Optical Fiber Communication - Solution Manual

Problem Solutions for Chapter 2 2-1. E=100cos

(

2π108t+30°

)

e_x +20 cos

(

2π108t −50°

)

e_y + 40cos

(

2π108_t+210°

)

_e z

2-2. The general form is:

y = (amplitude) cos(ωt - kz) = A cos [2π(νt - z/λ)]. Therefore (a) amplitude = 8 µm (b) wavelength: 1/λ = 0.8 µm-1 _{so that} λ _{= 1.25} µ_m (c) ω = 2πν = 2π(2) = 4π (d) At t = 0 and z = 4 µm we have y = 8 cos [2π(-0.8 µm-1₎₍₄ µ_m)] = 8 cos [2π(-3.2)] = 2.472

2-3. For E in electron volts and λ in µm we have E = 1.240

λ (a) At 0.82 µm, E = 1.240/0.82 = 1.512 eV At 1.32 µm, E = 1.240/1.32 = 0.939 eV At 1.55 µm, E = 1.240/1.55 = 0.800 eV (b) At 0.82 µm, k = 2π/λ = 7.662 µm-1 At 1.32 µm, k = 2π/λ = 4.760 µm-1 At 1.55 µm, k = 2π/λ = 4.054 µm-1

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 ωt Since the a's and the δ's are constants, we can set

a₁ sin δ₁ + a₂ sin δ₂ = A sin φ (2)

provided that constant values of A and φ exist which satisfy these equations. To verify this, first square both sides and add:

A2 (sin2 φ + cos2 φ) = a1 2 sin2δ1+cos2δ1

(

)

+ a₂2 _sin2_δ 2 +cos 2_δ 2

(

)

+ 2a₁a₂ (sin δ₁ sin δ₂ + cos δ₁ cos δ₂) or

A2 = a₁2+a₂2 + 2a₁a₂ cos (δ₁ - δ₂) Dividing (2) by (1) gives

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

Ex

E0 x

cos δ = cos (ωt - kz) cos δ to yield

E_y E_{0 y}

-Ex

E_0x cos δ = - sin (ωt - kz) sin δ (2.5-2)

sin2 (ωt - kz) = [1 - cos2 (ωt - kz)] = 1− Ex E_0x       2       (2.5-3)

Squaring both sides of Eq. (2.5-2) and substituting it into Eq. (2.5-3) yields

Ey E0 y − Ex E0x cosδ       2 = 1− Ex E0x       2       sin2δ

Expanding the left-hand side and rearranging terms yields

E_x E_0x       2 + Ey E_0y       2 - 2 Ex E_0x       Ey E_0y       cos δ = sin2δ 2-6. Plot of Eq. (2-7).

2-7. Linearly polarized wave. 2-8.

33 ° 33 °

90 ° Glass

Air: n = 1.0

(a) Apply Snell's law

n₁ cos θ₁ = n₂ cos θ₂ where n₁ = 1, θ₁ = 33°, and θ₂ = 90° - 33° = 57° ∴ n2 = cos 33° cos 57°= 1.540

(b) The critical angle is found from n_glass sin φ_glass = n_air sin φ_air

with φ_air = 90° and n_air = 1.0 ∴φcritical = arcsin 1 nglass = arcsin 1 1.540= 40.5° 2-9 Air Water 12 cm r θ

Find θ_c from Snell's law n₁ sin θ₁ = n₂ sin θ_c = 1 When n₂ = 1.33, then θ_c = 48.75°

Find r from tan θ_c =

r

12 cm , which yields r = 13.7 cm.

2-10.

45 °

Using Snell's law n_glass sin θ_c = n_alcohol sin 90° where θ_c = 45° we have

n_glass =

1.45

sin 45°= 2.05

2-11. (a) Use either NA =

(

n₁2−n₂2

)

1/ 2= 0.242 or

NA ≈ n1 2∆= n1 2(n1−n2)

n₁ = 0.243

(b) θ_0,max = arcsin (NA/n) = arcsin 0.242 1.0     = 14° 2-13. NA =

(

n₁2− n₂2

)

1/ 2=

[

n₁2 −n₁2(1− ∆)2

]

1/ 2 = n₁

(

2∆ − ∆2

)

1 / 2 Since ∆ << 1, ∆2 << ∆; ∴ NA ≈ n₁ 2∆

2-14. (a) Solve Eq. (2-34a) for jH_φ:

jH_φ = j εω

β Er -

1

βr

∂Hz

∂φ Substituting into Eq. (2-33b) we have

j β E_r + ∂Ez ∂r = ωµ j εω β Er − 1 βr ∂Hz ∂φ      

Solve for E_r and let q2 ₌ ω2

εµ

_- β2 _{to obtain Eq. (2-35a).}

(b) Solve Eq. (2-34b) for jH_r: jH_r = -j εω

β Eφ -

1

β ∂Hz

∂r Substituting into Eq. (2-33a) we have

j β E_φ+ 1 r ∂E_z ∂φ = -ωµ −j εω β Eφ− 1 β ∂Hz ∂r      

jE_r = 1 εω 1 r ∂Hz ∂φ +jrβHφ    

 Substituting into Eq. (2-33b) we have

β εω 1 r ∂Hz ∂φ +jrβHφ      + ∂ Ez ∂r = jωµ Hφ

Solve for H_φ and let q2 = ω2

εµ

- β2 to obtain Eq. (2-35d). (d) Solve Eq. (2-34b) for jE_φ

jE_φ = - 1

εω jβHr + ∂Hz

∂r



   Substituting into Eq. (2-33a) we have

1 r ∂E_z ∂φ - β εω jβHr + ∂Hz ∂r     = -jωµ Hr

Solve for H_r to obtain Eq. (2-35c).

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

- j q2 1 r ∂ ∂r β ∂H_z ∂φ + εωr ∂E_z ∂r      − ∂ ∂φ β ∂H_z ∂r − εω r ∂E_z ∂φ            = j

ε

ωEz

Upon differentiating and multiplying by jq2/

ε

ω we obtain Eq. (2-36).

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

- j q2 1 r ∂ ∂r β ∂Ez ∂φ − µωr ∂Hz ∂r      − ∂ ∂φ β ∂Ez ∂r + µω r ∂Hz ∂φ            = -j

µ

ωHz

Upon differentiating and multiplying by jq2/

ε

ω we obtain Eq. (2-37).

E_z = AJ₀(ur) ej(ωt− βz) and H_z = BJ₀(ur) ej(ωt− βz)

We want to find the coefficients A and B. From Eqs. (2-47) and (2-51), respectively, we have

C = Jν(ua)

K_ν(wa) A and D =

Jν(ua)

K_ν(wa) B

Substitute these into Eq. (2-50) to find B in terms of A:

A jβν a     u12 + 1 w2     = Bωµ _uJJ'ν(ua) ν(ua) + K'ν(wa) wK_ν(wa)      

For ν = 0, the right-hand side must be zero. Also for ν = 0, either Eq. (2-55a) or (2-56a) holds. Suppose Eq. (2-56a) holds, so that the term in square brackets on the right-hand side in the above equation is not zero. Then we must have that B = 0, which from Eq. (2-43) means that H_z = 0. Thus Eq. (2-56) corresponds to TM_0m modes.

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

0 = 1 u2 B

jβν

a Jν(ua)+Aωε1uJ'ν(ua)

    + 1 w2 B jβν a Jν(ua)+Aωε2w K'_ν(wa)J_ν(ua) Kν(wa)       With k₁2 = ω2µε1 and k2 2 = ω2µε2 rewrite this as Bν = ja βωµ 1 1 u2 + 1 w2         k1 2_J ν +k2 2_K ν

[

]

A

where J_ν and K_ν are defined in Eq. (2-54). If for ν = 0 the term in square brackets on the right-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 E_z = 0. Thus Eq. (2-55) corresponds to TE_0m modes.

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

∆ = n1 2_−n 2 2 2n1 2 = 1 2 1− n22 n1 2       ∆ << 1 implies n₁≈ n₂

Thus using Eq. (2-46), which states that n₂k = k₂≤ β ≤ k₁ = n₁k, we have

n2 2_k2

=k₂2 ≈n₁2k2 =k₁2 ≈ β2

2-17.

2-18. (a) From Eqs. (2-59) and (2-61) we have

M≈ 2π2a2 λ2 n1 2 _−n 2 2

(

)

=2π2a2 λ2

( )

NA 2 a = M 2π     1/ 2 λ NA= 1000 2     1/ 2 0.85µm 0.2π = 30.25µm Therefore, D = 2a =60.5 µm (b) M= 2π 2 30.25µm

(

)

2 1.32µm

(

)

2

( )

0.2 2 =414 (c) At 1550 nm, M = 300 2-19. From Eq. (2-58),

V = 2π (25 µm) 0.82 µm (1.48) 2 _−(1.46)2

[

]

1/ 2 = 46.5 Using Eq. (2-61) M ≈ V2/2 =1081 at 820 nm.

Similarly, M = 417 at 1320 nm and M = 303 at 1550 nm. From Eq. (2-72) P_clad

P



   _total≈ 4₃M-1/2 = 4_{3 1080}×100%= 4.1%

at 820 nm. Similarly, (P_clad/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

a = Vλ 2π n1 2 _−n 2 2

(

)

−1/ 2 = 2.40(1.32µm) 2π

[

(1.480)2−(1.478)2

]

1/ 2 = 6.55 µm (b) From Eq. (2-23) NA = n1 2₋ n2 2

(

)

1/ 2 =

[

(1.480)2−(1.478)2

]

1/ 2= 0.077 (c) From Eq. (2-23), NA = n sin θ_0,max. When n = 1.0 then

θ0,max = arcsin NA n     = arcsin   0.077_1.0   = 4.4° 2-22. n₂ = n1 2_−NA2 = (1.458)2 −(0.3)2 = 1.427 a = λV 2πNA= (1.30)(75) 2π(0.3) = 52 µm

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

λ n1 2∆

For a = 5 µm we have ∆ ≈ 0.002, so that at 0.82 µm V ≈

2π (5 µm)

0.82 µm 1.45 2(0.002) = 3.514

Thus the fiber is no longer single-mode. From Figs. 2-18 and 2-19 we see that the LP₀₁ and the LP₁₁ modes exist in the fiber at 0.82 µm.

2-24. 2-25. From Eq. (2-77) L_p = 2π β = λ n_y−n_x For Lp = 10 cm ny - nx = 1.3×10−6 _m 10−1_m = 1.3×10-5 For Lp = 2 m ny - nx = 1.3×10−6 _m 2m = 6.5×10-7 Thus 6.5×10-7 ≤ ny - nx ≤ 1.3×10-5

2-26. _{We want to plot n(r) from n2 to n1. From Eq. (2-78)}

n(r) = n1

[

1−2∆(r / a)α

]

1 / 2

= 1.48

[

1−0.02(r / 25)α

]

1 / 2 n2 is found from Eq. (2-79): n2 = n1(1 - ∆) = 1.465 2-27. From Eq. (2-81) M = α α +2 a 2_k2_n 1 2_{∆ =} α α +2 2πan₁ λ     2 ∆ where ∆ = n1−n2 n₁ = 0.0135

At λ = 820 nm, M = 543 and at λ = 1300 nm, M = 216. For a step index fiber we can use Eq. (2-61)

M_step≈ V 2 2 = 1 2 2πa λ     2 n₁2 _−n 2 2

(

)

At λ = 820 nm, M_step = 1078 and at λ = 1300 nm, M_step = 429. Alternatively, we can let α = ∞ in Eq. (2-81):

M_step = 2πan1 λ     2 ∆ = 1086 at 820 nm 432 at1300 nm   

2-28. Using Eq. (2-23) we have

(a) NA =

(

n₁2−n₂2

)

1/ 2=

[

(1.60)2−(1.49)2

]

1/ 2= 0.58 (b) NA =

[

(1.458)2−(1.405)2

]

1/ 2= 0.39

2-29. (a) From the Principle of the Conservation of Mass, the volume of a preform rod section of length Lpreform and cross-sectional area A must equal the volume of the fiber

drawn from this section. The preform section of length Lpreform is drawn into a fiber of

length Lfiber in a time t. If S is the preform feed speed, then Lpreform = St. Similarly, if s is the

fiber 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)2

and Fiber volume = Lfiber (d/2)2 = st (d/2)2

Equating these yields St D 2     2 = st d 2     2 or s = S D d     2 (b) S = s d D     2 = 1.2 m/s 0.125 mm 9 mm     2 = 1.39 cm/min

2-30. Consider the following geometries of the preform and its corresponding fiber: PREFORM FIBER 4 mm 3 mm R 25 µm 62.5 µm

We want to find the thickness of the deposited layer (3 mm - R). This can be done by comparing the ratios of the preform core-to-cladding cross-sectional areas and the fiber core-to-cladding cross-sectional areas:

Apreformcore Apreformclad = Afibercore A_fiberclad or π(32 _−R2₎ π(42 ₋₃2₎ = π(25)2 π

[

(62.5)2−(25)2

]

from which we have

R = 9− 7(25) 2 (62.5)2 −(25)2       1/ 2 = 2.77 mm Thus, thickness = 3 mm - 2.77 mm = 0.23 mm.

2-31. (a) The volume of a 1-km-long 50-µm diameter fiber core is V = πr2L = π (2.5×10-3 _cm)2 ₍₁₀5 _{cm) = 1.96 cm}3

The mass M equals the density ρ times the volume V: M = ρV = (2.6 gm/cm3)(1.96 cm3) = 5.1 gm

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

R =

5.1gm

0.5 gm / min = 10.2 min

2-32. Solving Eq. (2-82) for χ yields

χ = K

Yσ    

2

where Y = π for surface flaws. Thus

χ =

(20 N / mm3 / 2₎2

(70 MN / m2₎2 _π= 2.60×10-4 mm = 0.26 µm

2-33. (a) To find the time to failure, we substitute Eq. (2-82) into Eq. (2-86) and integrate (assuming that σ is independent of time):

χ−b / 2 χi χf

∫

dχ = AYbσb 0 t

∫

dt which yields 1 1−b 2 χf 1−b / 2_{− χ} i 1−b/ 2

[

]

= AYbσb_t or t = 2 (b−2)A(Yσ)b χi (2−b) / 2 _{− χ} f ( 2−b) / 2

[

]

(b) Rewriting the above expression in terms of K instead of χ yields

t = 2 (b−2)A(Yσ)b Ki Yσ     2−b − Kf Yσ     2−b       ≈ 2Ki 2−b (b−2)A(Yσ)b if Ki b−2 _{<< K} f b−2 or K_i2−b >>K_f2−b

2-34. Substituting Eq. (2-82) into Eq. (2-86) gives dχ

Integrating this from χ_i to χ_p where χ_i = K Yσi       2 _{and χp =} K Yσp       2

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

χ−b / 2 χi χp

∫

dχ = AYb σb 0 tp

∫

dt or 1 1−b 2 χp1−b / 2_{− χi}1−b / 2

[

]

= AYb σb_p t_p

for a constant stress σ_p. Substituting for χ_i and χ_p gives 2 b−2       K_Y  2−b σ_ib−2_{− σ} p b−2

[

]

= AYb σb_p tp or 2 b−2       K_Y  2−b 1 AYb σi b−2_{− σ} p b−2

[

]

= B

[

σ_ib−2 − σ_pb−2

]

= σp b t_p which is Eq. (2-87).

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

∫

dχ = AYb σs b 0 t_s

∫

dt

where χ_s is the crack depth at the fiber failure point. Integrating (as above) we get Eq. (2-89):

B

[

σ_pb−2 − σ_sb−2

]

= σ_sb t_s

2-35. (a) Substituting N_s as given by Eq. (2-92) and N_p as given by Eq. (2-93) into Eq. (2-94) yields F = 1 - exp − L L₀ σp b_t p + σs b_t s

(

)

/ B+ σs b−2

[

]

m b−2 σ0 m − σ_pb_t p/ B+ σp b−2

(

)

m b−2 σ0 m                   = 1 - exp − L L₀σ0 m σp b_t p / B+ σp b−2

[

]

bm−2 σp b_t p+ σs b_t s B       + σsb−2 σp b_t p/ B+ σp b−2           m b−2               −1             = 1 - exp −LN_p 1+σs b_t s σ_pb_t p + σs σ_p       b B σ_s2_t p       m b−2 1+ B σ_p2_t p −1                             ≈ 1 - exp −_{LNp 1}+ σs b_t s σ_pb tp       1 1+ B σ_p2 tp           m b−2 −1                         

(b) For the term given by Eq. (2-96) we have

σs σ_p       b B σ_s2_t p = (0.3)15 0.5 (MN / m2₎2 _s 0.3 (350 MN / m2₎

[

]

2 10 s= 6.5×10 -14

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

1 - exp − σ1c σ0       m L1 L0       = 1 - exp − σ2c σ0       m L2 L0      

which implies that

σ_1c σ₀       m L₁ L₀ = σ_2c σ₀       m L₂ L₀ or σ_1c σ_2c = L₂ L₁       1 / m If L1 = 20 m, then σ1c= 4.8 GN/m2 If L2 = 1 km, then σ2c= 3.9 GN/m2 Thus 4.8 3.9     m = 1000 20 = 50 gives m = log 50 log(4.8/ 3.9)= 18.8

Problem Solutions for Chapter 3 3-1. α

(

dB/ km

)

= 10 z log P(0) P(z)       = 10 z log e αpz

( )

=10αplog e=4.343 αp(1/ km)

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,

P(100 µW) = 10 log (100 µW/1.0 mW) = 10 log (0.10) = - 10.0 dBm P(150 µW) = 10 log (150 µW/1.0 mW) = 10 log (0.15) = - 8.24 dBm (a) At 8 km we have the following power levels:

P1300(8 km) = - 8.2 dBm – (0.6 dB/km)(8 km) = - 13.0 dBm = 50 µW

P1550(8 km) = - 10.0 dBm – (0.3 dB/km)(8 km) = - 12.4 dBm = 57.5 µW

(b) At 20 km we have the following power levels:

P1300(20 km) = - 8.2 dBm – (0.6 dB/km)(20 km) = - 20.2 dBm = 9.55 µW

P1550(20 km) = - 10.0 dBm – (0.3 dB/km)(20 km) = - 16.0 dBm = 25.1 µW

3-3. From Eq. (3-1c) with P_out = 0.45 P_in

α = (10/3.5 km) log (1/0.45) = 1.0 dB/km

3-4. (a) P_in = P_out 10αL/10 = (0.3 µW) 101.5(12)/10 = 18.9 µW (b) P_in = P_out 10αL/10 = (0.3 µW) 102.5(12)/10 = 300 µW

3-5. With λ in Eqs. (3-2b) and (3-3) given in µm, we have the following representative points for α_uv and α_IR:

λλ (µµm) αα_uv αα_IR 0.5 20.3 --0.7 1.44 --0.9 0.33 --1.2 0.09 _2.2×₁₀-6 1.5 0.04 0.0072 2.0 0.02 23.2 3.0 0.009 7.5×104

3-6. From Eq. (3-4a) we have

αscat = 8π 3 3λ4 (n 2 −1)2kBTfβT = 8π3 3(0.63µm)4 (1.46) 2 ₋₁

[

]

2 (1.38×10-16 dyne-cm/K)(1400 K) ×(6.8×10-12 cm2/dyne) = 0.883 km-1

To change to dB/km, multiply by 10 log e = 4.343: α_scat = 3.8 dB/km From Eq. (3-4b): αscat = 8π 3 3λ4 n8p2 kBTfβT = 1.16 km -1 = 5.0 dB/km 3-8. Plot of Eq. (3-7). 3-9. Plot of Eq. (3-9).

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

ννm P_clad/P αα_ννm = αα₁₁ + (αα_{2 2} - αα₁₁)P_clad/P 5 + 103Pclad/P

01 0.02 3.0 + 0.02 5 + 20 = 25 11 0.05 3.0 + 0.05 5 + 50 = 55 21 0.10 3.0 + 0.10 5 + 100 = 105 02 0.16 3.0 + 0.16 5 + 160 = 165 31 0.19 3.0 + 0.19 5 + 190 = 195 12 0.31 3.0 + 0.31 5 + 310 = 315

3-11. (a) We want to solve Eq. (3-12) for α_gi. With α = 2 in Eq. (2-78) and letting ∆ = n 2_(0)−n 2 2 2n2(0) we have α(r) = α₁ + (α₂ - α₁) n 2_(0)−n2_(r) n2_(0)−n 2 2 = α1 + (α2 - α1) r 2 a2 Thus αgi = α(r) p(r) r dr 0 ∞

∫

p(r) r dr 0 ∞

∫

= α1 + (α₂− α₁) a2 exp

(

−Kr2

)

_r3_dr 0 ∞

∫

exp

(

−Kr2

)

_{r dr} 0 ∞

∫

To evaluate the integrals, let x = Kr2_{, so that dx = 2Krdr. Then}

exp

(

−Kr2

)

_r3_dr 0 ∞

∫

exp

(

−Kr2

)

_{r dr} 0 ∞

∫

= 1 2K2 e −x _{x dx} 0 ∞

∫

1 2K e −x _dx 0 ∞

∫

= 1 K 1! 0! = 1 K Thus α_gi = α₁ + (α2− α1) Ka2 (b) p(a) = 0.1 P₀ = P₀ e−Ka 2 yields eKa 2 = 10. From this we have Ka2 = ln 10 = 2.3. Thus

αgi = α1 + (α2− α1)

2.3 = 0.57α1 + 0.43α2

n = 1+ 196.98 (13.4)2 _{−(1.24 /λ)}2       1/ 2

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

0.2 µm 1.548 1.550

0.6 µm 1.457 1.458

1.0 µm 1.451 1.450

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)

σ_mat

L = dτ

dλ σλ = [80 ps/(nm-km)](45 nm) = 3.6 ns/km

For a laser diode,

σ_mat

L = [80 ps/(nm-km)](2 nm) = 0.16 ns/km (b) From Fig. 3-13, dτmat

dλ = 22 ps/(nm-km)

Therefore, D_mat(λ) = [22 ps/(nm-km)](75 nm) = 1.65 ns/km 3-14. (a) Using Eqs. (2-48), (2-49), and (2-57), Eq. (3-21) becomes

b = 1 - ua V     2 = 1 - u 2_a2 u2_a2 _+w2_a2 = w2 u2_+w2 = β 2 _−k2_n 2 2 k2n1 2_{− β}2 _{+ β}2₋ k2n2 2 = β2_{/ k}2_−n 2 2 n1 2 ₋ n2 2 (b) Expand b as b = (β/ k+n2)(β/ k−n2) (n₁ +n₂)(n₁ −n₂)

β/ k+n2 n₁ +n₂ = n1(1−δ)+n2 n₁+n₂ = 1 - n1 n₁+n₂ δ Letting n2 = n1(1 - ∆) then yields

β/ k+n₂ n₁ +n₂ = 1 - δ 2− ∆≈ 1 since δ 2− ∆ << 1 Therefore, b ≈ β/ k−n2 n1−n2 or β = k[bn₁∆ + n₂] From n₂ = n₁(1 - ∆) we have 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 sin φ = x s = n2 n1 = n1(1− ∆) n1 = (1 - ∆)

The travel time of the highest order ray is thus Tmax= n1 c s L x           = n1L c 1 1− ∆

For the axial ray the travel time is T_min= Ln1 c

ϕ

θ

a

s

x

Therefore T_min- T_max= Ln1 c 1 1− ∆−1     = Ln_c1_{1− ∆}∆ ≈ Ln_c1∆

3-17. Since n2 =n1

(

1− ∆

)

, we can rewrite the equation as σ_mod L = n₁∆ c 1− π V    

where the first tern is Equation (3-30). The difference is then given by the factor 1− π V =1− πλ 2a 1 n₁2 _−n 2 2

(

)

1/ 2 ≈1− πλ 2a 1 n₁ 2∆ At 1300 nm this factor is 1− π(1.3) 2 62.5

(

)

1 1.48 2(0.015) =1−0.127=0.873 3-18. For ε = 0 and in the limit of α → ∞ we have

C₁ = 1, C₂ = 3 2, α α +1= 1, α +2 3α +2= 1 3, α +1 2α +1= 1 2, and (α +1) 2 (5α +2)(3α +2) = 1 15 Thus Eq. (3-41) becomes

σ_{int er mod al} = Ln1∆ 2 3c 1+3∆ + 12 5 ∆ 2     1/ 2 ≈ Ln1∆ 2 3c

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, ...)

C₁ = α −2 α +2 = 2 1− 6 5∆     −2 2 1−6 5∆     +2 = − 3 5∆ 1−3 5∆ ≈ −3 5∆ 1+ 3 5∆    

C₂ = 3α −2 2(α +2) = 3 2 1−6 5∆           −2 2 2 1−6 5∆     +2       = 1− 9 5∆ 2 1−3 5∆    

Evaluating the factors in Eq. (3-41) yields: (a) C₁2 ≈ 9 25 ∆2 (b) 4C1C2(α +1)∆ 2α +1 = 4∆ −3 5∆       1− 9₅∆  _{ }  2 1  −6₅∆  +1 _{ } 1−3 5∆     2 1  −₅3∆  _{ }  4 1  −₅6∆  +1 _{ } = −18∆ 1−11∆ +18 25     5 1 5∆     2 24 25∆  _  18 25 ∆2 (c) 16∆ 2_C 2 2_{(α +1)}2 (5α +2)(3α +2) = 16∆2 1− 9 5∆     2 2(1−6 5∆)+1     2 4 1−3 5∆     2 10(1−6 5∆)+2     6(1− 6 5∆)+2     = 16∆2 ₁₋9 5∆     2 9 1− 4 5∆     2 96(1− ∆)(1− 9 10∆)4 1− 3 5∆     2 ≈ 9 24 ∆2 Therefore, σint er mod al = Ln1∆ 2c α α +1       ₃α +_{α +}2₂  1/ 2 9 25∆ 2₋18 25∆ 2₊ 9 24∆ 2     1/ 2 = Ln1∆ 2 2c 2(1−6 5∆) 2(1−6 5∆)+1     2(1−6 5∆)+2 6(1−6 5∆)+2         1/ 2 3 10 6 ≈ n₁∆2_L 20 3c

3-20. We want to plot Eq. (3-30) as a function of σ_λ , 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 σ_{int er mod al} does not vary with σ_λ , we have

σ_{int er mod al} L = N₁∆ 2c α α +2       ₃α +_{α +}2₂  1/ 2 _4∆C 2(α +1) (5α +2)(3α +2)= 0.070 ns/km

With C₁ = 0 we have from Eq. (3-45)

σint ra mod al = 1 c σ_λ λ −λ 2d2n1 dλ2      = 0.098σ_λ ns / km at 850 nm 1.026×10−2_σ λns / km at1300 nm   

3-21. Using the same parameter values as in Prob. 3-18, except with ∆ = 0.001, we have from Eq. (3-41) σ_{int er mod al} /L = 7 ps/km, and from Eq. (3.45)

σint ra mod al L = 0.098σ_λ ns / km at850 nm 0.0103σ_λ ns / km at1300 nm    The plot of σ L= 1 L σint er 2 _{+ σ} int ra 2

(

)

1/ 2 vs σ_λ :

3-22. Substituting Eq. (3-34) into Eq. (3-33)

τ_g = L c dβ dk= L c 1 2β 2kn1 2_+2k2_n 1 dn₁ dk    - 2 α +2 α m a2     α α+2 2 α +2 n1 2 k2∆

(

)

α+22−1 × ∆ 2k2n1 dn1 dk +2kn1 2₊n1 2_k2 ∆ d∆ dk          = L c kn1 β N1− 4∆ α +2 α +2 α m a2 1 n1 2 k2∆       α α +2 N1 + n1k 2∆ d∆ dk            

= LN1 c kn1 β 1− 4∆ α +2 m M     α α +2 1+ ε 4           with N₁ = n₁ + k dn1

dk and where M is given by Eq. (2-97) and ε is defined in Eq. (3-36b).

3-23. From Eq. (3-39), ignoring terms of order ∆2,

λdτ dλ = L c dN1 dλ 1+ α − ε −2 α +2 m M     α α+2       + LN1 c α − ε −2 α +2 d dλ ∆ m M     α α +2       where N1 = n1 - λ dn1 dλ and M = α α +2a2k2n1 2 ∆ (a) dN1 dλ = d dλ n1− λ dn1 dλ     = - λ d 2_n 1 dλ2

Thus ignoring the term involving ∆ d

2_n 1

dλ2 , the first term in square brackets

becomes - L c λ2 d2_n 1 dλ2 (b) d dλ ∆ m M     α α+2       = m α α+2 d∆ dλ 1 M     α α +2 + ∆ −α α +2 dM dλ 1 M     α α +2+1       (c) dM dλ = α α +2a2 d dλ k 2 n1 2_∆

(

)

= α a2  _ d∆k2n2+2k2∆n dn1 +2kn2∆dk _

Ignoring d∆ dλand dn1 dλ terms yields dM dλ = 2α α +2a2k2n1 2_{∆ −}1 λ     = - 2M_λ so that d dλ ∆ m M     α α+2       = ∆ λ 2α α +2 m M     α α +2 . Therefore λdτ dλ = - L c λ2 d2_n 1 dλ2 + LN1 c α − ε −2 α +2 2α∆ α +2 m M     α α +2 3-24. Let a = λ2 d 2_n 1 dλ2 ; b = N1C1∆ _{α +}2α₂; γ = _{α +}α₂

Then from Eqs. (3-32), (3-43), and (3-44) we have

σ_{int ra mod al}2 = L2 σλ λ     2 1 M λ dτ_g dλ     m=0 M

∑

2 = L c     2 σλ λ     2 1 M −a+b m M     γ       m=0 M

∑

2 ≈ L c     2 σλ λ     2 1 M −a+b m M     γ       2 dm 0 M

∫

= L c     2 σλ λ     2 1 M a 2 −2ab m M     γ +b2 m M     2γ       dm 0 M

∫

= L c     2 _σ λ λ     2 a2 ₋ 2ab γ +1+ b2 2γ +1       = L c     2 _σ λ λ     2 −λ2 d2n1 dλ2      2   

- 2 λ2 d 2_n 1 dλ2      N1C1∆ _{α +}α ₁+ (N1C1∆) 2 4α2 (α +2)(3α +2)    3-25. Plot of Eq. (3-57). 3-26. (a) D= λ − λ

(

0

)

S0 = −50(0.07)= −3.5 ps /(nm−km) (b) D= 1500(0.09) 4 1− 1310 1500     4       =14.1 ps /(nm−km)

3-27. (a) From Eq. (3-48)

σstep L = n1∆ 2 3 c = 1.49(0.01) 2 3 (3×108) =14.4 ns / km (b) From Eq. (3-47) σopt L = n1∆2 20 3 c = 1.49(0.01)2 20 3 (3×108₎ =14.3 ps / km (c) 3.5 ps/km 3-28. (a) From Eq. (3-29)

σmod =Tmax−Tmin =

n1∆L c = (1.49)(0.01)(5×103_m) 3×108m / s =248 ns (b) From Eq. (3-48) σ_step = n1∆L 2 3 c = 248 2 3 =71.7 ns (c) BT = 0.2 σ_step =2.8 Mb / s (d) BT ⋅L=(2.8 MHz)(5 km)=13.9 MHz⋅km

3-29. For α =0.95αopt, we have σint er(α ≠ αopt) σint er(α = αopt) = (α − αopt) ∆(α +2) = − 0.05 (0.015)(1.95)= −170% For α =1.05αopt, we have

σint er(α ≠ αopt) σint er(α = αopt) = (α − αopt) ∆(α +2) = + 0.05 (0.015)(2.05) = +163%

Problem Solutions for Chapter 4 4-1. From Eq. (4-1), n_i = 2       2πkBT h2 3/2 (m_em_h)3/4 exp       - Eg 2k_BT = 2 2π(1.38×10 −23_{J/ K)} (6.63×10−34_J.s)2       3/ 2 T3 / 2

[

(.068)(.56)(9.11×10−31kg)2

]

3/ 4 × exp −(1.55−4.3×10 −4_T)eV 2(8.62×10−5_{eV / K)T}       = 4.15×1014 T3/2 exp − 1.55 2(8.62×10−5)T       exp 4.3×10−4 2(8.62×10−5)       = 5.03×1015 T3/2 exp _- 8991_ T

4-2. The electron concentration in a p-type semiconductor is n

P = ni = pi Since both impurity and intrinsic atoms generate conduction holes, the total conduction-hole concentration p

P is p_P = NA + ni = NA + n_P

From Eq. (4-2) we have that n P = n 2 i /pP . Then p P = NA + nP = NA + n 2 i /pP or p 2 P - NApP - n 2 i = 0 so that p_P = NA₂









1 + 4n2_i N2_A + 1

If ni << NA , which is generally the case, then to a good approximation p_P ≈ _{NA and nP} = n2_i /p_P ≈ n2_{i /NA}

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

x = 1₂ [- 4.759 + (4.759)2 + 4(.436) = 0.090]

The emission wavelength is λ = 1.240

1.540 = 805 nm.

(b) E_g = 1.424 + 1.266(0.15) + 0.266(0.15)2 = 1.620 eV, so that

λ = 1.240_1.620 = 766 nm

4-4. (a) The lattice spacings are as follows: a(BC) = a(GaAs) = 5.6536 Ao

a(BD) = a(GaP) = 5.4512 Ao a(AC) = a(InAs) = 6.0590 Ao a(AD) = a(InP) = 5.8696 Ao

a(x,y) = xy 5.6536 + x(1-y) 5.4512 + (1-x)y 6.0590 + (1-x)(1-y)5.8696 = 0.1894y - 0.4184x + 0.0130xy + 5.8696

(b) Substituting a(xy) = a(InP) = 5.8696 Ao into the expression for a(xy) in (a), we have

y = _{0.1894 - 0.0130x}0.4184x ≈ 0.4184x_0.1894 = 2.20x (c) With x = 0.26 and y = 0.56, we have

E_g = 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 eV 4-5. Differentiating the expression for E, we have

∆E= hc

λ2 ∆λ or ∆λ = λ2

hc∆E

For the same energy difference ∆E, the spectral width ∆λ is proportional to the wavelength squared. Thus, for example,

∆λ₁₅₅₀ ∆λ₁₃₁₀ = 1550 1310     2 =1.40

4-6. (a) From Eq. (4-10), the internal quantum efficiency is

η_int= 1

1+25/ 90 =0.783, and from Eq. (4-13) the internal power level is

P_int =(0.783) hc(35 mA) q(1310 nm)= 26 mW (b) From Eq. (4-16), P = 1 3.5 3.5

(

+1

)

226 mW =0.37 mW

4-7. Plot of Eq. (4-18). Some representative values of P/P0 are given in the table:

f in MHz P/P0 1 0.999 10 0.954 20 0.847 40 0.623 60 0.469 80 0.370 100 0.303

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,

10 log 1 1+

₍

2πfτ

₎

2

[

]

1/ 2       = −3

With a 5-ns lifetime, we find f = 1 2π

(

5 ns

)

10 0.6₋₁

(

)

=9.5 MHz 4-9. (a) Using Eq. (4-28) with Γ = 1

g_th = 1 0.05 cmln     1 0.32 2 + 10 cm-1 _{= 55.6 cm}-1 (b) With R1 = 0.9 and R2 = 0.32, g_th = 1 0.05 cm ln     1 0.9(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 E_g = 1.24 1.424 + 1.266(0.3) + 0.266(0.3)2 = 1.462 µm From Eq. (4-38) ηext = 0.8065 λ(µm) dP(mW) dI(mA)

Taking dI/dP = 0.5 mW/mA, we have η_ext = 0.8065 (1.462)(0.5) = 0.590 4-11. (a) From the given values, D = 0.74, so that ΓT = 0.216

Then n_eff2

= 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)2

2(400 µm)(3.6) = 0.22 nm

.85−.75 .22×10−3 =

.1

.22 ×103 = 455 modes

4-13. (a) From Eq. (4-44) we have g(λ) = (50 cm-1_{) exp}

      - (λ - 850 nm) 2 2(32 nm)2 = (50 cm-1) exp _- (λ - 850) _ 2 2048

(b) On the plot of g(λ) versus λ, drawing a horizontal line at g(λ) = α_t = 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 = λ 2 2Ln =

(850)2

2(3.6)(400 µm) = 0.25 nm

Therefore the number of modes in the range 820-to-880 nm is N = 880 - 820_0.25 = 240 modes

4-14. (a) Let N_m = n/λ = _2Lm be the wave number (reciprocal wavelength) of mode m. The difference ∆N between adjacent modes is then

∆N = N_m - N_m-1 = 1

2L (a-1)

We now want to relate ∆N to the change ∆λ in the free-space wavelength. First differentiate N with respect to λ:

dN dλ = d dλ    n λ = 1 λ dn dλ - n λ2 = - 1 λ2_    n - λ dn dλ

Thus for an incremental change in wavenumber ∆N, we have, in absolute values,

∆N = _λ1₂     n - λdn dλ ∆λ (a-2)

Equating (a-1) and (a-2) then yields ∆λ = λ 2 2L     n - λdn dλ

(b) The mode spacing is ∆λ = (.85 µm) 2

2(4.5)(400 µm) = 0.20 nm

4-15. (a) The reflectivity at the GaAs-air interface is

R₁ = R₂ = __n+1n-1_ 2 = __3.6+13.6-1_ 2 = 0.32 Then J_th = 1 β α + 1 2Lln 1 R1R2       = 2.65×103 A/cm2 Therefore I_th = J_th× l × w = (2.65×103 A/cm2)(250×10-4 cm)(100×10-4 cm) = 663 mA (b) I_th = (2.65×103 _A/cm2₎₍₂₅₀×₁₀-4 _cm)(10×₁₀-4 _{cm) = 66.3 mA}

4-16. From the given equation

∆E₁₁ =1.43 eV+

(

6.6256×10−34J⋅s

)

2 8 5 nm

(

)

2 1 6.19×10−32 _kg+ 1 5.10×10−31 _kg      =1.43 eV+0.25 eV =1.68 eV

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

n_e = mλB 2Λ =

2(1570 nm)

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

λ = λ_B± λ 2 B 2n_eL    1 2 = 1570 nm ± (1.57 µm)(1570 nm) 4(3.4)(300 µm) = 1570 nm ± 1.20 nm Therefore for m = 1, λ = λ_B± 3(1.20 nm) = 1570 nm ± 3.60 nm For m = 2, λ = λ_B± 5(1.20 nm) = 1570 nm ± 6.0 nm

4-19. (a) Integrate the carrier-pair-density versus time equation from time 0 to t_d (time for onset of stimulated emission). In this time the injected carrier pair density changes from 0 to n_th. t_d= dt 0 td

∫

= _J 1 qd − n τ 0 nth

∫

dn= −τ J qd− n τ      n=0 n=nth = τ ln J J−J_th      

where J = Ip/A and Jth = Ith/A. Therefore td = τ ln _ _ I_p I_p - I_th

(b) At time t = 0 we have n = nB, and at t = td we have n = nth. Therefore,

t_d = _⌡⌠ 0 t_d dt = 1 J qd− n τ dn nB n_th

∫

= τ ln J qd− nB τ J qd − nth τ        

In the steady state before a pulse is applied, n_B = J_B

τ

/qd. When a pulse is applied, the current density becomes I/A = J = J_B + J_p = (I_B + I_p)/A

Therefore, t_d = τ ln       I - I_B I - I th = τ ln       I_p I_{p +} I_B - I th

4-20. A common-emitter transistor configuration:

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 average value is

<

P(t) = P

>

₀[1 + 0.2m] = 1 mW

The minimum value is

P(t) = P₀[1 - 2.36m] ≥ 0 which implies m ≤ _2.361 = 0.42

Therefore for the average value we have

<

P(t) = P

>

₀[1 + 0.2(0.42)] ≤ 1 mW, which implies

P₀ = 1

1.084 = 0.92 mW so thatP(t) = 0.92[1 + 0.42x(t)] mW and i(t) = 10 P(t) = 9.2[1 + 0.42x(t)] mA

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

y(t) = a₁b₁ cos ω₁t + a₁b₂ cos ω₂t

+ a₂(b₁2cos2ω₁t + 2b₁b₂ cos ω₁t cos ω₂t + b₂2cos2ω₂t)

+ a₃(b₁3cos3ω₁t + 3b₁2b₂ cos2ω₁t cos ω₂t + 3b₁b₂2cos ω₁t cos2ω₂t+ b₂3cos3ω₂t) +a₄(b₁4cos4ω₁t + 4b₁3b₂ cos3ω₁t cos ω₂t + 6b₁2b₂2cos2ω₁t cos2ω₂t

+ 4b₁b₂3cos ω₁t cos3ω₂t + b₂4cos4ω₂t)

Use the following trigonometric relationships: i) cos2 _{x =} 1

2 (1 + cos 2x) ii) cos3 x = 1₄ (cos 3x + 3cos x) iii) cos4 _{x =} 1

8 (cos 4x + 4cos 2x + 3) iv) 2cos x cos y = cos (x+y) + cos (x-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)] then y(t) = 1 2    a₂b2₁ + a₂b2₂ + 3 4 a4b 4 1 + 3a4b 2 1b 2 2 + 3 4 a4b 4 2 constant terms

+ ₄3_a₃b₁3 + 2a₃b₁b2_2 cos ω₁t + 3₄ a_3b3₂ + 2b₁2b_{2 } cos ω₂t fundamental_terms

+ b2₁ 2 a2 + a4b  2 1 + 3a4b 2 2 cos 2ω1t + b2₂ 2 a2 + a4b  2 2 + 3a4b 2 1 cos 2ω2t 2nd-order harmonic terms

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

+ 1 8 a4b 4 1 cos 4ω1t + 1 8 a4b 4

2 cos 4ω2t 4th-order harmonic terms

+ _a₂b₁b₂ + 3₂ a₄b3₁b₂ + 3₂ a₄b₁b3₂_[cos (ω₁+ω₂)t + cos (ω₁-ω₂)t _]

2nd-order intermodulation terms

+ ₄3 a₃b2₁ b_{2 [}cos(2ω₁+ω₂)t + cos(2ω₁-ω₂)t + _] 3₄ a₃b₁b2_{2 [}cos(2ω₂+ω₁)t + cos(2ω₂-ω₁)t _] 3rd-order intermodulation terms

+ ₂1 a₄b3₁ b_{2 [}cos(3ω₁+ω₂)t + cos(3ω₁-ω₂)t _] + 3₄ a₄b2₁ b2_{2 [}cos(2ω₁+2ω₂)t + cos(2ω₁-2ω₂)t _]

+ 1₂ a₄b₁b3_{2 [}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 + A (ω ) cos 4ω t + A (ω ) cos ω t + A (ω ) cos 2ω t

+ A₃(ω₂) cos 3ω₂t + A₄(ω₂) cos 4ω₂t +

∑

m

∑

n

B_mn 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 τ_m = 2(5×104 hrs) = 105 hrs.

(a) 1 month = 720 hours. Therefore:

P(1 month) = (1 mW) exp(-720/105) = 0.99 mW (b) 1 year = 8760 hours. Therefore

P(1 year) = (1 mW) exp(-8760/105) = 0.92 mW

(c) 5 years = 5×8760 hours = 43,800 hours. Therefore P(5 years) = (1 mW) exp(-43800/105) = 0.65 mW

4-25. From Eq. (4-60) τ_s = K eEA/kBT or ln τ_{s = ln K + EA/kBT} where kB = 1.38×10-23 J/°K = 8.625×10-5 eV/°K At T = 60°C = 333°K, we have ln 4×104 = ln K + EA/[(8.625×10-5 eV)(333)] or 10.60 = ln K + 34.82 E_{A (1)} At T = 90°C = 363°K, we have ln 6500 = ln K + E_A/[(8.625×10-5 eV)(363)] or 8.78 = ln K + 31.94 E_{A (2)}

Solving (1) and (2) for EA and K yields E_A = 0.63 eV and k = 1.11×10-5 hrs Thus at T = 20°C = 293°K

τ

s = 1.11×10

Problem Solutions for Chapter 5 5-3. (a) cosL 30° = 0.5 cos 30° = (0.5)1/L = 0.8660 L = log 0.5/log 0.8660 = 4.82 (b) cosT 15° = 0.5 cos 15° = (0.5)1/T _{= 0.9659} T = log 0.5/log 0.9659 = 20.0

5-4. The source radius is less than the fiber radius, so Eq. (5-5) holds: P_LED-step = π2_r2

s B0(NA)2 = π2(2×10-3 cm)2(100 W/cm2)(.22)2 = 191 µW From Eq. (5-9)

P_LED-graded = 2π2(2×10-3 cm)2(100 W/cm2)(1.48)2(.01)_1 - 1_22_52_ = 159 µW 5-5. Using Eq. (5-10), we have that the reflectivity at the source-to-gel interface is

Rs−g= 3.600−1.305 3.600+1.305     2 =0.219

Similarly, the relfectivity at the gel-to-fiber interface is Rg−f = 1.465−1.305 1.465+1.305     2 =3.34×10−3

The total reflectivity then is R =R_s−gR_g−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

P =

⌡





⌠

0 r_m

⌡



⌠

0 2π













2π _⌡⌠ 0 θ0-max cos3θ _sin θ _dθ _dθ_{s r dr} Using ⌡ ⌠ 0 θ0 cos3θ _sin θ _dθ _{= ⌡}⌠ 0 θ0

(1 - sin2θ) _sin θ _d(sin θ_{) = ⌡}⌠ 0 sin θ₀ (x - x3) _dx we have P = 2π

⌡





⌠

0 r_m ⌡  ⌠ 0 2π       sin2θ_0-max 2 - sin4θ_0-max 4 dθs r dr = π

⌡





⌠

0 r_m ⌡  ⌠ 0 2π     NA2 - 1₂ NA4 dθ_s r dr = π 2 [2NA2 - NA4]⌡⌠ 0 rm r dr_⌡⌠ 0 2π dθ_s

5-7. (a) Let a = 25 µm and NA = 0.16. For r_s≥ a(NA) = 4 µm, Eq. (5-17) holds. For r_s≤ 4 µm, η = 1.

(b) With a = 50 µm and NA = 0.20, Eq. (5-17) holds for r_s≥ 10 µm. Otherwise, η = 1.

R_g−f = 1.485−1.305 1.485+1.305     2 =4.16×10−3

The power loss is (see Example 5-3)

L = −10 log (1−R)= −10 log (0.9958)=0.018 dB When there is no index-matching gel, the joint loss is Ra−f = 1.485−1.000 1.485+1.000     2 =0.038

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 sa - 1 2 cy s = aθ = a [2 arccos (y/a)] = 2a arccos __2ad_

c = 2     a2 - _d_ 2 2 1/2 Therefore

A_common = 2(shaded area) = sa – cy = 2a2 _arccos

    d 2a - d     a2 _-     d2 4 1/2 5-10.

Coupling loss (dB) for

Given axial misalignments (µµm) Core/cladding diameters (µµm) 1 3 5 10 50/125 0.112 0.385 0.590 1.266 62.5/125 0.089 0.274 0.465 0.985 100/140 0.056 0.169 0.286 0.590 5-11. arccos x = π 2 - arcsin x

For small values of x, arcsin x = x + x 3 2(3) +

x5

Therefore, for _2ad << 1, we have arccos _2ad ≈π₂ - _2ad

Thus Eq. (5-30) becomes P_T = 2_π P_π _ 2 - d 2a - 5d 6a = P _    1 - 8d 3πa d/a P_T/P (Eq.5-30) P_T/P (Eq.5-31)

0.00 1.00 1.00 0.05 0.9576 0.9576 0.10 0.9152 0.9151 0.15 0.8729 0.8727 0.20 0.8309 0.8302 0.25 0.7890 0.7878 0.30 0.7475 0.7454 0.35 0.7063 0.7029 0.40 0.6656 0.6605

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 of modes in the receiving fiber to the number of modes in the emitting fiber, where the number of modes M is found from Eq. (5-19). Therefore

η F = M_aR M_aE = k2_NA2₍₀₎     1 2 -1 α+2 a 2 R k2NA2(0)     1 2 -1 α+2 a 2 E = a2_R a2_E

Therefore from Eq. (5-21) the coupling loss for a_R≤ a_E is L_F = -10 log

        a2_R a2_E

5-14. For fibers with different NAs, where NA_R < NA_E

L_F = -10 log η F = -10 log M_R M_E = -10 log k2NA2_R(0)       α 2α+4 a2 k2_NA2 E(0) _     α 2α+4 a2

= -10 log         NA2_R(0) NA2_E(0)

5-15. For fibers with different α values, where α

R < αE LF = -10 log η_F = -10 log k2NA2(0)     α R 2α R + 4 a2 k2_NA2₍₀₎     α E 2α E + 4 a2 = -10 log     α R(αE + 2) α E(αR + 2)

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).

For fiber 1: NA₁(0)=n₁ 2∆ =1.46 2 0.01

(

)

=0.206 For fiber 2: NA₂(0)=n₁ 2∆ =1.48 2 0.015

(

)

=0.256

(a) The only loss is that from index-profile differences. From Eq. (5-37)

L_1→2

( )

α = −10 log 1.80(2.00+2)

2.00(1.80+2) =0.24 dB

(b) The losses result from core-size differences and NA differences.

L_2→1(a)= −20 log 50 62.5     =1.94 dB L_2→1(NA)= −20 log .206 .256     =1.89 dB

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

L_SM;ff = -10 log













4       W₁ W₂ + W₂ W₁ 2

For W₁ = 0.9W₂ , we then have L_SM;ff = -10 log __4.04464 _ = - 0.0482 dB 5-19. Plot of Eq. (5-44).

Problem Solutions for Chapter 6

6-1. From Eqs. (6-4) and (6-5) with R_f = 0, η = 1 - exp(-α_sw)

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

λλ (µµm) αα_s (cm-1) .60 _4.4×₁₀3 .65 2.9×103 .70 2.0×103 .75 1.4×103 .80 0.97×103 .85 630 .90 370 .95 190 1.00 70 6-2. I_p = qA _⌡⌠ 0 w G(x) dx = qA Φ₀α_s⌡⌠ 0 w e-αsx dx = qA Φ₀1 - e-αsw = qA P0(1 - Rf) hνA     1 - e-αsw 6-3. From Eq. (6-6), R = ηq hν = ηqλ hc = 0.8044 ηλ ( in µm) Plot R as a function of wavelength.

6-4. (a) Using the fact that V_a≈ V_B, rewrite the denominator as

1 -       V_a - I_MR_M V_B n = 1 -       V_B - V_B + V_a - I_MR_M V_B n = 1 -       1 - VB - Va + IMRM V_B n Since VB - V_Va + IMRM

1 -       1 - VB - V_Va + IMRM B n ≈ 1 -       1 - n( ) VB - Va + IMRM V_B = n( ) V_B - V_a + I_MR_M VB ≈ nI_MR_M VB Therefore, M₀ = IM I_p ≈ V_B n(VB - Va + IMRM) ≈ V_B nI_MR_M (b) M₀ = IM I_p = V_B nI_MR_M implies I 2 M = I_pV_B nR_M , so that M0 = _     V_B nI_pR_M 1/2 6-5.

<

i2_s(t)

>

= _T1_⌡⌠ 0 T i2_s(t)dt = ω 2π ⌡⌠ 0 2π/ω R2₀ P2(t) dt (where T = 2π/ω), = ω 2π R 2 0 P 2 0 ⌡⌠ 0 2π/ω (1 + 2m cos ωt + m2 cos2ωt) dt Using cos ωtdt= 0 2_π/_ω

∫

_ω1 sin ωt t₌0 t=2π/ω ₌₀ and _⌡⌠ 0 2π/ω cos2ωt dt = _ω1 ⌡  ⌠ 0 2π     1 2 + 1 2cos 2x dx = π ω we have

<

i2_s(t)

>

= R₀2 P2_01 + m _ 2 2

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 Then σQ 2 =2qIpB=2 1.6×10 −19 C

(

)

(0.593 µA)(150×106 Hz)=2.84×10−17 A2

(b) σ_DB2 =2qI_DB=2 1.6

(

×10−19 C

)

(1.0 nA)(150×106 Hz)=4.81×10−20 A2 (c) σ_T2 ₌4kBT RL B= 4 1.38×10 −23 _{J / K}

(

)

(293 K) 500 Ω 150×10 6 _Hz

(

)

=4.85×10−15 _A2 6-7. Using R0 = ηqλ

hc = 0.58 A/W, we have from Eqs. 4), 11b), 15), and (6-17)     S N Q = 1 2

(

R0P0m

)

2 M2 2qIpBM 1/ 2 M2 = R₀P₀m2 4qBM1/2 = 6.565×1012 P0     S N DB = ( ) R₀P₀m 2 4qIDBM1/2 = 3.798×10 22 _P2 0     S N DS = ( ) R0P0m 2M2 4qI_LB = 3.798×1026 P 2 0     S N T = 1 2 (R0P0m) 2 M2 4k_BTB/R_L = 7.333×1022 P 2 0

where P0 is given in watts. To convert P0 = 10-n W to dBm, use 10 log _ _ P₀ 10-3 = 10(3-n) dBm

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

S N = 1 2 (R0P0m) 2 M2 2qB(R₀P₀ + I_D)M5/2 + 2qI_LB + 4k_BTB/R_L = 1.215×10 −16_M2 2.176×10−23_M5/ 2_+1.656×10−19

The value of M for maximum S/N is found from Eq. (6-19), with x = 0.5: M_optimum = 62.1.

6-9. 0 = d dM    S N = d dM_    1 2 I 2 pM2 2q(I_p + I_D)M2+x + 2qI_L + 4k_BT/R_L 0 = I2_p M - (2+x)M1+x _{2q(Ip + ID)}1 2 I 2 pM2 2q(I_p + I_D)M2+x + 2qI_L + 4k_BT/R_L

Solving for M: M2+x_opt = 2qIL + 4kBT/RL xq(I_p + I_D)

6-10. (a) Differentiating pn, we have

∂p_n ∂x = 1 L_p    p_n0 + Be-αsw e(w-x)/Lp - α_sBe-αsx ∂2_p n ∂x2 = - 1 L2_p    p_n0 + Be-αsw e(w-x)/Lp + α2_s Be-αsx Substituting p_n and ∂ 2_p n

∂x2 into the left side of Eq. (6-23):

- Dp L2_p    p_n0 + Be-αsw e(w-x)/Lp + D_pα2_s Be-αsx + _τ1 p     p_n0 + Be-αsw e(w-x)/Lp - _τB p e-αsx + Φ₀α_s e-αsx =       B       D_pα2_s - _τ1 p + Φ₀α_s e-αsx

where the first and third terms cancelled because L2_p = D_pτ p . Substituting in for B: Left side =         Φ0 D_p αsL 2 p 1 - α2_s L2_p     D_pα2_s - _τ1 p + Φ₀α_s e-αsx

= Φ_D0 p         αsL2_p 1 - α2_s L2_p       α2_s L_p - 1 τ p + Dpαs e -α_sx = Φ_D0 p (-Dpαs + Dpαs) e -αsx

= 0 Thus Eq. (6-23) is satisfied.

(b) J_diff = qD_p ∂pn ∂x     _x=w = qD_p       1 L_p    p_n0 + Be-αsw - α_sBe-αsw = qD_p B     1 L_p - αs e -α_sw + qp_n0 Dp L_p = qΦ₀         αsL 2 p 1 - α2_s L2_p       1 - α_sL_p L_p e -α_sw + qp_n0Dp L_p = qΦ₀ αsLp 1 + α_sL_p e -α_sw + qp_n0Dp L_p c) Adding Eqs. (6-21) and (6-25), we have

J_total = J_drift + J_diffusion = qΦ₀

          1 - e-αsw +       αsLp 1 + α_sL_p e -α_sw + qp_n0Dp L_p = qΦ₀         1 - e -α_sw 1 + α_sL_p e -α_sw + qp_n0Dp L_p 6-11. (a) To find the amplitude, consider

    J_tot J*_tot sc 1/2 = qΦ0 (S S*)1/2 where S = 1 - e -jωtd jωt_d

We want to find the value of ωtd at which (S S*)1/2 = 1 2 . Evaluating (S S*)1/2 , we have

(S S*)1/2 =             1 - e-jωtd jωt_{d }    1 - e+jωtd -jωt_d 1/2 =     1 - e+jωtd + e-jωtd + 1 1/2 ωt_d = (2 - 2 cos ωtd) 1/2 ωt_d = [ ] (1 - cos ωtd)/2 1/2 ωt_d/2 = sin _ωtd_ 2 ωt_d 2 = sinc _ω₂td_

We want to find values of ωtd where (S S*)1/2 = 1 2 . x sinc x x sinc x 0.0 1.000 0.5 0.637 0.1 0.984 0.6 0.505 0.2 0.935 0.7 0.368 0.3 0.858 0.8 0.234 0.4 0.757 0.9 0.109

By extrapolation, we find sinc x = 0.707 at x = 0.442.

Thus ωtd

2 = 0.442 which implies ωtd = 0.884 (b) From Eq. (6-27) we have t_d = w

vd = 1 αsvd . Then ωt_d = 2πf_3-dB t_d = 2πf_{3-dB α}1 svd = 0.884 or f_3-dB = 0.884 α_sv_d/2π

6-12. (a) The RC time constant is

RC = R

ε

0KsA w = (104_{Ω)(8.85×10}−12_{F / m)(11.7)(5×10}−8_m2₎ 2×10−5_m = 2.59 ns

t_d = w v_d = 20×10−6_m 4.4×104_{m / s}= 0.45 ns c) _α1 s = 10 -3 _{cm = 10} µ_{m =} 1 2 w

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

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

(1) 1 - k1_{1 - k}(1 - k1) 2 = 1 - k₁ - k2₁ 1 - k₂ ≈ 1 - k₂ - k2₁ 1 - k₂ = 1 – keff (2) (1 - k1) 2 1 - k₂ = 1 - 2k1 + k2₁ 1 - k₂ ≈ 1 - 2k2 + k2₁ 1 - k₂ = 1 - k2 1 - k₂ - k₂ - k2₁ 1 - k₂ = 1 - keff Therefore Eq. (6-34) becomes Eq. (6-38): F_e = k_effM_e + 2(1 - k_eff) - 1 M_e(1 - keff) = keffMe + _    2 - 1 M_e (1 - keff) (b) With k₁≈ k₂ and k'_eff defined in Eq. (6-40), we have

(1) k2(1 - k1) k2₁(1 - k₂) ≈ k₂ - k2₁ k2₁(1 - k₂) = k' eff (2) (1 - k1) 2_k 2 k2₁(1 - k₂) = k₂ - 2k₁k₂ + k₂k2₁ k2₁(1 - k₂) ≈   k₂ - k2₁ - _k2₁ - k₂k2_1 k2₁(1 - k₂) = k' eff - 1

Therefore Eq. (6-35) becomes Eq. (6-39): F_h = k' eff Mh -     2 - 1 M_h (k ' eff - 1) 6-14. (a) If only electrons cause ionization, then β = 0, so that from Eqs. 36) and

(6-37), k₁ = k₂ = 0 and k_eff = 0. Then from Eq. (6-38) F_e = 2 - 1

M_e ≈ 2 for large Me

(b) If α = β, then from Eqs. (6-36) and (6-37), k₁ = k₂ = 1 so that k_eff = 1. Then, from Eq. (6-38), we have F_e = M_e.

Problem Solutions for Chapter 7

7-1. We want to compare F₁ = kM + (1 - k) 2− 1 M



   and F2 = Mx. For silicon, k = 0.02 and we take x = 0.3:

M F₁(M) F₂(M) % difference

9 2.03 1.93 0.60

25 2.42 2.63 8.7

100 3.95 3.98 0.80

For InGaAs, k = 0.35 and we take x = 0.7:

M F₁(M) F₂(M) % difference

4 2.54 2.64 3.00

9 4.38 4.66 6.4

25 6.86 6.96 1.5

100 10.02 9.52 5.0

For germanium, k = 1.0, and if we take x = 1.0, then F₁ = F₂. 7-2. The Fourier transform is

h_B(t) = H_B −∞ ∞

∫

(f )ej2πft _{df = R} ej2πft 1+j2πfRC df −∞ ∞

∫

Using the integral solution from Appendix B3:

ejpx (β +jx)n dx −∞ ∞

∫

= 2π(p) n−1_e− βp Γ(n) for p > 0, we have hB(t) = 1 2πC ej 2πft 1 2πRC +jf     df −∞ ∞

∫

= 1 Ce-t/RC 7-3. Part (a): hp −∞ ∞

∫

(t) dt = _α1_T b ⌡⌠ -αTb/2 αTb/2 dt = _α1 T_b     αT_b 2 + αT_b 2 = 1

Part (b): h_p −∞ ∞

∫

(t) dt = 1 2π 1 αTb exp − t 2 2(αTb) 2       −∞ ∞

∫

dt = 1 2π 1

αT_b π αTb 2 = 1 (see Appendix B3 for integral solution) Part (c): h_p −∞ ∞

∫

(t) dt = 1 αT_b exp − t αT_b       dt 0 ∞

∫

= -

[

e−∞−e−0

]

= 1 7-4. The Fourier transform is

F[p(t)*q(t)] = p(t)* q(t)e−j2πft _dt −∞ ∞

∫

= q(x) p(t−x) e−j2πft _{dt dx} −∞ ∞

∫

−∞ ∞

∫

= q(x) e−j2πfx p(t−x) e−j2πf( t−x) dt dx −∞ ∞

∫

−∞ ∞

∫

= q(x) e−j2πfx _{dx p(y) e}−j 2πfy _dy −∞ ∞

∫

−∞ ∞

∫

where y = t - x = F[q(t)] F[p(t)] = F[p(t)] F[q(t)] = P(f) Q(f)

7-5. From Eq. (7-18) the probability for unbiased data (a = b = 0) is P_e = 1

2

[

P0(vth)+P1(vth)

]

.

Substituting Eq. (7-20) and (7-22) for P₀ and P₁, respectively, we have

Pe = 1 2 1 2πσ2 e −v2_{/ 2σ}2 dv V / 2 ∞

∫

+ e−(v−V )2_{/ 2σ}2 dv −∞ V / 2

∫

     

In the first integral, let x = v/ 2σ2 so that dv = 2σ2 dx.

In the second integral, let q = v-V, so that dv = dq. The second integral then becomes

e−q2_{/ 2σ}2 dq −∞ V / 2−V

∫

= 2σ2 _e−x2 dx −∞ −V / 2 2σ2

∫

where x = q/ 2σ2 Then P_e = 1 2 2σ2 2πσ2 e −x2 dx V / 2 2σ2 ∞

∫

+ e−x2 dx −∞ −V / 2 2σ2

∫

      = 1 2 π e −x2 dx −∞ ∞

∫

−2 e−x2 dx 0 V / 2 2σ2

∫

     

Using the following relationships from Appendix B,

e−p2x2 dx −∞ ∞

∫

= π p and 2 π e −x2 dx 0 t

∫

= erf(t), we have P_e = 1 2 1−erf V 2σ 2          

7-6. (a) V = 1 volt and σ = 0.2 volts, so that V

2σ = 2.5. From Fig. 7-6 for V

2σ = 2.5, we find P_e≈ 7×10-3 errors/bit. Thus there are (2×105 bits/second)(7×10-3 errors/bit) = 1400 errors/second, so that

1

1400 errors/second = 7×10-4 seconds/error (b) If V is doubled, then V

2σ = 5 for which Pe≈ 3×10-7 errors/bit from Fig. 7-6.

Thus

1

(2×105bits / sec ond)(3×10−7errors / bit )= 16.7 seconds/error 7-7. (a) From Eqs. (7-20) and (7-22) we have

P₀(v_th) = 1 2πσ2 e −v2_{/ 2σ}2 dv V / 2 ∞

∫

= 1 2 1−erf V 2σ 2           and