Solutions Manual to
)LEHU2SWLFVDQG 2SWRHOHFWURQLFV
R.P. Khare Birla Institute of Technology and Science Pilani (Rajasthan)
© Oxford University Press 2004. All rights reserved. Not for sale or further circulation. Photocopying prohibited. For restricted use by faculty using Fiber Optics and Optoelectronics as a text in the classroom.
Preface This solutions manual has been written to aid the instructors in teaching this course. In the busy schedule it becomes difficult to find time for attempting newer problems and solving them. The author, therefore, felt that a readymade solutions manual should be made available to the instructors who are prescribing Fiber Optics and Optoelectronics as a textbook for their students. In fact a number of solved examples are already given in the textbook and the key to multiple choice questions is also given at the end of each set of such questions. However, the numerical review questions need elaborate solutions. These are presented in this manual. I hope the reader would now enjoy solving these problems. The author will welcome suggestions from instructors for improvement in the textbook or its presentation. They may mail to
[email protected].
R P Khare
1
Chapter 2: Ray Propagation in Optical Fibers 2.3
The parameter that varies with the surrounding medium is acceptance angle αm of the fiber. NA = na sin αm Since NA is a constant for a SI fiber, change in na will change the value of αm. In the present case, NA = 0.244 (see Example 2.1), and na = 1.33, we have na sin αm = 1.33 Sin αm = 0.244. This gives αm = 10.57o. Other parameters do not depend on na and hence they will not be affected.
2.4
NA = na sin αm = 1 × sin 20o ≈ 0.34 NA = n 1 2∆
However, NA is also given by: Therefore, n1 =
0.34
= 1.388
2 × 0.03
Now na sin αm = n1 sin θm = n1 cos φc = 0.34 This gives cos φc ≈ and 2.6
(a)
0.34 = 0.2449 1.388
φc ≈ 76o
NA = 0.17 = na sin αm = 1.33 × sin αm Therefore αm = 7.34o
(b)
(
Since NA = n − n 2 1
= [(0.17 )
)
1 2 2 2
= 0.17 and n2 = 1.46
]
1 2 2
n1 + (1.46 ) ≈ 1.47 Now na sin αm = n1cosφc = 1.47cosφc =0.17; which gives φc = 83.35o 2
2
2.7
(a)
( (
) )
c 3 × 10 8 ms −1 The r.i. of core = n1 = = = 1.5 v 2 × 10 8 ms −1 n2 , n1
and; since sin φc =
the r.i. of cladding = n2= n1 sin φc = 1.5 sin 75o ≈ 1.448 Therefore NA = [n
2.9
2 1
−n
] = [(1.5)
1 2 2 2
2
− (1.448)
∆T n 1 n 1 − n 2 = l c n2
(a)
The power p(t) in the pulse varies as follows:
(
)
p0 × t; 0 < t < 2τ 2τ
= 0; elsewhere Therefore, ε = ∫
2τ
0
2τ
p0 p t2 .t dt. = 0 = p 0 τ 2τ 2τ 2 0
1 2τ 1 2τ p 0 2 4 p( t ).t dt. = t dt = τ ∫ ∫ 0 0 p0τ 2τ 3 ε
(b)
t=
(c)
σ2 =
1 2τ 2 2 t p( t )dt − [t ] ∫ 0 ε
1 2τ 2 p 0 4 t . t dt − τ = ∫ p0τ 0 2τ 3 = 2τ2 –
=
2τ 2 9
σ=
τ 2 3
= 0.388
1.5(1.5 − 1.448) = = 1.79 × 10 −10 sm −1 8 −1 3 × 10 ms × 1.448
(b)
p(t) =
]
1 2 2
2
16τ 2 9
3
2.10
(a)
[
2
]
1 2 2
NA = (1.496 ) − (1.40 )
= 0.5272
For air, na = 1, Therefore NA = na sin αm = 0.5272 This gives αm = 31.81o (b)
For water, na = 1.33, Therefore na sin αm = 1.33 × sin αm = 0.5272, giving us αm = 23.35o
4
Chapter 3: Wave propagation in planar waveguides 3.8:
(a)
V=
2πa 2 n 1 − n 22 λ
(
)
1 2
=
π × 14(µm ) (1.46)2 − (1.455)2 1.30(µm )
[
]
1 2
V≈4 Referring to Fig. 3.4 of the textbook, the arc of circle of radius V = 4, intersects three ua versus wa curves corresponding to m = 0, 1 and 2. Therefore the guide supports these three modes. (The same information 2V and recalling that M is an integer can also be gathered by calculating π 2V greater than . π (b) The abscissae of the intersecting points of ua versus wa curves (see Fig. 3.4 of text book) for m = 0, 1 and 2 with the quadrant of circle of radius V = 4 give the values of uma and the corresponding ordinates give the value of wm a. bm and βm can also be calculated from these parameters. All these parameters for different values of m are listed in the table below. m
um a (rad)
um(m−1)
wma
wm(m−1)
w a bm = m V
0
1.25235
1.7890×105
3.79889
5.4269×105
0.90197
1
2.47458
3.5351×105
3.14269
4.4895×105
0.61109
2
3.59531
5.1361×105
1.75322
2.5046×105
0.19211
2
The phase propagation constants βm for different modes, m, can be calculated using the following relation:
{
(
β m = β 22 + b m β12 − β 22
)}
1 2
where β1 = kn1 =
2πn 1 2π × 1.46 = = 7.0565 × 10 6 m −1 λ 1.30 × 10 − 6
β2 = kn2 =
2πn 2 2π × 1.455 = = 7.0323 × 10 6 m −1 −6 λ 1.30 × 10
5
3.9
The maximum thickness of the guide layer that can support M modes is given by Mλ
2a =
(
2 n 12 − n 22
)
1 2
In the present case, M = 10, λ = 0.90 µm, n1 = 3.6 and n2 = 3.58. Therefore, 2a =
3.10
[
10 × 0.90(µm )
]
1 2 2
2 (3.6 ) − (3.58) 2
= 11.875µm
βmax = β1 = kn1 =
2πn 1 2π × 3.6 = = 2.513 × 10 5 m −1 λ 0.90 × 10 − 6
βmin = β2 = kn2 =
2πn2 2π × 3.58 = = 2.499 × 10 5 m −1 −6 λ 0.90 × 10
(a)
V=
2πa π × 10(µm ) n 1 2∆ = . 1.5 2 × 0.02 λ 1.55(µm )
(
)
(
)
V ≈ 6. From Fig. 3.4, it can be easily seen that for V = 6, the guide will support 4 modes: 2 symmetric modes corresponding to m = 0 and 2; and, 2 antisymmetric modes corresponding to m = 1 and 3. (b)
From Fig. 3.4 of textbook, the intersections of ua versus wa curves for different values of m with the circle of radius V = 6 gives the values of uma and w m a . The results are listed in the tabular form below. m
um a (rad)
um(m−1)
wma(rad)
wm(m−1)
w a bm = m V
0
1.34475
2.6895×105
5.84736
1.169472×106
0.94976
1
2.67878
5.35756×105
5.36881
1.073762×106
0.80067
2
3.98583
7.97166×105
4.48477
8.96954×105
0.558698
3
5.22596
1.045192×106
2.94776
5.89552×105
0.241369
{
(
β m = β 22 + b m β12 − β 22
)}
1 2
6
2
where β1 = kn1 =
β2 = kn2 =
2πn 1 2π × 1.5 = = 6.08 × 10 6 m −1 λ 1.55 × 10 −6 2πn 2 2π × 1.47 = = 5.9588 × 10 6 m −1 −6 λ 1.55 × 10
This gives, βm = [1.459 bm + 35.507]½ × 106 m−1 3.11
For antisymmetric modes, Ey(x) = B sin ux; | x | < a =
x D exp(− w | x |); | x |> a |x|
Power inside the guide layer, Pin =
=
1 β 2 ωµ 0
∫ (B sin ux ) dx a
2
−a
a β B 2 ∫ (1 − cos 2ux )dx 0 2ωµ 0
a
β 1 = B 2 x − sin 2ux 2ωµ 0 2u 0 =
β 1 B 2 a − sin 2ua 2ωµ 0 2u
Power outside the guide layer, Pout =
=
∞ 1 β −a 2 E y (x )dx + ∫ E 2y (x )dx ∫ a 2 ωµ 0 − ∞ ∞ 1 β 2 ∫ E 2y ( x )dx 2 ωµ 0 a
7
(1)
∞
=
β D 2 exp(− 2 wx )dx ωµ 0 ∫a ∞
β 1 − 2 wx = D 2 − e ωµ 0 2w a =
β 1 − 2 wa D 2 − e ωµ 0 2w
(2)
The confinement factor G=
Pin Pin + Pout
(3)
Substituting the values of Pin and Pout from (1) and (2) in (3), we get 1 β 1 .B 2 a − . sin 2ua 2 ωµ 0 2u G= 1 β 1 1 β 1 .B 2 a − . sin 2ua + .D 2 .e − 2 wa 2 ωµ 0 2u 2 ωµ 0 w 1 a − . sin 2ua 2 = 2 1 1 − 2 wa D sin 2ua + .e a − 2u w B At x = a,
B sin ua = D e−wa
Therefore,
D = e wa . sin ua B
Substituting for
(4)
(5)
D from (5) in (4) and manipulating, we get, B
1 a − . sin 2ua 2u G= 1 1 sin 2ua + . sin 2 ua a − 2u w
8
This result can also be expressed in the following form: 2 sin (ua ) G = 1 + sin ua. cos ua wa 1 − ua (b)
−1
(6)
Given that 2a = 6.523 µm, n1 = 1.50, n2 = 1.48 and λ = 1.0µm Therefore, V =
2πa 2 n 1 − n 22 λ
(
)
1 2
[
6.523 × π (1.5)2 − (1.48)2 = 1
]
1 2
≈5 So the guide supports 4 modes corresponding to m = 0, 1, 2 and 3. m = 0, 2 are symmetric modes m = 1, 3 are antisymmetric modes. From Fig. 3.4, the values of ua and wa may be written as follows. m
ua(rad)
wa(rad)
0
1.30644
4.8263
1
2.59574
4.27342
2
3.83747
3.20529
3
4.9063
0.963466
To calculate G-factors for symmetric modes, use formula (3.81) of text book; and for antisymmetric modes use formula (6) derived in part (a) of this question.
The respective calculations are given as follows: 9
2 cos (1.30644) Gm = 0 = 1 + cos(1.30644). sin(1.30644) 4.82631 + 1.30644
−1
= 0.98828 2 sin (2.5974) Gm = 1 = 1 + cos(2.5974). sin( 2.5974) 4.273421 − 2.5974
−1
= 0.94889
Gm = 2
2 cos (3.83747) = 1+ cos(3.83747). sin(3.83747) 3.205291 + 3.83747
−1
= 0.85992
Gm = 3
2 sin (4.9063) = 1 + cos(4.9063). sin( 4.9063) 0.9634661 − 4.9063 = 0.50960
10
−1
Chapter 4: Wave Propagation in Cylindrical Waveguides 4.2
(b)
(i)
∆=
n 12 − n 22 2n 12
Therefore n2 = n1 (1 − 2¨)½ = 1.46 (1 – 2 × 0.01)½ = 1.445 (ii)
V=
(iii)
Ms =
2πa π × 50(µm) n 1 2∆ = × 1.46 (2 × 0.01) ≈ 25 λ 1.3(µm) V 2 (25) = = 312 2 2 2
4.3
4.4
(b)
β max = kn 1 =
2π 2π × 1.48 n1 = = 10.93 × 10 6 m −1 −6 λ 0.85 × 10 (m)
β min = kn 2 =
2π 2π × 1.465 n2 = = 10.82 × 10 6 m −1 −6 λ 0.85 × 10 (m)
The cut off frequency for single mode operation is given by 1
2 2 Vc = 2.405 1 + α This should be equal to V given by V=
2πa n 0 2∆ λ
Therefore, for this case the following equality should be satisfied 1
2πa 2 2 n 0 2∆ = 2.405 1 + λ α 1
or
2 2 2.405λ1 + α 2a = πn 0 2∆
In the present problem, α = 1, n0 = 1.5 and ¨
11
Hence (i) for λ = 1.3 µm, 2.405 × 1.3(µm )(1 + 2 )2 1
2a =
π × 1.5 ×
= 7.1µm
(2 × 0.013)
and (ii) for λ = 1.55 µm, 2.405 × 1.55(µm )(1 + 2 )2 2a = = 8.5µm π × 1.5 × (2 × 0.013) 1
4.6
(M g ) eff Mg
2 1 (α + 2) 2a 3 3 = = 1 − + 2 2α∆ R 2n c kR
(1)
In the present problem α = 2, ¨ D µm, nc = 1.48 and k =
2π 2π = = 7.39(µm) −1 . λ 0.85(µm)
Substituting the valves of different parameters in eq. (1), we get 2 1 100 0.1371 3 = 1 − 100 + ; 2 R R
where R is measured in µm. Solving this equation, we get R ≈ 1.66 × 104 µm or 4.7
R ≈ 1.66 cm
The complete transverse electric field will be given by Ez = A Jl ur .eilφ.ei(ωt−βz); r < a a
12
(1)
and
Ez = B Kl wr .eilφ.ei(ωt−βz); r > a a
(2)
where A and B are arbitrary constants. The solutions for the transverse magnetic field are similar but the constants will be different. Thus
and
Hz = C Jl ur .eilφ.ei(ωt−βz); r < a a
(3)
Hz = D Kl wr .eilφ.ei(ωt−βz); r > a a
(4)
Let us derive the expression for Er first inside the core and then in the cladding. It is given that i k 2r
Er = −
∂E z 1 ∂H z β ∂r + ωµ r . ∂φ
(5)
Substituting the values of Ez and Hz from (1) and (3) in (5), we get i k 2r
Er = −
=−
or
Er = −
∂ ur ilφ i ( ωt −βz ) 1 ∂ ur + ωµ .e ilφ .CJ l .e i ( ωt −βz ) AB J l .e e r ∂φ a ∂r a
i βu ωµi ur ilφ i ( ωt −βz ) ur A .J ′l .e ilφ e i ( ωt −βz ) + C .J l e .e 2 r kr a a a i βu ur Cωµi ur ilφ i ( ωt −βz ) AJ ′l + J l e e 2 r kr a a a
) [
(
where k 2r = ω 2 εµ − β 2 = ω 2 .ε 0 ε r .µ 0 µ r − β 2
]
Inside the dielectric medium, such as glass fiber, µr = 1, εr = n2 and ε0µ0 =
1 . C2
13
(6)
Therefore n2 k 2r = ω 2 2 − β 2 = k 2 n 2 − β 2 c
[
where k =
]
(7)
ω 2π = free space propagation constant. = c λ
Note that for a SI fiber inside the core, n = n1 and in the cladding n = n2. Similarly, inside the cladding, (r > a). Er = −
i Bβ w wr Dωµi wr ilφ i ( ωt −βz ) K ′l + K l e .e r k 2r a a a
(8)
Expressions for Eφ, Hr and Hφ can be derived in a similar manner. 1
4.8
α 2 r n(r) = n0 1 − 2∆ ; r ≤ a a
Substituting ¨
n 02 − n c2 in the above equation and taking the square on both 2n 02
sides, we get n 02 − n c2 n (r ) = n 1 − 2 2 2n 0 2
2 0
α r . . a
α r = n 02 − n 02 − n c2 a
(
)
Therefore the local numerical aperture, NA(r) would be given by
{
NA(r) = n 2 (r ) − n c2
}
1 2
= n 02 − n 02 − n c2
(
− n c2
) ar
α
14
1 2
= n 02 − n c2
(
1
r α 2 1 − a
)
The mean square value of the numerical aperture taken over the core area may be given by the following relation:
Mean square value of NA =
∫
a
0
NA 2 (r ).2πrdr
∫
a
0
(n
2 0
−n
2 c
)∫
a
0
=
(n = =
(n
2πrdr
r α 1 − 2πrdr a πa 2
)
a
− n c2 2πr 2 2π r α + 2 − α. a (α + 2) 0 πa 2 2
2 0
)
− n c2 2 2πa 2 πa − α + 2 πa 2
2 0
α = n 02 − n c2 . α + 2
(
)
Therefore, root mean square value of numerical aperture for an α-profile would be given by 1
2 α 2 2 (NA)rms = . n 0 − n c α + 2
(
4.9
(a)
)
1
2 α 2 2 (NA)rms = . n 0 − n c α + 2
(
)
Given that α = 2, n0 = 1.460 and nc = 1.445
{
}
1
2 2 2 2 Therefore (NA)rms = (1.46 ) − (1.445) 2 + 2
15
= 0.1476 (b)
2π 2π n0 = × 1.46 = 7056mm −1 −3 λ 1.3 × 10 mm
β0 = kn0 =
(
)
n 02 − n c2 (1.46 )2 − (1.445)2 ¨ = = 0.0102 2 2n 02 2 × (1.46) V=
2πa 2 n 0 − n c2 λ
(
)
1 2
=
[
π × 50 (1.46)2 − (1.445)2 1.3
]
1 2
= 25.23
(c)
2 α V 2 (25.23) Mg = = = 159 2 α + 2 2 2+ 2
(a)
Given that α = 1, Mg = 500, n0 = 1.46, 2a = 75 µm and λ = 1.3 µm.
2
4.10
α V2 1 V2 V2 . = Therefore Mg = = . α + 2 2 1+ 2 2 6 or
6M g = 6 × 500 = 54.77
V =
As V can also be expressed as V≈
2πa n 0 2∆ , we have λ 2
1 λV 1 1.3 × 54.77 = 0.021 ¨ = 2 π(2a )n 0 2 π × 75 × 1.46 2
1
(b)
1
2 2 22 Vc = 2.405 1 + = 2.4051 + = 4.165 α 1
If we assume that the core diameter for single mode operation is 2a ′ , V can be written as V=
or
2πa ′ n 0 2∆ = 4.165 λ
2a ′ =
4.165 × 1.3(µm ) π × 1.46 2 × 0.021
16
= 5.76µm
4.11
(a)
(b)
Given that α = 2, 2a = 70µm n0 = 1.47, nc = 1.45 and λ = 1.3µm. Then β0 = kn0 =
2π 2π × 1.47 n0 = = 7.1048 × 10 6 m −1 −6 λ 1.3 × 10 (m)
βc = knc =
2π 2π × 1.45 nc = = 7.1008 × 10 6 m −1 −6 λ 1.3 × 10 (m)
∆=
n 02 − n c2 (1.47 )2 − (1.45)2 = = 0.01351 2 2n 02 2 × (1.47 )
V=
2πa π × 70 n 0 2∆ = × 1.47 2 × 0.01351 = 40.88 λ 1.3 V 2 (40.88) = = 418 . 4 4 2
Mg =
17
Chapter 5: Single Mode Fibers 5.2
(b)
Given that n1 = 1.46, 2a = 8µm, ¨ In general, V=
2πa n 1 2∆ λ
For single mode operation, in a SI fiber V = Vc = 2.405, Therefore, λc = or 5.4
V=
2πa π × 8(µm )× 1.46 2 × 0.0052 n 1 2∆ = Vc 2.405
λc = 1.556µm
2πa n 1 2∆ λ
At λ1 = 1.31µm, V1 =
2π × 4.1 × 1.4677 2 × 0.0036 = 2.449 1.31
Therefore, using eq. (5.3) of the text book we get, 1.619 2.879 w = 4.10.65 + + 3 6 ( 2.449) 2 ( 2 . 449 ) = 4.1 [0.65 + 0.4224 + 0.01334] or
w = 4.452 µm
Employing eq. (5.5), we get 1.567 wp = 4.452 – 4.1 0.016 + (2.449)7
18
or
wp = 4.374µm
At
λ2 = 1.55µm, V2 =
2π × 4.1 × 1.4682 2 × 0.0036 = 2.0705 1.55
1.619 2.879 w = 4.1 0.65 + + 3 (2.0705)6 2 ( 2 . 0705 )
= 4.1 [0.65 + 0.5434 + 0.0365] or
w = 5.0427 µm 1.567 wp = 5.0427 – 4.1 0.016 + (2.0705)7
5.5
or
wp = 4.9377µm
(a)
δβ1 =
2π 2π .δn 1 = × 10 − 6 = 4.833m −1 −6 λ 1.3 × 10 (m)
and δβ2 =
2π 2π .δn 2 = × 10 −5 = 48.33m −1 −6 λ 1.3 × 10 (m)
Range of δβ = 4.833 – 48.33 m−1 (b)
Lp1 =
2π 2π = (m) = 1.30m δβ1 4.833
Lp2 =
2π 2π = (m) = 0.13m δβ 2 48.33
Lp varies from 13 cm to 1.3 m 5.7
Given that n1 = 1.48, ¨ λ = 1.32 and a = 4.4 µm n2 ≈ n1(1 − ¨ V=
2π × 4.4 × 1.48 2 × 0.0027 = 2.2778 1.32
19
Using eq. (5.24) of the text book, we have Vd 2 2 (bV) ≈ 0.080 + 0.549(2.834 − V ) 2 dV = 0.2498 From eq. (5.23), we get Dw = −
1.476 × 0.0027 × 0.2498 3 × 10 8 ms −1 × 1.32(µm )
(
)
= −2.5138 × 10−12 s(µm−1) (km−1) = − 2.5138 ps (nm−1) (km−1) 5.9
D = Dm + Dw = 0 Since Dm = 7 ps nm−1 km−1, Dw = − 7 ps nm−1 km−1 From eq. (5.23) and (5.24), we get Dw = −
[
n 2∆ 2 0.080 + 0.549(2.834 − V ) cλ
]
n1 = 1.48, ¨ Q2 = n1(1 − ¨ − 0.01) or Dw = −
n2 = 1.4652
[
1.4652 × 0.01 2 0.080 + 0.549(2.834 − V ) 8 −1 3 × 10 ms × (1.55µm )
(
)
]
= − 31.509 (ps nm−1 km−1) [0.080 + 0.549 (2.834 − V)2] Thus, in order to get D = 0, we should have − 7 = − 31.509 [0.080 + 0.549 (2.834 − V)2] This gives V = 2.325 Since V =
2πa n 1 2∆ , we get λ
20
Core radius, a =
V×λ 2πn 1 2∆
=
2.325 × 1.55(µm ) 2π × 1.48 2 × 0.01
= 2.74 µm
5.11
8.2 µm 25 µm Fig. Q 5.11
Given that, n1 = 1.46, 2a = 8.2 µm, ¨ D1 = 25µm − 2r 2 P(r) = P0 exp 2 w At
λ1 = 1.30 µm, V1 =
2πa π × 8.2 n 1 2∆ = × 1.46 2 × 0.003 = 2.241 λ1 1.3
and at λ2 = 1.55 µm 21
V2 =
2πa π × 8.2 n 1 2∆ = × 1.46 2 × 0.003 = 1.879 λ2 1.55
Therefore at λ1 = 1.3 µm, (using eq. 5.3), 1.619 2.879 + w ≈ 4.1 0.65 + (2.241)1.5 (2.241)6 = 4.1 [0.65 + 0.48259 + 0.0227] w = 4.7368 µm and at λ2 = 1.55 µm, 1.619 2.879 w = 4.1 0.65 + + (1.879 )1.5 (1.879)6 = 4.1 [6.65 + 0.6285 + 0.0654] = 5.51 µm At
λ1 = 1.30 µm, the fractional power at r = a1 = 12.5 µm 12.5µm 2 r 2 P(r ) = exp − 2 = exp − 2 P0 w 4.7368µm P(r ) = 8.9387 × 10 −7 P0
(1)
and at λ2 = 1.55µm, the fractional power at r = a1 = 12.5µm 12.5µm 2 P(r ) = exp − 2 = 3.3869 × 10 −5 P0 5.51µm
(2)
Eq. (1) & (2) give the fraction of optical power reaching the inner-outer cladding interface and that may be lost by transmission into the outer parts of the cladding. In practice a fraction of this power may be reflected and some may be refracted.
22
Chapter 6: Optical fiber cables and connections 6.4
LF = − log10 (ηF) = 0.36 dB This gives ηF= 0.92 =
16k 2 (from eq. 6.3 of text book) (k + 1) 4
For air, n = 1, therefore k =
n1 = n1 n
Thus k2 − 2.17 k + 1 = 0 or
n 12 − 2.17n 1 + 1 = 0
Solving this we get n1 = 1.5. The second root of this equation is less than 1, which is not possible. 6.6
Using eq. (6.11) of the text book, Lang = − 10 log10 ηang = 0.6 dB
(1)
This is due to 4o angular misalignment. From eq. (1), ηang = 0.87 But from eq. (6.8), n∆θ ηang ≈ 1 − = 0.87 πNA In this problem, n = 1.46 and ¨θ = 4o = 0.0698 rad and hence NA =
6.8
Given that
1.46 × 0.0698 = 0.25 π × 0.13
∆y = 12% = 0.12 2a
n1 = 1.5 and n = 1.47
23
Therefore, k =
ηF =
And,
or
n 1 1.5 = = 1.02 n 1.47 16k 2 16(1.02) 2 16 × 1.041 = = = 0.99957 16.663 (k + 1) 4 (1.02 + 1) 4
ηlat
1 2 2 2 ∆y ∆y ∆y = cos −1 − × 1 − π 2a 2a 2a
ηlat
2 = 1.4505 − (0.12) 1 − (0.12) 2 π
{
} 1 2
= 0.8475 Hence the total coupling efficiency, ηT would be ηT = ηF × ηlat = 0.99957 × 0.84757 = 0.8472 This gives a total loss at the joint, LT = − 10 log10 ηT ≈ 0.72 dB. 6.10
The Fresnel reflection coefficient, R is given by n −n R = 1 n1 + n
2
Its first derivative w.r.t. n1 would be n − n 1 d dR (n 1 + n ) + (n 1 − n )− 1 2 × d (n 1 + n ) = 2 1 dn 1 n 1 + n n 1 + n dn 1 (n 1 + n ) dn 1 or
n − n 1 dR dn (n 1 − n ) 1 − − = 2 1 2 dn 1 n 1 + n (n 1 + n ) dn 1 (n 1 − n )
dn 1 + dn 1
In this expression n is a constant as it is the refractive index of the surrounding medium.
24
Hence
dn = 0 . This gives us dn 1
(n − n ) [ ( ) dR (n 1 + n ) − (n 1 − n )] = 4n n 1 − n3 =2 1 (n 1 + n ) dn 1 (n 1 + n ) For small variation in n1, ¨Q1, will, therefore, produce a small variation ¨5LQ5 given by: ¨5
4n (n 1 − n )
(n 1 + n )3
.∆n 1
25
Chapter 7: Optoelectronic Sources 7.3
The net rate of recombination per unit volume of excess minority electrons in the p-side =
− d∆n (t ) ∆n = dt τp
(1)
Solving (1) gives us ¨n(t) = ¨Q exp (−t/τp)
(2)
where ¨Q LVWKHH[FHVVFDUULHUFRQFHQWUDWLRQDWW 7KHPHDQlife time of the excess minority holes will then be given by ∞ t.∆n (0 )exp − t dt τ ∫0 p t = ∞ t − ( ) ∆ n 0 . exp ∫0 τ p or 7.4
t = τp
(3)
Given that Na = 1021 m−3 and Nd = 1023 m−3. It means that n-side is doped more heavily as compared with the p-side. Hence, in this case, (see eq. 7.49 of text book) ηinj =
Now
1 Dh 1 + De
De = µ e
Le Lh
(
Na Nd
) (
)
kT 0.85 m 2 V −1s −1 × 1.38 × 10 −23 JK −1 × (300K ) = e 1.6 × 10 −19 C
= 0.02199 m2 s−1 Similarly, Dh = h
kT = 1.035 × 10 −3 m 2 s −1 e
26
Therefore ηinj =
1 1.035 × 10 −3 1 10 21 × × 23 1 + 0.02199 1 10
= 0.9995 7.5
(a)
The power radiated by the p-n diode as a function of photon energy, P(Eph), may be written as P(E ph ) ∝ ∫ n (E 2 ).p(E 1 )dE 2 with the constraint that E2 – E1 = Eph
or
P(Eph) = α ∫
E v + E ph
E 2 =E c
(E − E c ) (E − E1 ) exp − 2 . exp − v dE 2 kT kT
where α is a constant. or
P(Eph) = α ∫
E v + E ph
E 2 =E c
(E − E v ) (E − E1 ) exp c . exp − 2 dE 2 kT kT
E ph − E g = α exp − kT
E v + E ph ∫ dE 2 E c
E ph − E g = α(E ph − E g )exp − kT
27
Electron Energy E2
Free Electron Density
Ev+Eph Ec
n (E2)
Eg
Eph Ev
p(E1)
Ec-Eph
Hole Density
E1
Fig. Q 7.5
Fig. Q.7.5
(b)
dP(E ph ) dE ph
(E ph − E g ) (E ph − E g ) = α exp − 1 = 0 , for maximum P. kT kT
Therefore (Eph)peak = Eg + kT; and P (at peak Eph) = 7.7
(a)
αkT . e
Using eq. (7.61) of text book 2 n a2 ( 1.50 ) F= 2 = = 0.0821 2n s 2(3.7 )2
28
(b)
Given that as = 10% = 0.1 t = 1 and ηinj = 0.40 ηext = ηinj (1 – as) × (1) × F. = 0.40 (1 – 0.1) × (1) × 0.0821 = 0.0295
(c)
The fraction of incident radiation collected and propagated by the fiber is given by (see eq. 7.113) φ (NA ) = (0.16) = 0.01137 = φm n a2 (1.50)2 2
2
φ = 0.0295 × 0.01137 = 3.35 × 10−4 φm
(d)
ηT = ηext
(e)
If the LED is emitting in air, φ (NA ) = (0.16) = 0.0256 = φm n a2 (1)2 2
7.8
2
1 1 = = 0.4545 τ rr 120 1 + 1+ 100 τ nr
(a)
ηint =
(b)
I φ int = ηint . E ph e 100 × 10 −3 (A ) [1.42 × e(J )] = 0.4545 × e = 0.0645 W = 64.5 mW Power supplied to LED = VI = 1.5 × 100 × 10−3 = 150 × 10−3W = 150 mW. Therefore internal power efficiency =
29
64.5 = 0.43. 150
(c)
If the diode is emitting in air, the external quantum efficiency would be given by ηext = ηint (1 − a s )
=
2
n s (n s + 1)
2
0.4545 × (1 − 0.1)× 2 2 3.7(3.7 + 1)
= 0.010 100 × 10 −3 I × (1.42 × e ) φ ext = η ext × × E ph = 0.010 × e e or
φext = 1.42 × 10−3W = 1.42 mW
Therefore, external power efficiency = (d)
ηT = ηext
2 ( NA ) .
n a2
= ηext
(n
1
2∆ n a2
)
1.42 = 0.0095 150
2
Given that na = 1.5, NA = n 1 2∆ = 1.46 2 × 0.02 =0.292 Using eq. (7.65), we can calculate ηext for this part as follows ηext = ηint (1 − as)
=
2n 3a
n s (n a + n s )
2
0.4545 × (1 − 0.1)2 × (1.5) 3 3.7(1.5 + 3.7) 2
= 0.0275 Therefore, ηT = 0.0275 ×
(0.292 )2 (1.5)2
= 1.042 × 10 −3
and
−3 I − 3 100 × 10 × (1.42 × e ) = × φT = ηT E ph 1.042 10 e e
or
φT = 1.4796 × 10−4W = 0.14796 mW.
30
Thus overall source-fiber power coupling efficiency =
0.14796mW = 9.86 × 10 − 4 150mW
and optical loss = − 10 log10 (9.86 × 10−4) = 30 dB. 7.9
In the presence of back enrission, the fraction, F, of the total optical power that can be collected at the semiconductor-air surface is given by n a2 F= 2n s2
(see eq. 7.61 of the text book)
In the absence of back enrission, this factor will be doubled and the new value of F will be F′ =
n a2 n s2
Further, R and t remain the same but as = 0 and hence T = 1. Therefore ηext = ηint F′tT = ηint ×
= ηint
n a2 4n a n s × ×1 2 n s (n a + n s )2 4n 3a
n s (n a + n s )
2
Given that na = 1, ns = 3.7, ηint = 0.60 ηext = (a)
0.60 × 4 × 1 = 0.029 2 3.7(1 + 3.7 )
Optical power emitted within the LED I = φint = ηint .E ph e 120 × 10 −3 A (1.43eV × e ) = 102.96 × 10−3W = 0.60 e 31
= 102.96 mW The total power consumed by the device = 120 × 10−3(A) × 1.5 (V) = 180 × 10−3 W = 180 mW. Therefore internal power efficiency = (b)
102.96 = 0.572 180
Optical power emitted in air by the LED I = φext = ηext .E ph e 120 × 10 −3 A (1.43eV × e ) = 0.029 e = 4.976 × 10−3W = 4.976 mW Therefore external power efficiency =
7.11
4.976 = 0.0276 180
Using eq. (7.91) of text book, we can calculate the cavity length, L as follows
(
)
c 3 × 10 8 ms −1 L= = = 1.62 × 10 − 4 m 9 2nδν 2 × 3.7 × 250 × 10 (Hz ) or
L = 162 µm
The number of longitudinal modes can be calculated using eq. (7.90) as follows: 2 × 3.7 × 162(µm ) ν 1 m = .2nL = 2nL = 0.85(µm ) c λ
32
or 7.12
m = 1410
In terms of wavelength m can be expressed as follows: 2nL 2 × 3.7 × 500(µm ) = = 4353 λ 0.85(µm )
m=
δm =
1 .2nLδλ λ2
Therefore mode separation in terms of wavelength will be given by (putting δm = 1), λ2 (0.85) = 1.95 × 10 − 4 µm δm = = 2nL 2 × 3.7 × 500 2
or
δλ = 0.195 nm
In terms of frequency, m is given by eq. (7.90), i.e. ν 2nL m = 2nL; and δm = δν . Therefore the mode separation in terms of c c frequency will be given by (with δm = 1) δm =
7.13
c 3 × 10 8 ms −1 = = 8.1 × 1010 Hz = 81 GHz. 2nL 2 × 3.7 × 500 × 10 − 6 (m)
The threshold gain coefficient for an ILD is given by (see eq. 7.92 of text book). gth =
1 1 1 ln α eff + Γ 2L R 1 R 2
Given that gth ≈ βJth. Therefore Jth =
1 1 1 ln α eff + βΓ 2L R 1 R 2
33
(1)
In the present problem, β = 0.02 cm A−1, αeff = 12 cm−1, L = 300 µm = 0.03 cm, width = W = 100 µm = 0.01 cm 2
2
n −1 3.7 − 1 n = 3.7 and hence R1 = = = 0.33 n +1 3.7 + 1 and
R2 = 1.
Substituting the values of given parameters in (1), for a strong carrier confinement (Γ = 1) , we get J th =
1 1 1 12 + ln 0.02 × 1 2 × 0.03 0.33
≈ 1524 A cm−2 The threshold current, Ith will be given by Ith = Jth LW = 1524 (A cm−2) × (0.03 cm) × (0.01 cm) = 0.4572 A 7.14
m=
2nL mλ 1 × 1.3(µm) . or L = = = 0.185µm λ 2n 2 × 3.5
Chapter 8: Optoelectronic Detectors 8.3
Given that η = 50% = 0.5 at λ = 0.90 µm. (a)
ℜ=
ηeλ 0.5 × 1.6 × 10 −19 (C) × 0.90 × 10 −6 (m) = hc 6.626 × 10 −34 (Js) × 3 × 10 8 (ms −1 )
= 0.362 AW−1 (b)
Ip = 10−6 A Using eq. (8.4) of the text book, we get 34
Pin = (c)
Ip ℜ
=
10 −6 A = 2.76 ×10−6 W = 2.76 µW −1 0.362 AW
(
)
The rate of photons received will be given by Pin Pin λ 2.76 × 10 −6 ( W ) × 0.90 × 10 −6 (m) rp = = = hν hc 6.626 × 10 −34 (Js) × 3 × 10 8 ms −1
(
= 1.25 × 1013 s−1 8.4
Given that η = 0.50 at λ = 1.3 µm. Its responsivity would be ℜ=
ηeλ 0.50 × 1.6 × 10 −19 × 1.3 × 10 6 = = 0.523 hc 6.626 × 10 −34 × 3 × 10 8
Therefore Ip = ℜ . Pin = 0.523 × 0.4 × 10−6 = 0.2092 × 10−6 A Ip = 0.2092 µA
or
The output photocurrent after avalanche gain is I = 8 µA. Thus the multiplication factor M =
8.5
I 8 = ≈ 38 . I p 0.2092
For an ideal p-n photodiode, η = 1. (a)
At λ = 0.85 µm ℜ=
(b)
At λ = 1.30 µm, ℜ=
(c)
1 × 1.6 × 10 −19 × 0.85 × 10 −6 = 0.684AW −1 8 − 34 6.626 × 10 × 3 × 10
1 × 1.6 × 10 −19 × 1.3 × 10 −6 = 1.046AW −1 6.626 × 10 −34 × 3 × 10 8
At λ = 1.55 µm, 35
)
ℜ=
8.6
1 × 1.6 × 10 −19 × 1.55 × 10 −6 = 1.248AW −1 8 −34 6.626 × 10 × 3 × 10
(
W Ip = ℜ × Pin = 0.40 (AW−1) × 100 × 10−6 × 2 mm 2 2 mm or
Ip = 80 × 10−6 A = 80 µA.
36
)
Chapter 9: Optoelectronic Modulators 9.2
x=
λ 589.3 × 10 −9 (m) = = 0.856 × 10 −6 m 4(n 0 − n e ) 4(1.658 − 1.486)
= 0.856 µm 9.3
λ 589.3 × 10 −9 (m) x= = = 32738.88 × 10 −9 m 2(n e − n 0 ) 2(1.553 − 1.544 ) = 0.0327 mm
9.5
From Table 9.1 of the text book, for lithium niobate (LiNbO3), n0 = 2.29, ne = 2.20, and r33 = 30.8 × 10−12 m/V. It is given that l = 5 mm, λ = 550 nm, and V = 100 volts (a)
Ez =
V 100 = = 2 × 10 4 V / m −3 l 5 × 10
∆n =
1 3 1 n 0 r33 .E z = (2.29) 3 × 30.8 × 10 −12 × 2 × 10 4 2 2
or ¨φ =
¨Q × 10−6 2π 2π × ∆n × l = × 3.698 × 10 − 6 × 5 × 10 −3 λ 550 × 10 −9
= 0.067π Therefore Φ = 2¨φ = 0.1344π (b)
Vπ = or
λ 550 × 10 −9 = 2n 30 r33 2 × (2.29) 3 × 30.8 × 10 −12 Vπ = 743.5 V
37
9.6
(a)
2πl (n 0 − n e ) λ Substituting the values of constants, we get ¨φ (due to intrinsic birefringence) =
π 2π × l = (2.29 − 2.20). 2 550 × 10 −9 m 550 × 10 −9 m Therefore, l = = 1.5277 × 10 −6 m ≈ 1.528µm. 4(0.09) (b)
¨φ (due to external field) =
π l r33 n 30 .V. λ d
−6 π π 3 1.528 × 10 −12 = × 30 . 8 × 10 × ( 2 . 29 ) 2 550 × 10 −9 m d
d=
(c)
9.8
× 20
2 × 30.8 × 10 −9 m × (2.29) 3 × (1.528) × 20 = 41.1 × 10 −9 m = 0.041µm. 550
Vπ =
550 × 10 −9 λ d 0.041 = × V = 40 V 3 3 −12 2n 0 r63 l (2.29) × 30.8 × 10 1.528
λ 633 × 10 −9 m = 1 Sin θm = m −6 n Λ × × 1 . 33 375 10 m 0 = 1.269 × 10 −3 Velocity of accoustic wave 1500ms −1 Λ = = = 375 × 10 −6 m 6 −1 frequency(Hz ) 4 × 10 s m=1 θm = 0.0727o
38
Chapter 10 : Optical Amplifiers 10.3
(a)
Psat =
hνA hcA = Γσ g τ c Γσ g τ c λ
From the given parameters, the cross-sectional area of the active region, A = 5 × 10−6(m) × 0.5 × 10−6(m) = 2.5 × 10−12 m2 Γ = 0.4, σg = 3 × 10−20 m2, τc = 1 ns = 1 × 10−9s and λ = 1.3 µm = 1.3 × 10−6 m. Therefore, Psat =
6.626 × 10 −34 (Js) × 3 × 10 8 (ms −1 ) × 2.5 × 10 −12 m 2 0.4 × 3 × 10 − 20 (m) 2 × 1 × 10 −9 (s) × 1.3 × 10 −6 (m)
= 3.185 × 10−2W = 31.85 mW. (b)
Zero (or small) signal gain coefficient, Iτ g0 = Γσg c − N tr eV = 0.4 × 3 × 10
−20
0.1(A) × 1 × 10 −19 (s) 24 −3 − × (m ) 1 . 0 10 ( m ) 3 −19 −16 1.6 × 10 (C) × 5 × 10 (m ) 2
where the volume of the active region has been calculated as follows: V = 5 × 10−6 (m) ×0.5 × 10−6(m) × 200× 10−6(m) = 5 × 10−16(m3) Thus (c)
g0 = 3000m−1.
Zero (or small) signal net gain over the length L of the active region will be given by G0 = exp (g0L) = exp (3000 × 2 × 10−4) = 1.82.
39
10.4
(i)
Using eq. (10.15) of the text book, for R1 = R2 = 0.01%, we get 2
1 + 10 10 − 4 × 10 − 4 1 + G s R 1 R 2 ¨* = 1 − 10 10 − 4 × 10 − 4 1 − G s R 1 R 2
2
2
1 + 10 −3 = 1.004 = −3 − 1 10 (ii)
For R1 = R2 = 1% 2
2 1 + 10 0.01 × 0.01 1.1 ¨* = 0.9 1 − 10 0.01 × 0.01
= 1.4938 10.6
If Ps, in and Ps, out are signal powers at the input and output ends of an EDFA at the signal wavelength, λs; and Pp, in is the input pump power at a wavelength of λp; using the principle of conservation of energy, the following inequality should hold true: λp
Ps, out ≤ Ps, in +
Therefore PCE ≡
λs
Pp, in
Ps ,out − Ps ,in Pp ,in
≤
λp λs
As λp < λs, PCE < 1. The maximum value that PCE can take is
Therefore (QCE )max =
λp λs
.
λs λ λp = 1. .(PCE )max = s . λp λp λs
40
10.7
(a)
The rate of absorption per unit volume from Er3+ level E1 to pump level E3 due to the pump at λp (assuming N2 = 0) =
=
σ pa Pp a p hν p
N1 ≈
σ pa Pp λ p a p hc
N t ; [N1 ≈ Nt as N2 ≈ 0]
2.17 × 10 −25 (m 2 ) × 30 × 10 −3 (Js −1 ) × 0.98 × 10 −6 (m) × 5 × 10 24 (m −3 ) 8.5 × 10 −12 (m 2 ) × 6.626 × 10 −34 (Js) × 3 × 10 8 (ms −1 )
= 1.8879 × 1028 m−3 s−1 (b)
The rate of absorption per unit volume from Er3+ level E1 to the metastable N level E2 due to the signal at λs (assuming N2 ≈ N1 ≈ t ) 2 =
σ sa Ps σ Pλ N N 1 = sa s s t a s hν s a s hc 2
=
2.57 × 10 −25 (m 2 ) × 200 × 10 −6 (Js −1 ) × 1.55 × 10 −6 (m) × 2.5 × 10 24 (m −3 ) 8.5 × 10 −12 (m 2 ) × 6.626 × 10 −34 (Js) × 3 × 10 8 (m −1 )
= 1.1788 × 1026 m−3 s−1 The rate of stimulated emission per unit volume from level E2 to level E1 due to the signal at λs (assuming again N2 ≈ N1 ≈ Nt/2) =
σ se Ps σ Pλ N N 2 = se s s t a s hν s a s hc 2
=
3.41 × 10 −25 (m 2 ) × 200 × 10 −6 (Js −1 ) × 1.55 × 10 −6 (m) × 2.5 × 10 24 (m −3 ) 8.5 × 10 −12 (m 2 ) × 6.626 × 10 −34 (Js) × 3 × 10 8 (m −1 )
= 1.564 × 1026 m−3 s−1 10.8
(a)
In the present problem λp = 0.98 µm, λs = 1.55 µm Ps, in = 0 dBm = 1 mW Ps, out = 20 dBm = 100 mW Therefore, the gain of the amplifier, G =
41
100mW = 100 1mW
G(dB) = 10 log10 100 = 20 dB. (b)
Assuming there is no spontaneous emission G=
Ps ,out Ps ,in
≤ 1+
λ p Pp ,in λ s Ps ,in
In order to achieve a specific maximum gain, G, the input pump power should be λ Pp ,in ≥ (G − 1) s λp
.Ps ,in
The limiting value of 1.55 −3 −3 Pp, in = (100 − 1)× × 1 × 10 = 156.58 × 10 W 0.98 or
Pp, in ≈ 156.6 mW.
42
Chapter 11: Wavelength Division Multiplexing 11.2
(b)
Using eq (11.2) and (11.3) of the text book P1 = P0 cos2 (κz) and
P2 = P0 sin2 (κz)
If the splitting ratio is 10 : 90,the required interaction length L will be given by P1 1 = = cot 2 (κL) P2 9 or
tan (κL) = 3
which gives L ≈ 1.25/κ 11.5
(b)
For demultiplexer with 32 – channels spaced at 50 GHz, the FSR should be at least 32 × 50 = 1600 GHz. Since the centre wavelength is 1.55 µm, corresponding frequency is νc =
C 3 × 10 8 = = 1.935 × 1014 Hz −6 λ c 1.55 × 10
Using eq. (11.15), we get m=
νc 1.935 × 1014 = ≈ 121 (∆ν )FSR 1600 × 10 9
43
Chapter 12: Fiber Optic Communication Systems 12.2
Given that Ptx = − 13 dBm, Prx = − 40 dBm, αf = 0.5 dB/km, αsplice = 0.1 dB/splice, αcon = 0.5 dB/connector, Ms = 7 dB and the continuous length of a single piece of fiber = 5 km. Assuming the repeaterless distance to be L, the number of splices would be L L L − 1 ≈ ; and the related splice loss would be ≈ × 0.1(dB) . 5 5 5
The attenuation within the fiber would be αf.L = 0.5L (dB) Using eq. (12.1) and (12.2) of the text book, we have Ptx = Prx + αfL + αcon + αsplice + Ms or
− 13 dBm = − 40 dBm + 0.5 L (dB) + 0.5 (dB) × 1 + 0.1 (dB) ×
Solving this, we get L ≈ 37.5 km. 12.3
Link Power Budget Power supplied by transmitter, Ptx = 0 dBm Channel loss, CL = (0.5 dB/km) × L + 1 dB × 2 (connectors) where L is link length or
CL = 0.5L + 2 (dB)
Safety margin, Ms = 6 dB Receiver sensitivity = Prx = − 35 dBm Therefore we should have Ptx = Prx + CL + Ms or
0 = − 35 dBm + (0.5 L + 2) + 6.
This gives L = 54 km.
44
L + 7 dB 5
12.4
Given that L = 12 km, B = 100 Mb/s ttx = 10 ns,
trx = 12 ns
The source (LED) has ∆λ = 30 nm, mean wavelength λ = 0.85µm = 850 nm. Fiber has core index n1 = 1.46, ∆ = 0.01 and Dm = 80 ps km−1 nm−1. Therefore the pulse broadening due to material dispersion would be given by tmat = Dm L ∆ λ = (80 ps km−1 nm−1) × (12 km) × (30 nm). = 28.8 ns. tintermodal =
n 1 (n 1 − n 2 ) L; n2 c
n 2 = n 1 (1 − 2∆ ) = 1.46 0.98 = 1.445
Thus, tintermodal =
1.46 1.46 − 1.445 × 12 × 10 3 (m) . 8 −1 1.445 3 × 10 ms
= 6.06 × 10−7s = 606 ns Using eq. (12.8) of the text book, we get
[
tsys = t 2tx + t f2 + t 2rx
[
]
1 2
{
}
2
2
2
]
1 2 2
= (10ns ) + (606ns ) + (28.8ns ) + (12ns ) = 607 ns = 0.607 µs For NRZ format, tsys ≤
0.70 0.70 = = 7.0 × 10 −9 s = 7 ns 6 B 100 × 10
The total rise time of the system is much greater than 0.70/B, and hence the system can not operate with NRZ format. What bit rate is possible?
45
B= 12.6
0.70 ≈ 1.15 × 10 6 bits/s = 1.15 Mb/s. −6 0.607 × 10 s
With reference to Fig. 12.6 (b) of the text book; if the optical loss within the Bus (the optical fiber) is not neglected, the power available at the Nth tap will be given by PN = PTC [(1 – C) (1 − δ) (1 – γ)]N−1
(1)
where PT is the transmitted power, C is the fraction of optical power coupled out at each tap, δ is the fractional loss (assumed to be same) at each tap, γ is the average fractional loss within the fiber between two taps (assumed to be same at equal distances of 50 m) and N is the number of subscribers. In the present case, PT = 1 mW, PN = − 40 dBm = 10−4 mW, and C = 0.05. The insertion loss at each tap = Ltap = 0.3 dB. Now Ltap = − 10 log10 (1 − δ) = 0.3 dB This gives δ = 0.06. The minimum distance between two taps is 50 m and the optical fiber is exhibiting a loss of 0.01 dB/m. Therefore the optical loss, Lf, within the fiber between two taps is 0.50 dB. Since, Lf = − 10 log10 (1 − γ) = 0.50 dB where γ is the fractional loss; we have γ = 0.10 Substituting the values of relevant parameters in eq. (1) above, we get 10−4 = 1 × 0.05 [(1 – 0.05) (1 – 0.06) (1 – 0.10)]N−1 This gives N = 29 12.7
(a)
Fiber attenuation between A and B
= (0.1 dB/km) × ((25 km) = 2.5 dB
Attenuation between B and C
= (0.12 dB/km) × (40 km) = 4.8 dB
Attenuation between C and D
= (0.1 dB/km) × (35 km) = 3.5 dB.
46
Total fiber attenuation between A and D
= 10.8 dB.
Losses at 2 couplers (one each at A & D)
= 0.5 dB × 2 = 1 dB.
In the forward direction (when the signal is traversing from A to D). There will be loss at the joint B due to mismatch of NA and profile parameter α. This loss can be calculated as follows: The total coupling efficiency at B when the signal is traversing in the forward direction. ηT = ηNA × ηα NA 2 = NA1
2
1 + 2 / α1 × 1+ 2 / α2
2 1+ 0.17 2 . 0 = × 0.20 1 + 2 1.9 2
= (0.7225) × (0.9743) = 0.70 Therefore loss at joint B = − 10 log10 ηT = 1.52 dB There will be no loss at joint C in the forward direction. Similarly, in the backward direction (when the signal is transversing from D to A), there will be 1.52 dB loss at joint C and no loss at B. Therefore, in either direction the total channel loss CL = 13.32 dB. The desired safety margin, Ms = 7 dB. The receiver sensitivity Prx = 0.5 µw = − 33 dBm. Therefore, Ptx = Pry + CL + Ms = −33 dBm + 13.32 dB + 7 dB = − 12.68 dBm
47
or (b)
Ptx = 54 µW (Both sides)
The pulse broadening due to material dispersion would be given by tmat = [Dm1 (Length of segments AB + CD) + Dm2 (Length of segment BC)) ∆λ or,
tmat = [(70 ps km−1 (nm−1) × (25 + 35) (km) + (80 ps km−1 (nm−1) × (40 km)] × (20 nm)
or
tmat = 0.148 µs
It is given that intermodal dispersion is negligible. Therefore total rise time of the system,
[
tsys = t 2tx + t 2mat + t 2rm
]
1 2
= 0.1488µs
Maximum bit rate for NRZ format =
48
0.70 = 4.7 × 10 6 bits / s = 4.7 Mb/s −6 0.1488 × 10
Chapter 13 : Fiber Optic Sensors 13.3
(a)
n∆θ Sang (normalized) = 1 − πNA 1.0 × (10π / 180) = 1 − π × (0.17) = 0.673
[(
)
]
1 2 2
NA = 1.46 2 − (1.45)
= 0.17
Range of Sang = (1 to 0.673) (b)
16(1.46) 2 (1 + 1.46) 4
ηang =
1.0 × (10π / 180) 1 − π(0.17)
= 0.627 Lang = − 10 log10 ηang = 2.027 dB Range of loss = 0 to 2.027 dB (c)
PT = Loss + PR PT (dBm) = (Lang) dB – 30 dBm = − 27.973 dBm PT = 0.0016 mw = 1.6 µw
13.7
We know that 2πLDΩ Cλ It is given that ¨φ =
L = 400m, D = 0.12 m, Ω = 5 × 10−4 rad/s λ = 0.633 × 10−6 m, C = 3 × 108 ms−1 Therefore ¨φ =
2π × 400 × 0.12 × 5 × 10 −4 rad = 0.066 rad 3 × 10 8 × 0.633 × 10 −6
49
Chapter 14 : Laser Based Systems 14.2
The pump energy that must fall on each cm2 of Ruby crystal surface to achieve threshold inversion is (see example 14.1 of the text book) equal to N 2 hν α (ν).η (ν) Given that ν = 5.1 × 1014 Hz, α (ν) = 2 cm−1, η(ν) = 0.95 and N0 = 1.5 × 1019 atoms of Cr3+ per cm3. Taking N2 ≈
N0 = 7.5 × 1018, we get 2
the required pump energy
=
7.5 × 1018 × 6.626 × 10 −34 × 5.1 × 1014 2 × 0.95
= 1.328 J cm−2
14.3
R2
R1 R1 P L
P0
R1R2P2L
L Fig. Q. 14.3 If we assume that the loss per unit length within the cavity is α, length of the cavity is L, R1 and R2 are the reflectivities of the two end mirrors and P0 is the incident power, then in one round trip the wave will return with power P2L given by P2L = P0 R1 R2 exp (−2 α L). 50
Therefore fractional loss in power per round trip is P0 − P2 L P0 − P0 R 1 R 2 exp(−2αL) = = 1 – R1 R2 exp (−2 α L). P0 P0 If n is the refractive index of the active medium and C speed of light in free space, 2Ln the time duration of this fractional loss is . This time corresponds to an C exponential decay time constant tc (of the power) approximately given by (Yariv 1997)* tc ≈
nL , assuming = 0 and R1 ≈ R2 ≈ R ≈ 1. C(1 − R ) )
and the population inversion density at threshold Nth = (N2 − N1)th =
8πn 3 ν 2 t spont ∆ν c3 t c
Given that R1 = 1, R2 = R = 0.99, L = 10 cm = 0.1 m, n = 1, ¨ν = 109 Hz, ν = c/λ = 4.74 × 1014 Hz and tspont = 0.1 µs. Substituting values of relevant parameters, we get tc =
nl 1 × 0.1(m) = = 3.33 × 10 −8 s c(1 − R ) 3 × 10 8 ms −1 (1 − 0.99 )
(
and N th = ( N 2 − N1 ) th =
)
8 × π × (1) 3 × (4.74 × 1014 s −1 ) 2 × 10 −7 (s)
(3 × 10
8
ms
) × 3.33 × 10
−1 3
−8
× 10 9 (s −1 )
(s)
= 6.28 × 1014 m−3 = 6.28 × 108 cm−3 14.8
The time difference between the two pulses is given by ¨t =
2dn , where the d is the depth of the object. c
Given that ¨W µs and n = 1.33. Therefore, *
A. Yariv : Optical Electronics in Modern Communications, (5th Ed. ), Oxford University Press, N.Y., p. 188 (1997)
51
d=
c∆t 3 × 10 8 (ms −1 ) × (0.53 × 10 −6 )(s) = 60 m = 2n 2 × 1.33
52
