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 f power f and f the f numerical f aperture f of f the f optical f fiber. f If f the f refractive f index f of
f the f core f is f 1.48, fdetermine fthe frefractive findex fof fthe fcladding.
Solution:
Transmit fpower fPT f = f1 fmW, fand fthe fpower fcoupled finto ffiber fPC f = f- f4 fdBm f= f10- f0.4 f W f=
2 f
f0.3981 fmW. f For fa fLambertian fsource, fthe fcoupled fpower fPC f = fNA × fPT fX f (for fderivation, fsee
fCherin f1983). fHence, 2 2 1
𝑃𝑃𝐶𝐶
1 implying fthat fn = fn – PC/PT f .
2f 2 2
NA = fPC/PT f = f0.3981. fFurther, f f 𝑁𝑁𝐴𝐴2 f f f f f
− f𝑛𝑛2 f 𝑃𝑃𝑇𝑇
f= f
𝑛𝑛2
Thus, fwe fobtain fn2 f as f=
𝑛𝑛2 f= f√1.482 f− f0.3981 f= f1.34.
2.4 Consider fa f20 fkm fsingle-mode foptical ffiber fwith fa floss fof f0.5 fdB/km fat f1330 fnm fand f0.2
fdB/km fat f 1550 fnm. fPresuming fthat fthe foptical ffiber fis ffed fwith fan foptical fpower fthat fis flarge
fenough fto fforce fthe f fiber f towards f exhibiting f nonlinear f effects, f determine f the f effective
f lengths f of f the f fiber f in f the f two f operating fconditions. fComment fon fthe fresults.
Solution:
With fL f= f20 fkm, ffirst fwe fconsider fthe fcase fwith ffiber floss fαdB f= f0.5 fdB/km. fSo, fthe floss fα fin
fneper/km
is fdetermined ffrom fαdB f= f10log10[exp(α)] fas
α f= fln f(10αdB/10) f= fln(100.05) f= f0.1151.
Hence, fwe fobtain fthe feffective ffiber flength fas
Lef ff f = f f[1 f– fexp(-αL)]/α
= f[(1 f– fexp(-0.1151 f× f20)]/0.1151 f= f7.82 fkm.
With f αdB f= f 0.2 f dB/km, f we f similarly f obtain f Lef f f f = f 13.06 f km, f which f is f expected f because
f with f lower f attenuation, f the f power f decays f slowly f along f the f fiber f and f thus f the f fiber
f nonlinearity f effects f can f take f place fover flonger ffiber flength.
2.5 Consider fan foptical fcommunication flink foperating fat f1550 fnm fover fa f60 fkm foptical ffiber
fhaving fa f loss fof f0.2 fdB/km. f Determine fthe fthreshold fpower ffor fthe f onset f of f SBS fin fthe
-11 f
ffiber. f Given: f SBS fgain f coefficient f gB f= f f5 f ×10 m/W, f feffective f farea f of f cross-section f
2
fof fthe f fiber f Aeff f = f 50 f µm , f fSBS f bandwidth f= f20 fMHz, flaser fspectral fwidth f= f200 fMHz.
Solution:
With fαdB f= f0.2 fdB/km fat f1550 fnm, fwe fobtain fα f= fln f(10αdB/10) f= fln(100.02) f=
f0.0461. f Hence, ffor fL f= f60 fkm, fwe fobtain fLef ff f f fas
Lef ff f f f= f[1 f– fexp(-αL)]/α
= f[1 f– fexp(-0.0461 f× f60)]/0.0461 f f= f20.327 fkm.
With fAeff f= f50 fμm2, fgB f= f5 f× f10-11 f m/W, fand fassuming fthe fpolarization-matching ffactor fto fbe
fηp f = f2, fwe f obtain fthe fSBS fthreshold fpower fas
21 f f f 𝜂𝜂𝑜𝑜 f 𝐴𝐴𝑛𝑛𝑓𝑓𝑓𝑓 𝛿𝛿𝛿𝛿 21 f × f 2 f × f 50 f f× 200
f 10−12
𝑃𝑃𝑡𝑡ℎ f (𝑆𝑆𝐵𝐵𝑆𝑆) f =
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 f power f and f the f numerical f aperture f of f the f optical f fiber. f If f the f refractive f index f of
f the f core f is f 1.48, fdetermine fthe frefractive findex fof fthe fcladding.
Solution:
Transmit fpower fPT f = f1 fmW, fand fthe fpower fcoupled finto ffiber fPC f = f- f4 fdBm f= f10- f0.4 f W f=
2 f
f0.3981 fmW. f For fa fLambertian fsource, fthe fcoupled fpower fPC f = fNA × fPT fX f (for fderivation, fsee
fCherin f1983). fHence, 2 2 1
𝑃𝑃𝐶𝐶
1 implying fthat fn = fn – PC/PT f .
2f 2 2
NA = fPC/PT f = f0.3981. fFurther, f f 𝑁𝑁𝐴𝐴2 f f f f f
− f𝑛𝑛2 f 𝑃𝑃𝑇𝑇
f= f
𝑛𝑛2
Thus, fwe fobtain fn2 f as f=
𝑛𝑛2 f= f√1.482 f− f0.3981 f= f1.34.
2.4 Consider fa f20 fkm fsingle-mode foptical ffiber fwith fa floss fof f0.5 fdB/km fat f1330 fnm fand f0.2
fdB/km fat f 1550 fnm. fPresuming fthat fthe foptical ffiber fis ffed fwith fan foptical fpower fthat fis flarge
fenough fto fforce fthe f fiber f towards f exhibiting f nonlinear f effects, f determine f the f effective
f lengths f of f the f fiber f in f the f two f operating fconditions. fComment fon fthe fresults.
Solution:
With fL f= f20 fkm, ffirst fwe fconsider fthe fcase fwith ffiber floss fαdB f= f0.5 fdB/km. fSo, fthe floss fα fin
fneper/km
is fdetermined ffrom fαdB f= f10log10[exp(α)] fas
α f= fln f(10αdB/10) f= fln(100.05) f= f0.1151.
Hence, fwe fobtain fthe feffective ffiber flength fas
Lef ff f = f f[1 f– fexp(-αL)]/α
= f[(1 f– fexp(-0.1151 f× f20)]/0.1151 f= f7.82 fkm.
With f αdB f= f 0.2 f dB/km, f we f similarly f obtain f Lef f f f = f 13.06 f km, f which f is f expected f because
f with f lower f attenuation, f the f power f decays f slowly f along f the f fiber f and f thus f the f fiber
f nonlinearity f effects f can f take f place fover flonger ffiber flength.
2.5 Consider fan foptical fcommunication flink foperating fat f1550 fnm fover fa f60 fkm foptical ffiber
fhaving fa f loss fof f0.2 fdB/km. f Determine fthe fthreshold fpower ffor fthe f onset f of f SBS fin fthe
-11 f
ffiber. f Given: f SBS fgain f coefficient f gB f= f f5 f ×10 m/W, f feffective f farea f of f cross-section f
2
fof fthe f fiber f Aeff f = f 50 f µm , f fSBS f bandwidth f= f20 fMHz, flaser fspectral fwidth f= f200 fMHz.
Solution:
With fαdB f= f0.2 fdB/km fat f1550 fnm, fwe fobtain fα f= fln f(10αdB/10) f= fln(100.02) f=
f0.0461. f Hence, ffor fL f= f60 fkm, fwe fobtain fLef ff f f fas
Lef ff f f f= f[1 f– fexp(-αL)]/α
= f[1 f– fexp(-0.0461 f× f60)]/0.0461 f f= f20.327 fkm.
With fAeff f= f50 fμm2, fgB f= f5 f× f10-11 f m/W, fand fassuming fthe fpolarization-matching ffactor fto fbe
fηp f = f2, fwe f obtain fthe fSBS fthreshold fpower fas
21 f f f 𝜂𝜂𝑜𝑜 f 𝐴𝐴𝑛𝑛𝑓𝑓𝑓𝑓 𝛿𝛿𝛿𝛿 21 f × f 2 f × f 50 f f× 200
f 10−12
𝑃𝑃𝑡𝑡ℎ f (𝑆𝑆𝐵𝐵𝑆𝑆) f =
2.2
