Link download solution manual optical fiber communications 4th edition by gerd keiser

To verify this, first we square both sides and add: 1... Use Snell’s Law and the relationship... 2.56a holds, so that the term in square brackets on the right-hand side in the above equa

Gerd Keiser, Optical Fiber Communications, McGraw-Hill, 4th

ed., 2011 Problem Solutions for Chapter 2

2.1

E  1 00co s 2  1 0 8 t  3 0  ex  2 0co s 2  1 0 8 t  5 0  ey

y = 8 cos [2  (-0.8  m-1

)(4  m)]

2.3 x1 = a1 cos (  t - 1) and x2 = a2 cos ( t - 2)

Adding x1 and x2 yields

equations To verify this, first we square both sides and add:

1

A2 (sin2 + cos2) = a

1 cos

2

1 

1 sin

+

2 2

2

2 + 2a1a2(sin1sin2

1 2 1 2

2 sin 

a cos   a cos 

Subtract from this the expression

E

0 y E

0x

(2.4-1)

(2.4-2)

Using the relation cos2 + sin2 = 1, we use Eq (2.2) to write

sin2 ( t - kz) = [1 - cos2 ( t - kz)] =

 E 

x 



(2.4-3)

Squaring both sides of Eq (2.4-2) and substituting it into Eq (2.4-3) yields

E y

E

0 y



2

cos 

1  x



2













sin2

Expanding the left-hand side and rearranging terms yields

E  E  E E

+

- 2

x  y

2.7

Air: n = 1.0









cos  = sin2

(a) Apply Snell's law

(b) The critical angle is found from

n

glass sin

glass = n

air sin

with air = 90 and nair = 1.0

n

2.8

Water



12 cm

12 cm

2.9

n

glass = 1.45

= 2.05

2.10 critical = arcsin npure

= arcsin

n

doped

1.450

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

2 2

or

n

1

1/ 2

= 0.242



)

= 0.243

 1.0  = 14

 1  n 2   1 1.00

2.13 (a) From Eq (2.21) the critical angle is  c  sin    sin   41 



n





2 2 1/ 2 2 2

= n1

2 1 / 2

Since  << 1, 2 <<  ; 

) 21/ 2

NA

 n 1 

j  Er + 

Er z = 

Solve for Er and let





q = 2 - 2 to obtain Eq (2.35a)

jHr = -j



Substituting into Eq (2.33a) we have

1 E  

Solve for E  and let q2 = 2

1 H 



 H

z

jEr =





= j  H 

j  H r

z



Substituting into Eq (2.33a) we have

z

2 



(f) Substitute Eqs (2.35a) and (2.35b) into Eq (2.33c)

= -j  Hz

- 2    r   

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

K





(ua)

K







(ua)

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





a

 2

 1

 



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):

+ 1  jJ (ua)  A w K' (wa)J (ua) 

w

2

a

2

2

= 22 rewrite this as

ja  1 

w

 u 2 2 

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

 = n

=

1  1



2





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

2 k 2

n

2

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

2 a2

2 a2

M 1/ 2  1000 1/ 2 0.85m



2

Therefore, D = 2a =60.5  m

(b)

(c) At 1550 nm, M = 300

2.19 From Eq (2.58),

(1.48) 2  2 1/

2

= 46.5

/2 =1081 at 820 nm

Similarly, M = 417 at 1320 nm and M = 303 at 1550 nm From Eq (2.72)

 P clad





total

3

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

Solving Eq (2.58) for the core radius a

(b) From Eq (2.23)

NA =



2 

n

1

2 1/ 2

1/ 2

A = arcsin  NA  =

 n  arcsin

 0.077

 1.0

2.22 n2 = 2

2

=

2 2

= 1.427

V 2 (5

0.82

m  1.45 2(0.002) = 3.514

Thus the fiber is no longer single-mode From Figs 2.18 and 2.19 we see that

2.25 From Eq (2.77), Lp = 2 

n

y

x

For Lp = 10 cm

For Lp = 2 m

Thus

ny - nx =

ny - nx =

2 m

m

m

= 1.3  10-5

= 6.5  10-7

2.26

n2 is found from Eq (2.79): n2 = n1(1 - ) =

1.465 2.27 From Eq (2.81)

M   2 a k n   2    1   

1

where

 =n 1  n 2

n

= 0.0135

For a step index fiber we can use Eq (2.61)

M

2

 n 2  n 2 

At  = 820 nm, Mstep = 1078 and at  = 1300 nm, Mstep = 429

M

step =

 =



2.28 Using Eq (2.23) we have

(a) NA =



2

n

1



2 1/ 2

2

= 0.58



= 0.39

2.29 (a) From the Principle of the Conservation of Mass, the volume of a

volume of the fiber drawn from this section The preform section of length

Thus, if D and d are the preform and fiber diameters, respectively, then

and

Equating these yields

d



2

(b) S = s  d  2

 D = 1.2 m/s

 2

R

4 mm

3 mm

FIBER PREFORM

done by comparing the ratios of the preform core-to-cladding cross-sectional areas and the fiber core-to-cladding cross-sectional areas:

A

A

preform core preform clad

A

fiber core

fiber clad

or

)

 (3  R

)

 (62.5) 2  (25) 2

from which we have

= 2.77 mm

9



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

(b) If R is the deposition rate, then the deposition time t is

Thus

3 / 2 ) 2

(70 MN/ m

2 2



) = 2.60 10-4 mm = 0.26  m

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

 f 

   b / 2 d 

 i

t

0

dt

which yields

1

or

t =

f  b / 21

2

 b)/ 2 2  b)/ 2 (2 (

2  2  b Kf 2  b 

b     

b  2