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SOLUTION MANUAL FOR OPTICAL FIBER COMMUNICATIONS 4TH

EDITION BY GERD KEISER

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

ed., 2011 Problem Solutions for Chapter 2

2.1

E 1 00co s 2 1 08 t 3 0 ex 2 0co s 21 08t 5 0 ey

4 0co s 21 08 t 2 10 ez

2.2 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

2 2

1 cos

2

1 1 sin

+

2 2

cos

2

2 + 2a1a2 (sin 1 sin 2 a2 sin 2

or

A2

= a 2 a 2 + 2a a cos ( - )

1 2 1 2 1 2

Dividing (2) by (1) gives

tan =

a 1

sin a 2 sin

1 2

a cos a cos

1 2

1 2

Thus we can write

x = x1 + x2 = A cos cos t + A sin

2.4 First expand Eq. (2.3) as

E y = cos (t - kz) cos - sin (t - kz) sin

E

0 y

Subtract from this the expression

E x cos = cos (t - kz) cos

E

0 x

to yield

Ey - E x

cos = - sin (t - kz) sin

E0 y

E0x

+ cos 1 cos 2)

sin t = A cos(t - )

(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

1 2 x

E

0x

(2.4-3)

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

2

E y

E

0 y

E x

E 0x

2

cos

=

E 1 x

E

0x

2

sin2

Expanding the left-hand side and rearranging terms yields

2

2

E E E E

x

+ y

- 2 x y

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

Glass 90

cos = sin2

(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

= arcsin 1

= 40.5n

g lass 1.540

3

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

Find r from tan c = r

, 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 n

glass = 1.45

= 2.05

sin 45

2.10 critical= arcsin

n

pure

= arcsin n

doped

1.450 1.460

= 83.3

2.11 Need to show that n1 cos 2 n 2 cos 1 0 . Use Snell’s Law and the relationship

tan sin

cos

5

2 2

2.12 (a) Use either NA = n1 n2

or

NA n1 2= n1 2(n1 n2

n1

1/ 2

= 0.242

) = 0.243

(b) A = arcsin (NA/n) = arcsin

0.242

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

1.50

1

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

2 2 1/ 2 2 2

(12.14 NA = n1 n2 = n1 n1

= n1

2 1 / 2

2

Since << 1, 2

<< ;

2.15 (a) Solve Eq. (2.34a) for jH:

)2 1/ 2

NA

n1

2

jH = j

Er -

1 Hz

r

Substituting into Eq. (2.33b) we have

j Er + E

rz =

Solve for Er and let

Er 1 H j z

r

q2

= 2 -

2 to obtain Eq. (2.35a).

(b) Solve Eq. (2.34b) for jHr:

jHr = -j

E -

1 Hz

Substituting into Eq. (2.33a) we have

r

6

1 E j E +

z = - j E

r

Solve for E and let q2

= 2

(c) Solve Eq. (2.34a) for jEr:

1 H z

r

- 2 to obtain Eq. (2.35b).

1 1 H

z

jEr =

jrH

Substituting into Eq. (2.33b) we have r

1 E H

z

z

jrH +

= jH

r

r

Solve for H and let q2

= 2 -

2 to obtain Eq. (2.35d).

(d) Solve Eq. (2.34b) for jE

jE= - 1

H z

jH r

r

1 E H - jHr

z

z

r r

Solve for Hr to obtain Eq. (2.35c).

Substituting into Eq. (2.33a) we have

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

j 1 Hz

Ez

Hz

Ez

- r

= jE

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

Ez

Hz Ez

Hz

= -jHz - 2 r 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

K

(ua) (wa)

A and D =

J

K

(ua) (wa)

B

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

A

j 1

a 2

u

1

w 2

= B

J' (ua) K' (wa)

(ua) wK (wa) uJ

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 B j J (ua) A uJ' (ua)

u2

a 1

8

+

1 jJ (ua) A w K' (wa)J (ua) B

w 2

a 2

K (wa)

With

k 2

1

= 21 and

k 2

2

= 22 rewrite this as

B =

ja 1

1

1

w

u 2 2

(k12J + k2

2K) 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. (2.56a) 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

=

n2 n

2

12n12

2

=

1 1

2

n 2 2

n 2

1

<< 1 implies n1 n2

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

2 k 2

k 2 2 k 2 k 2 2

n

2 n

2 1 1

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

M 2

2

2

a2 n12 n22

2

22

a2 NA

2

M 1/ 2 1000 1/ 2

0.85m a

30.25m

NA

2

0.22

9

Therefore, D = 2a =60.5 m

(b)

2 2

M 2 30.25m 2

414 0.2 1.32m 2

(c) At 1550 nm, M = 300

2.19 From Eq. (2.58),

V =

2 (25 m)

0.82 m

(1.48)2

2 1/

(1.46)

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

= 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

V 2 2 1/ 2 2.4 0(1.3 2m) m a = n n = = 6.55

1 2

2

2 2 (1.4 80) (1.4 78) 2 1/ 2

(b) From Eq. (2.23)

NA =

2 n

1

n 2 1/ 2

2

=

2

(1.480)

2

1/ 2

(1.478)

= 0.077

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

10

A = arcsin

NA =

n

arcsin

0.077

1.0

= 4.4

2.22 n2 =

2

NA 2

= 2 2

= 1.427 n1 (1.458) (0.3)

a = V

= (1.30)(75)

= 52 m 2 NA

2(0.3)

2.23 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

0.82

m)

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

the LP01 and the LP11 modes exist in the fiber at 0.82 m.

2.25 From Eq. (2.77), Lp =

2

=

n y

n x

For Lp = 10 cm

For Lp = 2 m

Thus

ny - nx =

ny - nx =

1.3 10 6

10 1

m

1.3 106

2 m

m

m

= 1.310-5

= 6.510-7

6.510-7

ny - nx 1.310-5

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): n2 = n1(1 - ) =

1.465 2.27 From Eq. (2.81)

2 2 2

2an 2 M

1

a k n

2 2

1

where

=

n 1 n 2

n 1

= 0.0135

At = 820 nm, M = 543 and at = 1300 nm, M = 216.

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

M

step

V 2

2

=

1 2a 2

n2 n2

2 1 2

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

Alternatively, we can let in Eq. (2-81):

M

step =

2an1 2

=

1086 at 820 nm

432 at 1300 nm

2.28 Using Eq. (2.23) we have

(a) NA =

2 n

1

n 2 1/ 2

2

=

2

(1.60)

2 1/

(1.49)

2

= 0.58

2 2

1/ 2

= 0.39 (b) NA = (1.458) (1.405)

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

and

Equating these yields

D St 2

Fiber volume = Lfiber (d/2)2

= st (d/2)2

2 d 2 D 2

= st or s = S d

2

(b) S = s

d 2

D

= 1.2 m/s

0.125 mm

9 mm

2

= 1.39 cm/min

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

A

preform core preform clad

=

A

A

fiber core

fiber clad

or 2 2

) (3 R 2 2

) (4 3

= (25)2

(62.5)2 (25)2

from which we have

R =

7(25)2

1/ 2

= 2.77 mm 9

2

2

(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 = r2

L = (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 time t is

t = M

= 5.1 gm

= 10.2 min

R

0.5 gm / min

2.32 Solving Eq. (2.82) for yields

K

2

=

where Y = for surface flaws. Y Thus

=

(20 N / mm 3 / 2 ) 2

(70 MN/ m 2 2

)

= 2.6010-4

mm = 0.26 m

2.33 (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

b / 2 d

i

t

= AYb

b

0

dt

which yields

1

1

or

t =

b 1f b / 2

1i b/ 2

= AYbbt

2

2 f b i

b)/ 2 2 b)/ 2 (2 (

(b 2)A(Y)

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

14

2 2 b Kf 2 b

t = Ki

(b 2)A(Y) b

Y Y

2K

2 b b 2

b 2

i

if K

or b i << K f

(b 2)A(Y)

K 2 b

K 2 b

i f

15