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 s power s and s the s numerical s aperture s of s the s optical s fiber. s If s the s refractive s index s of
s the s core s is s 1.48, sdetermine sthe srefractive sindex sof sthe scladding.
Solution:
Transmit spower sPT s = s1 smW, sand sthe spower scoupled sinto sfiber sPC s = s- s4 sdBm s= s10- s0.4 s W s=
2 s
s0.3981 smW. s For sa sLambertian ssource, sthe scoupled spower sPC s = sNA × sPT sX s (for sderivation,
ssee sCherin s1983). sHence, 2 2 1
𝑃𝑃𝐶𝐶
1 implying sthat sn = sn – PC/PT s .
2s 2 2
NA = sPC/PT s = s0.3981. sFurther, s s 𝑁𝑁𝐴𝐴2 s s s
− s𝑛𝑛2 s 𝑃𝑃𝑇𝑇
s s s = s 𝑛𝑛2
Thus, swe sobtain sn2 s as s=
𝑛𝑛2 s= s√1.482 s− s0.3981 s= s1.34.
2.4 Consider sa s20 skm ssingle-mode soptical sfiber swith sa sloss sof s0.5 sdB/km sat s1330 snm sand s0.2
sdB/km sat s 1550 snm. sPresuming sthat sthe soptical sfiber sis sfed swith san soptical spower sthat sis slarge
senough sto sforce sthe s fiber s towards s exhibiting s nonlinear s effects, s determine s the s effective
s lengths s of s the s fiber s in s the s two s operating sconditions. sComment son sthe sresults.
Solution:
With sL s= s20 skm, sfirst swe sconsider sthe scase swith sfiber sloss sαdB s= s0.5 sdB/km. sSo, sthe sloss sα
sin sneper/km
is sdetermined sfrom sαdB s= s10log10[exp(α)] sas
α s= sln s(10αdB/10) s= sln(100.05) s= s0.1151.
Hence, swe sobtain sthe seffective sfiber slength sas
Lef sf s = s s[1 s– sexp(-αL)]/α
= s[(1 s– sexp(-0.1151 s× s20)]/0.1151 s= s7.82 skm.
With s αdB s = s 0.2 s dB/km, s we s similarly s obtain s Lef s f s = s 13.06 s km, s which s is s expected s because
s with s lower s attenuation, s the s power s decays s slowly s along s the s fiber s and s thus s the s fiber
s nonlinearity s effects s can s take s place sover slonger sfiber slength.
2.5 Consider san soptical s communication slink soperating sat s1550 snm sover sa s60 skm s optical sfiber
shaving sa s loss sof s 0.2 sdB/km. s Determine s the sthreshold spower sfor sthe sonset s of s SBS sin sthe
-11 s
sfiber. s Given: s SBS sgain s coefficient s gB s= s s5 s ×10 m/W, s seffective s sarea s of s cross-section s
2
sof sthe s fiber s Aeff s= s 50 s µm , s sSBS s bandwidth s= s20 sMHz, slaser sspectral swidth s= s200 sMHz.
Solution:
With sαdB s= s0.2 sdB/km sat s1550 snm, swe sobtain sα s= sln s(10αdB/10) s= sln(100.02) s=
s0.0461. s Hence, sfor sL s= s60 skm, swe sobtain sLef sf s s sas
Lef sf s s s= s[1 s– sexp(-αL)]/α
= s[1 s– sexp(-0.0461 s× s60)]/0.0461 s s= s20.327 skm.
With sAeff s= s50 sμm2, sgB s= s5 s× s10-11 s m/W, sand sassuming sthe spolarization-matching sfactor sto sbe
sηp s = s2, swe s obtain sthe sSBS sthreshold spower sas
21 s s s 𝜂𝜂𝑜𝑜 s 𝐴𝐴𝑛𝑛𝑓𝑓𝑓𝑓 𝛿𝛿𝛿𝛿 21 s × s 2 s × s 50 s s× 200
s 10−12
𝑃𝑃𝑡𝑡ℎ s (𝑆𝑆𝐵𝐵𝑆𝑆) s =
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 s power s and s the s numerical s aperture s of s the s optical s fiber. s If s the s refractive s index s of
s the s core s is s 1.48, sdetermine sthe srefractive sindex sof sthe scladding.
Solution:
Transmit spower sPT s = s1 smW, sand sthe spower scoupled sinto sfiber sPC s = s- s4 sdBm s= s10- s0.4 s W s=
2 s
s0.3981 smW. s For sa sLambertian ssource, sthe scoupled spower sPC s = sNA × sPT sX s (for sderivation,
ssee sCherin s1983). sHence, 2 2 1
𝑃𝑃𝐶𝐶
1 implying sthat sn = sn – PC/PT s .
2s 2 2
NA = sPC/PT s = s0.3981. sFurther, s s 𝑁𝑁𝐴𝐴2 s s s
− s𝑛𝑛2 s 𝑃𝑃𝑇𝑇
s s s = s 𝑛𝑛2
Thus, swe sobtain sn2 s as s=
𝑛𝑛2 s= s√1.482 s− s0.3981 s= s1.34.
2.4 Consider sa s20 skm ssingle-mode soptical sfiber swith sa sloss sof s0.5 sdB/km sat s1330 snm sand s0.2
sdB/km sat s 1550 snm. sPresuming sthat sthe soptical sfiber sis sfed swith san soptical spower sthat sis slarge
senough sto sforce sthe s fiber s towards s exhibiting s nonlinear s effects, s determine s the s effective
s lengths s of s the s fiber s in s the s two s operating sconditions. sComment son sthe sresults.
Solution:
With sL s= s20 skm, sfirst swe sconsider sthe scase swith sfiber sloss sαdB s= s0.5 sdB/km. sSo, sthe sloss sα
sin sneper/km
is sdetermined sfrom sαdB s= s10log10[exp(α)] sas
α s= sln s(10αdB/10) s= sln(100.05) s= s0.1151.
Hence, swe sobtain sthe seffective sfiber slength sas
Lef sf s = s s[1 s– sexp(-αL)]/α
= s[(1 s– sexp(-0.1151 s× s20)]/0.1151 s= s7.82 skm.
With s αdB s = s 0.2 s dB/km, s we s similarly s obtain s Lef s f s = s 13.06 s km, s which s is s expected s because
s with s lower s attenuation, s the s power s decays s slowly s along s the s fiber s and s thus s the s fiber
s nonlinearity s effects s can s take s place sover slonger sfiber slength.
2.5 Consider san soptical s communication slink soperating sat s1550 snm sover sa s60 skm s optical sfiber
shaving sa s loss sof s 0.2 sdB/km. s Determine s the sthreshold spower sfor sthe sonset s of s SBS sin sthe
-11 s
sfiber. s Given: s SBS sgain s coefficient s gB s= s s5 s ×10 m/W, s seffective s sarea s of s cross-section s
2
sof sthe s fiber s Aeff s= s 50 s µm , s sSBS s bandwidth s= s20 sMHz, slaser sspectral swidth s= s200 sMHz.
Solution:
With sαdB s= s0.2 sdB/km sat s1550 snm, swe sobtain sα s= sln s(10αdB/10) s= sln(100.02) s=
s0.0461. s Hence, sfor sL s= s60 skm, swe sobtain sLef sf s s sas
Lef sf s s s= s[1 s– sexp(-αL)]/α
= s[1 s– sexp(-0.0461 s× s60)]/0.0461 s s= s20.327 skm.
With sAeff s= s50 sμm2, sgB s= s5 s× s10-11 s m/W, sand sassuming sthe spolarization-matching sfactor sto sbe
sηp s = s2, swe s obtain sthe sSBS sthreshold spower sas
21 s s s 𝜂𝜂𝑜𝑜 s 𝐴𝐴𝑛𝑛𝑓𝑓𝑓𝑓 𝛿𝛿𝛿𝛿 21 s × s 2 s × s 50 s s× 200
s 10−12
𝑃𝑃𝑡𝑡ℎ s (𝑆𝑆𝐵𝐵𝑆𝑆) s =
2.2
