Solution Manual Optical Networks 1st edition by Debasish Datta

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sd




ALL 15 CHAPTER COVERED
sd sd sd




SOLUTIONS MANUAL sd

, Errata


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




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




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


below Eq. 6.5.
sd sd




Page 572, Exercise 14.6.
sd sd sd Γ=
sd 0 24 40 50
sd sd sd sd sd sd Γ= 0
sd sd sd 50 25 60
sd sd sd


24 0 24 40
sd sd sd 25 0 50 60
sd sd sd


24 24 0 0
sd sd sd sd 25 30 0 30
sd sd sd


50 0 40 0.
sd sd sd sd 25 50 30 0.
sd sd sd




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




ii

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




2.1 A step-index multi-mode optical fiber has a refractive-index difference Δ =
s d s d s d s d s d s d s d s d s d s d


1% and a core refractive index of 1.5. If the core radius is 25 µm, find out the
s d s d s d s d s d sd sd sd sd sd sd sd sd sd sd sd sd sd


approximate number of propagating modes in the fiber, while operating with a wavelength of
sd sd sd sd s d sd sd sd sd sd sd sd sd sd


1300 nm.
sd sd



Solution:
Δ = 0.01, n1 = 1.5, a = 25 μm, w = 1300 nm,
sd sd sd and the number of modes Nmode
sd sd sd sd sd sd sd sd sd sd sd sd sd sd sd sd sd s d is given by
sd sd

𝐹𝐹 2 sd
sd

, with 𝐹𝐹 2𝜋𝜋𝑡𝑡 sd sd sd sd s d
𝑁𝑁𝐴𝐴.
𝑁𝑁𝑑𝑑𝑜𝑜𝑛𝑛𝑛𝑛 = sd sd
sd


𝑠𝑠
2
=
sd


The numerical aperture
2 sd 1 NA is obtained as sd sd sd sd sd




1
𝑁𝑁𝐴𝐴 s d = s d �𝑛𝑛2
− 𝑛𝑛 2 ≈ 𝑛𝑛 √2∆ = 1.5√0.02. sd
sd
sd sd sd sd


Hence, we obtain V parameter as,sd sd sd sd sd




2𝜋𝜋 × 25 × 10−6
sd sd sd sd
sd




𝐹𝐹 = 1300 × sd sd ×�
1.5√0.02�= 25.632,
sd sd sd




10−9 sd

leading to the number of modes Nmode , given
sd sd sd sd sd sd sd sd
by sd ≈ sd s d 329.
25.6322


𝑁𝑁𝑑𝑑𝑜𝑜𝑛𝑛𝑛𝑛 = s d 2
2.2 A step-index multi-mode optical fiber has a cladding with the refractive index of 1.45. If
sd sd sd sd sd sd sd sd sd sd sd sd sd sd

it has asd sd sd


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


sdthe maximum possible data transmission rate, that the fiber would support over a distance of
sd s d sd sd sd sd sd sd sd sd sd sd sd sd


5 km.
sd sd



Solution:
The cladding refractive index n2 =1.45, and the intermodal dispersion Dmod =
s d s d s d s d sd s d s d s d s d s d s d sd sd s d 35
ns/km. Dmod is expressed as
s d s d s d sd
𝑛𝑛1 − 𝑛𝑛2 sd sd
sd
sd
sd
𝑛𝑛1 Δ 𝑛𝑛1 𝑛𝑛1 − 𝑛𝑛2 = 35 ns/km. sd sd sd

�
sd sd sd s d sd


𝐷𝐷𝑑𝑑𝑜𝑜𝑛𝑛 ≈ sd

� = 𝑐𝑐 sd
= s d
sd


𝑐𝑐 𝑐𝑐 𝑛𝑛1
Hence, (n1 – n2) = cDmod = (3 × 10 ) × (35 × 10-9) = 0.0105, and n1 = n2 + 0.0105 =
sd
5
sd sd sd sd sd sd sd sd sd sd sd sd sd sd sd sd sd sd sd sd sd


1.4605. Therefore,
sd sd


we obtain NA as
sd sd sd



2 2 2 2

𝑁𝑁𝐴𝐴 = �𝑛𝑛1 − sd sd s d sd sd
sd sd sd
= sd − = sd s d 0.174815,

𝑛𝑛2 �1.4605 1.45 sd sd sd



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



The pulse spreading due to dispersion should remain ≤ 0.5/r, with r as the data-
s d s d sd s d s d s d sd s d s d sd s d sd s d s d




2.1

, transmission rate, implying that r ≤ 0.5/(Dmod L). Hence, we obtain the maximum
s d s d sd sd s d sd sd sd sd sd sd sd


sdpossible data transmission rate rmax over L = 5 km as
sd sd sd sd s d sd sd sd sd sd


0.5

𝑝𝑑𝑑𝑡𝑡𝑚𝑚 sd sd sd sd sd s d
=
= 2.86 Mbps. sd s d sd
35 × 10−9 × 5 sd sd
sd
sd sd

2.3 Consider that a step-index multi-mode optical fiber receives optical power from a
s d s d s d s d s d s d s d s d sd s d s d

Lambertian
s d


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


subtended by an
sd incident light ray from the source with the fiber axis. The total power
sd sd s d sd sd sd sd sd sd sd sd sd sd sd sd


emitted by the source is 1 mW while the power coupled into the fiber is found
sd sd sd sd sd sd sd s d s d s d s d s d s d s d s d s d


to be - 4 dBm. Derive the relation between the
s d s d s d s d s d s d s d s d s d s d




2.2