ALL 15 CHAPTER COVERED
SOLUTIONS MANUAL
, Errata
Context Present version in the book Corrected/changed version
Page 130, Exercise 2.7 5.27 nm 527 nm
Page 248, expression for Gd Gd = L/[2(M – 1) + L] Gd = L/[2(M – 1 + L)]
below Eq. 6.5.
Page 572, Exercise 14.6. Γ = 0 24 40 50 Γ = 0 50 25 60
24 0 24 40 25 0 50 60
24 24 0 0 25 30 0 30
50 0 40 0. 25 50 30 0.
Page 593, Exercise 15.7 0.1 µs 0.8 µs
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
𝐹𝐹 2
, with 𝐹𝐹 = 2𝜋𝜋𝑡𝑡
𝑁𝑁𝐴𝐴.
𝑁𝑁𝑑𝑑𝑜𝑜𝑛𝑛𝑛𝑛 =
The numerical aperture NA is obtained as 2 𝑠𝑠
1 2 1
𝑁𝑁𝐴𝐴= �𝑛𝑛2 2
− 𝑛𝑛 ≈ 𝑛𝑛 √2∆ = 1.5√0.02.
Hence, we obtain V parameter as,
2𝜋𝜋 × 25 × 10−6
𝐹𝐹 = 1300 × 10−9 × �1.5√0.02� = 25.632,
leading to the number of modes Nmode , given by
≈ 329.
25.6322
𝑁𝑁𝑑𝑑𝑜𝑜𝑛𝑛𝑛𝑛 =
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
𝑛𝑛1 − 𝑛𝑛2
𝑛𝑛1 Δ 𝑛𝑛1 𝑛𝑛1 − 𝑛𝑛2 = 35 ns/km.
= �=
𝐷𝐷𝑑𝑑𝑜𝑜𝑛𝑛 ≈ 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
2 2 2 2
𝑁𝑁𝐴𝐴 = �𝑛𝑛1 − 𝑛𝑛2 = �1.4605 − 1.45 = 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
0.5
𝑝𝑑𝑑𝑡𝑡𝑚𝑚 =
= 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,
𝑃𝑃𝐶𝐶
1 2 implying that n 22 = n12 – PC/PT .
2 2
NA = PC/PT = 0.3981. Further, 𝑁𝑁𝐴𝐴 = 𝑛𝑛 2
− 𝑛𝑛2 = 𝑃𝑃𝑇𝑇
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 μm2, gB = 5 × 10-11 m/W, and assuming the polarization-matching factor to be ηp = 2, we
obtain the SBS threshold power as
21 𝜂𝜂𝑜𝑜 𝐴𝐴𝑛𝑛𝑓𝑓𝑓𝑓 �1 + 𝛿𝛿𝛿𝛿 � = 21 × 2 × 50 × 10−12 200� = 22.73 mW.
𝑃𝑃𝑡𝑡ℎ (𝑆𝑆𝐵𝐵𝑆𝑆) = �1 +
𝑔𝑔𝐵𝐵 𝐵𝐵𝑛𝑛𝑓𝑓𝑓𝑓 Δ𝛿𝛿𝐵𝐵 5 × 10−11 × 20.327 × 103 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
SOLUTIONS MANUAL
, Errata
Context Present version in the book Corrected/changed version
Page 130, Exercise 2.7 5.27 nm 527 nm
Page 248, expression for Gd Gd = L/[2(M – 1) + L] Gd = L/[2(M – 1 + L)]
below Eq. 6.5.
Page 572, Exercise 14.6. Γ = 0 24 40 50 Γ = 0 50 25 60
24 0 24 40 25 0 50 60
24 24 0 0 25 30 0 30
50 0 40 0. 25 50 30 0.
Page 593, Exercise 15.7 0.1 µs 0.8 µs
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
𝐹𝐹 2
, with 𝐹𝐹 = 2𝜋𝜋𝑡𝑡
𝑁𝑁𝐴𝐴.
𝑁𝑁𝑑𝑑𝑜𝑜𝑛𝑛𝑛𝑛 =
The numerical aperture NA is obtained as 2 𝑠𝑠
1 2 1
𝑁𝑁𝐴𝐴= �𝑛𝑛2 2
− 𝑛𝑛 ≈ 𝑛𝑛 √2∆ = 1.5√0.02.
Hence, we obtain V parameter as,
2𝜋𝜋 × 25 × 10−6
𝐹𝐹 = 1300 × 10−9 × �1.5√0.02� = 25.632,
leading to the number of modes Nmode , given by
≈ 329.
25.6322
𝑁𝑁𝑑𝑑𝑜𝑜𝑛𝑛𝑛𝑛 =
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
𝑛𝑛1 − 𝑛𝑛2
𝑛𝑛1 Δ 𝑛𝑛1 𝑛𝑛1 − 𝑛𝑛2 = 35 ns/km.
= �=
𝐷𝐷𝑑𝑑𝑜𝑜𝑛𝑛 ≈ 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
2 2 2 2
𝑁𝑁𝐴𝐴 = �𝑛𝑛1 − 𝑛𝑛2 = �1.4605 − 1.45 = 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
0.5
𝑝𝑑𝑑𝑡𝑡𝑚𝑚 =
= 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,
𝑃𝑃𝐶𝐶
1 2 implying that n 22 = n12 – PC/PT .
2 2
NA = PC/PT = 0.3981. Further, 𝑁𝑁𝐴𝐴 = 𝑛𝑛 2
− 𝑛𝑛2 = 𝑃𝑃𝑇𝑇
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 μm2, gB = 5 × 10-11 m/W, and assuming the polarization-matching factor to be ηp = 2, we
obtain the SBS threshold power as
21 𝜂𝜂𝑜𝑜 𝐴𝐴𝑛𝑛𝑓𝑓𝑓𝑓 �1 + 𝛿𝛿𝛿𝛿 � = 21 × 2 × 50 × 10−12 200� = 22.73 mW.
𝑃𝑃𝑡𝑡ℎ (𝑆𝑆𝐵𝐵𝑆𝑆) = �1 +
𝑔𝑔𝐵𝐵 𝐵𝐵𝑛𝑛𝑓𝑓𝑓𝑓 Δ𝛿𝛿𝐵𝐵 5 × 10−11 × 20.327 × 103 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
