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 a power a and a the a numerical a aperture a of a the a optical a fiber. a If a the a refractive a index a of
a the a core a is a 1.48, adetermine athe arefractive aindex aof athe acladding.
Solution:
Transmit apower aPT a = a1 amW, aand athe apower acoupled ainto afiber aPC a = a- a4 adBm a= a10- a0.4 a W a=
2 a
a0.3981 amW. a For aa aLambertian asource, athe acoupled apower aPC a = aNA × aPT aX a (for aderivation,
asee aCherin a1983). aHence, 2 2 1
𝑃𝑃𝐶𝐶
1 implying athat an = an – PC/PT a .
2a 2 2
NA = aPC/PT a = a0.3981. aFurther, a a 𝑁𝑁𝐴𝐴2 a a
− a𝑛𝑛2 a 𝑃𝑃𝑇𝑇
a a a a = a 𝑛𝑛2
Thus, awe aobtain an2 a as a=
𝑛𝑛2 a= a√1.482 a− a0.3981 a= a1.34.
2.4 Consider aa a20 akm asingle-mode aoptical afiber awith aa aloss aof a0.5 adB/km aat a1330 anm aand
a0.2 adB/km aat a 1550 anm. aPresuming athat athe aoptical afiber ais afed awith aan aoptical apower athat ais
alarge aenough ato aforce athe a fiber a towards a exhibiting a nonlinear a effects, a determine a the
a effective a lengths a of a the a fiber a in a the a two a operating aconditions. aComment aon athe aresults.
Solution:
With aL a= a20 akm, afirst awe aconsider athe acase awith afiber aloss aαdB a= a0.5 adB/km. aSo, athe aloss aα
ain a neper/km
is adetermined afrom aαdB a= a10log10[exp(α)] aas
α a= aln a(10αdB/10) a= aln(100.05) a= a0.1151.
Hence, awe aobtain athe aeffective afiber alength aas
Lef af a = a a[1 a– aexp(-αL)]/α
= a[(1 a– aexp(-0.1151 a× a20)]/0.1151 a= a7.82 akm.
With a αdB a= a 0.2 a dB/km, a we a similarly a obtain a Lef a f a = a 13.06 a km, a which a is a expected
a because a with a lower a attenuation, a the a power a decays a slowly a along a the a fiber a and a thus a the
a fiber a nonlinearity a effects a can a take a place aover alonger afiber alength.
2.5 Consider aan aoptical a communication alink aoperating aat a 1550 anm aover aa a60 akm a optical afiber
a having aa a loss a of a 0.2 adB/km. a Determine athe athreshold a power afor athe a onset a of a SBS a in athe
-11 a
a fiber. a Given: a SBS a gain a coefficient a gB a= a a5 a ×10 m/W, a aeffective a aarea a of a cross-
2
section a aof athe a fiber a Aeff a= a 50 a µm , a aSBS a bandwidth a= a20 aMHz, alaser aspectral awidth a=
a200 aMHz.
Solution:
With aαdB a= a0.2 adB/km aat a1550 anm, awe aobtain aα a= aln a(10αdB/10) a= aln(100.02) a=
a0.0461. a Hence, afor aL a= a60 akm, awe aobtain aLef af a a aas
Lef af a a a= a[1 a– aexp(-αL)]/α
= a[1 a– aexp(-0.0461 a× a60)]/0.0461 a a= a20.327 akm.
With aAeff a= a50 aμm2, agB a= a5 a× a10-11 a m/W, aand aassuming athe apolarization-matching afactor ato
abe aηp a = a2, awe a obtain athe aSBS athreshold apower aas
21 a a a 𝜂𝜂𝑜𝑜 a 𝐴𝐴𝑛𝑛𝑓𝑓𝑓𝑓 𝛿𝛿𝛿𝛿 21 a × a 2 a × a 50 a 200
a× a 10−12
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 a power a and a the a numerical a aperture a of a the a optical a fiber. a If a the a refractive a index a of
a the a core a is a 1.48, adetermine athe arefractive aindex aof athe acladding.
Solution:
Transmit apower aPT a = a1 amW, aand athe apower acoupled ainto afiber aPC a = a- a4 adBm a= a10- a0.4 a W a=
2 a
a0.3981 amW. a For aa aLambertian asource, athe acoupled apower aPC a = aNA × aPT aX a (for aderivation,
asee aCherin a1983). aHence, 2 2 1
𝑃𝑃𝐶𝐶
1 implying athat an = an – PC/PT a .
2a 2 2
NA = aPC/PT a = a0.3981. aFurther, a a 𝑁𝑁𝐴𝐴2 a a
− a𝑛𝑛2 a 𝑃𝑃𝑇𝑇
a a a a = a 𝑛𝑛2
Thus, awe aobtain an2 a as a=
𝑛𝑛2 a= a√1.482 a− a0.3981 a= a1.34.
2.4 Consider aa a20 akm asingle-mode aoptical afiber awith aa aloss aof a0.5 adB/km aat a1330 anm aand
a0.2 adB/km aat a 1550 anm. aPresuming athat athe aoptical afiber ais afed awith aan aoptical apower athat ais
alarge aenough ato aforce athe a fiber a towards a exhibiting a nonlinear a effects, a determine a the
a effective a lengths a of a the a fiber a in a the a two a operating aconditions. aComment aon athe aresults.
Solution:
With aL a= a20 akm, afirst awe aconsider athe acase awith afiber aloss aαdB a= a0.5 adB/km. aSo, athe aloss aα
ain a neper/km
is adetermined afrom aαdB a= a10log10[exp(α)] aas
α a= aln a(10αdB/10) a= aln(100.05) a= a0.1151.
Hence, awe aobtain athe aeffective afiber alength aas
Lef af a = a a[1 a– aexp(-αL)]/α
= a[(1 a– aexp(-0.1151 a× a20)]/0.1151 a= a7.82 akm.
With a αdB a= a 0.2 a dB/km, a we a similarly a obtain a Lef a f a = a 13.06 a km, a which a is a expected
a because a with a lower a attenuation, a the a power a decays a slowly a along a the a fiber a and a thus a the
a fiber a nonlinearity a effects a can a take a place aover alonger afiber alength.
2.5 Consider aan aoptical a communication alink aoperating aat a 1550 anm aover aa a60 akm a optical afiber
a having aa a loss a of a 0.2 adB/km. a Determine athe athreshold a power afor athe a onset a of a SBS a in athe
-11 a
a fiber. a Given: a SBS a gain a coefficient a gB a= a a5 a ×10 m/W, a aeffective a aarea a of a cross-
2
section a aof athe a fiber a Aeff a= a 50 a µm , a aSBS a bandwidth a= a20 aMHz, alaser aspectral awidth a=
a200 aMHz.
Solution:
With aαdB a= a0.2 adB/km aat a1550 anm, awe aobtain aα a= aln a(10αdB/10) a= aln(100.02) a=
a0.0461. a Hence, afor aL a= a60 akm, awe aobtain aLef af a a aas
Lef af a a a= a[1 a– aexp(-αL)]/α
= a[1 a– aexp(-0.0461 a× a60)]/0.0461 a a= a20.327 akm.
With aAeff a= a50 aμm2, agB a= a5 a× a10-11 a m/W, aand aassuming athe apolarization-matching afactor ato
abe aηp a = a2, awe a obtain athe aSBS athreshold apower aas
21 a a a 𝜂𝜂𝑜𝑜 a 𝐴𝐴𝑛𝑛𝑓𝑓𝑓𝑓 𝛿𝛿𝛿𝛿 21 a × a 2 a × a 50 a 200
a× a 10−12
2.2
