Optical Fiber Communication - Solution Manual

1

Problem Solutions for Chapter 2

2-1.

E = 100cos 2π108 t + 30°( ) ex + 20 cos 2π108t − 50°( ) ey

+ 40cos 2π108 t + 210°( ) ez

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 µm

(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 µm)]

= 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 cos δ1 + a2 cos δ2 = A cos φ (1)

2

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) = a12 sin2 δ1 + cos2 δ1( )

+ a22 sin2 δ2 + cos2 δ2( ) + 2a1a2 (sin δ1 sin δ2 + cos δ1 cos δ2)

or

A2 = a12 + a2

2 + 2a1a2 cos (δ1 - δ2)

Dividing (2) by (1) gives

tan φ = a1 sinδ1 + a2 sinδ2

a1 cosδ1 + a2 cosδ2

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

E x

E0 x

cos δ = cos (ωt - kz) cos δ

to yield

Ey

E0 y

-Ex

E0x

cos δ = - sin (ωt - kz) sin δ (2.5-2)

Using the relation cos2 α + sin2 α = 1, we use Eq. (2-2) to write

3

sin2 (ωt - kz) = [1 - cos2 (ωt - kz)] = 1 −Ex

E0x

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

sin2 δ

Expanding the left-hand side and rearranging terms yields

Ex

E0x

2

+ Ey

E0y

2

- 2 Ex

E0x

Ey

E0y

cos δ = sin2 δ

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

nglass sin φglass = nair sin φair

4

with φair = 90° and nair = 1.0

∴ φcritical = arcsin 1

n glass

= arcsin 1

1.540= 40.5°

2-9Air

Water

12 cm

r

θ

Find θc from Snell's law n1 sin θ1 = n2 sin θc = 1

When n2 = 1.33, then θc = 48.75°

Find r from tan θc =

r12 cm

, which yields r = 13.7 cm.

2-10.

45 °

Using Snell's law nglass sin θc = nalcohol sin 90°

where θc = 45° we have

nglass =

1.45sin 45°

= 2.05

2-11. (a) Use either NA = n12 − n2

2( )1/ 2= 0.242

or

5

NA ≈ n1 2∆ = n12(n1 − n2)

n1

= 0.243

(b) θ0,max = arcsin (NA/n) = arcsin 0.2421.0

= 14°

2-13. NA = n12 − n2

2( )1/ 2= n1

2 − n12(1− ∆)2[ ]1/ 2

= n1 2∆ − ∆2( )1 / 2

Since ∆ << 1, ∆2 << ∆; ∴ NA ≈ n1 2∆

2-14. (a) Solve Eq. (2-34a) for jHφ:

jHφ = j εωβ

Er - 1βr

∂Hz

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

j β Er + ∂Ez

∂r= ωµ j

εωβ

Er −1

βr∂Hz

∂φ

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

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

jHr = -j εωβ

Eφ - 1β

∂Hz

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

j β Eφ + 1r

∂Ez

∂φ= -ωµ − j

εωβ

Eφ −1

β∂Hz

∂r

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

(c) Solve Eq. (2-34a) for jEr:

6

jEr = 1

εω1r

∂Hz

∂φ+ jrβHφ

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

β

εω1r

∂Hz

∂φ+ jrβHφ

+

∂Ez

∂r= jωµ Hφ

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

(d) Solve Eq. (2-34b) for jEφ

jEφ = - 1

εωjβHr +

∂Hz

∂r

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

1r

∂Ez

∂φ-

βεω

jβHr +∂Hz

∂r

= -jωµ Hr

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

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

-j

q 2

1r

∂∂r

β∂Hz

∂φ+ εωr

∂Ez

∂r

−

∂∂φ

β∂Hz

∂r−

εωr

∂Ez

∂φ

= jεωEz

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

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

-j

q2

1r

∂∂r

β∂Ez

∂φ− µω r

∂Hz

∂r

−

∂∂φ

β∂Ez

∂r+

µωr

∂Hz

∂φ

= -jµωHz

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

7

Ez = AJ0(ur) e j(ωt − βz) and Hz = BJ0(ur) e j(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 jβνa

1u2 +

1w2

= Bωµ

J' ν (ua)uJ ν (ua)

+K' ν (wa)wKν(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 Hz = 0. Thus Eq. (2-56) corresponds to TM0m modes.

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

0 = 1

u 2 Bjβνa

J ν(ua) + Aωε 1uJ'ν (ua)

+ 1

w2 Bjβνa

Jν(ua) + Aωε2wK' ν (wa)Jν(ua)

K ν(wa)

With k12 = ω2µε1 and k2

2 = ω2µε2 rewrite this as

Bν =

ja

βωµ

11

u 2+ 1

w2

k12Jν + k2

2K ν[ ] A

8

where Jν and Kν 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 Ez = 0. Thus Eq. (2-55) corresponds to TE0m

modes.

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

∆ = n 1

2 − n22

2n12 =

12

1 − n22

n12

∆ << 1 implies n1 ≈ n2

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

n22k2

= k22 ≈ n1

2k2= k1

2 ≈ β2

2-17.

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

M ≈2π2a2

λ2 n12 − n2

2( )=2π2a2

λ2 NA( )2

a =M2π

1/ 2 λNA

=1000

2

1/ 2 0.85µm0.2π

= 30.25µm

Therefore, D = 2a =60.5 µm

(b) M = 2π2 30.25µm( )2

1.32µm( )2 0.2( )2 = 414

(c) At 1550 nm, M = 300

2-19. From Eq. (2-58),

9

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)

Pclad

P

total

≈ 43

M-1/2 = 4 ×100%3 1080

= 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

a = Vλ2π

n12 − n2

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 = n12 − n2

2( )1/ 2= (1.480)2 − (1.478)2[ ]1/ 2

= 0.077

(c) From Eq. (2-23), NA = n sin θ0,max. When n = 1.0 then

θ0,max = arcsin NAn

= arcsin

0.0771.0

= 4.4°

2-22. n2 = n12 − NA2

= (1.458)2 − (0.3)2= 1.427

a = λV

2πNA=

(1.30)(75)2π(0.3)

= 52 µm

10

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) Lp = 2πβ

= λ

n y − nx

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

2 m= 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 =

αα + 2

a2k2n12∆ =

αα + 2

2πan1

λ

2

∆

where

∆ = n1 − n2

n1

= 0.0135

11

At λ = 820 nm, M = 543 and at λ = 1300 nm, M = 216.

For a step index fiber we can use Eq. (2-61)

Mstep ≈ V 2

2 =

12

2πa

λ

2

n12 − n2

2( )

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

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

Mstep = 2πan1

λ

2

∆ =

1086 at 820 nm

432 at 1300 nm

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

(a) NA = n12 − n2

2( )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 D2

2

= st d2

2

or s = S Dd

2

(b) S = s dD

2

= 1.2 m/s

0.125 mm9 mm

2

= 1.39 cm/min

12

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

PREFORMFIBER

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:

A preform core

Apreform clad

=

A fiber core

Afiber clad

orπ(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 cm3

The mass M equals the density ρ times the volume V:

M = ρV = (2.6 gm/cm3)(1.96 cm3) = 5.1 gm

13

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

t = MR

=

5.1 gm0.5 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

χ f

∫ dχ = AYbσb

0

t

∫ dt

which yields

1

1 − b2

χf1− b / 2 − χ i

1− b/ 2[ ] = AYbσbt

or

t = 2

(b − 2)A(Yσ)b χ i(2− b) / 2 − χf

(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 K ib− 2 << K f

b− 2 or K i2 −b >> Kf

2− b

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

dχdt

= AKb = AYbχb/2σb

14

Integrating this from χ i to χ p where

χ i = K

Yσ i

2

and χ p = K

Yσp

2

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

χ −b / 2

χ i

χ p

∫ dχ = AYb

σb

0

t p

∫ dt

or

1

1 − b2

χp1− b / 2 − χi

1−b / 2[ ]= AYb σ pb tp

for a constant stress σp. Substituting for χ i and χ p gives

2b − 2

KY

2−b

σ ib− 2 − σp

b− 2[ ]= AYb σ pb tp

or2

b − 2

KY

2−b 1AYb σ i

b− 2 − σpb− 2[ ]= B σ i

b−2 − σpb−2[ ]= σ p

b tp

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

χ p

χs

∫ dχ = AYb

σsb

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 σpb−2 − σs

b−2[ ]= σsb ts

Adding Eqs. (2-87) and (2-89) yields Eq. (2-90).

15

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 −LL0

σpbtp + σs

bts( )/ B + σsb−2[ ]

mb−2

σ0m −

σpbt p / B + σp

b−2( )m

b− 2

σ0m

= 1 - exp−

LL0σ0

m σpbtp / B + σp

b− 2[ ]m

b −2

σpbt p + σs

bts

B

+ σs

b −2

σpbtp / B + σp

b− 2

m

b− 2

− 1

= 1 - exp −LN p

1 +σs

b ts

σpbtp

+σs

σp

b

Bσs

2t p

m

b −2

1 + Bσp

2t p

−1

≈ 1 - exp −LN p 1+σs

bts

σpb tp

1

1+ Bσp

2 tp

m

b−2

− 1

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

σ s

σp

bB

σs2tp

= (0.3)15

0.5 (MN /m2 )2 s

0.3 (350 MN / m2)[ ]210 s

= 6.5×10-14

Thus this term can be neglected.

16

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

mL1

L0

= 1 - exp −

σ2c

σ0

mL2

L0

which implies that

σ1c

σ0

mL1

L0

= σ2c

σ0

mL2

L0

or

σ1c

σ2c

= L2

L1

1 / m

If L1 = 20 m, then σ1c= 4.8 GN/m2

If L2 = 1 km, then σ2c= 3.9 GN/m2

Thus4.83.9

m

= 1000

20= 50

gives

m = log 50

log(4.8/ 3.9)= 18.8

1


3-1.

α dB/ km( )=10z

log P(0)P(z)

=

10z

log eα pz( )

=10α p log e = 4.343 α p(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:



3-3. From Eq. (3-1c) with Pout = 0.45 Pin

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

3-4. (a) Pin = Pout 10αL/10 = (0.3 µW) 101.5(12)/10 = 18.9 µW

(b) Pin = Pout 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 representativepoints for αuv and αIR:

2

λλ (µµm) ααuv ααIR0.5 20.3 --0.7 1.44 --0.9 0.33 --1.2 0.09 2.2×10-6

1.5 0.04 0.00722.0 0.02 23.23.0 0.009 7.5×104

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

αscat = 8π3

3λ4 (n2 −1)2kBTfβT

=

8π3

3(0.63 µm)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: αscat = 3.8 dB/km

From Eq. (3-4b):

αscat = 8π3

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 Pclad/P ααννm = αα11 + (αα2 2 - αα11)Pclad/P 5 + 103Pclad/P01 0.02 3.0 + 0.02 5 + 20 = 2511 0.05 3.0 + 0.05 5 + 50 = 5521 0.10 3.0 + 0.10 5 + 100 = 10502 0.16 3.0 + 0.16 5 + 160 = 16531 0.19 3.0 + 0.19 5 + 190 = 19512 0.31 3.0 + 0.31 5 + 310 = 315

3

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

∆ = n 2 (0) − n2

2

2n2 (0)

we have

α(r) = α1 + (α2 - α1) n2 (0) − n2(r)

n 2(0) − n22 = α1 + (α2 - α1)

r2

a2

Thus

αgi =

α(r) p(r) r dr0

∞

∫

p(r) r dr0

∞

∫= α1 +

(α2 − α1)a2

exp −Kr2( ) r3dr0

∞

∫

exp −Kr2( ) r dr0

∞

∫

To evaluate the integrals, let x = Kr2, so that dx = 2Krdr. Then

exp −Kr 2( ) r3dr0

∞

∫

exp −Kr 2( ) r dr0

∞

∫=

12K2

e− x x dx0

∞

∫1

2K e−x dx

0

∞

∫=

1K

1!

0!=

1K

Thus αgi = 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

3-12. With λ in units of micrometers, we have

4

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. 3-120.2 µm 1.548 1.5500.6 µm 1.457 1.4581.0 µm 1.451 1.450

3-13. (a) From Fig. 3-13, dτdλ

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

have from Eq. (3-20)

σmat

L=

dτdλ

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

For a laser diode,

σmat

L = [80 ps/(nm-km)](2 nm) = 0.16 ns/km

(b) From Fig. 3-13, dτmat

dλ= 22 ps/(nm-km)

Therefore, Dmat(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 - uaV

2

= 1 - u2a2

u2a2 + w2a2 = w2

u 2 + w2

= β2 − k2n2

2

k2 n12 − β2 + β2 − k2 n2

2 = β2 / k2 − n2

2

n12 − n2

2

(b) Expand b as b = (β / k + n 2 )(β / k − n2 )

(n1 + n 2 )(n1 − n 2 )

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

5

β / k + n2

n1 + n2

= n1(1 −δ) + n2

n1 + n2

= 1 - n1

n 1 + n2

δ

Letting n2 = n1(1 - ∆) then yields

β / k + n2

n1 + n2

= 1 - δ

2 − ∆≈ 1 since

δ2 − ∆

<< 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 = xs

= n 2

n1

= n 1(1 − ∆)

n1

= (1 - ∆)

The travel time of the highest order ray is thus

Tmax =

n1

c s

Lx

=

n1Lc

11 − ∆

For the axial ray the travel time is Tmin = Ln1

c

ϕθa

s

x

6

Therefore

Tmin - Tmax = Ln1

c1

1 − ∆− 1

=

Ln1

c∆

1 − ∆≈

Ln1∆c

3-17. Since n2 = n1 1 − ∆( ), we can rewrite the equation as

σmod

L=

n1∆c

1 −πV

where the first tern is Equation (3-30). The difference is then given by the factor

1 −πV

= 1 −πλ2a

1

n12 − n2

2( )1/ 2 ≈ 1 −πλ2a

1n1 2∆

At 1300 nm this factor is 1 −π(1.3)2 62.5( )

11.48 2(0.015)

=1 − 0.127 = 0.873

3-18. For ε = 0 and in the limit of α → ∞ we have

C1 = 1, C2 = 32

, α

α + 1= 1,

α + 23α + 2

= 13

, α +1

2α +1= 1

2,

and (α + 1)2

(5α + 2)(3α + 2)=

115

Thus Eq. (3-41) becomes

σ int er mod al = Ln1∆2 3c

1+ 3∆ +125

∆2

1/ 2

≈ Ln1∆2 3c

3-19. For ε = 0 we have that α = 2(1 - 65

∆). 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 − 65

∆

+ 2

=− 3

5∆

1− 35

∆≈ − 3

5∆ 1+ 3

5∆

7

C2 = 3α − 2

2(α + 2)=

3 2 1 −65

∆

− 2

2 2 1− 65

∆

+ 2

=1 −

95

∆

2 1 − 35

∆

Evaluating the factors in Eq. (3-41) yields:

(a) C12 ≈

925

∆2

(b)4C1C2 (α + 1)∆

2α +1=

4∆ −35

∆

1 −

95

∆

2 1−

65

∆

+ 1

1 − 35

∆

2 1− 3

5∆

4 1 − 6

5∆

+ 1

= −

18∆1 −

11∆ +

1825

5 15

∆

2 2425

∆

1825

∆2

(c) 16∆2C2

2 (α + 1)2

(5α + 2)(3α + 2)

= 16∆2 1−

95

∆

2

2(1 −65

∆) +1

2

4 1 − 35

∆

2

10(1 − 65

∆) + 2

6(1 − 65

∆) + 2

= 16∆2 1−

95

∆

2

9 1 −45

∆

2

96(1 − ∆)(1 − 910

∆)4 1− 35

∆

2 ≈ 924

∆2

Therefore,

σ int er mod al = Ln1∆

2c

αα +1

α + 23α + 2

1/ 2 925

∆2 − 1825

∆2 + 924

∆2

1/ 2

= Ln1∆

2

2c

2(1 −65

∆)

2(1 − 65

∆) +1

2(1−65

∆) + 2

6(1− 65

∆) + 2

1/ 2

310 6

≈ n1∆

2 L20 3c

8

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

σ int 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 σλ , we have

σ int er mod al

L=

N1∆2c

αα + 2

α + 23α + 2

1/ 2 4∆C2(α + 1)(5α + 2)(3α + 2)

= 0.070 ns/km

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

σ int ra mod al = 1c

σλ

λ−λ2 d2n1

dλ2

=

0.098σλ ns / km at 850 nm

1.026 ×10−2 σλ ns / km at 1300 nm

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

σ int ra mod al

L=

0.098σλ ns / km at 850 nm

0.0103σλ ns / km at 1300 nm

The plot of σL

= 1L

σ int er2 + σ int ra

2( )1/ 2vs σλ :

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

τg = Lc

dβdk

= Lc

12β

2kn12 + 2k2n1

dn1

dk

- 2α + 2

αma 2

αα+ 2 2

α + 2n1

2k2∆( )2

α+2−1

× ∆ 2k2n1

dn 1

dk+ 2kn1

2 +n1

2k2

∆d∆dk

= Lc

kn1

βN1 −

4∆α + 2

α + 2

αma2

1n1

2k2∆

αα +2

N1 +n1k2∆

d∆dk

9

= LN1

ckn1

β1−

4∆α + 2

mM

αα +2

1 +ε4

with N1 = n1 + k dn1

dkand 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,

λdτdλ

= Lc

dN1

dλ1 +

α − ε − 2

α + 2mM

αα+ 2

+ LN1

cα − ε − 2

α + 2d

dλ∆

mM

αα +2

where

N1 = n1 - λ dn1

dλand M =

αα + 2

a2k2n12 ∆

(a)dN1

dλ=

ddλ

n1 − λdn1

dλ

= - λ

d2n1

dλ2

Thus ignoring the term involving ∆ d2n1

dλ2 , the first term in square brackets

becomes - Lc

λ2 d2n1

dλ2

(b)d

dλ∆

mM

αα+ 2

= m

αα+ 2 d∆

dλ1M

αα +2

+ ∆−α

α + 2dMdλ

1M

αα +2

+1

(c)dMdλ

= α

α + 2a2

ddλ

k2n12 ∆( )

= α

α + 2 a2

d∆dλ

k2n12 + 2k2∆n1

dn1

dλ+ 2kn1

2∆dkdλ

10

Ignoring d∆dλ

and dn1

dλterms yields

dMdλ

= 2α

α + 2a2k2n1

2 ∆ −1

λ

= -

2Mλ

so that

ddλ

∆mM

αα+ 2

=

∆λ

2αα + 2

mM

αα +2

. Therefore

λdτdλ

= - Lc

λ2 d2n1

dλ2 + LN1

cα − ε − 2

α + 22α∆α + 2

mM

αα +2

3-24. Let a = λ2 d2n1

dλ2 ; b = N1C1∆ 2α

α + 2; γ =

αα + 2

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

σ int ra mod al2 =

L2 σλ

λ

2 1M

λ dτg

dλ

m= 0

M

∑2

=

Lc

2 σλ

λ

21M

−a + b mM

γ

m= 0

M

∑2

≈ Lc

2 σλ

λ

21M

−a + bmM

γ

2

dm0

M

∫

=

Lc

2 σλ

λ

2

1M

a 2 − 2ab mM

γ

+ b2 mM

2γ

dm0

M

∫

= Lc

2 σλ

λ

2

a2 −2ab

γ + 1+

b2

2γ + 1

= Lc

2 σλ

λ

2

−λ2 d2n1

dλ2

2

11

- 2 λ2 d2n1

dλ2

N1C1∆

αα + 1

+ (N1C1∆)2 4α2

(α + 2)(3α + 2)

3-25. Plot of Eq. (3-57).

3-26. (a) D = λ − λ0( )S0 = −50(0.07) = −3.5 ps /(nm − km)

(b)

D =1500(0.09)

4 1 −

13101500

4

= 14.1 ps /(nm − km)

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

σ step

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∆Lc

=(1.49)(0.01)(5 ×103 m)

3 × 108 m / s= 248 ns

(b) From Eq. (3-48)

σstep =

n1∆L2 3 c

=2482 3

= 71.7 ns

(c)

BT = 0.2σstep

= 2.8 Mb / s

(d) BT ⋅ L = (2.8 MHz)(5 km) =13.9 MHz ⋅ km

12

3-29. For α = 0.95αopt , we have

σ int er (α ≠ αopt )

σ int er (α = αopt )=

(α − α opt)

∆(α + 2)= −

0.05(0.015)(1.95)

= −170%

For α =1.05αopt , we have

σ int er (α ≠ αopt )

σ int er (α = αopt )=

(α − α opt )

∆(α + 2)= +

0.05(0.015)(2.05)

= +163%

1


4-1. From Eq. (4-1), ni = 2

2πkBT

h2

3/2 (memh)

3/4 exp

- Eg

2kBT

= 2 2π(1.38 ×10−23 J / K)

(6.63 ×10 −34 J.s)2

3/ 2

T3 / 2 (.068)(.56)(9.11×10 −31 kg)2[ ]3/ 4

× exp −(1.55 − 4.3 × 10−4 T)eV2(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

-

8991T

4-2. The electron concentration in a p-type semiconductor is nP = ni = pi

Since both impurity and intrinsic atoms generate conduction holes, the total

conduction-hole concentration pP is

pP = NA + ni = NA + nP

From Eq. (4-2) we have that nP = n2i /pP . Then

pP = NA + nP = NA + n2i /pP or p

2P - NApP - n

2i = 0

so that

pP = NA2

1 + 4n

2i

N2A

+ 1

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

pP ≈ NA and nP = n2i /pP ≈ n

2i /NA

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

2

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 = 1.2401.540 = 805 nm.

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

λ = 1.2401.620 = 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 = 0.4184x

0.1894 - 0.0130x ≈ 0.4184x0.1894 = 2.20x

(c) With x = 0.26 and y = 0.56, we have

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 eV

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

3

∆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

= 15501310

2

= 1.40

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

ηint =1

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

Pint = (0.783)

hc(35 mA)q(1310 nm)

= 26 mW

(b) From Eq. (4-16),

P =

13.5 3.5 + 1( )2 26 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 0.99910 0.95420 0.84740 0.62360 0.46980 0.370100 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 log1

1 + 2πfτ( )2[ ]1/ 2

= −3

4

With a 5-ns lifetime, we find

f =1

2π 5 ns( )100.6 − 1( )= 9.5 MHz

4-9. (a) Using Eq. (4-28) with Γ = 1

gth =

10.05 cm

ln

1

0.32

2 + 10 cm-1 = 55.6 cm-1

(b) With R1 = 0.9 and R2 = 0.32,

gth = 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 = 1.24Eg

= 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 ηext = 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

Therefore the number of modes in the range 0.75-to-0.85 µm is

5

.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) = 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)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 Nm = n/λ = m2L be the wave number (reciprocal wavelength) of mode m.

The difference ∆N between adjacent modes is then

∆N = Nm - Nm-1 = 1

2L (a-1)

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

dNdλ =

ddλ

n

λ = 1λ

dndλ -

nλ2 = -

1λ2

n - λ

dndλ

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

∆N = 1λ2

n - λ

dndλ ∆λ (a-2)

6

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

2L

n - λ

dndλ

(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-1

n+1

2 =

3.6-1

3.6+1

2 = 0.32

Then Jth = 1

βα +

12L

ln1

R1R2

= 2.65×103 A/cm2

Therefore

Ith = Jth × l × w = (2.65×103 A/cm2)(250×10-4 cm)(100×10-4 cm) = 663 mA

(b) Ith = (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 × 10−34 J ⋅ s( )2

8 5 nm( )2

16.19 × 10−32 kg

+ 15.10 ×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

ne = mλB

2Λ = 2(1570 nm)2(460 nm) = 3.4

Then, from Eq. (4-48b), for m = 0

7

λ = λB ± λ

2B

2neL

1

2 = 1570 nm ± (1.57 µm)(1570 nm)

4(3.4)(300 µm) = 1570 nm ± 1.20 nm

Therefore for m = 1, λ = λB ± 3(1.20 nm) = 1570 nm ± 3.60 nm

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.

t d = dt0

t d

∫ = 1

Jqd

− nτ

0

n th

∫ dn = −τ Jqd

−nτ

n= 0

n= nth

= τ ln J

J − J th

where J = Ip/A and Jth = Ith/A. Therefore td = τ ln

Ip

Ip - Ith

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

td = ⌡⌠0

td dt =

1

Jqd

− nτ

dnnB

n th

∫ = τ ln

Jqd

− nB

τJ

qd− n th

τ

In the steady state before a pulse is applied, nB = JBτ/qd. When a pulse is applied,

the current density becomes I/A = J = JB + Jp = (IB + Ip)/A

Therefore, td = τ ln

I - IB

I - Ith

= τ ln

Ip

Ip + IB - Ith

4-20. A common-emitter transistor configuration:

4-21. Laser transmitter design.

8

4-22. Since the dc component of x(t) is 0.2, its range is -2.36 < x(t) < 2.76. The powerhas the form P(t) = P0[1 + mx(t)] where we need to find m and P0. The average

value is

< > P(t) = P0[1 + 0.2m] = 1 mW

The minimum value is

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

2.36 = 0.42

Therefore for the average value we have < > P(t) = P0[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(b12 cos2 ω1t + 2b1b2 cos ω1t cos ω2t + b2

2 cos2 ω2t)

+ a3(b13 cos3 ω1t + 3b1

2 b2 cos2 ω1t cos ω2t + 3b1b22 cos ω1t cos2 ω2t+ b2

3 cos3 ω2t)

+a4(b14 cos4 ω1t + 4b1

3 b2 cos3 ω1t cos ω2t + 6b12 b2

2 cos2 ω1t cos2 ω2t

+ 4b1b23 cos ω1t cos3 ω2t + b2

4 cos4 ω2t)

Use the following trigonometric relationships:

i) cos2 x = 12 (1 + cos 2x)

ii) cos3 x = 14 (cos 3x + 3cos x)

iii) cos4 x = 18 (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)]

9

vi) cos2 x cos2 y = 14 [1 + cos 2x+ cos 2y +

12 cos(2x+2y) +

12 cos(2x-2y)]

vii) cos3 x cos y = 18 [cos (3x+y) + cos (3x-y) + 3cos (x+y) + 3cos (x-y)]

then

y(t) = 12

a2b

21 + a2b

22 +

34 a4b

41 + 3a4b

21b

22 +

34 a4b

42

constantterms

+ 34

a3b

31 + 2a3b1b

22 cos ω1t +

34 a3

b

32 + 2b

21b2 cos ω2t

fundamentalterms

+ b

21

2

a2 + a4b

21 + 3a4b

22 cos 2ω1t +

b22

2

a2 + a4b

22 + 3a4b

21 cos 2ω2t

2nd-order harmonic terms

+ 14 a3b

31 cos 3ω1t +

14 a3b

32 cos 3ω2t 3rd-order harmonic terms

+ 18 a4b

41 cos 4ω1t +

18 a4b

42 cos 4ω2t 4th-order harmonic terms

+

a2b1b2 +

32 a4b

31b2 +

32 a4b1b

32 [ ]cos (ω1+ω2)t + cos (ω1-ω2)t

2nd-order intermodulation terms

+ 34 a3b

21 b2 [ ]cos(2ω1+ω2)t + cos(2ω1-ω2)t +

34 a3b1b

22 [ ]cos(2ω2+ω1)t + cos(2ω2-ω1)t

3rd-order intermodulation terms

+ 12 a4b

31 b2 [ ]cos(3ω1+ω2)t + cos(3ω1-ω2)t

+ 34 a4b

21 b

22 [ ]cos(2ω1+2ω2)t + cos(2ω1-2ω2)t

+ 12 a4b1b

32 [ ]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

+ A4(ω1) cos 4ω1t + A1(ω2) cos ω2t + A2(ω2) cos 2ω2t

10

+ 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) = 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) τs = K eEA/kBT

or ln τs = 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 EA (1)

At T = 90°C = 363°K, we have

ln 6500 = ln K + EA/[(8.625×10-5 eV)(363)]

or 8.78 = ln K + 31.94 EA (2)

Solving (1) and (2) for EA and K yields

EA = 0.63 eV and k = 1.11×10-5 hrs

Thus at T = 20°C = 293°K

11

τs = 1.11×10-5 exp0.63/[(8.625×10-5)(293)] = 7.45×105 hrs

1


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:

PLED-step = π2r2s B0(NA)2 = π2(2×10-3 cm)2(100 W/cm2)(.22)2 = 191 µW

From Eq. (5-9)

PLED-graded = 2π2(2×10-3 cm)2(100 W/cm2)(1.48)2(.01)

1 - 12

2

5

2 = 159 µW

5-5. Using Eq. (5-10), we have that the reflectivity at the source-to-gel interface is

R s− g =3.600 −1.3053.600 +1.305

2

= 0.219

Similarly, the relfectivity at the gel-to-fiber interface is

R g− f =1.465 −1.3051.465 +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

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

2

P =

⌡⌠

0

rm

⌡⌠

0

2π

2π ⌡⌠0

θ0-max

cos3 θ sin θ dθ dθs r dr

Using

⌡⌠0

θ0

cos3 θ sin θ dθ = ⌡⌠0

θ0

( )1 - sin2 θ sin θ d(sin θ) = ⌡⌠0

sin θ0

( )x - x3 dx

we have

P = 2π

⌡⌠

0

rm

⌡⌠

0

2π

sin2 θ0-max

2 - sin4 θ0-max

4 dθs r dr

= π

⌡⌠

0

rm

⌡⌠

0

2π

NA2 -

12 NA4 dθs r dr

= π2 [ ]2NA2 - NA4 ⌡⌠

0

rm

r dr ⌡⌠0

2π dθs

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

rs ≤ 4 µm, η = 1.

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

= 1.

5-8. Using Eq. (5-10), the relfectivity at the gel-to-fiber interface is

3

R g− f =1.485 −1.3051.485 +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

R a− f = 1.485 −1.0001.485 +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) = 12 sa -

12 cy

s = aθ = a [2 arccos (y/a)] = 2a arccos

d

2a

c = 2

a2 -

d

22 1/2

Therefore

Acommon = 2(shaded area) = sa – cy = 2a2 arccos

d

2a - d

a2 -

d2

4

1/2

5-10.Coupling loss (dB) forGiven axial misalignments (µµm)

Core/cladding diameters(µµm)

1 3 5 10

50/125 0.112 0.385 0.590 1.26662.5/125 0.089 0.274 0.465 0.985100/140 0.056 0.169 0.286 0.590

5-11. arccos x = π2 - arcsin x

For small values of x, arcsin x = x + x3

2(3) + x5

2(4)(5) + ...

4

Therefore, for d2a << 1, we have arccos

d2a ≈

π2 -

d2a

Thus Eq. (5-30) becomes PT = 2π P

π

2 - d2a -

5d6a = P

1 -

8d3πa

d/a PT/P (Eq.5-30) PT/P (Eq.5-31)0.00 1.00 1.000.05 0.9576 0.95760.10 0.9152 0.91510.15 0.8729 0.87270.20 0.8309 0.83020.25 0.7890 0.78780.30 0.7475 0.74540.35 0.7063 0.70290.40 0.6656 0.6605

5-12. Plots of mechanical misalignment losses.

5-13. From Eq. (5-20) the coupling efficiency ηF 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

ηF =

MaRMaE

= k2NA2(0)

1

2 -1

α+2a

2R

k2NA2(0)

1

2 -1

α+2a

2E

= a

2R

a2E

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

a

2R

a2E

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

LF = -10 log ηF = -10 log

MRME

= -10 log k2NA

2R(0)

α

2α+4a2

k2NA2E(0)

α

2α+4a2

5

= -10 log

NA

2R(0)

NA2E(0)

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

< αE

LF = -10 log ηF = -10 log

k2NA2(0)

α

R

2αR

+ 4a2

k2NA2(0)

α

E

2αE

+ 4a2

= -10 log

α

R(α

E + 2)

αE

(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)

L1→ 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.

L2→1(a) = −20 log

5062.5

= 1.94 dB

L2→1(NA) = −20 log

.206

.256

= 1.89 dB

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

6

LSM;ff = -10 log

4

W1

W2 +

W2W1

2

For W1 = 0.9W2 , we then have LSM;ff = -10 log

4

4.0446 = - 0.0482 dB

5-19. Plot of Eq. (5-44).

5-20. Plot of the throughput loss.

1


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 representative

values of the absorption coefficient:

λλ (µµm) ααs (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 1901.00 70

6-2. Ip = qA ⌡⌠0

w G(x) dx = qA Φ0 αs ⌡

⌠

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 = ηqhν =

ηqλhc = 0.8044 ηλ ( in µm)

Plot R as a function of wavelength.

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

1 -

Va - IMRM

VB

n = 1 -

VB - VB + Va - IMRM

VB

n

= 1 -

1 - VB - Va + IMRM

VB

n

Since VB - Va + IMRM

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

2

1 -

1 - VB - Va + IMRM

VB

n ≈ 1 -

1 - n( )VB - Va + IMRM

VB

= n( )VB - Va + IMRM

VB ≈

nIMRMVB

Therefore, M0 = IMIp

≈ VB

n( )VB - Va + IMRM ≈

VBnIMRM

(b) M0 = IMIp

= VB

nIMRM implies I

2M =

IpVBnRM

, so that M0 =

VB

nIpRM

1/2

6-5. < > i2s(t) =

1T ⌡

⌠

0

T

i2s(t)dt =

ω2π ⌡

⌠

0

2π/ω

R20 P2(t) dt (where T = 2π/ω),

= ω2π R2

0 P20 ⌡⌠

0

2π/ω (1 + 2m cos ωt + m2 cos2 ωt) dt

Using

cos ωt dt =0

2 π / ω

∫1

ω sin ωt t = 0

t = 2π / ω = 0

and ⌡⌠0

2π/ω cos2 ωt dt =

1ω

⌡⌠

0

2π

1

2 + 12cos 2x dx =

πω

we have < > i2s(t) = R

20 P2

0

1 + m2

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 =

ηqλhc

P0 = 0.593 µA

Then σQ2 = 2qIpB = 2 1.6 ×10−19

C( )(0.593 µA)(150 × 106 Hz) = 2.84 × 10−17

A2

3

(b) σDB2 = 2qIDB = 2 1.6 ×10−19 C( )(1.0 nA)(150 × 106 Hz) = 4.81× 10−20 A 2

(c)

σT2 = 4kBT

RL

B =4 1.38 × 10−23 J / K( )(293 K)

500 Ω150 ×106

Hz( )= 4.85 × 10−15 A 2

6-7. Using R0 = ηqλhc = 0.58 A/W, we have from Eqs. (6-4), (6-11b), (6-15), and (6-

17)

S

N Q =

12

R0 P0 m( )2M 2

2qIpBM1/ 2M2 = R0P0m2

4qBM1/2 = 6.565×1012 P0

S

N DB =

( )R0P0m2

4qIDBM1/2 = 3.798×1022 P20

S

N DS =

( )R0P0m2M2

4qILB = 3.798×1026 P20

S

N T =

12 ( )R0P0m

2M2

4kBTB/RL = 7.333×1022 P

20

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

SN =

12 ( )R0P0m

2M2

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

= 1.215 ×10−16M2

2.176 × 10−23 M 5/ 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.

4

6-9. 0 = d

dM

S

N = d

dM

1

2 I2pM2

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

0 = I2p M -

(2+x)M1+x 2q(Ip + ID)12 I

2pM2

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

Solving for M: M2+xopt =

2qIL + 4kBT/RLxq(Ip + ID)

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

∂pn

∂x=

1Lp

pn0 + Be-αsw

e(w-x)/Lp

- αsBe-αsx

∂2pn

∂x2 = - 1

L2p

pn0 + Be-αsw

e(w-x)/Lp

+ α2s Be

-αsx

Substituting pn and ∂2pn

∂x2 into the left side of Eq. (6-23):

- Dp

L2p

pn0 + Be-αsw

e(w-x)/Lp

+ Dp α2s Be

-αsx

+ 1τ

p

pn0 + Be-αsw

e(w-x)/Lp

- Bτ

p e

-αsx + Φ0 αs e

-αsx

=

B

Dp α

2s -

1τ

p + Φ0 αs e

-αsx

where the first and third terms cancelled because L2p = Dpτ

p .

Substituting in for B:

Left side =

Φ0

Dp

αsL2p

1 - α2s L

2p

Dp α

2s -

1τ

p + Φ0 αs e

-αsx

5

= Φ0Dp

αsL

2p

1 - α2s L

2p

α

2s Lp - 1

τp

+ Dpαs e-αsx

= Φ0Dp

( )-Dpαs + Dpαs e-αsx

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

(b) Jdiff = qDp ∂pn

∂x

x =w

= qDp

1

Lp

pn0 + Be-αsw

- αsBe-αsw

= qDp B

1

Lp - αs e

-αsw + qpn0

DpLp

= qΦ0

αsL

2p

1 - α2s L

2p

1 - αsLp

Lp e

-αsw + qpn0

DpLp

= qΦ0 αsLp

1 + αsLp e

-αsw + qpn0

DpLp

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

Jtotal = Jdrift + Jdiffusion = qΦ0

1 - e-αsw

+

αsLp

1 + αsLp e

-αsw + qpn0

DpLp

= qΦ0

1 - e-αsw

1 + αsLp e

-αsw + qpn0

DpLp

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

jωtd

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

= 12 .

Evaluating ( )S S*1/2

, we have

6

( )S S*1/2

=

1 - e-jωtd

jωtd

1 - e+jωtd

-jωtd

1/2

=

1 -

e+jωtd

+ e-jωtd

+ 11/2

ωtd =

( )2 - 2 cos ωtd1/2

ωtd

= [ ]( )1 - cos ωtd /2

1/2

ωtd/2 =

sin

ωtd

2ωtd2

= sinc

ωtd

2

We want to find values of ωtd where ( )S S*1/2

= 12 .

x sinc x x sinc x0.0 1.000 0.5 0.6370.1 0.984 0.6 0.5050.2 0.935 0.7 0.3680.3 0.858 0.8 0.2340.4 0.757 0.9 0.109

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

Thus ωtd2 = 0.442 which implies ωtd = 0.884

(b) From Eq. (6-27) we have td = wvd

= 1

αsvd . Then

ωtd = 2πf3-dB td = 2πf3-dB 1

αsvd = 0.884 or

f3-dB = 0.884 αsvd/2π

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

RC = Rε

0KsA

w = (104 Ω)(8.85 ×10−12F /m)(11.7)(5 × 10−8 m2 )

2 × 10−5m= 2.59 ns

(b) From Eq. (6-27), the carrier drift time is

7

td = wvd

= 20 ×10−6 m

4.4 × 104 m / s= 0.45 ns

c) 1αs

= 10-3 cm = 10 µm = 12 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 keff defined in Eq. (6-10), we have

(1) 1 - k1(1 - k1)

1 - k2 = 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 - k21 - k2

- k2 - k

21

1 - k2 = 1 - keff

Therefore Eq. (6-34) becomes Eq. (6-38):

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

Me(1 - keff) = keffMe +

2 -

1Me

(1 - keff)

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

(1)k2(1 - k1)

k21(1 - k2)

≈ k2 - k

21

k21(1 - k2)

= k'eff

(2)(1 - k1)2k2

k21(1 - k2)

= k2 - 2k1k2 + k2k

21

k21(1 - k2)

≈

k2 - k

21 -

k

21 - k2k

21

k21(1 - k2)

= k'eff - 1

8

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

2 -

1Mh

(k'eff - 1)

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

1


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

and F2 = Mx.

For silicon, k = 0.02 and we take x = 0.3:

M F1(M) F2(M) % difference9 2.03 1.93 0.6025 2.42 2.63 8.7100 3.95 3.98 0.80

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

M F1(M) F2(M) % difference4 2.54 2.64 3.009 4.38 4.66 6.425 6.86 6.96 1.5100 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

hB(t) =

HB−∞

∞

∫ (f )e j2πft df = R

e j2πft

1 + j2πfRC df

−∞

∞

∫

Using the integral solution from Appendix B3:

e jpx

(β + jx)n dx−∞

∞

∫ = 2π(p)n−1e− βp

Γ(n)for p > 0, we have

hB(t) =

12πC

e j 2πft

12πRC

+ jf

df−∞

∞

∫ = 1C

e-t/RC

7-3. Part (a):

hp−∞

∞

∫ (t) dt = 1

αTb ⌡⌠-αTb/2

αTb/2

dt = 1

αTb

αTb

2 + αTb

2 = 1

2

Part (b):

hp−∞

∞

∫ (t) dt =

12π

1

αTb

exp − t2

2(αTb )2

−∞

∞

∫ dt

= 12π

1αTb

π αTb 2 = 1 (see Appendix B3 for integral solution)

Part (c):

hp−∞

∞

∫ (t) dt =

1

αTb

exp −t

αTb

dt0

∞

∫ = - e−∞ − e−0[ ]= 1

7-4. The Fourier transform is

F[p(t)*q(t)] =

p(t)* q(t)e− j2 πft 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

Pe = 12

P0 (v th ) + P1(v th )[ ].

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

Pe =

12

1

2πσ 2 e−v 2 / 2σ 2

dvV / 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 thenbecomes

3

e−q2 / 2σ2

dq−∞

V / 2 −V

∫ =

2σ2 e− x2

dx−∞

−V / 2 2σ 2

∫ where x = q/ 2σ2

Then

Pe =

12

2σ2

2πσ 2 e− x2

dxV / 2 2σ 2

∞

∫ + e− x2

dx−∞

− V / 2 2σ2

∫

=

12 π

e−x 2

dx−∞

∞

∫ − 2 e−x2

dx0

V / 2 2σ2

∫

Using the following relationships from Appendix B,

e− p2x 2

dx−∞

∞

∫ = π

pand

2

π e−x 2

dx0

t

∫ = erf(t), we have

Pe = 12

1− erfV

2σ 2

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

2σ = 2.5. From Fig. 7-6 for

V2σ

= 2.5,

we find Pe ≈ 7×10-3 errors/bit. Thus there are (2×105 bits/second)(7×10-3

errors/bit) = 1400 errors/second, so that

11400 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 × 105 bits / sec ond)(3 ×10 −7errors / bit )

= 16.7 seconds/error

7-7. (a) From Eqs. (7-20) and (7-22) we have

P0(vth) =

1

2πσ2 e− v2 / 2σ 2

dvV / 2

∞

∫ = 12

1− erfV

2σ 2

and

4

P1(vth) = 1

2πσ2e− (v− V) 2 / 2σ 2

dv−∞

V / 2

∫ = 12

1− erfV

2σ 2

Then for V = V1 and σ = 0.20V1

P0(vth) = 12

1 - erf

1

2(.2) 2 =

12

1 - erf

2

0.8

= 12 [ ]1 - erf( )1.768 =

12 ( )1 - 0.987 = 0.0065

Likewise, for V = V1 and σ = 0.24V1

P1(vth) = 12

1 - erf

1

2(.24) 2 =

12

1 - erf

2

0.96

= 12 [ ]1 - erf( )1.473 =

12 ( )1 - 0.963 = 0.0185

(b) Pe = 0.65(0.0185) + 0.35(0.0065) = 0.0143

(c) Pe = 0.5(0.0185) + 0.5(0.0065) = 0.0125

7-8. From Eq. (7-1), the average number of electron-hole pairs generated in a time t is

N = ηEhν =

ηPthc/λ =

0.65(25 ×10−10W)(1 ×10−9 s)(1.3 × 10−6 m)(6.6256 ×10 −34 Js)(3 ×108 m / s)

= 10.6

Then, from Eq. (7-2)

P(n) = Nn e-N

n! = (10.6)5 e-10.6

5! = 133822

120 e-10.6 = 0.05 = 5%

7-9. v N = vout - vout

vN2 = vout − vout[ ]2

= vout2

-2 vout

2+ vout

2

= vout2 - vout

2

7-10. (a) Letting φ = fTb and using Eq. (7-40), Eq. (7-30) becomes

5

Bbae =

H p (0)

Hout (0)

2

Tb2

Hout' (φ)

Hp' (φ)

2

0

∞

∫ dφTb

= I2

Tb

since Hp(0) = 1 and Hout(0) = Tb. Similarly, Eq. (7-33) becomes

Be =

H p (0)

Hout (0)

2

Hout (f)Hp (f)

1 + j2πfRC( )2

0

∞

∫ df

=

1Tb

2 Hout (f)Hp (f)

2

0

∞

∫ 1 + 4π2f2 R2C2( ) df

=

1Tb

2 Hout (f)Hp (f)

2

0

∞

∫ df +

2πRC( )2

Tb2

Hout (f)Hp (f)

2

0

∞

∫ f2 df =

I2Tb

+ (2πRC)2

T3b

I3

(b) From Eqs. (7-29), (7-31), (7-32), and (7-34), Eq. (7-28) becomes

< > v2N = < > v

2s +< > v

2R +< > v

2I +< > v

2E

= 2q < >i0 < >m2 BbaeR2A2 + 4kBT

Rb BbaeR2A2 + SIBbaeR2A2 + SEBeA2

=

2q < >i0 M2+x + 4kBT

Rb + SI BbaeR2A2 + SEBeA2

= R2A2I2

Tb

2q < >i0 M2+x + 4kBT

Rb + SI +

SE

R2 + (2πRCA)2

T3b

SEI3

7-11. First let x = v − boff( )/ 2 σoff( ) with

dx = dv / 2 σoff( ) in the first part of Eq. (7-

49):

Pe =

2σoff

2πσoff

exp −x2( )vth − boff

2σoff

∞

∫ dx =

1

π exp −x2( )

Q / 2

∞

∫ dx

Similarly, let y = −v + bon( )/ 2 σon( ) so that

dy = -dv/ 2σon( )in the second part of Eq. (7-49):

6

Pe = -

1

π exp −y2( )

∞

− vth +b on

2σ on

∫ dy =

1

π exp −y2( )

Q / 2

∞

∫ dy

7-12. (a) Let x = V

2 2 σ =

K2 2

For K = 10, x = 3.536. Thus

Pe = e-x2

2 π x = 2.97×10-7 errors/bit

(b) Given that Pe = 10-5 = e-x2

2 π x then e-x2 = 2 π 10-5 x.

This holds for x ≈ 3, so that K = 2 2 x = 8.49.

7-13. Differentiating Eq. (7-54) with respect to M and setting dbon/dM = 0, we have

dbondM = 0

= -

Q(hν / η)Μ2 M2+ x η

hν

bonI2 + W

1/ 2

+ M 2+x ηhν

bonI2 (1 − γ

+ W

1/ 2

+

Q(hν / η)M(hν/ η)

12

(2 + x)M1+ x bonI2

M 2+ x ηhν

bonI2 + W

1/ 2 +

12

(2 + x)M1+x bonI2 (1− γ)

M 2+x ηhν

bonI2 (1− γ) + W

1/ 2

1 / 2

Letting G = M2+x

η

hν bonI2 for simplicity, yields

( )G + W1/2

+ [ ]G(1-γ) + W1/2

= G2 (2 + x)

1

(G + W)1/2 +

(1-γ)

[ ]G(1-γ) + W1/2

Multiply by (G + W)1/2

[ ]G(1-γ) + W1/2

and rearrange terms to get

(G + W)1/2

W -

Gx2 (1-γ) = [ ]G(1-γ) + W

1/2

Gx

2 - W

7

Squaring both sides and collecting terms in powers of G, we obtain the quadratic

equation

G2

x2γ

4 (1-γ) + G

x2γ

4 W(2-γ) - γW2(1+x) = 0

Solving this equation for G yields

G = -

x2

4 W(2-γ) ±

x4

16 W2(2-γ)2 + x2(1-γ)W2(1+x)

12

x2

2 (1-γ)

= W(2-γ)2(1-γ)

1 -

1 + 16

1+x

x2 1-γ

(2-γ)2

12

where we have chosen the "+" sign. Equation (7-55) results by letting

G = M2+xopt bonI2

η

hν

7-14. Substituting Eq. (7-55) for M2+xbon into the square root expressions in Eq. (7-

54) and solving Eq. (7-55) for M, Eq. (7-54) becomes

bon = Q

hν

η b1/(2+x)on

hν

η W(2-γ)2I2(1-γ)

K1/(2+x)

W(2-γ)

2(1-γ) K + W

12 +

W

2 (2-γ)K + W

12

Factoring out terms:

b(1+x)/(2+x)on = Q

hν

η

(1+x)/(2+x) W

x/2(2+x) I

1/(2+x)2

×

(2-γ)

2(1-γ) K + 1

12 +

1

2 (2-γ)K + 1

12

÷

(2-γ)K

2(1-γ)

1/(2+x)

or bon = Q(2+x)/(1+x)

hν

η Wx/2(1+x)

I1/(1+x)2 L

8

7-15. In Eq. (7-59) we want to evaluate

lim

γ →1(2 − γ)K2(1− γ )L

1+x

2+x

= lim

γ →1(2 − γ)K2(1− γ)

1+x

2+x lim

γ →11L

1+x

2+x

Consider first

lim

γ →1(2 − γ)K2(1− γ)

=

lim

γ →1

(2 − γ)2(1− γ )

−1 + 1 + B(1− γ )

(2 − γ)2

1

2

where B = 16(1+x)/x2 . Since γ→1, we can expand the square root term in a

binomial series, so that

lim

γ →1(2 − γ)K2(1− γ)

=

lim

γ →1

(2 − γ)2(1− γ )

−1 + 1 +12

B(1− γ)

(2 − γ)2 − Order(1− γ )2

=

lim

γ →1

B4(2 − γ )

= B4 = 4

1+xx2

Thuslim

γ →1(2 − γ)K2(1− γ)

1+x

2+x

=

4

1+xx2

1+x2+x

Next consider, using Eq. (7-58)

lim

γ →1

1L

1+ x

2+x

= lim

γ →1(2 − γ)K2(1− γ)

1/(2+ x)

÷ (2 − γ)2(1− γ)

K + 1

1

2

+12

(2 − γ )K +1

1

2

From the above result, the first square root term is

(2-γ)

2(1-γ) K + 1

12 =

4

1+xx2 + 1

12 =

x2 + 4x + 4

x2

12 =

x + 2x

From the expression for K in Eq. (7-55), we have that lim

γ →1 K = 0, so that

9

lim

γ →112

(2 − γ )K +1

1

2= 1 Thus

lim

γ →1(2 − γ )2(1− γ )

K + 1

1

2

+ 12

(2 − γ)K +1

1

2

=

x + 2x + 1 =

2(1 + x)x

Combining the above results yields

lim

γ →1(2 − γ)K2(1− γ )L

1+x

2+x

=

4

1+xx2

1+x2+x

4

1+xx2

12+x

2(1 + x)

x = 2x

so thatlim

γ →1 M

1+xopt =

W1/2

QI2

2x

7-16. Using H'p(f) = 1 from Eq. (7-69) for the impulse input and Eq. (7-66) for the

raised cosine output, Eq. (7-41) yields

I2 =

Hout' (φ)

2 dφ

0

∞

∫ =

12

Hout' (φ)

2 dφ

−∞

∞

∫

= 12 ⌡⌠

-1-β2

1-β2

dφ + ⌡⌠

1-β2

1+β2

18

1-sin

πφ

β - π2β

2dφ +

⌡⌠

- 1+β

2

- 1-β2

18

1-sin

πφ

β - π2β

2dφ

Letting y = πφβ -

π2β we have

I2 = 12 (1 - β) +

β4π⌡⌠

-π2

π2

[ ]1 - 2sin y + sin2y dy

= 12 (1 - β) +

β4π

π - 0 +

π2 =

12

1 - β4

10

Use Eq. (7-42) to find I3:

I3 =

Hout' (φ)

2φ2

0

∞

∫ dφ =12

Hout' (φ)

2φ2

−∞

∞

∫ dφ

= 12 ⌡⌠

-1-β2

1-β2

φ2 dφ + ⌡⌠

1-β2

1+β2

18

1-sin

πφ

β - π2β

2φ2dφ +

⌡⌠

- 1+β

2

- 1-β2

18

1-sin

πφ

β - π2β

2φ2dφ

Letting y = πφβ -

π2β

I3 = 13

1-β

2

3 +

β4π⌡

⌠

-π2

π2

[ ]1 - 2sin y + sin2y

β2y2

π2 + βyπ +

14 dy

= 13

1-β

2

3 +

β4π

⌡⌠

-π2

π2

β2y2

π2 + 14 ( )1 + sin2y dy -

2βπ ⌡⌠

-π2

π2

y sin y dy

where only even terms in "y" are nonzero. Using the relationships

⌡⌠0

π2

sin2y dy = π4 ; ⌡⌠

y sin ydy = -y cos y + sin y

and ⌡⌠

x2 sin x2 dx =

x3

6 -

x2

4 - 18 sin 2x -

x cos 2x4

we have

I3 = 13

1-β

2

3 +

β2π

β2

π2

1

3

π

2

3 +

16

π

2

3 +

π8 +

14

π

2 + π4 -

2βπ (1)

11

= β3

16

1

π2 - 16 - β2

1

π2 - 18 -

β32 +

124

7-17. Substituting Eq. (7-64) and (7-66) into Eq. (7-41), with s2 = 4π2α2 and β = 1, we

have

I2 =

⌡⌠

0

H

'out(φ)

H'p(φ)

2 dφ =

14⌡⌠

0

1

es2φ2

1-sin

πφ -

π2

2dφ

= ⌡⌠

0

1

es2φ2

1 + cos πφ

2

2dφ =

⌡⌠

0

1

es2φ2

cos4

πφ

2 dφ

Letting x = πφ2 yields I2 =

2π ⌡

⌠

0

π2

e16α2x2

cos4 x dx

Similarly, using Eqs. (7-64) and (7-66), Eq. (7-42) becomes

I3 = ⌡⌠

0

1

es2φ2

cos4

πφ

2 φ2dφ =

2π

3

x2e16α2 x2

cos4 x0

π2

∫ dx

7-18. Plot of I2 versus α for a gaussian input pulse:

7-19. Plot of I3 versus α for a gaussian input pulse:

7-20. Consider first lim

γ →1 K:

12

lim

γ →1 K =

lim

γ →1−1+ 1+16

1+ xx2

1− γ(2 − γ)2

1

2

= -1 + 1 = 0

Also lim

γ →1(1− γ) = 0. Therefore from Eq. (7-58)

lim

γ →1 L =

lim

γ →12(1− γ)(2 − γ)K

1/(1+ x)

(2-γ)

2(1-γ) K + 1

12 + 1

2+x1+x

Expanding the square root term in K yields

limγ →1

2 − γ1− γ

K = lim

γ →12 − γ1 − γ

−1+ 1+

162

1 + xx 2

1− γ(2 − γ)2 + order(1− γ)2

= lim

γ →18(1 + x)

x2

12 − γ

= 8(1+x)

x2

Therefore

lim

γ →1 L =

2x2

8(1+x)

1/(1+x)

4(1+x)

x2 + 1

12 + 1

2+x1+x

=

2x2

8(1+x)

1/(1+x)

x+2

x + 1

2+x1+x

= (1+x)

2

x

x1+x

7-21. (a) First we need to find L and L'. With x = 0.5 and γ = 0.9, Eq. (7-56) yields K =

0.7824, so that from Eq. (7-58) we have L = 2.89. With ε = 0.1, we have γ' = γ(1 -

ε) = 0.9γ = 0.81. Thus L' = 3.166 from Eq. (7-80). Substituting these values into

Eq. (7-83) yields

y(ε) = (1 + ε)

1

1 - ε

2+x1+x

L'L = 1.1

1

.9

5/3

3.1662.89 = 1.437

13

Then 10 log y(ε) = 10 log 1.437 = 1.57 dB

(b) Similarly, for x = 1.0, γ = 0.9, and ε = 0.1, we have L = 3.15 and L' = 3.35, so

that

y(ε) = 1.1

1

.9

3/2

3.353.15 = 1.37

Then 10 log y(ε) = 10 log 1.37 = 1.37 dB

7-22. (a) First we need to find L and L'. With x = 0.5 and γ = 0.9, Eq. (7-56) yields K =

0.7824, so that from Eq. (7-58) we have L = 2.89. With ε = 0.1, we have γ' = γ(1 -

ε) = 0.9γ = 0.81. Thus L' = 3.166 from Eq. (7-80). Substituting these values into

Eq. (7-83) yields

y(ε) = (1 + ε)

1

1 - ε

2+x1+x

L'L = 1.1

1

.9

5/3

3.1662.89 = 1.437

Then 10 log y(ε) = 10 log 1.437 = 1.57 dB

(b) Similarly, for x = 1.0, γ = 0.9, and ε = 0.1, we have L = 3.15 and L' = 3.35, so

that

y(ε) = 1.1

1

.9

3/2

3.353.15 = 1.37

Then 10 log y(ε) = 10 log 1.37 = 1.37 dB

7-23. Consider using a Si JFET with Igate = 0.01 nA. From Fig. 7-14 we have that α =

0.3 for γ = 0.9. At α = 0.3, Fig. 7-13 gives I2 = 0.543 and I3 = 0.073. Thus from

Eq. (7-86)

WJFET = 1B

2(.01nA )1.6 ×10−19C

+4(1.38×10 −23 J / K)(300K)

(1.6 × 10−19C)2105 Ω

0.543

+

1B

4(1.38×10−23 J / K)(300 K)(.7)

(1.6 × 10−19 C)2 (.005 S)(105

Ω)2

0.543

+

2π(10 pF)1.6 ×10−19 C

2 4(1.38 ×10−23 J / K)(300 K)(.7)(.005 S)

0.073 B

or

14

WJFET ≈ 3.51×1012

B+ 0.026B

and from Eq. (7-92) WBP = 3.39 ×1013

B+ 0.0049B

7-24. We need to find bon from Eq. (7-57). From Fig. 7-9 we have Q = 6 for a 10-9

BER. To evaluate Eq. (7-57) we also need the values of W and L. With γ = 0.9,Fig. 7-14 gives α = 0.3, so that Fig. 7-13 gives I2 = 0.543 and I3 = 0.073. Thus

from Eq. (7-86)

W = 3.51×1012

B+ 0.026B = 3.51×105 + 2.6×105 = 6.1×105

Using Eq. (7-58) to find L yields L = 2.871 at γ = 0.9 and x = 0.5. Substituting

these values into eq. (7-57) we have

bon = (6)5/3 (1.6×10-19/0.7) (6.1×105).5/3 (0.543)1/1.5 2.871 = 7.97×10-17 J

Thus Pr = bonB = (7.97×10-17 J)(107 b/s) = 7.97×10-10 W

orPr(dBm) = 10 log 7.97×10-10 = -61.0 dBm

7-25. From Eq. (7-96) the difference in the two amplifier designs is given by

∆W = 1

Bq2 2kBT

Rf I2 = 3.52×106 for I2 = 0.543 and γ = 0.9.

From Eq. (7-57), the change in sensitivity is found from

10 log

WHZ + ∆W

WHZ

x2(1+x)

= 10 log

1.0 + 3.52

1.0

.53

= 10 log 1.29 = 1.09 dB

7-26. (a) For simplicity, let

D = M2+x

η

hν I2 and F = QM

η

hν

so that Eq. (7-54) becomes, for γ = 1,

15

b = F[ ](Db + W)1/2 + W1/2

Squaring both sides and rearranging terms gives

b2

F2 - Db - 2W = 2W1/2 (Db + W)1/2

Squaring again and factoring out a "b2" term yields

b2 - (2DF2)b + (F4D2 - 4WF2) = 0

Solving this quadratic equation in b yields

b = 12 [ ]2DF2 ± 4F4D2 - 4F4D2 + 16WF2 = DF2 + 2F W

(where we chose the "+" sign) = hνη

MxQ2I2 +

2QM W1/2

(b) With the given parameter values, we have

bon = 2.286×10-19 39.1M 0.5 +1.7× 104

M

The receiver sensitivity in dBm is found from

Pr = 10 log bon(50 ×106 b / s)[ ]

Representative values of Pr for several values of M are listed in the table below:

M Pr(dBm) M Pr(dBm)

30 - 50.49 80 -51.92

40 -51.14 90 -51.94

50 -51.52 100 -51.93

60 -51.74 110 -51.90

70 -51.86 120 -51.86

16

7-27. Using Eq. (E-10) and the relationship

1

1+xa

2 dx0

∞

∫ = πa2

from App. B, we have from Eq. (7-97)

BHZ =

1(AR)2

(AR)2

1+ (2πRC)2 f 2 0

∞

∫ df = π2

12πRC

= 1

4RC

where H(0) = AR. Similarly, from Eq. (7-98)

BTZ =

11

1

1+2πRC

A

2

f2

0

∞

∫ df = π2

A2πRC

= A

4RC

7-28. To find the optimum value of M for a maximum S/N, differentiate Eq. (7-105)

with respect to M and set the result equal to zero:

d(S/N)dM =

(Ipm)2M

2q(Ip+ID)M2+x B + 4kBTB

Req FT

- q(Ip+ID) (2+x) M1+x B (Ipm)2M2

2q(Ip+ID)M2+x B + 4kBTB

Req FT

2 = 0

Solving for M,

M2+xopt =

4kBTBFT/Reqq(Ip+ID)x

7-29. (a) For computational simplicity, let K = 4kBTBFT/Req; substituting Mopt from

Problem 7-28 into Eq. (7-105) gives

17

SN =

12 (Ipm)2M

2opt

2q(Ip+ID)M2+xopt

B + KB =

12 (Ipm)2

K

q(Ip+ID)x

22+x

2q(Ip+ID)Kq(Ip+ID)x

B + KB

= xm2I

2p

2B(2+x) [ ]q(Ip+ID)x2/(2+x)

Req

4kBTFT

x/(2+x)

(b) If Ip >> ID, then

SN =

xm2I2p

2B(2+x) ( )qx2/(2+x)

I2/(2+x)p

Req

4kBTFT

x/(2+x)

= m2

2Bx(2+x)

( )xIp2(1+x)

q2(4kBTFT/Req)x

1/(2+x)

7-30. Substituting Ip = R0Pr into the S/N expression in Prob. 7-29a,

SN =

xm2

2B(2+x) ( )R0Pr

2

[ ]q(R0Pr+ID)x2/(2+x)

Req

4kBTFT

x/(2+x)

=

(0.8)2 (0.5 A / W)2 Pr2

2(5 × 106 / s) 3 [1.6 × 10−19 C(0.5Pr + 10−8 ) A]2 / 3

104 Ω / J1.656 × 10−20

1 / 3

= 1.530 ×1012 P

r 2

0.5Pr + 10−8( )2 / 3 where Pr is in watts.

18

We want to plot 10 log (S/N) versus 10 log Pr

1 mW . Representative values are

shown in the following table:

Pr (W) Pr (dBm) S/N 10 log (S/N) (dB)

2×10-9 - 57 1.237 0.92

4×10-9 - 54 4.669 6.69

1×10-8 - 50 25.15 14.01

4×10-8 - 44 253.5 24.04

1×10-7 - 40 998.0 29.99

1×10-6 - 30 2.4×104 43.80

1×10-5 - 20 5.2×105 57.18

1×10-4 - 10 1.13×107 70.52

1


8-1. SYSTEM 1: From Eq. (8-2) the total optical power loss allowed between the light

source and the photodetector is

PT = PS - PR = 0 dBm - (-50 dBm) = 50 dB

= 2(lc) + αfL + system margin = 2(1 dB) + (3.5 dB/km)L + 6 dB

which gives L = 12 km for the maximum transmission distance.

SYSTEM 2: Similarly, from Eq. (8-2)

PT = -13 dBm - (-38 dBm) = 25 dB = 2(1 dB) + (1.5 dB/km)L + 6 dB

which gives L = 11.3 km for the maximum transmission distance.

8-2. (a) Use Eq. (8-2) to analyze the link power budget. (a) For the pin photodiode,

with 11 joints

PT = PS - PR = 11(lc) + αfL + system margin

= 0 dBm - (-45 dBm) = 11(2 dB) + (4 dB/km)L + 6 dB

which gives L = 4.25 km. The transmission distance cannot be met with these

components.

(b) For the APD

0 dBm - (-56 dBm) = 11(2 dB) + (4 dB/km)L + 6 dB

which gives L = 7.0 km. The transmission distance can be met with these

components.

8-3. From g(t) = ( )1 - e-2πBt u(t) we have

1 - e-2πBt10

= 0.1 and

1 - e-2πBt90

= 0.9

so that

2

e-2πBt10

= 0.9 and e-2πBt90

= 0.1

Then

e2πBtr

= e2πB(t90-t10)

= .9.1 = 9

It follows that

2πBtr = ln 9 or tr = ln 92πB

= 0.35

B

8-4. (a) From Eq. (8-11) we have

1

2π σ exp

- t21/2

2σ2 = 12

1

2π σ which yields t1/2 = (2 ln 2)1/2 σ

(b) From Eq. (8-10), the 3-dB frequency is the point at which

G(ω) = 12 G(0), or exp

- (2πf3dB)2 σ2

2 = 12

Using σ as defined in Eq. (8-13), we have

f3dB = (2 ln 2)1/2

2πσ = 2 ln 2

π tFWHM =

0.44tFWHM

8-5. From Eq. (8-9), the temporal response of the optical output from the fiber is

g(t) = 1

2π σ exp

- t2

2σ2

If τe is the time required for g(t) to drop to g(0)/e, then

g(τe) = 1

2π σ exp

- τe2

2σ2 = g(0)

e = 1

2π σe

3

from which we have that τe = 2 σ. Since te is the full width of the pulse at the

1/e points, then te = 2τe = 2 2 σ.

From Eq. (8-10), the 3-dB frequency is the point at which

G(f3dB) = 12 G(0). Therefore with σ = te/(2 2 )

G(f3dB) = 1

2π exp

-

12(2πf3dB σ)2 =

12

1

2π

Solving for f3dB:

f3dB = 2 ln 22πσ =

2 ln 22π

2 2te

= 0.53

te

8-6. (a) We want to evaluate Eq. (8-17) for tsys.

Using Dmat = 0.07 ns/(nm-km), we have

tsys =

(2)2 + (0.07)2(1)2(7)2 +

440(7)0.7

800

2

+

350

90

2 1/2

= 4.90 ns

The data pulse width is Tb = 1B =

190 Mb/s = 11.1 ns

Thus 0.7Tb = 7.8 ns > tsys, so that the rise time meets the NRZ data requirements.

(b) For q = 1.0,

tsys =

(2)2 + (0.49)2 +

440(7)

800

2

+

350

90

2 1/2

= 5.85 ns

8-7. We want to plot the following 4 curves of L vs B = 1

Tb :

(a) Attenuation limit

PS - PR = 2(lc) + αfL + 6 dB, where PR = 9 log B - 68.5

so that L = (PS – 9 log B + 62.5 - 2lc)/αf

(b) Material dispersion

4

tmat = Dmat σλ L = 0.7Tb or

L = 0.7Tb

Dmat σλ =

0.7BDmat σλ

= 104

B (with B in Mb/s)

(c) Modal dispersion (one curve for q = 0.5 and one for q = 1)

tmod = 0.440L

q

800 = 0.7B or L =

800

0.44 0.7B

1/q

With B in Mb/s, L = 1273/B for q =1, and L = (1273/B)2 for q = .5.

8-8. We want to plot the following 3 curves of L vs B = 1

Tb :

(a) Attenuation limit

PS - PR = 2(lc) + αfL + 6 dB, where PR = 11.5 log B - 60.5, PS = -13 dBm, αf =

1.5 dB/km, and lc = 1 dB,

so that L = (39.5 - 11.5 log B)/1.5 with B in Mb/s.

(b) Modal dispersion (one curve for q = 0.5 and one for q = 1)

tmod = 0.440L

q

800 = 0.7B or L =

800

0.44 0.7B

1/q

With B in Mb/s, L = 1273/B for q =1, and L = (1273/B)2 for q = .5.

8-9. The margin can be found from

PS - PR = lc + 49(lsp) + 50αf + noise penalty + system margin

-13 - (-39) = 0.5 + 49(.1) + 50(.35) + 1.5 + system margin

from which we have

system margin = 1.6 dB

5

8-10. Signal bits

8-11. The simplest method is to use an exclusive-OR gate (EXOR), which can be

implemented using a single integrated circuit. The operation is as follows: when

the clock period is compared with the bit cell and the inputs are not identical, the

EXOR has a high output. When the two inputs are identical, the EXOR output is

low. Thus, for a binary zero, the EXOR produces a high during the last half of the

bit cell; for a binary one, the output is high during the first half of the bit cell.

A B C

L L L

L H H

H L H

H H L

1 0 1 1 1 1 0 0 100 1Signal bits

Baseband (NRZ-L) data

Clock signal

Optical Manchester

6

8-12.

8-13.

Original 010 001 111 111 101 000 000 001 111 110code

3B4B 0101 0011 1011 0100 1010 0010 1101 0011 1011 1100encoded

8-14. (a) For x = 0.7 and with Q = 6 at a 10-9 BER,

Pmpn = -7.94 log (1 - 18k2π4h4) where for simplicity h = BZDσλ

NRZ data

Freq. A

Freq. B

PSK data

7

(b) With x = 0.7 and k = 0.3, for an 0.5-dB power penalty at

140 Mb/s = 1.4×10-4 b/ps [to give D in ps/(nm.km)]:

0.5 = -7.94 log 1 - 18(0.3)2[π(1.4×10-4)(100)(3.5)]4D4

or

0.5 = -7.94 log 1 - 9.097×10-4D4 from which D = 2 ps/(nm.km)

B (Mb/s) D [ps/(nm.km)]

140 2

280 1

560 0.5

1


9-1.

9-2.

Received optical power (dBm)0 4 8 12 16

58

54

50

CNR

(dB)

RIN limit

Thermal noiselimit

QuantumNoiselimit

f1 f2 f3 f4 f5

∆Transmissionsystem

Triple-beatproducts

2-tone3rd order

2

9-3. The total optical modulation index is

m =

∑i

m2i

1/2

= 30(.03)2 + 30(.04)2[ ]1 / 2= 27.4 %

9-4 The modulation index is m =

∑i=1

120(.023)2

1/2

= 0.25

The received power is

P = P0 – 2(lc) - αfL = 3 dBm - 1 dB - 12 dB = -10 dBm = 100µW

The carrier power is

C = 12 (mR0P) 2 =

12

(15 ×10−6 A) 2

The source noise is, with RIN = -135 dB/Hz = 3.162×10-14 /Hz,

< > i2source = RIN (R0P)2 B = 5.69×10-13A2

The quantum noise is

< > i2Q = 2q(R0P + ID)B = 9.5×10-14A2

The thermal noise is

< > i2T =

4kBTReq

Fe = 8.25×10-13A2

3

Thus the carrier-to-noise ratio is

CN =

12

(15 ×10−6 A)2

5.69 × 10−13 A 2 + 9.5 ×10 −14 A2 + 8.25 ×10−13A2 = 75.6

or, in dB, C/N = 10 log 75.6 = 18.8 dB.

9-5. When an APD is used, the carrier power and the quantum noise change.


C = 12 (mR0MP) 2 =

12

(15 ×10−5 A)2


< > i2Q = 2q(R0P + ID)M2F(M)B = 2q(R0P + ID)M2.7B = 4.76×10-10A2


CN =

12

(15 ×10 −5 A)2

5.69 × 10−13 A 2 + 4.76 × 10−11 A2 + 8.25 ×10−13A2 = 236.3

or, in dB, CN = 23.7 dB

9-6. (a) The modulation index is m =

∑i=1

32(.044)2

1/2

= 0.25

The received power is P = -10 dBm = 100µW


C = 12 (mR0P) 2 =

12

(15 ×10−6 A) 2

4


< > i2source = RIN (R0P)2 B = 5.69×10-13A2


< > i2Q = 2q(R0P + ID)B = 9.5×10-14A2

The thermal noise is

< > i2T =

4kBTReq

Fe = 8.25×10-13A2


CN =

12

(15 ×10−6 A)2

5.69 × 10−13 A 2 + 9.5 ×10 −14 A2 + 8.25 ×10−13A2 = 75.6

or, in dB, C/N = 10 log 75.6 = 18.8 dB.

(b) When mi = 7% per channel, the modulation index is

m =

∑i=1

32(.07)2

1/2

= 0.396

The received power is P = -13 dBm = 50µW


C = 12 (mR0P) 2 =

12

(1.19 × 10−5 A) 2 = 7.06×10-11A2


< > i2source = RIN (R0P)2 B = 1.42×10-13A2

5


< > i2Q = 2q(R0P + ID)B = 4.8×10-14A2

The thermal noise is the same as in Part (a).


CN =

7.06 ×10−11A 2

1.42 × 10−13 A 2 + 4.8 ×10−14A 2 + 8.25 ×10−13 A2 = 69.6

or, in dB, C/N = 10 log 69.6 = 18.4 dB.

9-8. Using the expression from Prob. 9-7 with ∆ντ = 0.05, fτ = 0.05, and ∆ν = f = 10

MHz, yields

RIN(f) = 4R1R2

π ∆ν

f2 + ∆ν2

1 + e-4π∆ντ

- 2 e-2π∆ντ

cos(2πfτ)

= 4R1R2

π 1

20 MHz (.1442)

Taking the log and letting the result be less than -140 dB/Hz gives

-80.3 dB/Hz + 10 log R1R2 < -140 dB/Hz

If R1 = R2 then 10 log R1R2 = 20 log R1 < -60 dB

or 10 log R1 = 10 log R2 < -30 dB

1


10-1. In terms of wavelength, at a central wavelength of 1546 nm a 500-GHz channel

spacing is

∆λ =

λ2

c ∆f = 1546 nm( )2

3 ×108 m / s

500 ×10 9 s−1 = 4 nm

The number of wavelength channels fitting into the 1536-to-1556 spectral band

then is

N = (1556 – 1536 nm)/4 nm = 5

10-2. (a) We first find P1 by using Eq. (10-6):

10 log 200 µW

P1

= 2.7 dB yields P1 =10 (log 200 −0.27 ) = 107.4 µW

Similarly, P2 = 10(log 200 −0.47) = 67.8 µW

(b) From Eq. (10-5): Excess loss = 10 log

200107.4 + 67.8

= 0.58 dB

(c)P1

P1 + P2

=107.4175.2

= 61% andP2

P1 + P2

=67.8

175.2= 39%

10-3. The following coupling percents are are realized when the pull length is stopped at

the designated points:

Coupling percents from input fiber to output 2

Points A B C D E F1310 nm 25 50 75 90 100 01540 nm 50 88 100 90 50 100

10-4. From A out = s11Ain + s12Bin and Bout = s21Ain +s22 Bin = 0 , we have

Bin = −s21

s22

A in and

A out = s11 −s12s21

s22

A in

2

Then

T = Aout

Ain

2

= s11 − s12s21

s22

2

and R =Bin

Ain

2

=s21

s22

÷ s11 −

s12s21

s22

2

10-5. From Eq. (10-18)

P2

P0

= sin 2 0.4z( )exp −0.06z( ) = 0.5

One can either plot both curves and find the intersection point, or solve the

equation numerically to yield z = 2.15 mm.

10-6. Since βz ∝ n , then for nA > nB we have κA < κB. Thus, since we need to have

κALA = κBLB, we need to have LA > LB.

10-7. From Eq. (10-6), the insertion loss LIj for output port j is

L Ij = 10 logPi −in

Pj −out

Let

a j = Pi −in

Pj− out

= 10L Ij /10

, where the values of LIj are given in Table P10-7.

Exit port no. 1 2 3 4 5 6 7Value of aj 8.57 6.71 5.66 8.00 9.18 7.31 8.02

Then from Eq. (10-25) the excess loss is

10 log Pin

Pj∑

= 10 log

Pin

Pin

1a1

+1a2

+ ... +1an

=10 log 1

0.95

= 0.22 dB

10-8. (a) The coupling loss is found from the area mismatch between the fiber-core

endface areas and the coupling-rod cross-sectional area. If a is the fiber-core radius

and R is the coupling-rod radius, then the coupling loss is

3

Lcoupling = 10 log PoutPin

= 10 log 7πa2

πR2 = 10 log 7(25)2

(150)2 = -7.11 dB

(b) Similarly, for the linear-plate coupler

Lcoupling = 10 log 7πa2

l∞w = 10 log

7π(25)2

800(50) = -4.64 dB

10-9. (a) The diameter of the circular coupling rod must be 1000 µm, as shown in the

figure below. The coupling loss is

Lcoupling = 10 log 7πa2

πR2 = 10 log 7(100)2

(500)2 = -5.53 dB

(b) The size of the plate coupler must be 200 µm by 2600 µm.

The coupling loss is 10 log 7π(100)2

200(2600) = -3.74 dB

10-10. The excess loss for a 2-by-2 coupler is given by Eq. (10-5), where P1 = P2 for a 3-

dB coupler. Thus,

200 µm

400µm

Coupling roddiameter

4

Excess loss =

10 log P0

P1 + P2

= 10 log

P0

2P1

= 0.1 dB

This yields

P1 =

P0

2

÷10 0.01 = 0.977

P0

2

Thus the fractional power traversing the 3-dB coupler is FT = 0.977.

Then, from Eq. (10-27),

Total loss =

−10 log log FT

log 2− 1

log N = −10 log

log 0.977log 2

−1

log 2 n ≤ 30

Solving for n yields

n ≤−3

log 2 log 0.977

log 2− 1

= 9.64

Thus, n = 9 and N = 2n = 29 = 512

10-11. For details, see Verbeek et al., Ref. 34, p. 1012

For the general case, from Eq. (10-29) we find

M11 = cos 2κd( ) ⋅ cos k∆L / 2( )+ j sin k∆L / 2( )

M12 = M21 = j sin 2κd( )⋅ cos k∆L / 2( )

M 22 = cos 2κd( )⋅ cos k∆L / 2( )− j sin k∆L / 2( )

The output powers are then given by

Pout ,1 = cos2 2κd( )⋅ cos2 k∆L /2( )+ sin2 k∆L / 2( )[ ]Pin,1

+ sin 2 2κd( )⋅ cos2 k∆L / 2( )[ ]Pin,2

5

Pout ,2 = sin 2 2κd( ) ⋅cos2 k∆L / 2( )[ ]Pin,1

+ cos2 2κd( )⋅ cos2 k∆L /2( ) + sin2 k∆L / 2( )[ ]Pin,2

10-12. (a) The condition ∆ν = 125 GHz is equivalent to having ∆λ = 1 nm. Thus the

other three wavelengths are 1549, 1550, and 1551 nm.

(b) From Eqs. (10-42) and (10-43), we have

∆L1 =c

2n eff (2∆ν)= 0.4 mm and

∆L3 =c

2n eff ∆ν= 0.8 mm

10-13. An 8-to-1 multiplexer consists of three stages of 2 × 2 MZI multiplexers. The first

stage has four 2 × 2 MZIs, the second stage has two, and the final stage has one

2 × 2 MZI. Analogous to Fig. 10-14, the inputs to the first stage are (from top to

bottom) ν, ν + 4∆ν, ν + 2∆ν, ν + 6∆ν, ν + ∆ν, ν + 5∆ν, ν + 3∆ν, ν + 7∆ν.

In the first stage

∆L1 = c2n eff (4∆ν)

= 0.75 mm

In the second stage

∆L2 =

c2neff (2∆ν)

= 1.5 mm

In the third stage

∆L3 =c

2n eff (∆ν)= 3.0 mm

10-14. (a) For a fixed input angle φ, we differentiate both sides of the grating equation toget

cos θ dθ = k

n'Λ dλ ordθdλ =

kn'Λ cos θ

If φ ≈ θ, then the grating equation becomes 2 sin θ = kλn'Λ .

6

Solving this for k

n'Λ and substituting into the dθdλ equation yields

dθdλ =

2 sin θλ cos θ =

2 tan θλ

(b) For S = 0.01,

tan θ =

Sλ

2∆λ (1+m)

1/2 =

0.01(1350)2(26)(1+ 3)

1/ 2

= 0.2548

or θ = 14.3°

10-15. For 93% reflectivity

R = tanh 2(κL) = 0.93 yields κL = 2.0, so that L = 2.7 mm for κ = 0.75 mm-1.

10-16. See Bennion et al., Ref. 42, Fig. 2a.

10-17. Derivation of Eq. (10-49).

10-18. (a) From Eq. (10-45), the grating period is

Λ =λuv

2 sin θ2

=244 nm

2 sin(13.5°)=

2442(0.2334)

nm = 523 nm

(b) From Eq. (10-47), λ Bragg = 2Λn eff = 2(523 nm) 1.48 =1547 nm

(c) Using η = 1− 1/ 2 = 0.827 , we have from Eq. (10-51),

κ =π δn ηλBragg

=π 2.5 ×10 −4( )(0.827)

1.547 ×10−4 cm= 4.2 cm −1

(d) From Eq. (10-49), ∆λ = 1.547 µm( )2

π (1.48) 500 µm (2.1)2 + π2[ ]1 / 2

= 3.9 nm

7

(e) From Eq. (10-48), Rmax = tanh 2(κL) = tanh 2(2.1) = (0.97)2 = 94%

10-19. Derivation of Eq. (10-55).

10-20. (a) From Eq. (10-54),

∆L = m

λ 0

n c

= 118 1.554 µm

1.451= 126.4 µm

(b) From Eq. (10-57),

∆ν = xL f

n scdmλ2

n c

ng

=25 µm

9.36 × 103 µm 1.453 (3 ×10 8 m / s)(25 ×10 −6 m)

118 (1.554 ×10 −6 m)2 1.4511.475

= 100.5 GHz

∆λ =

λ2

c ∆ν =

(1.554 ×10−6 m)2

3 × 108 m / s 100.5 GHz = 0.81 nm

(c) From Eq. (10-60),

∆νFSR =c

ng∆L=

3 ×108 m / s1.475(126.4 µm)

= 1609 GHz

Then

∆λ =

λ2

c ∆νFSR =

(1.554 × 10−6 m)2

3 × 108 m / s

1609 GHz =12.95 nm

(d) Using the conditions

sin θ i ≈ θ i =

2(25 µm)9380 µm

= 5.33 ×10 −3 radians

and

sin θo ≈ θo = 21.3 ×10−3 radians

8

then from Eq. (10-59),

∆νFSR ≈ cng [∆L + d θ i + θo( )]

= 3 × 108 m /s1.475 (126.4 × 10−6 m) + (25 ×10−6 m) 5.33 + 21.3( )×10−3[ ]= 1601 GHz

10-21. The source spectral width is

∆λsignal =λ2 νc

=1550 nm( )2 (1.25 ×10 9 s−1)(3 ×10 8

m / s) 109 nm / m( ) =1 × 10−2

nm

Then from Eq. (10-61)

∆λtune = λ

∆n eff

n eff

= 1550 nm( ) 0.5%( ) = 7.75 nm

Thus, from Eq. (10-63)

N =∆λ tune

10 λ signal

=7.75 nm

10 0.01 nm( )= 77

10-22. (a) From Eq. (10-64), the grating period is

Λ =

λBragg

2neff

=1550 nm

2 3.2( )= 242.2 nm

(b) Again, from the grating equation,

∆Λ =

∆λ2n eff

=2.0 nm2 3.2( )

= 0.3 nm

10-23. (a) From Eq. (10-43)

∆L = c2n eff ∆ν

=λ2

∆λ1

2neff

= 4.0 mm

(b) ∆Leff = ∆n eff L implies that ∆n eff = 4 mm

100 mm= 0.04 = 4%

9

10-24. For example, see C. R. Pollock, Fundamentals of Optoelectronics, Irwin, 1995,

Fig. 15.11, p. 439.

10-25. (a) The driving frequencies are found from

fa = νo

v a∆nc

=va ∆n

λ

Thus we have

Wavelength (nm) 1300 1546 1550 1554Acousticfrequency (MHz)

56.69 47.67 47.55 47.43

(b) The sensitivity is (4 nm)/(0.12 MHz) = 0.033 nm/kHz

1


11-1. (a) From Eq. (11-2), the pumping rate is

R p = IqwdL

= 100 mA(1.6 ×10−19 C)(5 µm)(0.5 µm)(200 µm)

= 1.25 ×1027 (electrons / cm 3) / s

(b) From Eq. (11-8), the maximum zero-signal gain is

g0 = 0.3(1×10 −20 m2 )(1 ns) 1.25 ×1033 (electrons/ m3) / s −1.0 × 1024 /m3

1 ns

= 750 m−1 = 7.5 cm−1

(c) From Eq. (11-7), the saturation photon density is

N ph;sat =1

0.3 (1×10−20m2 ) 2 ×108 m / s( )(1 ns)= 1.67 ×1015 photons/ cm 3

(d) From Eq. (11-4), the photon density is

N ph =Pin λ

vg hc wd( )= 1.32 × 1010 photons / cm3

11-2. Carrying out the integrals in Eq. (11-14) yields

g0L = lnP(L)P(0)

+P(L) − P(0)

Pamp,sat

Then with P(0) = Pin, P(L) = Pout, G = Pout/Pin, and G0 = exp g0L( ) from Eq. (11-

10), we have

ln G 0 = g0L = ln G +GPin

Pamp,sat

−Pin

Pamp,sat

= ln G + 1 − G( ) Pin

Pamp,sat

Rearranging terms in the leftmost and rightmost parts then yields Eq. (11-15).

11-3. Plots of amplifier gains.

2

11-4. Let G = G0/2 and Pin = Pout / G = 2Pout,sat / G 0 . Then Eq. (11-15) yields

G0

2= 1+

G0Pamp.sat

2Pout.sat

ln 2

Solving for Pout,sat and with G0 >> 1, we have

Pout .sat =G 0 ln 2G0 − 2( ) Pamp.sat ≈ (ln 2) Pamp .sat = 0.693 Pamp.sat

11-5. From Eq. (11-10), at half the amplifier gain we have

G =12

G0 =12

exp g0L( )= exp gL( )

Taking the logarithm and substituting into the equation given in the problem,

g = g0 −1L

ln 2 =g0

1 + 4 ν3dB − ν0( )2

/ ∆ν( )2

From this we can find that

2 ν3dB − ν0( )∆ν

=g0

g0 − 1L

ln 2−1

1/ 2

=1

g0L / ln 2 −1

1/ 2

=1

log2G 0

2

1/ 2

11-6. Since

ln G = g λ( )L = g0 exp − λ − λ0( )2

/ 2 ∆λ( )2[ ]= ln G0 exp − λ − λ0( )2/ 2 ∆λ( )2[ ]

we have

lnln G 0

ln G

=λ − λ 0( )2

2 ∆λ( )2

The FWHM is given by 2(λ – λ0), so that from the above equation, with the 3-dB

gain G = 27 dB being 3 dB below the peak gain, we have

3

FWHM = 2 λ − λ0 = 2 2 ln ln G0

ln G

1/ 2

∆λ

= 2 2 ln ln 30ln 27

1/ 2

∆λ = 0.50 ∆λ

which is the expected result for a gaussian gain profile.

11-7. From Eq. (11-17), the maximum PCE is given by

PCE ≤λp

λ s

=980

1545= 63.4% for 980-nm pumping, and by

PCE ≤λp

λ s

= 14751545

= 95.5% for 1475-nm pumping

11-8. (a) 27 dBm = 501 mW and 2 dBm = 1.6 mW.

Thus the gain is

G = 10 log

5011.6

= 10 log 313 = 25 dB

(b) From Eq. (11-19),

313 ≤ 1+ 9801542

Pp,in

Ps, in

. With a 1.6-mW input signal, the pump power needed is

Pp,in ≥

312 1542( )980

1.6 mW( ) = 785 mW

11-9. (a) Noise terms:

From Eq. (6-17), the thermal noise term is

σT2 =

4kBTRL

B =4 1.38 ×10−23 J / K( ) 293 K( )

1000 Ω 1 GHz = 1.62 ×10−14 A2

From Eq. (11-26), we have

4

σshot− s2 = 2qR GPs, inB

= 2 1.6 ×10 −19 C( ) 0.73 A /W( ) 100( ) 1 µW( )1 GHz

= 2.34 ×10−14 A2

From Eqs. (11-26) and (11-24), we have

σshot− ASE2 = 2qR SASE∆νoptB = 2qR

hcλ

n spG∆νoptB

= 2 1.6 ×10−19 C( ) .73 A / W( ) 6.626 × 10−34 J/ K( )

× 3 ×108 m /s( )2(100)(3.77 THz)(1 GHz)/1550nm

= 2.26 ×10−14 A2

From Eq. (11-27) and (11-24), we have

σs− ASE2 = 4 0.73 A/ W( ) 100( ) 1 µW( )[ ]

× .73 A/ W( )6.626 ×10−34 J/ K( ) 3 ×108 m /s( )

1550nm2(100)(1 GHz)

= 5.47 ×10−12 A 2

From Eq. (11-28), we have

σASE − ASE2 = .73A / W( )2 6.626 × 10−34 J/ K( )3 × 108 m/ s( )

1550nm2(100)

2

× 2 1 THz( )− 1 GHz[ ]1 GHz( )

= 7.01×10−13 A 2

5

11-10. Plot of penalty factor from Eq. (11-36).

11-11. (a) Using the transparency condition Gexp(-αL) = 1 for a fiber/amplifier segment,

we have

P path =1L

P(z) dz0

L

∫ =Pin

L e− αz dz

0

L

∫

= Pin

αL 1 − e−αL[ ]= Pin

αL 1 − 1

G

= Pin

G

G −1ln G

since ln G = αL from the transparency condition.

(b) From Eq. (11-35) and using Eq. (11-24),

PASE path= NPASE

L e− αz

0

L

∫ dz = NPASE

αL 1 − e−αL( )

=α (NL)(αL)2 PASE 1−

1G

=

αL tot

(ln G)2 hνn sp(G −1)∆νopt 1 −1G

= αL tot hνnsp∆νopt 1G

G −1ln G

2

11-12. Since the slope of the gain-versus -input power curve is –0.5, then for a 6-dB drop

in the input signal, the gain increases by +3 dB.

1. Thus at the first amplifier, a –10.1-dBm signal now arrives and experiences a

+10.1-dB gain. This gives a 0-dBm output (versus a normal +3-dBm output).

2. At the second amplifier, the input is now –7.1 dBm (down 3 dB from the usual

–4.1 dBm level). Hence the gain is now 8.6 dB (up 1.5 dB), yielding an output

of

–7.1 dBm + (7.1 + 1.5) dB = 1.5 dBm

6

3. At the third amplifier, the input is now –5.6 dBm (down 1.5 dB from the usual

–4.1 dBm level). Hence the gain is up 0.75 dB, yielding an output of

–5.6 dBm + (7.1 + 0.75) dB = 2.25 dBm

4. At the fourth amplifier, the input is now –4.85 dBm (down 0.75 dB from the

usual –4.1 dBm level). Hence the gain is up 0.375 dB, yielding an output of

–4.85 dBm + (7.1 + 0.375) dB = 2.63 dBm

which is within 0.37 dB of the normal +3 dBm level.

11-13. First let 2πνi t + φ i = θ i for simplicity. Then write the cosine term as

cos θi =

e jθ i + e− jθ i

2, so that

P = E i(t)E i*(t) = 2Pi

i =1

N

∑ e jθ i + e− jθi

2

× 2Pkk =1

N

∑ e jθk + e− jθk

2

= 14

2Pik=1

N

∑ 2Pki =1

N

∑ e jθ i e− jθk + e jθk e− jθi + e jθ i e jθ k + e− jθ i e− jθ k[ ]

=14

2Pik=1

N

∑ 2Pki =1

N

∑ e j (θ i −θ k ) + e− j(θi − θk ) + e j(θi + θk ) + e− j(θ i +θ k )[ ]

= Pi +12i=1

N

∑ 2Pik≠ i

N

∑ 2Pki =1

N

∑ e j(θ i −θ k ) + e− j(θ i − θk )[ ]

where the last two terms in the second-last line drop out because they are beyond

the response frequency of the detector. Thus,

P = Pi +

i=1

N

∑ 2 PiPkk ≠i

N

∑i=1

N

∑ cos θ i − θk( )[ ]

11-14. (a) For N input signals, the output signal level is given by

Ps,out = G Ps,in

i =1

N

∑ (i) ≤1 mW .

The inputs are 1 µW (-30 dBm) each and the gain is 26 dB (a factor of 400).

7

Thus for one input signal, the output is (400)(1 µW) = 400 µW or –4 dBm.

For two input signals, the total output is 800 µW or –1 dBm. Thus the level of

each individual output signal is 400 µW or –4 dBm.

For four input signals, the total input level is 4 µW or –24 dBm. The output then

reaches its limit of 0 dBm, since the maximum gain is 26 dB. Thus the level of each

individual output signal is 250 µW or –6 dBm.

Similarly, for eight input channels the maximum output level is o dBm, so the level

of each individual output signal is 1/8(1 mW) = 125 µW or –9 dBm.

(b) When the pump power is doubled, the outputs for one and two inputs remains

at the same level. However, for four inputs, the individual output level is 500 µW

or –3 dBm, and for 8 inputs, the individual output level is 250 µW or –6 dBm.

11-15. Substituting the various expressions for the variances from Eqs. (11-26) through

(11-30) into the expression given for Q in the problem statement, we find

Q =AP

HP + D2( )1 / 2+ D

where we have defined the following terms for simplicity

A = 2R G

H = 4qR GB+ 8R2GSASEB and D2 = σoff

2

Rearrange terms in the equation for Q to get

Q2 HP + D2( )1/ 2= AP − QD

Squaring both sides and solving for P yields P =2QD

A+

Q2HA2

Substituting the expressions for A, H, and D into this equation, and recalling the

expression for the responsivity from Eq. (6-6), then produces the result stated in

the problem, where

8

F =1 + 2ηn sp(G −1)

ηG

1


12-1. We need to evaluate Pin using Eq. (12-11). Here Fc = 0.20,

CT = 0.05, Fi = 0.10, P0 = 0.5 mW, and A0 = e−2.3(3) / 10 = 0.933

Values of Pin as a function of N are given in the table below. Pin in

dBm is found from the relationship Pin(dBm) = 10 log

Pin (mW )1 mW

N Pin(nW) Pin(dBm) N Pin(nW) Pin(dBm)

2 387 -34.1 8 5.0 -53.0

3 188 -37.3 9 2.4 -56.2

4 91 -40.4 10 1.2 -59.2

5 44.2 -43.5 11 0.6 -62.2

6 21.4 -46.7 12 0.3 -65.5

7 10.4 -49.8

(b) Using the values in the above table, the operating margin for 8 stations is

-53 dBm - (-58 dBm) = 5 dB

(c) To have a 6-dB power margin, we can transmit over at most seven stations.

The dynamic range with N = 7 is found from Eq. (12-13):

DR = −10(N − 2) log .933(.8)2 (.95)2(.9)[ ]= −50 log (0.485) = 15.7 dBm

12-2. (a) Including a power margin, we have from Eq. (12-16)

PS − PR − power marg in = Lexcess + α(2L) + 2Lc +10log N

Thus

0 – (-38 dBm) – 6 dB = 3 dB + (0.3 dB/km)2(2 km) + 2(1.0 dB) + 10 log N

2

so that 10 log N = 25.8. This yields N = 380.1, so that 380 stations can be

attached.

(b) For a receiver sensitivity of –32 dBm, one can attach 95 stations.

12-3. (b) Let the star coupler be located in the ceiling in the wire room, as shown in the

figure below.

For any row we need seven wires running from the end of the row of offices to

each individual office. Thus, in any row we need to have (1+2+3+4+5+6+7)x15 ft

= 420 ft of optical fiber to connect the offices. From the wiring closet to the

second row of offices (row B), we need 8(10 + 15) ft = 200 ft; from the wiring

closet to the third row of offices (row C), we need 8(10 + 30) ft = 320 ft; and from

AWireroom B C D

3

the wiring closet to the fourth row of offices (row D), we need 8(20 + 45) ft = 520

ft of cable. For the 28 offices we also need 28x7 ft = 196 ft for wall risers.

Therefore for each floor we have the following cable needs:

(1) 4 x 420 ft for row runs

(2) 200 + 320 + 520 ft = 1040 for row connections

(3) 196 ft for wall risers

Thus, the total per floor = 2916 ft

Total cable in the building: 2x9 ft risers + 2916 ft x 2 floors = 5850 ft

12-4. Consider the following figure:

(a) For a bus configuration:

Cable length = N rows×(M-1)stations/row + (N-1) row interconnects

= N(M-1)d + (N-1)d = (MN-1)d

(b) The ring is similar to the bus, except that we need to close the loop with one

cable of length d. Therefore the cable length = MNd

(c) In this problem we consider the case where we need individual cables run

from the star to each station. Then the cable length is

L = cables run along the M vertical rows + cables run along the N horizontal

rows:

M

Nd

4

= Md ii =1

N −1

∑ + Nd jj=1

M −1

∑ = M N(N −1)

2d + N

M(M −1)2

d = MN2

(M + N − 2)d

12-5. (a) Let the star be located at the relative position (m,n). Then

L = N jj=1

m−1

∑ + N jj =1

M− m

∑ + M ii =1

n−1

∑ + M ii =1

N − n

∑

d

= Nm(m −1)

2+ (M − m)(M − m + 1)

2

+ Mn(n − 1)

2+ (N − n)(N − n +1)

2

d

= MN

2(M + N + 2) − Nm(M − m + 1) − Mn(N − n +1)

d

(b) When the star coupler is located in one corner of the grid, then

m = n= 1, so that the expression in (a) becomes

L = MN

2(M + N + 2) − NM − MN

d = MN

2(M + N − 2)d

(c) To find the shortest distance, we differentiate the expression for L given in (a)

with respect to m and n, and set the result equal to zero:

dLdm

= N(m - 1 - M) + Nm = 0 so that m = M +1

2

Similarly

dLdn

= M(n - 1 - N) + nM = 0 yields n = N + 1

2

Thus for the shortest cable runs the star should be located in the center of the grid.

12-6. (a) For a star network, one cannot reuse wavelengths. Thus, since each node must

be connected to N – 1 other nodes through a central point, we need N – 1

wavelengths.

5

For a bus network, these equations can easily be verified by drawing sample

diagrams with several even or odd stations.

For a ring network, each node must be connected to N – 1 other nodes. Without

wavelength reuse one thus needs N(N – 1) wavelengths. However, since each

wavelength can be used twice in the network, the number of wavelengths needed

is N(N-1)/2.

12-7. From Tables 12-4 and 12-5, we have the following:

OC-48 output for 40-km links: –5 to 0 dBm; α = 0.5 dB/km; PR = -18 dBm

OC-48 output for 80-km links: –2 to +3 dBm; α = 0.3 dB/km; PR = -27 dBm

The margin is found from: Margin = Ps − PR( )− αL − 2Lc

(a) Minimum power at 40 km:

Margin = [-2 – (-27)] – 0.5(40) –2(1.5) = +2 dB

(b) Maximum power at 40 km:

Margin = [0 – (-27)] – 0.5(40) –2(1.5) = +4 dB

(c) Minimum power at 80 km:

Margin = [-2 – (-27)] – 0.3(80) –2(1.5) = -2 dB

(d) Maximum power at 80 km:

Margin = [3 – (-27)] – 0.3(80) –2(1.5) = +3 dB

12-8. Expanding Table 12-6:

# of λs P1(dBm) P = 10 P1 / 10 (mW) Ptotal(mW) Ptotal(dBm)

1 17 50 50 17

2 14 25 50 17

3 12.2 16.6 49.8 17

4 11 12.6 50.4 17

5 10 10 50 17

6

6 9.2 8.3 49.9 17

7 8.5 7.1 49.6 17

8 8.0 6.3 50.4 17

12-9. See Figure 20 of ANSI T1.105.01-95.

12-10. See Figure 21 of ANSI T1.105.01-95.

12-11. The following wavelengths can be added and dropped at the three other nodes:

Node 2: add/drop wavelengths 3, 5, and 6



12-12. (b) From Eq. (12-18) we have

N λ = kpk +1 = 2(3)3 = 54

(c) From Eq. (12-20) we have

H =2(3)2 (3 − 1)(6 − 1) − 4(32 −1)

2(3 −1)[2(32 ) − 1]= 2.17

(d) From Eq. (12-21) we have

C =2 3( )3

2.17= 8.27

12-13. See Hluchyj and Karol, Ref. 25, Fig. 6, p. 1391 (Journal of Lightwave

Technology, Oct. 1991).

12-14. From Ref. 25:

In general, for a (p,k) ShuffleNet, the following spanning tree for assigning fixed

routes to packets generated by any given user can be obtained:

h Number of users h hops away from the source

1 p

7

2 p2

.

.

.

k – 1 pk-1

k Pk - 1

k + 1 Pk - p

k + 2 Pk - p2

.

.

.

2k – 1 Pk - pk-1

Summing these up results in Eq. (12-20).

12-15. See Li and Lee (Ref 40) for details.

12-16. The following is one possible solution:

(a) Wavelength 1 for path A-1-2-5-6-F

(b) Wavelength 1 for path B-2-3-C

(c) Wavelength 2 for the partial path B-2-5 and Wavelength 1 for path 5-6-F

(d) Wavelength 2 for path G-7-8-5-6-F

(e) Wavelength 2 for the partial path A-1-4 and Wavelength 1 for path 4-7-G

12-17. See Figure 4 of Barry and Humblet (Ref. 42).

12-18. See Shibata, Braun, and Waarts (Ref. 67).

(a) The following nine 3rd-order waves are generated due to FWM:

ν113 = 2(ν2 - ∆ν) – (ν2 + ∆ν) = ν2 - 3∆ν

ν112 = 2(ν2 - ∆ν) – ν2 = ν2 - 2∆ν

ν123 = (ν2 - ∆ν) + ν2 – (ν2 + ∆ν) = ν2 - 2∆ν

8

ν223 = 2ν2 – (ν2 + ∆ν) = ν2 - ∆ν = ν1

ν132 = (ν2 - ∆ν) + (ν2 + ∆ν) – ν2 = ν2

ν221 = 2ν2 – (ν2 - ∆ν) = ν2 + ∆ν = ν3

ν231 = ν2 + (ν2 + ∆ν) – (ν2 - ∆ν) = ν2 + 2∆ν

ν331 = 2(ν2 + ∆ν) – (ν2 - ∆ν) = ν2 + 3∆ν

ν332 = 2(ν2 + ∆ν) – ν2 = ν2 + 2∆ν

(b) In this case the nine 3rd-order waves are:

ν113 = 2(ν2 - ∆ν) – (ν2 + 1.5∆ν) = ν2 – 1.5∆ν

ν112 = 2(ν2 - ∆ν) – ν2 = ν2 - 2∆ν

ν123 = (ν2 - ∆ν) + ν2 – (ν2 + 1.5∆ν) = ν2 – 2.5∆ν

ν223 = 2ν2 – (ν2 + 1.5∆ν) = ν2 – 1.5∆ν

ν132 = (ν2 - ∆ν) + (ν2 + 1.5∆ν) – ν2 = ν2 + 0.5∆ν

ν221 = 2ν2 – (ν2 - ∆ν) = ν2 + ∆ν

ν231 = ν2 + (ν2 + 1.5∆ν) – (ν2 - ∆ν) = ν2 + 2.5∆ν

ν331 = 2(ν2 + 1.5∆ν) – (ν2 - ∆ν) = ν2 + 4∆ν

ν332 = 2(ν2 + 1.5∆ν) – ν2 = ν2 + 3∆ν

9

12-19. Plot: from Figure 2 of Y. Jaouën, J-M. P. Delavaux, and D. Barbier, “Repeaterless

bidirectional 4x2.5-Gb/s WDM fiber transmission experiment,” Optical Fiber

Technology, vol. 3, p. 239-245, July 1997.

12-20. (a) From Eq. (12-50) the peak power is

Ppeak = 1.7627

2π

2

A effλ

3

n 2c

DTs

2 =11.0 mW

(b) From Eq. (12-49) the dispersion length is

Ldisp = 43 km

(c) From Eq. (12-51) the soliton period is

L period =

π2

Ldisp = 67.5 km

(d) From Eq. (12-50) the peak power for 30-ps pulses is

10

Ppeak =

1.76272π

2

A effλ

3

n 2c

DTs

2 = 3.1 mW

12-21. Soliton system design.

12-22. Soliton system cost model.

12-23. (a) From the given equation, Lcoll = 80 km.

(b) From the given condition, Lamp ≤

12

Lcoll = 40 km

12-24. From the equation and conditions given in Prob. 12-23, we have that

∆λmax = Ts

DL amp

= 20 ps[0.4 ps /(nm ⋅ km)](25 km)

= 2 nm

Thus 2.0/0.4 = 5 wavelength channels can be accommodated.

12-25 Plot from Figure 3 of Ref. 103.

1


13-1 (a) From the given equation, nair = 1.000273. Thus,

λ vacuum = λairn air =1.000273(1550.0 nm) = 1550.42 nm

(b) From the given equation,

n T, P( )= 1 +(1.000273 − 1)(0.00138823)640

1+ 0.003671(0)= 1.000243

Then n T, P( )1550 nm( ) =1550.38 nm

13-2 Since the output voltage from the photodetector is proportional to the optical

power, we can write Eq. (13-1) as

α = 10L1 − L2

logV2

V1

where L1 is the length of the current fiber, L2 is the length cut off, and V1 and V2

are the voltage output readings from the long and short lengths, respectively. Then

the attenuation in decibels is

α =

101895 − 2

log 3.783.31

= 0.31 dB / km

13-3 (a) From Eq. (13-1)

α =10

LN − LF

log PN

PF

=10

LN − LF

log VN

VF

=10 log eLN − LF

ln VN

VF

From this we find

∆α = 10 log eL N − LF

∆VN

VN

+∆VF

VF

=4.343

LN − LF

±0.1% ± 0.1%( )= ± 8.686LN − LF

×10 −3

(b) If ∆α = 0.05 dB/km, then

L = LN − LF ≥

8.868 ×10−3

0.05 km = 176 m

2

13-4 (a) From Eq. (8-11) we have

1

2π σ exp

- t21/2

2σ2 = 12

1

2π σ which yields t1/2 = (2 ln 2)1/2 σ

(b) From Eq. (8-10), the 3-dB frequency is the point at which

G(ω) = 12 G(0), or exp

- (2πf3dB)2 σ2

2 = 12

Using σ as defined in Eq. (8-13), we have

f3dB = (2 ln 2)1/2

2πσ = 2 ln 2

π tFWHM =

0.44tFWHM

13-5 From Eq. (13-4), Pout (f) / Pin (f) = H(f) . To measure the frequency response, we

need a constant input amplitude, that is, Pin(f) = Pin(0). Thus,

P(f)P(0)

=Pout (f) / Pin (f)Pout (0) / Pin (0)

=H(f)H(0)

= H(f)

The following table gives some representative values of H(f) for different values of2σ:

f (MHz) 2σσ = 2 ns 2σσ = 1 ns 2σσ = 0.5 ns100 0.821 0.952 0.988200 0.454 0.821 0.952300 0.169 0.641 0.895500 0.0072 0.291 0.735700 0.089 0.5461000 0.0072 0.291

13-6 To estimate the value of D, consider the slope of the curve in Fig. P13-6 at λ =

1575 nm. There we have ∆τ = 400 ps over the wavelength interval from 1560 nm

to 1580 nm, i.e., ∆λ = 20 nm. Thus

D =

1L

∆τ∆λ

=1

10 km400 ps20 nm

= 2 ps /(nm ⋅ km)

3

Then, using this value of D at 1575 nm and with λ0 = 1548 nm, we have

S0 =

D λ( )λ − λ0

=2 ps /(nm ⋅ km)

(1575 − 1548) nm= 0.074 ps /(nm 2 ⋅ km)

13-7 With k = 1, λstart = 1525 nm, and λstop = 1575 nm, we have Ne = 17 extrema.

Substituting these values into Eq. (13-14) yields 1.36 ps.

13-8 At 10 Gb/s over a 100-km link, the given equation yields:

PISI ≈ 261 ps( )2 0.5 1− 0.5( )

100 ps( )2 = 6.5 ×10−4 dB

Similarly, at 10 Gb/s over a 1000-km link, PISI ≈ 0.065 dB .

This is the same result at 100 Gb/s over a 100-km link.

At 100 Gb/s over a 1000-km link, we have 6.5 dB.

13-9 For a uniform attenuation coefficient, β is independent of y. Thus, Eq. (13-16)

becomes

P(x) = P(0)exp −β dy0

x

∫

= P(0)e−βx

Writing this as exp(−βx) = P(0)/ P(x) and taking the logarithm on both sides

yields

βx log e = log

P(0)P(x)

. Since α = β(10 log e) , this becomes

αx =10 log

P(0)P(x)

For a fiber of length x = L with P(0) = PN being the near-end input power, this

equation reduces to Eq. (13-1).

4

13-10 Consider an isotropically radiating point source in the fiber. The power from this

point source is radiated into a sphere that has a surface area 4πr2. The portion of

this power captured by the fiber in the backward direction at a distance r from the

point source is the ratio of the area A = πa2 to the sphere area 4πr2. If θ is the

acceptance angle of the fiber core, then A = πa2 = π(rθ)2. Therefore S, as defined

in Eq. (13-18), is given by

S =A

4πr2 =πr 2θ2

4πr2 =θ2

4

From Eq. (2-23), the acceptance angle is

sin θ ≈ θ =

NAn

, so that S =θ2

4=

NA( )2

4n2

13-11 The attenuation is found from the slope of the curve, by using Eq. (13-22):

Fiber a:

α =10 log

PD(x1)PD(x2 )

2 x2 − x1( )=

10 log 7028

2 0.5 km( )= 4.0 dB / km

Fiber b:

α =10 log

2511

2 0.5 km( )= 3.6 dB/ km

Fiber c: α =

10 log 7

1.82 0.5 km( )

= 5.9 dB/ km

To find the final splice loss, let P1 and P2 be the input and output power levels,

respectively, at the splice point. Then for

For splice 1: Lsplice =10 log

P2

P1

= 10 log 2528

= −0.5 dB

For splice 2: Lsplice =10 log

711

= −2.0 dB

13-12 See Ref. 42, pp. 450-452 for a detailed and illustrated derivation.

5

Consider the light scattered from an infinitesimal interval dz that is located at L =

Tvgr. Light scattered from this point will return to the OTDR at time t = 2T. Upon

inspection of the pulse of width W being scattered form the point L, it can be

deduced that the back-scattered power seen by the OTDR at time 2T is the

integrated sum of the light scattered from the locations z = L – W/2 to z = L.

Thus, summing up the power from infinitesimal short intervals dz from the whole

pulse and taking the fiber attenuation into account yields

Ps L( ) = Sαs0

W

∫ P0 exp −2α L +z2

dz

= S αs

α P0 e

−2αL 1− e− αW( )

which holds for L ≥ W/2. For distances less than W/2, the lower integral limit gets

replaced by W – 2L.

13-13 For very short pulse widths, we have that αW << 1. Thus the expression in

parenthesis becomes

1α

1 − e−αW( )≈ 1α

1 − 1− αW( )[ ]= W

Thus

Ps L( ) ≈ S αs W P0 e−2αL

13-14 (a) From the given equation, for an 0.5-dB accuracy, the SNR is 4.5 dB.

The total loss of the fiber is (0.33 dB/km)(50 km) = 16.5 dB.

The OTDR dynamic range D is

D = SNR + αL + splice loss

= 4.5 dB +16.5 dB + 0.5 dB = 21.5 dB

6

Here the splice loss is added to the dynamic range because the noise that limits the

achievable accuracy shows up after the event.

(b) For a 0.05-dB accuracy, the OTDR dynamic range must be 26.5 dB.

13-15 To find the fault-location accuracy dL with an OTDR, we differentiate Eq. (13-23):

dL =c

2ndt

where is the accuracy to which the time difference between the original and

reflected pulses must be measured. For dL ≤ 1 m, we need

dt =

2nc

dL ≤2(1.5)

3 × 108 m / s

(0.5 m) = 5 ns

To measure dt to this accuracy, the pulse width must be ≤ 0.5dt (because we are

measuring the time difference between the original and reflected pulse widths).

Thus we need a pulse width of 2.5 ns or less to locate a fiber fault within 0.5 m of

its true position.

Optical Fiber Communication - Solution Manual

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Transcript of Optical Fiber Communication - Solution Manual