Solutions Manual to
)LEHU2SWLFVDQG
2SWRHOHFWURQLFV
R.P. Khare
Birla Institute of Technology and SciencePilani (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].
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
However, NA is also given by: NA = n1 2∆
Therefore, n1 = 03 . 0 2 34 . 0 × = 1.388
Now na sin αm = n1 sin θm = n1 cos φc = 0.34
This gives cos φc≈ 388 . 1 34 . 0 = 0.2449 and φc≈ 76o
2.6 (a) NA = 0.17 = na sin αm = 1.33 × sin αm
Therefore αm = 7.34o (b) Since NA =
(
)
2 1 2 2 2 1 n n − = 0.17 and n2 = 1.46 n1 =[
(
0.17) ( )
1.46]
2 1.47 1 2 2 ≈ +Now na sin αm = n1cosφc = 1.47cosφc =0.17;
2.7 (a) The r.i. of core = n1 =
( )
( )
1.5 ms 10 2 ms 10 3 v c 1 8 1 8 = × × = −−and; since sin φc =
1 2
n n
,
the r.i. of cladding = n2= n1 sin φc = 1.5 sin 75o≈ 1.448
Therefore NA =
[
n n]
[
( ) (
1.5 1.448)
]
2 0.388 1 2 2 2 1 2 2 2 1 − = − = (b)(
8( )
1)
10 1 2 2 1 1 sm 10 79 . 1 448 . 1 ms 10 3 448 . 1 5 . 1 5 . 1 n n n c n l T − − − × = × × − = − = ∆2.9 (a) The power p(t) in the pulse varies as follows: p(t) = t; 2 p0 × τ 0 < t < 2τ = 0; elsewhere Therefore, = τ τ = τ = ε τ τ
∫
0 2 0 2 0 2 0 0 p 2 t 2 p . dt t . 2 p (b) = τ τ τ = ε =∫
τ∫
τ 3 4 dt t 2 p p 1 . dt t ). t ( p 1 t 2 2 0 0 0 2 0 (c)∫
τ −[ ]
ε = σ 2 0 2 2 2 t dt ) t ( p t 1∫
τ τ − τ τ = 2 0 2 0 2 0 3 4 dt t 2 p . t p 1 = 2τ2 – 9 16τ2 = 9 2τ2 σ = τ 22.10 (a) NA =
[
(
1.496) ( )
1.40]
2 0.5272 1 2 2 − = 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,
Chapter 3: Wave propagation in planar waveguides 3.8: (a) V =
(
)
( )
( ) ( ) (
[
)
]
2 1 2 2 2 1 2 2 2 1 1.46 1.455 m 30 . 1 m 14 n n a 2 − µ µ × π = − λ π V ≈ 4Referring 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 can also be gathered by calculating
π
V 2
and recalling that M is an integer greater than π V 2 .
(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 ) 2 m m V a w b = 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
The phase propagation constants βm for different modes, m, can be calculated using the following relation:
(
)
{
}
2 1 2 2 2 1 m 2 2 m = β +b β −β β where β1 = kn1 = 1 6 6 1 m 10 0565 . 7 10 30 . 1 46 . 1 2 n 2 − − = × × × π = λ π β2 = kn2 = 2 6 7.0323 106m 1 10 30 . 1 455 . 1 2 n 2 − − = × × × π = λ π3.9 The maximum thickness of the guide layer that can support M modes is given by 2a =
(
)
2 1 2 2 2 1 n n 2 M − λIn the present case, M = 10, λ = 0.90 µm, n1 = 3.6 and n2 = 3.58.
Therefore, 2a =
( )
( ) ( )
[
3.6 3.58]
11.875 m 2 m 90 . 0 10 2 1 2 2 µ = − µ × βmax = β1 = kn1 = 1 5 6 1 m 10 513 . 2 10 90 . 0 6 . 3 2 n 2 − − = × × × π = λ π βmin = β2 = kn2 = 2 6 2.499 105 1 10 90 . 0 58 . 3 2 2 − − = × × × = m n π λ π 3.10 (a) V =(
)
( )
( )
.(
1.5 2 0.02)
m 55 . 1 m 10 2 n a 2 1 × µ µ × π = ∆ λ π 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 wma. The results are listed in the tabular form below.
m um a (rad) um(m− 1 ) wma(rad) wm(m− 1 ) m 2 m V a w b = 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
where β1 = kn1 = 1 6 6 1 m 10 08 . 6 10 55 . 1 5 . 1 2 n 2 − − = × × × π = λ π β2 = kn2 = 2 6 5.9588 106m 1 10 55 . 1 47 . 1 2 n 2 − − = × × × π = λ π This gives, βm = [1.459 bm + 35.507]½ × 106 m−1
3.11 For antisymmetric modes, Ey(x) = B sin ux; | x | < a = Dexp
(
w|x|)
;|x| a | x | x − >Power inside the guide layer,
(
)
∫
− ωµ β = a a 2 0 in Bsinux dx 2 1 P =∫
(
−)
ωµ β a 0 2 0 dx ux 2 cos 1 B 2 = a 0 2 0 ux 2 sin u 2 1 x B 2 − ωµ β = − ωµ β ua 2 sin u 2 1 a B 2 2 0 (1) Power outside the guide layer,( )
( )
+ ωµ β =∫
−∫
∞ − ∞ a a 2 y 2 y 0 out E x dx E x dx 2 1 P =∫
∞ ωµ β a 2 y 0 dx ) x ( E 2 2 1= D exp
(
2wx)
dx a 2 0∫
∞ − ωµ β = ∞ − − ωµ β a wx 2 2 0 e w 2 1 D = − ωµ β 2 −2wa 0 e w 2 1 D (2)The confinement factor G = out in in P P P + (3)
Substituting the values of Pin and Pout from (1) and (2) in (3), we get
ωµ β + − ωµ β − ωµ β = −2wa 2 0 2 0 2 0 e . w 1 D . 2 1 ua 2 sin . u 2 1 a B . 2 1 ua 2 sin . u 2 1 a B . 2 1 G = 2 wa 2 B D e . w 1 ua 2 sin u 2 1 a ua 2 sin . 2 1 a + − − − (4) At x = a, B sin ua = D e−wa Therefore, e .sinua B D = wa (5) Substituting for B D
from (5) in (4) and manipulating, we get,
G = + − − ua sin . 1 ua 2 sin 1 a ua 2 sin . u 2 1 a 2
This result can also be expressed in the following form: G =
( )
1 2 ua ua cos . ua sin 1 wa ua sin 1 − − + (6)(b) Given that 2a = 6.523 µm, n1 = 1.50, n2 = 1.48 and λ = 1.0µm
Therefore, V =
(
)
2 1 2 2 2 1 n n a 2 − λ π =[
( ) ( )
]
2 1 2 2 48 . 1 5 . 1 1 523 . 6 − π × ≈ 5So 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.
Gm = 0 = 1 2 30644 . 1 ) 30644 . 1 sin( ). 30644 . 1 cos( 1 8263 . 4 ) 30644 . 1 ( cos 1 − + + = 0.98828 Gm = 1 = 1 2 5974 . 2 ) 5974 . 2 sin( ). 5974 . 2 cos( 1 27342 . 4 ) 5974 . 2 ( sin 1 − − + = 0.94889 Gm = 2 = 1 2 83747 . 3 ) 83747 . 3 sin( ). 83747 . 3 cos( 1 20529 . 3 ) 83747 . 3 ( cos 1 − + + = 0.85992 Gm = 3 = 1 2 9063 . 4 ) 9063 . 4 sin( ). 9063 . 4 cos( 1 963466 . 0 ) 9063 . 4 ( sin 1 − − + = 0.50960
Chapter 4: Wave Propagation in Cylindrical Waveguides 4.2 (b) (i) 2 1 2 2 2 1 n 2 n n − = ∆ Therefore n2 = n1 (1 − 2¨)½ = 1.46 (1 – 2 × 0.01)½ = 1.445 (ii) 1.46 (2 0.01) 25 ) m ( 3 . 1 ) m ( 50 2 n a 2 V 1 × × ≈ µ µ × π = ∆ λ π = (iii) Ms =
( )
312 2 25 2 V2 2 = = 4.3 (b) max 1 1 6 10.93 106m 1 ) m ( 10 85 . 0 48 . 1 2 n 2 kn − = × − × × π = λ π = = β 1 6 6 2 2 min 10.82 10 m ) m ( 10 85 . 0 465 . 1 2 n 2 kn − = × − × × π = λ π = = β4.4 The cut off frequency for single mode operation is given by Vc = 2.405 2 1 2 1 α +
This should be equal to V given by
V = ∆ λ π 2 n a 2 0
Therefore, for this case the following equality should be satisfied
∆ λ π 2 n a 2 0 = 2.405 2 1 2 1 α + or 2a = ∆ π α + λ 2 n 2 1 405 . 2 0 2 1
Hence (i) for λ = 1.3 µm,
( )(
)
(
2 0.013)
7.1 m 5 . 1 2 1 m 3 . 1 405 . 2 a 2 2 1 µ = × × × π + µ × =and (ii) for λ = 1.55 µm,
( )(
)
(
2 0.013)
8.5 m 5 . 1 2 1 m 55 . 1 405 . 2 a 2 2 1 µ = × × × π + µ × = 4.6(
)
+ ∆ α + α − = = 3 2 c g eff g kR n 2 3 R a 2 2 2 1 2 1 M ) M ( (1)In the present problem α = 2, ¨ D µm, nc = 1.48 and k = 1 ) m ( 39 . 7 ) m ( 85 . 0 2 2 = µ − µ π = λ π . Substituting the valves of different parameters in eq. (1), we get + − = 3 2 R 1371 . 0 R 100 100 1 2 1 ; where R is measured in µm.
Solving this equation, we get R ≈ 1.66 × 104µm or R ≈ 1.66 cm
4.7 The complete transverse electric field will be given by Ez = A Jl a ur .eilφ.ei(ωt−βz); r < a (1)
and Ez = B Kl a wr .eilφ.ei(ωt−βz); r > 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 Hz = C Jl a ur .eilφ.ei(ωt−βz); r < a (3) and Hz = D Kl a wr .eilφ.ei(ωt−βz); r > a (4)
Let us derive the expression for Er first inside the core and then in the cladding. It
is given that φ ∂ ∂ ωµ + ∂ ∂ β − = z z 2 r r H . r 1 r E k i E (5)
Substituting the values of Ez and Hz from (1) and (3) in (5), we get
φ ∂ ∂ ωµ + ∂ ∂ − = φ ω−β φ i(ωt−βz) l il ) z t ( i il l 2 r r .e a ur CJ . e . r 1 e e . a ur J r AB k i E = ωµ + ′ β − φ ω−β ilφ i(ωt−βz) l ) z t ( i il l 2 r e . e a ur J . r i C e e . a ur J . a u A k i or 2 l l il i( t z) r r e e a ur J r i C a ur J A a u k i E φ ω−β ωµ + ′ β − = (6) where k2r =
(
ω2εµ−β2)
=[
ω2.ε0εr.µ0µr −β2]
Inside the dielectric medium, such as glass fiber,µr = 1, εr = n2 and
ε0µ0 = 2 C
1 .
Therefore
[
2 2 2]
2 2 2 2 2 r k n c n k = −β ω −β = (7) where k = λ π = ω 2c = free space propagation constant.
Note that for a SI fiber inside the core, n = n1 and in the cladding n = n2.
Similarly, inside the cladding, (r > a).
) z t ( i il l l 2 r r e .e a wr K r i D a wr K a w B k i E φ ω−β ωµ + ′ β − = (8)
Expressions for Eφ, Hr and Hφ can be derived in a similar manner.
4.8 n(r) = n0 ;r a a r 2 1 2 1 ≤ ∆ − α Substituting ¨ 2 0 2 c 2 0 n 2 n n −
in the above equation and taking the square on both sides, we get − − = α a r . n 2 n n 2 1 n ) r ( n 2 0 2 c 2 0 2 0 2 . =
(
)
− − α a r n n n 2c 2 0 2 0Therefore the local numerical aperture, NA(r) would be given by NA(r) =
{
}
2 1 2 c 2 n ) r ( n − =(
)
2 1 2 c 2 c 2 0 2 0 n a r n n n − − − α=
(
)
2 1 2 c 2 0 a r 1 n n − − αThe mean square value of the numerical aperture taken over the core area may be given by the following relation:
Mean square value of NA =
∫
∫
π π a 0 a 0 2 rdr 2 rdr 2 ). r ( NA =(
)
2 a 0 2 c 2 0 a rdr 2 a r 1 n n π π − −∫
α =(
)
a 0 2 2 2 2 c 2 0 ) 2 ( r . a 2 2 r 2 a n n + α π − π π − α+ α =(
)
+ α π − π π − 2 a 2 a a n n 2 2 2 2 c 2 0 =(
)
+ α α − 2 n n20 2c .Therefore, root mean square value of numerical aperture for an α-profile would be given by (NA)rms =
(
)
2 1 2 c 2 0 n n . 2 − + α α 4.9 (a) (NA)rms =(
)
2 1 2 c 2 0 n n . 2 − + α αGiven that α = 2, n0 = 1.460 and nc = 1.445
Therefore (NA)rms =
{
( ) (
)
}
2 1 2 2 445 . 1 46 . 1 2 2 2 − += 0.1476 (b) β0 = kn0 =
(
)
1 3 0 1.46 7056mm mm 10 3 . 1 2 n 2 − − × = × π = λ π ¨( ) (
)
( )
1.46 0.0102 2 445 . 1 46 . 1 n 2 n n 2 2 2 2 0 2 c 2 0 = × − = − V =(
)
[
( ) (
1.46 1.445)
]
25.23 3 . 1 50 n n a 2 2 2 21 2 1 2 c 2 0 − = × π = − λ π (c) Mg =(
)
159 2 23 . 25 2 2 2 2 V 2 2 2 = + = + α α4.10 (a) Given that α = 1, Mg = 500, n0 = 1.46, 2a = 75 µm and λ = 1.3 µm.
Therefore Mg = 6 V 2 V . 2 1 1 2 V . 2 2 2 2 = + = + α α or V = 6Mg = 6×500 =54.77 As V can also be expressed as
V ≈ ∆ λ π 2 n a 2 0 , we have ¨ 0.021 46 . 1 75 77 . 54 3 . 1 2 1 n ) a 2 ( V 2 1 2 2 0 = × × π × = π λ (b) Vc = 2.405 4.165 1 2 1 405 . 2 2 1 2 1 2 1 = + = α +
If we assume that the core diameter for single mode operation is 2a′, V can be written as V = 2 a n0 2∆ =4.165 λ ′ π
( )
m 3 . 1 165 . 4 × µ4.11 (a) Given that α = 2, 2a = 70µm n0 = 1.47, nc = 1.45 and λ = 1.3µm. Then β0 = kn0 = 0 6 7.1048 106m 1 ) m ( 10 3 . 1 47 . 1 2 n 2 − − = × × × π = λ π βc = knc = 1 6 6 c 7.1008 10 m ) m ( 10 3 . 1 45 . 1 2 n 2 − − = × × × π = λ π
( ) ( )
( )
1.47 0.01351 2 45 . 1 47 . 1 n 2 n n 2 2 2 2 0 2 c 2 0 = × − = − = ∆ (b) V = 1.47 2 0.01351 40.88 3 . 1 70 2 n a 2 0 × × = × π = ∆ λ π Mg =(
)
418 4 88 . 40 4 V2 = 2 = .Chapter 5: Single Mode Fibers 5.2 (b) Given that n1 = 1.46, 2a = 8µm, ¨ In general, V = ∆ λ π 2 n a 2 1
For single mode operation, in a SI fiber V = Vc = 2.405, Therefore,
( )
405 . 2 0052 . 0 2 46 . 1 m 8 2 n V a 2 1 c c × × µ × π = ∆ π = λ or λc = 1.556µm 5.4 V = ∆ λ π 2 n a 2 1 At λ1 = 1.31µm, V1 = 1.4677 2 0.0036 2.449 31 . 1 1 . 4 2 = × × × πTherefore, using eq. (5.3) of the text book we get, w =
(
) (
)
+ + 6 2 3 449 . 2 879 . 2 449 . 2 619 . 1 65 . 0 1 . 4 = 4.1 [0.65 + 0.4224 + 0.01334] or w = 4.452 µmEmploying eq. (5.5), we get wp = 4.452 – 4.1
(
)
+ 7 449 . 2 567 . 1 016 . 0or wp = 4.374µm At λ2 = 1.55µm, V2 = 1.4682 2 0.0036 2.0705 55 . 1 1 . 4 2π× × × = w = 4.1
(
) (
)
+ + 6 2 3 0705 . 2 879 . 2 0705 . 2 619 . 1 65 . 0 = 4.1 [0.65 + 0.5434 + 0.0365] or w = 5.0427 µm wp = 5.0427 – 4.1(
)
+ 7 0705 . 2 567 . 1 016 . 0 or wp = 4.9377µm 5.5 (a) δβ1 = 1 6 10 6 4.833m 1 ) m ( 10 3 . 1 2 n . 2 − − − × = × π = δ λ π and δβ2 = 2 6 10 5 48.33m 1 ) m ( 10 3 . 1 2 n . 2 − − − × = × π = δ λ π Range of δβ = 4.833 – 48.33 m−1 (b) Lp1 = (m) 1.30m 833 . 4 2 2 1 = π = δβ π Lp2 = (m) 0.13m 33 . 48 2 2 2 = π = δβ π Lp varies from 13 cm to 1.3 m5.7 Given that n1 = 1.48, ¨ λ = 1.32 and a = 4.4 µm
n2≈ n1(1 −¨ V = 1.48 2 0.0027 2.2778 32 . 1 4 . 4 2 = × × × π
Using eq. (5.24) of the text book, we have
(
)
2 2 2 V 834 . 2 549 . 0 080 . 0 ) bV ( dV Vd − + ≈ = 0.2498 From eq. (5.23), we get( )
( )
0.2498 m 32 . 1 ms 10 3 0027 . 0 476 . 1 Dw 8 1 × µ × × × − = − = −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−1From eq. (5.23) and (5.24), we get
Dw =
[
(
)
]
2 2 V 834 . 2 549 . 0 080 . 0 c n − + λ ∆ − n1 = 1.48, ¨ Q2 = n1(1 −¨ − 0.01) or n2 = 1.4652 Dw =( )
(
)
[
(
)
]
2 1 8 0.080 0.549 2.834 V m 55 . 1 ms 10 3 01 . 0 4652 . 1 + − µ × × × − − = − 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 = ∆ λ π 2 n a 2 1 , we get
Core radius, a =
( )
01 . 0 2 48 . 1 2 m 55 . 1 325 . 2 2 n 2 V 1 π× × µ × = ∆ π λ × = 2.74 µm 5.11 Fig. Q 5.11 Given that, n1 = 1.46, 2a = 8.2 µm, ¨ D1 = 25µm P(r) = P0 exp − 2 2 w r 2 At λ1 = 1.30 µm, V1 = 1.46 2 0.003 2.241 3 . 1 2 . 8 2 n a 2 1 1 = × × × π = ∆ λ π 8.2 µm 25 µmV2 = 1.46 2 0.003 1.879 55 . 1 2 . 8 2 n a 2 1 2 = × × × π = ∆ λ π
Therefore at λ1 = 1.3 µm, (using eq. 5.3), w ≈ 4.1
(
)
(
)
+ + 1.5 6 241 . 2 879 . 2 241 . 2 619 . 1 65 . 0 = 4.1 [0.65 + 0.48259 + 0.0227] w = 4.7368 µm and at λ2 = 1.55 µm, w = 4.1(
)
(
)
+ + 1.5 6 879 . 1 879 . 2 879 . 1 619 . 1 65 . 0 = 4.1 [6.65 + 0.6285 + 0.0654] = 5.51 µmAt λ1 = 1.30 µm, the fractional power at r = a1 = 12.5 µm
µ µ − = − = 2 2 0 4.7368 m m 5 . 12 2 exp w r 2 exp P ) r ( P 7 0 10 9387 . 8 ) ( = × − P r P (1) and at λ2 = 1.55µm, the fractional power at r = a1 = 12.5µm
5 2 0 10 3869 . 3 m 51 . 5 m 5 . 12 2 exp P ) r ( P = × − µ µ − = (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.
Chapter 6: Optical fiber cables and connections 6.4 LF = − log10 (ηF) = 0.36 dB This gives ηF= 0.92 = 4 2 ) 1 k ( k 16
+ (from eq. 6.3 of text book)
For air, n = 1, therefore k = 1 1
n n n = Thus k2− 2.17 k + 1 = 0 or n12 −2.17n1+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 log10ηang = 0.6 dB (1)
This is due to 4o angular misalignment. From eq. (1), ηang = 0.87
But from eq. (6.8),
87 . 0 NA n 1 ang = π θ ∆ − ≈ η
In this problem, n = 1.46 and ¨θ = 4o = 0.0698 rad and hence NA = 0.25 13 . 0 0698 . 0 46 . 1 = × π × 6.8 Given that 12% 0.12 a 2 y = = ∆ n1 = 1.5 and n = 1.47
Therefore, k = 1.02 47 . 1 5 . 1 n n1 = = ηF = 0.99957 663 . 16 041 . 1 16 ) 1 02 . 1 ( ) 02 . 1 ( 16 ) 1 k ( k 16 4 2 4 2 = × = + = + And, ∆ − × ∆ − ∆ π = η − 2 1 2 1 lat a 2 y 1 a 2 y a 2 y cos 2 or
{
}
− − π = η 2 1 2 lat 1.4505 (0.12)1 (0.12) 2 = 0.8475Hence 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 log10ηT≈ 0.72 dB.
6.10 The Fresnel reflection coefficient, R is given by R = 2 1 1 n n n n + −
Its first derivative w.r.t. n1 would be
(
) (
)
(
)
(
)
+ × + − − + + + + − = n n dn d n n 1 n n n n dn d n n 1 n n n n 2 dn dR 1 1 2 1 1 1 1 1 1 1 1 or(
)
(
(
)
)
+ − − − − + + − = 1 2 1 1 1 1 1 1 1 dn dn 1 n n n n dn dn 1 n n 1 n n n n 2 dn dRIn this expression n is a constant as it is the refractive index of the surrounding medium.
Hence 0 dn dn 1 = . This gives us
(
)
(
) (
[
) (
)
]
(
(
)
3)
1 1 1 1 1 1 1 n n n n n 4 n n n n n n n n 2 dn dR + − = − − + + − =For small variation in n1, ¨Q1, will, therefore, produce a small variation ¨5LQ5
given by: ¨5
(
)
(
)
3 1 1 1 n . n n n n n 4 ∆ + −Chapter 7: Optoelectronic Sources
7.3 The net rate of recombination per unit volume of excess minority electrons in the p-side =
( )
p n dt t n d τ ∆ = ∆ − (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
( )
( )
∫
∫
∞ ∞ τ − ∆ τ − ∆ = 0 p 0 p t exp . 0 n dt t exp 0 n . t t or t =τp (3)7.4 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) + = η d a h e e h inj N N L L D D 1 1 Now De =
(
) (
)
(
)
C 10 6 . 1 K 300 JK 10 38 . 1 s V m 85 . 0 e kT 19 1 23 1 1 2 e − − − − − × × × × = µ = 0.02199 m2 s−1 Similarly, Dh = 1 2 3 h 1.035 10 m s e kT = × − −Therefore ηinj = × × × + −3 2321 10 10 1 1 02199 . 0 10 035 . 1 1 1 = 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
( )
Eph ∝∫
n( ) ( )
E2 .pE1 dE2P
with the constraint that E2 – E1 = Eph
or P(Eph) =
(
)
(
)
2 1 v E E E E c 2 dE kT E E exp . kT E E exp ph v c 2 − − − − α∫
+ = where α is a constant. or P(Eph) =(
)
(
)
2 1 2 E E E E v c dE kT E E exp . kT E E exp ph v c 2 − − − α∫
+ = = α exp∫
+ − − v ph c E E E 2 g ph dE kT E E =(
)
− − − α kT E E exp E Eph g ph gElectron Energy E2 Ev+Eph Ec n (E2) Eg Ev p(E1) Ec-Eph E1 Fig. Q 7.5 Fig. Q.7.5 (b)
( )
(
)
(
)
0 kT E E -1 kT E E exp dE E dP ph g ph g ph ph = − − − α = , for maximum P. Therefore (Eph)peak = Eg + kT;and P (at peak Eph) = .
e kT
α
7.7 (a) Using eq. (7.61) of text book
( )
E
phFree Electron Density
(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)
( ) ( )
( )
1.50 0.01137 16 . 0 n NA 2 2 2 a 2 m = = = φ φ (d) ηT = ηext m φ φ = 0.0295 × 0.01137 = 3.35 × 10−4 (e) If the LED is emitting in air,( ) ( )
( )
1 0.0256 16 . 0 n NA 2 2 2 a 2 m = = = φ φ 7.8 (a) 0.4545 100 120 1 1 1 1 nr rr int = + = τ τ + = η (b) int int Eph e I . η = φ = 0.4545 ×( )
A[
1.42 e( )
J]
e 10 100 3 × × − = 0.0645 W = 64.5 mWPower supplied to LED = VI = 1.5 × 100 × 10−3 = 150 × 10−3W = 150 mW. Therefore internal power efficiency = 0.43.
150 5 . 64
(c) If the diode is emitting in air, the external quantum efficiency would be given by
(
)
(
)
2 s s s int ext 1 n n 2 a 1 + − η = η =(
)
(
)
2 1 7 . 3 7 . 3 2 1 . 0 1 4545 . 0 + × − × = 0.010(
1.42 e)
e 10 100 010 . 0 E e I 3 ph ext ext × × × × = × × η = φ − or φext = 1.42 × 10−3W = 1.42 mWTherefore, external power efficiency = 0.0095 150 42 . 1 = (d) ηT = ηext.
( )
(
2)
a 2 1 ext 2 a 2 n 2 n n NA =η ∆ Given that na = 1.5, NA = n1 2∆ = 1.46 2×0.02=0.292Using eq. (7.65), we can calculate ηext for this part as follows
ηext = ηint (1 − as)
(
)
2 s a s 3 a n n n n 2 + = 2 3 ) 7 . 3 5 . 1 ( 7 . 3 ) 5 . 1 ( 2 ) 1 . 0 1 ( 4545 . 0 + × − × = 0.0275 Therefore, ηT = 0.0275 ×(
)
( )
2 3 2 10 042 . 1 5 . 1 292 . 0 − × = and φT = ηT(
1.42 e)
e 10 100 10 042 . 1 E e I 3 3 ph × × × × = − −Thus overall source-fiber power coupling efficiency = 9.86 10 4 mW 150 mW 14796 . 0 = × −
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
F = 2 s 2 a n 2 n
(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 2 s 2 a n n F′=
Further, R and t remain the same but as = 0 and hence T = 1.
Therefore ηext = ηint F′tT
=
(
)
2 s a s a 2 s 2 a int n n n n 4 n n + × × η × 1 =(
)
2 s a s 3 a int n n n n 4 + ηGiven that na = 1, ns = 3.7, ηint = 0.60
(
1 3.7)
0.029 7 . 3 1 4 60 . 0 2 ext = + × × = η(a) Optical power emitted within the LED = φint = ηint .Eph e I = 0.60 120 10 A
(
1.43eV e)
3 × × − = 102.96 × 10−3W= 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 = 0.572 180
96 . 102
=
(b) Optical power emitted in air by the LED = φext = ηext .Eph e I = 0.029
(
1.43eV e)
e A 10 120 3 × × − = 4.976 × 10−3W = 4.976 mWTherefore external power efficiency = 0.0276 180
976 . 4
=
7.11 Using eq. (7.91) of text book, we can calculate the cavity length, L as follows
L =
( )
( )
Hz 1.62 10 m 10 250 7 . 3 2 ms 10 3 n 2 c 4 9 1 8 − − × = × × × × = δν or L = 162 µmThe number of longitudinal modes can be calculated using eq. (7.90) as follows:
( )
m 162 7 . 3 2 1 × × µ νor m = 1410
7.12 In terms of wavelength m can be expressed as follows:
( )
( )
m 85 . 0 m 500 7 . 3 2 nL 2 m µ µ × × = λ = = 4353 δm = δλ λ .2nL 1 2Therefore mode separation in terms of wavelength will be given by (putting
δm = 1),
( )
1.95 10 m 500 7 . 3 2 85 . 0 nL 2 m 4 2 2 µ × = × × = λ = δ − or δλ = 0.195 nmIn terms of frequency, m is given by eq. (7.90), i.e. m = ν c 2nL; and δm = δν c nL 2
. Therefore the mode separation in terms of frequency will be given by (with δm = 1)
10 6 1 8 10 1 . 8 ) m ( 10 500 7 . 3 2 ms 10 3 nL 2 c m = × × × × × = = δ −− Hz = 81 GHz.
7.13 The threshold gain coefficient for an ILD is given by (see eq. 7.92 of text book).
gth = + α Γ 1 2 eff R R 1 ln L 2 1 1
Given that gth≈βJth. Therefore
Jth = + α Γ β 1 2 eff R R 1 ln L 2 1 1 (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 n = 3.7 and hence R1 = 0.33 1 7 . 3 1 7 . 3 1 n 1 n 2 2 = + − = + − and R2 = 1.
Substituting the values of given parameters in (1), for a strong carrier confinement
(
Γ=1)
, we get × + × = 33 . 0 1 ln 03 . 0 2 1 12 1 02 . 0 1 Jth ≈ 1524 A cm−2The 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 0.185 m 5 . 3 2 ) m ( 3 . 1 1 2n m L or . nL 2 m = µ × µ × = λ = λ =
Chapter 8: Optoelectronic Detectors
8.3 Given that η = 50% = 0.5 at λ = 0.90 µm. (a) ) ms ( 10 3 ) Js ( 10 626 . 6 ) m ( 10 90 . 0 ) C ( 10 6 . 1 5 . 0 hc e 1 8 34 6 19 − − − − × × × × × × × = λ η = ℜ = 0.362 AW−1 −6
Pin = =
(
)
= ℜ − − 1 6 p AW 362 . 0 A 10 I 2.76 ×10−6 W = 2.76 µW (c) The rate of photons received will be given byrp = 34 8
( )
1 6 6 in in ms 10 3 ) Js ( 10 626 . 6 ) m ( 10 90 . 0 ) W ( 10 76 . 2 hc P h P − − − − × × × × × × = λ = ν = 1.25 × 1013 s−18.4 Given that η = 0.50 at λ = 1.3 µm. Its responsivity would be 523 . 0 10 3 10 626 . 6 10 3 . 1 10 6 . 1 50 . 0 hc e 8 34 6 19 = × × × × × × × = λ η = ℜ −− Therefore Ip = ℜ. Pin = 0.523 × 0.4 × 10−6 = 0.2092 × 10−6 A or Ip = 0.2092 µA
The output photocurrent after avalanche gain is I = 8 µA.
Thus the multiplication factor M = 38 2092 . 0 8 I I p ≈ = .
8.5 For an ideal p-n photodiode, η = 1. (a) At λ = 0.85 µm 1 8 34 6 19 AW 684 . 0 10 3 10 626 . 6 10 85 . 0 10 6 . 1 1 − − − − = × × × × × × × = ℜ (b) At λ = 1.30 µm, 1 8 34 6 19 AW 046 . 1 10 3 10 626 . 6 10 3 . 1 10 6 . 1 1 − − − − = × × × × × × × = ℜ λ µ
1 8 34 6 19 AW 248 . 1 10 3 10 626 . 6 10 55 . 1 10 6 . 1 1 − − − − = × × × × × × × = ℜ 8.6 Ip = ℜ × Pin = 0.40 (AW−1) × 100 × 10−6 2 2
( )
mm2 mm W × or Ip = 80 × 10−6 A = 80 µA.Chapter 9: Optoelectronic Modulators 9.2
(
) (
41.658 1.486)
0.856 10 m ) m ( 10 3 . 589 n n 4 x 6 9 e 0 − − × = − × = − λ = = 0.856 µm 9.3(
) (
)
32738.88 10 m 544 . 1 553 . 1 2 ) m ( 10 3 . 589 n n 2 x 9 9 0 e − − × = − × = − λ = = 0.0327 mm9.5 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 = 2 10 V/m 10 5 100 l V 4 3 = × × = − 4 12 3 z 33 3 0 (2.29) 30.8 10 2 10 2 1 E . r n 2 1 n= = × × × × ∆ − or ¨Q × 10−6 ¨φ = 9 3.698 10 6 5 10 3 10 550 2 l n 2 − − − × × × × × π = × ∆ × λ π = 0.067π Therefore Φ = 2¨φ = 0.1344π (b) Vπ = 3 12 9 33 3 0 2 (2.29) 30.8 10 10 550 r n 2 − − × × × × = λ or Vπ = 743.5 V
9.6 (a) ¨φ (due to intrinsic birefringence) = 2 l
(
n0 −ne)
λ π
Substituting the values of constants, we get ). 20 . 2 29 . 2 ( m 10 550 l 2 2 × 9 − × π = π − Therefore, 1.5277 10 m 1.528 m. ) 09 . 0 ( 4 m 10 550 l 6 9 µ ≈ × = × = − −
(b) ¨φ (due to external field) = .V. d l n r33 30 λ π
(
)
20 d 10 528 . 1 29 . 2 10 8 . 30 m 10 550 2 6 3 12 9 × × × × × × π = π − − − d = (2.29) (1.528) 20 41.1 10 m 0.041 m. 550 m 10 8 . 30 2 3 9 9 µ = × = × × × × × − − (c) Vπ = V 528 . 1 041 . 0 10 8 . 30 ) 29 . 2 ( 10 550 l d r n 2 3 12 9 63 3 0 × × × × = λ − − = 40 V 9.8 Sin θm = m × × × = Λ λ − − m 10 375 33 . 1 m 10 633 1 n 6 9 0 = 1.269 × 10 −3( )
× = × = = Λ − − − m 10 375 s 10 4 ms 1500 Hz frequency wave accoustic of Velocity 6 1 6 1 m = 1 θm = 0.0727oChapter 10 : Optical Amplifiers 10.3 (a) Psat = λ τ σ Γ = τ σ Γ ν c g c g hcA A h
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 = ) m ( 10 3 . 1 ) s ( 10 1 ) m ( 10 3 4 . 0 m 10 5 . 2 ) ms ( 10 3 ) Js ( 10 626 . 6 6 9 2 20 2 12 1 8 34 − − − − − − × × × × × × × × × × × = 3.185 × 10−2W = 31.85 mW.
(b) Zero (or small) signal gain coefficient, g0 = Γσg τ − tr c N eV I = 0.4 × 3 × 10−20(m2) × − × × × × × − − − − ) m ( 10 0 . 1 ) m ( 10 5 ) C ( 10 6 . 1 ) s ( 10 1 ) A ( 1 . 0 24 3 3 16 19 19
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 g0 = 3000m−
1
.
(c) 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
10.4 (i) Using eq. (10.15) of the text book, for R1 = R2 = 0.01%, we get ¨* 2 4 4 4 4 2 2 1 s 2 1 s 10 10 10 1 10 10 10 1 R R G 1 R R G 1 × − × + = − + − − − − = 1.004 10 1 10 1 2 3 3 = − + − − (ii) For R1 = R2 = 1% ¨* 2 2 9 . 0 1 . 1 01 . 0 01 . 0 10 1 01 . 0 01 . 0 10 1 = × − × + = 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: Ps, out≤ Ps, in + s p λ λ Pp, in Therefore PCE ≡ s in , p in , s out , s p P P P λ λ ≤ − As λp < λs, PCE < 1.
The maximum value that PCE can take is
s p λ λ
.
Therefore
(
QCE)
.(
PCE)
. 1.s p p s max p s max λ = λ λ λ = λ λ =
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) t p p p pa 1 p p p pa N hc a P N h a P σ λ ≈ ν σ = ; [N1≈ Nt as N2≈ 0] ) ms ( 10 3 ) Js ( 10 626 . 6 ) m ( 10 5 . 8 ) m ( 10 5 ) m ( 10 98 . 0 ) Js ( 10 30 ) m ( 10 17 . 2 1 8 34 2 12 3 24 6 1 3 2 25 − − − − − − − − × × × × × × × × × × × × = = 1.8879 × 1028 m−3 s−1
(b) The rate of absorption per unit volume from Er3+ level E1 to the metastable
level E2 due to the signal at λs (assuming N2≈ N1≈
2 Nt ) 2 N hc a P N h a P t s s s sa 1 s s s sa = σ λ ν σ = ) m ( 10 3 ) Js ( 10 626 . 6 ) m ( 10 5 . 8 ) m ( 10 5 . 2 ) m ( 10 55 . 1 ) Js ( 10 200 ) m ( 10 57 . 2 1 8 34 2 12 3 24 6 1 6 2 25 − − − − − − − − × × × × × × × × × × × × = = 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)
2 N hc a P N h a P t s s s se 2 s s s se = σ λ ν σ = ) m ( 10 3 ) Js ( 10 626 . 6 ) m ( 10 5 . 8 ) m ( 10 5 . 2 ) m ( 10 55 . 1 ) Js ( 10 200 ) m ( 10 41 . 3 1 8 34 2 12 3 24 6 1 6 2 25 − − − − − − − − × × × × × × × × × × × × = = 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
G(dB) = 10 log10 100 = 20 dB.
(b) Assuming there is no spontaneous emission G in , s s in , p p in , s out , s P P 1 P P λ λ + ≤ =
In order to achieve a specific maximum gain, G, the input pump power should be
(
)
s,in p s in , p G 1 .P P λ λ − ≥The limiting value of
Pp, in =
(
)
1 10 156.58 10 W 98 . 0 55 . 1 1 100 × × −3 = × −3 × − or Pp, in≈ 156.6 mW.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 2 2 1 cot 9 1 P P = = (κL) 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 Hz 10 935 . 1 10 55 . 1 10 3 C 14 6 8 c c = × × × = λ = ν −
Using eq. (11.15), we get m =
( )
121 10 1600 10 935 . 1 9 14 FSR c ≈ × × = ν ∆ νChapter 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
5 L 1 5 L ≈
− ; and the related splice loss would be 0.1
( )
dB 5L
×
≈ . 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) × 5 L
+ 7 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.
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 =
(
)
L; c n n n n 1 2 2 1 − n2 =n1(
1−2∆)
=1.46 0.98= 1.445 Thus, tintermodal = 12 10 (m) ms 10 3 445 . 1 46 . 1 445 . 1 46 . 1 3 1 8 × × × − − . = 6.06 × 10−7s = 606 nsUsing eq. (12.8) of the text book, we get tsys =
[
]
2 1 2 rx 2 f 2 tx t t t + + =[
(
) (
{
) (
)
}
(
)
]
2 1 2 2 2 2 ns 12 ns 8 . 28 ns 606 ns 10 + + + = 607 ns = 0.607 µsFor NRZ format, tsys≤ 7.0 10 s 7ns
10 100 70 . 0 B 70 . 0 9 6 = × = × = −
The total rise time of the system is much greater than 0.70/B, and hence the system can not operate with NRZ format.
B = 6 1.15 106 s 10 607 . 0 70 . 0 ≈ × × − bits/s = 1.15 Mb/s.
12.6 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
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 = α + α + × 2 1 2 1 2 / 2 1 / 2 1 NA NA = + + × 9 . 1 2 1 0 . 2 2 1 20 . 0 17 . 0 2 = (0.7225) × (0.9743) = 0.70
Therefore loss at joint B = − 10 log10ηT = 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
or Ptx = 54 µW (Both sides)
(b) 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 t t]
2 0.1488 s 1 2 rm 2 mat 2 tx + + = µMaximum bit rate for NRZ format = 4.7 10 bits/s 10 1488 . 0 70 . 0 6 6 = × × − = 4.7 Mb/s
Chapter 13 : Fiber Optic Sensors
13.3 (a) Sang (normalized) = π θ ∆ − NA n 1 =
(
)
× π π × − ) 17 . 0 ( 180 / 10 0 . 1 1 = 0.673 NA =[
( )
( )
]
2 1 2 2 45 . 1 46 . 1 − = 0.17 Range of Sang = (1 to 0.673) (b) ηang =(
)
π π × − + (0.17) 180 / 10 0 . 1 1 ) 46 . 1 1 ( ) 46 . 1 ( 16 4 2 = 0.627Lang = − 10 log10ηang = 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 ¨φ = λ Ω π C LD 2 It is given that 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 = rad 10 633 . 0 10 3 10 5 12 . 0 400 2 6 8 4 − − × × × × × × × π = 0.066 rad
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
) ( ). ( h N2 ν η ν α 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≈
2 N0
= 7.5 × 1018, we get the required pump energy =
95 . 0 2 10 1 . 5 10 626 . 6 10 5 . 7 18 34 14 × × × × × × − = 1.328 J cm−2 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 R1PL R1R2P2L L P0 R2 R1 Fig. Q. 14.3
Therefore fractional loss in power per round trip is 0 2 1 0 0 0 L 2 0 P ) L 2 exp( R R P P P P P − − α = − = 1 – R1 R2 exp (−2 α L).
If n is the refractive index of the active medium and C speed of light in free space, the time duration of this fractional loss is .
C Ln 2
This time corresponds to an exponential decay time constant tc (of the power) approximately given by (Yariv
1997)* tc≈
(
)
) R 1 C nL − , assuming = 0 and R1≈ R2≈ R ≈ 1.and the population inversion density at threshold Nth = (N2− N1)th = c 3 spont 2 3 t c t n 8π ν ∆ν 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
(
)
(
)
3.33 10 s 99 . 0 1 ms 10 3 ) m ( 1 . 0 1 ) R 1 ( c nl tc 8 1 8 − − − = × × × = − = and(
)
10 (s ) ) s ( 10 33 . 3 ms 10 3 ) s ( 10 ) s 10 74 . 4 ( ) 1 ( 8 ) N N ( N 9 1 8 3 1 8 7 2 1 14 3 th 1 2 th − − − − − × × × × × × × × π × = − = = 6.28 × 1014 m−3 = 6.28 × 108 cm−314.8 The time difference between the two pulses is given by
¨t = ,
c dn 2
where the d is the depth of the object. Given that ¨W µs and n = 1.33. Therefore,
d = 33 . 1 2 ) s )( 10 53 . 0 ( ) ms ( 10 3 n 2 t c 8 1 6 × × × × = ∆ − − = 60 m
