Optical fiber communication solution manual

Optical fiber communication solution manual

Problem Solutions for Chapter 2

2-1

E=100cos (2π108t+30°) ex +20 cos (2π108t −50°) ey

+ 40cos (2π108t+210°)ez

2-2 The general form is:

y = (amplitude) cos(ωt - kz) = A cos [2π(νt - z/λ)] Therefore

2-4 x1 = a1 cos (ωt - δ1) and x2 = a2 cos (ωt - δ2)

Adding x1 and x2 yields

x1 + x2 = a1 [cos ωt cos δ1 + sin ωt sin δ1]

+ a2 [cos ωt cos δ2 + sin ωt sin δ2]

= [a1 cos δ1 + a2 cos δ2] cos ωt + [a1 sin δ1 + a2 sin δ2] sin ωtSince the a's and the δ's are constants, we can set

a1 cos δ1 + a2 cos δ2 = A cos φ (1)

a1 sin δ1 + a2 sin δ2 = A sin φ (2)provided that constant values of A and φ exist which satisfy these equations Toverify this, first square both sides and add:

tan φ = a1sinδ1 +a2sinδ2

a1cosδ1 +a2cosδ2

Thus we can write

x = x1 + x2 = A cos φ cos ωt + A sin φ sin ωt = A cos(ωt - φ)

2-5 First expand Eq (2-3) as

Ey

E0 y= cos (ωt - kz) cos δ - sin (ωt - kz) sin δ (2.5-1)

Subtract from this the expression

E0x cos δ = - sin (ωt - kz) sin δ (2.5-2)

Using the relation cos2α + sin2α = 1, we use Eq (2-2) to write

(b) The critical angle is found from

nglass sin φglass = nair sin φair

with φair = 90° and nair = 1.0

∴φcritical = arcsin 1

nglass = arcsin

11.540= 40.5°

2-9

Air Water

2-11 (a) Use either NA = (n12−n22)1/ 2

= 0.242or

Solve for Er and let q2 = ω2εµ - β2 to obtain Eq (2-35a).

(b) Solve Eq (2-34b) for jHr:

jEr = 1

εω

1r

Solve for Hφ and let q2 = ω2εµ - β2 to obtain Eq (2-35d)

(d) Solve Eq (2-34b) for jEφ

Solve for Hr to obtain Eq (2-35c)

(e) Substitute Eqs (2-35c) and (2-35d) into Eq (2-34c)

Upon differentiating and multiplying by jq2/εω we obtain Eq (2-36)

(f) Substitute Eqs (2-35a) and (2-35b) into Eq (2-33c)

Upon differentiating and multiplying by jq2/εω we obtain Eq (2-37)

2-15 For ν = 0, from Eqs (2-42) and (2-43) we have

For the other case, substitute Eqs (2-47) and (2-51) into Eq (2-52):

where Jν and Kν are defined in Eq (2-54) If for ν = 0 the term in square brackets on theright-hand side is non-zero, that is, if Eq (2-56a) does not hold, then we must have that A

= 0, which from Eq (2-42) means that Ez = 0 Thus Eq (2-55) corresponds to TE0mmodes

2-16 From Eq (2-23) we have



   total≈ 43M-1/2 = 43 1080×100%= 4.1%

at 820 nm Similarly, (Pclad/P)total = 6.6% at 1320 nm and 7.8% at 1550 nm

2-20 (a) At 1320 nm we have from Eqs (2-23) and (2-57) that V = 25 and M = 312

(b) From Eq (2-72) the power flow in the cladding is 7.5%

2-21 (a) For single-mode operation, we need V ≤ 2.40

Solving Eq (2-58) for the core radius a

NA = n1

2− n2 2

2-23 For small values of ∆ we can write V ≈ 2πa

6.5×10-7 ≤ ny - nx ≤ 1.3×10-52-26 We want to plot n(r) from n2 to n1 From Eq (2-78)

At λ = 820 nm, Mstep = 1078 and at λ = 1300 nm, Mstep = 429.

Alternatively, we can let α = ∞ in Eq (2-81):

= 0.392-29 (a) From the Principle of the Conservation of Mass, the volume of a preform rodsection of length Lpreform and cross-sectional area A must equal the volume of the fiberdrawn from this section The preform section of length Lpreform is drawn into a fiber oflength Lfiber in a time t If S is the preform feed speed, then Lpreform = St Similarly, if s is thefiber drawing speed, then Lfiber = st Thus, if D and d are the preform and fiber diameters,respectively, then

Preform volume = Lpreform(D/2)2 = St (D/2)2and Fiber volume = Lfiber (d/2)2 = st (d/2)2

Equating these yields

St D2



  

2

= st d2

2-30 Consider the following geometries of the preform and its corresponding fiber:

Apreform core

Apreform clad

=

Afibercore

Afiberclador

(b) If R is the deposition rate, then the deposition time t is

t = M

R =

5.1gm0.5 gm / min = 10.2 min

2-32 Solving Eq (2-82) for χ yields

Integrating this from χi to χp where

are the initial crack depth and the crack depth after proof testing, respectively, yields

tp

∫ dtor

When a static stress σs is applied after proof testing, the time to failure is found from Eq.(2-86):

2-35 (a) Substituting Ns as given by Eq (2-92) and Np as given by Eq (2-93) into Eq.(2-94) yields

F = 1 - exp − L

L0

σpbtp + σsbts( )/ B+ σsb − 2

10 s= 6.5×10-14Thus this term can be neglected

2-36 The failure probability is given by Eq (2-85) For equal failure probabilities of thetwo fiber samples, F1 = F2, or

m = log 50log(4.8/ 3.9)= 18.8

Problem Solutions for Chapter 3

3-2 Since the attenuations are given in dB/km, first find the power levels in dBm for

100 µW and 150 µW These are, respectively,

3-5 With λ in Eqs (3-2b) and (3-3) given in µm, we have the following representative

points for αuv and αIR:

3-10 From Fig 2-22, we make the estimates given in this table:

ννm P clad /P ααννm = αα11 + (αα2 - αα11)P clad /P 5 + 10 3 P clad /P

3-11 (a) We want to solve Eq (3-12) for αgi With α = 2 in Eq (2-78) and letting

0

∞

∫

12K e

Thus αgi = α1 + (α2− α1)

Ka2(b) p(a) = 0.1 P0 = P0 e−Ka

To compare this with Fig 3-12, calculate three representative points, for example,

λ = 0.2, 0.6, and 1.0 µm Thus we have the following:

Wavelength λλ Calculated n n from Fig 3-12

3-13 (a) From Fig 3-13, dτ

dλ ≈ 80 ps/(nm-km) at 850 nm Therefore, for the LED we

have from Eq (3-20)

(b) Expand b as b = (β/ k+n2)(β/ k−n2)

(n1 +n2)(n1 −n2)Since n2< β/k < n1 , let β/k = n1(1 - δ) where 0 < δ < ∆ << 1 Thus,

n1 = n2(1 - ∆)-1 = n2(1 + ∆ + ∆2 + ) ≈ n2(1 + ∆)

Therefore, β = k[b n2(1 + ∆)∆ + n2] ≈ k n2(b∆ + 1)

3-16 The time delay between the highest and lowest order modes can be found from the

travel time difference between the two rays shown here

The travel time of each ray is given by

x

3-19 For ε = 0 we have that α = 2(1 - 6

5∆) Thus C1 and C2 in Eq (3-42) become(ignoring small terms such as ∆3, ∆4, )



2(1−6

5∆)+26(1−6

3-20 We want to plot Eq (3-30) as a function of σλ , where σint er mod al and

σint ra mod al are given by Eqs (3-41) and (3-45) For ε = 0 and α = 2, we have C1 =

0 and C2 = 1/2 Since σint er mod al does not vary with σλ , we have

With C1 = 0 we have from Eq (3-45)







3-21 Using the same parameter values as in Prob 3-18, except with ∆ = 0.001, we have

from Eq (3-41) σint er mod al /L = 7 ps/km, and from Eq (3.45)

σint ra mod al

L =

0.098σλ ns / km at850 nm0.0103σλ ns / km at1300 nm



  

α α+ 2

Thus ignoring the term involving ∆ d2n1

dλ2 , the first term in square brackets



  

α α+ 2



  

α α+ 2

2α

mM

L2 σλλ

3-29 For α =0.95αopt, we have

σint er(α ≠ αopt)

σint er(α = αopt) =(α − αopt)

(0.015)(1.95)= −170%For α =1.05αopt, we have

σint er(α ≠ αopt)

σint er(α = αopt) =(α − αopt)

(0.015)(2.05) = +163%

Problem Solutions for Chapter 4

N2A + 1

If ni << NA , which is generally the case, then to a good approximation

pP ≈ NA and nP = n2i /pP ≈ n2i /NA

4-3 (a) From Eq (4-4) we have 1.540 = 1.424 + 1.266x + 0.266x2 or

x2 + 4.759x - 0.436 = 0 Solving this quadratic equation yields (taking the plussign only)

Eg = 1.35 + 0.668(.26) - 1.17(.56) + 0.758(.26)2 + 0.18(.56)2

- 069(.26)(.56) - 322(.26)2(.56) + 0.03(.26)(.56)2 = 0.956 eV4-5 Differentiating the expression for E, we have

∆E= hc

λ2 ∆λ or ∆λ = λ2

hc∆EFor the same energy difference ∆E, the spectral width ∆λ is proportional to thewavelength squared Thus, for example,

∆λ1550

∆λ1310 =

15501310

4-8 The 3-dB optical bandwidth is found from Eq (4-21) It is the frequency f at

which the expression is equal to -3; that is,

With a 5-ns lifetime, we find

2π(5 ns) (100.6−1)=9.5 MHz4-9 (a) Using Eq (4-28) with Γ = 1

gth =

10.05 cmln  

10.32

2 + 10 cm-1 = 55.6 cm-1(b) With R1 = 0.9 and R2 = 0.32,

gth = 1

0.05 cm ln  

10.9(0.32) + 10 cm-1 = 34.9 cm-1(c) From Eq (4-37) ηext = ηi (gth - α )/gth ;

thus for case (a): ηext = 0.65(55.6 - 10)/55.6 = 0.53

For case (b): ηext = 0.65(34.9 - 10)/34.9 = 0.46

4-10 Using Eq (4-4) to find Eg and Eq (4-3) to find λ, we have for x = 0.03,

λ = 1.24

Eg =

1.241.424 + 1.266(0.3) + 0.266(0.3)2 = 1.462 µmFrom Eq (4-38)

ηext = 0.8065 λ(µm) dP(mW)

dI(mA)Taking dI/dP = 0.5 mW/mA, we have ηext = 0.8065 (1.462)(0.5) = 0.5904-11 (a) From the given values, D = 0.74, so that ΓT = 0.216

Then neff2

= 10.75 and W = 3.45, yielding ΓL = 0.856(b) The total confinement factor then is Γ = 0.185

4-12 From Eq (4-46) the mode spacing is

∆λ = λ2

2Ln =

(0.80 µm)22(400 µm)(3.6) = 0.22 nmTherefore the number of modes in the range 0.75-to-0.85 µm is

= (50 cm-1) exp - 

(λ - 850)2

2048 (b) On the plot of g(λ) versus λ, drawing a horizontal line at g(λ) = αt

= 32.2 cm-1 shows that lasing occurs in the region 820 nm < λ < 880 nm

(c) From Eq (4-47) the mode spacing is

∆λ = λ2

2Ln =

(850)22(3.6)(400 µm) = 0.25 nmTherefore the number of modes in the range 820-to-880 nm is

N = 880 - 8200.25 = 240 modes

4-14 (a) Let Nm = n/λ = 2Lm be the wave number (reciprocal wavelength) of mode m

The difference ∆N between adjacent modes is then

Equating (a-1) and (a-2) then yields ∆λ = λ2

2 =  

3.6-13.6+1

2 = 0.32

Then Jth = 1

12Lln

4-16 From the given equation

∆E11 =1.43 eV+(6.6256×10−34J⋅s)2

8 5 nm( )2

16.19×10−32 kg+ 1

Thus the emission wavelength is λ = hc/E = 1.240/1.68 = 739 nm

4-17 Plots of the external quantum efficiency and power output of a MQW laser

4-18 From Eq (4-48a) the effective refractive index is

ne = mλB

2Λ =

2(1570 nm)2(460 nm) = 3.4Then, from Eq (4-48b), for m = 0

For m = 2, λ = λB± 5(1.20 nm) = 1570 nm ± 6.0 nm

4-19 (a) Integrate the carrier-pair-density versus time equation from time 0 to td (time

for onset of stimulated emission) In this time the injected carrier pair densitychanges from 0 to nth

nth

∫ dn= −τ J

qd− nτ

qd−nτ

4-20 A common-emitter transistor configuration:

4-21 Laser transmitter design

4-22 Since the dc component of x(t) is 0.2, its range is -2.36 < x(t) < 2.76 The power

has the form P(t) = P0[1 + mx(t)] where we need to find m and P0 The averagevalue is

4-23 Substitute x(t) into y(t):

y(t) = a1b1 cos ω1t + a1b2 cos ω2t

+ a2(b12

cos2ω1t + 2b1b2 cos ω1t cos ω2t + b22

cos2ω2t)+ a3(b13

cos3ω1t + 3b12

b2 cos2ω1t cos ω2t + 3b1b22

cos ω1t cos2ω2t+ b23

cos3ω2t)+a4(b14

cos4ω1t + 4b13

b2 cos3ω1t cos ω2t + 6b12b22

cos2ω1t cos2ω2t+ 4b1b23

iii) cos4 x = 1

8 (cos 4x + 4cos 2x + 3) iv) 2cos x cos y = cos (x+y) + cos (x-y)

v) cos2 x cos y = 14 [cos (2x+y) + 2cos y + cos (2x-y)]

vi) cos2 x cos2 y = 14 [1 + cos 2x+ cos 2y + 12 cos(2x+2y) + 12 cos(2x-2y)]vii) cos3 x cos y = 1

8 [cos (3x+y) + cos (3x-y) + 3cos (x+y) + 3cos (x-y)]

+ 14 a3b31 cos 3ω1t + 14 a3b32 cos 3ω2t 3rd-order harmonic terms

2nd-order intermodulation terms

+ 34 a3b21 b2 [cos(2ω1+ω2)t + cos(2ω1-ω2)t + ] 34 a3b1b22 [cos(2ω2+ω1)t + cos(2ω2-ω1)t ]

3rd-order intermodulation terms

+ 12 a4b31 b2 [cos(3ω1+ω2)t + cos(3ω1-ω2)t ]

+ 34 a4b21 b22 [cos(2ω1+2ω2)t + cos(2ω1-2ω2)t ]

+ 12 a4b1b32 [cos(3ω2+ω1)t + cos(3ω2-ω1)t 4th-order intermodulation]

terms

This output is of the form

y(t) = A0 + A1(ω1) cos ω1t + A2(ω1) cos 2ω1t + A3(ω1) cos 3ω1t

+ A3(ω2) cos 3ω2t + A4(ω2) cos 4ω2t + ∑

m

∑n

Bmn cos(mω1+nω2)t where An(ωj) is the coefficient for the cos(nωj)t term

4-24 From Eq (4-58) P = P0 e-t/τm

where P0 = 1 mW and τm = 2(5×104 hrs) = 105hrs

(a) 1 month = 720 hours Therefore:

s = 1.11×10-5 exp{0.63/[(8.625×10-5)(293)]} = 7.45×105 hrs

Problem Solutions for Chapter 5

PLED-graded = 2π2(2×10-3 cm)2(100 W/cm2)(1.48)2(.01)1 - 1225 

2 = 159 µW5-5 Using Eq (5-10), we have that the reflectivity at the source-to-gel interface is

The total reflectivity then is R =Rs−gRg−f =7.30×10−4

The power loss in decibels is (see Example 5-3)

L = −10 log (1−R)= −10 log (0.999)=3.17×10−3 dB

5-6 Substituting B(θ) = B0 cosmθ into Eq (5-3) for B(θ,φ), we have

The power loss is L= −10 log (1−R)= −10 log (0.962)=0.17 dB

5-9 Shaded area = (circle segment area) - (area of triangle) = 1

2 1/2 Therefore

Acommon = 2(shaded area) = sa – cy = 2a2 arccos  

d2a - d a2 -  

d24

1/2

5-10

Coupling loss (dB) for Given axial misalignments (µµm) Core/cladding diameters

x52(4)(5) +

Therefore, for 2ad << 1, we have arccos 2ad ≈π2 - 2ad

Thus Eq (5-30) becomes PT = 2π Pπ2 - 

d2a -

5d6a = P 1 - 

5-12 Plots of mechanical misalignment losses

5-13 From Eq (5-20) the coupling efficiency η

F is given by the ratio of the number ofmodes in the receiving fiber to the number of modes in the emitting fiber, wherethe number of modes M is found from Eq (5-19) Therefore

a2R

a2E

Therefore from Eq (5-21) the coupling loss for aR≤ aE is LF = -10 log

5-14 For fibers with different NAs, where NAR < NAE

2α+4 a2

5-16 The splice losses are found from the sum of Eqs (5-35) through (5-37) First find

NA(0) from Eq (2-80b)

5-17 Plots of connector losses using Eq (5-43)

5-18 When there are no losses due to extrinsic factors, Eq (5-43) reduces to

5-20 Plot of the throughput loss.

Problem Solutions for Chapter 6

6-1 From Eqs (6-4) and (6-5) with Rf = 0, η = 1 - exp(-αsw)

To assist in making the plots, from Fig P6-1, we have the following representativevalues of the absorption coefficient:

6-4 (a) Using the fact that Va≈ VB, rewrite the denominator as

Since VB - VVa + IMRM

B << 1, we can expand the term in parenthesis:

6-6 Same problem as Example 6-6: compare Eqs (6-13), (6-14), and (6-17).

(a) First from Eq (6-6),

Ip = ηqλ

hc P0 =0.593 µA

σ2

=2qI B=2 1.6( ×10−19 C)(0.593 µA)(150×106 Hz)=2.84×10−17 A2

where P0 is given in watts To convert P0 = 10-n W to dBm, use 10 log

6-8 Using Eq (6-18) we have

S

N =

1

2 (R0P0m)2M22qB(R0P0 + ID)M5/2 + 2qILB + 4kBTB/RL

= 1.215×10− 16M2

2.176×10−23M5/ 2+1.656×10−19

The value of M for maximum S/N is found from Eq (6-19), with x = 0.5:

Moptimum = 62.1

0 = I2p M -

(2+x)M1+x 2q(Ip + ID)12 I2pM22q(Ip + ID)M2+x + 2qIL + 4kBT/RL

Solving for M: M2+xopt = 2qIL + 4kBT/RL

where the first and third terms cancelled because L2p = Dpτ

p Substituting in for B:

Lp

= qΦ0 αsLp

1 + αsLp e

-αsw + qpn0Dp

Lp c) Adding Eqs (6-21) and (6-25), we have

Jtotal = Jdrift + Jdiffusion = qΦ0

Lp 6-11 (a) To find the amplitude, consider

-jω td

jωtd

We want to find the value of ωtd at which (S S*)1/2 = 1

2 Evaluating (S S*)1/2 , we have

1/2

= 1 - e+jω td  

+ e-jω td

+ 11/2

ωtd2 = sinc 

td = w

vd =

20×10− 6m4.4×104m / s= 0.45 ns

c) α1s = 10-3 cm = 10 µm = 1

2 w

Thus since most carriers are absorbed in the depletion region, the carrier diffusiontime is not important here The detector response time is dominated by the RCtime constant

6-13 (a) With k1≈ k2 and keff defined in Eq (6-10), we have

Fe = keffMe + 2(1 - keff) - 1

Me(1 - keff) = keffMe + 2 - 

1

Me (1 - keff) (b) With k1≈ k2 and k'eff defined in Eq (6-40), we have

Therefore Eq (6-35) becomes Eq (6-39): Fh = k'

6-14 (a) If only electrons cause ionization, then β = 0, so that from Eqs 36) and

(6-37), k1 = k2 = 0 and keff = 0 Then from Eq (6-38)

Fe = 2 - 1

Me ≈ 2 for large Me(b) If α = β, then from Eqs (6-36) and (6-37), k1 = k2 = 1 so that

keff = 1 Then, from Eq (6-38), we have Fe = Me