1 Problem Solutions for Chapter 2 2-1. E = 100 cos 2 π 10 8 t 30° ( e x 20 cos 2π 10 8 t - 50° ( e y 40cos 2π 10 8 t 210° ( e z 2-2. The general form is: y = (amplitude) cos(ϖ t - 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) ϖ = 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. x 1 = a 1 cos (ϖ t - δ 1 ) and x 2 = a 2 cos (ϖ t - δ 2 ) Adding x 1 and x 2 yields x 1 + x 2 = a 1 [cos ϖ t cos δ 1 + sin ϖ t sin δ 1 ] + a 2 [cos ϖ t cos δ 2 + sin ϖ t sin δ 2 ] = [a 1 cos δ 1 + a 2 cos δ 2 ] cos ϖ t + [a 1 sin δ 1 + a 2 sin δ 2 ] sin ϖ t Since the a's and the δ's are constants, we can set a 1 cos δ 1 + a 2 cos δ 2 = A cos φ (1)
116
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Problem Solutions for Chapter 2 - SubodhTripathi · Problem Solutions for Chapter 2 ... 3 sin 2 (ωt - kz) = [1 ... fiber = )) = [] ...
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1
Problem Solutions for Chapter 2
2-1.
E = 100cos 2π108 t + 30°( ) ex + 20 cos 2π108t − 50°( ) ey
+ 40cos 2π108 t + 210°( ) ez
2-2. The general form is:
y = (amplitude) cos(ωt - kz) = A cos [2π(νt - z/λ)]. Therefore
(a) amplitude = 8 µm
(b) wavelength: 1/λ = 0.8 µm-1 so that λ = 1.25 µm
(c) ω = 2πν = 2π(2) = 4π
(d) At t = 0 and z = 4 µm we have
y = 8 cos [2π(-0.8 µm-1)(4 µm)]
= 8 cos [2π(-3.2)] = 2.472
2-3. For E in electron volts and λ in µm we have E = 1.240
λ
(a) At 0.82 µm, E = 1.240/0.82 = 1.512 eV
At 1.32 µm, E = 1.240/1.32 = 0.939 eV
At 1.55 µm, E = 1.240/1.55 = 0.800 eV
(b) At 0.82 µm, k = 2π/λ = 7.662 µm-1
At 1.32 µm, k = 2π/λ = 4.760 µm-1
At 1.55 µm, k = 2π/λ = 4.054 µm-1
2-4. x1 = a1 cos (ωt - δ1) and x2 = a2 cos (ωt - δ2)
Adding x1 and x2 yields
x1 + x2 = a1 [cos ωt cos δ1 + sin ωt sin δ1]
+ a2 [cos ωt cos δ2 + sin ωt sin δ2]
= [a1 cos δ1 + a2 cos δ2] cos ωt + [a1 sin δ1 + a2 sin δ2] sin ωt
Since the a's and the δ's are constants, we can set
a1 cos δ1 + a2 cos δ2 = A cos φ (1)
2
a1 sin δ1 + a2 sin δ2 = A sin φ (2)
provided that constant values of A and φ exist which satisfy these equations. To
verify this, first square both sides and add:
A2 (sin2 φ + cos2 φ) = a12 sin2 δ1 + cos2 δ1( )
+ a22 sin2 δ2 + cos2 δ2( ) + 2a1a2 (sin δ1 sin δ2 + cos δ1 cos δ2)
or
A2 = a12 + a2
2 + 2a1a2 cos (δ1 - δ2)
Dividing (2) by (1) gives
tan φ = a1 sinδ1 + a2 sinδ2
a1 cosδ1 + a2 cosδ2
Thus we can write
x = x1 + x2 = A cos φ cos ωt + A sin φ sin ωt = A cos(ωt - φ)
2-5. First expand Eq. (2-3) as
Ey
E0 y
= cos (ωt - kz) cos δ - sin (ωt - kz) sin δ (2.5-1)
Subtract from this the expression
E x
E0 x
cos δ = cos (ωt - kz) cos δ
to yield
Ey
E0 y
-Ex
E0x
cos δ = - sin (ωt - kz) sin δ (2.5-2)
Using the relation cos2 α + sin2 α = 1, we use Eq. (2-2) to write
3
sin2 (ωt - kz) = [1 - cos2 (ωt - kz)] = 1 −Ex
E0x
2
(2.5-3)
Squaring both sides of Eq. (2.5-2) and substituting it into Eq. (2.5-3) yields
Ey
E0 y
− Ex
E0x
cosδ
2
= 1 −Ex
E0x
2
sin2 δ
Expanding the left-hand side and rearranging terms yields
Ex
E0x
2
+ Ey
E0y
2
- 2 Ex
E0x
Ey
E0y
cos δ = sin2 δ
2-6. Plot of Eq. (2-7).
2-7. Linearly polarized wave.
2-8.
33 ° 33 °
90 °Glass
Air: n = 1.0
(a) Apply Snell's law
n1 cos θ1 = n2 cos θ2
where n1 = 1, θ1 = 33°, and θ2 = 90° - 33° = 57°
∴ n2 =
cos 33°cos 57°
= 1.540
(b) The critical angle is found from
nglass sin φglass = nair sin φair
4
with φair = 90° and nair = 1.0
∴ φcritical = arcsin 1
n glass
= arcsin 1
1.540= 40.5°
2-9Air
Water
12 cm
r
θ
Find θc from Snell's law n1 sin θ1 = n2 sin θc = 1
When n2 = 1.33, then θc = 48.75°
Find r from tan θc =
r12 cm
, which yields r = 13.7 cm.
2-10.
45 °
Using Snell's law nglass sin θc = nalcohol sin 90°
where θc = 45° we have
nglass =
1.45sin 45°
= 2.05
2-11. (a) Use either NA = n12 − n2
2( )1/ 2= 0.242
or
5
NA ≈ n1 2∆ = n12(n1 − n2)
n1
= 0.243
(b) θ0,max = arcsin (NA/n) = arcsin 0.2421.0
= 14°
2-13. NA = n12 − n2
2( )1/ 2= n1
2 − n12(1− ∆)2[ ]1/ 2
= n1 2∆ − ∆2( )1 / 2
Since ∆ << 1, ∆2 << ∆; ∴ NA ≈ n1 2∆
2-14. (a) Solve Eq. (2-34a) for jHφ:
jHφ = j εωβ
Er - 1βr
∂Hz
∂φSubstituting into Eq. (2-33b) we have
j β Er + ∂Ez
∂r= ωµ j
εωβ
Er −1
βr∂Hz
∂φ
Solve for Er and let q2 = ω2εµ - β2 to obtain Eq. (2-35a).
(b) Solve Eq. (2-34b) for jHr:
jHr = -j εωβ
Eφ - 1β
∂Hz
∂rSubstituting into Eq. (2-33a) we have
j β Eφ + 1r
∂Ez
∂φ= -ωµ − j
εωβ
Eφ −1
β∂Hz
∂r
Solve for Eφ and let q2 = ω2εµ - β2 to obtain Eq. (2-35b).
(c) Solve Eq. (2-34a) for jEr:
6
jEr = 1
εω1r
∂Hz
∂φ+ jrβHφ
Substituting into Eq. (2-33b) we have
β
εω1r
∂Hz
∂φ+ jrβHφ
+
∂Ez
∂r= jωµ Hφ
Solve for Hφ and let q2 = ω2εµ - β2 to obtain Eq. (2-35d).
(d) Solve Eq. (2-34b) for jEφ
jEφ = - 1
εωjβHr +
∂Hz
∂r
Substituting into Eq. (2-33a) we have
1r
∂Ez
∂φ-
βεω
jβHr +∂Hz
∂r
= -jωµ Hr
Solve for Hr to obtain Eq. (2-35c).
(e) Substitute Eqs. (2-35c) and (2-35d) into Eq. (2-34c)
-j
q 2
1r
∂∂r
β∂Hz
∂φ+ εωr
∂Ez
∂r
−
∂∂φ
β∂Hz
∂r−
εωr
∂Ez
∂φ
= jεωEz
Upon differentiating and multiplying by jq2/εω we obtain Eq. (2-36).
(f) Substitute Eqs. (2-35a) and (2-35b) into Eq. (2-33c)
-j
q2
1r
∂∂r
β∂Ez
∂φ− µω r
∂Hz
∂r
−
∂∂φ
β∂Ez
∂r+
µωr
∂Hz
∂φ
= -jµωHz
Upon differentiating and multiplying by jq2/εω we obtain Eq. (2-37).
2-15. For ν = 0, from Eqs. (2-42) and (2-43) we have
7
Ez = AJ0(ur) e j(ωt − βz) and Hz = BJ0(ur) e j(ωt − βz)
We want to find the coefficients A and B. From Eqs. (2-47) and (2-51),
respectively, we have
C = Jν (ua)
K ν (wa) A and D =
Jν (ua)K ν (wa)
B
Substitute these into Eq. (2-50) to find B in terms of A:
A jβνa
1u2 +
1w2
= Bωµ
J' ν (ua)uJ ν (ua)
+K' ν (wa)wKν(wa)
For ν = 0, the right-hand side must be zero. Also for ν = 0, either Eq. (2-55a) or (2-56a)
holds. Suppose Eq. (2-56a) holds, so that the term in square brackets on the right-hand
side in the above equation is not zero. Then we must have that B = 0, which from Eq. (2-
43) means that Hz = 0. Thus Eq. (2-56) corresponds to TM0m modes.
For the other case, substitute Eqs. (2-47) and (2-51) into Eq. (2-52):
0 = 1
u 2 Bjβνa
J ν(ua) + Aωε 1uJ'ν (ua)
+ 1
w2 Bjβνa
Jν(ua) + Aωε2wK' ν (wa)Jν(ua)
K ν(wa)
With k12 = ω2µε1 and k2
2 = ω2µε2 rewrite this as
Bν =
ja
βωµ
11
u 2+ 1
w2
k12Jν + k2
2K ν[ ] A
8
where Jν and Kν are defined in Eq. (2-54). If for ν = 0 the term in square brackets on the
right-hand side is non-zero, that is, if Eq. (2-56a) does not hold, then we must have that A
= 0, which from Eq. (2-42) means that Ez = 0. Thus Eq. (2-55) corresponds to TE0m
modes.
2-16. From Eq. (2-23) we have
∆ = n 1
2 − n22
2n12 =
12
1 − n22
n12
∆ << 1 implies n1 ≈ n2
Thus using Eq. (2-46), which states that n2k = k2 ≤ β ≤ k1 = n1k, we have
n22k2
= k22 ≈ n1
2k2= k1
2 ≈ β2
2-17.
2-18. (a) From Eqs. (2-59) and (2-61) we have
M ≈2π2a2
λ2 n12 − n2
2( )=2π2a2
λ2 NA( )2
a =M2π
1/ 2 λNA
=1000
2
1/ 2 0.85µm0.2π
= 30.25µm
Therefore, D = 2a =60.5 µm
(b) M = 2π2 30.25µm( )2
1.32µm( )2 0.2( )2 = 414
(c) At 1550 nm, M = 300
2-19. From Eq. (2-58),
9
V =
2π (25 µm)0.82 µm
(1.48)2 − (1.46)2[ ]1/ 2= 46.5
Using Eq. (2-61) M ≈ V2/2 =1081 at 820 nm.
Similarly, M = 417 at 1320 nm and M = 303 at 1550 nm. From Eq. (2-72)
Pclad
P
total
≈ 43
M-1/2 = 4 ×100%3 1080
= 4.1%
at 820 nm. Similarly, (Pclad/P)total = 6.6% at 1320 nm and 7.8% at 1550 nm.
2-20 (a) At 1320 nm we have from Eqs. (2-23) and (2-57) that V = 25 and M = 312.
(b) From Eq. (2-72) the power flow in the cladding is 7.5%.
2-21. (a) For single-mode operation, we need V ≤ 2.40.
Solving Eq. (2-58) for the core radius a
a = Vλ2π
n12 − n2
2( )−1/ 2=
2.40(1.32µm)
2π (1.480)2 − (1.478)2[ ]1/ 2 = 6.55 µm
(b) From Eq. (2-23)
NA = n12 − n2
2( )1/ 2= (1.480)2 − (1.478)2[ ]1/ 2
= 0.077
(c) From Eq. (2-23), NA = n sin θ0,max. When n = 1.0 then
θ0,max = arcsin NAn
= arcsin
0.0771.0
= 4.4°
2-22. n2 = n12 − NA2
= (1.458)2 − (0.3)2= 1.427
a = λV
2πNA=
(1.30)(75)2π(0.3)
= 52 µm
10
2-23. For small values of ∆ we can write V ≈ 2πa
λ n1 2∆
For a = 5 µm we have ∆ ≈ 0.002, so that at 0.82 µm
V ≈
2π (5 µm)0.82 µm
1.45 2(0.002) = 3.514
Thus the fiber is no longer single-mode. From Figs. 2-18 and 2-19 we see that the LP01
and the LP11 modes exist in the fiber at 0.82 µm.
2-24.
2-25. From Eq. (2-77) Lp = 2πβ
= λ
n y − nx
For Lp = 10 cm ny - nx =
1.3 ×10−6 m
10−1 m
= 1.3×10-5
For Lp = 2 m ny - nx =
1.3 ×10−6 m
2 m= 6.5×10-7
Thus
6.5×10-7 ≤ ny - nx ≤ 1.3×10-5
2-26. We want to plot n(r) from n2 to n1. From Eq. (2-78)
3-16. The time delay between the highest and lowest order modes can be found from the
travel time difference between the two rays shown here.
The travel time of each ray is given by
sin φ = xs
= n 2
n1
= n 1(1 − ∆)
n1
= (1 - ∆)
The travel time of the highest order ray is thus
Tmax =
n1
c s
Lx
=
n1Lc
11 − ∆
For the axial ray the travel time is Tmin = Ln1
c
ϕθa
s
x
6
Therefore
Tmin - Tmax = Ln1
c1
1 − ∆− 1
=
Ln1
c∆
1 − ∆≈
Ln1∆c
3-17. Since n2 = n1 1 − ∆( ), we can rewrite the equation as
σmod
L=
n1∆c
1 −πV
where the first tern is Equation (3-30). The difference is then given by the factor
1 −πV
= 1 −πλ2a
1
n12 − n2
2( )1/ 2 ≈ 1 −πλ2a
1n1 2∆
At 1300 nm this factor is 1 −π(1.3)2 62.5( )
11.48 2(0.015)
=1 − 0.127 = 0.873
3-18. For ε = 0 and in the limit of α → ∞ we have
C1 = 1, C2 = 32
, α
α + 1= 1,
α + 23α + 2
= 13
, α +1
2α +1= 1
2,
and (α + 1)2
(5α + 2)(3α + 2)=
115
Thus Eq. (3-41) becomes
σ int er mod al = Ln1∆2 3c
1+ 3∆ +125
∆2
1/ 2
≈ Ln1∆2 3c
3-19. For ε = 0 we have that α = 2(1 - 65
∆). Thus C1 and C2 in Eq. (3-42) become
(ignoring small terms such as ∆3, ∆4, ...)
C1 = α − 2
α + 2
=2 1 − 6
5∆
− 2
2 1 − 65
∆
+ 2
=− 3
5∆
1− 35
∆≈ − 3
5∆ 1+ 3
5∆
7
C2 = 3α − 2
2(α + 2)=
3 2 1 −65
∆
− 2
2 2 1− 65
∆
+ 2
=1 −
95
∆
2 1 − 35
∆
Evaluating the factors in Eq. (3-41) yields:
(a) C12 ≈
925
∆2
(b)4C1C2 (α + 1)∆
2α +1=
4∆ −35
∆
1 −
95
∆
2 1−
65
∆
+ 1
1 − 35
∆
2 1− 3
5∆
4 1 − 6
5∆
+ 1
= −
18∆1 −
11∆ +
1825
5 15
∆
2 2425
∆
1825
∆2
(c) 16∆2C2
2 (α + 1)2
(5α + 2)(3α + 2)
= 16∆2 1−
95
∆
2
2(1 −65
∆) +1
2
4 1 − 35
∆
2
10(1 − 65
∆) + 2
6(1 − 65
∆) + 2
= 16∆2 1−
95
∆
2
9 1 −45
∆
2
96(1 − ∆)(1 − 910
∆)4 1− 35
∆
2 ≈ 924
∆2
Therefore,
σ int er mod al = Ln1∆
2c
αα +1
α + 23α + 2
1/ 2 925
∆2 − 1825
∆2 + 924
∆2
1/ 2
= Ln1∆
2
2c
2(1 −65
∆)
2(1 − 65
∆) +1
2(1−65
∆) + 2
6(1− 65
∆) + 2
1/ 2
310 6
≈ n1∆
2 L20 3c
8
3-20. We want to plot Eq. (3-30) as a function of σλ , where σ int er mod al and
σ int ra mod al are given by Eqs. (3-41) and (3-45). For ε = 0 and α = 2, we have C1 =
0 and C2 = 1/2. Since σ int er mod al does not vary with σλ , we have
σ int er mod al
L=
N1∆2c
αα + 2
α + 23α + 2
1/ 2 4∆C2(α + 1)(5α + 2)(3α + 2)
= 0.070 ns/km
With C1 = 0 we have from Eq. (3-45)
σ int ra mod al = 1c
σλ
λ−λ2 d2n1
dλ2
=
0.098σλ ns / km at 850 nm
1.026 ×10−2 σλ ns / km at 1300 nm
3-21. Using the same parameter values as in Prob. 3-18, except with ∆ = 0.001, we havefrom Eq. (3-41) σ int er mod al /L = 7 ps/km, and from Eq. (3.45)
σ int ra mod al
L=
0.098σλ ns / km at 850 nm
0.0103σλ ns / km at 1300 nm
The plot of σL
= 1L
σ int er2 + σ int ra
2( )1/ 2vs σλ :
3-22. Substituting Eq. (3-34) into Eq. (3-33)
τg = Lc
dβdk
= Lc
12β
2kn12 + 2k2n1
dn1
dk
- 2α + 2
αma 2
αα+ 2 2
α + 2n1
2k2∆( )2
α+2−1
× ∆ 2k2n1
dn 1
dk+ 2kn1
2 +n1
2k2
∆d∆dk
= Lc
kn1
βN1 −
4∆α + 2
α + 2
αma2
1n1
2k2∆
αα +2
N1 +n1k2∆
d∆dk
9
= LN1
ckn1
β1−
4∆α + 2
mM
αα +2
1 +ε4
with N1 = n1 + k dn1
dkand where M is given by Eq. (2-97) and ε is defined in Eq.
(3-36b).
3-23. From Eq. (3-39), ignoring terms of order ∆2,
λdτdλ
= Lc
dN1
dλ1 +
α − ε − 2
α + 2mM
αα+ 2
+ LN1
cα − ε − 2
α + 2d
dλ∆
mM
αα +2
where
N1 = n1 - λ dn1
dλand M =
αα + 2
a2k2n12 ∆
(a)dN1
dλ=
ddλ
n1 − λdn1
dλ
= - λ
d2n1
dλ2
Thus ignoring the term involving ∆ d2n1
dλ2 , the first term in square brackets
becomes - Lc
λ2 d2n1
dλ2
(b)d
dλ∆
mM
αα+ 2
= m
αα+ 2 d∆
dλ1M
αα +2
+ ∆−α
α + 2dMdλ
1M
αα +2
+1
(c)dMdλ
= α
α + 2a2
ddλ
k2n12 ∆( )
= α
α + 2 a2
d∆dλ
k2n12 + 2k2∆n1
dn1
dλ+ 2kn1
2∆dkdλ
10
Ignoring d∆dλ
and dn1
dλterms yields
dMdλ
= 2α
α + 2a2k2n1
2 ∆ −1
λ
= -
2Mλ
so that
ddλ
∆mM
αα+ 2
=
∆λ
2αα + 2
mM
αα +2
. Therefore
λdτdλ
= - Lc
λ2 d2n1
dλ2 + LN1
cα − ε − 2
α + 22α∆α + 2
mM
αα +2
3-24. Let a = λ2 d2n1
dλ2 ; b = N1C1∆ 2α
α + 2; γ =
αα + 2
Then from Eqs. (3-32), (3-43), and (3-44) we have
σ int ra mod al2 =
L2 σλ
λ
2 1M
λ dτg
dλ
m= 0
M
∑2
=
Lc
2 σλ
λ
21M
−a + b mM
γ
m= 0
M
∑2
≈ Lc
2 σλ
λ
21M
−a + bmM
γ
2
dm0
M
∫
=
Lc
2 σλ
λ
2
1M
a 2 − 2ab mM
γ
+ b2 mM
2γ
dm0
M
∫
= Lc
2 σλ
λ
2
a2 −2ab
γ + 1+
b2
2γ + 1
= Lc
2 σλ
λ
2
−λ2 d2n1
dλ2
2
11
- 2 λ2 d2n1
dλ2
N1C1∆
αα + 1
+ (N1C1∆)2 4α2
(α + 2)(3α + 2)
3-25. Plot of Eq. (3-57).
3-26. (a) D = λ − λ0( )S0 = −50(0.07) = −3.5 ps /(nm − km)
4-8. The 3-dB optical bandwidth is found from Eq. (4-21). It is the frequency f at
which the expression is equal to -3; that is,
10 log1
1 + 2πfτ( )2[ ]1/ 2
= −3
4
With a 5-ns lifetime, we find
f =1
2π 5 ns( )100.6 − 1( )= 9.5 MHz
4-9. (a) Using Eq. (4-28) with Γ = 1
gth =
10.05 cm
ln
1
0.32
2 + 10 cm-1 = 55.6 cm-1
(b) With R1 = 0.9 and R2 = 0.32,
gth = 1
0.05 cm ln
1
0.9(0.32) + 10 cm-1 = 34.9 cm-1
(c) From Eq. (4-37) ηext = ηi (gth - α )/gth ;
thus for case (a): ηext = 0.65(55.6 - 10)/55.6 = 0.53
For case (b): ηext = 0.65(34.9 - 10)/34.9 = 0.46
4-10. Using Eq. (4-4) to find Eg and Eq. (4-3) to find λ, we have for x = 0.03,
λ = 1.24Eg
= 1.24
1.424 + 1.266(0.3) + 0.266(0.3)2 = 1.462 µm
From Eq. (4-38)
ηext = 0.8065 λ(µm) dP(mW)dI(mA)
Taking dI/dP = 0.5 mW/mA, we have ηext = 0.8065 (1.462)(0.5) = 0.590
4-11. (a) From the given values, D = 0.74, so that ΓT = 0.216
Then n eff2 = 10.75 and W = 3.45, yielding ΓL = 0.856
(b) The total confinement factor then is Γ = 0.185
4-12. From Eq. (4-46) the mode spacing is
∆λ = λ2
2Ln = (0.80 µm)2
2(400 µm)(3.6) = 0.22 nm
Therefore the number of modes in the range 0.75-to-0.85 µm is
5
.85 − .75.22 × 10−3 =
.1.22 ×103 = 455 modes
4-13. (a) From Eq. (4-44) we have g(λ) = (50 cm-1) exp
- (λ - 850 nm)2
2(32 nm)2
= (50 cm-1) exp
- (λ - 850)2
2048
(b) On the plot of g(λ) versus λ, drawing a horizontal line at g(λ) = αt
= 32.2 cm-1 shows that lasing occurs in the region 820 nm < λ < 880 nm.
(c) From Eq. (4-47) the mode spacing is
∆λ = λ2
2Ln = (850)2
2(3.6)(400 µm) = 0.25 nm
Therefore the number of modes in the range 820-to-880 nm is
N = 880 - 820
0.25 = 240 modes
4-14. (a) Let Nm = n/λ = m2L be the wave number (reciprocal wavelength) of mode m.
The difference ∆N between adjacent modes is then
∆N = Nm - Nm-1 = 1
2L (a-1)
We now want to relate ∆N to the change ∆λ in the free-space wavelength. Firstdifferentiate N with respect to λ:
dNdλ =
ddλ
n
λ = 1λ
dndλ -
nλ2 = -
1λ2
n - λ
dndλ
Thus for an incremental change in wavenumber ∆N, we have, in absolute values,
∆N = 1λ2
n - λ
dndλ ∆λ (a-2)
6
Equating (a-1) and (a-2) then yields ∆λ = λ2
2L
n - λ
dndλ
(b) The mode spacing is ∆λ = (.85 µm)2
2(4.5)(400 µm) = 0.20 nm
4-15. (a) The reflectivity at the GaAs-air interface is
R1 = R2 =
n-1
n+1
2 =
3.6-1
3.6+1
2 = 0.32
Then Jth = 1
βα +
12L
ln1
R1R2
= 2.65×103 A/cm2
Therefore
Ith = Jth × l × w = (2.65×103 A/cm2)(250×10-4 cm)(100×10-4 cm) = 663 mA
(b) Ith = (2.65×103 A/cm2)(250×10-4 cm)(10×10-4 cm) = 66.3 mA
4-16. From the given equation
∆E11 =1.43 eV +6.6256 × 10−34 J ⋅ s( )2
8 5 nm( )2
16.19 × 10−32 kg
+ 15.10 ×10−31 kg
=1.43 eV + 0.25 eV =1.68 eV
Thus the emission wavelength is λ = hc/E = 1.240/1.68 = 739 nm.
4-17. Plots of the external quantum efficiency and power output of a MQW laser.
4-18. From Eq. (4-48a) the effective refractive index is
ne = mλB
2Λ = 2(1570 nm)2(460 nm) = 3.4
Then, from Eq. (4-48b), for m = 0
7
λ = λB ± λ
2B
2neL
1
2 = 1570 nm ± (1.57 µm)(1570 nm)
4(3.4)(300 µm) = 1570 nm ± 1.20 nm
Therefore for m = 1, λ = λB ± 3(1.20 nm) = 1570 nm ± 3.60 nm
For m = 2, λ = λB ± 5(1.20 nm) = 1570 nm ± 6.0 nm
4-19. (a) Integrate the carrier-pair-density versus time equation from time 0 to td (time
for onset of stimulated emission). In this time the injected carrier pair densitychanges from 0 to nth.
t d = dt0
t d
∫ = 1
Jqd
− nτ
0
n th
∫ dn = −τ Jqd
−nτ
n= 0
n= nth
= τ ln J
J − J th
where J = Ip/A and Jth = Ith/A. Therefore td = τ ln
Ip
Ip - Ith
(b) At time t = 0 we have n = nB, and at t = td we have n = nth. Therefore,
td = ⌡⌠0
td dt =
1
Jqd
− nτ
dnnB
n th
∫ = τ ln
Jqd
− nB
τJ
qd− n th
τ
In the steady state before a pulse is applied, nB = JBτ/qd. When a pulse is applied,
the current density becomes I/A = J = JB + Jp = (IB + Ip)/A
Therefore, td = τ ln
I - IB
I - Ith
= τ ln
Ip
Ip + IB - Ith
4-20. A common-emitter transistor configuration:
4-21. Laser transmitter design.
8
4-22. Since the dc component of x(t) is 0.2, its range is -2.36 < x(t) < 2.76. The powerhas the form P(t) = P0[1 + mx(t)] where we need to find m and P0. The average
value is
< > P(t) = P0[1 + 0.2m] = 1 mW
The minimum value is
P(t) = P0[1 - 2.36m] ≥ 0 which implies m ≤ 1
2.36 = 0.42
Therefore for the average value we have < > P(t) = P0[1 + 0.2(0.42)] ≤ 1 mW,
which implies
P0 = 1
1.084 = 0.92 mW so thatP(t) = 0.92[1 + 0.42x(t)] mW and
i(t) = 10 P(t) = 9.2[1 + 0.42x(t)] mA
4-23. Substitute x(t) into y(t):
y(t) = a1b1 cos ω1t + a1b2 cos ω2t
+ a2(b12 cos2 ω1t + 2b1b2 cos ω1t cos ω2t + b2
2 cos2 ω2t)
+ a3(b13 cos3 ω1t + 3b1
2 b2 cos2 ω1t cos ω2t + 3b1b22 cos ω1t cos2 ω2t+ b2
3 cos3 ω2t)
+a4(b14 cos4 ω1t + 4b1
3 b2 cos3 ω1t cos ω2t + 6b12 b2
2 cos2 ω1t cos2 ω2t
+ 4b1b23 cos ω1t cos3 ω2t + b2
4 cos4 ω2t)
Use the following trigonometric relationships:
i) cos2 x = 12 (1 + cos 2x)
ii) cos3 x = 14 (cos 3x + 3cos x)
iii) cos4 x = 18 (cos 4x + 4cos 2x + 3)
iv) 2cos x cos y = cos (x+y) + cos (x-y)
v) cos2 x cos y = 14 [cos (2x+y) + 2cos y + cos (2x-y)]
9
vi) cos2 x cos2 y = 14 [1 + cos 2x+ cos 2y +
12 cos(2x+2y) +
12 cos(2x-2y)]
vii) cos3 x cos y = 18 [cos (3x+y) + cos (3x-y) + 3cos (x+y) + 3cos (x-y)]
7-10. (a) Letting φ = fTb and using Eq. (7-40), Eq. (7-30) becomes
5
Bbae =
H p (0)
Hout (0)
2
Tb2
Hout' (φ)
Hp' (φ)
2
0
∞
∫ dφTb
= I2
Tb
since Hp(0) = 1 and Hout(0) = Tb. Similarly, Eq. (7-33) becomes
Be =
H p (0)
Hout (0)
2
Hout (f)Hp (f)
1 + j2πfRC( )2
0
∞
∫ df
=
1Tb
2 Hout (f)Hp (f)
2
0
∞
∫ 1 + 4π2f2 R2C2( ) df
=
1Tb
2 Hout (f)Hp (f)
2
0
∞
∫ df +
2πRC( )2
Tb2
Hout (f)Hp (f)
2
0
∞
∫ f2 df =
I2Tb
+ (2πRC)2
T3b
I3
(b) From Eqs. (7-29), (7-31), (7-32), and (7-34), Eq. (7-28) becomes
< > v2N = < > v
2s +< > v
2R +< > v
2I +< > v
2E
= 2q < >i0 < >m2 BbaeR2A2 + 4kBT
Rb BbaeR2A2 + SIBbaeR2A2 + SEBeA2
=
2q < >i0 M2+x + 4kBT
Rb + SI BbaeR2A2 + SEBeA2
= R2A2I2
Tb
2q < >i0 M2+x + 4kBT
Rb + SI +
SE
R2 + (2πRCA)2
T3b
SEI3
7-11. First let x = v − boff( )/ 2 σoff( ) with
dx = dv / 2 σoff( ) in the first part of Eq. (7-
49):
Pe =
2σoff
2πσoff
exp −x2( )vth − boff
2σoff
∞
∫ dx =
1
π exp −x2( )
Q / 2
∞
∫ dx
Similarly, let y = −v + bon( )/ 2 σon( ) so that
dy = -dv/ 2σon( )in the second part of Eq. (7-49):
6
Pe = -
1
π exp −y2( )
∞
− vth +b on
2σ on
∫ dy =
1
π exp −y2( )
Q / 2
∞
∫ dy
7-12. (a) Let x = V
2 2 σ =
K2 2
For K = 10, x = 3.536. Thus
Pe = e-x2
2 π x = 2.97×10-7 errors/bit
(b) Given that Pe = 10-5 = e-x2
2 π x then e-x2 = 2 π 10-5 x.
This holds for x ≈ 3, so that K = 2 2 x = 8.49.
7-13. Differentiating Eq. (7-54) with respect to M and setting dbon/dM = 0, we have
dbondM = 0
= -
Q(hν / η)Μ2 M2+ x η
hν
bonI2 + W
1/ 2
+ M 2+x ηhν
bonI2 (1 − γ
+ W
1/ 2
+
Q(hν / η)M(hν/ η)
12
(2 + x)M1+ x bonI2
M 2+ x ηhν
bonI2 + W
1/ 2 +
12
(2 + x)M1+x bonI2 (1− γ)
M 2+x ηhν
bonI2 (1− γ) + W
1/ 2
1 / 2
Letting G = M2+x
η
hν bonI2 for simplicity, yields
( )G + W1/2
+ [ ]G(1-γ) + W1/2
= G2 (2 + x)
1
(G + W)1/2 +
(1-γ)
[ ]G(1-γ) + W1/2
Multiply by (G + W)1/2
[ ]G(1-γ) + W1/2
and rearrange terms to get
(G + W)1/2
W -
Gx2 (1-γ) = [ ]G(1-γ) + W
1/2
Gx
2 - W
7
Squaring both sides and collecting terms in powers of G, we obtain the quadratic
equation
G2
x2γ
4 (1-γ) + G
x2γ
4 W(2-γ) - γW2(1+x) = 0
Solving this equation for G yields
G = -
x2
4 W(2-γ) ±
x4
16 W2(2-γ)2 + x2(1-γ)W2(1+x)
12
x2
2 (1-γ)
= W(2-γ)2(1-γ)
1 -
1 + 16
1+x
x2 1-γ
(2-γ)2
12
where we have chosen the "+" sign. Equation (7-55) results by letting
G = M2+xopt bonI2
η
hν
7-14. Substituting Eq. (7-55) for M2+xbon into the square root expressions in Eq. (7-
54) and solving Eq. (7-55) for M, Eq. (7-54) becomes
bon = Q
hν
η b1/(2+x)on
hν
η W(2-γ)2I2(1-γ)
K1/(2+x)
W(2-γ)
2(1-γ) K + W
12 +
W
2 (2-γ)K + W
12
Factoring out terms:
b(1+x)/(2+x)on = Q
hν
η
(1+x)/(2+x) W
x/2(2+x) I
1/(2+x)2
×
(2-γ)
2(1-γ) K + 1
12 +
1
2 (2-γ)K + 1
12
÷
(2-γ)K
2(1-γ)
1/(2+x)
or bon = Q(2+x)/(1+x)
hν
η Wx/2(1+x)
I1/(1+x)2 L
8
7-15. In Eq. (7-59) we want to evaluate
lim
γ →1(2 − γ)K2(1− γ )L
1+x
2+x
= lim
γ →1(2 − γ)K2(1− γ)
1+x
2+x lim
γ →11L
1+x
2+x
Consider first
lim
γ →1(2 − γ)K2(1− γ)
=
lim
γ →1
(2 − γ)2(1− γ )
−1 + 1 + B(1− γ )
(2 − γ)2
1
2
where B = 16(1+x)/x2 . Since γ→1, we can expand the square root term in a
binomial series, so that
lim
γ →1(2 − γ)K2(1− γ)
=
lim
γ →1
(2 − γ)2(1− γ )
−1 + 1 +12
B(1− γ)
(2 − γ)2 − Order(1− γ )2
=
lim
γ →1
B4(2 − γ )
= B4 = 4
1+xx2
Thuslim
γ →1(2 − γ)K2(1− γ)
1+x
2+x
=
4
1+xx2
1+x2+x
Next consider, using Eq. (7-58)
lim
γ →1
1L
1+ x
2+x
= lim
γ →1(2 − γ)K2(1− γ)
1/(2+ x)
÷ (2 − γ)2(1− γ)
K + 1
1
2
+12
(2 − γ )K +1
1
2
From the above result, the first square root term is
(2-γ)
2(1-γ) K + 1
12 =
4
1+xx2 + 1
12 =
x2 + 4x + 4
x2
12 =
x + 2x
From the expression for K in Eq. (7-55), we have that lim
γ →1 K = 0, so that
9
lim
γ →112
(2 − γ )K +1
1
2= 1 Thus
lim
γ →1(2 − γ )2(1− γ )
K + 1
1
2
+ 12
(2 − γ)K +1
1
2
=
x + 2x + 1 =
2(1 + x)x
Combining the above results yields
lim
γ →1(2 − γ)K2(1− γ )L
1+x
2+x
=
4
1+xx2
1+x2+x
4
1+xx2
12+x
2(1 + x)
x = 2x
so thatlim
γ →1 M
1+xopt =
W1/2
QI2
2x
7-16. Using H'p(f) = 1 from Eq. (7-69) for the impulse input and Eq. (7-66) for the
raised cosine output, Eq. (7-41) yields
I2 =
Hout' (φ)
2 dφ
0
∞
∫ =
12
Hout' (φ)
2 dφ
−∞
∞
∫
= 12 ⌡⌠
-1-β2
1-β2
dφ + ⌡⌠
1-β2
1+β2
18
1-sin
πφ
β - π2β
2dφ +
⌡⌠
- 1+β
2
- 1-β2
18
1-sin
πφ
β - π2β
2dφ
Letting y = πφβ -
π2β we have
I2 = 12 (1 - β) +
β4π⌡⌠
-π2
π2
[ ]1 - 2sin y + sin2y dy
= 12 (1 - β) +
β4π
π - 0 +
π2 =
12
1 - β4
10
Use Eq. (7-42) to find I3:
I3 =
Hout' (φ)
2φ2
0
∞
∫ dφ =12
Hout' (φ)
2φ2
−∞
∞
∫ dφ
= 12 ⌡⌠
-1-β2
1-β2
φ2 dφ + ⌡⌠
1-β2
1+β2
18
1-sin
πφ
β - π2β
2φ2dφ +
⌡⌠
- 1+β
2
- 1-β2
18
1-sin
πφ
β - π2β
2φ2dφ
Letting y = πφβ -
π2β
I3 = 13
1-β
2
3 +
β4π⌡
⌠
-π2
π2
[ ]1 - 2sin y + sin2y
β2y2
π2 + βyπ +
14 dy
= 13
1-β
2
3 +
β4π
⌡⌠
-π2
π2
β2y2
π2 + 14 ( )1 + sin2y dy -
2βπ ⌡⌠
-π2
π2
y sin y dy
where only even terms in "y" are nonzero. Using the relationships
⌡⌠0
π2
sin2y dy = π4 ; ⌡⌠
y sin ydy = -y cos y + sin y
and ⌡⌠
x2 sin x2 dx =
x3
6 -
x2
4 - 18 sin 2x -
x cos 2x4
we have
I3 = 13
1-β
2
3 +
β2π
β2
π2
1
3
π
2
3 +
16
π
2
3 +
π8 +
14
π
2 + π4 -
2βπ (1)
11
= β3
16
1
π2 - 16 - β2
1
π2 - 18 -
β32 +
124
7-17. Substituting Eq. (7-64) and (7-66) into Eq. (7-41), with s2 = 4π2α2 and β = 1, we
have
I2 =
⌡⌠
0
H
'out(φ)
H'p(φ)
2 dφ =
14⌡⌠
0
1
es2φ2
1-sin
πφ -
π2
2dφ
= ⌡⌠
0
1
es2φ2
1 + cos πφ
2
2dφ =
⌡⌠
0
1
es2φ2
cos4
πφ
2 dφ
Letting x = πφ2 yields I2 =
2π ⌡
⌠
0
π2
e16α2x2
cos4 x dx
Similarly, using Eqs. (7-64) and (7-66), Eq. (7-42) becomes
I3 = ⌡⌠
0
1
es2φ2
cos4
πφ
2 φ2dφ =
2π
3
x2e16α2 x2
cos4 x0
π2
∫ dx
7-18. Plot of I2 versus α for a gaussian input pulse:
7-19. Plot of I3 versus α for a gaussian input pulse:
7-20. Consider first lim
γ →1 K:
12
lim
γ →1 K =
lim
γ →1−1+ 1+16
1+ xx2
1− γ(2 − γ)2
1
2
= -1 + 1 = 0
Also lim
γ →1(1− γ) = 0. Therefore from Eq. (7-58)
lim
γ →1 L =
lim
γ →12(1− γ)(2 − γ)K
1/(1+ x)
(2-γ)
2(1-γ) K + 1
12 + 1
2+x1+x
Expanding the square root term in K yields
limγ →1
2 − γ1− γ
K = lim
γ →12 − γ1 − γ
−1+ 1+
162
1 + xx 2
1− γ(2 − γ)2 + order(1− γ)2
= lim
γ →18(1 + x)
x2
12 − γ
= 8(1+x)
x2
Therefore
lim
γ →1 L =
2x2
8(1+x)
1/(1+x)
4(1+x)
x2 + 1
12 + 1
2+x1+x
=
2x2
8(1+x)
1/(1+x)
x+2
x + 1
2+x1+x
= (1+x)
2
x
x1+x
7-21. (a) First we need to find L and L'. With x = 0.5 and γ = 0.9, Eq. (7-56) yields K =
0.7824, so that from Eq. (7-58) we have L = 2.89. With ε = 0.1, we have γ' = γ(1 -
ε) = 0.9γ = 0.81. Thus L' = 3.166 from Eq. (7-80). Substituting these values into
Eq. (7-83) yields
y(ε) = (1 + ε)
1
1 - ε
2+x1+x
L'L = 1.1
1
.9
5/3
3.1662.89 = 1.437
13
Then 10 log y(ε) = 10 log 1.437 = 1.57 dB
(b) Similarly, for x = 1.0, γ = 0.9, and ε = 0.1, we have L = 3.15 and L' = 3.35, so
that
y(ε) = 1.1
1
.9
3/2
3.353.15 = 1.37
Then 10 log y(ε) = 10 log 1.37 = 1.37 dB
7-22. (a) First we need to find L and L'. With x = 0.5 and γ = 0.9, Eq. (7-56) yields K =
0.7824, so that from Eq. (7-58) we have L = 2.89. With ε = 0.1, we have γ' = γ(1 -
ε) = 0.9γ = 0.81. Thus L' = 3.166 from Eq. (7-80). Substituting these values into
Eq. (7-83) yields
y(ε) = (1 + ε)
1
1 - ε
2+x1+x
L'L = 1.1
1
.9
5/3
3.1662.89 = 1.437
Then 10 log y(ε) = 10 log 1.437 = 1.57 dB
(b) Similarly, for x = 1.0, γ = 0.9, and ε = 0.1, we have L = 3.15 and L' = 3.35, so
that
y(ε) = 1.1
1
.9
3/2
3.353.15 = 1.37
Then 10 log y(ε) = 10 log 1.37 = 1.37 dB
7-23. Consider using a Si JFET with Igate = 0.01 nA. From Fig. 7-14 we have that α =
0.3 for γ = 0.9. At α = 0.3, Fig. 7-13 gives I2 = 0.543 and I3 = 0.073. Thus from
Eq. (7-86)
WJFET = 1B
2(.01nA )1.6 ×10−19C
+4(1.38×10 −23 J / K)(300K)
(1.6 × 10−19C)2105 Ω
0.543
+
1B
4(1.38×10−23 J / K)(300 K)(.7)
(1.6 × 10−19 C)2 (.005 S)(105
Ω)2
0.543
+
2π(10 pF)1.6 ×10−19 C
2 4(1.38 ×10−23 J / K)(300 K)(.7)(.005 S)
0.073 B
or
14
WJFET ≈ 3.51×1012
B+ 0.026B
and from Eq. (7-92) WBP = 3.39 ×1013
B+ 0.0049B
7-24. We need to find bon from Eq. (7-57). From Fig. 7-9 we have Q = 6 for a 10-9
BER. To evaluate Eq. (7-57) we also need the values of W and L. With γ = 0.9,Fig. 7-14 gives α = 0.3, so that Fig. 7-13 gives I2 = 0.543 and I3 = 0.073. Thus
from Eq. (7-86)
W = 3.51×1012
B+ 0.026B = 3.51×105 + 2.6×105 = 6.1×105
Using Eq. (7-58) to find L yields L = 2.871 at γ = 0.9 and x = 0.5. Substituting