Optical Fiber Communication Solution Manual (1)

7/25/2019 Optical Fiber Communication Solution Manual (1)

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1

Problem Solutions for Chapter 2

2-1.

E = 100cos 2!108

t + 30 ex + 20 cos 2!108t" 50 ey

+ 40cos 2!108 t + 210( )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!(2) = 4!

(d) At t = 0 and z = 4 m we have

y = 8 cos [2!(-0.8 m-1)(4 m)]

= 8 cos [2!(-3.2)] = 2.472

2-3. For E in electron volts and %in m we have E =1.240

%

(a) At 0.82 m, E = 1.240/0.82 = 1.512 eV

At 1.32 m, E = 1.240/1.32 = 0.939 eV

At 1.55 m, E = 1.240/1.55 = 0.800 eV

(b) At 0.82 m, k = 2!/%= 7.662 m-1

At 1.32 m, k = 2!/%= 4.760 m-1

At 1.55 m, k = 2!/%= 4.054 m-1

2-4. x1= a1cos (#t - &1) and x2= a2cos (#t - &2)

Adding x1and x2yields

x1+ x2= a1[cos #t cos &1+ sin #t sin &1]

+ a2[cos #t cos &2+ sin #t sin &2]

= [a1cos &1+ a2cos &2] cos #t + [a1sin &1+ a2sin &2] sin #t

Since the a's and the &'s are constants, we can set

a1cos &1+ a2cos &2= A cos ' (1)


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2

a1sin &1+ a2sin &2= A sin ' (2)

provided that constant values of A and 'exist which satisfy these equations. To

verify this, first square both sides and add:

A2(sin2 '+ cos2 ') = a12

sin2

&1 +cos2

&1( )

+ a2

2 sin2 &2

+cos2 &2( )+ 2a1a2(sin &1sin &2+ cos &1cos &2)

or

A2= a1

2 +a2

2 + 2a1a2cos (&1- &2)

Dividing (2) by (1) gives

tan '=a

1sin&

1+a

2sin&

2

a1

cos&1

+a2cos&

2

Thus we can write

x = x1+ x2= A cos 'cos #t + A sin 'sin #t = A cos(#t - ')

2-5. First expand Eq. (2-3) as

Ey

E0 y= cos (#t - kz) cos &- sin (#t - kz) sin & (2.5-1)

Subtract from this the expression

E x

E0 xcos &= cos (#t - kz) cos &

to yield

Ey

E0 y

-ExE

0x

cos &= - sin (#t - kz) sin & (2.5-2)

Using the relation cos2(+ sin2(= 1, we use Eq. (2-2) to write


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3

sin2(#t - kz) = [1 - cos2(#t - kz)] = 1 " E

x

E0x

)

*+ ,

-.

2/

01

2

34 (2.5-3)

Squaring both sides of Eq. (2.5-2) and substituting it into Eq. (2.5-3) yields

E y

E0 y"

Ex

E 0xcos&

/

01

2

34

2

= 1 " Ex

E0x

)

*+ ,

-.

2/

01

2

34 sin

2&

Expanding the left-hand side and rearranging terms yields

Ex

E 0x

)

*+ ,

-.

2

+

Ey

E0y

)

*+

,

-.

2

- 2

Ex

E0x

)

*+ ,

-.

Ey

E0y

)

*+

,

-.

cos &= sin2

&

2-6. Plot of Eq. (2-7).

2-7. Linearly polarized wave.

2-8.

33 33

90 Glass

Air: n = 1.0

(a) Apply Snell's law

n1cos 51= n2cos 52

where n1= 1, 51= 33, and 52 = 90- 33= 57

6n2=cos 33

cos 57= 1.540

(b) The critical angle is found from

nglasssin 'glass = nairsin 'air


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4

with 'air= 90and nair= 1.0

6'critical= arcsin1

n glass= arcsin

1

1.540= 40.5

2-9

Air

Water

12 cm

r

5

Find 5cfrom Snell's law n1sin 51= n2sin 5c= 1

When n2= 1.33, then 5c= 48.75

Find r from tan 5c=r

12cm, which yields r = 13.7 cm.

2-10.

45

Using Snell's law nglasssin 5c = nalcoholsin 90

where 5c = 45we have

nglass=1.45

sin 45= 2.05

2-11. (a) Use either NA = n1

2 " n2

2( )1/ 2

= 0.242

or


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5

NA 7n1 28 = n12(n

1"n

2)

n1

= 0.243

(b) 50,max

= arcsin (NA/n) = arcsin0.242

1.0

)

*

,

-= 14

2-13. NA = n1

2 "n2

2( )1/ 2

= n1

2 "n1

2(1" 8)2[ ]1/ 2

= n1 28 " 82( )

1 / 2

Since 8


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6

jEr=1

9#

1

r

;Hz;'

+ jr:H')

*+ ,

-Substituting into Eq. (2-33b) we have

:9#

1r

;Hz;'

+ jr:H')

*+ ,

-+ ;E z

;r= j# H'

Solve for H'and let q2= #29- :2 to obtain Eq. (2-35d).

(d) Solve Eq. (2-34b) for jE'

jE'= -1

9# j:Hr +

;Hz;r

)*

,-

Substituting into Eq. (2-33a) we have

1

r

;Ez

;'-

:

9# j:Hr +

;Hz;r

)*

,-

= -j# Hr

Solve for Hrto obtain Eq. (2-35c).

(e) Substitute Eqs. (2-35c) and (2-35d) into Eq. (2-34c)

- j

q 21

r

;

;r:

;Hz;'

+9#r;Ez

;r

)

*+ ,

-"

;

;' :

;Hz;r

"9#

r

;Ez;'

)

*+ ,

-/

012

34= j9#Ez

Upon differentiating and multiplying by jq2/9#we obtain Eq. (2-36).

(f) Substitute Eqs. (2-35a) and (2-35b) into Eq. (2-33c)

-

j

q 2

1

r

;

;r :

;E z

;' " # r

;Hz

;r

)

*+ ,

-"

;

;' :

;Ez

;r +

#

r

;Hz

;'

)

*+ ,

-

/

01

2

34= -j#Hz

Upon differentiating and multiplying by jq2/9#we obtain Eq. (2-37).

2-15. For $= 0, from Eqs. (2-42) and (2-43) we have


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7

Ez= AJ0(ur) ej( #t " :z)

and Hz= BJ0(ur) ej( #t " :z)

We want to find the coefficients A and B. From Eqs. (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:

Aj:$

a

)* ,

-

1

u2 +

1

w2

)* ,-= B#

J'$

(ua)

uJ $ (ua)+

K'$

(wa)

wK$(wa)

/

012

34

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 TM0mmodes.

For the other case, substitute Eqs. (2-47) and (2-51) into Eq. (2-52):

0 =1

u2

Bj:$

aJ $ (ua) +A#91uJ'$ (ua)

/

0

2

3

+1

w2 B

j:$

aJ$ (ua) +A#92w

K' $ (wa)J$ (ua)

K $(wa)

/

01 34

With k12 = #291and k22 = #292 rewrite this as

B$=ja

:#

1

1

u2

+ 1

w2

/

0

1

1

2

3

4

4k

1

2J$ + k2

2K $[ ]A


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9

V =2! (25m)

0.82 m (1.48)2 "(1.46)2[ ]

1/ 2

= 46.5

Using Eq. (2-61) M 7V2/2 =1081 at 820 nm.

Similarly, M = 417 at 1320 nm and M = 303 at 1550 nm. From Eq. (2-72)

Pclad

P

)*

,-

total

74

3M-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


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10

2-23. For small values of 8we can write V 72!a

%n1 28

For a = 5 m we have 870.002, so that at 0.82 m

V 72! (5m)

0.82m1.45 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 LP11modes exist in the fiber at 0.82 m.

2-24.

2-25. From Eq. (2-77) Lp=2!

: =%

ny" nx

For Lp= 10 cm ny- nx=1.3 = 10"6 m

10"1

m= 1.3=10-5

For Lp= 2 m ny- nx=1.3 = 10"6 m

2 m= 6.5=10-7

Thus

6.5=10-7


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11

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

For a step index fiber we can use Eq. (2-61)

Mstep7V 2

2

=1

2

2!a

%

)

*

,

-

2

n1

2 " n2

2( )

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

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

Mstep=2!an1

%

)*

,-

2

8=1086at820nm

432at1300 nm

?@A

2-28. Using Eq. (2-23) we have

(a) NA = n12

"n22

( )1/ 2

= (1.60)2

" (1.49)2

[ ]1/ 2

= 0.58

(b) NA = (1.458)2 " (1.405)2[ ]

1/ 2

= 0.39

2-29. (a) From the Principle of the Conservation of Mass, the volume of a preform rod

section of length Lpreformand cross-sectional area A must equal the volume of the fiber

drawn from this section. The preform section of length Lpreformis drawn into a fiber of

length Lfiberin 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

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

Equating these yields

StD

2

)* ,-

2

= std

2

)* ,-

2

or s = SD

d

)* ,-

2

(b) S = sd

D

)*

,-

2

= 1.2 m/s0.125mm

9mm

)*

,-

2

= 1.39 cm/min


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12

2-30. Consider the following geometries of the preform and its corresponding fiber:

PREFORMFIBER

4 mm

3 mm

R

25 m

62.5 m

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=

Afibercore

Afiberclad

or

!(32 " R2 )

!(42 " 32 )=

!(25)2

! (62.5)2 "(25)

2[ ]

from which we have

R = 9" 7(25)

2

(62.5)2 "(25)

2

/

012

34

1/ 2

= 2.77 mm

Thus, 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=10-3cm)2(105 cm) = 1.96 cm3

The mass M equals the density Btimes the volume V:

M = BV = (2.6 gm/cm3)(1.96 cm3) = 5.1 gm


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13

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

t =M

R=

5.1gm

0.5gm / min= 10.2 min

2-32. Solving Eq. (2-82) for Cyields

C=K

YD

)*

,-

2

where Y = ! for surface flaws.

Thus

C=(20N / m m3 / 2 )2

(70MN/ m2 )2 != 2.60=10-4mm = 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 Dis independent of time):

C" b / 2 Ci

C f

E dC = AYbDb0

t

E dt

which yields

1

1 "b

2

Cf1" b / 2 " C i

1" b/ 2[ ]= AYbDbt

or

t =2

(b" 2)A(YD)b C i

(2 " b)/ 2 " Cf( 2" b)/ 2[ ]

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

t =2

(b" 2)A(YD)b

Ki

YD

)*

,-

2" b

" Kf

YD

)*

,-

2 "b/

012

34

72Ki

2" b

(b" 2)A(YD)b if K i

b" 2>Kf

2" b

2-34. Substituting Eq. (2-82) into Eq. (2-86) gives

dC

dt= AKb= AYbCb/2Db


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14

Integrating this from Cito C p where

Ci=

K

YD i

)

*

+ ,

-

.2

and Cp=

K

YDp

)

*

+,

-

.

2

are the initial crack depth and the crack depth after proof testing, respectively, yields

C "b / 2 C i

C p

E dC = AYb

Db

0

t p

E dt

or

1

1 "b

2

Cp1" b / 2

" Ci1"b / 2[ ]= AYb D pb tp

for a constant stress Dp. Substituting for C i and C p gives

2

b" 2

)*

,-

K

Y

)*

,-

2"b

Di

b" 2 " Dp

b" 2[ ]= AYb D pb tpor

2

b" 2

)*

,-

K

Y

)*

,-

2"b1

AYb D i

b" 2" Dp

b" 2[ ]= B D ib"2 " Dpb"2[ ]= D pb

tp

which is Eq. (2-87).

When a static stress Dsis applied after proof testing, the time to failure is found from Eq.(2-86):

C "b / 2 C p

Cs

E dC = AYb Dsb

0

t s

E dt

where C s is the crack depth at the fiber failure point. Integrating (as above) we get Eq. (2-

89):

B Dpb"2 " Dsb"2[ ]= Dsb ts

Adding Eqs. (2-87) and (2-89) yields Eq. (2-90).


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16

2-36. The failure probability is given by Eq. (2-85). For equal failure probabilities of the

two fiber samples, F1= F2, or

1 - exp " D1c

D0

)

*+ ,

-.

m

L1

L0

/

01

34= 1 - exp "

D2cD0

)

*+ ,

-.

m

L2

L0

/

01

34

which implies that

D1c

D0

)

*+ ,

-.

m

L1

L0

=D

2c

D0

)

*+ ,

-.

m

L2

L0

or

D1cD

2c

=L

2

L1

)

*+ ,

-.

1 / m

If L1= 20 m, then D1c= 4.8 GN/m2

If L2= 1 km, then D2c= 3.9 GN/m2

Thus

4.83.9

)* ,-

m

= 100020

= 50

gives

m =log50

log(4.8/ 3.9)= 18.8


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2

!!(m) ""uv ""IR0.5 20.3 --

0.7 1.44 --

0.9 0.33 --

1.2 0.09 2.2)10-6

1.5 0.04 0.00722.0 0.02 23.2

3.0 0.009 7.5)104

3-6. From Eq. (3-4a) we have

!scat=8*3

3(4 (n

2+ 1)

2kBTf,T

=

8*3

3(0.63m)4 (1.46)2

+ 1[ ]2

(1.38)10-16dyne-cm/K)(1400 K)

)(6.8)10-12cm2/dyne)

= 0.883 km-1

To change to dB/km, multiply by 10 log e = 4.343: !scat= 3.8 dB/km

From Eq. (3-4b):

!scat= 8*3

3(4

n8p2kBTf,T = 1.16 km-1

= 5.0 dB/km

3-8. Plot of Eq. (3-7).

3-9. Plot of Eq. (3-9).

3-10. From Fig. 2-22, we make the estimates given in this table:

##m Pclad/P ""##m= ""11+ (""22 - ""11)Pclad/P 5 + 103Pclad/P01 0.02 3.0 + 0.02 5 + 20 = 25

11 0.05 3.0 + 0.05 5 + 50 = 55

21 0.10 3.0 + 0.10 5 + 100 = 105

02 0.16 3.0 + 0.16 5 + 160 = 165

31 0.19 3.0 + 0.19 5 + 190 = 195

12 0.31 3.0 + 0.31 5 + 310 = 315


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3

3-11. (a) We want to solve Eq. (3-12) for !gi. With != 2 in Eq. (2-78) and letting

-=n

2(0)+n 2

2

2n2

(0)

we have

!(r) = !1+ (!2- !1)n 2 (0)+n 2(r)

n2(0) +n 2

2 = !1+ (!2- !1)r 2

a2

Thus

!gi=!(r)p(r)rdr

0

.

/

p(r) rdr

0

.

/= !1+

(!2+ !

1)

a2

exp+ Kr2( )r 3dr0

.

/

exp +Kr2( )rdr0

.

/

To evaluate the integrals, let x = Kr2, so that dx = 2Krdr. Then

exp+Kr 2( )r3dr0

.

/

exp+Kr 2( )rdr0

.

/=

1

2 K2 e+ xxdx

0

.

/

1

2K e+x dx

0

.

/=

1

K1!

0! =

1

K

Thus !gi = !1+(!2+ ! 1)

Ka2

(b) p(a) = 0.1 P0= P0e+ Ka2

yields eKa 2

= 10.

From this we have Ka2= ln 10 = 2.3. Thus

!gi = !1+ (!2+ ! 1)2.3

= 0.57!1+ 0.43!2

3-12. With (in units of micrometers, we have


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4

n = 1 + 196.98

(13.4)2 + (1.24 / ()2012

345

1/ 2

To compare this with Fig. 3-12, calculate three representative points, for example,

(= 0.2, 0.6, and 1.0 m. Thus we have the following:

Wavelength !! Calculated n n from Fig. 3-12

0.2 m 1.548 1.550

0.6 m 1.457 1.458

1.0 m 1.451 1.450

3-13. (a) From Fig. 3-13,d6

d(780 ps/(nm-km) at 850 nm. Therefore, for the LED we

have from Eq. (3-20)

8mat

L=

d6

d(8(= [80 ps/(nm-km)](45 nm) = 3.6 ns/km

For a laser diode,

8mat

L= [80 ps/(nm-km)](2 nm) = 0.16 ns/km

(b) From Fig. 3-13,d6mat

d(

= 22 ps/(nm-km)

Therefore, Dmat(() = [22 ps/(nm-km)](75 nm) = 1.65 ns/km

3-14. (a) Using Eqs. (2-48), (2-49), and (2-57), Eq. (3-21) becomes

b = 1 -ua

V

9:

;


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5

, / k +n 2n

1+n

2

=n

1(1 + =) +n 2n

1+n

2

= 1 -n

1

n1+n

2

=

Letting n2= n1(1 - -) then yields

, / k +n 2n

1+n

2

= 1 - =2+ -

71 since =2+ -

=x

s=

n 2

n1=

n 1(1 + -)

n1= (1 - -)

The travel time of the highest order ray is thus

Tmax =n

1

c s

L

x

9:

;


22/116

6

Therefore

Tmin

- Tmax

=Ln

1

c

1

1 + -+ 1

9:

;


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7

C2=3! + 2

2(! + 2)=

3 2 1 +6

5-

9:

;


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8

3-20. We want to plot Eq. (3-30) as a function of 8 ( , where 8 int er modal and

8 int ra modal are given by Eqs. (3-41) and (3-45). For A= 0 and != 2, we have C1=

0 and C2= 1/2. Since 8int er modal

does not vary with 8(

, we have

8int er modal

L=

N1-

2c

!

! +2

9:

;


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9

=LN

1

c

kn1

,1+

4-

! +2

m

M

9:

;


26/116

10

Ignoringd-

d(and

dn1

d(terms yields

dM

d(=

2!

! +2a2k2n

1

2 - +1

(

9

:

;


27/116

11

- 2 (2d 2n

1

d(29

:C ;


28/116

12

3-29. For ! =0.95!opt , we have

8 int er (! H !opt )

8 int er (! = !opt )=

(! + ! opt)

-(! +2)=+

0.05

(0.015)(1.95)=+170%

For ! =1.05!opt , we have

8 int er (! H !opt )

8 int er (! = !opt )=

(! + ! opt )

-(! +2)= +

0.05

(0.015)(2.05)= +163%


29/116

1


4-1. From Eq. (4-1), ni= 2!"#

$%&2'kBT

h2

3/2(memh)

3/4 exp

!"#

$%&

-Eg

2kBT

= 22'(1.38(10)23 J / K)

(6.63(10)34 J.s) 2*

+,-

./

3/ 2

T3 / 2

(.068)(.56)(9.11(10)31

kg)2[ ]

(exp )(1.55) 4.3(10)4 T)eV

2(8.62(10)5 eV/K)T

*

+,-

./

= 4.15(1014T3/2

exp ) 1.55

2(8.62(10)5

)T

*

+,-

./exp

4.3(10)4

2(8.62(10)5

)

*

+,-

./

= 5.03(1015T3/2

exp!"#

$%&

-8991

T

4-2. The electron concentration in a p-type semiconductor is nP = ni= pi

Since both impurity and intrinsic atoms generate conduction holes, the total

conduction-hole concentration pP is

pP = NA+ ni= NA+ nP

From Eq. (4-2) we have that nP = n

2

i /pP . Then

pP = NA+ nP = NA+ n

2i /p

P or p

2

P - NApP - n2i = 0

so that

pP =

NA

2

!""#

$%%&

1 +4n

2i

N2A

+ 1

If ni


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2

x2+ 4.759x - 0.436 = 0. Solving this quadratic equation yields (taking the plus

sign only)

x =1

2[ ]- 4.759 + (4.759)2+ 4(.436) = 0.090

The emission wavelength is 1=1.240

1.540 = 805 nm.

(b) Eg= 1.424 + 1.266(0.15) + 0.266(0.15)2= 1.620 eV, so that

1=1.240

1.620 = 766 nm

4-4. (a) The lattice spacings are as follows:

a(BC) = a(GaAs) = 5.6536 Ao

a(BD) = a(GaP) = 5.4512 Ao

a(AC) = a(InAs) = 6.0590 Ao

a(AD) = a(InP) = 5.8696 Ao

a(x,y) = xy 5.6536 + x(1-y) 5.4512 + (1-x)y 6.0590 + (1-x)(1-y)5.8696

= 0.1894y - 0.4184x + 0.0130xy + 5.8696

(b) Substituting a(xy) = a(InP) = 5.8696 Ao

into the expression for a(xy) in (a),

we have

y =0.4184x

0.1894 - 0.0130x 0

0.4184x

0.1894 = 2.20x

(c) With x = 0.26 and y = 0.56, we have

Eg= 1.35 + 0.668(.26) - 1.17(.56) + 0.758(.26)2+ 0.18(.56)2

- .069(.26)(.56) - .322(.26)2(.56) + 0.03(.26)(.56)2= 0.956 eV

4-5. Differentiating the expression for E, we have


31/116

3

2E = hc

1221 or 21 =

12

hc2E

For the same energy difference 2E, the spectral width 21is proportional to the

wavelength squared. Thus, for example,

211550

211310

= 1550

1310

#!

&$

2

=1.40

4-6. (a) From Eq. (4-10), the internal quantum efficiency is

3int

= 1

1 +25/ 90=0.783, and from Eq. (4-13) the internal power level is

Pint

=(0.783) hc(35mA)q(1310

nm)=26

mW

(b) From Eq. (4-16),

P = 1

3.5 3.5 + 1( )226 mW = 0.37mW

4-7. Plot of Eq. (4-18). Some representative values of P/P0are given in the table:

f in MHz P/P01 0.999

10 0.954

20 0.847

40 0.623

60 0.469

80 0.370

100 0.303

4-8. The 3-dB optical bandwidth is found from Eq. (4-21). It is the frequency f atwhich the expression is equal to -3; that is,

10 log 1

1 + 2'f4( )2[ ]

1/ 2

#

!"

&

$%=)3


32/116

4

With a 5-ns lifetime, we find f = 1

2' 5ns( )100.6 )1( )=9.5MHz

4-9. (a) Using Eq. (4-28) with 5= 1

gth=1

0.05

cmln!"#

$%&1

0.32

2 + 10 cm-1= 55.6 cm-1

(b) With R1= 0.9 and R2= 0.32,

gth=1

0.05 cm ln

+,*

./-1

0.9(0.32) + 10 cm-1= 34.9 cm-1

(c) From Eq. (4-37) 3ext= 3i(gth- 6)/gth;

thus for case (a): 3ext= 0.65(55.6 - 10)/55.6 = 0.53

For case (b): 3ext= 0.65(34.9 - 10)/34.9 = 0.46

4-10. Using Eq. (4-4) to find Egand Eq. (4-3) to find 1, we have for x = 0.03,

1=1.24

Eg =

1.24

1.424 + 1.266(0.3) + 0.266(0.3)2 = 1.462 m

From Eq. (4-38)

3ext= 0.8065 1(m)dP(mW)

dI(mA)

Taking dI/dP = 0.5 mW/mA, we have 3ext= 0.8065 (1.462)(0.5) = 0.590

4-11. (a) From the given values, D = 0.74, so that 5T= 0.216

Then neff

2= 10.75 and W = 3.45, yielding 5L= 0.856

(b) The total confinement factor then is 5= 0.185

4-12. From Eq. (4-46) the mode spacing is

21=12

2Ln =

(0.80 m)2

2(400 m)(3.6) = 0.22 nm

Therefore the number of modes in the range 0.75-to-0.85 m is


33/116

5

.85) .75

.22(10)3 =

.1

.22 (103= 455 modes

4-13. (a) From Eq. (4-44) we have g(1) = (50 cm-1) exp+,

*

./

--

(1- 850 nm)2

2(32 nm)2

= (50 cm-1) exp+,*

./-

-(1- 850)2

2048

(b) On the plot of g(1) versus 1, drawing a horizontal line at g(1) = 6t

= 32.2 cm-1shows that lasing occurs in the region 820 nm < 1< 880 nm.

(c) From Eq. (4-47) the mode spacing is

21=12

2Ln =

(850)2

2(3.6)(400 m) = 0.25 nm

Therefore the number of modes in the range 820-to-880 nm is

N =880 - 820

0.25 = 240 modes

4-14. (a) Let Nm= n/1= m2L be the wave number (reciprocal wavelength) of mode m.

The difference 2N between adjacent modes is then

2N = Nm- Nm-1=1

2L (a-1)

We now want to relate 2N to the change 21in the free-space wavelength. Firstdifferentiate N with respect to 1:

dN

d1 =

d

d1!"#

$%&n

1 =

1

1dn

d1 -

n

12 = -

1

12!"#

$%&

n - 1dn

d1

Thus for an incremental change in wavenumber 2N, we have, in absolute values,

2N =1

12!"#

$%&

n - 1dn

d1 21 (a-2)


34/116

6

Equating (a-1) and (a-2) then yields 21=12

2L!"#

$%&

n - 1dn

d1

(b) The mode spacing is 21=

(.85 m)2

2(4.5)(400 m) = 0.20 nm

4-15. (a) The reflectivity at the GaAs-air interface is

R1= R2= !"#

$%&n-1

n+1

2 =

!"#

$%&3.6-1

3.6+1

2 = 0.32

Then Jth=1

7 6+

1

2Lln

1

R1R2

*

+, ./= 2.65(103A/cm2

Therefore

Ith= Jth(l (w = (2.65(103A/cm2)(250(10-4cm)(100(10-4cm) = 663 mA

(b) Ith= (2.65(103A/cm2)(250(10-4cm)(10(10-4cm) = 66.3 mA

4-16. From the given equation

2E11 =1.43

eV + 6.6256( 10

)34

J 8 s( )

2

8 5nm( )2 1

6.19( 10)32 kg+ 1

5.10(10)31kg#!" &

$

=1.43 eV + 0.25eV =1.68eV

Thus the emission wavelength is 1= hc/E = 1.240/1.68 = 739 nm.

4-17. Plots of the external quantum efficiency and power output of a MQW laser.

4-18. From Eq. (4-48a) the effective refractive index is

ne=m1B29

=2(1570 nm)

2(460 nm) = 3.4

Then, from Eq. (4-48b), for m = 0


35/116

7

1= 1B1

2

B

2neL!"#$%&1

2 = 1570 nm

(1.57 m)(1570 nm)

4(3.4)(300 m) = 1570 nm 1.20 nm

Therefore for m = 1, 1= 1B3(1.20 nm) = 1570 nm 3.60 nm

For m = 2, 1= 1B5(1.20 nm) = 1570 nm 6.0 nm

4-19. (a) Integrate the carrier-pair-density versus time equation from time 0 to td(time

for onset of stimulated emission). In this time the injected carrier pair density

changes from 0 to nth.

td

= dt0

t d

: = 1

J

qd)

n

4

0

n th

: dn =)4J

qd)

n

4

#

!" &

$n= 0

n= nth

=4

ln

J

J ) Jth

#

!" &

$%

where J = Ip/A and Jth= Ith/A. Therefore td= 4ln!"#

$%&Ip

Ip- Ith

(b) At time t = 0 we have n = nB, and at t = tdwe have n = nth. Therefore,

td= ;P(t) = P0[1 + 0.2m] = 1 mW

The minimum value is

P(t) = P0[1 - 2.36m] =0 which implies m >1

2.36 = 0.42

Therefore for the average value we have < >P(t) = P0[1 + 0.2(0.42)] >1 mW,

which implies

P0= 11.084 = 0.92 mW so thatP(t) = 0.92[1 + 0.42x(t)] mW and

i(t) = 10 P(t) = 9.2[1 + 0.42x(t)] mA

4-23. Substitute x(t) into y(t):

y(t) = a1b1cos ?1t + a1b2cos ?2t

+ a2(b12cos2?1t + 2b1b2cos ?1t cos ?2t + b2

2cos2?2t)

+ a3(b13cos3?1t + 3b1

2b2cos2?1t cos ?2t + 3b1b2

2cos ?1t cos2?2t+ b2

3cos3?2t)

+a4(b14cos4?1t + 4b1

3b2cos3?1t cos ?2t + 6b1

2 b2

2cos2?1t cos2?2t

+ 4b1b23cos ?1t cos3?2t + b2

4cos4?2t)

Use the following trigonometric relationships:

i) cos2x =1

2(1 + cos 2x)

ii) cos3x =1

4(cos 3x + 3cos x)

iii) cos4x =1

8(cos 4x + 4cos 2x + 3)

iv) 2cos x cos y = cos (x+y) + cos (x-y)

v) cos2x cos y =1

4 [cos (2x+y) + 2cos y + cos (2x-y)]


37/116

9

vi) cos2x cos2y =1

4 [1 + cos 2x+ cos 2y +1

2 cos(2x+2y) +1

2 cos(2x-2y)]

vii) cos3x cos y =1

8 [cos (3x+y) + cos (3x-y) + 3cos (x+y) + 3cos (x-y)]

then

y(t) = 12+,*

./-a2b21+ a2b22+ 34a4b

41+ 3a4b

21b

22+ 34

a4b42 constantterms

+3

4+*

.-a3b

3

1+ 2a3b1b2

2 cos ?1t +3

4 a3+*

.-b

3

2+ 2b2

1b2 cos ?2tfundamental

terms

+b

2

1

2+*

.-a2 + a4b

2

1+ 3a4b2

2 cos 2?1t +b

2

2

2+*

.-a2 + a4b

2

2+ 3a4b2

1 cos 2?2t

2nd-order harmonic terms

+1

4 a3b

3

1 cos 3?1t +1

4 a3b

3

2 cos 3?2t 3rd-order harmonic terms

+1

8 a4b

4

1 cos 4?1t +1

8 a4b

4

2 cos 4?2t 4th-order harmonic terms

+ +*

.-a2b1b2+

3

2a4b3

1b2+3

2a4b1b3

2 [ ]cos (?1+?2)t + cos (?1-?2)t

2nd-order intermodulation terms

+3

4 a3b2

1 b2[ ]cos(2?1+?2)t + cos(2?1-?2)t +3

4 a3b1b2

2[ ]cos(2?2+?1)t + cos(2?2-?1)t

3rd-order intermodulation terms

+1

2 a4b3

1 b2[ ]cos(3?1+?2)t + cos(3?1-?2)t

+3

4 a4b2

1 b2

2[ ]cos(2?1+2?2)t + cos(2?1-2?2)t

+1

2 a4b1b3

2[ ]cos(3?2+?1)t + cos(3?2-?1)t 4th-order intermodulation

terms

This output is of the form

y(t) = A0+ A1(?1) cos ?1t + A2(?1) cos 2?1t + A3(?1) cos 3?1t

+ A4(?1) cos 4?1t + A1(?2) cos ?2t + A2(?2) cos 2?2t


38/116

10

+ A3(?2) cos 3?2t + A4(?2) cos 4?2t + @m

@n

Bmncos(m?1+n?2)t

where An(?j) is the coefficient for the cos(n?j)t term.

4-24. From Eq. (4-58) P = P0 e-t/4m

where P0= 1 mW and4m= 2(5(104hrs) = 105

hrs.

(a) 1 month = 720 hours. Therefore:

P(1 month) = (1 mW) exp(-720/105) = 0.99 mW

(b) 1 year = 8760 hours. Therefore

P(1 year) = (1 mW) exp(-8760/105) = 0.92 mW

(c) 5 years = 5(8760 hours = 43,800 hours. Therefore

P(5 years) = (1 mW) exp(-43800/105) = 0.65 mW

4-25. From Eq. (4-60) 4s= K eEA/kBT

or ln4s= ln K + EA/kBT

where kB= 1.38(10-23J/K = 8.625(10-5eV/K

At T = 60C = 333K, we have

ln 4(104= ln K + EA/[(8.625(10-5eV)(333)]

or 10.60 = ln K + 34.82 EA (1)

At T = 90C = 363K, we have

ln 6500 = ln K + EA/[(8.625(10-5eV)(363)]

or 8.78 = ln K + 31.94 EA (2)

Solving (1) and (2) for EAand K yields

EA= 0.63 eV and k = 1.11(10-5hrs

Thus at T = 20C = 293K


39/116

11

4s = 1.11(10-5exp{0.63/[(8.625(10-5)(293)]} = 7.45(105hrs


40/116

1


5-3. (a) cosL30 = 0.5

cos 30 = (0.5)1/L= 0.8660

L = log 0.5/log 0.8660 = 4.82

(b) cosT15 = 0.5

cos 15 = (0.5)1/T= 0.9659

T = log 0.5/log 0.9659 = 20.0

5-4. The source radius is less than the fiber radius, so Eq. (5-5) holds:

PLED-step= !2r2s B0(NA)

2= !2(2"10-3cm)2(100 W/cm2)(.22)2= 191 W

From Eq. (5-9)

PLED-graded= 2!2(2"10-3cm)2(100 W/cm2)(1.48)2(.01)#$%

&'(

1 -1

2)*+,2

5

2 = 159 W

5-5. Using Eq. (5-10), we have that the reflectivity at the source-to-gel interface is

R s- g = 3.600-1.3053.600+1.305*) ,+

2

= 0.219

Similarly, the relfectivity at the gel-to-fiber interface is

R g- f = 1.465 -1.305

1.465 +1.305

*)

,+

2

= 3.34"10-3

The total reflectivity then is R =Rs- g

Rg-f = 7.30"10

-4

The power loss in decibels is (see Example 5-3)

L =-10 log (1-R) =-10 log (0.999)=3.17 "10-3

dB

5-6. Substituting B(.) = B0cosm. into Eq. (5-3) for B(.,/), we have


41/116

2

P =

01

12

0

rm

012

0

2!

#$%

&'(

2! 02

0

.0-maxcos3.sin .d. d.sr dr

Using

02

0

.0cos3.sin .d. = 02

0

.0

( )1 - sin2. sin .d(sin .) = 02

0

sin .0

( )x - x3 dx

we have

P = 2!

0112

0

rm

012

0

2!

#$%

&'(sin2.0-max

2-

sin4.0-max4

d.sr dr

= !

0112

0

rm

012

0

2!

#$% &'(NA2- 12NA

4 d.sr dr

=!2

[ ]2NA2- NA4 02

0

rm

r dr 02

0

2! d.s

5-7. (a) Let a = 25 m and NA = 0.16. For rs3a(NA) = 4 m, Eq. (5-17) holds. For

rs44 m, 5= 1.

(b) With a = 50 m and NA = 0.20, Eq. (5-17) holds for rs310 m. Otherwise, 5

= 1.

5-8. Using Eq. (5-10), the relfectivity at the gel-to-fiber interface is


42/116

3

Rg- f =

1.485 -1.305

1.485 +1.305

*)

,+

2

= 4.16"10-3

The power loss is (see Example 5-3)

L =-10

log

(1-R) =-10

log

(0.9958) =0.018

dB

When there is no index-matching gel, the joint loss is

R a- f = 1.485 -1.000

1.485 +1.000

*)

,+

2

=0.038

The power loss is L =-10 log (1-R) =-10 log (0.962) = 0.17dB

5-9. Shaded area = (circle segment area) - (area of triangle) =1

2 sa -

1

2 cy

s = a.= a [2 arccos (y/a)] = 2a arccos)6*

+7,d

2a

c = 2#$%

&'(

a2-)6*+7,d

22 1/2

Therefore

Acommon

= 2(shaded area) = sa cy = 2a2arccos)6

*

+7

,d

2a - d

#$

%

&'

(a2-

)6

*

+7

,d2

4

1/2

5-10.

Coupling loss (dB) for

Given axial misalignments (m)

Core/cladding diameters

(m)

1 3 5 10

50/125 0.112 0.385 0.590 1.266

62.5/125 0.089 0.274 0.465 0.985

100/140 0.056 0.169 0.286 0.590

5-11. arccos x =!2

- arcsin x

For small values of x, arcsin x = x +x3

2(3) +

x5

2(4)(5) + ...


43/116

4

Therefore, ford

2a


44/116

5

= -10 log

#$$%

&''(NA

2

R(0)

NA2E(0)

5-15. For fibers with different 9values, where 9R

< 9E

LF= -10 log 5F = -10 log

k2NA2(0))66*

+77,9

R

29R

+ 4a2

k2NA2(0))66*

+77,9

E

29E

+ 4a2

= -10 log#$$%

&''(9

R(9

E+ 2)

9E

(9R

+ 2)

5-16. The splice losses are found from the sum of Eqs. (5-35) through (5-37). First find

NA(0) from Eq. (2-80b).

For fiber 1: NA1(0) =n

1 2: = 1.46 2 0.01( ) =0.206

For fiber 2: NA2

(0) = n1

2: = 1.48 2 0.015( ) = 0.256

(a)

The only loss is that from index-profile differences. From Eq. (5-37)

L1; 2 9( ) =-10 log

1.80(2.00 + 2)

2.00(1.80 + 2)= 0.24

dB

(b)

The losses result from core-size differences and NA differences.

L2;1

(a) =-20

log

50

62.5

*)

,+=

1.94

dB

L2;1

(NA) =-20

log

.206

.256

%#

(&= 1.89

dB

5-17. Plots of connector losses using Eq. (5-43).

5-18. When there are no losses due to extrinsic factors, Eq. (5-43) reduces to


45/116

6

LSM;ff= -10 log

#$%

&'(4

)6*

+7,W1

W2+

W2W1

2

For W1= 0.9W2, we then have LSM;ff= -10 log #$%

&'(44.0446 = - 0.0482 dB

5-19. Plot of Eq. (5-44).

5-20. Plot of the throughput loss.


46/116

1


6-1. From Eqs. (6-4) and (6-5) with Rf= 0, != 1 - exp(-"sw)

To assist in making the plots, from Fig. P6-1, we have the following representativevalues of the absorption coefficient:

!!(m) ""s(cm-1)

.60 4.4#103

.65 2.9#103

.70 2.0#103

.75 1.4#103

.80 0.97#103

.85 630

.90 370

.95 190

1.00 70

6-2. Ip= qA $%

0

w

G(x) dx = qA &0"s$%

0

w

e-"sx

dx

= qA &0'( )*1 - e-"sw = qA P0(1 - Rf)h+A'( )*1 - e-"sw

6-3. From Eq. (6-6), R =!qh+

=!q,hc

= 0.8044 !, ( in m)

Plot R as a function of wavelength.

6-4. (a) Using the fact that Va-VB, rewrite the denominator as

1 - ./

0

12

3Va- IMRMVB

n

= 1 - ./

0

12

3VB- VB+ Va- IMRMVB

n

= 1 -./0

123

1 -VB- Va+ IMRM

VB

n

SinceVB- Va+ IMRM

VB


47/116

2

1 -./0

123

1 -VB- Va+ IMRM

VB

n -1 -

'4(

)5*

1 -n( )VB- Va+ IMRM

VB

=

n( )VB- Va+ IMRM

VB -nIMRM

VB

Therefore, M0=IMIp

-VB

n( )VB- Va+ IMRM -

VBnIMRM

(b) M0=IMIp

=VB

nIMRM implies I

2M =

IpVBnRM

, so that M0=./0

123VB

nIpRM

1/2

6-5. < >i2

s(t) =

1

T$%0

T

i

2

s(t)dt =

627 $%

0

27/6

R

2

0P2(t) dt (where T = 27/6),

=627

R2

0 P2

0 $%

0

27/6

(1 + 2m cos 6t + m2cos26t) dt

Using

cos 6tdt =0

7 6

8 1

6

sin 6t t = 0t = 27 / 6

=0

and $%

0

27/6

cos26t dt =1

6

$9%

0

27

./0

1231

2+

1

2cos 2x dx =

76

we have < >i2s(t) = R2

0 P2

0./0

123

1 +m2

2

6-6. Same problem as Example 6-6: compare Eqs. (6-13), (6-14), and (6-17).

(a)First from Eq. (6-6), I p =!q,

hcP0 =0.593A

Then :Q2= 2qI pB =2 1.6# 10

;19C( )(0.593A)(150# 106 Hz) = 2.84# 10;17 A2


48/116


49/116

4

6-9. 0 =d

dM./0

123S

N =

d

dM'44(

)55*1

2I2

pM2

2q(Ip+ ID)M2+x+ 2qIL+ 4kBT/RL

0 = I2p M -

(2+x)M1+x 2q(Ip+ ID)1

2I2pM

2

2q(Ip+ ID)M2+x+ 2qIL+ 4kBT/RL

Solving for M: M2+x

opt =2qIL+ 4kBT/RL

xq(Ip+ ID)

6-10. (a) Differentiating pn, we have

=pn

=x =

1

Lp.

/0

1

23

pn0+ Be

-"sw

e

(w-x)/Lp

- "sBe

-"sx

= 2pn=x

2= -

1

L2

p

./0

123

pn0+ Be-"sw

e(w-x)/Lp

+ "2

s Be-"sx

Substituting pnand= 2p

n

=x2into the left side of Eq. (6-23):

-Dp

L2

p

./0

123

pn0+ Be-"sw

e(w-x)/Lp

+ Dp"2

s Be-"sx

+1

>p

./0

123

pn0+ Be-"sw

e(w-x)/Lp

-B

>p

e-"sx

+ &0"se-"sx

=

'4(

)5*

B

./0

123

Dp"2

s -1

>p

+ &0"s e-"sx

where the first and third terms cancelled because L2

p = Dp>p .

Substituting in for B:

Left side =

'44(

)55*&0

Dp

"sL2p

1 - "2

s L2

p

./0

123

Dp"2

s -1

>p

+ &0"s e-"sx


50/116

5

=&0Dp

'44(

)55*"sL

2

p

1 - "2s L

2p

.//0

1223"

2

s Lp- 1

>p

+ Dp"s e-"sx

=

&0Dp( )-Dp"s+ Dp"s e

-"sx

= 0 Thus Eq. (6-23) is satisfied.

(b) Jdiff= qDp=pn=x

0.

31

x =w

= qDp'4(

)5*1

Lp./0

123

pn0+ Be-"sw

- "sBe-"sw

= qDpB

.

/0

1

231

Lp- "s e

-"sw + qpn0

DpLp

= q&0

.//0

1223"sL

2p

1 - "2

s L2

p

./0

1231 - "sLp

Lp e

-"sw + qpn0

DpLp

= q&0"sLp

1 + "sLp e

-"sw + qpn0

DpLp

c) Adding Eqs. (6-21) and (6-25), we have

Jtotal= Jdrift+ Jdiffusion = q&0'4(

)5*

.0

13

1 - e-"sw

+./0

123"sLp

1 + "sLpe

-"sw + qpn0

DpLp

= q&0.//0

1223

1 -e-"sw

1 + "sLp e

-"sw + qpn0

DpLp

6-11. (a) To find the amplitude, consider

.0 13JtotJ*totsc

1/2 = q&0( )S S* 1/2 where S = 1 - e

-j6td

j6td

We want to find the value of 6tdat which ( )S S*1/2

=1

2 .

Evaluating ( )S S*1/2

, we have


51/116


52/116

7

td=w

vd =

20 # 10;6 m

4.4# 104 m / s= 0.45 ns

c)1

"s = 10-3cm = 10 m =

1

2 w

Thus since most carriers are absorbed in the depletion region, the carrier diffusion

time is not important here. The detector response time is dominated by the RC

time constant.

6-13. (a) With k1-k2and keffdefined in Eq. (6-10), we have

(1) 1 -k1(1 - k1)

1 - k2 = 1 -

k1- k2

1

1 - k2 -1 -

k2- k2

1

1 - k2 = 1 keff

(2)(1 - k1)2

1 - k2 =

1 - 2k1+ k2

1

1 - k2 -

1 - 2k2+ k2

1

1 - k2

=1 - k21 - k2

-k2- k

2

1

1 - k2 = 1 - keff

Therefore Eq. (6-34) becomes Eq. (6-38):

Fe= keffMe+ 2(1 - keff) -1

Me(1 - keff) = keffMe+ ./0 1232 -

1

Me (1 - keff)

(b) With k1-k2and k'eff

defined in Eq. (6-40), we have

(1)k2(1 - k1)

k2

1(1 - k2) -

k2- k2

1

k2

1(1 - k2) = k

'eff

(2)(1 - k1)2k2

k2

1(1 - k2) =

k2- 2k1k2+ k2k2

1

k2

1(1 - k2)

-.0

13k2- k

2

1 - .0

13k

2

1- k2k2

1

k2

1(1 - k2) = k

'eff

- 1


53/116

8

Therefore Eq. (6-35) becomes Eq. (6-39): Fh= k'eff

Mh-./0

123

2 -1

Mh(k

'eff

- 1)

6-14. (a) If only electrons cause ionization, then @= 0, so that from Eqs. (6-36) and (6-

37), k1= k2= 0 and keff= 0. Then from Eq. (6-38)

Fe= 2 -1

Me - 2 for large Me

(b) If "= @, then from Eqs. (6-36) and (6-37), k1= k2= 1 so that

keff= 1. Then, from Eq. (6-38), we have Fe= Me.


54/116

1


7-1. We want to compare F1= kM + (1 - k) 2 ! 1

M

"#

$%and F2= M

x.

For silicon, k = 0.02 and we take x = 0.3:

M F1(M) F2(M) % difference

9 2.03 1.93 0.60

25 2.42 2.63 8.7

100 3.95 3.98 0.80

For InGaAs, k = 0.35 and we take x = 0.7:

M F1(M) F2(M) % difference

4 2.54 2.64 3.009 4.38 4.66 6.4

25 6.86 6.96 1.5

100 10.02 9.52 5.0

For germanium, k = 1.0, and if we take x = 1.0, then F1= F2.

7-2. The Fourier transform is

hB(t) = HB!&

&

' (f)ej2(ft

df = Rej2(ft

1+j2(fRC

df!&

&

'

Using the integral solution from Appendix B3:

ejpx

() +jx)n dx

!&

&

' =2((p) n !1e! )p

*(n)for p > 0, we have

hB(t) =1

2(C

ej 2(ft

1

2(RC+ jf

"#

$%

df!&

&

' =1

Ce-t/RC

7-3. Part (a):

hp!&

&

' (t) dt =1

+Tb ,-

-+Tb/2

+Tb/2

dt =1

+Tb#."

%/$+Tb

2+

+Tb2

= 1


55/116

2

Part (b):

hp

!&

&

' (t) dt =1

2(

1

+Tb exp !

t2

2(+Tb )2

0

123

45!&

&

' dt

= 12(

1

+Tb

( +Tb 2 = 1 (see Appendix B3 for integral solution)

Part (c):

hp

!&

&

' (t) dt =1

+Tb

exp ! t

+Tb

0

123

45dt

0

&

' = - e!& ! e

!0[ ]= 1

7-4. The Fourier transform is

F[p(t)*q(t)] = p(t)* q(t)e!j2(ft

dt!&

&

' = q(x) p(t!x)e!j2(ft

dtdx!&

&

'!&

&

'

= q(x)e!j2(fx

p(t!x)e!j2(f( t ! x)

dtdx!&

&

'!&

&

'

= q(x)

e!j2(fx

dx

p(y)

e!j 2(fy

dy!&

&

'!&

&

' where y = t - x

= F[q(t)] F[p(t)] = F[p(t)] F[q(t)] = P(f) Q(f)

7-5. From Eq. (7-18) the probability for unbiased data (a = b = 0) is

Pe=1

2P0 (v th ) + P1(v th )[ ].

Substituting Eq. (7-20) and (7-22) for P0and P1, respectively, we have

Pe=1

2

1

2(6 2 e

!v 2 / 26 2dv

V / 2

&

' + e! (v !V ) 2 / 26 2

dv!&

V / 2

'0

12

45

In the first integral, let x = v/ 262 so that dv = 262 dx.

In the second integral, let q = v-V, so that dv = dq. The second integral thenbecomes


56/116

3

e!q 2 / 26 2

dq!&

V / 2 !V

' = 262

e! x2

dx!&

!V / 2 26 2

' where x = q/ 2 62

Then

Pe=1

2

262

2(6 2 e

! x 2dx

V / 2 262

&

' + e! x 2

dx!&

! V / 2 26 2

'0

12

3

45

=1

2 ( e

!x 2

dx!&

&

' ! 2 e!x 2

dx0

V / 2 26 2

'0

12

3

45

Using the following relationships from Appendix B,

e

! p2 x 2

dx!&

&

' =(

p and2

( e!x 2

dx0

t

' = erf(t), we have

Pe=1

21! erf

V

26 2

"#

$%

0

123

45

7-6. (a) V = 1 volt and 6= 0.2 volts, so thatV

26= 2.5. From Fig. 7-6 for

V

26= 2.5,

we find Pe77810-3errors/bit. Thus there are (28105bits/second)(7810-3

errors/bit) = 1400 errors/second, so that

1

1400 errors/second = 7810-4seconds/error

(b) If V is doubled, thenV

26= 5 for which Pe73810-7errors/bit from Fig. 7-6.

Thus

1

(28 105

bits / sec ond)(3 8 10!7

errors/bit)= 16.7 seconds/error

7-7. (a) From Eqs. (7-20) and (7-22) we have

P0(vth) =1

2(62

e! v 2 / 26 2

dvV / 2

&

' =1

21! erf

V

26 2

"#

$%

0

12 45

and


57/116


58/116

5

Bbae=H p (0)

Hout (0)

2

Tb2

Hout' (v2N = < >v2

s +< >v2

R +< >v2

I +< >v2

E

= 2q < >i0 < >m2 BbaeR2A2+4kBT

Rb BbaeR2A2+ SIBbaeR2A2+ SEBeA2

=#." %/$2q < >i0 M2+x+

4kB

T

Rb+ SI BbaeR2A2+ SEBeA2

=R2A2I2

Tb#."

%/$

2q < >i0 M2+x+4kBT

Rb+ SI+

SE

R2 +

(2(RCA)2

T3

b

SEI3

7-11. First let x = v! boff( )/ 2 6 off with dx = dv / 2 6off in the first part of Eq. (7-

49):

Pe=26off

2(6off

exp! x2( )vth ! b off

26off

&

' dx =1

( exp !x2( )

Q / 2

&

' dx

Similarly, let y = !v +bon

( )/ 2

6on( )so that

dy = -dv/ 26on( )in the second part of Eq. (7-49):


59/116

6

Pe= -1

( exp !y

2( )&

! vth +b on26 on

' dy =1

( exp!y

2( )Q / 2

&

' dy

7-12. (a) Let x = V2 2 6

= K2 2

For K = 10, x = 3.536. Thus

Pe=e-x

2

2 (x = 2.97810-7errors/bit

(b) Given that Pe= 10-5=e-x

2

2 (x then e-x

2 = 2 ( 10-5x.

This holds for x 73, so that K = 2 2 x = 8.49.

7-13. Differentiating Eq. (7-54) with respect to M and setting dbon/dM = 0, we have

dbondM

= 0

= -Q(h: /9)

=2 M

2+ x 9

h:

"#

$% bonI 2 +W

0

123

45

1/ 2

+ M2+x 9

h:

"#

$% bonI2 (1 ! >

0

123

45+W

?@A

BCD

1/ 2

+Q(h: /9)

M( h:/9)

1

2(2 + x)M

1+ xb

onI

2

M 2+ x 9

h:

"#

$%

bon

I2 +W

0

123

45

1/ 2 +

1

2(2 +x)M

1+xb

onI

2(1! >)

M 2+x 9

h:

"#

$%

bon

I2

(1! > ) +W0

123

45

1/ 2

?

@E

AE

B

CE

DE

1 / 2

Letting G = M2+x#."

%/$9

h: bonI2for simplicity, yields

A@?

DCB

( )G + W1/2

+ [ ]G(1->) + W1/2

=G

2(2 + x)

A

@?

D

CB1

(G + W)

1/2+(1->)

[ ]G(1->) + W

1/2

Multiply by (G + W)1/2

[ ]G(1->) + W1/2

and rearrange terms to get

(G + W)1/2

120

453

W -Gx

2(1->) = [ ]G(1->) + W

1/2#."

%/$Gx

2- W


60/116

7

Squaring both sides and collecting terms in powers of G, we obtain the quadratic

equation

G2120

453x2>

4(1->) + G

120

453x2>

4W(2->) - >W2(1+x) = 0

Solving this equation for G yields

G =

-x2

4W(2->)

AE@E?

DECEBx4

16W2(2->)2+ x2(1->)W2(1+x)

1

2

x2

2(1->)

=

W(2->)2(1->) AE

@E?

DECEB

1 - 120

453

1 + 16 #."

%/$1+x

x2

1->

(2->)2

1

2

where we have chosen the "+" sign. Equation (7-55) results by letting

G = M2+x

opt bonI2 #."

%/$9

h:

7-14. Substituting Eq. (7-55) for M2+xbon into the square root expressions in Eq. (7-

54) and solving Eq. (7-55) for M, Eq. (7-54) becomes

bon=

Q#."

%/$h:

9b

1/(2+x)

on

120

453h:

9W(2->)2I2(1->)

K1/(2+x)AE

@E?

DECEB

120

453W(2->)

2(1->)K + W

1

2+

120

453W

2(2->)K + W

1

2

Factoring out terms:

b(1+x)/(2+x)

on = Q#."

%/$h:

9

(1+x)/(2+x)

Wx/2(2+x)

I1/(2+x)

2

8AE@E?

DECEB

120

453(2->)

2(1->)K + 1

12+

120

4531

2(2->)K + 1

12

120

453(2->)K

2(1->)

1/(2+x)

or bon= Q(2+x)/(1+x)

#."

%/$h:

9 W

x/2(1+x)

I1/(1+x)

2 L


61/116

8

7-15. In Eq. (7-59) we want to evaluate

lim

> F 1

(2! > )K

2(1! > )L

0

123

45

1+x

2+x

=lim

> F 1

(2 ! > )K

2(1! > )

0

123

45

1+x

2+x lim

> F 1

1

L

"#

$%

1+x

2+x

Consider first

lim

> F 1

(2 ! >)K

2(1! >)

0

12 45=

lim

> F1

(2 ! >)

2(1! > )!1+ 1 +B

(1! > )

(2! > )2

0

123

45

1

2?

@E

AE

B

CE

DE

where B = 16(1+x)/x2. Since >F1, we can expand the square root term in a

binomial series, so that

lim> F 1

(2 ! >)K2(1! >)

012

345= lim

> F1

(2 ! >)2(1! > )

!1+ 1 +12

B (1! > )(2 ! > )2

! Order(1 ! > )2012

345

?@A

BCD

=lim

> F 1

B

4(2 ! > )=

B

4 = 4

1+x

x2

Thuslim

> F1

(2 ! > )K

2(1! >)

0

123

45

1+x

2+x

=#."

%/$

41+x

x2

1+x

2+x

Next consider, using Eq. (7-58)

lim

> F 1

1

L

"#

$%

1+ x

2+x

=lim

> F1

(2 ! > )K

2(1! >)

0

123

45

1/(2+ x)

(2! >)

2(1! >)K +1

0

123

45

1

2

+ 1

2(2 ! > )K +10

134

1

2?

@E

AE

B

CE

D

From the above result, the first square root term is

120

453(2->)

2(1->)K + 1

1

2 =

120

453

41+x

x2+ 1

1

2 =

#."

%/$x2+ 4x + 4

x2

1

2 =

x + 2

x

From the expression for K in Eq. (7-55), we have thatlim

> F 1K = 0, so that


62/116

9

lim

> F 1

1

2(2 ! > )K +10

134

1

2

= 1 Thus

lim> F 1

(2 ! > )2(1! > )

K +1012

345

1

2+ 1

2(2 ! >)K +10

134

1

2

?

@E

AE

B

C

DE

= x + 2x

+ 1 = 2(1 + x)x

Combining the above results yields

lim

> F 1

(2! > )K

2(1! > )L

0

123

45

1+x

2+x

=#."

%/$

41+x

x2

1+x

2+x

#."

%/$

41+x

x2

1

2+x

2(1 + x)

x =

2

x

so that

lim

> F1 M1+x

opt =W1/2

QI2 2

x

7-16. Using H'p(f) = 1 from Eq. (7-69) for the impulse input and Eq. (7-66) for the

raised cosine output, Eq. (7-41) yields

I2= Hout'

(


63/116

10

Use Eq. (7-42) to find I3:

I3= Hout'

(


64/116


65/116

12

lim

> F 1K =

lim

> F1 !1+ 1+16

1+x

x2"#

$%

1! >

(2 ! >) 20

123

45

1

2?

@E

AE

B

CE

DE

= -1 + 1 = 0

Also

lim

> F 1(1! >) = 0. Therefore from Eq. (7-58)

lim

> F 1L =

lim

> F1

2(1! >)

(2 ! > )K

0

123

45

1/(1+ x)

AE@E?

DECEB

120

453(2->)

2(1->)K + 1

1

2+ 1

2+x

1+x

Expanding the square root term in K yields

lim

> F 1

2 ! >

1! >K =

lim

> F1

2! >

1 ! >

"

#. $

%/ !1+ 1+

16

2

1 + x

x 2"#

$%

1! >

(2 ! > ) 2+order(1! >) 2

0

123

45

?@

A

BC

D

=lim

> F 1

8(1 + x)

x2

1

2 ! >

0

123

45=

8(1+x)

x2

Therefore

lim

> F 1

L =

1

20

4

532x2

8(1+x)

1/(1+x)

AE

@E?

DE

CEB

1

20

4

534(1+x)

x2 + 1

1

2+ 1

2+x

1+x

=120

4532x2

8(1+x)

1/(1+x)

AE@E?

DECEBx+2

x+ 1

2+x

1+x

= (1+x)#."

%/$2

x

x

1+x

7-21. (a) First we need to find L and L'. With x = 0.5 and >= 0.9, Eq. (7-56) yields K =

0.7824, so that from Eq. (7-58) we have L = 2.89. With H= 0.1, we have >' = >(1 -

H) = 0.9

>= 0.81. Thus L' = 3.166 from Eq. (7-80). Substituting these values into

Eq. (7-83) yields

y(H) = (1 + H)#."

%/$1

1 - H

2+x

1+x

L'

L = 1.1

#."

%/$1

.9

5/3

3.166

2.89 = 1.437


66/116

13

Then 10 log y(H) = 10 log 1.437 = 1.57 dB

(b) Similarly, for x = 1.0, >= 0.9, and H= 0.1, we have L = 3.15 and L' = 3.35, so

that

y(H) = 1.1#."

%/$1

.9

3/2

3.35

3.15 = 1.37

Then 10 log y(H) = 10 log 1.37 = 1.37 dB

7-22. (a) First we need to find L and L'. With x = 0.5 and >= 0.9, Eq. (7-56) yields K =

0.7824, so that from Eq. (7-58) we have L = 2.89. With H= 0.1, we have >' = >(1 -

H) = 0.9>= 0.81. Thus L' = 3.166 from Eq. (7-80). Substituting these values into

Eq. (7-83) yields

y(H) = (1 + H)#."

%/$1

1 - H

2+x

1+x

L'

L = 1.1

#."

%/$1

.9

5/3

3.166

2.89 = 1.437

Then 10 log y(H) = 10 log 1.437 = 1.57 dB

(b) Similarly, for x = 1.0, >= 0.9, and H= 0.1, we have L = 3.15 and L' = 3.35, so

that

y(H) = 1.1#."

%/$1

.9

3/2

3.35

3.15 = 1.37

Then 10 log y(H) = 10 log 1.37 = 1.37 dB

7-23. Consider using a Si JFET with Igate= 0.01 nA. From Fig. 7-14 we have that +=

0.3 for >= 0.9. At += 0.3, Fig. 7-13 gives I2= 0.543 and I3= 0.073. Thus from

Eq. (7-86)

WJFET=1

B

2(.01nA )

1.6 8 10!19C+

4(1.388 10 !23 J / K)(300K)

(1.6 8 10!19C)2105 I

0

123

450.543

+1

B

4(1.38 8 10!23 J/ K)(300 K)(.7)

(1.6 8 10!19 C)2 (.005 S)(10 5 I) 20

123

450.543

+2((10pF)

1.6 8 10!19 C01

34

2 4(1.38 8 10!23 J / K)(300K)(.7)

(.005S)0.073 B

or


67/116

14

WJFET73.518 1012

B+ 0.026B

and from Eq. (7-92) WBP=3.39 8 1013

B+ 0.0049B

7-24. We need to find bonfrom Eq. (7-57). From Fig. 7-9 we have Q = 6 for a 10-9

BER. To evaluate Eq. (7-57) we also need the values of W and L. With >= 0.9,

Fig. 7-14 gives += 0.3, so that Fig. 7-13 gives I2= 0.543 and I3= 0.073. Thus

from Eq. (7-86)

W =3.518 1012

B+ 0.026B = 3.518105+ 2.68105= 6.18105

Using Eq. (7-58) to find L yields L = 2.871 at >= 0.9 and x = 0.5. Substitutingthese values into eq. (7-57) we have

bon= (6)5/3(1.6810-19/0.7) (6.18105).5/3(0.543)1/1.52.871 = 7.97810-17J

Thus Pr= bonB = (7.97810-17J)(107b/s) = 7.97810-10W

or

Pr(dBm) = 10 log 7.97810-10= -61.0 dBm

7-25. From Eq. (7-96) the difference in the two amplifier designs is given by

JW =1

Bq22kBT

Rf I2= 3.528106for I2= 0.543 and >= 0.9.

From Eq. (7-57), the change in sensitivity is found from

10 log120

453WHZ+ JW

WHZ

x

2(1+x)

= 10 log120

4531.0 + 3.52

1.0

.5

3

= 10 log 1.29 = 1.09 dB

7-26. (a) For simplicity, let

D = M2+x#."

%/$9

h: I2 and F =

Q

M#."

%/$9

h:

so that Eq. (7-54) becomes, for >= 1,


68/116

15

b = F[ ](Db + W)1/2+ W1/2

Squaring both sides and rearranging terms gives

b2

F2 - Db - 2W = 2W1/2

(Db + W)1/2

Squaring again and factoring out a "b2" term yields

b2- (2DF2)b + (F4D2- 4WF2) = 0

Solving this quadratic equation in b yields

b =1

2[ ]2DF2 4F4D2- 4F4D2+ 16WF2 = DF2+ 2F W

(where we chose the "+" sign) =h:9

#."

%/$

MxQ2I2+2Q

MW1/2

(b) With the given parameter values, we have

bon= 2.286810-19 39.1M0.5

+1.78 104

M

"

#. $

%

The receiver sensitivity in dBm is found from

Pr= 10 log bon(508106 b / s )

Representative values of Prfor several values of M are listed in the table below:

M Pr(dBm) M Pr(dBm)

30 - 50.49 80 -51.92

40 -51.14 90 -51.94

50 -51.52 100 -51.93

60 -51.74 110 -51.90

70 -51.86 120 -51.86


69/116

16

7-27. Using Eq. (E-10) and the relationship

1

1+ x

a

"#

$%

2 dx0

&

' =(a2

from App. B, we have from Eq. (7-97)

BHZ=1

(AR) 2

(AR)2

1+ (2(RC)2 f2

0

&

' df=(2

1

2(RC =

1

4RC

where H(0) = AR. Similarly, from Eq. (7-98)

BTZ

=1

1

1

1+ 2(RCA

"#

$%

2

f2

0

&

' df=

(

2

A

2(RC =

A

4RC

7-28. To find the optimum value of M for a maximum S/N, differentiate Eq. (7-105)

with respect to M and set the result equal to zero:

d(S/N)

dM =

(Ipm)2M

2q(Ip+ID)M2+x B +4kBTB

ReqFT

-q(Ip+ID) (2+x) M1+x B (Ipm)2M2

120

453

2q(Ip+ID)M2+x B +4kBTB

ReqFT

2 = 0

Solving for M,

M2+x

opt =4kBTBFT/Req

q(Ip+ID)x

7-29. (a) For computational simplicity, let K = 4kBTBFT/Req; substituting Moptfrom

Problem 7-28 into Eq. (7-105) gives


70/116

17

S

N =

1

2(Ipm)2M

2

opt

2q(Ip+ID)M2+x

optB + KB

=

1

2(Ipm)2

120

453K

q(Ip+ID)x

2

2+x

2q(Ip+ID)K

q(Ip+ID)xB + KB

=xm2I

2

p

2B(2+x) [ ]q(Ip+ID)x2/(2+x)#

."

%/$Req

4kBTFT

x/(2+x)

(b) If Ip>> ID, then

S

N =

xm2I2p

2B(2+x) ( )qx2/(2+x)

I2/(2+x)

p

#."

%/$Req

4kBTFT

x/(2+x)

=m2

2Bx(2+x)1220

4553

( )xIp

2(1+x)

q2(4kBTFT/Req)x

1/(2+x)

7-30. Substituting Ip= R0Printo the S/N expression in Prob. 7-29a,

S

N =

xm2

2B(2+x)

( )R0Pr

2

[ ]q(R0Pr+ID)x2/(2+x)#

."

%/$Req

4kBTFT

x/(2+x)

=(0.8)2 (0.5A / W )2 P

r2

2(5 8106 / s) 3[1.6 810!19 C(0.5Pr+10

!8 )A]2 / 3104 I / J

1.656 8 10!20"

#. $

%

1 / 3

=1.5308 1012 P

r2

0.5Pr+ 10

!8( )2 / 3 where Pris in watts.


71/116

18

We want to plot 10 log (S/N) versus 10 logPr

1 mW . Representative values are

shown in the following table:

Pr (W) Pr (dBm) S/N 10 log (S/N) (dB)

2810-9 - 57 1.237 0.92

4810-9 - 54 4.669 6.69

1810-8 - 50 25.15 14.01

4810-8 - 44 253.5 24.04

1810-7 - 40 998.0 29.99

1810-6 - 30 2.48104 43.80

1810-5 - 20 5.28105 57.18

1810-4 - 10 1.138107 70.52


72/116

1


8-1. SYSTEM 1: From Eq. (8-2) the total optical power loss allowed between the light

source and the photodetector is

PT= PS- PR = 0 dBm - (-50 dBm) = 50 dB

= 2(lc) + !fL + system margin = 2(1 dB) + (3.5 dB/km)L + 6 dB

which gives L = 12 km for the maximum transmission distance.

SYSTEM 2: Similarly, from Eq. (8-2)

PT= -13 dBm - (-38 dBm) = 25 dB = 2(1 dB) + (1.5 dB/km)L + 6 dB

which gives L = 11.3 km for the maximum transmission distance.

8-2. (a) Use Eq. (8-2) to analyze the link power budget. (a) For thepin photodiode,

with 11 joints

PT= PS- PR= 11(lc) + !fL + system margin

= 0 dBm - (-45 dBm) = 11(2 dB) + (4 dB/km)L + 6 dB

which gives L = 4.25 km. The transmission distance cannot be met with these

components.

(b) For the APD

0 dBm - (-56 dBm) = 11(2 dB) + (4 dB/km)L + 6 dB

which gives L = 7.0 km. The transmission distance can be met with these

components.

8-3. From g(t) = ( )1 - e-2"Bt u(t) we have

#$

%&

1 - e-2"Bt10

= 0.1 and #$

%&

1 - e-2"Bt90

= 0.9

so that


73/116

2

e-2"Bt10

= 0.9 and e-2"Bt90

= 0.1

Then

e2"Btr

= e2"B(t90-t10)

=.9

.1

= 9

It follows that

2"Btr= ln 9 or tr=ln 9

2"B =

0.35

B

8-4. (a) From Eq. (8-11) we have

1

2"' exp#(

($

%))&

-

t2

1/2

2'2 =1

21

2"' which yields t1/2= (2 ln 2)1/2'

(b) From Eq. (8-10), the 3-dB frequency is the point at which

G(*) =1

2 G(0), or exp

+,-

./0

-(2"f3dB)2'2

2 =

1

2

Using 'as defined in Eq. (8-13), we have

f3dB=(2 ln 2)1/2

2"' =

2 ln 2

"tFWHM =

0.44

tFWHM

8-5. From Eq. (8-9), the temporal response of the optical output from the fiber is

g(t) =1

2"'

exp

#

($

%

)&

-t2

2'2

If1eis the time required for g(t) to drop to g(0)/e, then

g(1e) =1

2"' exp

#($

%)&

-1e2

2'2 =

g(0)

e =

1

2"'e


74/116

3

from which we have that 1e= 2 '. Since teis the full width of the pulse at the

1/e points, then te = 21e= 2 2 '.

From Eq. (8-10), the 3-dB frequency is the point at which

G(f3dB) = 12 G(0). Therefore with '= te/(2 2 )

G(f3dB) =1

2" exp

+,-

./0

-1

2(2"f3dB')2 =

1

2

1

2"

Solving for f3dB:

f3dB=2 ln 2

2"'

=2 ln 2

2"

2 2

te

=0.53

te

8-6. (a) We want to evaluate Eq. (8-17) for tsys.

Using Dmat= 0.07 ns/(nm-km), we have

tsys= 234

567

(2)2+ (0.07)2(1)2(7)2++,-

./0440(7)0.7

800

2

++,-

./0350

90

2 1/2

= 4.90 ns

The data pulse width is Tb=

1

B =

1

90 Mb/s = 11.1 ns

Thus 0.7Tb= 7.8 ns > tsys, so that the rise time meets the NRZ data requirements.

(b) For q = 1.0,

tsys= 234

567

(2)2+ (0.49)2++,-

./0440(7)

800

2

++,-

./0350

90

2 1/2

= 5.85 ns

8-7. We want to plot the following 4 curves of L vs B =1

Tb :

(a) Attenuation limit

PS- PR= 2(lc) + !fL + 6 dB, where PR= 9 log B - 68.5

so that L = (PS 9 log B + 62.5 - 2lc)/!f

(b) Material dispersion


75/116

4

tmat= Dmat'8L = 0.7Tb or

L =0.7Tb

Dmat'8=

0.7

BDmat'8=

104

B (with B in Mb/s)

(c) Modal dispersion (one curve for q = 0.5 and one for q = 1)

tmod =0.440L

q

800 =

0.7

B or L =

+,-

./0

#($

%)&800

0.440.7

B

1/q

With B in Mb/s, L = 1273/B for q =1, and L = (1273/B)2for q = .5.

8-8. We want to plot the following 3 curves of L vs B =1

Tb :

(a) Attenuation limit

PS- PR= 2(lc) + !fL + 6 dB, where PR= 11.5 log B - 60.5, PS= -13 dBm, !f=

1.5 dB/km, and lc= 1 dB,

so that L = (39.5 - 11.5 log B)/1.5 with B in Mb/s.

(b) Modal dispersion (one curve for q = 0.5 and one for q = 1)

tmod =0.440Lq

800 =

0.7

B or L =

+,-

./0

#($

%)&800

0.440.7

B

1/q

With B in Mb/s, L = 1273/B for q =1, and L = (1273/B)2for q = .5.

8-9. The margin can be found from

PS- PR= lc+ 49(lsp) + 50!f+ noise penalty + system margin

-13 - (-39) = 0.5 + 49(.1) + 50(.35) + 1.5 + system margin

from which we have

system margin = 1.6 dB


76/116

5

8-10. Signal bits

8-11. The simplest method is to use an exclusive-OR gate (EXOR), which can be

implemented using a single integrated circuit. The operation is as follows: when

the clock period is compared with the bit cell and the inputs are not identical, the

EXOR has a high output. When the two inputs are identical, the EXOR output is

low. Thus, for a binary zero, the EXOR produces a high during the last half of the

bit cell; for a binary one, the output is high during the first half of the bit cell.

A B C

L L L

L H H

H L H

H H L

1 0 1 1 1 1 0 0 100 1Signal bits

Baseband (NRZ-L) data

Clock signal

Optical Manchester


77/116

6

8-12.

8-13.

Original 010 001 111 111 101 000 000 001 111 110

code

3B4B 0101 0011 1011 0100 1010 0010 1101 0011 1011 1100

encoded

8-14. (a) For x = 0.7 and with Q = 6 at a 10-9BER,

Pmpn= -7.94 log (1 - 18k2"4h4) where for simplicity h = BZD'8

NRZ data

Freq. A

Freq. B

PSK data


78/116

7

(b) With x = 0.7 and k = 0.3, for an 0.5-dB power penalty at

140 Mb/s = 1.4910-4b/ps [to give D in ps/(nm.km)]:

0.5 = -7.94 log {1 - 18(0.3)2["(1.4910-4)(100)(3.5)]4D4}

or

0.5 = -7.94 log {1 - 9.097910-4D4} from which D = 2 ps/(nm.km)

B (Mb/s) D [ps/(nm.km)]

140 2

280 1

560 0.5


79/116

1


9-1.

9-2.

Received optical power (dBm)

0 4 8 12 16

58

54

50

CNR

(dB)

RIN limit

Thermal noise

limit

Quantum

Noise

limit

f1 f2 f3 f4 f5

!

Transmissionsystem

Triple-beatproducts

2-tone3rd order


80/116

2

9-3. The total optical modulation index is

m =

"#$

%&'

(i

m2

i

1/2

= 30(.03)2+ 30(.04)

2[ ]1 / 2

= 27.4 %

9-4 The modulation index is m =

"##$

%&&'

(i=1

120

(.023)21/2

= 0.25

The received power is

P = P0 2(lc) - )fL = 3 dBm - 1 dB - 12 dB = -10 dBm = 100W

The carrier power is

C =1

2(mR0P) 2=

1

2(15*10

+6A) 2

The source noise is, with RIN = -135 dB/Hz = 3.162*10-14/Hz,

< >i2

source

= RIN (R0P)2 B = 5.69*10-13A2

The quantum noise is

< >i2Q = 2q(R0P + ID)B = 9.5*10-14A2

The thermal noise is

< >i2

T =

4kBT

Req Fe= 8.25*10-13

A2


81/116

3

Thus the carrier-to-noise ratio is

C

N =

1

2(15 *10+6 A)2

5.69*10+13

A2+ 9.5 *10

+14A

2+8.25 *10

+13A

2 = 75.6

or, in dB, C/N = 10 log 75.6 = 18.8 dB.

9-5. When an APD is used, the carrier power and the quantum noise change.


C = 12(mR0MP) 2= 1

2(15 *10+5 A) 2


< >i2Q = 2q(R0P + ID)M2F(M)B = 2q(R0P + ID)M2.7B = 4.76*10-10A2


C

N =

12

(15 *10 +5 A)2

5.69*10+13 A 2 + 4.76* 10+11 A2 +8.25 *10+13A2= 236.3

or, in dB,C

N = 23.7 dB

9-6. (a) The modulation index is m =

"

##$

%

&&'

(

i=1

32

(.044)21/2

= 0.25

The received power is P = -10 dBm = 100W


C =1

2(mR0P) 2=

1

2(15*10+6 A) 2


82/116

4


< >i2source = RIN (R0P)2 B = 5.69*10-13A2


< >i2Q = 2q(R0P + ID)B = 9.5*10-14A2

The thermal noise is

< >i2T =4kBT

Req Fe= 8.25*10-13A2


C

N =

1

2(15 *10+6 A)2

5.69*10+13

A2+ 9.5 *10

+14A

2+8.25 *10

+13A

2 = 75.6

or, in dB, C/N = 10 log 75.6 = 18.8 dB.

(b) When mi= 7% per channel, the modulation index is

m =

"##$

%&&'

(i=1

32

(.07)21/2

= 0.396

The received power is P = -13 dBm = 50W


C = 12(mR0P) 2=

1

2(1.19*10

+5A) 2= 7.06*10-11A2


< >i2source = RIN (R0P)2 B = 1.42*10-13A2


83/116

5


< >i2Q = 2q(R0P + ID)B = 4.8*10-14A2

The thermal noise is the same as in Part (a).


C

N =

7.06*10+11A 2

1.42* 10+13 A 2 +4.8 *10+14A 2 +8.25 *10+13 A2= 69.6

or, in dB, C/N = 10 log 69.6 = 18.4 dB.

9-8. Using the expression from Prob. 9-7 with !,-= 0.05, f-= 0.05, and !,= f = 10

MHz, yields

RIN(f) =4R1R2

.

!,f2+ !,2

"$

%'

1 + e-4.!,-

- 2 e-2.!,-

cos(2.f-)

=4R1R2

. 1

20 MHz(.1442)

Taking the log and letting the result be less than -140 dB/Hz gives

-80.3 dB/Hz + 10 log R1R2< -140 dB/Hz

If R1= R2then 10 log R1R2= 20 log R1 < -60 dB

or 10 log R1= 10 log R2< -30 dB


84/116

1


10-1. In terms of wavelength, at a central wavelength of 1546 nm a 500-GHz channel

spacing is

!" ="2

c

!f = 1546

nm( )2

3 # 108m / s

500 # 109s

$1 =4nm

The number of wavelength channels fitting into the 1536-to-1556 spectral band

then is

N = (1556 1536 nm)/4 nm = 5

10-2. (a) We first find P1by using Eq. (10-6):

10 log 200

W

P1

%

&' (

)*=2.7dB yields P1 = 10

(log

200$0.27 )=107.4W

Similarly, P2

=10(log 200$0.47) =67.8W

(b) From Eq. (10-5): Excess loss = 10

log

200

107.4 + 67.8

%&

()=

0.58

dB

(c) P1P

1+ P

2

=107.4175.2

=61% and P2P

1+P

2

= 67.8175.2

=39%

10-3. The following coupling percents are are realized when the pull length is stopped at

the designated points:

Coupling percents from input fiber to output 2

Points A B C D E F

1310 nm 25 50 75 90 100 0

1540 nm 50 88 100 90 50 100

10-4. From A out =s11Ain +s12Bin and Bout = s21Ain + s22 Bin = 0 , we have

Bin =$s21

s22A in and

Aout

= s11$

s12

s21

s22

+

,-.

/0A

in


85/116

2

Then

T = Aout

Ain

2

= s11$s12s21

s22

2

and R = B

in

Ain

2

= s

21

s22

%

&' (

)* s

11$

s12

s21

s22

%

&' (

)*

2

10-5. From Eq. (10-18)

P2

P0=sin

20.4z( )exp$0.06z( ) =0.5

One can either plot both curves and find the intersection point, or solve the

equation numerically to yield z = 2.15 mm.

10-6. Since 1z2n , then for n

A> n

Bwe have 3

A< 3

B. Thus, since we need to have

3ALA= 3BLB, we need to have LA> LB.

10-7. From Eq. (10-6), the insertion loss LIjfor output port j is

LIj

= 10 log P

i $in

Pj $out

%

&'

(

)*

Let

a j = Pi $in

Pj$ out=10

L Ij /10

, where the values of LIjare given in Table P10-7.

Exit port no. 1 2 3 4 5 6 7

Value of aj 8.57 6.71 5.66 8.00 9.18 7.31 8.02

Then from Eq. (10-25) the excess loss is

10

log

Pin

Pj4

%

&'(

)*=10 log

Pin

Pin

1

a1

+ 1

a2

+... + 1

an

%

&' (

)*

%

&

'

''

(

)

*

**

= 10

log

1

0.95

%& () = 0.22

dB

10-8. (a) The coupling loss is found from the area mismatch between the fiber-core

endface areas and the coupling-rod cross-sectional area. If a is the fiber-core radius

and R is the coupling-rod radius, then the coupling loss is


86/116

3

Lcoupling= 10 logPoutPin

= 10 log75a2

5R2 = 10 log

7(25)2

(150)2 = -7.11 dB

(b) Similarly, for the linear-plate coupler

Lcoupling= 10 log75a2

l6w = 10 log

75(25)2

800(50) = -4.64 dB

10-9. (a) The diameter of the circular coupling rod must be 1000 m, as shown in the

figure below. The coupling loss is

Lcoupling= 10 log

75a2

5R2 = 10 log7(100)2

(500)2 = -5.53 dB

(b) The size of the plate coupler must be 200 m by 2600 m.

The coupling loss is 10 log

75(100)2

200(2600) = -3.74 dB

10-10. The excess loss for a 2-by-2 coupler is given by Eq. (10-5), where P1= P2for a 3-

dB coupler. Thus,

200 m

400

m

Coupling rod

diameter


87/116

4

Excess loss = 10 log P

0

P1

+P2

%

&' (

)*= 10 log

P0

2P1

%

&' (

)* = 0.1dB

This yields

P1

= P0

2%&

()

10 0.01 =0.977P02

%&

()

Thus the fractional power traversing the 3-dB coupler is FT= 0.977.

Then, from Eq. (10-27),

Total loss = $10 log log F

T

log 2$1

%

&' (

)log N = $10 log

log 0.977

log 2$ 1

%

&' (

)log 2

n 730

Solving for n yields

n7 $3

log 2log 0.977

log 2$ 1

%

&' (

)

=9.64

Thus, n = 9 and N =2n =2

9 = 512

10-11. For details, see Verbeek et al., Ref. 34, p. 1012

For the general case, from Eq. (10-29) we find

M

11= cos

23d( ) 8 cos

k!L / 2( )+ j

sin

k!L / 2( )

M

12=M

21=j sin 23d( )8 cos k!L / 2( )

M 22 =cos 23d( )8 cos k!L / 2( )$j sin k!L / 2( )

The output powers are then given by

Pout ,1

= cos2 23d( )8 cos2 k!L / 2( )+ sin2 k!L / 2( )[ ]Pin,1

+sin 2 23d( )8 cos2 k!L / 2( )[ ]Pin,2


88/116

5

Pout ,2

= sin2 23d( ) 8cos2 k!L / 2( )[ ]Pin,1

+ cos 2 23d( )8 cos2 k!L / 2( ) + sin2 k!L / 2( )[ ]Pin,2

10-12. (a) The condition !9= 125 GHz is equivalent to having !"= 1 nm. Thus the

other three wavelengths are 1549, 1550, and 1551 nm.

(b) From Eqs. (10-42) and (10-43), we have

!L1 = c

2neff

(2!9)=0.4 mm and

!L3 = c

2neff

!9=0.8mm

10-13. An 8-to-1 multiplexer consists of three stages of 2# 2MZI multiplexers. The first

stage has four 2# 2MZIs, the second stage has two, and the final stage has one

2# 2MZI. Analogous to Fig. 10-14, the inputs to the first stage are (from top to

bottom) 9, 9+ 4!9, 9+ 2!9, 9+ 6!9, 9+ !9, 9+ 5!9, 9+ 3!9, 9+ 7!9.

In the first stage

!L1 = c

2neff(4!9)= 0.75mm

In the second stage

!L2 = c2neff(2!9)

=1.5

mm

In the third stage

!L 3 = c

2neff

(!9)=3.0

mm

10-14. (a) For a fixed input angle :, we differentiate both sides of the grating equation toget

cos ;d;=k

n'< d" or

d;d"

=k

n'


89/116

6

Solving this fork

n'< and substituting into the

d;d"

equation yields

d;

d" =

2 sin ;

"cos ; =

2 tan ;

"

(b) For S = 0.01,

tan ;=,-+

/0.S"

2!"(1+m)

1/2

=0.01(1350)

2(26)(1+3)

+

,-.

/0

1/ 2

= 0.2548

or ;= 14.3

10-15. For 93% reflectivity

R =tanh2(3L) = 0.93 yields 3L = 2.0, so that L = 2.7 mm for 3= 0.75 mm-1.

10-16. See Bennion et al., Ref. 42, Fig. 2a.

10-17. Derivation of Eq. (10-49).

10-18. (a) From Eq. (10-45), the grating period is

< = "uv

2 sin;

2

= 244nm

2 sin(13.5)=

244

2(0.2334)nm =523nm

(b) From Eq. (10-47), "Bragg

= 2 =1$1/ 2 =0.827 , we have from Eq. (10-51),

3 =5?n>

"Bragg

=5 2.5 # 10 $4( )(0.827)

1.547 # 10$4 cm=4.2cm

$1

(d) From Eq. (10-49), !" = 1.547m( )

2

5(1.48)500 m

(2.1)2

+52[ ]1 / 2

= 3.9 nm


90/116

7

(e) From Eq. (10-48), R max = tanh2(3L) =tanh

2(2.1) =(0.97)

2=94%

10-19. Derivation of Eq. (10-55).

10-20. (a) From Eq. (10-54),

!L =m" 0n c

=1181.554m

1.451= 126.4m

(b) From Eq. (10-57),

!9 = x

L f

nscd

m"2

nc

ng

= 25m

9.36# 103 m

1.453(3 # 10 8 m/s)(25 # 10 $6 m)

118(1.554 # 10 $6 m)2

1.451

1.475=100.5GHz

!" ="2

c!9 =

(1.554 # 10$6 m)2

3 # 108 m / s100.5GHz =0.81nm

(c) From Eq. (10-60),

!9FSR = c

n g!L=

3 # 108 m / s

1.475(126.4m)= 1609GHz

Then

!" =

"2

c!9FSR =

(1.554 # 10$6 m)2

3 # 108m / s

1609GHz =12.95nm

(d) Using the conditions

sin;i= ;

i= 2(25m)

9380m=5.33 # 10 $3 radians

and

sin;o = ;o =21.3 # 10$3

radians


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then from Eq. (10-59),

!9FSR = c

ng [!L + d; i +;o( )]

= 3 # 108 m / s

1.475 (126.4 # 10$6 m) + (25 # 10$6 m) 5.33 +21.3( ) # 10$3[ ]=1601GHz

10-21. The source spectral width is

!"signal

="2 9

c=

1550

nm( )2 (1.25 # 10 9

s$1)

(3 # 10 8

m/s) 109

nm / m( )=1 # 10$2

nm

Then from Eq. (10-61)

!"tune ="!neff

neff= 1550nm( ) 0.5%( ) =7.75nm

Thus, from Eq. (10-63)

N = !" tune10 " signal

= 7.75nm

10 0.01nm( )= 77

10-22. (a) From Eq. (10-64), the grating period is

< ="Bragg

2neff=

1550 nm

2 3.2( )= 242.2nm

(b) Again, from the grating equation,

!< = !"

2neff

=2.0nm

2 3.2( )= 0.3nm

10-23. (a) From Eq. (10-43)

!L = c

2neff!9=

"2

!"

1

2 neff=4.0mm

(b) !Leff =!n effL implies that !neff = 4mm

100mm=0.04 =4%


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10-24. For example, see C. R. Pollock, Fundamentals of Optoelectronics, Irwin, 1995,

Fig. 15.11, p. 439.

10-25. (a) The driving frequencies are found from

fa =9ov a !n

c=

va !n

"

Thus we have

Wavelength (nm) 1300 1546 1550 1554

Acoustic

frequency (MHz)

56.69 47.67 47.55 47.43

(b) The sensitivity is (4 nm)/(0.12 MHz) = 0.033 nm/kHz


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11-1. (a) From Eq. (11-2), the pumping rate is

Rp=

I

qwdL=

100mA

(1.6! 10"19 C)(5m)(0.5 m)(200 m)

= 1.25 ! 1027 (electrons / cm 3) / s

(b) From Eq. (11-8), the maximum zero-signal gain is

g0= 0.3(1! 10 "20 m2 )(1ns) 1.25 ! 1033 (electrons/ m3) / s "

1.0 ! 1024 / m3

1ns

#

$%&

'(

= 750 m"1 =7.5cm"1

(c) From Eq. (11-7), the saturation photon density is

Nph;sat

= 1

0.3 (1! 10"20 m2 ) 2! 108 m / s( )(1ns)= 1.67 ! 1015 photons/ cm3

(d) From Eq. (11-4), the photon density is

N ph = Pin

)v

ghc wd( )

= 1.32 ! 1010 photons / cm3

11-2. Carrying out the integrals in Eq. (11-14) yields

g0L =lnP(L)

P(0)+

P(L) " P(0)

Pamp,sat

Then with P(0) = Pin, P(L) = Pout, G = Pout/Pin, and G0 =exp g 0L( )from Eq. (11-

10), we have

ln G 0 =g 0L =ln G + GP

in

Pamp,sat

" P

in

Pamp,sat

=ln G + 1 " G( ) P

in

Pamp,sat

Rearranging terms in the leftmost and rightmost parts then yields Eq. (11-15).

11-3. Plots of amplifier gains.


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11-4. Let G = G0/2 and Pin =Pout/ G = 2Pout,sat / G 0 . Then Eq. (11-15) yields

G0

2

=1+G

0P

amp.sat

2Pout.satln 2

Solving for Pout,satand with G0>> 1, we have

Pout .sat

=G

0ln 2

G0"2( )

Pamp.sat *(ln 2)Pamp.sat =

Optical Fiber Communication Solution Manual (1)

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