Problem Solutions for Chapter 2 2-1. 2-2.
E = 100cos (2 π108 t + 30°) e x + 20 cos (2π10 8t − 50°) e y + 40cos (2π10 8 t + 210°) e z 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
(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 verify this, first square both sides and add: A2 (sin2 φ + cos2 φ) = a 1 (sin δ1 + cos δ1 ) 2
2
2
+ a 22 (sin 2 δ 2 + cos2 δ 2 ) + 2a1a2 (sin δ1 sin δ2 + cos δ1 cos δ2) or A2 = a 12 + a 22 + 2a1a2 cos (δ1 - δ2) Dividing (2) by (1) gives
tan φ =
a 1 sin δ1 + a 2 sin δ2 a 1 cosδ1 + a 2 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 = cos (ωt - kz) cos δ - sin (ωt - kz) sin δ E0 y
(2.5-1)
Subtract from this the expression Ex cos δ = cos (ωt - kz) cos δ E0 x to yield Ey E x cos δ = - sin (ωt - kz) sin δ E 0 y E 0x
(2.5-2)
Using the relation cos2 α + sin2 α = 1, we use Eq. (2-2) to write
2
E 2 sin2 (ωt - kz) = [1 - cos2 (ωt - kz)] = 1 − x E 0x
(2.5-3)
Squaring both sides of Eq. (2.5-2) and substituting it into Eq. (2.5-3) yields 2 2 Ey Ex Ex 2 1 E − E cos δ = − E 0x sin δ 0y 0x
Expanding the left-hand side and rearranging terms yields 2
2 Ex Ey E E + - 2 x y cos δ = sin2 δ E 0x E 0y E 0x E 0y
2-6.
Plot of Eq. (2-7).
2-7.
Linearly polarized wave.
2-8. Air: n = 1.0
33 °
33 ° 90 °
Glass
(a) Apply Snell's law n1 cos θ1 = n2 cos θ2 where n1 = 1, θ1 = 33°, and θ2 = 90° - 33° = 57° ∴ n2 =
cos 33° = 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 ∴ φcritical = arcsin
1 n glass
= arcsin
1 = 40.5° 1.540
2-9 Air
r
Water
θ 12 cm
Find θc from Snell's law
n1 sin θ1 = n2 sin θc = 1
When n2 = 1.33, then θc = 48.75° r , which yields r = 13.7 cm. Find r from tan θc = 12 cm
2-10.
45 °
Using Snell's law
nglass sin θc = nalcohol sin 90°
where θc = 45° we have 1.45 nglass = = 2.05 sin 45° 2-11. (a) Use either NA = (n12 − n22 ) = 0.242 1/ 2
or
4
NA ≈ n1 2∆ = n1
2(n1 − n 2 ) = 0.243 n1
0.242 (b) θ0,max = arcsin (NA/n) = arcsin = 14° 1.0 2-13. NA = (n12 − n22 ) = [n12 − n12 (1− ∆)2 ] 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
εω 1 ∂H z Er β βr ∂φ ∂E z = ωµ ∂r
j β Er +
Substituting into Eq. (2-33b) we have
εω 1 ∂H z j E − β r βr ∂φ
Solve for Er and let q2 = ω2εµ - β2 to obtain Eq. (2-35a). (b) jHr = -j
Solve Eq. (2-34b) for jHr: εω 1 ∂H z Eφ β β ∂r
j β Eφ +
1 ∂E z = -ωµ r ∂φ
Substituting into Eq. (2-33a) we have
1 ∂Hz εω − j E − φ β β ∂r
Solve for Eφ and let q2 = ω2εµ - β2 to obtain Eq. (2-35b).
(c)
Solve Eq. (2-34a) for jEr: 5
jEr =
β εω
1 1 ∂Hz + jrβH φ εω r ∂φ
Substituting into Eq. (2-33b) we have
1 ∂Hz ∂E z + jrβH φ + = jωµ Hφ r ∂φ ∂r
Solve for Hφ and let q2 = ω2εµ - β2 to obtain Eq. (2-35d). (d)
Solve Eq. (2-34b) for jEφ jEφ = -
∂H z 1 jβ H r + εω ∂r
Substituting into Eq. (2-33a) we have
1 ∂E z β ∂H z = -jωµ Hr jβ H r + r ∂φ εω ∂r Solve for Hr to obtain Eq. (2-35c).
(e)
Substitute Eqs. (2-35c) and (2-35d) into Eq. (2-34c)
-
j 1 ∂ ∂Hz ∂E z ∂ ∂H z εω ∂E z = jεωEz − β + εωr β − 2 r ∂φ q r ∂r ∂φ ∂r ∂φ ∂r
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 2 q
∂H z ∂ ∂E z µω ∂Hz 1 ∂ ∂E z = -jµωHz β − µω − β + r ∂r ∂φ ∂r r ∂r ∂φ r ∂φ
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
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
C=
J ν (ua) A and K ν (wa)
D=
J ν (ua) B K ν (wa)
Substitute these into Eq. (2-50) to find B in terms of A:
A
J' ν (ua) K' ν (wa) 1 jβν 1 + = Bωµ uJ ν (ua) + wK (wa) a u2 w 2 ν
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. (243) 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 jβν B J ν (ua) + Aωε 1uJ' ν (ua) 2 u a
+
1 2 w
jβν K' (wa)J ν (ua) B J ν (ua) + Aωε 2 w ν a K ν (wa)
With k21 = ω2µε1 and k22 = ω2µε2 rewrite this as ja 1 [k 2 J + k 2 K ] A Bν = 1 ν 2 ν 1 1 βωµ + 2 2 u w
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. 2-16. From Eq. (2-23) we have n 21 − n22 1 n 22 ∆ = = 1− 2 2 2n1 2 n1 ∆ << 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.
2-18. (a) From Eqs. (2-59) and (2-61) we have
M≈
2π 2 a 2 2 2π 2 a 2 2 n n − (NA )2 = ( ) 2 2 1 2 λ λ
M a= 2π
1/ 2
λ 1000 = NA 2
1/ 2
0.85µm = 30.25µm 0.2π
Therefore, D = 2a =60.5 µm 2π (30.25µm ) (0.2)2 = 414 2 (1.32µm ) 2
(b) M =
2
(c) At 1550 nm, M = 300 2-19. From Eq. (2-58),
8
V=
1/ 2 2π (25 µm) (1.48)2 − (1.46)2 ] = 46.5 [ 0.82 µm
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) 4 -1/2 4 × 100% Pclad = = 4.1% P total ≈ 3 M 3 1080 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=
−1/ 2 Vλ 2 2.40(1.32µm) n1 − n22 ) = = 6.55 µm ( 2 2 1/ 2 2π 2 π[(1.480) − (1.478) ]
(b) From Eq. (2-23) NA = (n1 − n2 ) = [(1.480) − (1.478) 2
2 1/ 2
2
]
2 1/ 2
= 0.077
(c) From Eq. (2-23), NA = n sin θ0,max. When n = 1.0 then NA 0.077 = arcsin θ0,max = arcsin 1.0 = 4.4° n
2-22. n2 =
2
2
n1 − NA =
a=
2
2
(1.458) − (0.3) = 1.427
(1.30)(75) λV = = 52 µm 2 πNA 2 π (0.3)
9
2-23. For small values of ∆ we can write V ≈
2 πa n1 λ
2∆
For a = 5 µm we have ∆ ≈ 0.002, so that at 0.82 µm V≈
2 π (5 µm) 1.45 0.82 µm
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 =
2π λ = β n y − nx
For Lp = 10 cm
ny - nx =
1.3 × 10 −6 m = 1.3×10-5 10 −1 m
For Lp = 2 m
ny - nx =
1.3 × 10 −6 m = 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) n(r) = n1 [1− 2∆(r / a)α ] = 1.48 [1− 0.02(r / 25)α ] 1 /2
n2 is found from Eq. (2-79):
1 /2
n2 = n1(1 - ∆) = 1.465
2-27. From Eq. (2-81) α α 2πan1 2 2 2 2 M= a k n1 ∆ = ∆ α +2 λ α+2 where ∆=
n1 − n2 = 0.0135 n1
10
At λ = 820 nm, M = 543 and at λ = 1300 nm, M = 216. For a step index fiber we can use Eq. (2-61) V2 1 2πa 2 2 n1 − n 22 ) Mstep ≈ = ( 2 2 λ At λ = 820 nm, Mstep = 1078 and at λ = 1300 nm, Mstep = 429. Alternatively, we can let α = ∞ in Eq. (2-81): 2
2πan1 ∆= Mstep = λ
1086 at 820 nm 432 at 1300 nm
2-28. Using Eq. (2-23) we have (a) NA = (n12 − n22 ) = [(1.60)2 − (1.49)2 ] = 0.58 1/ 2
1/ 2
(b) NA = [(1.458)2 − (1.405)2 ] = 0.39 1/ 2
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 Fiber volume = Lfiber (d/2)2 = st (d/2)2
and Equating these yields
2
D d St = st 2 2 (b)
2
D s = S d
or
2 2 d 0.125 mm S = s = 1.2 m/s = 1.39 cm/min D 9 mm
11
2
2-30. Consider the following geometries of the preform and its corresponding fiber:
25 µm R
4 mm 62.5
3 mm
µm
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 core-to-cladding cross-sectional areas: A preform core A fiber core = Apreform clad A fiber clad or
π(32 − R2 ) π (25)2 = π(42 − 32 ) π [(62.5)2 − (25)2 ]
from which we have 7(25)2 R = 9 − 2 2 (62.5) − (25)
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
12
(b) If R is the deposition rate, then the deposition time t is t=
M 5.1 gm = = 10.2 min R 0.5 gm / min
2-32. Solving Eq. (2-82) for χ yields 2 K χ= Yσ
where Y =
π
for surface flaws.
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 integrate (assuming that σ is independent of time): χf
∫
χi
t
χ − b / 2 dχ = AYbσb ∫ dt 0
which yields 1
b [χ f 1− 2 or t=
1− b / 2
b/ 2 − χ1− ] = AYbσbt i
2 (2− b) / 2 − χ (f 2− b) / 2 ] b [χ i (b − 2)A(Yσ)
(b) Rewriting the above expression in terms of K instead of χ yields Ki 2− b K f 2 −b 2 t= − b (b − 2)A(Yσ ) Yσ Yσ 2− b
2Ki ≈ b (b − 2)A(Yσ )
2 2 if K b− << K b− or i f
2-34. Substituting Eq. (2-82) into Eq. (2-86) gives dχ = AKb = AYbχb/2σb dt
13
K i2 −b >> Kf 2− b
Integrating this from χ i to χ p where K χi = Yσ i
2
and
K χp = Yσ p
2
are the initial crack depth and the crack depth after proof testing, respectively, yields tp
χp
∫χ
−b / 2
χi
dχ = AYb ∫ σ dt b
0
or 1
[
1− b / 2
b χp 1− 2
1−b / 2
− χi
]= AYb σ
b p
tp
for a constant stress σp. Substituting for χ i and χ p gives 2−b 2 K 2 2 σ b− − σ b− = AYb σ bp tp i p b − 2 Y
]
[
or 2 K b − 2 Y
2−b
1 b− 2 b− 2 b − σ p = B σ b−2 − σ b−2 = σ p tp b σi i p AY
[
] [
]
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): χs
∫
χ −b / 2 dχ = AYb σ bs
χp
ts
∫
dt
0
where χ s is the crack depth at the fiber failure point. Integrating (as above) we get Eq. (289):
[
]
B σ b−2 − σsb−2 = σ bs ts p 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 L F = 1 - exp − L0
b (σ p t p + σsb t s )/ B + σsb−2 σ m0
[
− L b 2 t / B + σ b− m σp p p = 1 - exp L 0 σ 0
[
m b−2
]
(σ t −
b p p
m b− 2 / B + σ b−2 ) p m σ0
m b− 2 σ bp t p + σ bs t s b −2 + σs m B b −2 −1 2 σ bp t p / B + σ b− p
]
m σ b t σ b B b −2 1+ s s + s σb 2 σ σ t t p s p p p = 1 - exp − LN p − 1 B 1+ 2 σpt p
m b−2 σ bs t s 1 − 1 ≈ 1 - exp − LN p 1 + b σ p t p 1 + B σ 2p t p
(b) For the term given by Eq. (2-96) we have b
σ B s 2 = (0.3)15 σ p σs t p
0.5 (MN / m 2 )2 s
[0.3 (350 MN / m )] 10 s 2
2
Thus this term can be neglected.
15
= 6.5×10-14
2-36. The failure probability is given by Eq. (2-85). For equal failure probabilities of the two fiber samples, F1 = F2, or σ m L σ m L 1c 1 2 1 - exp − = 1 - exp − 2c σ0 L0 σ0 L0 which implies that m
m
σ1c L1 σ 2c L 2 = σ 0 L 0 σ0 L0 or σ1c L 2 = σ 2c L1
1/m
If L1 = 20 m, then σ1c= 4.8 GN/m2 If L2 = 1 km, then σ2c= 3.9 GN/m2 Thus m 1000 4.8 = = 50 3.9 20
gives m=
log 50 = 18.8 log(4.8/ 3.9)
16
Problem Solutions for Chapter 3 3-1. α (dB/ km ) =
10 P(0) 10 α z log = log e p z P(z) z
( )
= 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: P1300(20 km) = - 8.2 dBm – (0.6 dB/km)(20 km) = - 20.2 dBm = 9.55 µW P1550(20 km) = - 10.0 dBm – (0.3 dB/km)(20 km) = - 16.0 dBm = 25.1 µW
3-3.
From Eq. (3-1c) with Pout = 0.45 Pin α = (10/3.5 km) log (1/0.45) = 1.0 dB/km
3-4.
3-5.
(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
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) 0.5 0.7 0.9 1.2 1.5 2.0 3.0
3-6.
αuv 20.3 1.44 0.33 0.09 0.04 0.02 0.009
αIR ---2.2×10-6 0.0072 23.2 7.5×104
From Eq. (3-4a) we have αscat =
=
8π 3 2 2 4 (n − 1) kBTfβT 3λ
2 8π 3 (1.46)2 − 1] (1.38×10-16 dyne-cm/K)(1400 K) 4[ 3(0.63 µm)
×(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 =
8π 3 8 2 -1 4 n p kBTfβT = 1.16 km = 5.0 dB/km 3λ
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: νm 01 11 21 02 31 12
Pclad/P 0.02 0.05 0.10 0.16 0.19 0.31
αν m = α1 + (α α2 - α1 )Pclad/P 3.0 + 0.02 3.0 + 0.05 3.0 + 0.10 3.0 + 0.16 3.0 + 0.19 3.0 + 0.31
2
5 + 103Pclad/P 5 + 20 = 25 5 + 50 = 55 5 + 100 = 105 5 + 160 = 165 5 + 190 = 195 5 + 310 = 315
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 α(r) = α1 + (α2 - α1)
n 2 (0) − n 2(r) r2 = α + (α α ) 1 2 1 2 2 2 n (0) − n 2 a
Thus ∞
αgi =
∫ 0
∞
α(r) p(r) r dr = α1 +
∞
∫
p(r) r dr
(α2 − α1 ) a2
0
∫
exp(− Kr 2 ) r 3 dr
∫
exp(−Kr 2 ) r dr
0 ∞
0
To evaluate the integrals, let x = Kr2, so that dx = 2Krdr. Then ∞
∫
0 ∞
∫ 0
1 2K2 = 1 2 exp(−Kr ) r dr 2K
exp (−Kr 2 ) r 3dr
Thus αgi = α1 +
∞
∫
e − x x dx
∫
e −x
0 ∞
1 1! 1 = K = K 0! dx
0
(α2 − α1 ) 2 Ka
(b) p(a) = 0.1 P0 = P0 e
− Ka 2
yields e
Ka 2
= 10.
From this we have Ka2 = ln 10 = 2.3. Thus αgi = α1 +
(α2 − α1 ) = 0.57α1 + 0.43α2 2.3
3-12. With λ in units of micrometers, we have
3
196.98 n = 1 + 2 2 (13.4) − (1.24 / λ)
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 λ 0.2 µm 0.6 µm 1.0 µm
Calculated n 1.548 1.457 1.451
n from Fig. 3-12 1.550 1.458 1.450
dτ ≈ 80 ps/(nm-km) at 850 nm. Therefore, for the LED we dλ have from Eq. (3-20)
3-13. (a) From Fig. 3-13,
σ mat dτ = σλ = [80 ps/(nm-km)](45 nm) = 3.6 ns/km L dλ For a laser diode, σ mat = [80 ps/(nm-km)](2 nm) = 0.16 ns/km L (b) From Fig. 3-13,
dτ mat = 22 ps/(nm-km) dλ
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 2 u 2a2 w2 ua = b = 1 - = 1 - 2 2 u a + w2 a 2 u 2 + w 2 V
=
β 2 − k 2 n 22 β 2 / k 2 − n 22 = 2 2 2 2 2 2 2 2 k n1 − β + β − k n2 n1 − n 2
(b) Expand b as
b=
(β / k + n 2 )(β / k − n 2 ) (n1 + n 2 )(n1 − n 2 )
Since n2< β/k < n1 , let β/k = n1(1 - δ) where 0 < δ < ∆ << 1. Thus,
4
n1 β / k + n 2 n 1 (1 − δ) + n 2 = =1δ n 1 + n2 n1 + n 2 n 1 + n2 Letting n2 = n1(1 - ∆) then yields β / k + n2 δ δ =1≈ 1 since << 1 n1 + n 2 2 −∆ 2 −∆ Therefore, b ≈
β / k − n2 or β = k[bn1∆ + n2] n1 − n 2
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.
θ s
ϕ a
x
The travel time of each ray is given by sin φ =
x n n (1 − ∆ ) = 2 = 1 = (1 - ∆) s n1 n1
The travel time of the highest order ray is thus T max =
n 1 L n 1L 1 s = c x c 1−∆
For the axial ray the travel time is T min =
5
Ln 1 c
Therefore T min - T max =
Ln 1 1 Ln 1 ∆ Ln 1 ∆ 1 = − ≈ c 1 − ∆ c 1−∆ c
3-17. Since n 2 = n1 (1 − ∆ ), we can rewrite the equation as σ mod n1 ∆ π 1− = L c V where the first tern is Equation (3-30). The difference is then given by the factor 1−
1 π πλ πλ 1 1 =1− ≈ − 1/ 2 V 2a (n12 − n 22 ) 2a n 1 2∆
At 1300 nm this factor is 1 −
1 π(1.3) = 1 − 0.127 = 0.873 2(62.5) 1.48 2(0.015)
3-18. For ε = 0 and in the limit of α → ∞ we have C1 = 1, C2 =
3 , 2
α α+2 1 α +1 1 = 1, = , = , α+1 3α + 2 3 2α + 1 2
1 (α + 1)2 = and (5α + 2)(3α + 2) 15 Thus Eq. (3-41) becomes σ int er mod al
Ln1 ∆ 12 2 1/ 2 Ln1 ∆ 1+ 3∆ + ∆ = ≈ 2 3c 5 2 3c
6 ∆). Thus C1 and C2 in Eq. (3-42) become 5 (ignoring small terms such as ∆3, ∆4, ...)
3-19. For ε = 0 we have that α = 2(1 -
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 9 32 1 − ∆ − 2 1− ∆ 3α − 2 5 5 C2 = = = 3 6 2(α + 2) 2 2 1− ∆ + 2 2 1 − ∆ 5 5 Evaluating the factors in Eq. (3-41) yields: (a) C12 ≈
9 2 ∆ 25
9 6 3 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 − =
(c)
18∆ 11 18 1− ∆ + 25 51
∆ 5
2
24 ∆ 25
18 2 ∆ 25
16∆2 C22 (α + 1)2 (5α + 2)(3α + 2) 2
2
6 9 2 16∆ 1− ∆ 2(1 − ∆) + 1 5 5 = 2 6 6 3 4 1 − ∆ 10(1 − ∆) + 2 6(1 − ∆) + 2 5 5 5 2 4 2 9 16∆ 1− ∆ 9 1 − ∆ 9 2 5 5 = ∆ 2 ≈ 9 3 24 96(1 − ∆)(1 − ∆)4 1− ∆ 5 10 2
Therefore, σ int er mod al =
1/ 2 1/ 2 Ln 1 ∆ α α + 2 9 2 18 2 9 2 ∆ − ∆ + ∆ 2c α + 1 3α + 2 25 25 24
1/ 2 6 6 2(1 − ∆) 2(1− ∆) + 2 3 n1∆2 L Ln 1 ∆2 5 5 = ≈ 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 σ int er mod al N 1 ∆ α α + 2 1/ 2 4∆C 2 (α + 1) = = 0.070 ns/km 2c α + 2 3α + 2 (5α + 2)(3α + 2) L With C1 = 0 we have from Eq. (3-45) σ int ra mod al =
1 σ λ 2 d 2 n1 −λ = c λ dλ2
0.098σ λ ns / km at 850 nm −2 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) 0.098σ λ ns / km at 850 nm σ int ra mod al = L 0.0103σ λ ns / km at 1300 nm The plot of
1/ 2 σ 1 2 = (σ int er + σ 2int ra ) vs σ λ : L L
3-22. Substituting Eq. (3-34) into Eq. (3-33) τg =
L dβ L 1 dn 2 2 = 2kn1 + 2k n1 1 c dk c 2β dk α
2 −1 α + 2 m α+ 2 2 2 2 α+2 n1 k ∆ ) - 2 ( 2 α a α+2
n 2 k 2 d∆ 2 dn 2 × ∆ 2k n1 1 + 2kn1 + 1 dk ∆ dk α α +2 L kn1 4∆ α + 2 m 1 n1 k d∆ = N1 + N1 − 2 2 2 c β 2∆ dk α + 2 α a n1 k ∆
8
α LN1 kn1 4∆ m α +2 ε 1− 1 = + c β α + 2 M 4
with
N1 = n1 + k
dn 1 and where M is given by Eq. (2-97) and ε is defined in Eq. dk
(3-36b). 3-23. From Eq. (3-39), ignoring terms of order ∆2, α dτ L dN 1 α − ε − 2 m α+ 2 1+ λ = dλ c dλ α + 2 M
α LN1 α − ε − 2 d m α +2 + ∆ c α + 2 dλ M
where N1 = n1 - λ
(a)
α 2 2 2 dn1 and M = a k n1 ∆ α+2 dλ
d dn d 2 n1 dN 1 n1 − λ 1 = - λ = 2 dλ dλ dλ dλ
Thus ignoring the term involving ∆ L 2 d 2 n1 becomes - λ 2 c dλ
d 2 n1 2 , the first term in square brackets dλ
α
α
α
(b)
α +1 α +2 d m α+ 2 −α dM 1 α +2 α+ 2 d∆ 1 m ∆ = + ∆ dλ M dλ M α + 2 dλ M
(c)
dM α 2 d 2 2 = a (k n1 ∆ ) dλ α + 2 dλ =
α dn dk d∆ 2 2 2 2 k n1 + 2k ∆n1 1 + 2kn1 ∆ a2 dλ dλ dλ α+2 9
Ignoring
d∆ dn and 1 terms yields dλ dλ
dM 2α 2 2 2 1 2M = a k n1 ∆ − = λ λ dλ α + 2
so that
α α d m α+ 2 ∆ 2α m α +2 ∆ = . dλ M λ α + 2 M
Therefore
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 3-24. Let
a = λ2
d 2 n1 2α ; 2 ; b = N1C1∆ dλ α+2
γ=
α α+2
Then from Eqs. (3-32), (3-43), and (3-44) we have 2
σ
2 int ra mod al
dτ g λ dλ
2
γ m b −a + M
2
1 σ = L λ λ M
2
2
2
L σλ 1 = c λ M
M
∑
m= 0
M
∑
m= 0
2
2
2 γ L σλ 1 M m ≈ b dm c λ M ∫0 −a + M
L σλ = c λ
2
2 L σλ = c λ
2
2 2ab b2 a − + γ + 1 2γ + 1
2 L σ = λ λ c
2
2 d 2 n 2 1 −λ dλ2
2
1 M
∫
M
0
2γ γ 2 m 2 m a − 2ab b dm + M M
10
2 d 2 n1 4α 2 α 2 λ (N C ∆) 2 N1C1∆ + 1 1 2 dλ (α + 2)(3α + 2) α+1
-
3-25. Plot of Eq. (3-57). 3-26. (a) D = (λ − λ 0 )S0 = −50(0.07) = −3.5 ps /(nm − km) 1500(0.09) (b) D = 4
1310 4 1− 14.1 ps /(nm − km) 1500 =
3-27. (a) From Eq. (3-48) σ step L
=
n1 ∆ 1.49(0.01) = 8 = 14.4 ns / km 2 3 c 2 3 (3 × 10 )
(b) From Eq. (3-47) σ opt n ∆2 1.49(0.01)2 = 1 = = 14.3 ps / km L 20 3 c 20 3 (3 × 108 ) (c) 3.5 ps/km 3-28. (a) From Eq. (3-29) σ mod = Tmax − Tmin =
n1 ∆L (1.49)(0.01)(5 × 103 m) = = 248 ns 8 c 3 × 10 m / s
(b) From Eq. (3-48) σ step =
n 1∆L 248 = = 71.7 ns 2 3c 2 3
(c) BT =
0.2 = 2.8 Mb / s σ step
(d) BT ⋅ L = (2.8 MHz)(5 km) = 13.9 MHz ⋅ km
11
3-29. For α = 0.95αopt , we have σ int er (α ≠ α opt ) (α − α opt ) 0.05 = =− = −170% (0.015)(1.95) σ int er (α = α opt ) ∆(α + 2) For α = 1.05αopt , we have σ int er (α ≠ αopt ) (α − α opt ) 0.05 = =+ = +163% (0.015)(2.05) σ int er (α = αopt ) ∆(α + 2)
12
Problem Solutions for Chapter 4
4-1.
3/4 2πkBT3/2 Eg From Eq. (4-1), ni = 2 2 (memh) exp - 2k T B h
2π(1.38 × 10−23 J / K) =2 −34 2 (6.63 × 10 J.s)
3/ 2
T
3 /2
[(.068)(.56)(9.11× 10
kg)
]
4.3 10 −4
−31
2 3/ 4
(1.55 − 4.3 × 10 −4 T)eV × exp − −5 2(8.62 × 10 eV / K)T
× 3/2 = 4.15×1014 T exp − exp −5 −5 2(8.62 × 10 )T 2(8.62 × 10 ) = 5.03×1015 T 4-2.
3/2
1.55
8991 exp - T
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 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 pP = 2
2 4ni 1+ 2 NA
+1
If ni << NA , which is generally the case, then to a good approximation 2 2 pP ≈ NA and nP = ni /pP ≈ ni /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) ]
1.240 The emission wavelength is λ = 1.540
= 0.090
= 805 nm.
(b) Eg = 1.424 + 1.266(0.15) + 0.266(0.15)2 = 1.620 eV, so that 1.240 λ = 1.620 4-4.
= 766 nm
(a) The lattice spacings are as follows: o
a(BC) = a(GaAs) = 5.6536 A o
a(BD) = a(GaP) = 5.4512 A o
a(AC) = a(InAs) = 6.0590 A o
a(AD) = a(InP) = 5.8696 A 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 o
(b) Substituting a(xy) = a(InP) = 5.8696 A into the expression for a(xy) in (a), we have 0.4184x y = 0.1894 - 0.0130x
0.4184x ≈ 0.1894
= 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
2
∆E =
hc λ2 ∆λ or ∆λ = ∆E hc λ2
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 4-6.
(a) From Eq. (4-10), the internal quantum efficiency is ηint =
1 = 0.783, and from Eq. (4-13) the internal power level is 1 + 25/ 90
Pint = (0.783)
hc(35 mA ) = 26 mW q(1310 nm)
(b) From Eq. (4-16), P= 4-7.
Plot of Eq. (4-18). Some representative values of P/P0 are given in the table:
f in MHz 1 10 20 40 60 80 100
4-8.
1 2 26 mW = 0.37 mW 3.5(3.5 + 1)
P/P0 0.999 0.954 0.847 0.623 0.469 0.370 0.303
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 2 1 + (2πfτ )
[
1/ 2 = −3
]
3
With a 5-ns lifetime, we find f =
4-9.
1 10 0.6 − 1) = 9.5 MHz ( 2π (5 ns )
(a) Using Eq. (4-28) with Γ = 1 gth =
1 1 2 ln 0.32 + 10 cm-1 = 55.6 cm-1 0.05 cm
(b) With R1 = 0.9 and R2 = 0.32, 1 1 gth = 0.05 cm ln 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, 1.24 1.24 λ= E = = 1.462 µm 1.424 + 1.266(0.3) + 0.266(0.3)2 g From Eq. (4-38) dP(mW) ηext = 0.8065 λ(µm) 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 2eff = 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 λ2 ∆λ = 2Ln =
(0.80 µm)2 = 0.22 nm 2(400 µm)(3.6)
Therefore the number of modes in the range 0.75-to-0.85 µm is 4
.1 .85 − .75 3 −3 = .22 ×10 = 455 modes .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 (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 λ2 ∆λ = 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 = 240 modes 0.25
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)
We now want to relate ∆N to the change ∆λ in the free-space wavelength. First differentiate N with respect to λ: dN d n 1 dn n 1 dn = = - 2 = - 2 n - λ dλ λ λ dλ λ dλ dλ λ Thus for an incremental change in wavenumber ∆N, we have, in absolute values, 1 dn ∆N = 2 n - λ dλ λ
∆λ
(a-2)
5
Equating (a-1) and (a-2) then yields ∆λ =
(b) The mode spacing is
∆λ =
λ2 dn 2Ln - λ dλ
(.85 µm)2 2(4.5)(400 µm)
= 0.20 nm
4-15. (a) The reflectivity at the GaAs-air interface is n-1 2 3.6-1 2 R1 = R2 = n+1 = 3.6+1 = 0.32 Then Jth =
1 1 1 ln = 2.65×103 A/cm2 α + β 2L R1R2
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
(6.6256 × 10 = 1.43 eV +
J ⋅ s) 1 1 + 2 −32 −31 6.19 × 10 kg 5.10 × 10 kg 8(5 nm ) −34
2
= 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 2(1570 nm) = 2(460 nm) = 3.4 2Λ
Then, from Eq. (4-48b), for m = 0
6
2
λB 1 (1.57 µm)(1570 nm) = 1570 nm ± 1.20 nm λ = λB ± 2n L 2 = 1570 nm ± 4(3.4)(300 µm) e 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 density changes from 0 to nth. td
n th
0
0
t d = ∫ dt =
J dn = −τ − J n qd − qd τ 1
∫
n τ
n= n th
n= 0
J = τ ln J − J th
Ip where J = Ip/A and Jth = Ith/A. Therefore td = τ ln I - I p th (b) At time t = 0 we have n = nB, and at t = td we have n = nth. Therefore, td td = ⌠ ⌡ dt = 0
n th
∫
nB
J − nB qd τ dn = τ ln J n J n − − th qd τ qd τ 1
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 Ip I - IB Therefore, td = τ ln I - I = τ ln I I - I th p + B th 4-20. A common-emitter transistor configuration:
4-21. Laser transmitter design.
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 The minimum value is P(t) = P0[1 - 2.36m] ≥ 0
1 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
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 + 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 + 4b1 b 32 cos ω1t cos3 ω2t + b 42 cos4 ω2t)
Use the following trigonometric relationships: 1 i) cos2 x = 2 (1 + cos 2x) 1 ii) cos3 x = 4 (cos 3x + 3cos x) 1 iii) cos4 x = 8 (cos 4x + 4cos 2x + 3) iv) v)
2cos x cos y = cos (x+y) + cos (x-y) 1 cos2 x cos y = 4 [cos (2x+y) + 2cos y + cos (2x-y)]
8
1
1
vi) vii)
1
cos2 x cos2 y = 4 [1 + cos 2x+ cos 2y + 2 cos(2x+2y) + 2 cos(2x-2y)] 1 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 y(t) = 2 a2b1 + a2b2 + 4 a4b1 + 3a4b1b2 + 4 a4b2
constant terms
3 2 3 3 2 3 + 4a3b1 + 2a3b1b2 cos ω1t + 4 a3b2 + 2b1b2 cos ω2t 2
fundamental terms
2
b1 b2 2 2 2 2 + 2 a2 + a4b1 + 3a4b2 cos 2ω1t + 2 a2 + a4b2 + 3a4b1 cos 2ω2t 2nd-order harmonic terms 1 1 3 3 + 4 a3b1 cos 3ω1t + 4 a3b2 cos 3ω2t
3rd-order harmonic terms
1 1 4 4 + 8 a4b1 cos 4ω1t + 8 a4b2 cos 4ω2t
4th-order harmonic terms
3 3 3 3 + a2b1b2 + 2 a4b1b2 + 2 a4b1b2[ cos (ω1+ω2)t + cos (ω1-ω2)t]
2nd-order intermodulation 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
3
2
+ 2 a4b1 b2 [ cos(3ω1+ω2)t + cos(3ω1-ω2)t] 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 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
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.
(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 E /k T 4-25. From Eq. (4-60) τs = K e A B
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
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
10
(2)
τ
s
= 1.11×10-5 exp{0.63/[(8.625×10-5)(293)]} = 7.45×105 hrs
11
Problem Solutions for Chapter 5
5-3.
(a) cosL 30° = 0.5 cos 30° = (0.5)1/L = 0.8660 L = log 0.5/log 0.8660 = 4.82 (b) cosT 15° = 0.5 cos 15° = (0.5)1/T = 0.9659 T = log 0.5/log 0.9659 = 20.0
5-4.
The source radius is less than the fiber radius, so Eq. (5-5) holds: 2
PLED-step = π2rs B0(NA)2 = π2(2×10-3 cm)2(100 W/cm2)(.22)2 = 191 µW From Eq. (5-9) 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 2
3.600 − 1.305 R s− g = 0.219 3.600 + 1.305 = Similarly, the relfectivity at the gel-to-fiber interface is 2
R g− f
1.465 − 1.305 −3 = = 3.34 × 10 1.465 + 1.305
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
1
rm
⌠ ⌠ θ0-max P = 2π ⌠ ⌡ cos3 θ sin θ dθ dθs r dr ⌡ 0 ⌡0 2π
0 Using θ0 ⌠ ⌡ cos3 θ sin θ dθ 0
sin θ0 θ0 =⌠ ⌡ ( x - x3) dx ⌡ ( 1 - sin2 θ) sin θ d(sin θ) = ⌠ 0 0
we have rm
⌠ 2π ⌠ sin2 θ0-max sin4 θ0-max dθs r dr P = 2π 4 ⌡ 2 ⌡0 0 rm
⌠ 2π 1 ⌠ = π NA2 - 2 NA4 dθs r dr ⌡ ⌡0 0 rm 2π π 2 4 = 2 [ 2NA - NA ] ⌠ ⌡ r dr ⌠ ⌡ 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.
5-8.
Using Eq. (5-10), the relfectivity at the gel-to-fiber interface is
2
2 1.485 − 1.305 R g− f = 4.16 × 10 −3 1.485 + 1.305 =
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
=
2 1.485 − 1.000 1.485 + 1.000 = 0.038
The power loss is L = −10 log (1 − R) = −10 log (0.962) = 0.17 dB 5-9.
1 1 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 d21/2 d Acommon = 2(shaded area) = sa – cy = 2a2 arccos 2a - d a2 - 4
5-10.
Core/cladding diameters (µ µ m) 50/125 62.5/125 100/140
Coupling loss (dB) for Given axial misalignments (µ µ m) 1 3 5 10 0.112 0.089 0.056
0.385 0.274 0.169
π 5-11. arccos x = 2 - arcsin x x3 x5 For small values of x, arcsin x = x + 2(3) + 2(4)(5) + ...
3
0.590 0.465 0.286
1.266 0.985 0.590
d d π arccos 2a ≈ 2 - 2a
d Therefore, for 2a << 1, we have Thus Eq. (5-30) becomes d/a 0.00 0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.40
PT =
8d 2 π d 5d P2 - 2a - 6a = P 1 π 3πa
PT/P (Eq.5-30) 1.00 0.9576 0.9152 0.8729 0.8309 0.7890 0.7475 0.7063 0.6656
PT/P (Eq.5-31) 1.00 0.9576 0.9151 0.8727 0.8302 0.7878 0.7454 0.7029 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
MaR ηF = M aE
1 1 2 k2NA2(0) 2 a α+2 R = 1 1 2 k2NA2(0) 2 a α+2 E
2
=
aR 2
aE
a2 R Therefore from Eq. (5-21) the coupling loss for aR ≤ aE is LF = -10 log 2 aE 5-14. For fibers with different NAs, where NAR < NAE
MR LF = -10 log ηF = -10 log M E
2 α 2 k2NAR(0) a 2α+4 = -10 log 2 α 2 k2NAE(0) a 2α+4
4
NA2 (0) R = -10 log 2 NAE(0) 5-15. For fibers with different α values, where αR < αE αR a2 k2NA2(0) 2αR + 4 LF = -10 log ηF = -10 log αE a2 k2NA2(0) 2αE + 4
αR(αE + 2) = -10 log 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: NA 1 (0) = n1 2∆ = 1.46 2(0.01) = 0.206 For fiber 2: NA 2 (0) = n1 2∆ = 1.48 2(0.015) = 0.256 (a) The only loss is that from index-profile differences. From Eq. (5-37) L 1→ 2 (α) = −10 log
1.80(2.00 + 2) = 0.24 dB 2.00(1.80 + 2)
(b) The losses result from core-size differences and NA differences. L 2→1 (a) = −20 log
50 1.94 dB 62.5 =
.206 L 2→1 (NA) = −20 log = 1.89 dB .256 5-17. Plots of connector losses using Eq. (5-43). 5-18. When there are no losses due to extrinsic factors, Eq. (5-43) reduces to
5
LSM;ff = -10 log
4
2 W1 + W2 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). 5-20. Plot of the throughput loss.
6
Problem Solutions for Chapter 6
6-1.
From Eqs. (6-4) and (6-5) with Rf = 0, η = 1 - exp(-αsw) To assist in making the plots, from Fig. P6-1, we have the following representative values of the absorption coefficient: λ (µ µ m) .60 .65 .70 .75 .80 .85 .90 .95 1.00
6-2.
αs (cm-1) 4.4×103 2.9×103 2.0×103 1.4×103 0.97×103 630 370 190 70
w w ⌠ -αsx dx Ip = qA ⌠ ⌡ G(x) dx = qA Φ0 αs ⌡ e 0 0 -αsw -αsw P0(1 - Rf) = qA Φ0 1 - e = qA 1 - e hνA
6-3.
From Eq. (6-6),
R=
ηqλ ηq = hc = 0.8044 ηλ ( in µm) hν
Plot R as a function of wavelength. 6-4.
(a) Using the fact that Va ≈ VB, rewrite the denominator as VB - VB + Va - IMRMn Va - IMRMn 1- = 1 VB VB VB - Va + IMRMn = 1 - 1 VB Since
VB - Va + IMRM VB
<< 1, we can expand the term in parenthesis:
1
n( VB - Va + IMRM) VB - Va + IMRMn 1 ≈ 1 1 - 1 VB VB =
n( VB - Va + IMRM) nIMRM ≈ VB VB
IM Therefore, M0 = I p
≈
VB VB ≈ nI R n( VB - Va + IMRM) M M
IM VB (b) M0 = I = nI R p M M
6-5.
IpVB 2 VB 1/2 implies IM = nR , so that M0 = nI R M p M
< >
T 2π/ω 1⌠ 2 ω ⌠ 2 2 R P (t) dt = T ⌡ is (t)dt = 2π ⌡ 0 0 0
ω = 2π
2 P0
2 is (t)
2 R0
(where T = 2π/ω),
2π/ω ⌠ ⌡ (1 + 2m cos ωt + m2 cos2 ωt) dt 0
Using 2 π/ω
∫
cos ωt dt =
0
and
t = 2π / ω t=0
=0
2π 2π/ω 1 π ⌠ 1 1 ⌠ 2 + 2cos 2x dx = ⌡ cos2 ωt dt = ω ω ⌡ 0 0
we have
6-6.
1 sin ωt ω
< i2s(t)>
2 m2 2 = R0 P0 1 + 2
Same problem as Example 6-6: compare Eqs. (6-13), (6-14), and (6-17). ηqλ P0 = 0.593 µA hc
(a) First from Eq. (6-6), I p = Then σ Q = 2qI p B = 2(1.6 × 10 2
−19
C)(0.593 µA)(150 × 10 Hz) = 2.84 × 10 6
2
−17
A
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 4(1.38 × 10 −23 J / K)(293 K) 4k B T 6 −15 2 (c) σ = B= 150 × 10 Hz) = 4.85 × 10 A ( RL 500 Ω 2 T
6-7.
ηqλ Using R0 = hc = 0.58 A/W, we have from Eqs. (6-4), (6-11b), (6-15), and (617) 1 2 R0 P0 m) M 2 ( R0P0m2 S 2 = = N 1/ 2 2 4qBM1/2 Q 2qI p BM M 2 ( R0P0m) S = N 4qIDBM1/2 DB
= 6.565×1012 P0
2
= 3.798×1022 P0
2 2 R P m M 0 0 ( ) S = N 4qILB DS
= 3.798×1026 P0
2 1 R0P0m) M2 ( 2 S N = 4k TB/R T B L
= 7.333×1022 P0
2
2
P0 where P0 is given in watts. To convert P0 = 10-n W to dBm, use 10 log -3 = 10 10(3-n) dBm 6-8.
Using Eq. (6-18) we have 2 2 1 R P m M 0 0 ( ) 2 S 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
6-9.
1 2 2 2 IpM d S d 0 = dM N = dM 2+x + 2qIL + 4kBT/RL 2q(Ip + ID)M 1 2 (2+x)M1+x 2q(Ip + ID)2 IpM2 2 0 = Ip M 2q(Ip + ID)M2+x + 2qIL + 4kBT/RL
2+x
Solving for M: Mopt
=
2qIL + 4kBT/RL xq(Ip + ID)
6-10. (a) Differentiating pn, we have -αsw (w-x)/Lp 1 ∂p n e = L pn0 + Be p ∂x -αsw (w-x)/Lp 1 ∂ 2 pn e 2 = - 2 pn0 + Be ∂x Lp Substituting pn and
-αsx 2 + αs Be
∂ 2 pn into the left side of Eq. (6-23): ∂x2
-αsw (w-x)/Lp Dp e - 2 pn0 + Be Lp +
-αsx - αsBe
-αsw (w-x)/Lp 1 e pn0 + Be τp
-αsx 2 + Dp αs Be
-
-αsx B -αsx e + Φ0 αs e 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: 2 Φ -α x s 0 αsLp 2 1 Left side = D Dp αs - + Φ0 αs e 2 2 τ p p 1 - αs Lp
4
2 2 -α x Φ0 αsLp αs Lp - 1 s =D + Dpαs e 2 2 τ p 1-α L p s p
-αsx Φ0 =0 = D ( -Dpαs + Dpαs) e p
(b) Jdiff = qDp
Thus Eq. (6-23) is satisfied.
∂pn ∂x x =w
-αsw -α w 1 - αsBe s = qDp L pn0 + Be p -αsw Dp 1 = qDp BL - αs e + qpn0 L p p αsL2 1 - α L Dp p s p -αsw e + qp = qΦ0 n0 2 2 Lp Lp 1 - αs Lp = qΦ0
-αsw αsLp Dp e + qpn0 L 1 + αsLp p
c) Adding Eqs. (6-21) and (6-25), we have -αsw αsLp -αsw Dp Jtotal = Jdrift + Jdiffusion = qΦ0 1 - e + qpn0 L e + p 1 + αsLp -αsw Dp -αsw e = qΦ0 1 e + qpn0 L p 1 + αsLp 6-11. (a) To find the amplitude, consider J J* 1/2 = qΦ ( S S*) 1/2 0 tot tot sc
-jωtd 1-e where S = jωtd
We want to find the value of ωtd at which ( S S*)
Evaluating ( S S*)
1/2
, we have 5
1/2
=
1 . 2
( S S*)
1/2
+jωtd1/2 -jωtd 1 - e 1 - e = jωtd -jωtd 1/2
-jωtd +jωtd +e 1 - e + 1 = ωtd [ ( 1 - cos ωtd) /2] = ωtd/2
1/2
( 2 - 2 cos ωtd) = ωtd
ωtd sin 2 ωtd = = sinc 2 ωtd 2
We want to find values of ωtd where ( S S*) x 0.0 0.1 0.2 0.3 0.4
1/2
sinc x 1.000 0.984 0.935 0.858 0.757
1/2
=
1 . 2
x 0.5 0.6 0.7 0.8 0.9
sinc x 0.637 0.505 0.368 0.234 0.109
By extrapolation, we find sinc x = 0.707 at x = 0.442. ωtd Thus 2 = 0.442 which implies
ωtd = 0.884
w 1 (b) From Eq. (6-27) we have td = v = . Then αsvd d ωtd = 2πf3-dB td = 2πf3-dB
1 = 0.884 αsvd
or
f3-dB = 0.884 αsvd/2π 6-12. (a) The RC time constant is
RC =
Rε0KsA w
=
(10 4 Ω)(8.85 × 10 −12 F / m)(11.7)(5 × 10 −8 m 2 ) = 2.59 ns 2 × 10 −5 m
(b) From Eq. (6-27), the carrier drift time is
6
w 20 × 10 −6 m td = v = = 0.45 ns d 4.4 × 10 4 m / s c)
1 1 = 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. 6-13. (a) With k1 ≈ k2 and keff defined in Eq. (6-10), we have 2
2
(1)
k1 - k1 k2 - k1 k1(1 - k1) 1- 1-k =1- 1-k ≈1- 1-k 2 2 2
(2)
1 - 2k1 + k1 1 - 2k2 + k1 (1 - k1)2 = ≈ 1 - k2 1 - k2 1 - k2
2
= 1 – keff
2
2
k2 - k1 1 - k2 =1-k - 1-k 2 2
= 1 - keff
Therefore Eq. (6-34) becomes Eq. (6-38): 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
(1)
(2)
k2(1 - k1) 2 k1(1 - k2)
(1 - k1)2k2 2
2
≈
k2 - k1
2
=
k2 - 2k1k2 + k2k1
k1(1 - k2)
≈
'
= keff 2 k1(1 - k2)
2
k1(1 - k2)
k2 - k2 - k2 - k2k2 1 1 1 2 k1(1 - k2)
7
' = keff - 1
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 (637), k1 = k2 = 0 and keff = 0. Then from Eq. (6-38) 1 Fe = 2 - M
e
≈ 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.
8
Problem Solutions for Chapter 7
7-1.
1 We want to compare F1 = kM + (1 - k) 2 − and F2 = Mx. M For silicon, k = 0.02 and we take x = 0.3: M 9 25 100
F1(M) 2.03 2.42 3.95
F2(M) 1.93 2.63 3.98
% difference 0.60 8.7 0.80
For InGaAs, k = 0.35 and we take x = 0.7: M 4 9 25 100
F1(M) 2.54 4.38 6.86 10.02
F2(M) 2.64 4.66 6.96 9.52
% difference 3.00 6.4 1.5 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) =
j2πft ∫ HB (f )e df = R
−∞
∞
e j2πft ∫ 1 + j2πfRC df −∞
Using the integral solution from Appendix B3: ∞
2π(p) n−1 e − βp e jpx = ∫ β + jx) n dx Γ(n) −∞ ( 1 hB(t) = 2πC
7-3.
∞
∫
−∞
for p > 0, we have
e j 2πft 1 -t/RC df = e 1 C + jf 2πRC
Part (a): ∞
1 ∫ h p (t) dt = αTb
−∞
αTb/2
⌠ dt ⌡ -αTb/2
=
1 αTb αTb + 2 αTb 2
1
=1
Part (b): ∞
∫h
p
(t) dt =
−∞
1 1 2 π αTb
1 1 π αTb 2π αTb
=
∞
∫
−∞
t2 dt exp − 2 2( αTb )
2 = 1 (see Appendix B3 for integral solution)
Part (c): ∞
1 ∫ h p (t) dt = αTb −∞ 7-4.
∞
∫ 0
t exp − dt = - [e −∞ − e −0 ]= 1 αT b
The Fourier transform is ∞
F[p(t)*q(t)] =
∫
p(t)* q(t)e
− j2 πft
∞
∫
dt =
−∞ ∞
=
∫
q(x) e
−∞ ∞
=
∫
− j2πfx
∞
q(x)
−∞ ∞
∫
p(t − x) e
− j2πf( t − x)
∫
p(t − x) e − j2πft dt dx
−∞
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 =
1 [P0 (v th ) + P1 (v th )]. 2
Substituting Eq. (7-20) and (7-22) for P0 and P1, respectively, we have 1 Pe = 2
V /2 ∞ −v 2 / 2σ 2 1 − (v−V ) 2 / 2σ 2 e dv e dv + ∫ 2πσ 2 V∫/ 2 −∞
In the first integral, let x = v/ 2σ2 so that dv = 2σ2 dx. In the second integral, let q = v-V, so that dv = dq. The second integral then becomes
2
V / 2 −V −q 2 / 2σ 2
∫e
dq =
2
2σ
−V / 2 2σ 2 − x2
∫
−∞
e
where x = q/ 2σ2
dx
−∞
Then ∞ − V / 2 2σ 2 2σ 2 − x2 − x2 e dx e dx + ∫ 2πσ 2 V / 2∫ 2σ 2 −∞
1 Pe = 2
∞ −x 2 = ∫ e dx − 2 2 π −∞ 1
V / 2 2σ 2
∫
e
−x 2
0
dx
Using the following relationships from Appendix B, ∞
∫
e
− p2 x 2
−∞
Pe =
7-6.
π dx = p
2 π
and
t
∫
e
−x 2
dx = erf(t), we have
0
1 V 1− erf 2σ 2 2
V V = 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 (a) V = 1 volt and σ = 0.2 volts, so that
errors/bit) = 1400 errors/second, so that 1 1400 errors/second
= 7×10-4 seconds/error
(b) If V is doubled, then
V = 5 for which Pe ≈ 3×10-7 errors/bit from Fig. 7-6. 2σ
Thus 1 = 16.7 seconds/error −7 (2 × 10 bits / sec ond)(3 × 10 errors / bit ) 5
7-7.
(a) From Eqs. (7-20) and (7-22) we have P0(vth) =
1 2πσ2
∞
∫
V /2
e− v
2
/ 2σ 2
dv =
1 V 1− erf 2σ 2 2
and
3
1 2πσ2
P1(vth) =
V /2
− ( v− V) ∫e
2
/ 2σ 2
dv =
−∞
1 V 1− erf 2σ 2 2
Then for V = V1 and σ = 0.20V1 2 1 = 2 1 - erf0.8
1 1 P0(vth) = 2 1 - erf 2(.2) 2 1 = 2 [ 1 - erf( 1.768) ]
1 = 2 ( 1 - 0.987)
= 0.0065
Likewise, for V = V1 and σ = 0.24V1 1 1 P1(vth) = 2 1 - erf 2(.24) 2 1 = 2 [ 1 - erf( 1.473) ]
2 1 = 2 1 - erf0.96
1 = 2 ( 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 ηE N= hν
ηPt = hc/λ
0.65(25 × 10 −10 W)(1 × 10 −9 s)(1.3 × 10 −6 m) = = 10.6 34 8 (6.6256 × 10 − Js)(3 × 10 m / s)
Then, from Eq. (7-2) e-N P(n) = Nn n! 7-9.
e-10.6 = (10.6)5 5!
133822 = 120
e-10.6 = 0.05 = 5%
v N = v out - v out v2N =
[v
out
− vout
] 2
2
2
= vout -2 vout + = v2out -
vout
vout
2
2
7-10. (a) Letting φ = fTb and using Eq. (7-40), Eq. (7-30) becomes 4
H p (0) Bbae = Hout (0)
2
∞
H'out (φ) T ' H p (φ)
∫
2
dφ I 2 = Tb Tb
2 b
0
since Hp(0) = 1 and Hout(0) = Tb. Similarly, Eq. (7-33) becomes H p (0) Be = Hout (0) 1 = 2 Tb 1 = 2 Tb
∞
∫ 0
∞
∫ 0
2
2
∞
Hout (f) (1 + j2πfRC ) df H p (f)
∫ 0
Hout (f) H p (f)
2
(1 + 4π f
2 2
2
RC
2
2
Hout (f) (2πRC)2 df + H p (f) Tb2
) df ∞
∫ 0
2
I2 Hout (f) 2 f df = Tb H p (f)
+
(2πRC)2 3
Tb
I3
(b) From Eqs. (7-29), (7-31), (7-32), and (7-34), Eq. (7-28) becomes
< v2N >
=
< v2s >
= 2q
< v2R >
+
4kBT BbaeR2A2 + R b
< v2I >
< v2E >
+
+
BbaeR2A2 + SIBbaeR2A2 + SEBeA2
4kBT = 2q M2+x + R + SI BbaeR2A2 + SEBeA2 b =
4k T R2A2I2 S 2+x + B + S + E 2q i M 0 I < > Rb Tb R2
+
(2πRCA)2 3
Tb
SE I 3
7-11. First let x = (v − boff )/ ( 2 σ off ) with dx = dv / ( 2 σoff ) in the first part of Eq. (749):
Pe =
2σ off 2πσ off
∞
1 exp(− x ) dx = ∫ π v th − b off 2
∞
∫
Q/
exp (−x 2 ) dx 2
2σ off
Similarly, let y = (−v + b on ) / ( 2 σon ) so that dy = -dv/ ( 2σon )in the second part of Eq. (7-49): 5
− v th +b on 2σ on
1 π
Pe = -
7-12. (a) Let x =
∫
∞
V 2 2σ
∞
1 exp(−y ) dy = π
∫
2
=
K 2 2
Q/ 2
exp (−y
2
) dy
For K = 10, x = 3.536. Thus
2
Pe =
e-x 2 πx
= 2.97×10-7 errors/bit
2 e-x -5 (b) Given that Pe = 10 =
then
This holds for x ≈ 3, so that K = 2 2
x = 8.49.
2 πx
e-x
2
=2 π
10-5 x.
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 + + + − γ =M W M b I W b I (1 hν on 2 hν on 2 Μ 2
1/ 2
1 /2
1 1 1+ x 1+x (2 + x)M b I (2 + x)M b I (1− γ) on 2 on 2 Q(hν / η) 2 2 + + 1/ 2 1/ 2 M(hν/ η) 2+ x η 2+x η M b I W M hν bon I 2 (1− γ) + W hν on 2 + η Letting G = M2+x bonI2 for simplicity, yields hν ( G
+ W)
1/2
+ [ G(1-γ) + W]
1/2
1/2
Multiply by (G + W) 1/2
(G + W)
1 G (1-γ) = 2 (2 + x) 1/2 + 1/2 [ G(1-γ) + W] (G + W)
[ G(1-γ) + W]
Gx W - 2 (1-γ)
1/2
and rearrange terms to get
= [ G(1-γ) + W]
6
1/2 Gx
2 - W
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
Solving this equation for G yields 1
x2
G=
- 4 W(2-γ) ±
x4 16
2 x2(1-γ)W2(1+x)
W2(2-γ)2 + x2 2 (1-γ)
1 1+x 1-γ W(2-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
G = Mopt
η bonI2 hν
7-14. Substituting Eq. (7-55) for M2+xbon into the square root expressions in Eq. (754) and solving Eq. (7-55) for M, Eq. (7-54) becomes hν 1/(2+x) 1 1 Q bon W(2-γ) W 2 2 η bon = K + W + 2 (2-γ)K + W hν W(2-γ) 1/(2+x) 2(1-γ) η 2I (1-γ) K 2 Factoring out terms: (1+x)/(2+x)
bon
hν(1+x)/(2+x) = Q η
x/2(2+x)
1/(2+x)
W
I2
1 1 (2-γ) 1 2 2 × K + 1 + 2 (2-γ)K + 1 2(1-γ)
or
(2+x)/(1+x)hν
bon = Q
η
x/2(1+x)
W
1/(1+x)
I2
7
(2-γ)K1/(2+x) ÷ 2(1-γ) L
7-15. In Eq. (7-59) we want to evaluate 1+x
1+x
1+x
lim (2 − γ)K 2+x lim 1 2+x lim (2 − γ)K 2+x = γ →1 2(1− γ) γ →1 L γ →1 2(1− γ )L Consider first 1 lim (2 − γ)K lim (2 − γ) (1− γ ) 2 = −1 + 1 + B 2 γ →1 2(1− γ) γ →1 2(1− γ ) (2 − γ)
where B = 16(1+x)/x2 . Since γ→1, we can expand the square root term in a binomial series, so that lim (2 − γ)K lim (2 − γ) 1 (1− γ) 2 = 1 B Order(1− ) −1 + + − γ 2 2 (2 − γ) γ →1 2(1− γ) γ →1 2(1− γ ) =
lim B B =4 γ →1 4(2 − γ )
=4
1+x
1+x
Thus
1+x x2
lim (2 − γ )K 2+x 1+x2+x = 4 2 γ → 1 2(1 − γ ) x
Next consider, using Eq. (7-58) 1+ x
lim 1 2+x γ →1 L lim (2 − γ)K = γ →1 2(1− γ)
1/(2+ x)
1 1 (2 − γ) 2 1 2 ÷ K + 1 + (2 − γ )K + 1 2 2(1− γ)
From the above result, the first square root term is 1
(2-γ) 2 K + 1 2(1-γ)
1
1+x 2 = 4 2 + 1 x
1
x2 + 4x + 42 = x2
From the expression for K in Eq. (7-55), we have that
8
=
x+2 x
lim K = 0, so that γ →1
1
lim 1 2 (2 − γ )K + 1 = 1 γ →1 2
Thus
1 1 lim (2 − γ ) 2 1 2 x+2 K + 1 + (2 − γ )K + 1 = x 2 γ → 1 2(1 − γ )
+1=
2(1 + x) x
Combining the above results yields 1+x
1+x
1
lim (2 − γ)K 2+x 1+x2+x 1+x2+x 2(1 + x) = 4 2 4 x2 x γ →1 2(1− γ )L x lim 1+x Mopt γ →1
so that
2 =x
W1/2 2 = QI x 2
' 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 ∞
I2 =
∫ 0
1 H (φ) dφ = 2
1-β 2
1 =2 ⌠ ⌡ dφ -
2
' out
1-β 2
Letting y =
∫
2
H'out (φ) dφ
−∞
1+β 2
-
⌠ 1 πφ π 2 + 81-sin - dφ β 2β ⌡
πφ β
-
π 2β
1-β 2
⌠ 1 πφ π 2 + 81-sin - dφ β 2β ⌡
1-β 2
1+β 2
we have
π 2 β + ⌠ [ 1 - 2sin y + sin2y] dy 4π ⌡ π -2
1 I2 = 2 (1 - β)
1 = 2 (1 - β)
∞
+
β π π - 0 + 2 4π
1 β = 2 1 - 4
9
Use Eq. (7-42) to find I3: ∞
I3 =
∫ 0
2
∞
∫
2
'
2
Hout (φ) φ dφ
−∞
1+β 2
1-β 2
1 =2 ⌠ ⌡ φ2 dφ -
1 H (φ) φ dφ = 2 2
' out
1-β 2
-
⌠ 1 πφ π 2 + 81-sin - φ2dφ β 2β ⌡
⌠ 1 πφ π 2 + 81-sin - φ2dφ β 2β ⌡
1-β 2
Letting y =
-
πφ β
1 1-β3 I3 = 3 2
1 1-β3 =3 2
-
1-β 2
1+β 2
π 2β
π 2 β⌠ β2y2 βy 1 + [ 1 - 2sin y + sin2y] 2 + + dy 4π ⌡ π 4 π π -2
β + 4π
π π 2 2 2β ⌠ β2y2 1 2 y sin y dy 2 + 4( 1 + sin y) dy - ⌠ π⌡ ⌡ π π π -2 2
where only even terms in "y" are nonzero. Using the relationships π 2 ⌠ ⌡ sin2y 0
and
π dy = 4
⌠ ⌡ y sin ydy
;
⌠ ⌡ x2 sin x2 dx
x3 = 6
x2 1 - 4 - 8
= -y cos y + sin y
sin 2x -
x cos 2x 4
we have 1 1-β3 I3 = 3 2
+
β β2 1π3 1π3 π 1π π 2β + + + + - (1) 62 8 42 4 π 2π π2 32 10
β 3 1 1 = 16 2 - 6 π
β - 32
1 1 - β2 2 - 8 π
1 + 24
7-17. Substituting Eq. (7-64) and (7-66) into Eq. (7-41), with s2 = 4π2α2 and β = 1, we have
⌠ Hout(φ)2 I2 = ' H (φ) ⌡ p '
1 1⌠ s2φ2 π2 dφ = 4e 1-sin πφ - 2 dφ ⌡ 0
0 1 1 2 2 2 ⌠ s φ 1 + cos πφ ⌠ s2φ2 4 πφ dφ = cos 2 dφ = e e 2 ⌡ ⌡ 0 0
πφ Letting x = 2 yields
π 2 2 ⌠ 16α2x2 4 I2 = ⌡e cos x dx π 0
Similarly, using Eqs. (7-64) and (7-66), Eq. (7-42) becomes 1 3 ⌠ s2φ2 4 πφ 2 2 I3 = e cos 2 φ dφ = π ⌡ 0
π 2
∫
2 16α 2 x 2
xe
0
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 K: γ →1
11
4
cos x dx
1 2 lim lim 1+ x 1− γ K= 1+ 16 = -1 + 1 = 0 −1+ x 2 (2 − γ) 2 γ →1 γ →1
Also
lim (1− γ) = 0. Therefore from Eq. (7-58) γ →1
lim 2(1− γ) lim L= γ →1 γ →1 (2 − γ)K
1/(1+ x)
2+x
1 1+x (2-γ) 2 K + 1 + 1 2(1-γ)
Expanding the square root term in K yields lim 2 − γ lim 2 − γ −1+ 1+ 16 1 + x 1− γ + order(1− γ) 2 K = 2 2 γ →1 1− γ γ →1 1 − γ 2 x (2 − γ) =
lim 8(1 + x) 1 8(1+x) = x2 γ → 1 x 2 2 − γ
Therefore 2+x 1 1/(1+x) 1+x 2 lim 2x 4(1+x) + 12 + 1 L = 8(1+x) γ →1 x2
2+x 1+x 2x2 1/(1+x) x+2 = 8(1+x) + 1 x
x
21+x = (1+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'
y(ε) = (1 + ε) 1 - ε
1 5/3 3.166 = 1.1 .9 L 2.89 = 1.437
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'
y(ε) = (1 + ε) 1 - ε
1 5/3 3.166 = 1.1 .9 L 2.89 = 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 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)
WJFET =
+
+
1 2(.01nA ) 4(1.38× 10 −23 J / K)(300K) + 0.543 B 1.6 × 10 −19 C (1.6 × 10 −19 C)2 10 5 Ω 1 4(1.38× 10 −23 J / K)(300 K)(.7) 0.543 B (1.6 × 10 −19 C)2 (.005 S)(10 5 Ω) 2
2 −23 2π(10 pF) 4(1.38 × 10 J / K)(300 K)(.7) 0.073 B 1.6 × 10 −19 C (.005 S)
or
13
WJFET ≈
3.51× 1012 + 0.026B B
and from Eq. (7-92)
WBP =
3.39 × 1013 + 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) W=
3.51× 1012 + 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 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
or Pr(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 2kBT 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 HZ 7-26. (a) For simplicity, let Q η η D = M2+x I2 and F = M hν hν so that Eq. (7-54) becomes, for γ = 1,
14
= 10 log 1.29 = 1.09 dB
b = F[ (Db + W)1/2 + W1/2] Squaring both sides and rearranging terms gives b2 - Db - 2W = 2W1/2 (Db + W)1/2 F2 Squaring again and factoring out a "b2" term yields b2 - (2DF2)b + (F4D2 - 4WF2) = 0 Solving this quadratic equation in b yields 1 b = 2 [ 2DF2 ±
4F4D2 - 4F4D2 + 16WF2] = DF2 + 2F W
(where we chose the "+" sign)
=
2Q hν x 2 M Q I2 + M W1/2 η
(b) With the given parameter values, we have 1.7× 10 0.5 bon = 2.286×10-19 39.1M + 4
M
The receiver sensitivity in dBm is found from Pr = 10 log [b on (50 ×10 b / s)] 6
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
15
7-27. Using Eq. (E-10) and the relationship
∫
∞
0
πa 1 2 dx = 2 x 1+ a
from App. B, we have from Eq. (7-97)
BHZ =
1 (AR) 2
∫
∞
0
(AR) 2 1 π 1 df = 2 = 4RC 2 2 2πRC 1+ (2πRC) f
where H(0) = AR. Similarly, from Eq. (7-98)
BTZ =
1 ∞ 1 ∫0
A 1 π A df = 2 = 4RC 2 2πRC 2πRC 2 f 1+ 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) dM =
(Ipm)2M 4kBTB 2q(Ip+ID)M2+x B + R FT eq
-
q(Ip+ID) (2+x) M1+x B (Ipm)2M2 2 =0 4kBTB FT 2q(Ip+ID)M2+x B + R eq
Solving for M, 2+x
Mopt
=
4kBTBFT/Req 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
S N =
2 2+x
1 2M2 (I m) p opt 2 2+x
2q(Ip+ID)Mopt B + KB
2
xm2Ip
=
2B(2+x) [ q(Ip+ID)x]
K 1 2 2 (Ipm) q(Ip+ID)x = 2q(I +I )K p D q(Ip+ID)x B + KB
Req x/(2+x) 2/(2+x) 4kBTFT
(b) If Ip >> ID, then 2
S N =
xm2Ip
Req x/(2+x) 2/(2+x) 2/(2+x) 4kBTFT 2B(2+x) ( qx) Ip
2(1+x) 1/(2+x) ( xIp) m2 = 2Bx(2+x) 2( x q 4kBTFT/Req)
7-30. Substituting Ip = R0Pr into the S/N expression in Prob. 7-29a, 2 xm2 S = N 2B(2+x)
( R0Pr) [ q(R0Pr+ID)x]
Req x/(2+x) 2/(2+x) 4kBTFT
10 4 Ω / J (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
(0.5P + 10 )
−8 2 / 3
where Pr is in watts.
r
17
1 /3
Pr We want to plot 10 log (S/N) versus 10 log 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
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 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) -2πBt10 1 - e = 0.1 and
we have
-2πBt90 1 - e = 0.9
so that
1
-2πBt10
e
-2πBt90
= 0.9 and e
= 0.1
Then 2πBtr
e
2πB(t90-t10)
.9 = .1 = 9
=e
It follows that 2πBtr = ln 9 8-4.
or
tr =
ln 9 0.35 = B 2πB
(a) From Eq. (8-11) we have t2 1/2 1 1 exp- 2 = 2 2σ 2π σ
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 1 G(ω) = 2 G(0),
or
(2πf3dB)2 σ2 1 = exp 2 2
Using σ as defined in Eq. (8-13), we have
f3dB =
8-5.
2 ln 2 0.44 (2 ln 2)1/2 = =t π tFWHM 2πσ FWHM
From Eq. (8-9), the temporal response of the optical output from the fiber is
g(t) =
1 t2 exp- 2 2σ 2π σ
If τe is the time required for g(t) to drop to g(0)/e, then
g(τe) =
τe2 1 g(0) exp- 2 = e = 2σ 2π σ
2
1 2π σe
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 1 G(f3dB) = 2 G(0). Therefore with σ = te/(2 2 )
G(f3dB) =
1 2π
1 1 1 exp - 2(2πf3dB σ)2 = 2 2π
Solving for f3dB:
f3dB = 8-6.
2 ln 2 2πσ
=
0.53 2 ln 2 2 2 = t t 2π e e
(a) We want to evaluate Eq. (8-17) for tsys. Using Dmat = 0.07 ns/(nm-km), we have 2 2 1/2 440(7)0.7 350 2 2 2 2 tsys = (2) + (0.07) (1) (7) + 800 + 90
= 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. (b) For q = 1.0, 2 21/2 440(7) 350 2 2 tsys = (2) + (0.49) + 800 + 90
8-7.
= 5.85 ns
1 We want to plot the following 4 curves of L vs B = T
b
(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
3
:
tmat = Dmat σλ L = 0.7Tb
L=
0.7Tb Dmat σλ
=
0.7 BDmat σλ
or 104 = B (with B in Mb/s)
(c) Modal dispersion (one curve for q = 0.5 and one for q = 1) q
0.440L tmod = 800
0.7 = B
1/q
or
800 0.7 L = 0.44 B
With B in Mb/s, L = 1273/B for q =1, and L = (1273/B)2 for q = .5.
8-8.
1 We want to plot the following 3 curves of L vs B = T
b
:
(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) q
0.440L tmod = 800
0.7 = B
1/q
or
800 0.7 L = 0.44 B
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
4
8-10. Signal bits
0 0 1 1 0 1 1 1 1 0 0 1
Signal bits
Baseband (NRZ-L) data
Clock signal
Optical Manchester
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
5
8-12.
NRZ data
Freq. A
Freq. B
PSK data
8-13. Original 010 code
001 111 111 101 000 000 001 111 110
3B4B 0101 0011 1011 0100 1010 0010 1101 0011 1011 1100 encoded
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
6
(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
7
Problem Solutions for Chapter 9 9-1.
RIN limit 58
CNR
Quantum Noise limit
54
Thermal noise limit
(dB) 50
0
4
12
8
Received optical power (dBm) 9-2.
f1 Transmission system
f2
f3
∆
Triplebeat products
2-tone 3rd order
1
f4
f5
16
9-3.
The total optical modulation index is
21/2 m = ∑mi i 9-4
= [30(.03)2 + 30(.04)2 ] = 27.4 % 1 /2
120 1/2 2 The modulation index is m = ∑(.023) i=1
= 0.25
The received power is P = P0 – 2(lc) - αfL = 3 dBm - 1 dB - 12 dB = -10 dBm = 100µW The carrier power is 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,
< 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 2 iT
< >
4kBT Fe = 8.25×10-13A2 = R eq
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
or, in dB,
9-5.
C/N = 10 log 75.6 = 18.8 dB.
When an APD is used, the carrier power and the quantum noise change. The carrier power is 1 1 C = 2 (mR0MP) 2 = (15 × 10−5 A) 2 2 The quantum noise is
< i2Q>
= 2q(R0P + ID)M2F(M)B = 2q(R0P + ID)M2.7B = 4.76×10-10A2
Thus the carrier-to-noise ratio is 1 (15 × 10 −5 A)2 C 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
9-6.
32 1/2 2 (a) The modulation index is m = ∑(.044) i=1 The received power is
= 0.25
P = -10 dBm = 100µW
The carrier power is 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,
< 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>
4kBT = R Fe = 8.25×10-13A2 eq
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
or, in dB,
C/N = 10 log 75.6 = 18.8 dB.
(b) When mi = 7% per channel, the modulation index is 32 1/2 2 m = ∑(.07) i=1
= 0.396
The received power is
P = -13 dBm = 50µW
The carrier power is 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,
< i2source>
= RIN (R0P)2 B = 1.42×10-13A2
4
The quantum noise is
< i2Q>
= 2q(R0P + ID)B = 4.8×10-14A2
The thermal noise is the same as in Part (a). Thus the carrier-to-noise ratio 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
or, in dB,
9-8.
C/N = 10 log 69.6 = 18.4 dB.
Using the expression from Prob. 9-7 with ∆ντ = 0.05, fτ = 0.05, and ∆ν = f = 10 MHz, yields
RIN(f) =
=
4R1R2 ∆ν -4π∆ντ -2π∆ντ -2e cos(2πfτ) 1 + e 2 2 π f + ∆ν 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 If R1 = R2 then 10 log R1R2 = 20 log R1 < -60 dB or 10 log R1 = 10 log R2 < -30 dB
5
Problem Solutions for Chapter 10 10-1. In terms of wavelength, at a central wavelength of 1546 nm a 500-GHz channel spacing is λ2 (1546 nm )2 9 −1 ∆λ = ∆f = 500 × 10 s = 4 nm 8 c 3 × 10 m / s 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): 200 µW = 2.7 dB yields 10 log P1 Similarly, P2 = 10(log
200 −0.47)
(log P1 = 10
200 −0.27 )
= 107.4 µW
= 67.8 µW
200 = 0.58 dB (b) From Eq. (10-5): Excess loss = 10 log 107.4 + 67.8
(c)
P1 107.4 = = 61% P1 + P2 175.2
and
P2 67.8 = = 39% P1 + P2 175.2
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 1310 nm 1540 nm
A 25 50
B 50 88
10-4. From A out = s 11A in + s 12Bin Bin = −
s 21 A in s 22
and
C 75 100
and
D 90 90
E 100 50
F 0 100
Bout = s 21A in + s 22 Bin = 0 , we have
s s A out = s 11 − 12 21 A in s 22
1
Then A T = out A in
2
s s = s11 − 12 21 s 22
2
2
and
s B s s R = in = 21 ÷ s11 − 12 21 Ain s 22 s 22
2
10-5. From Eq. (10-18) 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 aj = Exit port no. Value of aj
Pi −in L /10 = 10 Ij , where the values of LIj are given in Table P10-7. P j− out 1 8.57
2 6.71
3 5.66
4 8.00
5 9.18
6 7.31
7 8.02
Then from Eq. (10-25) the excess loss is P Pin 1 10 log in = 10 log = 0.22 dB = 10 log 1 1 1 0.95 ∑ Pj Pin + + ... + an a1 a 2 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 Lcoupling = 10 log P in
= 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
7π(25)2 = 10 log 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
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 3dB coupler. Thus,
3
P0 = 10 log P0 = 0.1 dB Excess loss = 10 log P1 + P2 2P1 This yields P1 =
P0 P0 10 0.01 = 0.977 ÷ 2 2
Thus the fractional power traversing the 3-dB coupler is FT = 0.977. Then, from Eq. (10-27), log FT log 0.977 − 1 log N = −10 log − 1 log 2 n ≤ 30 Total loss = −10 log log 2 log 2 Solving for n yields n≤
−3 = 9.64 log 0.977 log 2 −1 log 2
n 9 Thus, n = 9 and N = 2 = 2 = 512
10-11. For details, see Verbeek et al., Ref. 34, p. 1012 For the general case, from Eq. (10-29) we find M 11 = cos (2κd) ⋅ cos (k∆L / 2) + j sin (k∆L / 2) M 12 = M 21 = 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 ) + sin 2 ( k∆L / 2 )]Pin,1 +[sin 2 (2κd ) ⋅ cos2 (k∆L / 2)]Pin,2
4
Pout ,2 = [sin 2 (2κd) ⋅cos 2 ( k∆L / 2)]Pin,1 +[cos 2 (2κd ) ⋅ cos2 (k∆L / 2) + sin 2 (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 = 0.4 mm and 2n eff (2∆ν)
∆L 3 =
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 =
c = 0.75 mm 2n eff (4 ∆ν)
In the second stage ∆L 2 =
c = 1.5 mm 2neff (2∆ν)
In the third stage
∆L 3 =
c = 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 cos θ dθ =
k n'Λ
dλ
dθ dλ
or
=
If φ ≈ θ, then the grating equation becomes 2 sin θ =
5
k n'Λ cos θ kλ . n'Λ
Solving this for
dθ dλ
=
k n'Λ
and substituting into the
2 sin θ λ cos θ
=
dθ dλ
equation yields
2 tan θ λ
(b) For S = 0.01, Sλ 1/2 tan θ = 2∆λ (1+m)
0.01(1350) = 2(26)(1+ 3)
1/ 2
= 0.2548
or θ = 14.3° 10-15. For 93% reflectivity 2
R = tanh (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=
λ uv
244 nm 244 nm = 523 nm θ = 2 sin(13.5°) = 2(0.2334) 2 sin 2
(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), −4 π δn η π (2.5 × 10 )(0.827) κ= = = 4.2 cm −1 1.547 × 10−4 cm λ Bragg
(d) From Eq. (10-49), ∆λ =
(1.547 µm )2 π (1.48) 500 µm
6
[(2.1)
2
+ π2 ]
1 /2
= 3.9 nm
2
2
2
(e) From Eq. (10-48), R max = tanh (κL) = tanh (2.1) = (0.97) = 94%
10-19. Derivation of Eq. (10-55).
10-20. (a) From Eq. (10-54), ∆L = m
1.554 µm λ0 = 118 = 126.4 µm nc 1.451
(b) From Eq. (10-57), ∆ν =
=
∆λ =
x n s cd n c L f mλ2 n g 1.453 (3 × 10 8 m / s)(25 × 10 −6 m) 1.451 25 µm = 100.5 GHz 118 (1.554 × 10 −6 m)2 1.475 9.36 × 103 µm (1.554 × 10 −6 m) 2 λ2 100.5 GHz = 0.81 nm ∆ν = c 3 × 108 m / s
(c) From Eq. (10-60), ∆νFSR = Then
c 3 × 108 m / s = = 1609 GHz n g ∆L 1.475(126.4 µm) (1.554 × 10 −6 m)2 λ2 1609 GHz = 12.95 nm ∆λ = ∆νFSR = 8 c 3 × 10 m / s
(d) Using the conditions sin θ i ≈ θ i =
2(25 µm) = 5.33 × 10 −3 radians 9380 µm
and sin θ o ≈ θo = 21.3 × 10
−3
radians
7
then from Eq. (10-59), ∆νFSR ≈
=
c ng [∆L + d(θ i + θ o )]
1.475[(126.4 × 10−6
3 × 10 8 m / s = 1601 GHz m) + (25 × 10 −6 m)( 5.33 + 21.3) × 10−3 ]
10-21. The source spectral width is ∆λ signal =
λ2 ν (1550 nm )2 (1.25 × 10 9 s −1 ) = = 1 × 10 −2 nm (3 × 10 8 m / s)(109 nm / m ) c
Then from Eq. (10-61) ∆λ tune = λ
∆n eff = (1550 nm )(0.5%) = 7.75 nm n eff
Thus, from Eq. (10-63) N=
7.75 nm ∆λ tune = = 77 10 λ signal 10(0.01 nm )
10-22. (a) From Eq. (10-64), the grating period is Λ=
λ Bragg 1550 nm = = 242.2 nm 2neff 2(3.2)
(b) Again, from the grating equation, ∆Λ =
2.0 nm ∆λ = = 0.3 nm 2n eff 2(3.2 )
10-23. (a) From Eq. (10-43) λ2 1 ∆L = = = 4.0 mm 2n eff ∆ν ∆λ 2neff c
(b) ∆L eff = ∆n eff L implies that ∆ n eff =
8
4 mm = 0.04 = 4% 100 mm
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 ∆n va ∆n = c λ
Thus we have Wavelength (nm) Acoustic frequency (MHz)
1300 56.69
1546 47.67
1550 47.55
(b) The sensitivity is (4 nm)/(0.12 MHz) = 0.033 nm/kHz
9
1554 47.43
Problem Solutions for Chapter 11
11-1. (a) From Eq. (11-2), the pumping rate is Rp =
I 100 mA = −19 qwdL (1.6 × 10 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 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 (c) From Eq. (11-7), the saturation photon density is N ph;sat =
0.3 (1× 10
−20
1 = 1.67 × 1015 photons/ cm 3 8 m )(2 × 10 m / s )(1 ns) 2
(d) From Eq. (11-4), the photon density is N ph =
Pin λ = 1.32 × 1010 photons / cm 3 vg hc (wd )
11-2. Carrying out the integrals in Eq. (11-14) yields g 0 L = ln
P(L) P(L) − P(0) + P(0) Pamp,sat
Then with P(0) = Pin, P(L) = Pout, G = Pout/Pin, and G 0 = exp(g 0 L ) from Eq. (1110), we have ln G 0 = g 0 L = ln G +
GPin P P − 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).
11-3. Plots of amplifier gains. 1
11-4. Let G = G0/2 and Pin = Pout / G = 2Pout,sat / G 0 . Then Eq. (11-15) yields GP G0 = 1+ 0 amp.sat ln 2 2 2Pout.sat Solving for Pout,sat and with G0 >> 1, we have Pout .sat = 11-5.
G 0 ln 2 P ≈ (ln 2) Pamp .sat = 0.693 Pamp.sat (G0 − 2) amp.sat
From Eq. (11-10), at half the amplifier gain we have G=
1 1 G 0 = exp(g 0 L ) = exp(gL) 2 2
Taking the logarithm and substituting into the equation given in the problem, g = g0 −
1 g0 ln 2 = 2 2 L 1 + 4(ν3dB − ν0 ) / (∆ν)
From this we can find that 2(ν3dB − ν 0 ) g0 = − 1 1 ∆ν g0 − ln 2 L
1/ 2
1 = g 0L / ln 2 − 1
1/ 2
=
1 G0 log 2 2
1/ 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 ) ln = 2 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
ln G0 FWHM = 2 λ − λ 0 = 2 2 ln ln G ln 30 = 2 2 ln ln 27
1/ 2
∆λ
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 ≤
PCE ≤
λp λs
=
980 = 63.4% for 980-nm pumping, and by 1545
λ p 1475 = = 95.5% for 1475-nm pumping λ s 1545
11-8. (a) 27 dBm = 501 mW and 2 dBm = 1.6 mW. Thus the gain is G = 10 log
501 10 log 313 = 25 dB 1.6 =
(b) From Eq. (11-19), 313 ≤ 1 +
Pp,in ≥
980 Pp,in . With a 1.6-mW input signal, the pump power needed is 1542 Ps, in
312(1542 ) (1.6 mW ) = 785 mW 980
11-9. (a) Noise terms: From Eq. (6-17), the thermal noise term is 4 (1.38 × 10 −23 J / K)(293 K ) 4k B T B= 1 GHz = 1.62 × 10−14 A2 σ = RL 1000 Ω 2 T
From Eq. (11-26), we have
3
σ 2shot− s = 2qR GPs, in B = 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 σ 2shot− ASE = 2qR SASE ∆ν opt B = 2qR
hc n spG∆ν opt B λ
= 2(1.6 × 10−19 C)(.73 A / W)(6.626 × 10−34 J/ K ) ×(3 × 10 8 m / s )2(100)(3.77 THz)(1 GHz)/ 1550nm = 2.26 × 10−14 A2 From Eq. (11-27) and (11-24), we have σ 2s− ASE = 4[(0.73 A/ W )(100)(1 µW )] 6.626 × 10 −34 J / K)(3 × 108 m / s ) ( × (.73 A/ W ) 2(100)(1 GHz) 1550nm = 5.47 × 10 −12 A 2
From Eq. (11-28), we have σ
2 ASE − ASE
(6.626 × 10 −34 J/ K )(3 × 10 8 m / s ) 2(100) = (.73A / W ) 1550nm 2
× [2(1 THz) − 1 GHz ](1 GHz) = 7.01× 10 −13 A 2 4
2
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
1 = L
=
L
∫ 0
P P(z) dz = in L
L
∫
e − αz dz
0
1 P G − 1 Pin P 1 − e −αL ] = in 1 − = in [ αL αL G G ln G
since ln G = αL from the transparency condition. (b) From Eq. (11-35) and using Eq. (11-24), PASE
path
NPASE = L
=
L
∫
e − αz dz =
0
NPASE αL
(1 − e ) −αL
1 1 α (NL) αL tot PASE 1− = 2 2 hνn sp (G − 1)∆νopt 1 − (αL) (ln G) G G
= αL tot hνnsp ∆ν opt
1 G − 1 2 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 cos θi =
e jθ i + e − jθ i , so that 2
jθ − jθ N e i +e i P = E i (t)E *i (t) = ∑ 2Pi 2 i =1
jθ − jθ N e k +e k × 2P k ∑ 2 k =1
1 N N 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 ∑ ∑ 4 i =1 k=1
]
1 N N − j( θ − θ ) − j( θ +θ ) j (θ −θ ) j( θ + θ ) = ∑ ∑ 2Pi 2Pk e i k + e i k + e i k + e i k 4 i =1 k=1
]
=
[
[
N
1 2
= ∑ Pi + i=1
N
N
∑∑
[
2Pi 2Pk e
i =1 k≠ i
j(θ i −θ k )
+e
− j(θ i − θ k )
]
where the last two terms in the second-last line drop out because they are beyond the response frequency of the detector. Thus, N
N
N
i=1
i=1 k ≠i
P = ∑ Pi + ∑ ∑ 2 Pi Pk [cos(θ i − θ k )] 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 Q=
AP
(HP + D )
2 1 /2
+D
where we have defined the following terms for simplicity A = 2R G 2 H = 4qR GB + 8R GSASE B and
2 2 D = σoff
Rearrange terms in the equation for Q to get Q (HP + D 2
)
2 1/ 2
= AP − QD
Squaring both sides and solving for P yields P =
2QD Q 2H + 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
F=
1 + 2ηn sp (G − 1) ηG
8
Problem Solutions for Chapter 12 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) (.95) (.9)] = −50 log (0.485) = 15.7 dBm 2
2
12-2. (a) Including a power margin, we have from Eq. (12-16) PS − PR − power m arg in = L excess + α(2L) + 2L c + 10log N Thus 0 – (-38 dBm) – 6 dB = 3 dB + (0.3 dB/km)2(2 km) + 2(1.0 dB) + 10 log N
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
B C
A
D
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
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. 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:
M
d
N
(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:
3
= Md
N −1
M −1
i =1
j=1
∑ i + Nd
∑j=
M
M(M − 1) N(N − 1) d+N d = 2 2
MN (M + N − 2)d 2
12-5. (a) Let the star be located at the relative position (m,n). Then M− m n−1 N−n m−1 L = N ∑ j + N ∑ j + M∑ i + M ∑ i d j=1 j =1 i =1 i =1
n(n − 1) (N − n)(N − n + 1) m(m − 1) (M − m)(M − m + 1) = N + + M + d 2 2 2 2 MN (M + N + 2) − Nm(M − m + 1) − Mn(N − n + 1) d = 2
(b) When the star coupler is located in one corner of the grid, then m = n= 1, so that the expression in (a) becomes MN MN (M + N + 2) − NM − MN d = (M + N − 2)d L= 2 2 (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: dL = N(m - 1 - M) + Nm = 0 dm
so that
m=
yields
n=
M +1 2
Similarly dL = M(n - 1 - N) + nM = 0 dn
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.
4
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 − 2L c
(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
1
17
2
P1 / 10
(mW)
Ptotal(mW)
Ptotal(dBm)
50
50
17
14
25
50
17
3
12.2
16.6
49.8
17
4
11
12.6
50.4
17
5
10
10
50
17
5
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 Node 3: add/drop wavelengths 1, 2, and 3 Node 4: add/drop wavelengths 1, 4, and 5 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) = 2.17 2(3 − 1)[2(32 ) − 1]
(d) From Eq. (12-21) we have 2(3) C= = 8.27 2.17 3
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
6
p2
2 . . . k–1
pk-1
k
Pk - 1
k+1
Pk - p
k+2
Pk - p2
. . . Pk - pk-1
2k – 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
7
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
8
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 eff λ3 D 2 = 11.0 mW n 2 c Ts
(b) From Eq. (12-49) the dispersion length is Ldisp = 43 km (c) From Eq. (12-51) the soliton period is L period =
π L = 67.5 km 2 disp
(d) From Eq. (12-50) the peak power for 30-ps pulses is
9
Ppeak =
1.7627 2π
2
A eff λ3 D = 3.1 mW n 2 c Ts2
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, L amp ≤
1 L = 40 km 2 coll
12-24. From the equation and conditions given in Prob. 12-23, we have that ∆λ max =
Ts 20 ps = = 2 nm DL amp [0.4 ps /(nm ⋅ km)](25 km)
Thus 2.0/0.4 = 5 wavelength channels can be accommodated.
12-25 Plot from Figure 3 of Ref. 103.
10
Problem Solutions for Chapter 13 13-1
(a) From the given equation, nair = 1.000273. Thus, λ vacuum = λ air n air = 1.000273(1550.0 nm) = 1550.42 nm (b) From the given equation, n (T, P) = 1 +
(1.000273 − 1)(0.00138823)640 = 1.000243 1+ 0.003671(0)
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=
10 V 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 α= 13-3
10 3.78 log = 0.31 dB / km 1895 − 2 3.31
(a) From Eq. (13-1) α=
10 P 10 V 10 log e V log N = log N = ln N LN − LF PF L N − L F VF L N − LF VF
From this we find ∆α =
4.343 10 log e ∆ VN ∆VF 8.686 + = ± ± = ± × 10 −3 ( 0.1% 0.1% ) − − − L N L F VN VF L N L F LN L F
(b) If ∆α = 0.05 dB/km, then L = LN − LF ≥
8.868 × 10 −3 km = 176 m 0.05
1
13-4
(a) From Eq. (8-11) we have t2 1/2 1 1 exp- 2 = 2 2σ 2π σ
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 1 G(ω) = 2 G(0),
(2πf3dB)2 σ2 1 = exp 2 2
or
Using σ as defined in Eq. (8-13), we have
f3dB = 13-5
2 ln 2 0.44 (2 ln 2)1/2 = =t π tFWHM 2πσ FWHM
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) Pout (f) / Pin (f) H(f) = = = H(f) P(0) Pout (0) / Pin (0) H(0) The following table gives some representative values of H(f) for different values of 2σ: f (MHz) 100 200 300 500 700 1000
13-6
2σ σ = 2 ns 0.821 0.454 0.169 0.0072
2σ σ = 1 ns 0.952 0.821 0.641 0.291 0.089 0.0072
2σ σ = 0.5 ns 0.988 0.952 0.895 0.735 0.546 0.291
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=
1 ∆τ 1 400 ps = = 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 S0 = 13-7
2 ps /(nm ⋅ km) D(λ ) 2 = = 0.074 ps /(nm ⋅ km) λ − λ 0 (1575 − 1548) nm
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 ≈ 26
(1 ps)2 0.5(1− 0.5) 6.5 10−4 = × 2 (100 ps)
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 x −βx P(x) = P(0)exp −β ∫ dy = P(0)e 0
Writing this as exp(−βx) = P(0)/ P(x) and taking the logarithm on both sides yields βx log e = log αx = 10 log
P(0) . Since α = β(10 log e), this becomes P(x)
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).
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 From Eq. (2-23), the acceptance angle is NA sin θ ≈ θ = , so that n
θ2 ( NA) 2 S= = 4 4n2
13-11 The attenuation is found from the slope of the curve, by using Eq. (13-22): PD (x1 ) 70 PD (x 2 ) 10 log 28 = = 4.0 dB / km 2(x 2 − x1 ) 2(0.5 km )
10 log Fiber a: α =
25 11 = 3.6 dB/ km Fiber b: α = 2(0.5 km ) 10 log
7 1.8 = 5.9 dB/ km Fiber c: α = 2(0.5 km ) 10 log
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: L splice = 10 log
P2 25 = 10 log = −0.5 dB P1 28
For splice 2: L splice = 10 log
7 = −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 W z Ps (L ) = ∫ Sαs P0 exp −2α L + dz 2 0
=S
αs 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 1 −αW 1 − e )≈ [1 − (1 − αW )]= W ( 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. The OTDR dynamic range D is D = SNR + αL + splice loss = 4.5 dB + 16.5 dB + 0.5 dB = 21.5 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. (1323): dL =
c 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 dt =
2n 2(1.5) dL ≤ (0.5 m) = 5 ns 8 c 3 × 10 m / s
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.
6
