Optical Networks (Solutions, Instructor Solution Manual) [1 ed.] 0198834225, 9780198834229

Citation preview

Solution Manual (March 2021)

Optical Networks Debasish Datta

Oxford University Press

Preface

This solution manual has been prepared chapter-wise starting from Chapter 2, after submission of the book manuscript. Special attention has been given to the holistic design issues and performance features in all segments of optical networks, calling for the crosslayer aspects therein, and hence in most of the chapters some of the exercise problems need adequate knowledge in the technologies (Chapter 2) and transmission impairments in optical networks (Chapter 10). The instructors may please address this issue and encourage the students to appreciate this aspect for developing a practical perspective of optical networks. Moreover, while solving the problems after submission of the book manuscript, some changes on the numerical data have been incorporated in the exercises of the book to make them more worthwhile, along with a correction in an expression in Chapter 6. All these modifications are included in the errata presented in the following page. The instructors are most welcome to contact me by email ([email protected]), if they need any clarifications, and they may please write to me if they come across any error in the book. Debasish Datta

i

Errata

Context

Present version in the book

Corrected/changed version

5.27 nm

527 nm

Gd = L/[2(M – 1) + L]

Gd = L/[2(M – 1 + L)]

Page 572, Exercise 14.6.

Γ = 0 24 40 50 24 0 24 40 24 24 0 0 50 0 40 0.

Γ = 0 50 25 60 25 0 50 60 25 30 0 30 25 50 30 0.

Page 593, Exercise 15.7

0.1 µs

0.8 µs

Page 130, Exercise 2.7 Page 248, expression for Gd below Eq. 6.5.

ii

Exercise Problems and Solutions for Chapter 2 (Technologies for Optical Networks) 2.1 A step-index multi-mode optical fiber has a refractive-index difference Δ = 1% and a core refractive index of 1.5. If the core radius is 25 µm, find out the approximate number of propagating modes in the fiber, while operating with a wavelength of 1300 nm. Solution: Δ = 0.01, n1 = 1.5, a = 25 μm, w = 1300 nm, and the number of modes Nmode is given by 𝑁𝑁𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚 =

The numerical aperture NA is obtained as

𝑉𝑉 2 , with 2

𝑉𝑉 =

2𝜋𝜋𝜋𝜋 𝑤𝑤

𝑁𝑁𝑁𝑁.

𝑁𝑁𝑁𝑁 = �𝑛𝑛12 − 𝑛𝑛22 ≈ 𝑛𝑛1 √2∆ = 1.5√0.02. Hence, we obtain V parameter as, 2𝜋𝜋 × 25 × 10−6 × �1.5√0.02� = 25.632, 1300 × 10−9 leading to the number of modes Nmode , given by 𝑉𝑉 =

𝑁𝑁𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚 =

25.6322 ≈ 329. 2

2.2 A step-index multi-mode optical fiber has a cladding with the refractive index of 1.45. If it has a limiting intermodal dispersion of 35 ns/km, find its acceptance angle. Also calculate the maximum possible data transmission rate, that the fiber would support over a distance of 5 km. Solution: The cladding refractive index n2 =1.45, and the intermodal dispersion Dmod = 35 ns/km. Dmod is

expressed as 𝐷𝐷𝑚𝑚𝑚𝑚𝑚𝑚 ≈

𝑛𝑛1 Δ 𝑛𝑛1 𝑛𝑛1 − 𝑛𝑛2 𝑛𝑛1 − 𝑛𝑛2 = � �= = 35 ns/km. 𝑐𝑐 𝑛𝑛1 𝑐𝑐 𝑐𝑐

Hence, (n1 – n2) = cDmod = (3 × 105) × (35 × 10-9) = 0.0105, and n1 = n2 + 0.0105 = 1.4605. Therefore, we obtain NA as 𝑁𝑁𝑁𝑁 = �𝑛𝑛12 − 𝑛𝑛22 = �1.46052 − 1.452 = 0.174815,

and the acceptance angle is obtained as θA = sin-1(NA) = sin-1(0.174815) = 10.068o. The pulse spreading due to dispersion should remain ≤ 0.5/r, with r as the data-transmission rate, implying that r ≤ 0.5/(Dmod L). Hence, we obtain the maximum possible data transmission rate rmax over L = 5 km as 𝑟𝑟𝑚𝑚𝑚𝑚𝑚𝑚 =

0.5 = 2.86 Mbps. 35 × 10−9 × 5

2.3 Consider that a step-index multi-mode optical fiber receives optical power from a Lambertian source with the emitted intensity pattern given by I(θ) = I0 cosθ, where θ is the angle subtended by an incident light ray from the source with the fiber axis. The total power emitted by the source is 1 mW while the power coupled into the fiber is found to be - 4 dBm. Derive the relation between the

2.1

launched power and the numerical aperture of the optical fiber. If the refractive index of the core is 1.48, determine the refractive index of the cladding. Solution: Transmit power PT = 1 mW, and the power coupled into fiber PC = - 4 dBm = 10- 0.4 W = 0.3981 mW. For a Lambertian source, the coupled power PC = NA2 × PT X (for derivation, see Cherin 1983). Hence, 𝑃𝑃 NA2 = PC/PT = 0.3981. Further, 𝑁𝑁𝑁𝑁2 = 𝑛𝑛12 − 𝑛𝑛22 = 𝑃𝑃 𝐶𝐶 implying that n22 = n12 – PC/PT . 𝑇𝑇

Thus, we obtain n2 as

𝑛𝑛2 = √1.482 − 0.3981 = 1.34. 2.4 Consider a 20 km single-mode optical fiber with a loss of 0.5 dB/km at 1330 nm and 0.2 dB/km at 1550 nm. Presuming that the optical fiber is fed with an optical power that is large enough to force the fiber towards exhibiting nonlinear effects, determine the effective lengths of the fiber in the two operating conditions. Comment on the results. Solution: With L = 20 km, first we consider the case with fiber loss αdB = 0.5 dB/km. So, the loss α in neper/km is determined from αdB = 10log10[exp(α)] as

α = ln (10αdB/10) = ln(100.05) = 0.1151. Hence, we obtain the effective fiber length as Lef f = [1 – exp(-αL)]/α = [(1 – exp(-0.1151 × 20)]/0.1151 = 7.82 km. With αdB = 0.2 dB/km, we similarly obtain Lef f = 13.06 km, which is expected because with lower attenuation, the power decays slowly along the fiber and thus the fiber nonlinearity effects can take place over longer fiber length. 2.5 Consider an optical communication link operating at 1550 nm over a 60 km optical fiber having a loss of 0.2 dB/km. Determine the threshold power for the onset of SBS in the fiber. Given: SBS gain coefficient gB = 5 ×10-11 m/W, effective area of cross-section of the fiber Aeff = 50 µm2, SBS bandwidth = 20 MHz, laser spectral width = 200 MHz. Solution: With αdB = 0.2 dB/km at 1550 nm, we obtain α = ln (10αdB/10) = ln(100.02) = 0.0461. Hence, for L = 60 km, we obtain Lef f as Lef f = [1 – exp(-αL)]/α = [1 – exp(-0.0461 × 60)]/0.0461 = 20.327 km. With Aeff = 50 μm , gB = 5 × 10 m/W, and assuming the polarization-matching factor to be ηp = 2, we obtain the SBS threshold power as 2

𝑃𝑃𝑡𝑡ℎ (𝑆𝑆𝑆𝑆𝑆𝑆) =

-11

21 𝜂𝜂𝑝𝑝 𝐴𝐴𝑒𝑒𝑒𝑒𝑒𝑒 𝛿𝛿𝛿𝛿 21 × 2 × 50 × 10−12 200 �1 + �= �1 + � = 22.73 mW. −11 3 𝑔𝑔𝐵𝐵 𝐿𝐿𝑒𝑒𝑒𝑒𝑒𝑒 5 × 10 × 20.327 × 10 Δ𝜔𝜔𝐵𝐵 20

2.6 Consider an optical communication link operating at 1550 nm over a 60 km optical fiber having a loss of 0.2 dB/km. The effective area of cross-section of the fiber Aeff = 50 µm2, where an optical power of 0 dBm is launched. Determine the nonlinear phase shift introduced by SPM in the fiber. Given: ñ(ω) = 2.6 × 10-20 m2/W.

2.2

Solution: The transmit power PT = 1 mW and the effective area of cross-section of the fiber is Aeff = 50 μm2 Hence, the intensity of launched light I = PT/Aeff = 10-3/(50 × 10-12) = 2 × 107 W/m2. With αdB = 0.2 dB/km., α = 0.0461 and Leff = 20.327 km (see Exercise 2.5). Further, the nonlinearity parameter 𝜉𝜉of the fiber is given by 𝜉𝜉 =

𝑛𝑛� (𝜔𝜔)𝜔𝜔 𝑐𝑐

=

2𝜋𝜋 �𝑛𝑛 (𝜔𝜔) 𝑤𝑤

with ñ(ω) = 2.6 × 10-20 m2/W and w = 1550 nm.

Therefore, the phase shift Δ𝜙𝜙𝑆𝑆𝑆𝑆𝑆𝑆 introduced by SPM in the optical fiber is obtained as Δ𝜙𝜙𝑆𝑆𝑆𝑆𝑆𝑆 = 𝐿𝐿𝑒𝑒𝑒𝑒𝑒𝑒 𝜉𝜉 𝐼𝐼 = (20.327 × 103 ) ×

2𝜋𝜋×2.6×10 −20 1.55×10 −6

× (2 × 107 ) = 0.04285 radian = 2.455o

2.7 Determine the Bragg wavelength of an integrated-optic grating with a period of 527 nm (earlier it was 5.27 nm in the book) and an effective refractive index of 1.47. Sketch a block schematic using Bragg grating and other relevant components for an optical add-drop multiplexer for three wavelengths at a given node in a WDM ring network. Solution: The period of grating is Λ = 527 nm, and the effective refractive index is neff = 1.47. Hence the Bragg wavelength wb of the integrated-optic grating is obtained as 𝑤𝑤𝑏𝑏 = 2𝑛𝑛𝑒𝑒𝑒𝑒𝑒𝑒 Λ = 2 × 1.47 × 0.527 = 1549.38 nm.

Using Bragg grating one can realize an add-drop multiplexer (OADM) for three wavelengths w1, w2,

w3 (w2 as the drop wavelength) as shown in the following. Circulator Bragg grating

w1, w2, w3 w2

Coupler

w1, w2, w3

w2

OADM using Bragg grating

2.8 A laser cavity (i.e., an active layer in DH configuration) has a length L with a medium loss α dB per unit length, which is pumped with a gain g per unit length using an appropriate forward bias current. The walls on two ends of the cavity are designed with reflectivities r1 and r2, and the cavity has a confinement factor of Γ. Determine the condition to be satisfied by the pumped cavity to function as a laser. Give a typical sketch of g as a function of wavelength, and explain its impact on the laser spectrum. Solution: The DH laser cavity operates as shown in the following diagram with the two reflecting walls (shaded parts).

r1

Cavity (α, g, Γ)

r2

L Total loss-cum-gain ρLG in the cavity must be ≥ 1, with ρLG given by 𝜌𝜌𝐿𝐿𝐿𝐿 = 𝑟𝑟1 𝑟𝑟2 𝑒𝑒𝑒𝑒𝑒𝑒{−(𝛼𝛼 − 𝑔𝑔Γ)𝐿𝐿)}. 2.3

Hence, the condition that the cavity must satisfy to function as a laser is expressed as 𝑟𝑟1 𝑟𝑟2 𝑒𝑒𝑒𝑒𝑒𝑒{−(𝛼𝛼 − 𝑔𝑔Γ)𝐿𝐿)} ≥ 1,

leading to the expression for the minimum cavity gain, given by 𝑔𝑔𝑚𝑚𝑚𝑚𝑚𝑚 =

𝛼𝛼 Γ

ln (𝑟𝑟1 𝑟𝑟2 ) . 𝐿𝐿Γ

−

g (w)

w

g(w) shapes the spectrum and thus reduces the number of modes in a laser. 2.9 What is the fundamental difference between the spectral spreads in lasers due to the phase noise and modulation? Determine the spectral spread of a 1550 nm laser transmitting at 10 Gbps with the unmodulated linewidths of (i) 200 MHz and (ii) 0.08 nm. Solution: The spectral spread due to phase noise is an internal phenomenon in a laser, arising from the underlying spontaneous noise and finite bandwidth of the laser cavity. On the other hand, the spectral spread due to the modulation is an external effect from the phase, frequency, amplitude, intensity modulation (governed by the modulation scheme) by the input data stream and hence is related to the bit rate of the data stream. (i) Laser linewidth BL = 200 MHz, modulation bandwidth BM = 10 GHz (considering intensity modulation). Hence the total spectral spread BT of the laser after modulation (3-dB bandwidth) is obtained as 𝐵𝐵𝑇𝑇 = �𝐵𝐵𝐿𝐿2 + 𝐵𝐵𝑀𝑀2 = �(200 × 106 )2 + (10 × 109 )2 = √100.04 × 1018 ≈ 10 GHz.

(ii) Laser linewidth is given in terms of wavelength spread δw = 0.08 nm, which leads to a frequency spread BL, given by 𝐵𝐵𝐿𝐿 =

𝑐𝑐𝑐𝑐𝑐𝑐 𝑤𝑤 2

=

3×10 8 ×0.08×10 −9 (1.55×10 −6 )2

= 9.99 GHz ≈ 10 GHz.

Using this value for BL, we obtain the total laser spectral spread as

𝐵𝐵𝑇𝑇 = �𝐵𝐵𝐿𝐿2 + 𝐵𝐵𝑀𝑀2 = �(10 × 109 )2 + (10 × 109 )2 = 14.14 GHz.

2.10 The threshold current density of a stripe-geometry AlGaAs laser is 3000 A/cm2 at 15oC. Estimate the required threshold current at 50oC, when the laser characteristic temperature T0 = 170oK and the contact stripe of the laser has a size (area) of 20 μm × 100 μm. Solution: The threshold current Ith (T) in laser has a strong dependence on temperature, given by 𝑇𝑇 𝑇𝑇0

𝐼𝐼𝑡𝑡ℎ (𝑇𝑇) ∝ exp � �.

Let T1 = 50oC and T2 = 15oC.

At T = T2, the current density Jth (T2) = 3000 A/cm2 = 3 × 107 A/m2. Using the above relation between Ith with T (and hence between Jth with T), we obtain 𝐼𝐼𝑡𝑡ℎ (𝑇𝑇1 ) 𝐼𝐼𝑡𝑡ℎ (𝑇𝑇2 )

=

𝐽𝐽 𝑡𝑡ℎ (𝑇𝑇1 ) 𝐽𝐽 𝑡𝑡ℎ (𝑇𝑇2 )

exp (𝑇𝑇 /𝑇𝑇 )

𝑇𝑇1 −𝑇𝑇2 � 𝑇𝑇0

= exp (𝑇𝑇1 /𝑇𝑇0 ) = exp � 2.4

2

0

Therefore, we obtain Jth (T1) as 𝑇𝑇1 −𝑇𝑇2 � 𝑇𝑇0

𝐽𝐽𝑡𝑡ℎ (𝑇𝑇1 ) = 𝐽𝐽𝑡𝑡ℎ (𝑇𝑇2 )exp �

50−15 � A/m2 . 170

= 3 × 107 × exp �

Using the above expression, we therefore obtain Ith (T1) as the product of Jth (T1) and the stripe area Ast (= 20 μm × 100 μm), given by 50−15 �× 170

𝐼𝐼𝑡𝑡ℎ (𝑇𝑇1 ) = 𝐽𝐽𝑡𝑡ℎ (𝑇𝑇1 )𝐴𝐴𝑠𝑠𝑠𝑠 = 3 × 107 × exp �

(20 × 100 × 10−12 ) = 73.72 A.

2.11 Consider that a binary optical signal is incident from a fiber onto a photodetector. Presuming that the incident light has a duality in its nature (particle and wave), give examples for the manifestation of both the forms of light on the performance of a digital optical receiver. Solution: One of the manifestations of the wave nature of light takes place in the form of fiber dispersion mechanisms leading to the pulse spreading at the receiving end. On the other hand, the particle nature of light is reflected from the fact that, the photocurrent (number of electrons per unit time) is proportional to the number of photons (light particles) arriving per unit time at the receiver. 2.12 An APD operates at a wavelength of 900 nm with 95% quantum efficiency. Consider that an incident light with a power of -30 dBm has produced a photocurrent of 15 μA at the APD output. Determine the mean avalanche gain of the APD. Solution: The APD operates at 900 nm with a quantum efficiency of η = 0.95. The incident power Pin = -30 dBm = 10-6 W, which has generated a photocurrent IAPD = 15 μA. Hence, the primary photocurrent Ip in the APD is obtained as 𝐼𝐼𝑝𝑝 = 𝑅𝑅𝑤𝑤 𝑃𝑃𝑖𝑖𝑖𝑖 =

𝜂𝜂𝜂𝜂 ℎ𝑓𝑓

𝑃𝑃𝑖𝑖𝑖𝑖 =

𝜂𝜂𝜂𝜂𝜂𝜂 ℎ𝑐𝑐

𝑃𝑃𝑖𝑖𝑖𝑖 =

0.95×1.6×10 −19 ×900×10 −9 6.63×10 −34 ×3×10 8

Using the above result, we obtain the APD mean gain MAPD as 𝑀𝑀𝐴𝐴𝐴𝐴𝐴𝐴 =

× 10−6 = 0.6878 μA.

𝐼𝐼𝐴𝐴𝐴𝐴𝐴𝐴 15 = = 21.8. 𝐼𝐼𝑝𝑝 0.6878

2.13 If the received optical power in a PIN-based optical receiver is -20 dBm and the preamplifier needs a voltage swing of 1 mV at its output, calculate the value of the feedback resistance needed for the preamplifier. Given: optical transmitter uses a laser with perfect extinction, PIN diode responsivity = 0.8 A/W. Solution: The received optical power PR = -20 dBm = 10-5 W and the PIN responsivity = 0.8 A/W. Hence, the photocurrent Ip = 0.8 × 10-5 A. The preamplifier output voltage swing needs to be Vs = 10-3 V which is the product of the input photocurrent Ip and the transimpedance RTI, implying Vs = Ip RTI. Hence we obtain the transimpedance as 𝑅𝑅𝑇𝑇𝑇𝑇 =

𝑉𝑉𝑠𝑠 10−3 = = 125 ohm. 𝐼𝐼𝑝𝑝 0.8 × 10−5

2.14 Draw a block schematic for a strictly nonblocking 8 × 8 optical switch, employing 2 × 2 electrooptic switching elements as the building block in Spanke's architecture. Estimate the total insertion loss of a similar N × N (with N = 2k) optical switch in terms of its number of stages and the losses incurred in all the passive devices during the traversal of a lightpath from an input port to an output port. Solution: 2.5

Considering the Spanke’s switch architecture in Fig. 2.58, we note that the first (i.e., input) column will have N (1:N) electro-optic switches, with k = log2N. Similarly, the second (output) column will have N (N:1) electro-optic switches. Using this observation, the readers are instructed to draw the full switch configuration. Hence, the insertion loss Lsw incurred by a lightpath in the above switch will be the sum of the losses in the two columns of the switches along with the connector losses Lc the input and output sides of each 1:N and N:1 switch. Thus, we express Lsw as 𝐿𝐿𝑠𝑠𝑠𝑠 = 2𝑘𝑘𝐿𝐿𝑒𝑒𝑒𝑒 + 4𝐿𝐿𝑐𝑐 = 2(log 2 𝑁𝑁 × 𝐿𝐿𝑒𝑒𝑒𝑒 + 2𝐿𝐿𝑐𝑐 ),

with Leo as the insertion loss in each (2 × 2) electro-optic switch, used as (1 × 2) and (2 × 1) switches at the input and output columns, respectively. 2.15 Using the results on gain saturation in EDFA (Eq. 2.119), estimate the decrease in EDFA gain when its input power increases from -20 dBm to -10 dBm. Given: Q = 10 dBm. Discuss how the gain saturation in EDFAs can affect the performance of an optical link in a WDM network and suggest some possible remedy. Solution: Using Eq. 2.119, EDFA gains with the input power levels of -20 dBm and -10 dBm are estimated as 27.5 dB and 22.4 dB respectively, implying that the gain falls by 27.5 – 22.4 = 5.1 dB with the 10 dB increase in the input power level, thereby affecting the optical SNR at the destination receivers. Gain control: By using a separate wavelength in each fiber link and controlling its own power, one can bring down the variation in the total optical power (from all wavelengths) handled by each EDFA along the fiber links across the network.

2.6

Exercise Problems and Solutions for Chapter 3 (Optical LANs/MANs and SANs) 3.1 Discuss with suitable illustrations why the fiber-optic bus or ring topology cannot use traditional one-packet-at-a-time MAC protocols, such as CSMA/CD or token-ring protocol, as used in IEEE standards for copper-based LANs. Also indicate the difficulty in using CSMA/CD protocol in passive-star-based optical LANs. Solution: In optical LANs, the packet sizes shrink with high speed while the network size increases. This leaves a large space (and hence bandwidth) in the optical fiber as unused resource. Further, in bus, the progressive power losses of each packet along the optical fiber, makes collision detection difficult when the packets overlap with dissimilar power levels. Passive-star-based optical LANs don’t suffer from the progressive loss, but the ratio of the packet duration to the round-trip propagation delay becomes much smaller with large network size. 3.2 Obtain an expression for the progressive power loss in an optical fiber bus connecting N nodes to set up a LAN, with each node tapping power from and transmitting (coupling) power into the fiber bus using a transmit-receive coupler (Fig.3.2). Using this formulation, evaluate the dynamic range needed in the receivers used for the network nodes connected to the fiber bus with N = 5 and 10. If the receiver dynamic range cannot be more than 10 dB, comment on the feasibility of the LANs in the two cases. Given: connector loss = 1 dB, power tapping for the receiver in the transmit-receive coupler = 10%, insertion loss in the waveguide of each transmit-receive coupler = 0.3 dB, loss at the transmitter-coupling point in each transmit-receive coupler = 0.2 dB, fiber loss = 0.2 dB/km, distance between two adjacent nodes = 1 km. Solution: RX

TX Each node employs a passive transmit-receive coupler, as shown above, with two taps for transmit and receive operations, and four connectors on input/output and transmit/receive ports. The necessary parameters associated with the nodes and the network are defined in the following for the analysis. • • • • • • • • •

Ptx: transmitted power from each node in mW (including the loss of isolator/circulator). d: distance between two successive nodes in km. All nodes are assumed to be equispaced. lf : loss in a fiber segment (d km) in neper/km between each adjacent node pair with, lf = exp(αf d) and lf (dB) = 10 log10(lf ). lc: connector loss (ratio). lcin: insertion loss (loss) inside the coupler. lrtap: tapping ratio for the receiving port. ltxc: loss (ratio) at the transmitting port. lrxc: loss (ratio) at the receiving port due to power tapping (= 1 – lrtap). ltcp: loss (ratio) while coupling the optical signal from the transmitter into the transmit-receive coupler

The power at node 1 output is given by 𝑃𝑃1 = 𝑃𝑃𝑡𝑡𝑡𝑡 𝑙𝑙𝑐𝑐 𝑙𝑙𝑡𝑡𝑡𝑡𝑡𝑡 𝑙𝑙𝑐𝑐 = 𝑃𝑃𝑡𝑡𝑡𝑡 𝑙𝑙𝑐𝑐2 𝑙𝑙𝑡𝑡𝑡𝑡𝑡𝑡 ,

and the power received at node 2 from node 1 is given by 3.1

𝑃𝑃𝑅𝑅12 = 𝑃𝑃𝑡𝑡𝑡𝑡 𝑙𝑙𝑐𝑐2 𝑙𝑙𝑡𝑡𝑡𝑡𝑡𝑡 × 𝑙𝑙𝑓𝑓 𝑙𝑙𝑐𝑐 𝑙𝑙𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟 𝑙𝑙𝑐𝑐 = 𝑃𝑃𝑡𝑡𝑡𝑡 𝑙𝑙𝑐𝑐4 𝑙𝑙𝑡𝑡𝑡𝑡𝑡𝑡 𝑙𝑙𝑓𝑓 𝑙𝑙𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟 .

Following the same approach, we obtain the power received at node N as the power received at the input of node N multiplied by 𝑙𝑙𝑐𝑐2 𝑙𝑙𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟 . Moreover, the power loss at each transmit-receive coupler is given by 𝑙𝑙 𝑇𝑇𝑇𝑇𝑇𝑇 = 𝑙𝑙𝑐𝑐2 (𝑙𝑙𝑟𝑟𝑟𝑟𝑟𝑟 𝑙𝑙𝑡𝑡𝑡𝑡𝑡𝑡 𝑙𝑙𝑐𝑐𝑖𝑖𝑛𝑛 ), and the optical signal from node 1 to node N has to pass through N-2 transmit-receive couplers with this loss. Therefore, we obtain the power received at the receiver of node N as 𝑁𝑁−1 2 𝑃𝑃𝑅𝑅1𝑁𝑁 = 𝑃𝑃𝑡𝑡𝑡𝑡 𝑙𝑙𝑐𝑐2 𝑙𝑙𝑡𝑡𝑡𝑡𝑡𝑡 𝑙𝑙 𝑁𝑁−2 𝑙𝑙𝑐𝑐 𝑙𝑙𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟 . 𝑇𝑇𝑇𝑇𝑇𝑇 𝑙𝑙𝑓𝑓

Substituting for lTRC in the above expression, we obtain

𝑃𝑃𝑅𝑅1𝑁𝑁 = 𝑃𝑃𝑡𝑡𝑡𝑡 𝑙𝑙𝑐𝑐2 𝑙𝑙𝑡𝑡𝑡𝑡𝑡𝑡 {𝑙𝑙𝑐𝑐2 (𝑙𝑙𝑟𝑟𝑟𝑟𝑟𝑟 𝑙𝑙𝑡𝑡𝑡𝑡𝑡𝑡 𝑙𝑙𝑐𝑐𝑐𝑐𝑐𝑐 )}𝑁𝑁−2 𝑙𝑙𝑓𝑓𝑁𝑁−1 𝑙𝑙𝑐𝑐2 𝑙𝑙𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟

𝑃𝑃𝑅𝑅1𝑁𝑁 = 𝑃𝑃𝑡𝑡𝑡𝑡 𝑙𝑙𝑐𝑐2𝑁𝑁 𝑙𝑙𝑡𝑡𝑡𝑡𝑡𝑡 𝑙𝑙𝑓𝑓𝑁𝑁−1 (𝑙𝑙𝑟𝑟𝑟𝑟𝑟𝑟 𝑙𝑙𝑡𝑡𝑡𝑡𝑡𝑡 𝑙𝑙𝑐𝑐𝑐𝑐𝑐𝑐 )𝑁𝑁−2 𝑙𝑙𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟 .

or,

From the above results, the dynamic range DR of the bus can be expressed as 𝑃𝑃𝑡𝑡𝑡𝑡 𝑙𝑙𝑐𝑐4 𝑙𝑙𝑡𝑡𝑡𝑡𝑡𝑡 𝑙𝑙𝑓𝑓 𝑙𝑙𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟 𝑃𝑃𝑅𝑅12 � 𝐷𝐷𝐷𝐷 = 10log10 � 1𝑁𝑁 � = 10log10 � 𝑃𝑃𝑅𝑅 𝑃𝑃𝑡𝑡𝑡𝑡 𝑙𝑙𝑐𝑐2𝑁𝑁 𝑙𝑙𝑡𝑡𝑡𝑡𝑡𝑡 𝑙𝑙𝑓𝑓𝑁𝑁−1 (𝑙𝑙𝑟𝑟𝑟𝑟𝑟𝑟 𝑙𝑙𝑡𝑡𝑡𝑡𝑡𝑡 𝑙𝑙𝑐𝑐𝑐𝑐𝑐𝑐 )𝑁𝑁−2 𝑙𝑙𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟

= 10log10 �

1

𝑁𝑁−2 � �𝑙𝑙𝑐𝑐2 𝑙𝑙𝑓𝑓 𝑙𝑙𝑟𝑟𝑟𝑟𝑟𝑟 𝑙𝑙𝑡𝑡𝑡𝑡𝑡𝑡 𝑙𝑙𝑐𝑐𝑐𝑐𝑐𝑐 �

𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 = (𝑁𝑁 − 2)�2𝑙𝑙𝑐𝑐𝑑𝑑𝑑𝑑 + 𝑙𝑙𝑓𝑓𝑑𝑑𝑑𝑑 + 𝑙𝑙𝑟𝑟𝑟𝑟𝑟𝑟 + 𝑙𝑙𝑡𝑡𝑡𝑡𝑡𝑡 + 𝑙𝑙𝑐𝑐𝑐𝑐𝑐𝑐 �.

For N = 10, d = 1 km, αfdB = 0.2 dB/km, lcdB = 1 dB, ltxcdB= 0.2 dB, lcindB = 0.3 dB, and lrxc = 1 – lrtap = 1 – 0.1 = 0.9 with lrxcdB = - log10(0.9) = 0.46 dB, we obtain DR as 𝐷𝐷𝐷𝐷 = 8[2 × 1 + 0.2 + 0.46 + 0.2 + 0.3] = 8 × 3.16 = 25.28 dB,

which is not acceptable for the given optical receiver. However with N = 5, DR comes down to 3 × 3.16 = 9.48 dB ( > 1), i.e., the LAN should satisfy 𝜏𝜏 𝑇𝑇𝑇𝑇 (max) ≤

𝐷𝐷𝑀𝑀𝑀𝑀𝑀𝑀 𝐾𝐾

.

Using the above criterion, one can obtain the conditions needed for the respective PC-based singlehop WDM LANs as follows. Aloha/Aloha:

Slotted Aloha/Aloha:

𝜏𝜏 𝑇𝑇𝑇𝑇 (max) ≤

Slotted Aloha/M-server switch:

𝜏𝜏 𝑇𝑇𝑇𝑇 (max) ≤

1 2𝐺𝐺𝐺𝐺 𝑀𝑀 − 1 exp � �1 + ��. 𝐾𝐾 𝑀𝑀 𝐿𝐿 1 2𝐺𝐺𝐺𝐺 𝑀𝑀 − 2 exp � �1 + ��. 𝐾𝐾 𝑀𝑀 2𝐿𝐿 1

𝜏𝜏 𝑇𝑇𝑇𝑇 (max) ≤ 𝐾𝐾

exp (𝐺𝐺)

[𝐿𝐿𝐿𝐿 exp (−G )]𝑀𝑀 /𝑀𝑀 ! 1− 𝑀𝑀 ∑𝑖𝑖=0 [𝐿𝐿𝐿𝐿 exp (−G )]𝑖𝑖 /𝑖𝑖!

.

Note that, the above conditions need to be fulfilled for the entire range of G expected in the LAN. Moreover, a large value for K (say 10) might make the design infeasible for the LANs, and in such cases, the used optical spectrum can be divided into multiple segments, and each segment can be assigned one tunable laser and filter in each node.

6.2

6.3 Consider a spanning tree representing the minimum-hop paths from a source to all the destinations in a multihop 31-node WDM-LAN with a logical nodal degree of two. Evaluate the full-tree height and the expected number of hops. Solution: The number N of nodes in a spanning tree with a nodal degree p and a height H is given by 2𝐻𝐻+1 − 1 𝑁𝑁 ≤ , 𝑝𝑝 > 1. 𝑝𝑝 − 1 Hence with N = 31 and p = 2, we obtain H as 2𝐻𝐻+1 − 1 31 ≤ , 2−1 leading to the estimate of H as 𝐻𝐻 = ⌈𝑙𝑙𝑙𝑙𝑔𝑔2 (31 + 1) − 1 ⌉ = 4. The expected number of hops is therefore obtained as 2 − 24+1 + 32 × 4 + 4 𝐸𝐸[ℎ] = = 3.29. 32 − 1

6.4 Consider a six-node passive-star multihop WDM LAN. For this network, form a unidirectional logical ring topology and determine the required number of wavelengths. Estimate the expected number of hops needed in this network to communicate between any two nodes. If each wavelength operates at 10 Gbps, evaluate the network throughput in Gbps. Solution: The 6-node multihop WDM LAN with the unidirectional (clockwise) ring as its logical topology is shown in the following. 1 6

2

3

5 4

With 6 nodes, the LAN needs M = 6 wavelengths, wherein the connections are set up with node 1 as the source node (note that hop requirements will be the same for each node due to the circular topology). The number of hops (h) for each connection would be as follows • Node 1 to 2: h = 1 • Node 1 to 3: h = 2 • Node 1 to 4: h = 3 • Node 1 to 5: h = 4 • Node 1 to 6: h = 5. Hence, the expected number of hops is obtained as 1+2+3+4+5 = 3. 𝐸𝐸[ℎ] = 6−1 Using E[h], we obtain the network throughput in Gbps (for 10 Gbps as the transmission speed) as 6.3

𝑀𝑀 × 10 Gbps 6 × 10 = = 20 Gbps 𝐸𝐸[ℎ] 3 6.5 Consider a (p,k) Shufflenet with p = 3, k = 2. a) Draw the (3,2) Shufflenet topology using the node numbering convention used for self routing. b) How many wavelengths should this network have to set up concurrent transmissions for all possible connections? c) Find out the number of hops from node 0 to all the other nodes in the network. d) Calculate the expected number of hops in the network. e) Trace a shortest-path route from node 0 to the highest-numbered node in the network. How many such shortest-path routes exist? Identify all of these paths. 𝜂𝜂 =

Solution: a) The number of nodes in the (3,2) ShuffleNet is N = kpk = 2×32 = 18. So the first column of the ShuffleNet has 9 nodes numbered from top to bottom as (using p = 3 as the radix for the row numbering): (1,00), (1,01), ... , (1,22). Similarly the 9 nodes in the second column are numbered as: (2,00), (2,01), ... , (2,22), while the wrapped-up last column is the replica of the first column. Using this numbering scheme, sketch the interconnection for the ShuffleNet as illustrated in Section 6.4.1. b) The number of wavelengths M = kpk+1 = 2×33 = 54. c) The numbers of hops needed to reach from node 0 to all other nodes are presented below in tabular form. h = 1: Route from node 0 Number of hops (h) Node 0 → 9, 10,11 1 h = 2: Route from node 0 Node 0 → 9 → 1,2 Node 0 → 10 → 3,4,5 Node 0 → 11 → 6,7,8

Number of hops (h) 2 2 2

Route from node 0 Node 0 → 9 → 1 → 12,13,14 Node 0 → 9 → 2 → 15,16,17

Number of hops (h) 3 3

Alternate paths with h = 3: Route from node 0 Node 0 → 10 → 3 → 9,10,11 Node 0 → 10 → 4 → 9,10,11 Node 0 → 10 → 5 → 12,13,14 Node 0 → 11 → 6 → 12,13,14 Node 0 → 11 → 7 → 15,16,17 Node 0 → 11 → 8 → 15,16,17

Number of hops (h) 3 3 3 3 3 3

h = 3:

d) The expected number of hops E[h] is obtained as 6.4

𝑘𝑘𝑝𝑝𝑘𝑘 (𝑘𝑘 − 1)(3𝑘𝑘 − 1) − 2𝑘𝑘(𝑝𝑝𝑘𝑘 − 1) 𝐸𝐸[ℎ] = 2(𝑝𝑝 − 1)(𝑘𝑘𝑝𝑝𝑘𝑘 − 1) =

2×32 (3−1)(3×2−1)−2×2�32 −1� 2(3−1)(2×32 −1)

= 2.1765.

e) The shortest path between node 0 to node 17 is: 0 → 9 → 2 → 17. (other options: 0 → 11→ 7 → 17 and 0 → 11 → 8 → 17).

6.6 Consider a (3,4,5) GHCNet and find out the following: a) number of nodes in the network, b) number of wavelengths in the network for collision-free transmission, c) logical nodal degree, d) nodes reachable in one hop from node (3,2,1) (in mixed-radix representation), e) path from node (0,0,0) to (4,3,2) using the GHCNet routing algorithm, f) expected number of hops for this network. Solution: (3,4,5) GHCNet: a) Number of nodes N: 𝑟𝑟

𝑁𝑁 = � 𝑚𝑚𝑘𝑘 = 3 × 4 × 5 = 60. 𝑘𝑘=1

b) Logical nodal degree p: 𝑟𝑟

𝑝𝑝 = �(𝑚𝑚𝑘𝑘 − 1) = (3 − 1) + (4 − 1) + (5 − 1) = 9. 𝑘𝑘=1

c) Number of wavelengths M:

𝑀𝑀 = 𝑁𝑁𝑁𝑁 = 60 × 9 = 540. d) Nodes reachable in one hop from node (3,2,1): i. Decrease or increase the third digit: (3,2,1) → (3,2,0), (3,2,2). ii. Decrease or increase the second digit: (3,2,1) → (3,0,1), (3,1,1), (3,3,1). iii. Decrease or increase the first digit: (3,2,1) → (0,2,1), (1,2,1), (2,2,1), (4,2,1). e) Path from node (0,0,0) to (4,3,2): (0,0,0) → (0,0,2) → (0,3,2) → (4,3,2). f) Expected number of hops E[h]: ∑𝑟𝑟ℎ=1 ℎ𝑛𝑛ℎ . 𝑁𝑁 − 1 In the given GHCNet, r = 3. Hence, we need to evaluate n1, n2 and n3. 𝑛𝑛1 = ∑3𝑖𝑖=1(𝑚𝑚𝑖𝑖 − 1) = (3 − 1) + (4 − 1) + (5 − 1) = 9. Similarly, 𝐸𝐸[ℎ] =

𝑛𝑛2 = ∑3𝑖𝑖=1 ∑3𝑗𝑗 ≠𝑖𝑖=1(𝑚𝑚𝑖𝑖 − 1)�𝑚𝑚𝑗𝑗 − 1� = 26,

3

Hence,

3

𝑛𝑛3 = � �

3

�

(𝑚𝑚𝑖𝑖 − 1)�𝑚𝑚𝑗𝑗 − 1�(𝑚𝑚𝑘𝑘 − 1) = 24.

𝑖𝑖=1 𝑗𝑗 =1 𝑘𝑘=1(𝑖𝑖≠𝑗𝑗 ≠𝑘𝑘)

6.5

𝐸𝐸[ℎ] =

∑𝑟𝑟ℎ=1 ℎ𝑛𝑛ℎ 1 × 9 + 2 × 26 + 3 × 24 = = 2.2542. 𝑁𝑁 − 1 60 − 1

6.7 Consider a (2,4) dBNet and draw its topology. Find out the following: a) number of nodes in the network, b) number of wavelengths in the network for collision-free transmission, c) logical nodal degree, d) route between two extreme nodes according to the self-routing algorithm, e) number of hops from node (0000) to all other nodes, f) expected number of hops for this network. Solution: (p,D) dBNet with p = 2, D = 4. a) Number of nodes N = pD = 16. b) Number of wavelengths M = Np – p = 32 – 2 = 30. c) Logical nodal degree is p + 1 for N – p nodes, and p for p nodes. Hence 14 nodes have a logical degree of 3 and 2 nodes have a logical degree of 2 due to self-looping. d)Route between two extreme nodes (node (0,0,0,0) and node (1,1,1,1): Using the routing scheme (shift and merge), we get Adds Shift Merge Routing 0000 0000 1111 1111 00001111 0000 → 0001 → 0011 → 0111 → 1111. e) Hop counts from node (0000) : 1 hop 0000 to 0001: Adds Shift Merge Routing 0000 0000 0001 0001 00001 0000 → 0001. 2 hops 0000 to 0010: Adds Shift 0000 0000 0010 0010 0000 to 0010: Adds Shift 0000 0000 0011 0011

Merge

Routing

000010

0000 → 0001. → 0010.

Merge

Routing

000011

0000 → 0001. → 0011..

Similarly, for 3 and 4 hops, we obtain the following results: 3 hops 6.6

0000 to 0101: 0000 → 0001→ 0010 → 0101, 0000 to 0110: 0000 → 0001→ 0011 → 0110, 0000 to 0111: 0000 → 0001→ 0011 → 0111. 4 hops 0000 to 1000: 0000 → 0001→ 0010 → 0100 → 1000, and so on for the remaining 7 (below) 0000 to 1001 0000 to 1010 0000 to 1011 0000 to 1100 0000 to 1101 0000 to 1110 0000 to 1111 f) Expected number of hops E[p] is obtained as 𝐸𝐸[ℎ] =

𝑝𝑝 − 𝑝𝑝𝐷𝐷+1 + 𝑝𝑝𝐷𝐷 𝐷𝐷(𝑝𝑝 − 1)2 + 𝐷𝐷(𝑝𝑝 − 1) 2 − 24+1 + 24 × 4(2 − 1)2 + 4(2 − 1) = = 2.5333. (𝑝𝑝𝐷𝐷 − 1)(𝑝𝑝 − 1)2 (24 − 1)(2 − 1)2

6.8 Draw the topology of a (2,3) ShuffleNet with the number of nodes starting from 0. For this network, a) trace the route from node 0 to node 23 following the self-routing scheme of ShuffleNet, b) if both the outgoing links from the nodes 9 and 11 fail, then check whether node 23 would be reachable from node 0 - justify your answer by tracing the routes, c) mention one possible situation, when node 18 would be unreachable from node 0 and justify your answer from the topology. Solution: a, b) The (2,3) ShuffleNet with column-row-based addressing for the 24 nodes (0 to 23) is shown below: Columns → 0 1 2 0 • Following routes/paths are shown by the respective Rows ↓ --------------------------arrows in the network diagram on the left side. 000 | 0 8 16 0 • Actual path obtained using the self-routing scheme based on the column-row-based addresses of the 001 | 1 9 17 1 nodes: 010 | 2 10 18 2 0 → 9 → 19 → 7 → 15 → 23 011 | 3 11 19 3 • If outgoing links of node 9 are down, still there will 100 | 4 12 20 4 exist the following alternate shortest paths: 101 | 5 13 21 5 0 → 8 → 17→ 3 → 15 → 23, 110 | 6 14 22 6 0 → 8 → 16→ 1 → 11 → 23. 111 | 7 15 23 7 c) Having drawn the full topology of the network, it becomes evident that, node 18 will become unreachable from node 0, if the outgoing links from the nodes 9 and 13 are down.

6.7

Exercise Problems and Solutions for Chapter 7 (WDM Access Networks) 7.1 Consider three types of PONs using WDM transmission: (a) WDM PONs with single-stage RN using AWG, (b) TWDM PON using two stages of RN with AWG as the first-stage RN and multiple optical splitters at the second stage, and (c) TWDM PON using splitters only at RNs. Make an exhaustive comparison between these three options in respect of the power budget, cycle time, and hardware complexity of the OLT/ONUs. Solution: See the discussions in Sections 7.2, 7.3, and 7.9. 7.2 A long-reach WDM PON has to be set up to serve 64 ONUs over a span of 100 km by using AWG-based RN. If the transmit power of the optical source in the OLT is 3 dBm for each wavelength and the per-channel receiver sensitivity of the ONUs is -28 dBm for 1 Gbps downstream transmission, examine the feasibility of the power budget for downstream connection. Given: fiber loss = 0.2 dB/km, AWG insertion loss = 4 dB, connector loss = 1 dB. Solution: OLT transmit power PT = 3 dBm, receiver sensitivity PR (min) = -28 dBm, fiber loss = 0.2 dB/km, AWG loss lawg = 4 dB, and connector loss lc = 1 dB. Power budget: PR = PT – (100 × αf + lawg + 4 × lc) = 3 – (100 × 0.2 + 4 + 4 × 1) dBm = 3 – (20 + 4 + 4) = - 25 dBm > PR (min) (= -28 dBm). Hence, the power budget is feasible, though there will be some variation of power at the AWG output ports, which should be taken care of by the dynamic range of the receivers at the ONUs. 7.3 A TWDM PON has to be set up to serve 64 ONUs over a span of 40 km at 1 Gbps using 8 wavelengths, and the network designer is given two options: (a) a static TWDM PON using AWGOSP configuration, and (b) dynamic TWDM PON using OSP-only configuration. The OSPs are realized with multiple stages of 2 × 2 optical splitters as building block, with each of them having an insertion loss of 0.3 dB. If the transmit power of the optical source at the OLT is 3 dBm over each wavelength and the per-channel receiver sensitivity of the ONUs is -28 dBm, examine the feasibility of the downstream power budget in each case. Given: fiber loss = 0.2 dB/km, AWG insertion loss = 4 dB, connector loss = 1 dB. Solution: Number of ONUs N = 64, maximum distance L = 40 km, transmit power PT = 3 dBm, receiver sensitivity PR (min) = -28 dBm, bit rate r = 1 Gbps, connector loss lc = 1 dB, AWG insertion loss = 4 dB. Power budget with AWG and power splitter: Power splitter loss = 10 log10N + 0.3 × log2N. The number of wavelengths M = 8, and hence the AWG (with the insertion loss of 4 dB) needs 1:8 configuration, with each wavelength needing 64/8 = 8 splitters, each with a split ratio of 1:8. Between the OLT and each ONU, 6 connectors are needed with a loss of 6lc = 6 dB. Hence, the power budget is obtained as PR = PT – [40 × 0.2 + 4 + (10log108 + 0.3 × log28) + 6] dBm = PT - [8 + 4 + 9 + 0.9 + 6] dBm = 3 – 27.9 = -24.9 dBm,

7.1

which is larger than the receiver sensitivity (= - 28dBm) with a 3.1 dB as system margin. Hence, this design offers a feasible power budget. Power budget with two stages of power splitters: In this case, end-to-end connection between the OLT and each ONU requires 6 connectors with a loss of 6 dB. Hence, we obtain PR as PR = PT – [40 × 0.2 + (10 × log108 + 0.3 × log28) + (10 × log108 + 0.3 × log28) + 6] = 3 – [8 + (9 + 0.9) + (9 + 0.9) + 6] = 3 - 33.8 = -30.8 dBm, which is less than the receiver sensitivity. Hence this design doesn’t offer a feasible power budget. 7.4 Consider a static TWDM PON with AWG-OSP configuration. The PON transmits at 1 Gbps (both ways) and supports 128 ONUs over 40 km with a fixed grant size of 10,000 bytes with a guard time of 1 μs. There are two options for the PON design in respect of the AWG and OSPs: a) 1 × 16 AWG, 1 × 8 OSP, b) 1 × 8 AWG, 1 × 16 OSP. Determine the cycle time of a TDMA group for each design. Compare the two designs. Given: propagation time in optical fiber = 5 μs/km. Solution: Maximum distance L = 40 km, grant size G =10,000 bytes, bit rate r = 1 Gbps RTT = 40 × 2 × 5 = 400 μs, guard time τg = 1 μs. a) 16 × 8 (1:16 AWG, 1:8 power splitter) Number of wavelengths M = 16, number of ONUs in each TDM group = 8 Cycle time TC = RTT + 8τg + 8G/r = 400 × 10-6 + 8 × 10-6 + 8 × 80,000 × 10-9 s = 400 + 8 + 640 = 1048 μs = 1.048 ms. b) 8 × 16 (1:8 AWG, 1:16 power splitter) Number of wavelengths M = 8, number of ONUs in each TDM group = 16 Cycle time TC = RTT + 16τg + 16G/r = 400 × 10-6 + 16 × 10-6 + 16 × 80,000 × 10-9 sec = 400 + 16 + 1280 = 1696 μs = 1.696 ms. In case (a), the power budget is (obviously) better, as well as, the cycle time is smaller, however needing AWG. In case (b), more flexible DBA scheme can be realized due to larger number of ONUs sharing each wavelength, thereby offering higher bandwidth utilization. 7.5 Consider the design of a 128 ONU TWDM PON using two stages of RN using OSPs only. The TWDM PON needs to cover a distance of 40 km and transmit at 1 Gbps both ways, with a transmit power of 3 dBm. The receivers have a sensitivity of -28 dBm while operating at 1 Gbps, and the OLT allocates a fixed grant size of 10,000 bytes for all ONUs and uses a guard time of 1 μs. If the design needs to ensure that the cycle time for the TDMA groups of ONUs remains confined within 1.5 ms, then estimate the number of wavelengths to be used and check whether any EDFA is needed to realize a feasible power budget. Given: fiber loss = 0.2 dB/km, connector loss = 1 dB. Solution: The design uses splitters only, while the cycle time TC < 1∙5 ms. The number of ONUs N = 128, distance L = 40 km, and receiver sensitivity PR (min) = - 28dBm, fiber loss αf = 0.2 dB/km, connector loss lc = 1 dB. For the ease of deployment (fiber laying), we assume that, the 1:128 splitting would be realized in two stages, e.g, with M × K = 128, with the second stage of splitting having K splitters. This configuration 7.2

will divide the ONUs into M groups, each group of K ONUs sharing typically one of the M wavelengths. However, large K will lead to long cycle time. Thus, one needs to design judiciously the values for K and M. Cycle time in each group of ONUs: Grant size duration τ = 10,000 × 8 × 10-9 s = 80 μs. Hence, the cycle time is expressed as TC ≥ RTT + Kτg + Kτ. with τg as the guard time. Hence, 1500 ≥ 400 + K × 1 + K × 80 or, K ≤ 1100/81 = 13.58 → 13, thereby needing 1:16 splitters at the second stage (with 2n-splitting using 2 × 2 splitters as the building block), with two ports remaining unused. With K = 13, we therefore obtain 𝑀𝑀 = �

128 � 13

= 10.

Hence, the first-stage splitter will also employ the nearest available, i.e., the 1:16 splitter configuration, wherein 6 ports will remain unused. Thus, with both stages of power splitting needing the 1:16 configuration, overall power-splitting will increase from 1:128 to 1:162 (= 1:256), due to the unused ports in both stages, thereby making the power budget more critical. Power budget: We obtain the power budget with total splitting ratio of 1:256 (with 6 connectors) as PR = PT – (αfL + 6lc + 10log10256 + log2256 × 0.3) = 3 – [0.2 × 40 + 6 × 1 + (24 + 2.4)] = -37.4 dBm. Hence, the TWDM PON needs to employ EDFAs in both directions with a gain Gedfa given by Gedfa ≥ {-28 – (-37.4)} + PM = 9.4 dB + PM, with PM as the system power margin. With a PM ≈ 5 dB, an EDFA with a gain of 15 dB would make the design feasible in the physical layer. 7.6 A WDM PON uses one 1:N AWG, which has a non-uniform power distribution P(θ) across its output ports, following a Gaussian distribution, given by 𝑃𝑃(𝜃𝜃) =

1

exp �

−θ2 �, 2σ2

√2𝜋𝜋𝜎𝜎 2 where θ is the angle made by the output ports with respect to the longitudinal axis passing through the input port. The angle θ takes up discrete values across the output ports and σ is determined by the overall AWG construction. Determine the expression for the power ratio PR (in dB) of the signal power levels at the central output port (with N assumed to be odd) and the outermost output port on one of two sides. Using this result, the network provider decides to connect the farthest ONU to the central AWG port, while the closer ONUs are connected to the outer ports of the AWG. If dmax and dmin are the distances of the farthest and closest ONUs from the RN respectively, determine the expression for the revised PR (as seen at the ONUs) with the above connection strategy for the ONUs. Solution: Power received at the outermost output port of AWG (i.e., θ = θmax) expressed as 𝑃𝑃(𝜃𝜃𝑚𝑚𝑚𝑚𝑚𝑚 ) =

1

√2𝜋𝜋𝜎𝜎 2

−θ2max 2σ2

exp �

�.

Similarly, the power received at the central port with θ = 0 is given by 𝑃𝑃(0) =

1

√2𝜋𝜋𝜎𝜎 2

.

The ratio of the two power levels at the central and outermost ports can be expressed in dB as PR, given by

7.3

θ2max 𝑃𝑃(0) 𝑃𝑃𝑃𝑃 = 10log10 = 10log10 exp � 2 �. 𝑃𝑃(𝜃𝜃𝑚𝑚𝑚𝑚𝑚𝑚 ) 2σ

Hence, the ONU receivers should have a dynamic range DR in dB, such that the combined effect of PR and the varying distances of ONUs from the OLT keeps the received-power ratio at ONUs below the receiver DR. In order to address this issue, the farthest ONU (at a distance d = dmax) from the AWG is connected to the central AWG output port and the nearest ONU (at a distance d = dmin) is connected to the outermost AWG output port (one of the two on two sides), such that the receivedpower ratio PR′ at the ONUs gets reduced as dmax �� 𝑃𝑃𝑃𝑃 ′ = �𝑃𝑃𝑅𝑅 − 10log10 � dmin θ2

d

max min = �10log10 exp �� 2σ ��� 2 � �d

with α = dmin/dmax.

=

αθ2max �10log10 exp �� 2σ 2

max

���,

7.7 Develop a pseudo code to simulate the EFT and EFT-VF schemes for DWBA in a dynamic TWDM PON with N ONUs and compare the performance of the two schemes using computer simulation for a set of practical network specifications. Solution: Consider Poisson arrivals of data traffic at ONUs from the user-ends and implement the respective DWBA schemes (EFT and EFT-VF) as described in Section 7.9.5 for a TWDM PON with typical system/device parameters.

7.4

Exercise Problems and Solutions for Chapter 8 (WDM Metro Networks)

8.1. Discuss using an appropriate layered architecture, how the packet- and circuit-switched traffic flows are circuit-multiplexed in SONET-over-WDM ring networks. What is the limitation of the circuit-switched transmission of packet-switched traffic and why does this practice continue to exist? Solution: See the discussions in Section 8.2. 8.2. Consider an N-node mesh-configured WRON having bidirectional fiber links with a given set of lightpaths, where a node pair is allocated only one lightpath in each direction. Assume that a cut-set Ci bisects the network into two parts with N-ni and ni nodes. If the cut-set consists of fi fiber pairs and li lightpaths, then find out the expression for the lower-bound for the number of wavelengths Mmin needed in the entire network. Using this model, show that for an N-node ring (i.e., two-connected mesh) with N as an odd integer, Mmin = (N2 – 1)/8. Solution: The N-node WRON mesh is bisected into two parts with N-ni and ni nodes with a cut-set Ci, as shown below.

N - ni nodes

ni

...

nodes

Cut-set Ci Cut-set Ci consists of fi fiber pairs and li lightpaths. Hence to maintain the connectivity between the two sides, one needs at least M wavelengths, given by 𝑙𝑙𝑖𝑖 𝑀𝑀 ≥ . 𝑓𝑓𝑖𝑖

Hence, the lower-bound on M for all possible cut-sets is obtained as 𝑙𝑙𝑖𝑖 𝑀𝑀𝐿𝐿𝐿𝐿 = �𝑚𝑚𝑚𝑚𝑚𝑚𝑖𝑖 �. 𝑓𝑓𝑖𝑖

For a WDM ring with N as an odd number, (N-1)/2 and (N+1)/2 are almost the half of N, with fi = 2. Hence, we obtain MLB for the WDM ring as 𝑁𝑁 2 − 1 1 (𝑁𝑁 − 1) (𝑁𝑁 + 1) 𝑀𝑀𝐿𝐿𝐿𝐿 = � � × . �� = 2 2 2 8

Note that, N being odd ≥ 3, N is also odd and hence N -1 is always divisible by 8. 2

2

8.1

8.3. Repeat the above exercise for WRON rings with even number of nodes and determine the lowerbound for the number of wavelengths in the network. Solution: Following the same approach as in the previous exercise, we divide the network into two sets for N as an even number, each having N/2 nodes. Hence, we express MLB as 𝑁𝑁 2 1 𝑁𝑁 𝑁𝑁 𝑀𝑀𝐿𝐿𝐿𝐿 = � � × �� = � �. 8 2 2 2

In this case, the ceiling function is needed to estimate MLB as N 2 is not always divisible by 8.

8.4. Consider a six-node bidirectional ring with the following traffic matrix (integer multiples of OC12 connection). Γ=

077256 905976 740546 886049 375403 967540

Find out the numbers of ADMs and wavelengths by using a multihop heuristic design to support the traffic demand with a grooming capacity of 4 in each wavelength. Solution: Use the heuristic method presented in Section 8.3.3 to design the WDM ring. 8.5. Repeat the above exercise using a multihop LP-based method. Compare the results obtained from the LP formulation and the heuristic scheme. Solution: Use the LP-based design presented in Section 8.3.3 to design the WDM ring, and compare the results from the heuristic and LP-based methods, as illustrated using the case-study results (Fig. 8.14).

8.6. Consider that a WRON-based ring with five nodes employs single-hub configuration, where the numbers of OC-48 lightpaths needed from the various nodes to the hub node have been estimated from the traffic matrix for the WRON as follows (node 0 is the hub): a) b) c) d)

from node 4 to hub: 2, from node 3 to hub: 4, from node 2 to hub: 1, from node 1 to hub: 3.

Determine the number of wavelengths needed in the network. Also calculate the number of OC-48 transceivers to be provisioned in the hub.

8.2

Solution: Consider the 5-node single-hub WRON ring, as shown below with node 0 as the hub. Hub node

0 1

4 WRON ring

2

3

The number of lightpaths li0 needed from a node to the hub on either side of the hub are as follows: • • • •

l40 = 2, l30 = 4, l20 = 1, l10 = 3.

Hence, the number of OC-48 lightpaths needed on each side of the ring (lightpaths on two sides of the hub being link-disjoint) can be expressed as the sum of the lightpaths needed from each node to the hub on one of the two sides. Thus, we obtain the number M of wavelengths necessary for the ring as 1 1 𝑀𝑀 = � (𝑙𝑙10 + 𝑙𝑙20 + 𝑙𝑙30 + 𝑙𝑙40 )� = � (3 + 1 + 4 + 2)� = 5. 2 2

The number of transceivers required in the hub is 2M = 10.

8.7. Consider a WRON-based ring with six nodes, whose virtual topology looks like a circle with several arcs as lightpaths representing the end-to-end connections between all non-adjacent node pairs. With this network-setting, sketch the schematic diagram of an OADM for the add-drop and passthrough operations. Solution: The 6-node WRON ring is shown below, wherein the dashed lines are used to show some end-to-end connections between non-adjacent nodes. However, all such connections are not shown to avoid 0 5

1 2

4 3

8.3

clumsiness. The number of nodes being even, we express the number M of wavelengths as 𝑀𝑀 = �

𝑁𝑁 2 𝑁𝑁 𝜌𝜌 + �� �, 8 4 𝑁𝑁 − 1

with ρ as the ingress traffic at each node (OADM), which is distributed uniformly between the

remaining (N-1) nodes. Assuming ⌈𝜌𝜌/(𝑁𝑁 − 1)⌉ = 1 (which can be increased proportionately corresponding to the actual traffic) and N = 6, we obtain M as 62

6 + � × 1 = 6. 8 4

𝑀𝑀 = �

Consider one half of the node operating for the clockwise traffic over the shortest paths. In order to construct the configuration of node 0 for the clockwise traffic, we consider the relevant transit as well as add/drop lightpaths (with shortest-path routing) as follows: 1. 2. 3. 4. 5. 6. 7. 8. 9.

node 4 to node 1 (transit lightpath on wavelength w1), node 5 to node 2 (transit lightpath on wavelength w2), node 5 to node 1 (transit lightpath on wavelength w3), node 3 to node 0 (drop lightpath on wavelength w4), node 4 to node 0 (drop lightpath on wavelength w5), node 5 to node 0 (drop lightpath on wavelength w6), node 0 to node 3 (add lightpath on wavelength w4), node 0 to node 2 (add lightpath on wavelength w5), node 0 to node 1 (add lightpath on wavelength w6).

Using the above add/drop/transit lightpaths, the clockwise part of the OADM at node 0 is constructed as shown below. Optical demux w1

w1,w2, ..., w6

Optical mux

w2 w3

w1,w2, ..., w6

w4 w5 w6 5-0 4-0 3-0

0-1

0-2

0-3

RX TX

RX TX

8.4

RX TX

8.8. In the six-node WRON ring of the above exercise, assume that the optical transmitters transmit 15 dBm of optical power at 2.5 Gbps for each lightpath and the receivers have a sensitivity of -23 dBm, and the ring has a circumference of 30 km with equal distances between the adjacent nodes. Trace the longest lightpath in the ring and check the feasibility of the power budget for the same. Given: connector loss = 1 dB, demultiplexer loss = 3 dB, multiplexer loss = 3 dB, fiber loss = 0.2 dB/km. Solution: As shown in the WRON ring in the previous exercise, one of the longest lightpaths runs from node 3 to node 0.

0 With equal distance between the adjacent nodes, the distance between any adjacent node pair is 30/6 = 5 km.

1 2 3

With the above observation, we therefore consider the lightpath from node 3 to node 0, and estimate its power budget. For this purpose, we enumerate in the following the losses incurred by the lightpath from node 3 to node 0. •

Loss at node 3: Connector loss = 2lc = 2 × 1 = 2 dB, optical multiplexer (OMUX) loss lmx = 3 dB, Hence, total loss = 2 + 3 = 5 dB. • Losses at nodes 2 and 1 (same loss at both nodes): Connector loss = 4lc = 4 dB, optical demultiplexer (ODMUX) loss ldm = 3 dB, OMUX loss lmx = 3 dB. Hence, total loss at node (and at node 2) = 4 + 3 + 3 = 10 dB. • Loss at node 0: Connector loss = 2lc = 2 dB, ODMUX loss ldm = 3 dB, Hence, total loss = 2 + 3 = 5 dB. Further, the transmit power PT = 15 dBm and loss in each 5-km link = 0.2 ×5 = 1 dB. We therefore express the received power PR as 𝑃𝑃𝑅𝑅 = 𝑃𝑃𝑇𝑇 − (losses in 4 nodes + losses in 3 links)

= 15 − {(5 + 10 + 10 + 5) + 3 × 1} = −18 dBm,

which is larger than the receiver sensitivity (= -23 dBm) with a 5-dB power margin. Hence, the design is feasible. 8.9. Discuss critically the realizability of the various packet-switched WDM metro networks, particularly in respect of their scalability of physical size, node hardware and latency.

8.5

Solution: See the discussion on packet-switched WDM rings in Section 8.4. 8.10. Consider a five-node packet-switched WDM metro ring based on HORNET architecture. With the HORNET-node configuration presented in Fig. 8.28, formulate the power budget for the data packets with a ring circumference of 15 km. If the transmit power from the lasers is 15 dBm and the receiver sensitivity for the data packets is -27 dBm at 1 Gbps, estimate the worst-case power budget and comment on its feasibility. Given: insertion and connecting loss in the input splitter (in addition to the 10% tapping loss) = 1 dB, insertion loss in FBG (including circulator) = 2 dB, insertion loss in the output combiner = 2 dB, coupling and insertion loss in the output combiner = 1.5 dB, connector loss = 1 dB. Assume equal distance between adjacent nodes. Solution: In order to examine the power budget of HORNET, we consider an example scenario as shown in the figure below. Circulator

Circulator Con

Splitter

Con

FBG Combiner

10%

Splitter

FBG Combiner

3 intermediate nodes (without details Add CP Drop HORNET node (destination node)

Add CP Drop HORNET node (source node)

The figure shows the source and destination nodes along with three intermediate nodes for the longest connection span in the 5-node ring, with the add and drop ports (shown in bold face) of the source and destination nodes, being under consideration for the transmission and reception, respectively, between a given farthest node pair. In the figure, CP represents the tapped signal for the control packets, FBG stands for fiber Bragg grating, Con represents the connectors, while Add and Drop ports are for the data packets. For evaluation of the path loss, we define the necessary parameters as follows: lf = fiber loss in dB/km, lc = connector loss, lcom = combiner loss, lcoup = coupling loss in combiner, lfbg = FBG loss (including circulator), lsp = 1 + 10log10(10/9) = 1.456 dB = splitter loss for passthrough signal including the connector losses integrated with the device. Using these physical-layer settings, we express the received power in the farthest node as (e.g., node 0 to node 3 with 3 intermediate nodes with 4 fiber links each with a length = 15/5 = 3 km) 𝑃𝑃𝑅𝑅 = 𝑃𝑃𝑇𝑇 − ��𝑙𝑙𝑐𝑐 + 𝑙𝑙𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 + 𝑙𝑙𝑐𝑐 � + 3�𝑙𝑙𝑐𝑐 + 𝑙𝑙𝑠𝑠𝑠𝑠 + 𝑙𝑙𝑓𝑓𝑓𝑓𝑓𝑓 + 𝑙𝑙𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 + 𝑙𝑙𝑐𝑐 � + 4 × 3 × 𝑙𝑙𝑓𝑓 + �𝑙𝑙𝑐𝑐 + 𝑙𝑙𝑠𝑠𝑠𝑠 + 𝑙𝑙𝑓𝑓𝑓𝑓𝑓𝑓 + 𝑙𝑙𝑐𝑐 ��

= 15 − [(1 + 2 + 1) + 3{1 + 1.456 + 2 + 2 + 1} + 12 × 0.2 + {1 + 1.456 + 2 + 1}]

= 15 − [4 + 3 × 7.456 + 2.4 + 5.456] = −19.224 dBm,

which is larger than the receiver sensitivity (= - 27 dBm), thereby making the design feasible with a system margin of 7.776 dB.

8.6

Exercise Problems and Solutions for Chapter 9 (WDM Long-haul Networks) 9.1 As illustrated in the example shown in Fig.9.1, sketch some other candidate topological interconnections between multiple WDM rings to form mesh-like topology and specify the requirements of the interconnecting nodes in the proposed interconnected-ring networks. Solution: MDR(4)

MDR(4)

MDR(4)

MDR(4)

MDR(4)

MDR(4)

MDR (3)

MDR(6)

MDR(5) MDR(3)

MDR(3)

MDR(5)

MDR(5) MDR(4)

MDR(4)

• • • •

Shaded boxes: multi-degree ROADMs with degree n (MDR(n)). White boxes: ROADMs with a degree of 2. Dashed lines: to interconnect separate rings. In-line EDFAs are needed when these links exceed a limit (≈ 60 km, typically). All fiber links are bidirectional using WDM transmission and all nodes (ROADMs) are capable of wavelength routing.

9.1

9.2 Consider an MD-ROADM used in Fig. 9.1 at node nAB and sketch its configuration using optical demultiplexers, multiplexers, and switches. Sketch alternate configurations for the same MDROADM by using (i) optical splitters and WSSs and (ii) WSSs only. Comment on the basic differences between these configurations. Solution: See the figures and the related discussions in Sections 2.16 and 9.2. 9.3 Draw the configuration for a MEMS-based OXC that can support three input and three output fibers with 16 wavelengths on each fiber, such that any four wavelengths from/to each fiber can be dropped/added through tunable transceivers (i.e., with logical nodal degree of four). Specify the sizes of the optical demultiplexers and multiplexers, optical switch (MEMS) and the number of tunable transceivers needed. Solution: 3 × 3 OXC with Δ = 4 (tunable transceivers), number of wavelengths M = 16:

ODMUX

w1,w2, ..., w16

w1,w2, ..., w16

w1,w2, ..., w16

OMUX

.. .

1 2

1 2

16

16

.. .

17 18

17 18

.. .

32

52 × 52 MEMS switch

32

33 34

33 34

48

48

49 50 51 52

49 50 51 52

4 tunable TXs

.. . .. . .. .

w1,w2, ..., w16

w1,w2, ..., w16

w1,w2, ..., w16

4 tunable RXs

9.4. Consider an N-node mesh-configured WRON offering a set C of connections {cs,d} with s and d representing the source and destination nodes for the connection cs,d. Assume that the connection cs,d employs l(s,d) fiber links. If the WRON has E bidirectional fiber links, determine the expression for the lower-bound on number of wavelengths needed in the network in terms of the number of nodes,

9.2

total inter-nodal fiber links used for the connections in C and the average physical degree of a node in the network. Solution: The N-node WRON mesh employs C = {cs,d} connections, each connection having l(s,d) fiber links. Hence, the total number of intermodal fiber links LT can be expressed as 𝐿𝐿 𝑇𝑇 = � 𝑙𝑙(𝑠𝑠, 𝑑𝑑). 𝑠𝑠,𝑑𝑑

On the other hand, the network with E edges has 2E links (each edge being bidirectional), and each link carries M wavelengths. Hence, total capacity of the network is 2EM, which turns out to be the upper-bound for LT, i.e., 2𝐸𝐸𝐸𝐸 ≥ 𝐿𝐿 𝑇𝑇,

leading to the lower limit for the number of wavelengths to support the network as 𝑀𝑀𝑚𝑚𝑚𝑚𝑚𝑚 =

𝐿𝐿 𝑇𝑇 ∑𝑠𝑠,𝑑𝑑 𝑙𝑙(𝑠𝑠, 𝑑𝑑) = . 2𝐸𝐸 2𝐸𝐸

Using E and N, the average physical degree Dav of the nodes in the WRON can be expressed as

Thus, Mmin can be expressed using Dav as

𝐷𝐷𝑎𝑎𝑎𝑎 =

𝑀𝑀𝑚𝑚𝑚𝑚𝑚𝑚 =

2𝐸𝐸 . 𝑁𝑁

∑𝑠𝑠,𝑑𝑑 𝑙𝑙(𝑠𝑠, 𝑑𝑑) . 𝐷𝐷𝑎𝑎𝑎𝑎 𝑁𝑁

9.5 Consider a 3 × 3 OXC for four wavelengths using MEMS in Spanke’s switch architecture and sketch the OXC configuration. Estimate the gain of an EDFA as the loss-compensating amplifier placed after the OXC. Given: connector loss = 1 dB, loss in optical demultiplexer/multiplexer = 3 dB, loss in a MEMS element = 1 dB. Solution: Consider the OXC shown in Fig. 2.80 with MEMS-based optical switch fabric using Spanke’s architecture, wherein we assume that there are 3 input/output ports operating with 4 wavelengths. Further, we define the loss parameters as follows: loxc = OXC loss, lc = connector loss, LM = MEMS loss in one stage (it uses two stages in Spanke’s architecture), ldm = demultiplexer loss, lmx = multiplexer loss. Using these definitions, we obtain the OXC loss for a passthrough lightpath as a sum of three components (demultiplexer, switch, and multiplexer units) as 𝑙𝑙𝑜𝑜𝑜𝑜𝑜𝑜 = (𝑙𝑙𝑐𝑐 + 𝑙𝑙𝑑𝑑𝑑𝑑 + 𝑙𝑙𝑐𝑐 ) + (𝑙𝑙𝑐𝑐 + 2𝑙𝑙𝑀𝑀 + 𝑙𝑙𝑐𝑐 ) + (𝑙𝑙𝑐𝑐 + 𝑙𝑙𝑚𝑚𝑚𝑚 + 𝑙𝑙𝑐𝑐 ) = 5 + 4 + 5 = 14 dB.

Hence the OXC should be followed by an EDFA with a gain of 14 dB to fully compensate for the OXC loss for a passthrough lightpath.

9.6 Consider an N-node WRON mesh with a given traffic matrix. The network vendor can afford ΔT transceivers for the entire network and decides to provision them for the nodes in accordance with the traffic handled by each node. Using this provisioning scheme, develop an expression for the number 9.3

of transceivers Δi for each node, while ensuring that each node has at least one transceiver to remain connected with the network. Solution: Each node in the WRON must have at least one incoming and one outgoing lightpath to remain connected with the rest of the network. Besides this constraint, the logical in/out degree of a node is made proportional to the traffic handled by the node. Hence, if ΔT represents the total number of transceivers available in the WRON, then (ΔT - 1× N) transceivers should be proportionately distributed to all nodes determined by the traffic handled therein. Assuming the network traffic matrix is mostly symmetric, and denoting λ(s,d) as the traffic from node s to node d, we therefore obtain the number Δi of transmitters/receivers in node i as ∆𝑖𝑖 = (∆ 𝑇𝑇 − 𝑁𝑁)

∑𝑑𝑑 𝜆𝜆(𝑖𝑖, 𝑑𝑑) + ∑𝑠𝑠 𝜆𝜆(𝑠𝑠, 𝑖𝑖) + 1. ∑𝑠𝑠 ∑𝑑𝑑≠𝑠𝑠 𝜆𝜆(𝑠𝑠, 𝑑𝑑)

9.7 Consider the seven-node WDM mesh network shown in Fig. 9.19 with two tunable transmitters and two tunable receivers per node and two wavelengths per fiber. Assume that the nodes don’t have wavelength converters. The traffic matrix Γ for node pairs in the network is given below (in arbitrary units): Γ=

0344521 3042210 3308641 2210220 5735077 1484102 0 1 4 2 2 3 0.

Determine a set of lightpaths to be constructed using an algorithm maximizing the single-hop traffic with the largest traffic considered first, where if two traffic elements have the same value, the first traffic element obtained in the row-wise sequence (i.e., beginning with the topmost row from left to right and then moving downward to the next row) is selected. Show the RWA for the lightpaths obtained by using the shortest-path and first-fit schemes for routing and WA, respectively. Solution: Using the heuristic scheme on the WRON with the given traffic matrix, the RWA solution is obtained and presented in a tabular form as follows.

Serial No.

Lightpath route

Cost (distance)

Wavelength (wi)

1. 2.

3→ 4 6 → 2 →3

100 390

w1 w1

3.

5 →3 →2

420

w1

4.

5→6

400

w1

5.

3→5

120

w1

6.

1 →2 → 3 → 5

620

w2

9.4

7.

4 →5 → 6 → 7

650

w2

8.

1→2→3

500

w2

9.

6→2

90

w2

10.

2→ 1

200

w1

11.

7→6

160

w1

12.

2→6→7

250

w1

13.

4→3→2→1

600

w2

14.

7→6→5→4

650

w2

The direct lightpaths for other (s,d) pairs cannot be allocated, and the connections for these node pairs need to be accommodated with concatenated (multihop) lightpaths by using traffic grooming. 9.8 Evaluate the wavelength utilization ratio in a WRON mesh using 10 wavelengths with full and without wavelength conversion for five-hop connections for a blocking probability of 0.001. Using these results, estimate also the gain in wavelength utilization that can be achieved by using full wavelength conversion in a network which currently doesn't employ any wavelength conversion. Solution: Number of wavelengths M = 10, number of hops H = 5, and the probability of blocking Pb due to resource exhaustion should be ≤ 0.001. Hence, the wavelength utilization without wavelength conversion is obtained as 1 𝐻𝐻

1/𝑀𝑀

𝛾𝛾𝑛𝑛𝑛𝑛 = − ln �1 − 𝑃𝑃𝑏𝑏

1 5

� = − ln�1 − 0.0011/10 � = 0.13910.

With full wavelength conversion at all nodes, the wavelength utilization improves to 𝑃𝑃𝑏𝑏 1/𝑀𝑀 0.001 1/10 𝛾𝛾𝑓𝑓𝑓𝑓 = =� � = 0.42668. 𝐻𝐻 5 Thus, with full wavelength conversion, one can obtain a gain in wavelength utilization given by 𝐺𝐺𝑤𝑤𝑤𝑤 =

0.42668 = 3.06743. 0.13910

9.9 In a given WRON, two client nodes (IP routers) are connected to two OXCs at two different WRON nodes, with four intermediate nodes between them. In the WRON, each fiber between any two nodes supports four wavelengths. Consider that 0.7 is the probability that a wavelength is being used on a fiber for an ongoing connection. a) Find the probability that a new connection request is blocked when the nodes are not equipped with wavelength converters. b) Find the probability that a new connection request is blocked when the nodes are equipped with wavelength converters, having full wavelength-conversion capability. Solution: OXC

OXC

OXC

OXC

OXC

OXC

OXC Client

Client 9.5

Number of wavelengths M = 4, number of hops H = 5, wavelength utilization γ = 0.7. a) Without wavelength conversion: 𝑃𝑃𝑏𝑏 = [1 − (1 − 𝛾𝛾)𝐻𝐻 ]𝑀𝑀 = [1 − (1 − 0.7)5 ]4 = 0.99032.

b) With full wavelength conversion:

𝑃𝑃𝑏𝑏 = 1 − [1 − 𝛾𝛾 𝑀𝑀 ]𝐻𝐻 = 1 − [1 − 0.74 ]5 = 0.74661.

9.10 Figure 9.20 shows a five-node linear WRON with six all-optical connection requests in the form of VPs, needing WA. Form an auxiliary graph using the VPs (VPi's) shown in the given network, where these VPs would be represented by the equivalent nodes. Using the auxiliary graph, assign wavelengths (i.e., colors) to the VPs by using a heuristic scheme, where the nodes in the auxiliary graph with the largest degree are considered first in every step to assign the wavelengths. Solution: Using the given linear network and the set of VPs assigned therein (i.e., VP1, VP2, VP3, VP4, VP5, and VP6), the auxiliary graph (with its nodes representing the VPs) is drawn, as shown below. VP1

VP2

VP3

VP6 VP4

VP5

With the above auxiliary graph, the wavelengths indexed as w1, w2, ... and so on, we start coloring the the nodes, in descending order of the nodal degrees Di’s and obtain the wavelength assignment to the VPs (transforming them into lightpaths) as follows (nodes with the same degree are chosen arbitrarily in the assignment process). • • • • • •

VP6 (D6 = 4): assigned wavelength = w1, VP2 (D2 = 4): having a common edge with VP6, cannot be assigned w1. Hence, VP2 is assigned the next available wavelength in the list, i.e., w2. VP5 (D5 = 4): having common edges with VP2 and VP6, VP5 shouldn’t be assigned w1, w2. Hence, VP5 is assigned the next wavelength w3. VP4 (D4 = 4): having common edges with VP2, VP3, VP5 and VP6, VP4 shouldn’t be assigned w1, w2, and w3. Hence VP4 is assigned the next wavelength w4. VP1 (D1 = 2): having common edges with VP2 and VP5, VP1 shouldn’t be assigned w2, w3. Hence, VP1 is assigned the wavelength available with lowest index from the list, i.e., w1. VP3 (D3 = 2): similarly, having common edges with VP6 and VP4, VP3 is assigned w2.

The readers are encouraged to repeat the above WA problem using ILP as well.

9.6

Exercise Problems and Solutions for Chapter 10 (Transmission Impairments and Power Consumption in Optical Networks) 10.1 How in practice is the power/energy-consumption efficiency defined in telecommunication networks? Discuss how the efficiency has evolved over the last three centuries. Also, discuss the conflicting roles between the ways to enhance power-consumption efficiency and spectrum utilization in the high-speed WDM networks. Solution: See the discussion in Section 10.1. 10.2. Indicate the major differences between the shot and thermal noise components in optical receivers. Write down the expression for the shot-noise variance in an APD-based optical receiver and discuss how one can optimize the receiver performance (in respect of SNR) in the presence of APD excess noise. Solution: Shot noise: signal-dependent noise, with its route cause lying in the optical source. Thermal noise: doesn’t depend on signal and is typically a localized noise process in the receiver. The shot-noise variance in an APD is expressed as 2 𝜎𝜎𝐴𝐴𝐴𝐴𝐴𝐴 = 2𝑞𝑞𝑅𝑅𝑤𝑤 𝑃𝑃𝑅𝑅 𝑀𝑀2 𝐹𝐹(𝑀𝑀)𝐵𝐵𝑒𝑒 ,

with the excess noise factor F(M) ≈ M x, x being in the range of [0,1]. Hence, the signal-to-noise ratio (SNR) in an APD-based receiver is expressed as 𝑆𝑆𝑆𝑆𝑆𝑆 =

(𝑅𝑅𝑤𝑤 𝑃𝑃𝑅𝑅 𝑀𝑀)2 . 2𝑞𝑞𝑅𝑅𝑤𝑤 𝑃𝑃𝑅𝑅 𝑀𝑀2+𝑥𝑥 𝐵𝐵𝑒𝑒 + 4𝑘𝑘𝑘𝑘𝐵𝐵𝑒𝑒 /𝑅𝑅

To maximize the SNR with an optimum APD gain Mopt, we equate the two terms in the denominator with M = Mopt as or

leading to Hence, we obtain Mopt as

2+𝑥𝑥 𝐵𝐵𝑒𝑒 = 2𝑞𝑞𝑅𝑅𝑤𝑤 𝑃𝑃𝑅𝑅 𝑀𝑀𝑜𝑜𝑜𝑜𝑜𝑜

2+𝑥𝑥 𝑀𝑀𝑜𝑜𝑜𝑜𝑜𝑜 =

4𝑘𝑘𝑘𝑘𝐵𝐵𝑒𝑒 , 𝑅𝑅

2𝑘𝑘𝑘𝑘 1 × = 𝑍𝑍 (say), 𝑅𝑅 𝑞𝑞𝑅𝑅𝑤𝑤 𝑃𝑃𝑅𝑅

(2 + 𝑥𝑥)log10 𝑀𝑀𝑜𝑜𝑜𝑜𝑜𝑜 = log10 𝑍𝑍. 𝑀𝑀𝑜𝑜𝑜𝑜𝑜𝑜 = 10{log 10 𝑍𝑍/(2+𝑥𝑥)}

10.3 Consider that, a PIN-based binary optical receiver, operating at 10 Gbps with an incoming optical signal at 1.55 μm, offers a BER of 10-9 with a received optical power of -18 dBm. Determine how inferior it is (in dB) with respect to an equivalent receiver operating in quantum limit. Discuss how one can improve the BER of this receiver. (Given: Planck's constant = 6.63 × 10-34 Joulesecond).

10.1

Solution: In quantum limit, the received optical power at 10 Gbps and 1550 nm for a BER = 10-9 is obtained as

Λ 𝑎𝑎𝑎𝑎 ℎ𝑓𝑓 Λ 𝑎𝑎𝑎𝑎 ℎ𝑐𝑐 10.5 × 6.63 × 10−34 × 3 × 108 𝑃𝑃𝑅𝑅 = = = = 13.474 nW = −48.73 dBm, 𝑇𝑇 𝑇𝑇𝑇𝑇 10−10 × 1550 × 10−9

with Λav as the average photon count received and T = 1/r is the bit interval. The given PIN-based receiver requires -18 dBm, and hence the receiver is inferior to the receiver in quantum limit by -18 – (-48.73) = 30.73 dB. In order to improve the receiver sensitivity, one needs to employ APD as the photodetector with appropriate APD gain. 10.4. Consider an optical communication link operating between two cities using WDM transmission over C band of a single-mode fiber. If the link uses 32 wavelengths, each operating at 10 Gbps, calculate the effective transmission capacity (in Gbps) of the link. Estimate the maximum allowable value of chromatic dispersion for each wavelength, if the link length is 100 km and the laser linewidth is 100 MHz. Ignore polarization-mode dispersion. Solution: Number of wavelengths M = 32 and per-channel bit rate r = 10 Gbps. Hence, the effective transmission capacity of the WDM link is 𝐶𝐶𝑊𝑊𝑊𝑊𝑊𝑊 = 𝑀𝑀𝑀𝑀 = 32 × 10 = 320 Gbps.

The effective spectral width of the laser due to the inherent linewidth and modulation is estimated as (Eq. B.57) 𝐵𝐵𝐿𝐿𝑀𝑀 = �𝑟𝑟 2 + 𝛿𝛿𝛿𝛿 2 = �(1010 )2 + (107 )2 ≈ 10 GHz

The corresponding value for wavelength spread δw (with the central wavelength being ≈ 1550 nm in C band), is obtained as 𝛿𝛿𝛿𝛿 =

𝑤𝑤 2 𝑀𝑀 (1550 × 10−9 )2 𝐵𝐵 = × 1010 = 0.08 nm. 𝑐𝑐 𝐿𝐿 3 × 108

With Dch as the chromatic dispersion, the pulse spread over L = 100 km link is expressed as Δ𝜏𝜏𝑐𝑐ℎ = 𝐷𝐷𝑐𝑐ℎ 𝛿𝛿𝛿𝛿𝐿𝐿 = 𝐷𝐷𝑐𝑐ℎ × 0.08 × 100 ps,

which should be ≤ T with T = 100 ps (for per-channel 10 Gbps transmission). Thus, the link should satisfy the condition, given by

thereby needing to satisfy

𝐷𝐷𝑐𝑐ℎ × 0.08 × 100 < 100, 𝐷𝐷𝑐𝑐ℎ
1. Hence, this is a non-Nyquist transmission system, b) r = 40 Gbps: ρ = Δf/r = 50/40 = 1.25. In this case, ρ being greater than but close to 1, this is a quasi-Nyquist transmission system. c) r = 100 Gbps: ρ = Δf/r = 50/100 = 0.5