Optical Fiber Communication Solution Man
Optical Fiber Communication Solution Man
Optical Fiber Communication Solution Man
(a) amplitude = 8 µm
= [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)
1
a1 sin δ1 + a2 sin δ2 = A sin φ (2)
provided that constant values of A and φ exist which satisfy these equations. To
or
A2 = a 12 + a 22 + 2a1a2 cos (δ1 - δ2)
a 1 sin δ1 + a 2 sin δ2
tan φ =
a 1 cosδ1 + a 2 cosδ2
Ey
= cos (ωt - kz) cos δ - sin (ωt - kz) sin δ (2.5-1)
E0 y
Ex
cos δ = cos (ωt - kz) cos δ
E0 x
to yield
Ey E x
- cos δ = - sin (ωt - kz) sin δ (2.5-2)
E 0 y E 0x
2
E 2
sin2 (ωt - kz) = [1 - cos2 (ωt - kz)] = 1 − x (2.5-3)
E 0x
Squaring both sides of Eq. (2.5-2) and substituting it into Eq. (2.5-3) yields
Ex
2 2
Ey Ex
E
− cos δ = 1 − sin2 δ
0y E 0x E 0x
Ex Ey E E
2 2
+ - 2 x y cos δ = sin2 δ
E 0x E 0y E 0x E 0y
2-8.
Air: n = 1.0
33 ° 33 °
Glass 90 °
cos 33°
∴ n2 = = 1.540
cos 57°
(b) The critical angle is found from
nglass sin φglass = nair sin φair
3
with φair = 90° and nair = 1.0
1 1
∴ φcritical = arcsin = arcsin = 40.5°
n glass 1.540
2-9
Air r
Water
θ
12 cm
2-10.
45 °
or
4
2(n1 − n 2 )
NA ≈ n1 2∆ = n1 = 0.243
n1
0.242
(b) θ0,max = arcsin (NA/n) = arcsin = 14°
1.0
= n1 (2∆ − ∆2 )
1 /2
εω 1 ∂H z
jHφ = j Er - Substituting into Eq. (2-33b) we have
β βr ∂φ
∂E z εω 1 ∂H z
j β Er + = ωµ j E −
β r βr ∂φ
∂r
εω 1 ∂H z
jHr = -j Eφ - Substituting into Eq. (2-33a) we have
β β ∂r
1 ∂E z εω 1 ∂Hz
j β Eφ + = -ωµ −j Eφ −
r ∂φ β β ∂r
5
1 1 ∂Hz
jEr = + jrβH φ Substituting into Eq. (2-33b) we have
εω r ∂φ
β 1 ∂Hz ∂E z
+ jrβH φ + = jωµ Hφ
εω r ∂φ ∂r
1 ∂H z
jEφ = - jβ H r + Substituting into Eq. (2-33a) we have
εω ∂r
1 ∂E z β ∂H z
- jβ H r + = -jωµ Hr
r ∂φ εω ∂r
Solve for Hr to obtain Eq. (2-35c).
j 1 ∂ ∂Hz ∂E ∂ ∂H z εω ∂E z
- β + εωr z − β − = jεωEz
q r ∂r ∂φ
2
∂r ∂φ ∂r r ∂φ
1 ∂ ∂E z ∂H ∂ ∂E µω ∂Hz
− µω r z − β z +
j
- β = -jµωHz
r ∂r ∂φ ∂r ∂φ ∂r r ∂φ
2
q
6
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
J ν (ua) J ν (ua)
C= A and D= B
K ν (wa) K ν (wa)
For ν = 0, the right-hand side must be zero. Also for ν = 0, either Eq. (2-55a) or (2-56a)
holds. Suppose Eq. (2-56a) holds, so that the term in square brackets on the right-hand
side in the above equation is not zero. Then we must have that B = 0, which from Eq. (2-
43) means that 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):
1 jβν
0= B J ν (ua) + Aωε 1uJ' ν (ua)
u a
2
Bν =
ja 1 [k 2 J + k 2 K ] A
1 ν ν
βωµ +
1 1
2
u 2
w
2
7
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.
n 21 − n22 1 n 22
∆ = = 1− 2
2 n1
2
2n1
∆ << 1 implies n1 ≈ n2
Thus using Eq. (2-46), which states that n2k = k2 ≤ β ≤ k1 = n1k, we have
n 22 k 2 = k 22 ≈ n 21 k 2 = k12 ≈ β 2
2-17.
2π 2 a 2 2 2π 2 a 2
M≈
λ 2 (n 1 − n2
2 ) =
λ 2 (NA )2
λ
1/ 2 1/ 2
M 1000 0.85µm
a= = = 30.25µm
2π NA 2 0.2π
Therefore, D = 2a =60.5 µm
2π (30.25µm )
2 2
8
2π (25 µm)
[(1.48)2 − (1.46)2 ] = 46.5
1/ 2
V=
0.82 µm
Similarly, M = 417 at 1320 nm and M = 303 at 1550 nm. From Eq. (2-72)
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%.
2.40(1.32µm)
(n1 − n22 ) =
Vλ 2 −1/ 2
a= = 6.55 µm
2π 2 π[(1.480) − (1.478) ]
2 2 1/ 2
NA 0.077
θ0,max = arcsin = arcsin 1.0 = 4.4°
n
λV (1.30)(75)
a= = = 52 µm
2 πNA 2 π (0.3)
9
2πa
2-23. For small values of ∆ we can write V ≈ n1 2∆
λ
For a = 5 µm we have ∆ ≈ 0.002, so that at 0.82 µm
2 π (5 µm)
V≈ 1.45 2(0.002) = 3.514
0.82 µm
Thus the fiber is no longer single-mode. From Figs. 2-18 and 2-19 we see that the LP01
2-24.
2π λ
2-25. From Eq. (2-77) Lp = =
β n y − nx
1.3 × 10 −6 m
For Lp = 10 cm ny - nx = = 1.3×10-5
10 −1 m
1.3 × 10 −6 m
For Lp = 2 m ny - nx = = 6.5×10-7
2m
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)
α α 2πan1 2
M= a k n1 ∆ =
2 2 2
∆
α+2 α +2 λ
where
n1 − n2
∆= = 0.0135
n1
10
At λ = 820 nm, M = 543 and at λ = 1300 nm, M = 216.
V2 1 2πa 2 2
Mstep ≈
2
=
2 λ (n1 − n 22 )
2πan1
2
1086 at 820 nm
λ ∆=
Mstep =
432 at 1300 nm
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
D d D
2 2 2
St = st or s = S
2 2 d
d 0.125 mm
2 2
(b) S = s = 1.2 m/s = 1.39 cm/min
D 9 mm
11
2-30. Consider the following geometries of the preform and its corresponding fiber:
25 µm
R
4 mm
62.5 µm
3 mm
FIBER
PREFORM
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
or
π(32 − R2 ) π (25)2
=
π(42 − 32 ) π [(62.5)2 − (25)2 ]
12
(b) If R is the deposition rate, then the deposition time t is
M 5.1 gm
t= = = 10.2 min
R 0.5 gm / min
K
2
χ= where Y = π for surface flaws.
Yσ
Thus
(20 N / mm 3 / 2 ) 2
χ= = 2.60×10-4 mm = 0.26 µm
(70 MN / m 2 )2 π
2-33. (a) To find the time to failure, we substitute Eq. (2-82) into Eq. (2-86) and
∫ χ − b / 2 dχ = AYbσb ∫ dt
χi 0
which yields
b [χ f ] = AYbσbt
1 1− b / 2
− χ1−
i
b/ 2
1−
2
or
b [χ i − χ (f 2− b) / 2 ]
2 (2− b) / 2
t=
(b − 2)A(Yσ)
2 Ki 2− b K f 2 −b
t= −
(b − 2)A(Yσ ) Yσ Yσ
b
2− b
2Ki
≈ if K b− 2
<< K b− 2
or K i2 −b >> Kf 2− b
(b − 2)A(Yσ )
b i f
dχ
= AKb = AYbχb/2σb
dt
13
Integrating this from χ i to χ p where
K
2
K
2
χi = and χp =
Yσ i Yσ p
are the initial crack depth and the crack depth after proof testing, respectively, yields
χp tp
∫χ dχ = AYb ∫ σ dt
−b / 2 b
χi 0
or
[χ ]= AYb σ
1
− χi
1− b / 2 1−b / 2 b
b p p tp
1−
2
2 K
2−b
b − 2 Y σ b−
i
2
[
− σ b−
p
2
= AYb σ bp tp ]
or
2−b
2 K
[ ] [ ]
1
b σi − σ p = B σ b−2 − σ b−2 = σ p tp
b− 2 b− 2 b
b − 2 Y AY i p
When a static stress σs is applied after proof testing, the time to failure is found from Eq.
(2-86):
χs ts
∫ χ −b / 2 dχ = AYb σ bs ∫ dt
χp 0
where χ s is the crack depth at the fiber failure point. Integrating (as above) we get Eq. (2-
89):
[
B σ b−2
p − σsb−2 = σ bs ts ]
Adding Eqs. (2-87) and (2-89) yields Eq. (2-90).
14
2-35. (a) Substituting Ns as given by Eq. (2-92) and Np as given by Eq. (2-93) into Eq.
(2-94) yields
b m
[ ]
m
(σ p t p + σsb t s )/ B + σsb−2 b−2
(σ t
b
/ B + σ b−2 )b− 2
F = 1 - exp −
L
−
p p p
L0 σ m0 σ0
m
m
σ bp t p + σ bs t s b− 2
+ σsb −2
m
[
− L σ b t / B + σ b− 2
]
B
−1
b −2
= 1 - exp L 0 σ m0 p p p
σ bp t p / B + σ b− 2
p
m
σ b t σ b B b −2
1+ s s + s
σ b
t
p p σ
p s p σ 2
t
= 1 - exp − LN p − 1
B
1 +
σ 2p t p
m
b−2
σ bs t s 1
≈ 1 - exp − LN p 1 + b − 1
σ p t p 1 + B
σ 2p t p
σ B
b
0.5 (MN / m 2 )2 s
s 2 = (0.3)15 = 6.5×10-14
σ p σs t p [0.3 (350 MN / m )] 10 s 2 2
15
2-36. The failure probability is given by Eq. (2-85). For equal failure probabilities of the
two fiber samples, F1 = F2, or
σ m L σ m L
1 - exp − 1c
1
= 1 - exp − 2c 2
σ0 L0 σ0 L0
σ1c L1 σ 2c L 2
m m
=
σ 0 L 0 σ0 L0
or
σ1c L 2
1/m
=
σ 2c L1
Thus
4.8
m
1000
= = 50
3.9 20
gives
log 50
m= = 18.8
log(4.8/ 3.9)
16
Problem Solutions for Chapter 3
3-1.
P(0) 10
α (dB/ km ) =
10
z
log =
P(z) z
α z
log e p ( )
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
P1550(8 km) = - 10.0 dBm – (0.3 dB/km)(8 km) = - 12.4 dBm = 57.5 µW
P1300(20 km) = - 8.2 dBm – (0.6 dB/km)(20 km) = - 20.2 dBm = 9.55 µW
P1550(20 km) = - 10.0 dBm – (0.3 dB/km)(20 km) = - 16.0 dBm = 25.1 µW
3-5. With λ in Eqs. (3-2b) and (3-3) given in µm, we have the following representative
points for αuv and αIR:
1
λ (µ
µ m) αuv αIR
0.5 20.3 --
0.7 1.44 --
0.9 0.33 --
1.2 0.09 2.2×10-6
1.5 0.04 0.0072
2.0 0.02 23.2
3.0 0.009 7.5×104
8π 3 2
αscat = 4 (n − 1) kBTfβT
2
3λ
4[
(1.46)2 − 1] (1.38×10-16 dyne-cm/K)(1400 K)
8π 3 2
=
3(0.63 µm)
×(6.8×10-12 cm2/dyne)
= 0.883 km-1
8π 3 8 2
αscat = 4 n p kBTfβT = 1.16 km = 5.0 dB/km
-1
3λ
3-10. From Fig. 2-22, we make the estimates given in this table:
νm Pclad/P αν m = α1 + (α
α2 - α1 )Pclad/P 5 + 103Pclad/P
01 0.02 3.0 + 0.02 5 + 20 = 25
11 0.05 3.0 + 0.05 5 + 50 = 55
21 0.10 3.0 + 0.10 5 + 100 = 105
02 0.16 3.0 + 0.16 5 + 160 = 165
31 0.19 3.0 + 0.19 5 + 190 = 195
12 0.31 3.0 + 0.31 5 + 310 = 315
2
3-11. (a) We want to solve Eq. (3-12) for αgi. With α = 2 in Eq. (2-78) and letting
n 2 (0) − n 22
∆= 2
2n (0)
we have
n 2 (0) − n 2(r) r2
α(r) = α1 + (α2 - α1) = α1 + (α2 - α1 )
n (0) − n 2
2 2 2
a
Thus
∞ ∞
∞ ∞
exp (−Kr 2 ) r 3dr
1
∫ 2K2 ∫ e − x x dx 1
1! 1
0
∞ = 0
∞ = K =
exp(−Kr ) r dr
1 0! K
∫ ∫ e −x
2
dx
0
2K 0
(α2 − α1 )
Thus αgi = α1 + 2
Ka
− Ka 2 Ka 2
(b) p(a) = 0.1 P0 = P0 e yields e = 10.
(α2 − α1 )
αgi = α1 + = 0.57α1 + 0.43α2
2.3
3
1/ 2
196.98
n = 1 + 2
(13.4) − (1.24 / λ)
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:
dτ
3-13. (a) From Fig. 3-13, ≈ 80 ps/(nm-km) at 850 nm. Therefore, for the LED we
dλ
have from Eq. (3-20)
σ mat dτ
= σλ = [80 ps/(nm-km)](45 nm) = 3.6 ns/km
L dλ
σ mat
= [80 ps/(nm-km)](2 nm) = 0.16 ns/km
L
dτ mat
(b) From Fig. 3-13, = 22 ps/(nm-km)
dλ
3-14. (a) Using Eqs. (2-48), (2-49), and (2-57), Eq. (3-21) becomes
ua
2
u 2a2 w2
b=1- = 1 - =
V u 2 a 2 + w2 a 2 u 2 + w 2
β 2 − k 2 n 22 β 2 / k 2 − n 22
= =
k n1 − β + β − k n2 n1 − n 2
2 2 2 2 2 2 2 2
(β / k + n 2 )(β / k − n 2 )
(b) Expand b as b=
(n1 + n 2 )(n1 − n 2 )
Since n2< β/k < n1 , let β/k = n1(1 - δ) where 0 < δ < ∆ << 1. Thus,
4
β / k + n 2 n 1 (1 − δ) + n 2 n1
= =1- δ
n1 + n 2 n 1 + n2 n 1 + n2
β / k + n2 δ δ
=1- ≈ 1 since << 1
n1 + n 2 2 −∆ 2 −∆
β / k − n2
Therefore, b ≈ or β = k[bn1∆ + n2]
n1 − n 2
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.
θ ϕ a
s
x n n (1 − ∆ )
sin φ = = 2 = 1 = (1 - ∆)
s n1 n1
n 1 L n 1L 1
T max = s =
c x c 1−∆
Ln 1
For the axial ray the travel time is T min =
c
5
Therefore
Ln 1 1 Ln 1 ∆ Ln 1 ∆
T min - T max = − 1 = ≈
c 1 − ∆ c 1−∆ c
σ mod n1 ∆ π
= 1−
L c V
where the first tern is Equation (3-30). The difference is then given by the factor
π πλ 1 πλ 1
1− =1− ≈ 1 −
2a (n12 − n 22 )
1/ 2
V 2a n 1 2∆
π(1.3) 1
At 1300 nm this factor is 1 − = 1 − 0.127 = 0.873
2(62.5) 1.48 2(0.015)
3 α α+2 1 α +1 1
C1 = 1, C2 = , = 1, = , = ,
2 α+1 3α + 2 3 2α + 1 2
(α + 1)2 1
and =
(5α + 2)(3α + 2) 15
Ln1∆ 12 2 1/ 2 Ln1 ∆
σ int er mod al = 1+ 3∆ + ∆ ≈
2 3c 5 2 3c
6
3-19. For ε = 0 we have that α = 2(1 - ∆). Thus C1 and C2 in Eq. (3-42) become
5
(ignoring small terms such as ∆3, ∆4, ...)
− ∆ −
6 3
α − 2 2 1 5 2 − ∆
3 3
C1 = = = 5 ≈ − ∆ 1 + ∆
α + 2 2 1 − 6 ∆ + 2 1 − 3 ∆ 5 5
5 5
6
6
32 1 − ∆ − 2
9
1− ∆
3α − 2 5 5
C2 = = =
2(α + 2) 2 2 1− 6
2 1 − 3 ∆
5 ∆ + 2 5
9 2
(a) C12 ≈ ∆
25
3 9 6
4∆ − ∆ 1 − ∆ 2 1− ∆ + 1
4C1 C2 (α + 1)∆ 5 5 5
(b) =
2α + 1 1 3 ∆ 2 1− 3 ∆ 4 1 6 ∆ 1
− − +
5 5 5
18∆ 11 18
− 1− ∆ +
25 18 2
= ∆
24
2
∆
25
51 ∆
5 25
9
2
4 2
16∆ 1− ∆ 9 1 − ∆
2
5 5 9 2
= 2 ≈ ∆
9 3 24
96(1 − ∆)(1 − ∆)4 1− ∆
10 5
Therefore,
Ln 1∆ α α + 2 9 2 18 2 9 2
1/ 2 1/ 2
σ int er mod al = ∆ − ∆ + ∆
2c α + 1 3α + 2 25 25 24
1/ 2
6
2(1 − ∆) 6
2(1− ∆) + 2
Ln 1 ∆2 5 5 3 n1∆2 L
= ≈
2c 2(1 − 6 ∆) + 1 6(1− 6 ∆) + 2 10 6 20 3c
5 5
7
3-20. We want to plot Eq. (3-30) as a function of σ λ , 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 σ λ , we have
1 σ λ 2 d 2 n1 0.098σ λ ns / km at 850 nm
σ int ra mod al = −λ = −2
c λ dλ2 1.026 × 10 σ λ ns / km at 1300 nm
3-21. Using the same parameter values as in Prob. 3-18, except with ∆ = 0.001, we have
from Eq. (3-41) σ int er mod al /L = 7 ps/km, and from Eq. (3.45)
σ 1 2
= (σ int er + σ 2int ra ) vs σ λ :
1/ 2
The plot of
L L
L dβ L 1 dn
τg = = 2kn1 + 2k n1 1
2 2
c dk c 2β dk
α
α + 2 m α+ 2 2
2
(n1 k ∆ )
2 2 −1
α+2
-2
α a 2
α+2
2 dn n 2 k 2 d∆
× ∆ 2k n1 1 + 2kn1 + 1
2
dk ∆ dk
α
L kn1
4∆ α + 2 m 1 α +2
n1 k d∆
= N1 − N1 +
c β α + 2 α a n1 k ∆ 2∆ dk
2 2 2
8
α
LN1 kn1 4∆ m α +2 ε
+
c β α + 2 M 4
= 1− 1
dn 1
with N1 = n1 + k and where M is given by Eq. (2-97) and ε is defined in Eq.
dk
(3-36b).
α
dτ L dN 1 α − ε − 2 m α+ 2
λ 1+
c dλ
=
dλ α + 2 M
α
LN1 α − ε − 2 d m α +2
+ ∆
c α + 2 dλ M
where
dn1 α 2 2 2
N1 = n1 - λ and M = a k n1 ∆
dλ α+2
d d 2 n1
n1 − λ 1 = - λ
dN 1 dn
(a) =
dλ dλ
2
dλ dλ
d 2 n1
Thus ignoring the term involving ∆ 2 , the first term in square brackets
dλ
L 2 d 2 n1
becomes - λ 2
c dλ
α α α
d m α+ 2
α
−α dM 1 α +2
+1
α+ 2 d∆ 1
α +2
∆ + ∆
dλ M dλ M
(b) = m
α + 2 dλ M
α 2 d 2 2
(k n1 ∆ )
dM
(c) = a
dλ α + 2 dλ
α d∆ 2 2
k n1 + 2k ∆n1 1 + 2kn1 ∆
2 dn 2 dk
= a2
α+2 dλ dλ dλ
9
d∆ dn
Ignoring and 1 terms yields
dλ dλ
dM 2α 2 2 2 1 2M
= a k n1 ∆ − = - so that
dλ α + 2 λ λ
α α
d m α+ 2 ∆ 2α m α +2
∆
dλ M
= . Therefore
λ α + 2 M
α
dτ L d 2 n1 LN1 α − ε − 2 2α∆ m α +2
λ = - λ2 +
dλ c dλ2 c α + 2 α + 2 M
d 2 n1 2α α
3-24. Let a = λ2 2 ; b = N1C1∆ ; γ=
dλ α+2 α+2
2
σλ 1 dτ g
2 M
σ 2
int ra mod al =L 2
λ M ∑ λ
dλ
m= 0
γ 2
m
2
L σλ 1
2 M
=
c λ M ∑
m= 0
−a + b
M
γ 2
L σλ 1 M m
2 2
c λ M ∫0 −a + M
≈ b dm
γ
2
2
L σλ
2 2γ
1 M
m 2 m
=
c λ M ∫0
a − 2ab
M + b
M
dm
L σλ
2 2
2 2ab b2
= a − +
c λ γ + 1 2γ + 1
L σλ
2 2
2 d 2 n 2
−λ dλ2
1
=
c λ
10
2 d 2 n1 α 2 4α 2
- 2 λ N1C1∆ (N C
+ 1 1 ∆)
dλ 2
α+1 (α + 2)(3α + 2)
1500(0.09) 1310 4
(b) D =
− 1500 =
1 14.1 ps /(nm − km)
4
σ step n1 ∆ 1.49(0.01)
= = 8 = 14.4 ns / km
L 2 3 c 2 3 (3 × 10 )
σ opt n ∆2 1.49(0.01)2
= 1 = = 14.3 ps / km
L 20 3 c 20 3 (3 × 108 )
n1 ∆L (1.49)(0.01)(5 × 103 m)
σ mod = Tmax − Tmin = = = 248 ns
3 × 10 m / s
8
c
n 1∆L 248
σ step = = = 71.7 ns
2 3c 2 3
0.2
(c) BT = = 2.8 Mb / s
σ step
11
3-29. For α = 0.95αopt , we have
12
Problem Solutions for Chapter 4
2πkBT3/2 3/4 Eg
4-1. From Eq. (4-1), ni = 2 2 (memh) exp - 2k T
h B
3/ 2
2π(1.38 × 10−23 J / K)
=2 T
3 /2
[(.068)(.56)(9.11× 10 −31
kg) ]
2 3/ 4
(6.63 × 10 J.s)
−34 2
3/2 8991
= 5.03×1015 T exp - T
Since both impurity and intrinsic atoms generate conduction holes, the total
conduction-hole concentration pP is
pP = NA + ni = NA + nP
2
From Eq. (4-2) we have that nP = ni /pP . Then
2 2 2
pP = NA + nP = NA + ni /pP or pP - NApP - ni = 0
so that
NA
2
pP = 2
4ni
1+ 2 +1
NA
4-3. (a) From Eq. (4-4) we have 1.540 = 1.424 + 1.266x + 0.266x2 or
1
x2 + 4.759x - 0.436 = 0. Solving this quadratic equation yields (taking the plus
sign only)
1
x = 2 [ - 4.759 + (4.759)2 + 4(.436) ] = 0.090
1.240
The emission wavelength is λ = 1.540 = 805 nm.
1.240
λ = 1.620 = 766 nm
0.4184x 0.4184x
y = 0.1894 - 0.0130x ≈ 0.1894 = 2.20x
2
hc λ2
∆E = ∆λ or ∆λ = ∆E
λ2 hc
For the same energy difference ∆E, the spectral width ∆λ is proportional to the
wavelength squared. Thus, for example,
∆λ 1550 1550 2
= = 1.40
∆λ 1310 1310
1
ηint = = 0.783, and from Eq. (4-13) the internal power level is
1 + 25/ 90
hc(35 mA )
Pint = (0.783) = 26 mW
q(1310 nm)
1
P= 2 26 mW = 0.37 mW
3.5(3.5 + 1)
4-7. Plot of Eq. (4-18). Some representative values of P/P0 are given in the table:
f in MHz P/P0
1 0.999
10 0.954
20 0.847
40 0.623
60 0.469
80 0.370
100 0.303
4-8. The 3-dB optical bandwidth is found from Eq. (4-21). It is the frequency f at
which the expression is equal to -3; that is,
1
10 log 1/ 2 = −3
[
1 + (2πfτ )
2
]
3
(
10 0.6 − 1) = 9.5 MHz
1
With a 5-ns lifetime, we find f =
2π (5 ns )
1 1 2
gth = ln 0.32 + 10 cm-1 = 55.6 cm-1
0.05 cm
1 1
gth = 0.05 cm ln 0.9(0.32) + 10 cm-1 = 34.9 cm-1
4-10. Using Eq. (4-4) to find Eg and Eq. (4-3) to find λ, we have for x = 0.03,
1.24 1.24
λ= E = = 1.462 µm
g 1.424 + 1.266(0.3) + 0.266(0.3)2
dP(mW)
ηext = 0.8065 λ(µm) dI(mA)
λ2 (0.80 µm)2
∆λ = 2Ln = = 0.22 nm
2(400 µm)(3.6)
4
.85 − .75 .1
−3 = .22 ×10 = 455 modes
3
.22 × 10
(λ - 850 nm)2
4-13. (a) From Eq. (4-44) we have g(λ) = (50 cm-1) exp -
2(32 nm)2
(λ - 850)2
= (50 cm-1) exp - 2048
= 32.2 cm-1 shows that lasing occurs in the region 820 nm < λ < 880 nm.
λ2 (850)2
∆λ = 2Ln = = 0.25 nm
2(3.6)(400 µm)
880 - 820
N= 0.25 = 240 modes
m
4-14. (a) Let Nm = n/λ = 2L be the wave number (reciprocal wavelength) of mode m.
The difference ∆N between adjacent modes is then
1
∆N = Nm - Nm-1 = 2L (a-1)
dN d n 1 dn n 1 dn
= = - 2 = - 2 n - λ
dλ dλ λ λ dλ λ λ dλ
1 dn
∆N = 2 n - λ ∆λ (a-2)
λ dλ
5
λ2
Equating (a-1) and (a-2) then yields ∆λ =
dn
2Ln - λ
dλ
(.85 µm)2
(b) The mode spacing is ∆λ = = 0.20 nm
2(4.5)(400 µm)
n-1 2 3.6-1 2
R1 = R2 = n+1 = 3.6+1 = 0.32
1 1 1
Then Jth = α + ln = 2.65×103 A/cm2
β 2L R1R2
Therefore
∆E11 = 1.43 eV +
(6.6256 × 10 −34
J ⋅ s)
2
1
+
1
8(5 nm ) 6.19 × 10 kg 5.10 × 10 kg
−32 −31
2
4-17. Plots of the external quantum efficiency and power output of a MQW laser.
6
2
λB 1 (1.57 µm)(1570 nm)
λ = λB ± 2n L 2 = 1570 nm ± = 1570 nm ± 1.20 nm
e 4(3.4)(300 µm)
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 density
changes from 0 to nth.
J
n= n th
J
td n th
1 n
t d = ∫ dt = ∫ dn = −τ − = τ ln
0 0
J n
− qd τ J − J th
n= 0
qd τ
Ip
where J = Ip/A and Jth = Ith/A. Therefore td = τ ln I - I
p th
J − nB
td n th
1 qd τ
td = ⌠
⌡ dt = ∫ J n dn
= τ ln
J − n th
0 nB −
qd τ qd 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
I - IB Ip
Therefore, td = τ ln I - I = τ ln I I - I
th p + B th
7
4-22. Since the dc component of x(t) is 0.2, its range is -2.36 < x(t) < 2.76. The power
has the form P(t) = P0[1 + mx(t)] where we need to find m and P0. The average
value is
< P(t)> = P0[1 + 0.2m] = 1 mW
1
P(t) = P0[1 - 2.36m] ≥ 0 which implies m ≤ 2.36 = 0.42
Therefore for the average value we have < P(t)> = P0[1 + 0.2(0.42)] ≤ 1 mW,
which implies
1
P0 = 1.084 = 0.92 mW so thatP(t) = 0.92[1 + 0.42x(t)] mW and
+ a2(b12 cos2 ω1t + 2b1b2 cos ω1t cos ω2t + b 22 cos2 ω2t)
+ a3(b13 cos3 ω1t + 3b12 b2 cos2 ω1t cos ω2t + 3b1 b 22 cos ω1t cos2 ω2t+ b 32 cos3 ω2t)
+a4(b14 cos4 ω1t + 4b13 b2 cos3 ω1t cos ω2t + 6b12 b 22 cos2 ω1t cos2 ω2t
8
1 1 1
vi) cos2 x cos2 y = 4 [1 + cos 2x+ cos 2y + 2 cos(2x+2y) + 2 cos(2x-2y)]
1
vii) cos3 x cos y = 8 [cos (3x+y) + cos (3x-y) + 3cos (x+y) + 3cos (x-y)]
then
1 2 2 3 4 2 2 3 4 constant
y(t) = 2 a2b1 + a2b2 + 4 a4b1 + 3a4b1b2 + 4 a4b2
terms
fundamental
+ 4a3b1 + 2a3b1b2 cos ω1t + 4 a3b2 + 2b1b2 cos ω2t
3 3 2 3 3 2
terms
2 2
b1 b2
+ 2 a2 + a4b1 + 3a4b2 cos 2ω1t + 2 a2 + a4b2 + 3a4b1 cos 2ω2t
2 2 2 2
1 3 1 3
+ 4 a3b1 cos 3ω1t + 4 a3b2 cos 3ω2t 3rd-order harmonic terms
1 4 1 4
+ 8 a4b1 cos 4ω1t + 8 a4b2 cos 4ω2t 4th-order harmonic terms
3 2 3 2
+ 4 a3b1 b2 [ cos(2ω1+ω2)t + cos(2ω1-ω2)t] + 4 a3b1b2 [ cos(2ω2+ω1)t + cos(2ω2-ω1)t]
3rd-order intermodulation terms
1 3
+ 2 a4b1 b2 [ cos(3ω1+ω2)t + cos(3ω1-ω2)t]
3 2 2
+ 4 a4b1 b2 [ cos(2ω1+2ω2)t + cos(2ω1-2ω2)t]
1 3
+ 2 a4b1b2 [ cos(3ω2+ω1)t + cos(3ω2-ω1)t] 4th-order intermodulation
terms
y(t) = A0 + A1(ω1) cos ω1t + A2(ω1) cos 2ω1t + A3(ω1) cos 3ω1t
9
+ A3(ω2) cos 3ω2t + A4(ω2) cos 4ω2t + ∑ ∑ Bmn cos(mω1+nω2)t
m n
where An(ωj) is the coefficient for the cos(nωj)t term.
-t/τ
4-24. From Eq. (4-58) P = P0 e m where P0 = 1 mW and τm = 2(5×104 hrs) = 105
hrs.
E /k T
4-25. From Eq. (4-60) τs = K e A B or ln τs = ln K + EA/kBT
10
τ = 1.11×10-5 exp{0.63/[(8.625×10-5)(293)]} = 7.45×105 hrs
s
11
Problem Solutions for Chapter 5
5-4. The source radius is less than the fiber radius, so Eq. (5-5) holds:
2
PLED-step = π2rs B0(NA)2 = π2(2×10-3 cm)2(100 W/cm2)(.22)2 = 191 µW
1 2 2
PLED-graded = 2π2(2×10-3 cm)2(100 W/cm2)(1.48)2(.01)1 - 2 5 = 159 µW
5-5. Using Eq. (5-10), we have that the reflectivity at the source-to-gel interface is
3.600 − 1.305
2
R s− g = = 0.219
3.600 + 1.305
1.465 − 1.305
2
−3
R g− f = = 3.34 × 10
1.465 + 1.305
5-6. Substituting B(θ) = B0 cosm θ into Eq. (5-3) for B(θ,φ), we have
1
rm
⌠ ⌠ θ0-max
2π
P = 2π ⌠
⌡ cos3 θ sin θ dθ dθs r dr
⌡ 0
⌡0
0
Using
θ0 θ0 sin θ0
⌠
⌡ cos3 θ sin θ dθ =⌠
⌡ ( 1 - sin2 θ) sin θ d(sin θ) = ⌠⌡ ( x - x3) dx
0 0 0
we have
rm
⌠ 2π
P = 2π
⌠ sin2 θ0-max sin4 θ0-max
- dθs r dr
⌡ 2 4
⌡0
0
rm
⌠ 2π
= π NA2 - 2 NA4 dθs r dr
⌠ 1
⌡
⌡0
0
rm 2π
π 2 4
⌡ r dr ⌠
= 2 [ 2NA - NA ] ⌠ ⌡ dθs
0 0
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.
2
1.485 − 1.305
2
R g− f = 4.16 × 10 −3
1.485 + 1.305 =
1.485 − 1.000
2
R a− f = = 0.038
1.485 + 1.000
1 1
5-9. Shaded area = (circle segment area) - (area of triangle) = 2 sa - 2 cy
d
s = aθ = a [2 arccos (y/a)] = 2a arccos 2a
d21/2
c = 2 a2 - 2
Therefore
d d21/2
Acommon = 2(shaded area) = sa – cy = 2a2 arccos 2a - d a2 - 4
5-10.
Coupling loss (dB) for
µ m)
Given axial misalignments (µ
Core/cladding diameters 1 3 5 10
µ m)
(µ
50/125 0.112 0.385 0.590 1.266
62.5/125 0.089 0.274 0.465 0.985
100/140 0.056 0.169 0.286 0.590
π
5-11. arccos x = 2 - arcsin x
x3 x5
For small values of x, arcsin x = x + 2(3) + 2(4)(5) + ...
3
d d π d
Therefore, for 2a << 1, we have arccos 2a ≈ 2 - 2a
2 π d 5d 8d
Thus Eq. (5-30) becomes PT = P2 - 2a - 6a = P 1 -
π 3πa
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
1 1 2
k2NA2(0) 2 - a 2
MaR α+2 R aR
ηF = M = =
aE 1 1 2 2
k2NA2(0) 2 - a aE
α+2 E
a2
Therefore from Eq. (5-21) the coupling loss for aR ≤ aE is LF = -10 log 2
R
aE
5-14. For fibers with different NAs, where NAR < NAE
2 α 2
k2NAR(0) a
MR 2α+4
LF = -10 log ηF = -10 log M = -10 log
E 2 α 2
k2NAE(0) a
2α+4
4
NA2 (0)
= -10 log 2
R
NAE(0)
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).
(a) The only loss is that from index-profile differences. From Eq. (5-37)
1.80(2.00 + 2)
L 1→ 2 (α) = −10 log = 0.24 dB
2.00(1.80 + 2)
50
L 2→1 (a) = −20 log
62.5 =
1.94 dB
5-18. When there are no losses due to extrinsic factors, Eq. (5-43) reduces to
5
LSM;ff = -10 log
4
W1 + W2
2
W2 W1
4
For W1 = 0.9W2 , we then have LSM;ff = -10 log 4.0446 = - 0.0482 dB
5-19. Plot of Eq. (5-44).
6
Problem Solutions for Chapter 6
To assist in making the plots, from Fig. P6-1, we have the following representative
values of the absorption coefficient:
λ (µ
µ m) αs (cm-1)
.60 4.4×103
.65 2.9×103
.70 2.0×103
.75 1.4×103
.80 0.97×103
.85 630
.90 370
.95 190
1.00 70
w
w ⌠ -αsx
6-2. Ip = qA ⌠
⌡ G(x) dx = qA Φ0 αs ⌡ e dx
0 0
ηq ηqλ
6-3. From Eq. (6-6), R= = hc = 0.8044 ηλ ( in µm)
hν
6-4. (a) Using the fact that Va ≈ VB, rewrite the denominator as
VB - Va + IMRMn
= 1 - 1 -
VB
VB - Va + IMRM
Since VB << 1, we can expand the term in parenthesis:
1
VB - Va + IMRMn n( VB - Va + IMRM)
1 - 1 - ≈ 1 - 1 -
VB VB
n( VB - Va + IMRM) nIMRM
= VB ≈ VB
IM VB VB
Therefore, M0 = I ≈ ≈ nI R
p n( VB - Va + IMRM) M M
IM VB 2 IpVB VB 1/2
(b) M0 = I = nI R implies IM = nR , so that M0 = nI R
p M M M p M
T 2π/ω
2 1⌠ 2 ω ⌠ 2 2
6-5. < > is (t) = T ⌡ is (t)dt = R P (t) dt
2π ⌡ 0
(where T = 2π/ω),
0 0
2π/ω
ω 2 2
= R 0 P0 ⌠
⌡ (1 + 2m cos ωt + m2 cos2 ωt) dt
2π
0
Using
2 π/ω
1
∫
t = 2π / ω
cos ωt dt = sin ωt t=0 =0
0 ω
2π
2π/ω
1 ⌠ 1 1 π
and ⌠
⌡ cos2 ωt dt = ω 2 + 2cos 2x dx =
⌡ ω
0 0
2 2 m2
we have < i2s(t)> = R0 P0 1 + 2
6-6. Same problem as Example 6-6: compare Eqs. (6-13), (6-14), and (6-17).
ηqλ
(a) First from Eq. (6-6), I p = P0 = 0.593 µA
hc
2
(b) σ 2DB = 2qID B = 2(1.6 × 10−19 C)(1.0 nA)(150 × 10 6 Hz) = 4.81× 10 −20 A 2
ηqλ
6-7. Using R 0 = hc = 0.58 A/W, we have from Eqs. (6-4), (6-11b), (6-15), and (6-
17)
1
S (R0 P0 m) M 2
2
R0P0m2
N = 2 = = 6.565×1012 P0
Q 2qI p BM M
1/ 2 2
4qBM1/2
2
S ( R0P0m) 2
N = = 3.798×1022 P0
DB 4qIDBM1/2
2 2
S ( R0 P 0m) M 2
N = = 3.798×1026 P0
DS 4qILB
1 2
( R0P0m) M2
S 2 2
N = 4k TB/R = 7.333×1022 P0
T B L
P0
where P0 is given in watts. To convert P0 = 10-n W to dBm, use 10 log -3 =
10
10(3-n) dBm
1 2 2
S 2 ( R0 P 0m) M
N = 2qB(R0P0 + ID)M5/2 + 2qILB + 4kBTB/RL
1.215 × 10 −16 M 2
=
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.
3
1 2 2
d S d 2 IpM
0 = dM N = dM
+ 2qIL + 4kBT/RL
6-9.
2q(Ip + ID)M2+x
1 2
2 (2+x)M1+x 2q(Ip + ID)2 IpM2
0 = Ip M -
2q(Ip + ID)M2+x + 2qIL + 4kBT/RL
∂ 2 pn 1 -αsw (w-x)/Lp
e 2 -αsx
2 = - 2 pn0 + Be + αs Be
∂x Lp
∂ 2 pn
Substituting pn and into the left side of Eq. (6-23):
∂x2
1 -αsw (w-x)/Lp
e B -αsx -αsx
+ pn0 + Be - e + Φ0 αs e
τp τp
2 1 -αsx
= BDp αs - + Φ0 αs e
τp
2
where the first and third terms cancelled because Lp = Dpτp .
Substituting in for B:
Φ 2 -α x
0 αsLp 2 1
Dp αs - + Φ0 αs e
s
Left side = D 2 2 τ
1 - αs Lp
p
p
4
Φ0 αsLp αs Lp - 1 -α x
2 2
+ Dpαs e
s
=D
p 1-α L2 2 τ
s p p
Φ0 -αsx
= D ( -Dpαs + Dpαs) e =0 Thus Eq. (6-23) is satisfied.
p
∂pn
(b) Jdiff = qDp
∂x x =w
1 -αsw -α w
= qDp L pn0 + Be - αsBe s
p
-αsw
= qDp BL - αs e
1 Dp
+ qpn0 L
p p
αsL2 1 - α L
p s p -αsw Dp
= qΦ0
2 2 Lp e + qp n0 Lp
1 - αs Lp
αsLp -αsw Dp
= qΦ0 e + qpn0 L
1 + αsLp p
-αsw
e -αsw Dp
= qΦ0 1 - e + qpn0 L
1 + αsLp p
-jωtd
J J* 1/2 = qΦ ( S S*) 1/2 where S =
1-e
tot tot 0 jωtd
sc
1/2 1
We want to find the value of ωtd at which ( S S*) = .
2
1/2
Evaluating ( S S*) , we have
5
-jωtd +jωtd1/2
1/2 1 - e 1 - e
( S S*) =
jωtd -jωtd
1/2
+jωtd -jωtd 1/2
1 - e +e + 1 ( 2 - 2 cos ωtd)
= =
ωtd ωtd
1/2 ωtd
[ ( 1 - cos ωtd) /2] sin 2
ωtd
= = = sinc 2
ωtd/2 ωtd
2
1/2 1
We want to find values of ωtd where ( S S*) = .
2
x sinc x x sinc x
0.0 1.000 0.5 0.637
0.1 0.984 0.6 0.505
0.2 0.935 0.7 0.368
0.3 0.858 0.8 0.234
0.4 0.757 0.9 0.109
ωtd
Thus 2 = 0.442 which implies ωtd = 0.884
w 1
(b) From Eq. (6-27) we have td = v = . Then
d αsvd
1
ωtd = 2πf3-dB td = 2πf3-dB = 0.884 or
αsvd
6
w 20 × 10 −6 m
td = v = = 0.45 ns
d 4.4 × 10 4 m / s
1 1
c) = 10-3 cm = 10 µm = 2 w
αs
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.
2 2
k1(1 - k1) k1 - k1 k2 - k1
(1) 1- 1-k =1- 1-k ≈1- 1-k = 1 – keff
2 2 2
2 2
(1 - k1)2 1 - 2k1 + k1 1 - 2k2 + k1
(2) 1 - k2 = 1 - k2 ≈ 1 - k2
2
1 - k2 k2 - k1
=1-k - 1-k = 1 - keff
2 2
1 1
Fe = keffMe + 2(1 - keff) - M (1 - keff) = keffMe + 2 - M (1 - keff)
e e
'
(b) With k1 ≈ k2 and keff defined in Eq. (6-40), we have
2
k2(1 - k1) k2 - k1 '
(1) 2 ≈ 2 = keff
k1(1 - k2) k1(1 - k2)
2
(1 - k1)2k2 k2 - 2k1k2 + k2k1
(2) 2 = 2
k1(1 - k2) k1(1 - k2)
7
' 1 '
Therefore Eq. (6-35) becomes Eq. (6-39): Fh = keff Mh - 2 - M (keff - 1)
h
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)
1
Fe = 2 - M ≈ 2 for large Me
e
8
Problem Solutions for Chapter 7
We want to compare F1 = kM + (1 - k) 2 − and F2 = Mx.
1
7-1.
M
∞
1 e j 2πft 1 -t/RC
hB(t) = ∫
df = e
+ jf
2πC 1 C
−∞
2πRC
αTb/2
1 αTb αTb
∞
1
∫ h p (t) dt = αTb ⌠ dt
⌡ =
αTb 2
+ 2
=1
−∞ -αTb/2
1
Part (b):
∞ ∞
1 1 t2
∫h p (t) dt =
2 π αTb ∫ exp −
2( αTb )
2 dt
−∞ −∞
1 1
= π αTb 2 = 1 (see Appendix B3 for integral solution)
2π αTb
Part (c):
∞ ∞
t
dt = - [e −∞ − e −0 ]= 1
1
∫ h p (t) dt = αTb ∫ exp −
αT
−∞ 0 b
∫ ∫ ∫
− j2 πft
F[p(t)*q(t)] = p(t)* q(t)e dt = q(x) p(t − x) e − j2πft dt dx
−∞ −∞ −∞
∞ ∞
∫ ∫
− j2πfx − j2πf( t − x)
= q(x) e p(t − x) e dt dx
−∞ −∞
∞ ∞
Pe =
1
[P0 (v th ) + P1 (v th )].
2
Substituting Eq. (7-20) and (7-22) for P0 and P1, respectively, we have
1 1 ∞ −v 2 / 2σ 2 V /2
2πσ 2 V∫/ 2 ∫
− (v−V ) 2 / 2σ 2
Pe = e dv + e dv
2 −∞
In the second integral, let q = v-V, so that dv = dq. The second integral then
becomes
2
V / 2 −V −V / 2 2σ 2
∫e ∫
−q 2 / 2σ 2 − x2
dq = 2σ 2
e dx where x = q/ 2σ2
−∞ −∞
Then
1 2σ 2 ∞ − V / 2 2σ 2
∫ ∫
− x2 − x2
Pe = e dx + e dx
2 2πσ 2 V / 2 2σ 2 −∞
∞ −x 2
1
V / 2 2σ 2
∫ e dx − 2 ∫
−x 2
= e dx
2 π −∞ 0
1 V
Pe = 1− erf
2 2σ 2
V V
7-6. (a) V = 1 volt and σ = 0.2 volts, so that = 2.5. From Fig. 7-6 for = 2.5,
2σ 2σ
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
1
1400 errors/second = 7×10-4 seconds/error
V
(b) If V is doubled, then = 5 for which Pe ≈ 3×10-7 errors/bit from Fig. 7-6.
2σ
Thus
1
−7 = 16.7 seconds/error
(2 × 10 bits / sec ond)(3 × 10 errors / bit )
5
and
3
1 V
V /2
1
∫e
− ( v− V) 2
/ 2σ 2
P1(vth) = dv = 1− erf
2πσ2 −∞
2 2σ 2
1 1 1 2
P0(vth) = 2 1 - erf = 2 1 - erf0.8
2(.2) 2
1 1
= 2 [ 1 - erf( 1.768) ] = 2 ( 1 - 0.987) = 0.0065
1 1 1 2
P1(vth) = 2 1 - erf = 2 1 - erf0.96
2(.24) 2
1 1
= 2 [ 1 - erf( 1.473) ] = 2 ( 1 - 0.963) = 0.0185
7-8. From Eq. (7-1), the average number of electron-hole pairs generated in a time t is
[v ]
2
v2N = out − vout
2 2 2
= vout -2 vout + vout
2
= v2out - vout
7-10. (a) Letting φ = fTb and using Eq. (7-40), Eq. (7-30) becomes
4
2 ∞ 2
H p (0) H'out (φ) dφ I 2
∫
2
Bbae = Tb ' =
Hout (0) 0 H p (φ) Tb Tb
2 ∞ 2
H p (0) Hout (f)
Be = ∫ (1 + j2πfRC ) df
Hout (0) 0
H p (f)
∞ 2
(1 + 4π f ) df
1 Hout (f)
∫
2 2 2 2
= 2 RC
Tb 0 H p (f)
2 2
1
∞
Hout (f) (2πRC)2 ∞
Hout (f) 2 I2 (2πRC)2
= 2
Tb ∫ H p (f)
df +
Tb2 ∫ H p (f)
f df = T
b
+
Tb
3 I3
0 0
(b) From Eqs. (7-29), (7-31), (7-32), and (7-34), Eq. (7-28) becomes
< v2N > = < v2s > + < v2R > + < v2I > +< v2E >
4kBT
= 2q <i0 > <m2 > BbaeR2A2 + R BbaeR2A2 + SIBbaeR2A2 + SEBeA2
b
4kBT
= 2q <i0 > M2+x + R + SI BbaeR2A2 + SEBeA2
b
R2A2I2
2+x + B + SI + E
4k T S (2πRCA)2
= Tb 2q < i0 > M + SE I 3
Rb R2 3
Tb
7-11. First let x = (v − boff )/ ( 2 σ off ) with dx = dv / ( 2 σoff ) in the first part of Eq. (7-
49):
∞ ∞
5
− v th +b on
2σ on ∞
V K
7-12. (a) Let x = = For K = 10, x = 3.536. Thus
2 2σ 2 2
2
e-x
Pe = = 2.97×10-7 errors/bit
2 πx
2
e-x 2
-5
(b) Given that Pe = 10 = then e-x =2 π 10-5 x.
2 πx
7-13. Differentiating Eq. (7-54) with respect to M and setting dbon/dM = 0, we have
dbon
dM =0
1/ 2
Q(hν / η) 2+ x η 2+x η
1/ 2
=- + + − γ +
Μ 2
M b I
hν on 2 W M b I
hν on 2 (1 W
1 /2
1 1
(2 + x)M 1+ x
b on I 2 (2 + x)M 1+x
bon I 2 (1− γ)
Q(hν / η) 2 2
+ + 1/ 2
M(hν/ η) 2+ x η
1/ 2
2+x η
M b I + W
hν on 2 M hν bon I 2 (1− γ) + W
η
Letting G = M2+x bonI2 for simplicity, yields
hν
G 1/2 1/2
G 1 (1-γ)
( + W) + [ G(1-γ) + W] = 2 (2 + x) 1/2 + 1/2
(G + W) [ G(1-γ) + W]
1/2 1/2
Multiply by (G + W) [ G(1-γ) + W] and rearrange terms to get
6
Squaring both sides and collecting terms in powers of G, we obtain the quadratic
equation
x2γ x2γ
G2 4 (1-γ) + G 4 W(2-γ) - γW2(1+x) = 0
1
x2 x4 2
- 4 W(2-γ) ± W2(2-γ)2 + x2(1-γ)W2(1+x)
16
G=
x2
2 (1-γ)
1
W(2-γ) 1+x 1-γ 2
= 1 - 1 + 16 2
2(1-γ) x (2-γ)2
where we have chosen the "+" sign. Equation (7-55) results by letting
2+x η
G = Mopt bonI2
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
hν 1/(2+x)
W(2-γ)
Q bon 1 1
η
2 W 2
bon = K + W + 2 (2-γ)K + W
hν W(2-γ) 1/(2+x) 2(1-γ)
η K
2I2(1-γ)
(2-γ)
1 1
(2-γ)K1/(2+x)
2 1 2
× K + 1 + 2 (2-γ)K + 1 ÷
2(1-γ) 2(1-γ)
7
7-15. In Eq. (7-59) we want to evaluate
1+x 1+x 1+x
lim (2 − γ)K 2+x lim (2 − γ)K 2+x lim 1 2+x
=
γ →1 2(1− γ )L γ →1 2(1− γ) γ →1 L
Consider first
lim (2 − γ)
1
lim (2 − γ)K (1− γ ) 2
= −1 + 1 + B
γ →1 2(1− γ) γ →1 2(1− γ ) (2 − γ)
2
where B = 16(1+x)/x2 . Since γ→1, we can expand the square root term in a
binomial series, so that
lim B B 1+x
= =4 =4
γ →1 4(2 − γ ) x2
1+x 1+x
lim (2 − γ )K 2+x 1+x2+x
Thus = 4 2
γ → 1 2(1 − γ ) x
1 1
2
1/(2+ x)
lim (2 − γ)K (2 − γ) 2 1
= ÷ K + 1 + (2 − γ )K + 1
γ →1 2(1− γ) 2(1− γ) 2
1 1 1
(2-γ) 2 x2 + 4x + 42
= 4 2 + 1
1+x 2 x+2
K + 1 = =
2(1-γ) x x2 x
lim
From the expression for K in Eq. (7-55), we have that K = 0, so that
γ →1
8
1
lim 1
(2 − γ )K + 1 = 1
2
Thus
γ →1 2
1 1
lim (2 − γ ) 2 1 2 x+2 2(1 + x)
K + 1 + (2 − γ )K + 1 = x +1=
γ → 1 2(1 − γ ) 2 x
1+x 1+x 1
lim (2 − γ)K 2+x 1+x2+x 1+x2+x 2(1 + x) 2
= 4 2 4 2 =x
γ →1 2(1− γ )L x x x
'
7-16. Using Hp(f) = 1 from Eq. (7-69) for the impulse input and Eq. (7-66) for the
raised cosine output, Eq. (7-41) yields
∞ ∞
1
∫ ∫
' 2 2
I2 = H (φ) dφ =
out H'out (φ) dφ
0
2 −∞
1+β 1-β
1-β 2 - 2
2
1 ⌠ 1 πφ π 2 ⌠ 1 πφ π 2
=2 ⌠
⌡ dφ + 81-sin - dφ + 81-sin - dφ
1-β ⌡ β 2β ⌡ β 2β
- 2 1-β 1+β
2 - 2
πφ π
Letting y = - we have
β 2β
π
2
1 β
I2 = 2 (1 - β) + ⌠ [ 1 - 2sin y + sin2y] dy
4π ⌡
π
-2
1 β π 1 β
= 2 (1 - β) + π - 0 + 2 = 2 1 - 4
4π
9
Use Eq. (7-42) to find I3:
∞ ∞
1
2 2
∫ H (φ) φ dφ = ∫ Hout (φ) φ dφ
' 2 ' 2
I3 = out
0 2 −∞
1+β 1-β
1-β 2 - 2
2
1 ⌠ 1 πφ π 2 ⌠ 1 πφ π 2
=2 ⌠
⌡ φ2 dφ + 81-sin - φ2dφ + 81-sin - φ2dφ
1-β
⌡ β 2β ⌡ β 2β
- 2 1-β 1+β
2 - 2
πφ π
Letting y = -
β 2β
π
2
1 1-β3 β⌠ β2y2 βy 1
I3 = 3 2 + [ 1 - 2sin y + sin2y] 2 + + dy
4π ⌡ π π 4
π
-2
π
2 π
β
2
1 1-β3 ⌠ β2y2 1 2β
=3 2 2 + 4( 1 + sin y) dy - ⌠
2
π⌡
+ y sin y dy
4π
⌡ π
π
π
-2 - 2
where only even terms in "y" are nonzero. Using the relationships
π
2
π
⌠
⌡ sin2y dy = 4 ; ⌠
⌡ y sin ydy = -y cos y + sin y
0
x3 x2 1 x cos 2x
and ⌠
⌡ x2 sin x2 dx = 6 - 4 - 8 sin 2x -
4
we have
10
β 3 1 1 1 1 β 1
= 16 2 - 6 - β2 2 - 8 - 32 + 24
π π
7-17. Substituting Eq. (7-64) and (7-66) into Eq. (7-41), with s2 = 4π2α2 and β = 1, we
have
1
⌠ Hout(φ)2
'
1⌠ s2φ2 π2
I2 = ' dφ = 4e 1-sin πφ - 2 dφ
⌡
⌡ p
H (φ)
0
0
1 1
⌠ s φ 1 + cos πφ
2 2 2 ⌠ s2φ2 4 πφ
= e dφ = e cos 2 dφ
⌡ 2 ⌡
0 0
π
2
πφ 2 ⌠ 16α2x2 4
Letting x = 2 yields I2 = ⌡e cos x dx
π
0
1 π
⌠ s2φ2 4 πφ 2
3 2
2
∫
2 16α 2 x 2
I3 = e cos 2 φ dφ =
4
xe cos x dx
⌡ π
0
0
lim
7-20. Consider first K:
γ →1
11
lim
1
γ 2
lim 1+ x 1−
K= −1+ 1+ 16
x 2 (2 − γ) 2
= -1 + 1 = 0
γ →1 γ →1
lim
Also (1− γ) = 0. Therefore from Eq. (7-58)
γ →1
2+x
lim 2(1− γ)
1/(1+ x) 1 1+x
lim (2-γ)
K + 1 + 1
2
L=
γ →1 γ →1 (2 − γ)K 2(1-γ)
lim 2 − γ lim 2 − γ
K = −1+ 1+ 16 1 + x 1− γ + order(1− γ) 2
γ →1 1− γ γ →1 1 − γ 2 x (2 − γ)
2 2
Therefore
2+x
1
lim 2x
2 1/(1+x)
4(1+x) + 12 + 1
1+x
L = 8(1+x)
γ →1 x2
2+x x
2x2 1/(1+x) x+2 1+x 21+x
= 8(1+x) + 1 = (1+x) x
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
2+x
1 1+x L' 1 5/3 3.166
y(ε) = (1 + ε) = 1.1 .9 2.89 = 1.437
1 - ε L
12
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
1 3/2 3.35
y(ε) = 1.1.9 3.15 = 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
2+x
1 1+x L' 1 5/3 3.166
y(ε) = (1 + ε) = 1.1 .9 2.89 = 1.437
1 - ε L
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
1 3/2 3.35
y(ε) = 1.1.9 3.15 = 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)
13
3.51× 1012
WJFET ≈ + 0.026B
B
3.39 × 1013
and from Eq. (7-92) WBP = + 0.0049B
B
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)
3.51× 1012
W= + 0.026B = 3.51×105 + 2.6×105 = 6.1×105
B
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
7-25. From Eq. (7-96) the difference in the two amplifier designs is given by
1 2kBT
∆W = I2 = 3.52×106 for I2 = 0.543 and γ = 0.9.
Bq2 Rf
From Eq. (7-57), the change in sensitivity is found from
x .5
WHZ + ∆W2(1+x) 1.0 + 3.52 3
10 log W = 10 log 1.0 = 10 log 1.29 = 1.09 dB
HZ
η Q η
D = M2+x I2 and F = M
hν hν
14
b = F[ (Db + W)1/2 + W1/2]
b2
- Db - 2W = 2W1/2 (Db + W)1/2
F2
1
b = 2 [ 2DF2 ± 4F4D2 - 4F4D2 + 16WF2] = DF2 + 2F W
hν x 2 2Q
(where we chose the "+" sign) = M Q I2 + M W1/2
η
1.7× 10 4
bon = 2.286×10-19 39.1M +
0.5
M
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
15
7-27. Using Eq. (E-10) and the relationship
∞ 1 πa
∫ x
2 dx =
2
1+
0
a
1 ∞ (AR) 2 π 1 1
BHZ =
(AR) 2 ∫0 1+ (2πRC) f2 2
df =
2 2πRC
= 4RC
1 ∞ 1 π A A
1 ∫0
BTZ = df = = 4RC
2πRC 2
2 2 2πRC
1+ f
A
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) (Ipm)2M
dM = 4kBTB
2q(Ip+ID)M2+x B + R FT
eq
Solving for M,
2+x 4kBTBFT/Req
Mopt = q(Ip+ID)x
7-29. (a) For computational simplicity, let K = 4kBTBFT/Req; substituting Mopt from
Problem 7-28 into Eq. (7-105) gives
16
2
2+x
2
1 1 K
2M2
S 2 (I p m) opt 2 (Ipm) q(Ip+ID)x
N = 2+x = 2q(I +I )K
p D
2q(Ip+ID)Mopt B + KB
q(Ip+ID)x B + KB
2
xm2Ip Req x/(2+x)
= 2/(2+x) 4kBTFT
2B(2+x) [ q(Ip+ID)x]
2
S xm2Ip Req x/(2+x)
N = 2/(2+x) 2/(2+x) 4kBTFT
2B(2+x) ( qx) Ip
2(1+x) 1/(2+x)
m2 ( xIp)
= 2Bx(2+x) 2( x
q 4kBTFT/Req)
2
S xm2 ( R0Pr) Req x/(2+x)
N = 2B(2+x) 2/(2+x) 4kBTFT
[ q(R0Pr+ID)x]
10 4 Ω / J
1 /3
(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 1.656 × 10 −20
1.530 × 1012 Pr 2
= where Pr is in watts.
(0.5P + 10 )
r
−8 2 / 3
17
Pr
We want to plot 10 log (S/N) versus 10 log 1 mW . Representative values are
18
Problem Solutions for Chapter 8
8-1. SYSTEM 1: From Eq. (8-2) the total optical power loss allowed between the light
source and the photodetector is
8-2. (a) Use Eq. (8-2) to analyze the link power budget. (a) For the pin photodiode,
with 11 joints
which gives L = 4.25 km. The transmission distance cannot be met with these
components.
(b) For the APD
which gives L = 7.0 km. The transmission distance can be met with these
components.
-2πBt10 -2πBt90
1 - e = 0.1 and 1 - e = 0.9
so that
1
-2πBt10 -2πBt90
e = 0.9 and e = 0.1
Then
2πBtr 2πB(t90-t10) .9
e =e = .1 = 9
It follows that
ln 9 0.35
2πBtr = ln 9 or tr = = B
2πB
t2
1 1/2 1 1
exp- 2 = 2 which yields t1/2 = (2 ln 2)1/2 σ
2π σ 2σ 2π σ
(b) From Eq. (8-10), the 3-dB frequency is the point at which
1 (2πf3dB)2 σ2 1
G(ω) = 2 G(0), or exp - =
2 2
(2 ln 2)1/2 2 ln 2 0.44
f3dB = = =t
2πσ π tFWHM FWHM
8-5. From Eq. (8-9), the temporal response of the optical output from the fiber is
1 t2
g(t) = exp- 2
2π σ 2σ
1 τe2 g(0) 1
g(τe) = exp- 2 = e =
2π σ 2σ 2π σe
2
from which we have that τe = 2 σ. Since te is the full width of the pulse at the
1 1 1 1
G(f3dB) = exp - 2(2πf3dB σ)2 = 2
2π 2π
Solving for f3dB:
2 ln 2 2 ln 2 2 2 0.53
f3dB = = t = t
2πσ 2π e e
440(7)0.7 350
2 2 1/2
tsys = (2) + (0.07) (1) (7) + 800 + 90
2 2 2 2
= 4.90 ns
1 1
The data pulse width is Tb = B = 90 Mb/s = 11.1 ns
Thus 0.7Tb = 7.8 ns > tsys, so that the rise time meets the NRZ data requirements.
1
8-7. We want to plot the following 4 curves of L vs B = T :
b
3
tmat = Dmat σλ L = 0.7Tb or
(c) Modal dispersion (one curve for q = 0.5 and one for q = 1)
q 1/q
0.440L 0.7 800 0.7
tmod = 800 = B or L = 0.44 B
With B in Mb/s, L = 1273/B for q =1, and L = (1273/B)2 for q = .5.
1
8-8. We want to plot the following 3 curves of L vs B = T :
b
(b) Modal dispersion (one curve for q = 0.5 and one for q = 1)
q 1/q
0.440L 0.7 800 0.7
tmod = 800 = B or L = 0.44 B
With B in Mb/s, L = 1273/B for q =1, and L = (1273/B)2 for q = .5.
4
8-10. Signal bits
Signal bits 0 0 1 1 0 1 1 1 1 0 0 1
Clock signal
Optical Manchester
8-11. The simplest method is to use an exclusive-OR gate (EXOR), which can be
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
5
8-12.
NRZ data
Freq. A
Freq. B
PSK data
8-13.
Original 010 001 111 111 101 000 000 001 111 110
code
3B4B 0101 0011 1011 0100 1010 0010 1101 0011 1011 1100
encoded
6
(b) With x = 0.7 and k = 0.3, for an 0.5-dB power penalty at
or
B (Mb/s) D [ps/(nm.km)]
140 2
280 1
560 0.5
7
Problem Solutions for Chapter 9
9-1.
RIN limit
58
Quantum
Noise
CNR limit
54
Thermal noise
(dB) limit
50
0 4 8 12 16
Received optical power (dBm)
9-2.
f1 f2 f3 f4 f5
Transmission
system
∆
Triple-
beat
products
2-tone
3rd order
1
9-3. The total optical modulation index is
21/2
m = ∑mi = [30(.03)2 + 30(.04)2 ] = 27.4 %
1 /2
i
120 1/2
9-4
The modulation index is m = ∑(.023)
2 = 0.25
i=1
The received power is
P = P0 – 2(lc) - αfL = 3 dBm - 1 dB - 12 dB = -10 dBm = 100µW
1 1 −6
C = 2 (mR0P) 2 = (15 × 10 A) 2
2
The source noise is, with RIN = -135 dB/Hz = 3.162×10-14 /Hz,
2 4kBT
< >iT = R
eq
Fe = 8.25×10-13A2
2
Thus the carrier-to-noise ratio is
1
(15 × 10 −6 A)2
C 2
N = 5.69 × 10−13 A 2 + 9.5 × 10 −14 A2 + 8.25 × 10 −13 A 2 = 75.6
9-5. When an APD is used, the carrier power and the quantum noise change.
1 1
C = 2 (mR0MP) 2 = (15 × 10−5 A) 2
2
The quantum noise is
1
C (15 × 10 −5 A)2
2
N = 5.69 × 10−13 A 2 + 4.76 × 10−11 A2 + 8.25 × 10 −13 A 2 = 236.3
C
or, in dB, N = 23.7 dB
32 1/2
9-6.
(a) The modulation index is m = ∑(.044)
2 = 0.25
i=1
The received power is P = -10 dBm = 100µW
1 1
C = 2 (mR0P) 2 = (15 × 10−6 A) 2
2
3
The source noise is, with RIN = -135 dB/Hz = 3.162×10-14 /Hz,
4kBT
< i2T> = R
eq
Fe = 8.25×10-13A2
32 1/2
m = ∑(.07)
2 = 0.396
i=1
The received power is P = -13 dBm = 50µW
1 1
C = 2 (mR0P) 2 = (1.19 × 10 −5 A) 2 = 7.06×10-11A2
2
The source noise is, with RIN = -135 dB/Hz = 3.162×10-14 /Hz,
4
The quantum noise is
C 7.06 × 10 −11 A 2
N = 1.42 × 10−13 A 2 + 4.8 × 10 −14 A 2 + 8.25 × 10 −13 A 2 = 69.6
9-8. Using the expression from Prob. 9-7 with ∆ντ = 0.05, fτ = 0.05, and ∆ν = f = 10
MHz, yields
4R1R2 1
= (.1442)
π 20 MHz
Taking the log and letting the result be less than -140 dB/Hz gives
-80.3 dB/Hz + 10 log R1R2 < -140 dB/Hz
5
Problem Solutions for Chapter 10
λ2 (1546 nm )2 9 −1
∆λ = ∆f = 500 × 10 s = 4 nm
3 × 10 m / s
8
c
The number of wavelength channels fitting into the 1536-to-1556 spectral band
then is
200 µW
10 log = 2.7 dB yields P1 = 10
(log 200 −0.27 )
= 107.4 µW
P1
200 −0.47)
Similarly, P2 = 10(log = 67.8 µW
200 = 0.58 dB
(b) From Eq. (10-5): Excess loss = 10 log
107.4 + 67.8
P1 107.4 P2 67.8
(c) = = 61% and = = 39%
P1 + P2 175.2 P1 + P2 175.2
10-3. The following coupling percents are are realized when the pull length is stopped at
the designated points:
Points A B C D E F
1310 nm 25 50 75 90 100 0
1540 nm 50 88 100 90 50 100
10-4. From A out = s 11A in + s 12Bin and Bout = s 21A in + s 22 Bin = 0 , we have
s 21 s s
Bin = − A in and A out = s 11 − 12 21 A in
s 22 s 22
1
Then
2
s
2 2 2
A s s B s s
T = out = s11 − 12 21 and R = in = 21 ÷ s11 − 12 21
A in s 22 Ain s 22 s 22
P2
= sin 2 (0.4z )exp(− 0.06z ) = 0.5
P0
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 β z ∝ 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
P
L Ij = 10 log i −in
Pj −out
Let
Pi −in
aj = = 10 Ij , where the values of LIj are given in Table P10-7.
L /10
P j− out
P Pin 1
10 log in = 10 log = 10 log
0.95 =
0.22 dB
∑ Pj Pin
1 1 1
+ + ... +
a1 a 2 an
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
2
Pout 7πa2 7(25)2
Lcoupling = 10 log P = 10 log = 10 log = -7.11 dB
in πR2 (150)2
7πa2 7π(25)2
Lcoupling = 10 log = 10 log 800(50) = -4.64 dB
l∞w
10-9. (a) The diameter of the circular coupling rod must be 1000 µm, as shown in the
figure below. The coupling loss is
7πa2 7(100)2
Lcoupling = 10 log = 10 log = -5.53 dB
πR2 (500)2
200 µm
400
µm
Coupling rod
diameter
(b) The size of the plate coupler must be 200 µm by 2600 µm.
7π(100)2
The coupling loss is 10 log 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,
3
P0
Excess loss = 10 log = 10 log P0 = 0.1 dB
P1 + P2 2P1
This yields
P0 P0
P1 = ÷ 10 0.01 = 0.977
2 2
−3
n≤ = 9.64
log 0.977
log 2 −1
log 2
4
Pout ,2 = [sin 2 (2κd) ⋅cos 2 ( k∆L / 2)]Pin,1
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.
c c
∆L1 = = 0.4 mm and ∆L 3 = = 0.8 mm
2n eff (2∆ν) 2n eff ∆ν
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
c
∆ L1 = = 0.75 mm
2n eff (4 ∆ν)
c
∆L 2 = = 1.5 mm
2neff (2∆ν)
c
∆L 3 = = 3.0 mm
2n eff (∆ν)
10-14. (a) For a fixed input angle φ, we differentiate both sides of the grating equation to
get
k dθ k
cos θ dθ = dλ or =
n'Λ dλ n'Λ cos θ
kλ
If φ ≈ θ, then the grating equation becomes 2 sin θ = .
n'Λ
5
k dθ
Solving this for and substituting into the equation yields
n'Λ dλ
dθ 2 sin θ 2 tan θ
= =
dλ λ cos θ λ
1/ 2
Sλ 1/2 0.01(1350)
tan θ = = = 0.2548
2∆λ (1+m) 2(26)(1+ 3)
or θ = 14.3°
R = tanh (κL) = 0.93 yields κL = 2.0, so that L = 2.7 mm for κ = 0.75 mm-1.
2
λ uv 244 nm 244
Λ= θ = 2 sin(13.5°) = 2(0.2334 )
nm = 523 nm
2 sin
2
(b) From Eq. (10-47), λ Bragg = 2Λn eff = 2(523 nm) 1.48 = 1547 nm
π δn η π (2.5 × 10 )(0.827)
−4
−1
κ= = = 4.2 cm
λ Bragg 1.547 × 10−4 cm
(1.547 µm )2
[(2.1) + π2 ]
1 /2
(d) From Eq. (10-49), ∆λ = 2
= 3.9 nm
π (1.48) 500 µm
6
(e) From Eq. (10-48), R max = tanh (κL) = tanh (2.1) = (0.97) = 94%
2 2 2
λ0 1.554 µm
∆L = m = 118 = 126.4 µm
nc 1.451
x n s cd n c
∆ν =
L f mλ2 n g
λ2 (1.554 × 10 −6 m) 2
∆λ = ∆ν = 100.5 GHz = 0.81 nm
c 3 × 108 m / s
c 3 × 108 m / s
∆νFSR = = = 1609 GHz
n g ∆L 1.475(126.4 µm)
Then
λ2 (1.554 × 10 −6 m)2
∆λ = ∆νFSR = 1609 GHz = 12.95 nm
3 × 10 m / s
8
c
2(25 µm)
sin θ i ≈ θ i = = 5.33 × 10 −3 radians
9380 µm
and
−3
sin θ o ≈ θo = 21.3 × 10 radians
7
then from Eq. (10-59),
c
∆νFSR ≈
ng [∆L + d(θ i + θ o )]
3 × 10 8 m / s
= = 1601 GHz
1.475[(126.4 × 10−6 m) + (25 × 10 −6 m)( 5.33 + 21.3) × 10−3 ]
λ2 ν (1550 nm )2 (1.25 × 10 9 s −1 )
∆λ signal = = = 1 × 10 −2 nm
c (3 × 10 8 m / s)(109 nm / m )
∆n eff
∆λ tune = λ = (1550 nm )(0.5%) = 7.75 nm
n eff
∆λ tune 7.75 nm
N= = = 77
10 λ signal 10(0.01 nm )
λ Bragg 1550 nm
Λ= = = 242.2 nm
2neff 2(3.2)
∆λ 2.0 nm
∆Λ = = = 0.3 nm
2n eff 2(3.2 )
c λ2 1
∆L = = = 4.0 mm
2n eff ∆ν ∆λ 2neff
4 mm
(b) ∆L eff = ∆n eff L implies that ∆ n eff = = 0.04 = 4%
100 mm
8
10-24. For example, see C. R. Pollock, Fundamentals of Optoelectronics, Irwin, 1995,
Fig. 15.11, p. 439.
v a ∆n va ∆n
fa = νo =
c λ
Thus we have
9
Problem Solutions for Chapter 11
I 100 mA
Rp = =
qwdL (1.6 × 10 C)(5 µm)(0.5 µm)(200 µm)
−19
1.0 × 1024 / m 3
g 0 = 0.3(1× 10 −20 m 2 )(1 ns) 1.25 × 1033 (electrons/ m 3) / s −
1 ns
= 750 m −1 = 7.5 cm −1
1
N ph;sat = = 1.67 × 1015 photons/ cm 3
0.3 (1× 10 −20
m )(2 × 10 m / s )(1 ns)
2 8
Pin λ
N ph = = 1.32 × 1010 photons / cm 3
vg hc (wd )
Then with P(0) = Pin, P(L) = Pout, G = Pout/Pin, and G 0 = exp(g 0 L ) from Eq. (11-
10), we have
GPin P P
ln G 0 = g 0 L = ln G + − in = ln G + (1 − G ) in
Pamp,sat Pamp,sat Pamp,sat
Rearranging terms in the leftmost and rightmost parts then yields Eq. (11-15).
1
11-4. Let G = G0/2 and Pin = Pout / G = 2Pout,sat / G 0 . Then Eq. (11-15) yields
G0 GP
= 1+ 0 amp.sat ln 2
2 2Pout.sat
G 0 ln 2
Pout .sat = P ≈ (ln 2) Pamp .sat = 0.693 Pamp.sat
(G0 − 2) amp.sat
11-5. From Eq. (11-10), at half the amplifier gain we have
1 1
G= G 0 = exp(g 0L ) = exp(gL)
2 2
Taking the logarithm and substituting into the equation given in the problem,
1 g0
g = g0 − ln 2 =
1 + 4(ν3dB − ν0 ) / (∆ν)
2 2
L
L log 2 2
11-6. Since
[ ]
ln G = g(λ )L = g0 exp − (λ − λ 0 ) / 2(∆λ ) = ln G0 exp −(λ − λ 0 ) / 2(∆λ )
2 2
[ 2 2
]
we have
ln G 0 (λ − λ 0 )
2
ln =
ln G 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
2
1/ 2
FWHM = 2 λ − λ 0 = 2 2 ln
ln G0
∆λ
ln G
1/ 2
ln 30
= 2 2 ln ∆λ = 0.50 ∆λ
ln 27
λp 980
PCE ≤ = = 63.4% for 980-nm pumping, and by
λs 1545
λp 1475
PCE ≤ = = 95.5% for 1475-nm pumping
λs 1545
501
G = 10 log
1.6 =
10 log 313 = 25 dB
980 Pp,in
313 ≤ 1 + . With a 1.6-mW input signal, the pump power needed is
1542 Ps, in
312(1542 )
Pp,in ≥ (1.6 mW ) = 785 mW
980
3
σ shot−
2
s = 2q GPs, in B
R
= 2.34 × 10−14 A2
hc
σ shot−
2
ASE = 2qR S ASE ∆ν opt B = 2qR n spG∆ν opt B
λ
= 2.26 × 10−14 A2
× (.73 A/ W )
(6.626 × 10 −34 J / K)(3 × 108 m / s )
2(100)(1 GHz)
1550nm
= 5.47 × 10 −12 A 2
= 7.01× 10 −13 A 2
4
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
L L
1 P
P path =
L ∫ P(z) dz = in
L ∫ e − αz dz
0 0
P G − 1
1 − e −αL ] = in 1 − = in
[
Pin P 1
=
αL αL G G ln G
(1 − e )
NPASE NPASE
= ∫ e − αz dz = −αL
PASE path
L 0 αL
α (NL) αL tot
PASE 1− =
1 1
= 2 hνn sp (G − 1)∆νopt 1 −
(αL) 2 G (ln G) G
1 G − 1 2
= αL tot hνnsp ∆ν opt
G ln G
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
5
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 2 πνi t + φ i = θ i for simplicity. Then write the cosine term as
e jθ i + e − jθ i
cos θi = , so that
2
− jθ − jθ
N e i +e i
jθ
N e k +e k
jθ
P = E i (t)E *i (t) = ∑ 2Pi
∑
× 2P
i =1 2 k =1
k
2
1 N N
= ∑ ∑
4 i =1 k=1
[
2Pi 2Pk e jθ i e − jθ k + e jθ k e − jθi + e jθ i e jθ k + e − jθ i e − jθ k ]
1 N N
4 i =1 k=1
[
= ∑ ∑ 2Pi 2Pk e i k + e i k + e i k + e i k
j (θ −θ ) − j( θ − θ ) j( θ + θ ) − j( θ +θ )
]
N N N
[ ]
1
= ∑ Pi + ∑∑
j(θ i −θ k ) − j(θ i − θ k )
2Pi 2Pk e +e
i=1 2 i =1 k≠ i
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 + ∑ ∑ 2 Pi Pk [cos(θ i − θ k )]
N N N
i=1 i=1 k ≠i
11-14. (a) For N input signals, the output signal level is given by
N
Ps,out = G∑ Ps,in (i) ≤ 1 mW .
i =1
The inputs are 1 µW (-30 dBm) each and the gain is 26 dB (a factor of 400).
6
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
AP
Q=
(HP + D )
2 1 /2
+D
A = 2R G
Q (HP + D )
2 1/ 2
= AP − QD
2
2QD Q 2 H
Squaring both sides and solving for P yields P = + 2
A A
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
7
1 + 2ηn sp (G − 1)
F=
ηG
8
Problem Solutions for Chapter 12
Pin (mW )
dBm is found from the relationship Pin(dBm) = 10 log
1 mW
7 10.4 -49.8
(b) Using the values in the above table, the operating margin for 8 stations is
(c) To have a 6-dB power margin, we can transmit over at most seven stations.
Thus
1
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.
Wire
room
A B C D
For any row we need seven wires running from the end of the row of offices to
= 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
2
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.
d
N
(b) The ring is similar to the bus, except that we need to close the loop with one
(c) In this problem we consider the case where we need individual cables run
L = cables run along the M vertical rows + cables run along the N horizontal
rows:
3
N −1 M −1
N(N − 1) M(M − 1) MN
= Md ∑ i + Nd ∑j= M
2
d+N
2
d =
2
(M + N − 2)d
i =1 j=1
12-5. (a) Let the star be located at the relative position (m,n). Then
(b) When the star coupler is located in one corner of the grid, then
MN (M N 2) NM MN MN
L= + + − − d= (M + N − 2)d
2 2
(c) To find the shortest distance, we differentiate the expression for L given in (a)
dL M +1
= N(m - 1 - M) + Nm = 0 so that m=
dm 2
Similarly
dL N +1
= M(n - 1 - N) + nM = 0 yields n=
dn 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
wavelengths.
4
For a bus network, these equations can easily be verified by drawing sample
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.
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 − 2L c
# of λs P = 10
P1 / 10
P1(dBm) (mW) Ptotal(mW) Ptotal(dBm)
1 17 50 50 17
2 14 25 50 17
4 11 12.6 50.4 17
5 10 10 50 17
5
6 9.2 8.3 49.9 17
12-11. The following wavelengths can be added and dropped at the three other nodes:
2(3)
3
C= = 8.27
2.17
12-13. See Hluchyj and Karol, Ref. 25, Fig. 6, p. 1391 (Journal of Lightwave
In general, for a (p,k) ShuffleNet, the following spanning tree for assigning fixed
1 p
6
2 p2
k–1 pk-1
k Pk - 1
k+1 Pk - p
k+2 Pk - p2
2k – 1 Pk - pk-1
(c) Wavelength 2 for the partial path B-2-5 and Wavelength 1 for path 5-6-F
(e) Wavelength 2 for the partial path A-1-4 and Wavelength 1 for path 4-7-G
(a) The following nine 3rd-order waves are generated due to FWM:
7
ν223 = 2ν2 – (ν2 + ∆ν) = ν2 - ∆ν = ν1
8
12-19. Plot: from Figure 2 of Y. Jaouën, J-M. P. Delavaux, and D. Barbier, “Repeaterless
1.7627 A eff λ3 D
2
Ppeak = 2 = 11.0 mW
2π n 2 c Ts
Ldisp = 43 km
π
L period = L = 67.5 km
2 disp
(d) From Eq. (12-50) the peak power for 30-ps pulses is
9
1.7627
2
A eff λ3 D
Ppeak = = 3.1 mW
2π n 2 c Ts2
1
(b) From the given condition, L amp ≤ L = 40 km
2 coll
12-24. From the equation and conditions given in Prob. 12-23, we have that
Ts 20 ps
∆λ max = = = 2 nm
DL amp [0.4 ps /(nm ⋅ km)](25 km)
10
Problem Solutions for Chapter 13
(1.000273 − 1)(0.00138823)640
n (T, P) = 1 + = 1.000243
1+ 0.003671(0)
13-2 Since the output voltage from the photodetector is proportional to the optical
power, we can write Eq. (13-1) as
10 V
α= log 2
L1 − L 2 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
10 3.78
α= log = 0.31 dB / km
1895 − 2 3.31
10 P 10 V 10 log e V
α= log N = log N = ln N
LN − LF PF L N − L F VF L N − LF VF
8.868 × 10 −3
L = LN − LF ≥ km = 176 m
0.05
1
13-4 (a) From Eq. (8-11) we have
t2
1 1/2 1 1
exp- 2 = 2 which yields t1/2 = (2 ln 2)1/2 σ
2π σ 2σ 2π σ
(b) From Eq. (8-10), the 3-dB frequency is the point at which
1 (2πf3dB)2 σ2 1
G(ω) = 2 G(0), or exp - =
2 2
(2 ln 2)1/2 2 ln 2 0.44
f3dB = = =t
2πσ π tFWHM FWHM
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,
The following table gives some representative values of H(f) for different values of
2σ:
f (MHz) σ = 2 ns
2σ σ = 1 ns
2σ σ = 0.5 ns
2σ
100 0.821 0.952 0.988
200 0.454 0.821 0.952
300 0.169 0.641 0.895
500 0.0072 0.291 0.735
700 0.089 0.546
1000 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
1 ∆τ 1 400 ps
D= = = 2 ps /(nm ⋅ km)
L ∆λ 10 km 20 nm
2
Then, using this value of D at 1575 nm and with λ0 = 1548 nm, we have
λ − λ 0 (1575 − 1548) nm
13-7 With k = 1, λstart = 1525 nm, and λstop = 1575 nm, we have Ne = 17 extrema.
PISI ≈ 26
(1 ps)2 0.5(1− 0.5) 6.5 10−4
= × dB
(100 ps)
2
x
P(x) = P(0)exp −β ∫ dy = P(0)e
−βx
0
Writing this as exp(−βx) = P(0)/ P(x) and taking the logarithm on both sides
yields
P(0)
βx log e = log . Since α = β(10 log e), this becomes
P(x)
P(0)
αx = 10 log
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).
3
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
A πr 2 θ2 θ 2
S= 2 = 2 =
4πr 4πr 4
NA θ2 ( NA) 2
sin θ ≈ θ = , so that S= =
n 4 4n2
13-11 The attenuation is found from the slope of the curve, by using Eq. (13-22):
PD (x1 ) 70
10 log
PD (x 2 ) 10 log 28
Fiber a: α = = = 4.0 dB / km
2(x 2 − x1 ) 2(0.5 km )
25
10 log
Fiber b: α = 11 = 3.6 dB/ km
2(0.5 km )
7
10 log
Fiber c: α = 1.8 = 5.9 dB/ km
2(0.5 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
P2 25
For splice 1: L splice = 10 log = 10 log = −0.5 dB
P1 28
7
For splice 2: L splice = 10 log = −2.0 dB
11
13-12 See Ref. 42, pp. 450-452 for a detailed and illustrated derivation.
4
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
z
W
Ps (L ) = ∫ Sαs P0 exp −2α L + dz
2
0
αs
=S P0 e −2αL (1− e − αW )
α
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
(1 − e )≈ [1 − (1 − αW )]= W
1 −αW 1
α α
Thus
Ps (L ) ≈ S α s 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.
5
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):
c
dL = dt
2n
where is the accuracy to which the time difference between the original and
reflected pulses must be measured. For dL ≤ 1 m, we need
2n 2(1.5)
dt = dL ≤ (0.5 m) = 5 ns
3 × 10 m / s
8
c
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.