*****INSTANT DOWNLOAD*****Solution Manual Optical Networks 1st edition by Debasish Datta

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c c c




SOLUTIONS MANUAL c

, Errata


Context Present version in the book
c c c c Corrected/changed version c




Page 130, Exercise 2.7
c c c 5.27 nm c 527 nm c




Page 248, expression for Gd
c c c c Gd = L/[2(M – 1) + L]
c c c c c c Gd = L/[2(M – 1 + L)]
c c c c c c




below Eq. 6.5.
c c




Page 572, Exercise 14.6.
c c c Γ = 0 24 40 50
c ccc c c c Γ = 0 50 25 60
c cc c c c




24 0 24 40 c c c 25 0 50 60 c c c




24 24 0 0 c c cc 25 30 0 30 c c c




50 0 40 0. cc c c 25 50 30 0. c c c




Page 593, Exercise 15.7
c c c 0.1 µs c 0.8 µs c




ii

, Exercise Problems and Solutions for Chapter 2 (Technologies for Optical Networks)
c c c c c c c c c c




2.1 A step-index multi-mode optical fiber has a refractive-index difference Δ = 1% and a
c c c c c c c c c c c c c




ccore refractive index of 1.5. If the core radius is 25 µm, find out the approximate number of
c c c c c c c c c c c c c c c c c




propagating modes in the fiber, while operating with a wavelength of 1300 nm.
c c c c c c c c c c c c c




Solution:
Δ = 0.01, n1 = 1.5, a = 25 μm, w = 1300 nm, and the2 number of modes Nmode is given by
c c c c c c c c c c c c c c c c c c c c c c c


𝐹𝐹 c



, with 𝐹𝐹 = 2𝜋𝜋𝑡𝑡
c




𝑁𝑁𝐴𝐴.
c c c cc c




𝑁𝑁𝑑𝑑𝑜𝑜𝑛𝑛𝑛𝑛 = cc c




The numerical aperture NA is obtained as
c
2 𝑠𝑠 c c c c c




1 2 1
2
𝑁𝑁𝐴𝐴 c = �𝑛𝑛2
−𝑛𝑛 ≈𝑛𝑛 √2∆ =1.5√0.02.c c
c

c c c c




Hence, we obtain V parameter as, c c c c c




2𝜋𝜋 × 25 × 10−6 c c c c
c




𝐹𝐹 = 1300 × 10−9 ×�1.5√0.02�= 25.632, c c c c c c




leading to the number of modes Nmode , given by c c c c c c c


≈ 329.
c c

c c



25.6322

𝑁𝑁𝑑𝑑𝑜𝑜𝑛𝑛𝑛𝑛 =
2
c




2.2 A step-index multi-mode optical fiber has a cladding with the refractive index of 1.45. If it has a
c c c c c c c c c c c c c c c c c




limiting intermodal dispersion of 35 ns/km, find its acceptance angle. Also calculate the maximum
c c c c c c c c c c c c c




c possible data transmission rate, that the fiber would support over a distance of 5 km.
c c c c c c c c c c c c c c




Solution:
The cladding refractive index n2 =1.45, and the intermodal dispersion Dmod = 35 ns/km. Dmod
c c c c c c c c c c c cc c c c c




is expressed as
𝑛𝑛1 − 𝑛𝑛2
c c c


c



𝑛𝑛1 𝑛𝑛1 − 𝑛𝑛2 = 35 ns/km. 𝑛𝑛1 Δ
c c

c c c c


= � �=
c c cc c c

c


𝐷𝐷𝑑𝑑𝑜𝑜𝑛𝑛 ≈ 𝑐𝑐 𝑐𝑐 𝑐𝑐
𝑛𝑛1 c




5 -9
Hence, (n1 – n2) = cDmod = (3 × 10 ) × (35 × 10 ) = 0.0105, and n1 = n2 + 0.0105 = 1.4605. Therefore,
c c c c c c c c c c c c c c c c c c c c c c c c




we obtain NA as
c c c




2 2 2 2

𝑁𝑁𝐴𝐴 = �𝑛𝑛1 − 𝑛𝑛2 = �1.4605 − 1.45 = 0.174815, cc c cc
ccc
c cc c c c




and the acceptance angle is obtained as θA = sin-1(NA) = sin-1(0.174815) = 10.068o.
c c c c c c c c c c c c c




The pulse spreading due to dispersion should remain ≤ 0.5/r, with r as the data-transmission rate,
c c c c c c c c c c c c c c c




cimplying that r ≤ 0.5/(Dmod L). Hence, we obtain the maximum possible data transmission rate rmax
c c c c c c c c c c c c c c c




cover L= 5 km as c c c c c




0.5

𝑝𝑑𝑑𝑡𝑡𝑚𝑚 = ccccc c


= 2.86 Mbps.
35 × 10−9 × 5
c c c
c
c c c c




2.3 Consider that a step-index multi-mode optical fiber receives optical power from a Lambertian c c c c c c c c c c c c




source with the emitted intensity pattern given by I(θ) = I0 cosθ, where θ is the angle subtended by an
c c c c c c c c c c c c c c c c c c c




c incident light ray from the source with the fiber axis. The total power emitted by the source is 1 mW
c c c c c c c c c c c c c c c c c c c




cwhile the power coupled into the fiber is found to be - 4 dBm. Derive the relation between
c c c c c c c c c c c c c c c c c




cthe

2.1

, launched power and the numerical aperture of the optical fiber. If the refractive index of the core is
c c c c c c c c c c c c c c c c c




c1.48, determine the refractive index of the cladding.
c c c c c c c




Solution:
Transmit power PT = 1 mW, and the power coupled into fiber PC = - 4 dBm = 10- 0.4 W = 0.3981 mW. For a
c c c c c c c c c c c c c c c c c c
c c

c c c c c




Lambertian source, the coupled power PC = NA2 × PT X (for derivation, see Cherin 1983). Hence,
c c c c c c c c
c

c c c c c c c c



𝑃𝑃𝐶𝐶
1 2 implying that n 22 = n 12 – PC/PT . c c c c c c


2 2 2
NA = PC/PT = 0.3981. Further, 𝑁𝑁𝐴𝐴 = 𝑛𝑛
c
ccc c



− 𝑛𝑛2 = 𝑃𝑃𝑇𝑇
c c c c c c c
cc



Thus, we obtain n2 as
c

c c c c




𝑛𝑛2 =√1.482 −0.3981=1.34. c
c
c
c c c




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 c c c c c c c c c c c c c c c c c c c c




nm. Presuming that the optical fiber is fed with an optical power that is large enough to force the fiber
c c c c c c c c c c c c c c c c c c c c




ctowards exhibiting nonlinear effects, determine the effective lengths of the fiber in the two
c c c c c c c c c c c c c




coperating conditions. Comment on the results. c c c c c




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
c c c c c c c c c c c c c c c c c c c c c c




is determined from αdB = 10log10[exp(α)] as
c c c c c c




α = ln (10αdB/10) = ln(100.05) = 0.1151.
c c c c c c c




Hence, we obtain the effective fiber length as
c c c c c c c




Lef f = [1 – exp(-αL)]/α c c cc c c




= [(1 – exp(-0.1151 × 20)]/0.1151 = 7.82 km.
c c c c c c c c




With αdB = 0.2 dB/km, we similarly obtain Lef f = 13.06 km, which is expected because with lower
c c c c c c c c c c c c c c c c c c




attenuation, the power decays slowly along the fiber and thus the fiber nonlinearity effects can take
c c c c c c c c c c c c c c c c




place over longer fiber length.
c c c c c




2.5 Consider an optical communication link operating at 1550 nm over a 60 km optical fiber having a c c c c c c c c c c c c c c c c




closs of 0.2 dB/km. Determine the threshold power for the onset of SBS in the fiber. Given: SBS gain
c c c c c c c c c c c c c c c c c c




ccoefficient gB = 5 ×10-11 m/W, effective area of cross-section of the fiber Aeff = 50 µm2, SBS c c c c c
c

c c c c c c c c c c c c c c c c




cbandwidth = 20 MHz, laser spectral width = 200 MHz. c c c c c c c c c




Solution:
With αdB = 0.2 dB/km at 1550 nm, we obtain α = ln (10αdB/10) = ln(100.02) = 0.0461. Hence, for
c c c c c c c c c c c c c c c c c c c




c L = 60 km, we obtain Lef f as
c c c c c c c ccc




Lef f = [1 – exp(-αL)]/α
c ccc c c c




= [1 – exp(-0.0461 × 60)]/0.0461 = 20.327 km.
c c c c c cc c c




With Aeff = 50 μm2, gB = 5 × 10-11 m/W, and assuming the polarization-matching factor to be ηp = 2, we
c c c c c c c c c
c

c c c c c c c c c c c




obtain the SBS threshold power as
c c c c c c




21 𝜂𝜂𝑜𝑜 𝐴𝐴𝑛𝑛𝑓𝑓𝑓𝑓 cc c
c
𝛿𝛿𝛿𝛿 21 × 2 × 50 × c c c c cc 200
�1 + c �= c
c10−12 �=22.73mW.
c c c




𝑃𝑃𝑡𝑡ℎ (𝑆𝑆𝐵𝐵𝑆𝑆) �1 +
𝑔𝑔𝐵𝐵 𝐵𝐵𝑛𝑛𝑓𝑓𝑓𝑓 5 × 10−11 × 20.327 × 103
c
c

Δ𝛿𝛿𝐵𝐵 c c
c
c c c
cc
20

c =
2.6 Consider an optical communication link operating at 1550 nm over a 60 km optical fiber having a c c c c c c c c c c c c c c c c




closs of 0.2 dB/km. The effective area of cross-section of the fiber Aeff = 50 µm2, where an optical
c c c c c c c c c c c c c c c c c c




cpower of 0 dBm is launched. Determine the nonlinear phase shift introduced by SPM in the fiber.
c c c c c c c c c c c c c c c c




2.2