Solution manual for optical fiber communications 4th edition by gerd keiser

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/>Gerd Keiser, Optical Fiber Communications, McGraw-Hill, 4th ed., 2011
Problem Solutions for Chapter 2
E  100cos 210 t  30 e x  20cos 210 t  50 e y
8

2.1
2.2

8

 40cos 2108 t  210 e z

The general form is:
y = (amplitude) cos(t - kz) = A cos [2(t - z/)]. Therefore
(a) amplitude = 8 m
(b) wavelength: 1/ = 0.8 m-1 so that  = 1.25 m
(c)  = 2(2) = 4
(d) At time t = 0 and position z = 4 m we have
y = 8 cos [2(-0.8 m-1)(4 m)]
= 8 cos [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 ) = a 21 sin2 1  cos2 1 
+ a 22 sin2  2  cos2  2  + 2a1a2 (sin 1 sin 2 + cos 1 cos 2)
or
A2 = a 12  a 22 + 2a1a2 cos (1 - 2)
Dividing (2) by (1) gives

tan  =

a 1 sin1  a 2 sin2
a 1 cos 1  a 2 cos 2

Thus we can write
x = x1 + x2 = A cos  cos t + A sin  sin t = A 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
cos  = cos (t - kz) cos 
E0 x
to yield
Ey
E0 y

-

Ex
cos  = - sin (t - kz) sin 
E 0x

(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

E y

E
 x cos  =
E

 0 y E 0x


  E 2 
2
 x 
1  E   sin 
0x



Expanding the left-hand side and rearranging terms yields
2

2
 E 

 E x   E y 
  +   - 2  E 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: n = 1.0

33 

33 
90 

Glass

(a) Apply Snell's law
n1 cos 1 = n2 cos 2
where n1 = 1, 1 = 33, and 2 = 90 - 33 = 57

 n2 =

cos 33
= 1.540
cos 57

(b) The critical angle is found from
nglass sin glass = nair sin air
with air = 90 and nair = 1.0
 critical = arcsin

1
n glass

= arcsin

3

1
= 40.5
1.540


4


2.8
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
r
Find r from tan c =
, which yields r = 13.7 cm.
12 cm

2.9

45 

Using Snell's law nglass sin c = nalcohol sin 90
where c = 45 we have
1.45
nglass =
= 2.05
sin 45
n pure

1.450
= 83.3
1.460

2.10

critical = arcsin

2.11

Need to show that n1 cos 2  n 2 cos 1  0 . Use Snell’s Law and the relationship
sin 
tan  
cos 

n doped

= arcsin

5


2.12

(a) Use either NA = n12  n22  = 0.242
1/ 2

or
NA  n1 2 = n1

2(n1  n 2 )
= 0.243
n1

(b) A = arcsin (NA/n) = arcsin

2.13

0.242 
= 14
 1.0 

n 
 1.00 
(a) From Eq. (2.21) the critical angle is  c  sin 1  2   sin 1 
  41
 1.50 
 n1 




(c) The number of angles (modes) gets larger as the wavelength decreases.

2.14

NA = n12  n22  = n12  n12(1 )2 
1/ 2

1/ 2

= n1 2  2 

1 /2

Since  << 1, 2 << ;  NA  n1 2

2.15

(a) Solve Eq. (2.34a) for jH:

jH = j


1 Hz
Er 
r 

j  Er +

Substituting into Eq. (2.33b) we have

 
E z
1 Hz 
=  j

Er 

r
r  
 


Solve for Er and let q2 = 2 - 2 to obtain Eq. (2.35a).
(b)
jHr = -j

Solve Eq. (2.34b) for jHr:

1 Hz
E 
 r

Substituting into Eq. (2.33a) we have

6


j  E +

1 E z
 
1 Hz 
= -  j

E 

r 
 r 
 


Solve for E and let q2 = 2 - 2 to obtain Eq. (2.35b).

(c)

Solve Eq. (2.34a) for jEr:

jEr =




1


1 
Hz  jrH 


r  

Substituting into Eq. (2.33b) we have

1 
Hz  jrH + E z = jH

 r
r  

Solve for H and let q2 = 2 - 2 to obtain Eq. (2.35d).
(d)

Solve Eq. (2.34b) for jE
jE= -

1 
Hz 
jHr 


r 

Substituting into Eq. (2.33a) we have

Hz 
1 E z  
= -jHr
jHr 

r  
r 
Solve for Hr to obtain Eq. (2.35c).

(e)

Substitute Eqs. (2.35c) and (2.35d) into Eq. (2.34c)

-




j 1  
 Hz  r E z    Hz   E z  = jE
z
2
q 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  r Hz    E z   Hz  = -jHz
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(t  z ) and Hz = BJ0(ur) e 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)
A and
K  (wa)

D=

J (ua)
B
K  (wa)

Substitute these into Eq. (2.50) to find B in terms of A:

A

J'  (ua) K'  (wa) 
j  1

1
 2 = B

2

 a u
w 
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  j

J  (ua)  A 1uJ'  (ua)
2 B
u  a


8


+

1
2
w

 j
K' (wa)J (ua) 
B
J (ua)  A 2 w 


K (wa)
 a



With k21 = 21 and k22 = 22 rewrite this as




ja  1
2
2
B =
1
 (k1 J + k2 K) A
   1 
 u 2 w2 
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. (2.56a) 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.
2.17

From Eq. (2.23) we have
 =

2
1 
n 21  n22
1  n 2 

=
2
2
2  n1 
2n1

 << 1

implies n1  n2

Thus using Eq. (2.46), which states that n2k = k2  k1 = n1k, we have

2

n 2k

2.18

2



2

2

k 2  n 1k

2



2

k1  

2

(a) From Eqs. (2.59) and (2.61) we have

M

2 2 a 2 2
2 2 a 2
2
n

n


NA 2

2
1
2
2


1/ 2

 M
a   
2

1/ 2


1000
 
NA
2 

0.85m
 30.25m
0.2

9

Therefore, D = 2a =60.5 m
2 30.25m
2
(b) M 
0.2  414
2
1.32m
2

2

(c) At 1550 nm, M = 300
2.19

From Eq. (2.58),

V=

2 (25 m)
2
2 1/ 2
(1.48)  (1.46)  = 46.5

0.82 m

Using Eq. (2.61) M  V2/2 =1081 at 820 nm.
Similarly, M = 417 at 1320 nm and M = 303 at 1550 nm. From Eq. (2.72)

4  100%
4

Pclad 
 M-1/2 =
= 4.1%
 P total 3
3 1080
at 820 nm. Similarly, (Pclad/P)total = 6.6% at 1320 nm and 7.8% at 1550 nm.

2.20 (a) At 1320 nm we have from Eqs. (2.23) and (2.57) that V = 25 and M = 312.
(b) From Eq. (2.72) the power flow in the cladding is 7.5%.
2.21

(a) For single-mode operation, we need V  2.40.

Solving Eq. (2.58) for the core radius a

a=

2.40(1.32m)
V 2
2 1/ 2
= 6.55 m
n1  n2  =

2
2 1/ 2
2
2(1.480)  (1.478) 

(b) From Eq. (2.23)
NA = n1  n2  = (1.480)  (1.478)

2

2 1/ 2

2



2 1/ 2

= 0.077

(c) From Eq. (2.23), NA = n sin A. When n = 1.0 then

10


A = arcsin

2.22

n2 =

2
2
n1  NA =

a=

2.23

0.077 
NA 
= arcsin
= 4.4
 1.0 
 n 

2
2
(1.458) (0.3) = 1.427

(1.30)(75)
V
=
= 52 m
2(0.3)
2NA

For small values of  we can write V 

2a
n1


2

For a = 5 m we have  0.002, so that at 0.82 m
V 

2 (5 m)
1.45
0.82 m

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 m.

2.25

From Eq. (2.77), Lp =


2
=
n y  nx


1.3  10 6 m
= 1.310-5
1
10 m

For Lp = 10 cm

ny - nx =

For Lp = 2 m

1.3  10 6 m
ny - nx =
= 6.510-7
2m

Thus
6.510-7 ny - nx 1.310-5
2.26

We want to plot n(r) from n2 to n1. From Eq. (2.78)
n(r) = n1 1 2(r / a)



 1 /2

= 1.48 1 0.02(r / 25)



 1 /2

11


n2 is found from Eq. (2.79):
2.27

n2 = n1(1 - ) = 1.465

From Eq. (2.81)
2


 2an1 
2 2 2
M
a k n1  

2
  2   

where
=

n1  n2
= 0.0135
n1

At  = 820 nm, M = 543 and at  = 1300 nm, M = 216.
For a step index fiber we can use Eq. (2.61)
2

1 2a  2
V2
2
Mstep 
=
n1  n 2 




2 
2

At  = 820 nm, Mstep = 1078 and at  = 1300 nm, Mstep = 429.
Alternatively, we can let  in Eq. (2-81):
2

Mstep =
2.28

2an1 
=
  

1086 at 820 nm

432 at 1300 nm

Using Eq. (2.23) we have
(a) NA = n1  n2  = (1.60)  (1.49)
2

2 1/ 2



2 1/ 2

2

= 0.58

(b) NA = (1.458)2  (1.405)2  = 0.39
1/ 2

2.29

(a) From the Principle of the Conservation of Mass, the volume of a preform rod

section of length Lpreform and cross-sectional area A must equal the volume of the fiber
drawn from this section. The preform section of length Lpreform is drawn into a fiber of
length Lfiber in a time t. If S is the preform feed speed, then Lpreform = St. Similarly, if s is
the fiber drawing speed, then Lfiber = st. Thus, if D and d are the preform and fiber
diameters, respectively, then
Preform volume = Lpreform(D/2)2 = St (D/2)2

12


Fiber volume = Lfiber (d/2)2 = st (d/2)2

and
Equating these yields

2

2

D 
d 
St
= st
2 
2 

2

D 
s=S
d 

or

2

2

0.125 mm 
d 
= 1.2 m/s
= 1.39 cm/min
 9 mm 
D 

(b)

S=s

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
Apreform clad

or

=

A fiber core
Afiber clad

(32  R2 )

 (25)2
=
2
2
2
2
(4  3 )  (62.5)  (25) 

from which we have
1/ 2


7(25)2

R = 9 
2
2  = 2.77 mm
 (62.5)  (25) 

13


Thus, the thickness = 3 mm - 2.77 mm = 0.23 mm.

2.31

(a) The volume of a 1-km-long 50-m diameter fiber core is
V = r2L =  (2.510-3 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 time t is
t=

2.32

5.1 gm
M
=
= 10.2 min
R 0.5 gm / min

Solving Eq. (2.82) for  yields
2

 K 
=
Y 
Thus
=

2.33

where Y =



for surface flaws.

(20 N / mm 3 / 2 ) 2

= 2.6010-4 mm = 0.26 m
2 2
(70 MN/ m ) 

(a) To find the time to failure, we substitute Eq. (2.82) into Eq. (2.86) and

integrate (assuming that  is independent of time):
f



t

b /2

d

= AYbb

i



dt

0

which yields

1

or


b f
1
2
t=

1 b / 2

1 b/ 2

 i

 = AYbbt

2
(2 b)/ 2
(2 b)/ 2
 f

b  i
(b  2)A(Y)

(b) Rewriting the above expression in terms of K instead of yields

14

 Ki 2 b  Kf 2 b 
2
t=

b 
Y  
(b  2)A(Y) 
 Y 

2 b

2Ki

b
(b  2)A(Y)

2
2
if K b
<< K b
or
i
f

15

2 b

Ki

 Kf

2 b