Download Full
Solution Manual for Optical Fiber Communications 4th Edition by Gerd
Keiser
https://getbooksolutions.com/download/solution-manual-optical-fibercommunications-4th-edition-by-keiser Gerd Keiser, Optical Fiber Communications, McGraw-Hill, 4th ed., 2011 Problem Solutions for Chapter 2 E
2.1
2.2
100cos 2 1 0 t 3 0 e x 20cos 21 0 t 5 0 e y 8
8
40cos 21 08 t 2 10 e z
The general form is: y = (amplitude) cos( cos(t - kz) = A cos [2 [2(t - z/ z/)]. Therefore (a) amplitude = 8 m (b) wavelength: 1/ 1/ = 0.8 m-1 so that = 1.25 m (c) = = 2 2(2) = 4 4 (d) At time t = 0 and position z = 4 m we have y = 8 cos [2 [2(-0.8 m-1)(4 m)] = 8 cos [2 [2 (-3.2)] = 2.472
2.3
x1 = a1 cos ( (t - 1) and x2 = a2 cos ( (t - 2) Adding x1 and x2 yields x1 + x2 = a1 [cos t cos 1 + sin t sin 1] + a2 [cos t cos 2 + sin t sin 2] = [a1 cos 1 + a2 cos 2] cos t + [a1 sin 1 + a2 sin 2] sin t Since the a's and the 's are constants, we can set a1 cos 1 + a2 cos 2 = A cos
(1)
a1 sin 1 + a2 sin 2 = A sin
(2)
provided that constant values of A and exist which satisfy these equations. To verify this, first we square both sides and add:
1
A2 (sin2 + cos2 ) =
+
2
2
a 2 sin
2
2
a 1 sin
1 cos cos 1 2
2
2 cos cos 2 + 2a1a2 (sin 1 sin 2 + cos 1 cos 2)
or A2 =
2
a1
2
a2
+ 2a1a2 cos ( (1 - 2)
Dividing (2) by (1) gives
tan =
a 1 sin 1 a 2 sin 2 a 1 cos cos 1 a 2 cos 2
Thus we can write x = x1 + x2 = A cos cos t + A sin sin t = A cos( cos(t - )
2.4
First expand Eq. (2.3) as Ey E0 y
= cos ( (t - kz) cos - sin ( (t - kz) sin
(2.4-1)
Subtract from this the expression Ex E0x
cos = cos ( (t - kz) cos
to yield Ey E0 y
-
Ex E 0x
cos = - sin ( ( t - kz) sin
(2.4-2)
Using the relation cos2 + sin2 = 1, we use Eq. (2.2) to write
E 2 sin2 ( (t - kz) = [1 - cos2 ( (t - kz)] = 1 x E 0x
(2.4-3)
Squaring both sides of Eq. (2.4-2) and substituting it into Eq. (2.4-3) yields
2
2 2 Ey E x Ex 2 E 0 y E 0x cos = 1 E 0x sin
Expanding the left-hand side and rearranging terms yields 2
2 E E E x E y + - 2 x y cos = sin2 E 0x E 0y E 0x E 0y
2.5
Plot of Eq. (2.7).
2.6
Linearly polarized wave.
2.7 Air: Air: n = 1.0
33
33
Glass
90
(a) Apply Snell's law n1 cos 1 = n2 cos 2 where n1 = 1, 1 = 33 33, and 2 = 90 90 - 33 33 = 57 57 n2 =
cos 33 cos 57
= 1.540
(b) The critical angle is found from nglass sin glass = nair sin air with air = 90 90 and nair = 1.0 critical = arcsin
1 n glass
= arcsin
3
1 1.540
= 40.5 40.5
4
2.8 Air Air
r
Water
12 cm
Find c from Snell's law
n1 sin 1 = n2 sin c = 1
When n2 = 1.33, then c = 48.75 48.75 Find r from tan c =
r 12 cm
, which yields r = 13.7 cm.
2.9
45
Using Snell's law
nglass sin c = nalcohol sin 90 90
where c = 45 45 we have nglass =
1.45 sin 45
n pu re
2.10
critical = arcsin
2.11
Need to show that tan
n doped
= 2.05
= arcsin
1.450 1.460
= 83.3 83.3
n1 cos cos 2 n 2 cos cos 1 0 .
cos sin
5
Use Snell’s Law and the relationship
2.12
(a) Use either NA =
n
2
1
1/ 2
n 2 = 0.242 2
or NA n1
2 =
n1
2(n 2( n1 n 2 ) n1
= 0.243
(b) A = arcsin (NA/n) = arcsin
2.13
0.242 = 14 14 1.0
(a) From Eq. (2.21) the critical angle is
n sin n 1
2
c
1
1.00 sin 41 41 1 . 50 1
(c) The number of angles (modes) gets larger as th e wavelength decreases.
2.14
NA =
n
1/ 2
2 1/ 2
n 2 = n1 n1 (1 )
2
2
1
= n1
2
2
1 /2
2 2
Since << 1, 2 << ; NA n1
2.15
2
(a) Solve Eq. (2.34a) for jH:
jH = j
j Er +
Er -
E z r
1 H z
Substituting into Eq. (2.33b) we have
r
= j
E r
Hz r 1
Solve for Er and let q2 = 2 - 2 to obtain Eq. (2.35a). (2.35a). (b) jHr = -j
Solve Eq. (2.34b) for jHr :
E -
H r 1
z
Substituting into Eq. (2.33a) we have
6
E = - - r
1
j E +
z
1 Hz j E r
Solve for E and let q2 = 2 - 2 to obtain obtain Eq. (2.35b).
(c)
Solve Eq. (2.34a) for jEr :
jEr =
1 1 Hz
jr H r
1 Hz jr H + r
E z r
Substituting into Eq. (2.33b) we have
= j j H
Solve for H and let q2 = 2 - 2 to obtain Eq. (2.35d). (d)
Solve Eq. (2.34b) for jE jE = -
1
E z
r
-
1
jHr
Hz r
Substituting into Eq. (2.33a) we have
Hz jHr = -j -j Hr r
Solve for Hr to obtain Eq. (2.35c).
(e)
Substitute Eqs. (2.35c) and (2.35d) into Eq. (2.34c)
-
j 1 q
2
Hz E Hz E z r z = jEz r r r r r
Upon differentiating and multiplying by jq2/ we obtain Eq. (2.36).
7
(f)
Substitute Eqs. (2.35a) and (2.35b) into Eq. (2.33c)
-
j 1
E z Hz E z Hz = -jHz r 2 q r r r r r
Upon differentiating and multiplying by jq2/ we obtain Eq. (2.37).
2.16
For = 0, from Eqs. (2.42) and (2.43) we have
Ez = AJ0(ur) e
j( j( t z )
and Hz = BJ0(ur) e
j( j( t z )
We want to find the coefficients A and B. From Eq. (2.47) and (2.51), respectively, we have
C=
J (ua) K (wa)
A and
D=
J (ua)
K (wa)
B
Substitute these into Eq. (2.50) to find B in terms of A:
A
j 1 a
u 2
1
w 2
= B B
J' (ua) K' (wa) uJ (ua) wK (wa)
For = 0, the right-hand side must be zero. Also for = 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
j B J (ua) A1uJ' (ua) a
8
+
j K' (wa)J (ua) B J (ua) A2 w w a K (wa) 1
With
2
2
k 1 =
2
21 and
22 rewrite this as
k 2 =
ja 1 B = 1 1 w u 2
2
2 2 (k 1 J + k 2 K) A
where J and K are defined in Eq. (2.54). If for = 0 the term in square brackets on the right-hand side is non-zero, that is, if Eq. (2.56 a) does not hold, then we must have that A = 0, which from Eq. (2.42) (2.42 ) means that Ez = 0. Thus Eq. (2.55) corresponds to TE0m modes. 2.17
From Eq. (2.23) we have n 21 n 22
=
2 1
2 2 2
2n
<< 1
n = 1 2 n 1
1
implies n1 n2
Thus using Eq. (2.46), which states that n2k = k 2 k k 1 = n1k, we have
2
2
n 2k
2.18
2
k 2
2
2
n 1 k
2
k 1
2
(a) From Eqs. (2.59) and (2.61) we have 2
M
2 a 2
M a 2
2
2
n
2
1
1/ 2
2
n2
2 a
2
2
1000 NA 2
1/ 2
2
NA
0.85m 0.2
9
30.25m
Therefore, D = 2a =60.5 m
(b)
2
2
30.25m 0.2 4 14 1.32m
2
M
2
2
(c) At 1550 nm, M = 300 2.19
From Eq. (2.58), 2 (25 m)
V=
(1.48)
2
0.82 m
(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
4 3
M-1/2 =
4
100%
3 1080 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=
V
n 2
2
1
1/
n2 2
2
=
2.40(1.32m)
2 (1.480)
2
(1.478)2
1/ 2
= 6.55 m
(b) From Eq. (2.23) NA =
n
2
1
1/ 2
n 2 = (1.480) (1.478) 2
2
2 1/ 2
= 0.077
(c) From Eq. (2.23), NA = n sin A. When n = 1.0 then
10
A = arcsin
2.22
n2 =
2
n1
n
2
NA N A =
a=
2.23
NA
= arcsin
0.077 1.0
2
= 4.4 4.4
2
(1.458) (0.3) = 1.427
(1.30)(75) V = = 52 m 2 (0.3) 2 NA 2 a
For small values of we can write write V
n1
2
For a = 5 m we have 0.002, 0.002, so that at 0.82 m V
2 (5 m) 0.82
m
2(0.002) = 3.514
1.45
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 m.
2.25
From Eq. (2.77), L p =
For L p = 10 cm
For L p = 2 m
2
=
n y nx
ny - nx =
ny - nx =
1.3 10 10
1
1.3 10
6
m
m
6
2m
m
= 1.3 1.310-5
= 6.5 6.510-7
Thus 6.5 6.510-7 ny - nx 1.3 1.310-5 2.26
We want to plot n(r) from n 2 to n1. From Eq. (2.78) n(r) = n1 1 2 (r/a)
1/2
= 1.48 1 0.02(r 0.02(r / 25) 25)
11
1/2
n2 is found from Eq. (2.79): 2.27
n2 = n1(1 - ) = 1.465
From Eq. (2.81)
M 2
a
2
2an1 2 k n1 2 2
2
where =
n1
n
2
n1
= 0.0135
At = 820 nm, M = 543 and at = 1300 nm, M = 216. For a step index fiber we can use Eq. (2.61) Mstep
V
2
=
2
1 2 a 2
2
n
2 1
n 22
At = 820 nm, Mstep = 1078 and at = 1300 nm, Mstep = 429. Alternatively, we can let in in Eq. (2-81): Mstep = 2.28
2an
1086 at 820 432 at 1300
2
=
1
nm
Using Eq. (2.23) we have (a) NA =
n
2
1
1/ 2
n 2 = (1.60) (1.49) 2
2
2 1/ 2
(b) NA = (1.458) (1.405) 2
2.29
nm
2 1/ 2
= 0.58
= 0.39
(a) From the the Principle of the Conservation of Mass, Mass, the volume of a preform rod rod
section of length L preform and cross-sectional area A must equal the volume of the fiber drawn from this section. The preform section of length L preform is drawn into a fiber of length Lfiber in a time t. If S is the preform feed speed, then L preform = St. Similarly, if s is the fiber drawing speed, then Lfiber = st. Thus, if D and d are the preform and fiber diameters, respectively, then Preform volume = L preform(D/2)2 = St (D/2)2
12
Fiber volume = Lfiber (d/2)2 = st (d/2)2
and Equating these yields
2
D d St = st 2 2
2
D s=S d
or
2
2
2
(b)
d 0.125 mm S=s = 1.2 m/s = 1.39 cm/min D 9 mm
2.30
Consider the following geometries of the preform and its corresponding fiber:
25 m R
4 mm 62.5 m
3 mm
FIBER
PREFORM
We want to find the thickness of the deposited layer (3 mm - R). This can be done by comparing the ratios of the preform core-to-cladding cross-sectional areas and the fiber core-to-cladding cross-sectional areas: A preform core A preform clad
=
A fiber core fiber core Afiber clad fiber clad
or
(32 R 2 ) 2
(4
2
3 )
=
(25)2 (62.5) (25) 2
2
from which we have R=
7(25)2 9 2 2 (62.5) (25)
1/ 2
= 2.77 mm
13
Thus, the thickness = 3 mm - 2.77 mm = 0.23 mm.
2.31
(a) The volume of a 1-km-long 50- 50-m diameter fiber core is V = r 2L = (2.5 (2.510-3 cm)2 (105 cm) = 1.96 cm3
The mass M equals the density times the volume V: M = V = (2.6 gm/cm3)(1.96 cm3) = 5.1 gm (b) If R is the deposition rate, then the deposition d eposition time t is M
t=
2.32
R
5.1 gm
=
0.5 gm / min min
= 10.2 min
Solving Eq. (2.82) for yields
K = Y
2
where Y =
for surface flaws.
Thus ( 20 N / mm3 / 2 ) 2
=
2.33
2 2
(70 MN/ m )
= 2.60 2.6010-4 mm = 0.26 m
(a) To find find the time to to failure, we substitute Eq. (2.82) into Eq. (2.86) and
integrate (assuming that is independent of time): f
t
b / 2
d =
AY b b
i
dt
0
which yields 1 1
b
1 b / 2 f
1 b/ 2
i
= AY b bt
2
or t=
2 b
(b 2)A(Y )
(2 b) / 2 i
(2 b) / 2
f
(b) Rewriting the above expression in terms of K instead of yields
14
t=
K i 2 b K f 2 b b (b 2)A(Y ) Y Y 2
2K i
2 b b
(b (b 2)A(Y 2)A(Y )
if
b 2
K i
<<
b 2
K f
15
or
2 b
K i
K f
2 b
