber is minimized. Are the pulses longer or shorter than the initial pulses? (Exam. 980529) 9.7 The group-velocity dispersion needs to be compensated in a ten-channel WDM system, each channel operating at 10 Gbit/s and separated in wavelength by 100 GHz. The single-mode fiber is of standard type with D=17 ps/(km nm), approximately equal for all WDM channels. The compensation should be done with chirped fiber gratings inserted every 50 km. Assume that the mode index equals 1.5. (a) How long should the gratings be to compensate all channels simultaneously? (b) How long should the gratings be if each channel is compensated separately (i.e. when ten gratings are used every 50 km)? Assume that the grating bandwidth must cover at least a bandwidth around each channel of 25 GHz. Exam(000526)

9.8 A long-haul WDM link is using 8000 km of SMF with dispersion D = 17 ps/(nm km) and dispersion slope S = 0.06 ps/(nm2 km) at 1550 nm. The dispersion is compensated for using DCF with D = â&#x2C6;&#x2019;104 ps/(nm km) and dispersion slope S = â&#x2C6;&#x2019;0.25 ps/(nm2 km) at 1550 nm. a) How much DCF is needed to perfectly compensate the dispersion at 1550 nm? b) Using this length of DCF, the dispersion will, due to the dispersion slope, not be perfectly compensated at wavelengths other than 1550 nm. How much accumulated dispersion (in ps/nm) remains at the edges of the C-band (1528 and 1563 nm)? How long SMF does these two values correspond to? (Exam. 120111)

Coherent systems

10.1 Derive the expression for the intensity of the combined signal and local oscillator optical fields on the surface of the photodetector that produces a photocurrent in a coherent optical receiver. The signal field is given by Es = As cos[ωs t + φs (t)] and the local oscillator field is given by ELO = ALO cos[ωLO t + φLO (t)]. 10.2 Derive an expression for the SNR of a homodyne receiver by taking into account both the shot noise and the thermal noise. 10.3 Prove that the SNR of an ideal PSK homodyne receiver (perfect phase-locking and 100% quantum efficiency) approaches 4 < Np >, where < Np > is the average number of photons/bit. Assume that the receiver bandwidth equals half the bit rate and that the receiver operated in the shot-noise limit. 10.4 Calculate the sensitivity (in dBm units) of a homodyne ASK receiver operating at 1.55 µm in the shot-noise limit by using the SNR expression obtained in Problem 10.2. Assume η = 0.8 and ∆f = 1 GHz. What is the receiver sensitivity when the PSK format is used in place of ASK? 10.5 In a coherent receiver, you can use either balanced or single-ended detection. If balanced detection is used, the photocurrent can be written as ibal ∼ |Esig + nsig + ELO |2 − |Esig + nsig − ELO |2 where Esig is the electric field of the signal, ELO the electric field of the local oscillator (LO) laser and nsig is the signal ASE noise. On the other hand, if single-ended detection is used, only one of the two terms will remain. Show that with a single-ended coherent receiver, you need to have high LO power to reach the same SNR as for a balanced receiver. What will happen to the spontaneous-spontaneous beat-noise when you use balanced coherent detection? 10.6 Make a sketch in the complex plane of what a QPSK constellation would look like as the time pass by if there is a fixed frequency separation between the local oscillator laser and the signal laser. What needs to be done to solve this problem?

Part II

2.6 No. Since the field vector is fixed in one of the arms, you can only go between complete constructive and complete destructive interference by changing the refractive index in the other arm.

Answers 1

2.7 85.4 %

Introduction, optical fibers −7

2.8 9.3 Gsymbol/s

−1

2.9 3 dB for QPSK and 10 dB for 16-QAM.

1.1 0.2 dB/km: α = 4.61 · 10 cm , L = 50 km 20 dB/km: α = 4.61 · 10−5 cm−1 , L = 0.5 km 2000 dB/km: α = 4.61 · 10−3 cm−1 , L = 5 m

2.10 -

1.2 781 and 3.02 · 107

Signal propagation in optical fibers

1.3 0.42 Mbit, 420 µs 1.4 156

3.1

1.5 0011 0110 1011 0101 (3, 6, 11, 5 in binary form) 1.6 1562.5 photons 3.2

1.7 24.5 km 1.8

(a) One mode (b) Material dispersion and waveguide dispersion

(a) 126 µW (b) 1 pW

(a) ∆T = (LN A2 )/(2n2 c)

1.9

(a) 8.1 µm (b) λ =0.85 µm: 4 modes. λ = 0.63 µm: 7 modes

(b) NA=0.173

3.3 D = −83.3 ps/(nm km), β2 = 28.3 ps2 /km

3.4 -

1.10 NA = 0.093, B = 10 Mbit/s

3.5 -

1.11 0.227 mW

3.6 3.2 µm, 3.5 µm, 0.81 3.7 22.6 Mbit/s

1.12 25 ps

3.8 1300 nm: 11.8 Gbit/s, 1550 nm: 7.1 Gbit/s

1.13 500 GHz

3.9 -

1.14 For ∆=1.5% we have: BLmax =13.1 Mbit km/s Pout =67 µW

3.10 150 Gbit/s 3.11 C = −6: L = 5.3 km C = 0: L = 112 km

For ∆=0.5% we have: BLmax =39.8 Mbit km/s Pout =22 µW The fiber with ∆=1.5% works, but not the one with ∆=0.5%!

3.12 Core radius = 2.0 µm, core-cladding index difference = 6.1·10-3. 3.13 3.14 76 ps 3.15 L = 100 km: System 2 L = 500 km: System 1

1.15 BLmax = 90.7 Mb/(s km) for NA=0.1 BLmax = 3.3 Mb/(s km) for NA=0.5

3.16 51.67 ps/m

3.17 Transmitter Transmitter Transmitter Transmitter

Signal generation, transmitters 2.1 -

Fiber Fiber Fiber Fiber

1: 2: 1: 2:

19.6 87.0 95.2 87.0

km km km km

(disp. limited) (att. limited) (att. limited) (att. limited)

3.18 (Hint: Use the pulse widths in the figure to get the initial chirp, and then the dispersion. Assuming negative β2 for the high wavelength and positive for the low wavelength leads to two sets of solutions. Assuming a linear change of the dispersion between the two wavelengths, it is possible to get the delay time difference by integrating dispersion between the two frequencies) Two solutions, Td =125 ps or 2.27 ps

2.2 100 ps and 400 ps, respectively 2.3 2.62 · 107 2.4 Id (t) = Rd P (t) · 0.5 · [1 ± cos(∆φ)] 2.5

1, 1, 2, 2,

(a) 40 ps (b) 80 ps (c) It will increase with a factor of 16 and a factor of 64, respectively.

3.19 The delay is 8 µs smaller when operating at 1700 nm.

Nonlinear impairments and solitons 4.1 4.2 4.3 4.4 4.5 4.6 4.7

6.5

745 km 65.2 mW 18.2 GHz Modulation formats with multiple amplitude levels are more sensitive to nonlinear phase shifts since symbols with different amplitudes will obtain phase shifts of different magnitude. This makes it more difficult to detect the bits accurately and for large phase shifts, information can even be lost.

(b) 70 km (c) 35.5 mV 6.6 The dispersion limit gives the criterion Ldisp < 1/(16β2 B 2 ), and the loss limit gives Latt < 10 log(Ptr /Prec )/α where Prec = Np hνB is the minimum received power. Plot or numerics then show that the system is dispersion limited for bit rates above 17.1 Gbit/s, and loss limited for bit rates below 17.1 Gbit/s. 6.7 Infinite penalty means a BER-floor exists and it occurs when rI = (1 − rex )/Q.

Optical tansmitters and receivers 5.1 5.2 5.3 5.4 5.5 5.6 5.7 5.8

0.84 A/W, 1.0 A/W 0.496, 7.94 nA σs =3.66 nA, σT =18.2 nA, σA = σT =18.2 nA, SNR=38.1 dB 5.9 Shot noise limit: 2.8 nW, Thermal noise limit: 0.616 µW B T Fn 5.10 Mopt = ( qRL4k )1/(2+x) x(RPin +Id ) σT )1/(1+x) = 28.4 5.11 Mopt = ( q∆f Qx 2

5.12 δI = −10 log10 [(1 − rex ) −

(a) Assume thermal limit (which can be shown to be true a posteriori), and the attenuation limited distance then becomes 70 km (17.4 dB loss can be accepted). The rise time limit corresponds to 499 ps dispersion or Lmax = 356 km. Thus the system is attenuation limited.

6.8

(a) 2.8 % (b) 2.5·10−5 (c) 1.5 nW, 15 µW

6.9 0.217 6.10 Overhead: 12.5%, Redundancy 0.111 6.11 7.36 dB 6.12

(a) BER=1.4·10−5 (b) BER=9.3·10−12

6.13

(a) Np ≥ 22.3 (b) Ptrans = 60nW

6.14

(a) 1.25 V (b) 0.21 V (c) 1.68 V

6.15

rI2 Q2 ]

(a) BER = 1.6 · 10−10 (b) BER = 3 · 10−3 (c) BER = 1.6 · 10−11

6.16 Lightwave systems (Error probability, power penalties) 6.1 6.2 6.3 6.4

Q < 6 corresponds to ∆f < 2.45·1012 Hz. As this is much larger than the bandwidth of the electri circuits, no filter is needed. ∼ 3.4µW. 6.17 Prec =

20 km 56 km δI = −10 log10 [(1 − rex )2 − (rI2 + N/M SR2 )Q2 ] The power budget is loss power level [dB] [dBm] transmitter +10 input coupling 7 +3 fiber 16 -13 connectors (2 each) 2 -15 splices (5 each) 1 -16 output coupling 1 -17 receiver sensitivity -25 margin 8 dB

Multichannel systems 7.1 34.6 Tb/s 7.2 390 channels. BL = 7800 (Tbit/s)· km 7.3 1250 channels, 375 TBit km/s p 7.4 ∆SP M < ∆2Available − ∆2P ulse = 8.48 GHz which gives P0 < 82.7mW . 7.5

(a) All signals are suffering from FWM induced crosstalk. The generated frequencies within the signal bandwidth can not be suppressed by filtering.

(b) To reduce the crosstalk it is necessary to reallocate the signals so the new frequencies do not coincide with the signal frequencies. Now the signals can be filtered out by a narrow optical filter.

9.3 9.4 -

9.5 The longest possible distance is Lmax = 1.5LD = 43.3 km, and it is obtained for Copt = 1/1.5 = 0.667. (c) Other ways to reduce the FWM-induced crosstalk: 9.6 (a) TF W HM (L1 ) = 20.07, TF W HM (L2 ) = 27.16, • Increase the channel spacing enough so TF W HM (L3 ) = 52.06 ps phase matching is no longer satisfied. (b) TF W HM (L1 + L2 + L3 ) = 37.16 ps • Increase the local dispersion so the phase (c) L2,opt = 5.23 km, and the compensated pulse matching condition is no longer satishas the same width as the initial one. fied. • Keep the signal power low. 9.7 (a) 62.9 cm (b) 1.7 cm

Optical amplifiers 10

8.1 421 GHz, 334 GHz 8.2 34.6 mW

10.1 I(t) = 0.5(A2S +A2LO +AS ALO cos[ωIF +φ(t)] cos(θ)

8.3 g = σs Nt (Wp τsp − 1)/(Wp τsp + 1) 8.4 -

10.2 The SNR is twice that of the heterodyne case, i.e. 2 P̄s PLO SN R = 2q(RPLO +I4R d )∆f +4kB T Fn ∆f /RL

8.5 The sensitivities are -41.4 dBm and -40.4 dBm.

10.3 -

8.6 -

10.4 5.8 nW, 2.9 nW

8.7 -

10.5 The spontaneous-spontaneous beat-noise will cancel in the detection process.

8.8 -

10.6 The QPSK constellation will obtain the shape of a ring if there is a fixed frequency separation between the lasers. This is because there will be a linear phase increase as a function of time, which rotates the symbols in the complex plane. To handle this problem, the intermediate frequency needs to be tracked.

8.9 8.10 SNRopt = 2.23 or 3.5 dB 8.11 8.12 Configuration I: Fn,ef f = 6 dB, configuration II: Fn,ef f = 7.5 dB Thus configuration I is the best one. 8.13 System output power: –9.5 dBm ( if operating correctly –9 dBm) the output power degrades with 0.5 dB which probably not deteriorates the system very much. The effective noise figure is 41.7 dB (if operating correctly 36.5 dB), i.e. the effective noise figure increases 5.2 dB due to the defective amplifier. 8.14 Hint: SNR degradation is equal to effective noise figure. SNRDegradation =17.4 dB. 8.15 Optimum configuration: Amp1 : 50 km, G=22 dB, Pout =10 dBm Amp2 : 140 km, G=15-21.9 dB, Pout =3.25-10 dBm The signal should be kept as high as possible in the beginning of the system and the lowest level should be located (if possible) at the receiver. 8.16 8.17 -

Coherent systems

Dispersion management 9.1 Lmax ∼ =7.4 km ∼ km 9.2 Lmax =30

Part III

Exams The following exams are provided: • 081023 • 090114 • 121217

(2p)

Assume fiber loss = 0.2 dB/km and thus an EDFA gain = 20 dB in the binary PSK case, and that the optical filter bandwidth is the same in all cases. (10p)

If a binary PSK system needs ten amplifiers to satisfy the OSNR requirement, how many would be needed when instead using QPSK or 16-QAM modulation formats?

3. In a DWDM system using PSK modulation one wishes to transport data over 1000 km with as few as possible wavelength channels. In the system one also wishes to minimize the number of in-line EDFAs used in the link. The required OSNR for a given total bit rate scales approximately linearly with number of bits/symbols, e.g. QPSK (2 bits/symbol) needs 3 dB higher OSNR than binary PSK (1 bit/symbol) even though the baud rate (and noise bandwidth) is lower in the case of QPSK, etc.

Under what condition will the EDFA improve the electrical SNR? Express the condition in terms of the EFDA noise figure and the quantum efficiency of the detector used in the receiver. (10p)

2. A coherent homodyne ASK receiver is considered to be modified by adding a low noise EDFA preamplifier (gain >> 1) to boost the signal before it is mixed with the local oscillator and detected. Assume that the noise is entirely dominated by the ASE-local oscillator beat noise with the EDFA, that the LO power is much larger than the amplified signal, and that shot noise dominates the noise in the case without EDFA.

OOK, 15 Gbit/s; 64-QAM, 40 Gbit/s; QPSK, 20 GSymbols/s

c) Rank these three signals regarding their dispersion tolerance.

ps/(km.nm). Describe how the pulse width (in both time domain and spectral domain) will be affected by this. (3p)

b) A negatively chirped (its frequency decreases as a function of time) pulse centered at 1546 nm enters a dispersion compensating fiber (D = -104

1. a) You need a multi-mode fiber that can guide all incoming light (via total internal reflection) within a cone of 40 degrees and transmit a 10 Mbit/s signal 500 m. The refractive index of the core is 1.5000 and the medium outside fiber is air. Choose a value for the refractive index of the cladding and show that you fulfill the criterions (5p)

Questions during the examination: Peter Andrekson, 070-3088 606. ___________________________________________________________

Results will be posted by November 4, 2008. The marking can be examined on personal visit to Peter Andrekson, ext 1606. Room C428 at MC.

Note: Home assignments, their solutions, old exams, and “Problems and answers to Fiber Optics” are not allowed.

Allowed material: • "Lightwave Technology: Telecommunication Systems", by G.P. Agrawal • Lecture notes • Two manuals for the laboratory exercises. • Mathematical and physical tables, e.g.: Tefyma, Physics Handbook, Standard Mathematical Tables, Beta Mathematical Handbook. • English-Swedish Wordbook • Table of Gaussian pulses • Calculator of choice

Note: Only one problem per page. Motivate all equations used and make reasonable assumptions when necessary. Unreadable solutions will not be corrected!

Examination in Fiber Optical Communication (MCC100) October 23, 2008, pm, Hörsalslängan

Department of Microtechnology and Nanoscience, Photonics Laboratory

Solutions for examination of Fiber Optical communication, October 23, 2008.

6. You want to transmit data over an old fiber optic system. It uses 120 km of dispersion-shifted fiber with a loss of 0.23 dB/km and an old transmitter laser that is only providing -10 dBm of output power. You have one upgrade to the system however, an EDFA that can provide up to 25 dB small-signal gain. But since the low dispersion makes the system less tolerant to nonlinear effects, you cannot allow an input power of more than +3 dBm. Should you use the EDFA as a booster at the fiber input or at the fiber output as an optical preamplifier to the receiver? See the figure below. Since you can only allow 13 dB gain if you are using it as a booster, you must lower the EDFA pump power, which will give a higher noise figure, 8 dB. When the EDFA is operating at maximum gain, the noise figure is 4 dB. What system has the best receiver SNR? You do not have to compute the values of the SNR, only the relative difference. Assume that signal-spontaneous beat noise dominates in both cases. (10p)

c) Now you are considering another way to improve your system. You want to add an EDFA preamplifier to the receiver that has a gain of 30 dB and a noise figure of 6 dB, followed by an optical filter with a bandwidth of 0.5 nm. Find the sensitivity and power penalty for this case. (4p)

b) To avoid the problem of broadening due to GVD you are considering adding some DCF to the system. The DCF module compensates the dispersion perfectly, but has a loss of 3.5 dB. Find the power penalty due to the GVD, i.e. find how much the sensitivity has degraded compared to the case without GVD. Is adding the DCF a good idea? (2p)

change the average transmitter power (up or down) from its current setting to operate the system at a BER of 10-9? (4p)

a) Determine the sensitivity of the receiver with the dispersion-degraded extinction ratio as above. The sensitivity is defined as the optical received power required for a bit-error rate of 10-9. By how much (in dB) should you

5. You have an RZ-transmitter at a wavelength of 1550 nm operating at 10 Gbit/s and a 40 km standard SMF with a loss of 0.2 dB/km and a dispersion coefficient of 16 ps/(nm km). The receiver consists of a p-i-n photodiode with a quantum efficiency of 0.7 and a negligible dark current, as well as an electrical amplifier with a load resistance of 50 â&#x201E;Ś, noise figure of 3 dB and an noise equivalent bandwidth of 8 GHz. Due to GVD, the pulses have broadened into the neighboring bit slot after transmission. By looking at the eye-diagram with an oscilloscope at the output, the average optical power in 1-bits and 0-bits is measured to be -8 dBm and -13.5 dBm, respectively. Assume that 1-bits and 0bits are equally likely and that the decision threshold is such that the probability of errors due to 1-bits and 0-bits are equal. Make sure to show that your assumptions are valid!

Determine the maximum transmission reach in the four cases and also explain why you might expect this result intuitively. The effect of laser linewidth can be neglected and assume that the system operates at 1550 nm. The attenuation of the fiber can also be neglected.

4. Two single-mode fiber transmission systems (1 and 2) use a laser launching a non-chirped Gaussian pulses. System 1 and System 2 are operating at 10 Gbit/s and 100 Gbit/s, respectively. The RMS-width of the input pulse is equal to 0.1 times the bit-slot and the RMS-width of the output pulse cannot exceed 0.25 times of the bit-slot. The fiber for each system can be chosen to be either Fiber 1 or Fiber 2, the dispersion characteristics of which are shown by the graph below.

ln G4 (G1 − 1) OSNR (4 bits/symbo l) =4 = (G4 − 1) ln G1 OSNR (1 bit/symbol )

System 1 {

Linewidth is neglected

Pulse broadening formula

}

; Non‐chirp

Fiber2: @1550 nm is 0,3 ps/(km – nm) @1550 nm is the slope of D@1550 nm which equals to 0,05 ps/(km – nm^2) Therefore, @1550 nm:

Fiber1: @1550 nm is 0,4 ps/(km – nm) @1550 nm is the slope of D@1550 nm which equals to 1 ps/(km – nm^2) Therefore, @1550 nm:

4. D and S parameters can be found from the graph.

Thus the number of amplifiers needed with QPSK is 13 and for 16-QAM is 17.

The corresponding amplifier spacing will need to be no more than 80.5 km (QPSK) and 60 km (16-QAM), respectively.

G1 = 20 dB Î (G1-1)/lnG1 = 21.5 Î G2 = 16.1 dB and G4 = 12 dB

OSNR (2 bits/symbo l) ln G2 (G1 − 1) = =2 OSNR (1bit/symbol ) (G2 − 1) ln G1

Since QPSK has 2 bits/symbol and 16-QAM has 4 bits/symbol, it follows that the relative required OSNR is:

n 2c n − n2 n c 1 , where Δ = 1 . We get that L < 2 2 ⋅ = 739 m n1 n 12 ⋅ Δ n1 ⋅ Δ B

2q ( Rd PL )Δf

4 Rd2 P0 PL

3. Eq. 6.2.2 in the textbook (and on p.10-7 in the notes) states the OSNR for a given total reach Ps ln G OSNR = 2nsp ⋅ hν 0 ⋅ Δν oαLT (G − 1)

An improvement is obtained when Fn < 1/η. The EDFA can only improve the sensitivity if the quantum efficiency is lower than 50% since Fn is always > 2.

SNREDFA 1 = SNR ηFn

Use SASE = (G-1)hνnsp ≅ GhνFn/2 and Rd = qη/hν and the relative difference is found to be:

2 With an EDFA, the ASE-LO beat noise dominates: σ 2 LO − sp = 4 Rd PL S ASE Δf 2 4 Rd2 G P0 PL I SNREDFA = = σ2 4 Rd2 PL S ASE Δf

SNR =

2. Using the lecture notes # 12 – p. 23-25 we have without the EDFA:

c) QPSK with 20 GSymbols/s occupies a bandwidth of about 20 GHz. The 64-QAM signal transmits six bits per symbol. The bandwidth is therefore about 40/6 = 6.7 GHz. The OOK signal has a bandwidth of 15 GHz. Since dispersion tolerance decreases when bandwidth increases, the most resilient signal is the 64-QAM, the OOK signal comes in second place, and the QPSK signal has the worst dispersion tolerance.

b) The dispersion is normal since D < 0 and since C < 0, the pulse will first be compressed because of its frequency variation. Then, if the DCF is long enough, it will start to broaden. The spectral shape is unaffected since the transmission is linear.

BL <

To verify that we can transmit a 10 Mbit/s signal 500 m, use the BL product:

n 2 = n 1 sin (90 − θ r ) =1.4605

In the interface between the core and the cladding, we can apply Snell’s law again and set the refraction angle to 90˚ to satisfy the condition for total internal reflection:

sin (θ in ) = n 1 sin(θ r )

1. a) The refractive index needs to be set so that incoming light can have incident angle θ in of 20 degrees. The medium surrounding the fiber is air (refractive index 1) which means that the refraction angle θ r , given by Snell’s law, is:

P1 + P0 ηqλ , P0 = P1 ⋅ rex and Rd = . 2 hc

(1 + rex ) 6 ⋅ 2.302 ⋅ 10 −6 ⋅ 6.626 ⋅ 10 −34 ⋅ 3 ⋅ 108 1.282 Qσ T ⋅ = ⋅ = 2.82 ⋅ 10 −5 W = −15.5 dBm 0.7 ⋅ 1.602 ⋅ 10 −19 ⋅ 1550 ⋅ 10 −9 0.718 ηqλ / hc (1 − rex )

Prec (rex = 0.282) (1 + rex ) = = 1.786 = 2.52 dB (1 − rex ) Prec (rex = 0)

Rd GP1 (1 − rex ) Rd GP1 (1 − rex ) = = σ 0 + σ1 2 Rd GS sp ΔfP1 (1 + rex )

and the receiver sensitivity is found as:

The Q-value of the system is formed as:

2 2 σ sig − sp , 0 / 1 = 4 Rd GP0 / 1 S sp Δf .

2 S sp Δf (1 + rex ) (1 + rex )

GPrec (1 − rex )

c) EDFA: Assume signal-spontaneous beat noise dominates the noise. Since both 1-bits and 0-bits contain optical power due to the dispersion-induced non-ideal extinction ratio, both will have beat-noise contributions. The signal-spontaneous beat noise is:

In other words, the DCF would not improve the system!

PP =

b) The power penalty is defined as the receiver sensitivity divided by the receiver sensitivity in the case of rex = 0:

So we can lower the output power of the transmitter by 5.57 dB and still have a BER of 10-9.

Pin = 12 (10 −0.8 + 10 −1.35 ) = 0.101 mW = - 9.93 dBm .

The power received power is -8 dBm in 1-bits and -13.5 dBm in 0-bits. Thus,

σ s2,1 = 2qRd P1 Δf = 2qRd Δf ⋅ 2 Prec /(1 + rex ) = 9.86 ⋅10 −14 A 2 << σ T2 OK!

Now we can check our assumption of a thermally limited system:

Prec =

Solving for the sensitivity yields:

Prec =

where we have made use of the fact that

I −I R P (1 − rex ) Rd P rec (1 − rex ) ηqλP rec (1 − rex ) = = , Q= 1 0 = d 1 σ1 + σ 0 2σ T σ T (1 + rex ) hcσ T (1 + rex )

The Q-value is formed as:

, , and in Eq.1. The RMS-width f the pulse after fiber would be (after

, , and in Eq.1. The maximal length of fiber would be (after solving the

, , and in Eq.1. The RMS-width f the pulse after fiber would be (after

, , and in Eq.1. The maximal length of fiber would be (after solving the

}

σ T2 =

4k BTFn Δf 4 ⋅1.38 ⋅10 −23 ⋅ 300 ⋅ 2 ⋅ 8 ⋅10 9 = = 5.30 ⋅10 −12 A 2 RL 50

To compute the sensitivity Prec, start by assuming that the system is thermally limited (and check this later!). The thermal noise variance is found to be:

5. a) A BER of 10-9 corresponds to Q = 6. A power difference of -5.5 dB between 1-bits and 0-bits means that the extinction ratio is rex = 10-0.55 = 28.2%.

The reason that fiber 1 is a better choice at high bit rate even though the β2 is higher, is that the impact of higher-order dispersion increases with shorter pulses.

Thus, Fiber 1 should be chosen at 100 Gbit/s.

solving the equation)

Fiber2: Put

equation)

Fiber1: Put

System 2 {

Thus, Fiber 2 should be chosen at 10 Gbit/s.

solving the equation)

Fiber2: Put

equation)

Fiber1: Put

So in case a the signal is 12 dB lower than in case b, and the noise power is 35.8 dB lower than in case b meaning that the SNR is 35.8 – 12 = 23.8 dB better in case a.

So the noise spectral density decreases with the gain, but increases with the noise figure. The difference between case a and case b is 10 log ((20-1)/(316-1)) + (8-4)= -8.2 dB. Furthermore, in case a, the noise power spectral density will experience the 0.23·120 = 27.6 dB of loss due to the fiber propagation, meaning that at the receiver it is 35.8 dB lower. G (1 − rex ) 2

2Q 2 S sp Δf (1 + rex )(1 + rex ) 2 =

2 ⋅ 36 ⋅ 2.562 ⋅10 −16 ⋅ 8 ⋅ 10 9 ⋅ (1 + 0.282)(1 + 0.282 ) 2 = 1000 ⋅ (1 − 0.282) 2

2 Prec (rex = 0.282) (1 + rex )(1 + rex ) = = 5.83 = 7.66 dB 2 Prec (rex = 0) (1 − rex )

2 s 2 sig − sp

( Rd Ps ) 2 Ps = 4 Rd2 Ps S sp Δf 4 S sp Δf

S sp = (G − 1) ⋅ nsp ⋅ h ⋅ c / λ = (G − 1) ⋅ 0.5 ⋅ Fo ⋅ h ⋅ c / λ .

Between the two configurations, the only quantities that changes are the signal power and the noise power spectral density. With the EDFA at the fiber input (case a), the detected signal power will be 12 dB lower than when using it at the output (case b), since the gain is only 13 dB due to the nonlinearity limitation of the fiber. The noise power spectral density is given by:

SNR =

6) The SNR is given by the ratio between the average detected signal power and noise power, in this case only signal-spontaneous beat noise:

The power penalty is much larger now, since the noise is dependant on the power. Thus, the DCF would be advantageous to use in this case.

PP =

The power penalty is found in the same way as before:

and thermal noise is the same as before, as it is not dependent on signal power. Thus our assumption that signal-spontaneous beat noise dominates holds.

σ s2, 0 = σ s2,1 ⋅ rex = 8.47 ⋅10 −16 A 2 ,

σ s2,1 = 2qRd P1 Δf = 3.00 ⋅ 10 −15 A 2

σ sp2 − sp = 4 Rd2 S sp 2 Δν o Δf = 1.00 ⋅10 −10 A 2 , where Δν o = 0.5 nm = 62.5 GHz

2 2 −9 σ sig A2 − sp , 0 = σ sig − sp ,1 ⋅ rex = 2.37 ⋅ 10

2 2 2 −9 σ sig A2 − sp ,1 = 4 Rd GP1 S sp Δf = 4 Rd G ( 2 ⋅ Prec /(1 + rex )) S sp Δf = 8.40 ⋅ 10

We must now check the noise assumption:

S sp = (G − 1) ⋅ n sp ⋅ h ⋅ c / λ = 2.562 ⋅ 10 −16 .

where we have made use of the the fact that nsp = Fo/2 = 2 and

= 8.59 ⋅10 −7 W = - 30.66 dBm

Prec =

(5p)

Among the possible solutions, which:

(5p)

4. A non-chirped Gaussian pulse with RMS width of 80ps is launched into an ideal single mode fiber made of pure silica. This fiber has no waveguide dispersion and the dispersion slope [S] is 0.07 ps/(nm2-km) @operating

Which system will experience the largest nonlinear phase shift?

B) NRZ transmitter with 5 mW average optical power. The wavelength is 1560 nm where the fiber attenuation is 0.18 dB/km.

A) NRZ transmitter with 40 mW average optical power. The wavelength is 1530 nm where the fiber attenuation is 0.22 dB/km.

3. In a 10 Gb/s transmission link without in-line optical amplifiers, which of the following systems will have the longest reach, and how long will it be? Both use standard SMF with full dispersion compensation at the receiver and an EDFA preamplifier with -35 dBm sensitivity at BER=10-9 and a bit rate of 10 Gb/s. (5p)

- is most tolerant to chromatic dispersion/polarization-mode dispersion? - is experiencing most presence of fiber nonlinearities?

(5p)

Which of the above twelve combinations are possible?

2. A 1 Tb/s DWDM EDFA amplified transmission system should be designed that occupies no more than about 10 nm of optical bandwidth between 1550 – 1560 nm. The data format is NRZ and you can choose bit rates in all channels of 10 Gb/s, 40 Gb/s, or 100 Gb/s. The available frequency grid spacings are 25 GHz, 50 GHz, 100 GHz or 200 GHz and the maximum allowable bitrate/frequency ratio (i.e. spectral efficiency) is 1 b/s/Hz.

d) You transmit two OOK-signals through 55 km standard single-mode fiber. At the input of the fiber, the average signal power is 3 dBm for one of the signals, and 12 mW for the other. What is the combined signal power at the output of the fiber? α = 0.2 dB/km. (2p)

c) The bit-rate is 5 Gbit/s for OOK and 8 Gbit/s for 4-PAM. Explain which of these signals that is most dispersion-tolerant, and why. (3p)

What is the ratio between the average signal powers for the two signals shown in the figure? (Assume that all states occur equally often.) (3p)

b) The amplitude of the electric field is shown below for two signals using the modulation formats on-off-keying (OOK) and four-pulse-amplitude-modulation (4-PAM).

a) What is the diameter of the core? If you want single-mode operation, how much smaller diameter do you need? (2p)

1. A particular optical fiber has core index of 1.500 and cladding index of 1.496. A signal with λ = 1.55 μm that travels in this fiber has seven modes. The seventh mode is almost cut-off.

Questions during the examination: Carl Lundström, 0736 – 49 18 19

Results will be posted by January 28, 2009. The marking can be examined on personal visit to Peter Andrekson, ext. 1606. Room C428 at MC2

Note: Home assignments, their solutions, old exams, and “Problems and answers to Fiber Optics” are not allowed.

Note: Only one problem per page. Motivate all equations used and make reasonable assumptions when necessary. Unreadable solutions will not be corrected!

Examination in Fiber Optical Communication (MCC100) January 14, 2009, am, Halls at "Väg och vatten”

Department of Microtechnology and Nanoscience, Photonics Laboratory

(6p)

b) Use the figure below, which shows the relation between OSNR and Q-value and determine the BER with the two span lengths. To be able to use the longer span length, forward error correction coding (FEC) is required. Assuming that the added redundancy bits do not impose a significant dispersion penalty, what coding gain is needed to have the same BER with 75 km span length as in the case of 50 km spans and no FEC? (4p)

a) Calculate the OSNR (at the receiver) in the two cases.

(4p)

Each span consists of either 50 or 75 km of SMF (loss: 0.2 dB/km) but due to the dispersion compensation, the average loss is 0.3 dB/km. Each span is followed by an EDFA with a gain equal to the span loss. The EDFA noise figure is 5 dB and the average input signal power into each span is 1 mW. The wavelength is 1550 nm.

6. You are designing a 3000 km long fully dispersion-compensated fiber-optic link. The link is to be periodically amplified, and you are considering either 50 km or 75 km spacing between the amplifiers. The dilemma is that while an increased span length saves you a lot of money on EDFAs, the OSNR will be worse.

c) If you replace the p-i-n with an APD, would the sensitivity improve? Why/why not? Explain (briefly) how an APD can improve sensitivity despite the fact that it adds more noise. (2p)

b) The receiver sensitivity is defined as the average received optical power required for a BER of 10-9. Determine the sensitivity of this receiver. (4p)

a) Determine the bit-error-rate (BER).

Assume that 1-bits and 0-bits are equally likely and that the decision threshold is ideal.

5. You have a 10 Gbit/s NRZ transmitter at 1550 nm wavelength that outputs an average of 9 dBm of optical power. The extinction ratio is ideal. The transmitter is connected to a fiber-optic link that has a total loss of 28 dB, including dispersion compensation, coupling losses and fiber loss. There are no in-line amplifiers. At the receiving end, you have a p-i-n photodetector with a quantum efficiency of 80%, and negligible dark current. The photocurrent is amplified by an amplifier with a load resistance of 50 â&#x201E;Ś, a noise equivalent bandwidth of 5 GHz and a noise figure of 3 dB.

b) If the same pulse (non-chirped/80ps RMS width) is generated from another light source with linewidth (RMS) of 0.67 nm, determine the maximal length of this fiber that satisfies the same condition as in a).

a) Determine the maximal length of this fiber so that the RMS width of the pulse does not exceed 100ps after propagation.

wavelength (1550nm). The laser linewidth can be considered as a Gaussian with 100MHz FWHM spectral width.

−6

) ⋅ 0.8 ⋅ 1550 ⋅ 10 ⋅ 25 ⋅ 10 ⋅ 5 ⋅ 10 = 4 ⋅ 10−14 A 2 6.626 ⋅ 10− 34 ⋅ 3 ⋅ 108

4k BTΔfFn 4 ⋅ 1.38 ⋅ 10−23 ⋅ 300 ⋅ 2 ⋅ 5 ⋅ 109 = = 3.31 ⋅ 10−12 A 2 . RL 50

2 ⋅ (1.602 ⋅ 10

−9

⎛ 0.8 ⋅1.602 ⋅10−19 ⋅1550 ⋅ 10−9 ⎞ −6 ⎜ ⎟⎟ ⋅ 25 ⋅ 10 6.626 ⋅ 10− 34 ⋅ 3 ⋅ 108 I1 − I 0 I1 Rd P1 ⎜⎝ ⎠ Q= = = = = 6.87 σ 1 + σ 0 σ T + σ T 2σ T 2 3.31 ⋅ 10−12

The thermal noise dominates and we can neglect shot noise. The Q-value is now formed as:

σ T2 =

The thermal noise variance is:

σ s2,1 = 2qRd P1Δf = 2q 2ηλP1Δf / hc =

−19 2

5. a) Since the extinction ratio is ideal, there is no optical power in a 0-bit. Therefore the shot noise is also 0 in a 0-bit. The average transmitted power is 9 dBm, meaning that the transmitted power in 1-bits is 12 dBm, and after the 28 dB of loss, the received power in a 1bit is -16 dBm, i.e. 25 μW. The shot noise variance in a 1-bit is computed as:

Ö Lmax = 4.15 km

Pulse broadening formula: σ2/σ02 = 1 + (1+ Vw2)(β2·L/2σ02)2 + (1+ Vw2)2·(β3·L/4·1.414·σ03)2 Ö Lmax2 = [(100/80)2 – 1] / [84.12·(-27.48/2·802)2 + 84.14·(0.1247 /4·1.414·803)2]

Ö Vw = 2σwσ0 = 2·526·109·80·10-12 = 84.1 >> 1 (can not be neglected)

b) σw= 2πC/λ2·σλ = 526 Grad/s

Ö Lmax = 349.24 km

Pulse broadening formula: σ2/σ02 = 1 + (β2·L/2σ02)2 + (β3·L/4·1.414·σ03)2 Ö Lmax2 = [(100/80)2 – 1] / [(-27.48/2·802)2 + (0.1247 /4·1.414·803)2]

Ö Vw = 2σwσ0 = 2·267·106·80·10-12 = 0.042 << 1 (can be neglected)

a) σw= 2π·100/(8·ln2)1/2 = 267 Mrad/s

Dispersion Slope, S = dD/dλ = (2πC/ λ2)2·β3 + (4π C/ λ3)·β2 Ö β3 = 0.1589 ps3/km

!!! 1276 nm is zero-dispersion wavelength which can be found in the book and the lecture note.

4. Dispersion Parameter, D = DMaterial + Dwaveguide = 122·(1 - 1276/1550) + 0 = 21.56 ps2/(kmnm) Ö β2 = Dλ2/2∏C = -27.48 ps2/km Vλ 2π n 12 − n 22

= 9.0μm ⇒ d = 18.0μm

PRatio = 0 + E 12

0 + E 12 + 4E 12 + 9E 12 = 14

B) is thus a slightly better choice.

The nonlinear phase shift is proportional to PinLeff, where Leff is defined in eq (4.1.8). Thus, Leff (A) = 19.7 km and Leff (B) = 24.2 km, and the nonlinear phase shift will be 40/5x19.7/24.2 = 6.5 times larger in system A than in system B.

3. A) Lmax = (16+35)/0.22 = 232 km B) Lmax = (7+35)/0.18 = 233 km.

The 100 Gb/s solution will be more sensitive to dispersion (shorter pulses). Also, since the bandwidth of a 100 Gb/s signal is larger, it will need a higher signal power to reach a certain SNR/Q/BER. Therefore, the 100 Gb/s system will accumulate more nonlinear effects.

The two possible solutions are thus: 25x40 Gb/s at 50 GHz grid with SE = 0.8 b/s/Hz 10x100 Gb/s on a 100 GHz grid with SE = 1 b/s/Hz

At 10 Gb/s one needs 100 channels, which with the smallest available grid (25 GHz & spectral efficiency of 0.4) occupies 100x25 GHz = 2.5 THz is not possible. At 40 Gb/s (25 channels) and 50 GHz grid (SE = 0.8) the spectral width will be 1.25 GHz. The other frequency grids will not work. At 100 Gb/s (10 channels) both 100 and 200 GHz grid work but the spectral width will be sufficiently small only for a 100 GHz grid.

2. The available frequency bandwidth is about 1.25 THz.

d) The combined power at the input is 14 mW. It is attenuated 12.6 times ( 55km ⋅ 0.2dB / km = 11 dB = 12.6 ), so the output power is 14 mW / 12.6 = 1.1mW .

c) The 4-PAM-signal has higher dispersion tolerance since its symbol rate is smaller. Its bandwidth is therefore smaller and this is beneficial from a “resilience towards dispersion” point-of-view”.

The 4-PAM signal has 14 times higher average power than the OOK-signal.

b) The ratio is:

We want V = 2.405. At this normalized frequency, a is equal to 5.4 μm and d = 10.8 μm. This means that d must be reduced by 7.2 μm.

a (V = 4) =

1. a) V ≈ 4 when the seventh mode almost is cutoff. The diameter d can be calculated to be

Solutions for examination of Fiber Optical communication, January 14, 2009

) exp(−Q 2 / 2) = 3.3 ⋅10 −12 Q 2π 1

I1 − I 0

σ1 + σ 0

Rd P1 Rd P rec ηqλP rec = = 2σ T σT hcσ T

Qσ T 6 ⋅ 3.31 ⋅ 10 −12 ⋅ 6.626 ⋅ 10 −34 ⋅ 3 ⋅ 108 = 1.09 ⋅ 10 − 5 W = −19.6dBm = ηqλ / hc 0.8 ⋅ 1.602 ⋅ 10 −19 ⋅ 1550 ⋅ 10 − 9

G c = 20 ⋅ log(7 / 3.3) = 6.5 dB

so we need a coding gain of

Gc = 20 ⋅ log(Qcode / Q) ,

b) From the figure we determine that an OSNR of 20 dB gives a Q-value of 7, corresponding to a BER of 10-12. An OSNR of 14.5 dB gives a Q-value of about 3.3, which corresponds to a BER of 5.3·10-4. The coding gain is defined (in dB) as

Case 2: OSNRdB = 0 – 16 – 5 – 22.5 + 58 = 14.5 dB

Case 1: OSNRdB = 0 – 18 – 5 – 15 + 58 = 20 dB

The EDFA gain needs to be 15 dB in case 1 (50 km · 0.3 dB/km) and 22.5 dB in case 2. The number of EDFAs needed (NA) is 60 (~18 dB) in case 1 and 40 (~16 dB) in case 2. Thus:

6. a) From the lecture notes: OSNRdB = Ps(dBm)-NA(dB)-Fo(dB)-G(dB)+58(dBm)

c) An APD would improve the sensitivity, since the system is thermally limited. An APD can improve the sensitivity if the system is dominated by thermal noise, despite adding more (shot) noise because the signal current can increase more than the total noise, since the thermal noise remains unchanged.

Prec =

Setting Q = 6 and solving for the sensitivity yields:

b) Since the system is thermally limited at a BER of 3.3·10-12 it will be thermally limited at a lower optical power level giving a BER of 10-9 (corresponding to a Q-value of 6) as well. The Q-value is formed as:

BER = 12 erfc Q / 2 ≈

(

The BER can now be computed using the relation:

2(4) 2012-12-17

EAM

180 ο

0ο

(b)

CW ? ?

EAM EAM

EAM

(c) How does the impact of dispersion (and therefore also the number of taps in the compensation filter) scale with the symbol rate? Use your answer to estimate the number of filter taps needed to compensate for 1000 ps/nm for a 28 Gbaud polarizationmultiplexed QPSK signal. (3p)

(b) The link in Fig. 2b has a DCF module after each SMF section, but the DCF is a bit too short and there is a residual CD after transmission. How many taps are required for the compensation? (3p)

(a) How many filter taps are required to compensate for the CD in the link shown in Fig. 2a, which consists of 10 identical sections with single-mode fiber? (4p)

2. In a digital coherent receiver, chromatic dispersion (CD) can be compensated after detection with a finite-impulse response filter whose number of taps depends on the transmission distance, the dispersion parameter, and the symbol rate. You are designing a coherent receiver for a 40 Gbit/s polarization-multiplexed QPSK system (λ = 1550 nm) and plan to use a rule-of-thumb which says that four filter taps are required per 1000 ps/nm of accumulated CD.

Figure 1: (a) A configuration that can generate OOK and BPSK. (b) A configuration capable of generating, e.g., QPSK.

(a)

EAM

(b) In Fig. 1b, two devices such as the one in Fig. 1a are connected in parallel. What should the phase shifts in the different branches be to make it possible to generate quadrature phase-shift keying (QPSK) symbols? Motivate your answer. Notice that the value of the phase shifter is an “effective value” including any possible phase shift in the splitters/combiners. (It is only necessary to state one of the different possible combinations.) (4p)

(a) Explain in words and illustrate your explanation using constellation diagrams how the transmitter can be used to generate both OOK and BPSK. (6p)

1. The transmitter in Fig. 1a can be used to generate both on–off keying (OOK) and binary phase-shift keying (BPSK). An incoming continuous wave (CW) signal is split equally into two branches. In each branch there is an electro-absorption modulator (EAM) which can turn the light on and off, and a phase shifter which provides a constant phase shift. The signals in the two branches are then recombined at the transmitter output.

Examination, Fiber Optical Communication, MCC100 Photonics Laboratory, Department of Microtechnology and Nanoscience

1(4) 2012-12-17

• Calculator of choice (laptops and other devices capable of connecting to communication networks are not allowed)

• English dictionary

• Mathematical and physical tables, e.g.: Tefyma, Physics Handbook, Standard Mathematical Tables, Beta Mathematical Handbook

• The two laboratory exercise manuals

• The printed lecture notes

• ”Fiber-Optic Communication Systems”, by G. P. Agrawal

Questions during the examination: Samuel Olsson will visit after approximately one and three hours. Alternatively, use the phone number 0705-391492.

The correction can be examined on personal visit to Pontus Johannisson, room C427 in the MC2 building. Phone: 031-772 1626. E-mail address: pontus.johannisson@chalmers.se.

The solution will be posted on the board in the “Canyon” in MC2 (just outside the entrance to the Photonics Laboratory) directly after the exam. The correction will be finished no later than 20 days after the exam. Results will then be announced via e-mail.

The maximum score is 60 points. The examination contains 6 questions that each can give up to 10 points. The questions are not given in order of increasing difficulty.

No personal communication is allowed. No use of any kind of communication device is allowed.

Allowed material include:

Note: Solve only one problem per page. Mark each page clearly with your personal identification code. Motivate all equations used and make reasonable assumptions when necessary. Unreadable solutions will not be corrected!

14:00–18:00, December 17, 2012, HA, HB, HC

Examination, Fiber Optical Communication, MCC100

Examination, Fiber Optical Communication, MCC100 Photonics Laboratory, Department of Microtechnology and Nanoscience

Figure 3: EDFA setup

4(4) 2012-12-17

(d) Assume the use of dual-polarization 12.5 Gbaud 16-QAM data using a channel spacing of 25 GHz. (3p)

(b) Assume the use of single-polarization 10 Gbaud OOK channels using a channel spacing of 100 GHz. (3p)

(a) First calculate the bandwidth corresponding to 30 nm. (1p)

6. A typical EDFA has a gain bandwidth of about 30 nm. You plan to make a WDM system that uses this entire bandwidth. Calculate the total data throughput using the below given channel specifications.

(b) The p-i-n photodetector is replaced by an APD with FA (M ) = M 0.7 and M = 300. The quantum efficiency, bandwidth, and load impedance are the same as above. What is the BER or Q factor at −20 dBm received optical power? (5p)

(a) Calculate the bit error rate (BER) or the Q factor at −20 dBm received optical power. (5p)

5. A 1550 nm receiver with 5 GHz bandwidth is operating at 5 Gbit/s. The quantum efficiency of the p-i-n photodiode is 90 %, the load impedance is 50 Ω, and the noise figure of the electrical amplifier following the photodiode is 5 dB. The ones and zeros are transmitted with equal probabilities with 0.1 extinction ratio. The temperature is 300 K.

EDFA

Examination, Fiber Optical Communication, MCC100 Photonics Laboratory, Department of Microtechnology and Nanoscience

L = 100 km D = 17 ps/km nm

10 sections Rx Tx

(b) L = 100 km D = 17 ps/km nm

L = 15 km D = −100 ps/km nm

10 sections Rx

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(b) During the experiments one of the EDFAs breaks down and is replaced with another one with a noise figure of 10 dB. Which noise component(s) increase(s) and by how much? Does it matter where in the system the replacement EDFA is placed? Motivate the answer. (3p)

(a) Which noise source dominates in the system? Motivate the answer by calculating the relevant noise variances. (7p)

4. A 10 Gbit/s, amplified optical channel operating at λ = 1550 nm is configured as in Fig. 3. The average launch power is −7 dBm, the modulation format is on-off keying with −7 dB extinction ratio. The noise figure of the EDFAs is 5 dB. Loss in each fiber span is 20 dB, which is compensated by the following EDFA. The dispersion compensation can be assumed to be ideal. To detect the signal you use a photoreceiver with a responsivity of 1 A/W, 10 GHz bandwidth, and with an electrical amplifier with 6 dB noise figure. In front of the receiver there is a tunable optical filter to reject the out-of-band ASE noise. Assume that the system operates at normal room temperature.

(c) Assume that we change from nonlinear to “almost linear” propagation by lowering the signal input power significantly. We will then have two (main) problems at the receiver. What are these two problems and how can you solve them? (4p)

(b) Consider a fundamental soliton that is losing power during propagation through a fiber. What will happen to the soliton? How will this affect the OOK data? (3p)

(a) What is the reason that the fundamental soliton was used for this purpose? Describe why the fundamental soliton has this special property. (3p)

3. Modern systems are designed to keep the nonlinear effects small but for a long time soliton systems were considered to be a practical solution for optical communication. In these systems, the fundamental soliton was used to implement transmission using on–off keying (OOK).

Figure 2: (a) A system without hardware dispersion compensation. (b) A system with imperfect hardware dispersion compensation.

(a)

Examination, Fiber Optical Communication, MCC100 Photonics Laboratory, Department of Microtechnology and Nanoscience

6(4) 2012-12-17

thermal noise: 1.32E-11 shot noise for level 1: 1.1E-12 shot noise for level 0: 2.2E-13 sig-sp noise, level 1: 1.8E-9 sig-sp noise, level 0: 2.1E-10

and σs2 = 2qI∆f.

(4)

(3)

P1 = Pavg

(6)

I1 − I0 σ1 + σ0

BER =

√ 1 erfc(Q/ 2) 2

and with the given parameters it is 3.6. The BER is

The Q factor, assuming that P (0|1) = P (1|0) is

(8)

(7)

where B is bandwidth and RL is the load resistance. A similar expression is found in the same way for the zeros.

σ12 = 2qI1 B + 4kB T BFn /RL ,

2 , (5) 1+r where Pavg is the average power and r is the extinction ratio. The optical power in zeros: P0 = P1 r. The photodiode responsivity is Rd = ηλ/1.24, where η is the quantum efficiency and λ is the wavelength in µm. The photocurrents corresponding to one and zeros are I1 = Rd P1 and I0 = Rd P0 , respectively. The noise current variance for ones is

5. The optical power in ones is

• sig-sp noise, level 1, second case: 1.7E-9 • sig-sp noise, level 0, second case: 3.3E-10

where the indices 1, 2, 3, and 4 correspond to the different amplifiers. When one of the EDFAs has worse performance, the order of the EDFAs does not matter. As seen above, the population inversion factors are simply added. The noise components are calculated is in (a). The signal-spontaneous noise is now:

SASE = (nsp,1 + nsp,2 + nsp,3 + nsp,4 )hν0 (G − 1),

(b) The total ASE spectral density will now be be

• • • • •

The results are as follows:

σT2 = 4kB T FN,el ∆f /RL

where Ps is the average signal power after the final amplification. Tuning the optical filter, the spontaneous-spontaneous noise can be neglected. The thermal and shot noise variances are

Examination, Fiber Optical Communication, MCC100 Photonics Laboratory, Department of Microtechnology and Nanoscience

5(4) 2012-12-17

2 σsig-sp = 4Rd2 Ps SASE ∆f,

(2)

since the gain G is equal to the loss A in each span. At the receiver, the signalspontaneous noise variance is

(a) The loss of each span is 20 dB, so the gain of each EDFA must compensate for it and the EDFA gain G = 20 dB. Given the noise figure Fn = 5 dB, one can find the nsp = Fn G/[2(G − 1)]. Since nsp is known and the same for each amplifier, we can calculate the ASE spectral density 3 G2 G G + + + 1 = 4 × nsp hν0 (G − 1) (1) SASE = nsp hν0 (G − 1) A3 A2 A

(c) The two main problems are (i) an increased pulse width (as above) that leads to ISI and (ii) a decreased power level that leads to a low SNR. These problems were typically solved by using DCF to compensate for the dispersion and EDFAs as in Fig. 3 to compensate for the losses.

(b) When the pulse is losing power, the balance between the dispersion and the Kerr nonlinearity is disturbed. The lower power level implies that the dispersion becomes the stronger effect and the pulse starts to broaden. As usual, this dispersive broadening leads to ISI and problems to detect the signal.

(a) The fundamental soliton can propagate in an optical fiber without changing its shape. This is possible due to the fact that the dispersion and the Kerr nonlinearity set up a robust balance. In particular, the phase shift due to the dispersion that would lead to the pulse broadening is perfectly canceled by the nonlinear phase shift.

(c) The impact from CD has a quadratic dependence on the symbol rate (pulse width), meaning that the number of taps required to compensate for a CD of 1000 ps/nm is 4 × (28/10)2 = 31.4, which we round upwards to 32.

(b) The residual dispersion is 17000 − 10 × 15 × 100 = 2000 ps/nm, and the number of required filter taps is 4 × 2000/1000 = 8.

(a) With 10 sections the total accumulated CD in the link is 17000 ps/nm, and the number of required filter taps is 4 × 17000/1000 = 68.

(b) The simplest solution is to set up the phases to be 0◦ , 90◦ , 180◦ , 270◦ . Then, electrical signals are generating from the data in such a way that only one EAM is in the on state at every given instant.

(a) For OOK it is sufficient to turn the power on and off. For example, the lower EAM can be in off state and the data signal is then fed to the upper EAM. BPSK signals have either 0 or π phase, and to generate this both of the EAMs need to be switched. This is done by making sure that the upper EAM is in the on state when the lower is in the off state and vice versa.

Examination, Fiber Optical Communication, MCC100 Photonics Laboratory, Department of Microtechnology and Nanoscience

7(4) 2012-12-17

(9)

c 2.998 × 108 dλ = (30 × 10−9 ) ≈ 3.7 THz. 2 λ (1.55 × 10−6 )2

(10)

(11)

(12)

3.7 × 1012 100 × 109 = 15 Tbit/s 25 × 109

(13)

(d) Dual-polarization 12.5 Gbaud 16-QAM data using a channel spacing of 25 GHz means 100 Gbit/s/ch and a total throughput of

3.7 × 1012 50 × 109 = 3.7 Tbit/s 50 × 109

(c) Single-polarization 25 Gbaud DQPSK data using a channel spacing of 50 GHz means 50 Gbit/s/ch and a total throughput of

3.7 × 1012 10 × 109 = 370 Gbit/s 100 × 109

(b) Single-polarization 10 Gbaud OOK channels using a channel spacing of 100 GHz means 10 Gbit/s/ch and a total throughput of

dν =

(a) The corresponding bandwidth is

The Q factor and BER are calculated in the same way as in the previous case. The Q factor is 9.9 and BER is 2 × 10−23 .

and it is calculated in the same way for zeros.

σ12 = 2qRd P1 M 2 B + 4kB T BFn /RL ,

With the APD, the photocurrents are I1 = Rd P1 M and I0 = Rd P0 M , where M is the multiplication factor. The noise current variance for ones is

and with the given parameters it is equal to 1.7 × 10−4 .

Examination, Fiber Optical Communication, MCC100 Photonics Laboratory, Department of Microtechnology and Nanoscience