88758088 Solution Manual Optical Fiber Communication Gerd Keiser 3rd Ed 1

7/31/2019 88758088 Solution Manual Optical Fiber Communication Gerd Keiser 3rd Ed 1

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Problem Solutions for Chapter 2

2-1.

E = 100cos 2108 t + 30( )ex + 20 cos 210 8t 50( )ey+ 40cos 210 8 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 = 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)


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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 square both sides and add:

A2 (sin2 + cos2 ) = a12

sin2 1 + cos

2 1( )

+ a2

2sin

2 2

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

or

A2 = a1

2 + a2

2+ 2a1a2 cos (1 - 2)

Dividing (2) by (1) gives

tan =a

1sin1 + a2 sin2

a1

cos1

+ a2

cos2

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

Ex

E0 xcos = cos (t - kz) cos

to yield

Ey

E0 y

-ExE

0 x

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

x

E0x

2

(2.5-3)

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

Ey

E0 y Ex

E0xcos

2

= 1 Ex

E0x

2

sin2

Expanding the left-hand side and rearranging terms yields

Ex

E0x

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

n1 cos 1 = n2 cos 2

where n1 = 1, 1 = 33, and 2 = 90 - 33 = 57

n2 =cos 33cos 57

= 1.540

(b) The critical angle is found from

nglass sin glass = nair sin air


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with air = 90 and nair = 1.0

critical = arcsin1

nglass= arcsin

1

1.540= 40.5

2-9

Air

Water

12 cm

r

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

12 cm, which yields r = 13.7 cm.

2-10.

45

Using Snell's law nglass sin c = nalcohol sin 90

where c = 45 we have

nglass =1.45

sin 45= 2.05

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

2 n2

2( )1/ 2 = 0.242or


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NA n1 2 = n12( n

1 n

2)

n1

= 0.243

(b) 0,max

= arcsin (NA/n) = arcsin0.242

1. 0

= 14

2-13. NA = n1

2 n2

2( )1/ 2 = n12 n12(1 )2[ ]1/ 2

= n1 2 2( )1 / 2

Since


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jEr =1

1

r

Hz

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

1r

Hz

+ jrH + Ez

r= j H

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

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

jE = -1

jHr +

Hzr

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

1

r

Ez

-

jHr +

Hzr

= -j Hr

Solve for Hr to obtain Eq. (2-35c).

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

-j

q2

1

rr

Hz

+ r Ezr

Hzr

r

Ez

= jEz

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

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

-

j

q 2

1

r

r

Ez r

Hzr

Ezr +

r

Hz

= -jHz

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

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


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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(w a)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)

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

u2

Bja

J(ua) + A 1uJ' (ua)

+1

w2 B

ja

J(ua) + A 2wK' (wa)J( ua)

K( wa)

With k12 = 21 and k22 = 22 rewrite this as

B =ja

1

1

u2

+ 1w

2

k1

2J + k 2

2K[ ]A


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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.

2-16. From Eq. (2-23) we have

=n 1

2 n22

2n12 =

1

21

n 22

n12


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V =2(25m )

0.82m (1.48)

2 (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

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 2.40.

Solving Eq. (2-58) for the core radius a

a =V2

n1

2 n2

2( )1/ 2 = 2.40(1.32m )2 (1.480)2 (1. 478)2[ ]1/ 2

= 6.55 m

(b) From Eq. (2-23)

NA = n12 n2

2( )1/ 2 = (1. 480)2 (1.478) 2[ ]1/ 2 = 0.077

(c) From Eq. (2-23), NA = n sin 0,max. When n = 1.0 then

0,max = arcsinNA

n

= arcsin 0.0771. 0

= 4.4

2-22. n2 = n12 NA 2 = (1. 458)2 ( 0.3) 2 = 1.427

a =V

2NA=

(1.30)( 75)

2(0. 3)= 52 m


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

n1 2

For a = 5 m we have 0.002, so that at 0.82 m

V 2(5m )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 the LP01

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

2-24.

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

n

y nx

For Lp = 10 cm ny - nx =1. 3 10 6m

101

m= 1.310-5

For Lp = 2 m ny - nx =1. 3 10 6m

2 m= 6.510-7

Thus

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

n2 is found from Eq. (2-79): n2 = n1(1 - ) = 1.465

2-27. From Eq. (2-81)

M =

+ 2

a2k

2n

1

2 =

+ 2

2an1

2

where

=n

1 n 2n

1

= 0.0135


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

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

MstepV

2

2

=1

2

2a

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 =2an1

2

=1086at820 nm

432 at1300 nm

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 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

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

9 mm

2

= 1.39 cm/min


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(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 yields

=K

Y

2

where Y = for surface flaws.

Thus

=( 20N /mm 3 / 2 )2

(70 MN / m2

)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):

b / 2 i

f

d = AYbb0

t

dt

which yields

1

1 b2

f1 b /2 i

1 b/ 2[ ]= AYbbt

or

t =2

(b 2)A(Y)b i

(2 b ) / 2 f( 2 b ) / 2[ ]

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

t =2

(b 2)A(Y)bKi

Y

2 b

KfY

2 b

2Ki

2 b

(b 2)A(Y)b if K i

b 2> Kf

2 b

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

ddt

= AKb = AYbb/2b


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Integrating this from ito p where

i=

K

Y

i

2

and p =K

Y

p

2

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

b / 2 i

p

d = AYb

b 0

tp

dtor

1

1 b

2

p1 b /2 i

1b / 2[ ]= AYb pb tp

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

2

b 2

K

Y

2b

i

b 2 p

b 2[ ]= AYb pb tpor

2

b 2

K

Y

2b1

A Yb i

b 2 pb 2[ ]= B ib2 pb2[ ]= pb tp

which is Eq. (2-87).

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

b / 2 p

s

d = AYb sb0

ts

dt

where s is the crack depth at the fiber failure point. Integrating (as above) we get Eq. (2-89):

B pb2 sb2[ ]= sb ts

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


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2-35. (a) Substituting Ns as given by Eq. (2-92) and Np as given by Eq. (2-93) into Eq.

(2-94) yields

F = 1 - exp LL

0

pbtp + sbts( )/ B + sb2

[ ]m

b2

0

m pbtp / B + pb2( )

m

b 2

0

m

= 1 - exp

L

L0

0

m

p

bt

p/B +

p

b 2[ ]m

b 2

pbt

p + sbt

s

B

+ s

b 2

p

bt

p

/B + p

b 2

m

b 2

1

= 1 - exp LNp

1 +

s

bts

pbtp

+

s

p

b

B

s2tp

m

b 2

1 + B

p2t

p

1

1 - exp LNp 1+

s

bts

p

btp

1

1+ B

p

2tp

m

b2

1

(b) For the term given by Eq. (2-96) we have

sp

b

B

s2t

p

= (0.3)150.5 (MN /m 2 )2 s

0. 3(350 MN/ m2)[ ]210 s

= 6.510-14

Thus this term can be neglected.


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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 1c0

m

L 1

L 0

= 1 - exp

2c0

m

L 2

L 0

which implies that

1c

0

m

L1

L0

=

2c

0

m

L2

L0

or

1c

2c

=L

2

L1

1 /m

If L1 = 20 m, then 1c= 4.8 GN/m2

If L2 = 1 km, then 2c= 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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3-1.

dB/km( )= 10

z

log P(0)

P(z )

=

10

z

log e pz

( )=10p log e = 4.343 p(1/ km )

3-2. Since the attenuations are given in dB/km, first find the power levels in dBm for

100 W and 150 W. These are, respectively,

P(100 W) = 10 log (100 W/1.0 mW) = 10 log (0.10) = - 10.0 dBm

P(150 W) = 10 log (150 W/1.0 mW) = 10 log (0.15) = - 8.24 dBm

(a) At 8 km we have the following power levels:

P1300(8 km) = - 8.2 dBm (0.6 dB/km)(8 km) = - 13.0 dBm = 50 W

P1550(8 km) = - 10.0 dBm (0.3 dB/km)(8 km) = - 12.4 dBm = 57.5 W

(b) At 20 km we have the following power levels:



3-3. From Eq. (3-1c) with Pout = 0.45 Pin

= (10/3.5 km) log (1/0.45) = 1.0 dB/km

3-4. (a) Pin = Pout 10L/10 = (0.3 W) 101.5(12)/10 = 18.9 W

(b) Pin = Pout 10L/10 = (0.3 W) 102.5(12)/10 = 300 W

3-5. With in Eqs. (3-2b) and (3-3) given in m, we have the following representativepoints for uv and IR:


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(m) uv IR0.5 20.3 --

0.7 1.44 --

0.9 0.33 --

1.2 0.09 2.210-6

1.5 0.04 0.00722.0 0.02 23.2

3.0 0.009 7.5104

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

scat =83

34(n

2 1)2 kBTfT

=

83

3(0.63m )4 (1.46)2

1[ ]2

(1.3810-16 dyne-cm/K)(1400 K)

(6.810-12 cm2/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 = 83

34n8p2 kBTfT = 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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3-11. (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( )r3dr

0

exp Kr2( )rdr

0

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

exp Kr2( )r3dr0

exp Kr2( )rdr0

=

1

2 K2 exxdx

0

1

2K ex dx

0

=

1

K1!

0! =

1

K

Thus gi = 1 +(2 1)

Ka2

(b) p(a) = 0.1 P0 = P0eKa2

yields eKa 2

= 10.

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

gi = 1 + (2 1)2.3

= 0.571 + 0.432

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


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n = 1 +196.98

(13.4)2 (1. 24 /)2

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 1.5500.6 m 1.457 1.4581.0 m 1.451 1.450

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

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

have from Eq. (3-20)

mat

L=

dd

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

For a laser diode,

mat

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

(b) From Fig. 3-13,dmat

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

2

= 1 -u 2a2

u2a

2 + w 2a2=

w 2

u2 + w 2

=2 k 2n 2

2

k2

n12 2 + 2 k2 n2

2=

2/ k 2 n 22

n12 n 2

2

(b) Expand b as b =(/ k + n 2 )(/k n 2 )

( n1

+ n2

)(n1

n2

)

Since n2< /k < n1 , let /k = n1(1 - ) where 0 < <


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/ k + n 2n

1+ n

2

=n

1(1 ) + n 2

n1+ n

2

= 1 -n

1

n1+ n

2

Letting n2 = n1(1 - ) then yields

/ k + n 2n

1+ n

2

= 1 - 2

1 since 2


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Therefore

Tmin

- Tmax

=Ln

1

c

1

1 1

=

Ln1

c

1

Ln

1

c

3-17. Since n 2 = n1 1 ( ), we can rewrite the equation as

modL

=n

1c

1 V

where the first tern is Equation (3-30). The difference is then given by the factor

1 V

= 1 2a

1

n1

2 n 22( )1/ 2

1 2a

1

n1

2

At 1300 nm this factor is 1 (1. 3)

2 62.5( )1

1.48 2(0.015)= 1 0.127= 0.873

3-18. For = 0 and in the limit of we have

C1 = 1, C2 =3

2,

+ 1

= 1, + 2

3 + 2=

1

3,

+12 +1

= 12

,

and( + 1)2

(5 + 2)(3 + 2)=

1

15

Thus Eq. (3-41) becomes

int ermodal =Ln

1

2 3c1+ 3 +

12

52

1/ 2

Ln12 3c

3-19. For = 0 we have that = 2(1 -6

5). Thus C1 and C2 in Eq. (3-42) become

(ignoring small terms such as 3, 4, ...)

C1 = 2 + 2

=2 1

6

5

2

2 1 65

+ 2

= 3

5

1 35

3

5 1+

3

5


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C2 =3 2

2( + 2)=

3 2 1 6

5

2

2 2 1 65

+ 2

=1

9

5

2 1 35

Evaluating the factors in Eq. (3-41) yields:

(a) C12

9

252

(b)4 C

1C

2( + 1)

2 +1=

4 3

5

1

9

5

2 1

6

5

+ 1

1 35

2 1

3

5

4 1

6

5

+ 1

=

18 1 11 + 1825

5 15

224

25

18

252

(c)162C2

2( + 1)2

(5 + 2)(3 + 2)

=

162 19

5

2

2(1 6

5) + 1

2

4 1 35

2

10(1 65

) + 2

6(1 65

) + 2

=

162 19

5

2

9 1 4

5

2

96(1 )(1 910

)4 1 35

2 9

242

Therefore,

int ermodal = Ln1

2c

+1 +

23 + 2

1/ 2

925

2 1825

2 + 924

2

1/ 2

=Ln

12

2c

2(1 6

5)

2(1 65

) +1

2(1 6

5) + 2

6(1 65

) + 2

1/ 2

3

10 6

n1

2L

20 3c


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8

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

int ra modal are given by Eqs. (3-41) and (3-45). For = 0 and = 2, we have C1 =0 and C2 = 1/2. Since

int ermodal

does not vary with

, we have

int ermodal

L=

N1

2 c

+ 2

+ 23 + 2

1/ 24C

2( + 1)

(5 + 2)(3 + 2 )= 0.070 ns/km

With C1 = 0 we have from Eq. (3-45)

int ra modal =1

c

2d2n

1

d2

=

0.098 ns/km at850 nm1.026102 ns/km at1300 nm

3-21. Using the same parameter values as in Prob. 3-18, except with = 0.001, we havefrom Eq. (3-41) int ermodal /L = 7 ps/km, and from Eq. (3.45)

int ra modalL

=0.098ns /km at850 nm

0.0103 ns/kmat1300 nm

The plot ofL

=1

L

int er

2 + int ra

2( )1/ 2 vs :

3-22. Substituting Eq. (3-34) into Eq. (3-33)

g =L

c

ddk

=L

c

1

22kn

1

2 + 2k 2n1dn

1

dk

- 2 + 2

m

a2

+ 2 2

+ 2n1

2k

2( )2

+ 21

2k 2n1 dn1

dk+ 2kn1

2 + n12k 2

ddk

=L

c

kn1

N1

4 + 2

+ 2

m

a2

1

n12k

2

+2

N1 +n1k

2ddk


25/116

9

=LN

1

c

kn1

1

4 + 2

m

M

+2

1 +4

with N1 = n1 + k dn1

dkand where M is given by Eq. (2-97) and is defined in Eq.

(3-36b).

3-23. From Eq. (3-39), ignoring terms of order 2,

dd

=L

c

dN1

d1 +

2 + 2

m

M

+ 2

+LN

1

c

2 + 2

d

d

m

M

+2

where

N1 = n1 - dn1

dand M =

+ 2

a2k2n 12

(a)dN1

d=

d

dn 1

dn1

d

= -

d2n1

d2

Thus ignoring the term involving d

2n1

d2, the first term in square brackets

becomes -L

c2

d2n1

d2

(b)d

d

m

M

+ 2

= m

+ 2 d

d1

M

+2

+

+ 2dM

d1

M

+2

+1

(c)dM

d=

+ 2

a2d

dk

2n1

2 ( )

=

+ 2a2

dd

k2n1

2 + 2k 2n1dn1

d+ 2kn1

2dk

d


26/116

10

Ignoringdd

anddn1

dterms yields

dM

d=

2

+ 2a2k2n

1

2

1

= -

2 M

so that

d

d

m

M

+ 2

=

2 + 2

m

M

+2

. Therefore

dd

= -L

c2

d2n1

d2+

LN1

c

2 + 2

2 + 2

m

M

+2

3-24. Let a = 2 d2

n1d2

; b = N1C1 2 + 2; = + 2

Then from Eqs. (3-32), (3-43), and (3-44) we have

int ra modal

2 = L2

21

M

dg

d

m = 0

M

2

=L

c

2

21

M a + b

m

M

m = 0

M

2

L

c

2

21

Ma + b

m

M

2

dm0

M

=L

c

2

2

1

M a

2 2abm

M

+ b 2m

M

2

dm

0

M

= Lc

2

2

a2 2ab + 1 +b

2

2 + 1

=L

c

2

2

2 d2n1

d2

2


27/116


28/116


29/116

1


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

2kBT

h2

3/2(memh)

3/4exp

-Eg

2kBT

= 22(1. 38 1023J/K)

(6. 63 10 34J. s)2

3/ 2

T3 / 2

(. 068)(. 56)( 9. 11 10 31 kg)2[ ]3/ 4

exp (1.55 4.3 10 4 T)eV2(8.62 105 eV /K)T

= 4.151014 T3/2

exp 1.55

2(8.62 10 5 )T

exp

4.3 10 4

2(8.62 105 )

= 5.031015 T3/2

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 + n2i /pP or p

2

P - NApP - n2i = 0

so that

pP

=NA

2

1 +4n

2i

N2A

+ 1

If ni


30/116

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 =1.240

1.540= 805 nm.

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

=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

0.4184x

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

E =hc

2 or =

2

hcE

For the same energy difference E, the spectral width is proportional to thewavelength squared. Thus, for example,

1550

1310

= 15501310

2

= 1.40

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

int =1

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

Pint = (0.783) hc(35mA )

q(1310 nm) = 26mW

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

P =1

3.5 3.5 + 1( )226mW = 0.37mW

4-7. Plot of Eq. (4-18). Some representative values of P/P0 are 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 log1

1 + 2f( )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 = 1

gth =1

0.05cm ln

1

0.32

2+ 10 cm-1 = 55.6 cm-1

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

gth =1

0.05 cmln

1

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

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

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

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

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

=1.24

Eg=

1.24

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

From Eq. (4-38)

ext = 0.8065 (m)dP(mW)

dI(mA)

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

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

Then neff

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

(b) The total confinement factor then is = 0.185

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

=2

2Ln=

(0.80 m)22(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(

) = (50 cm-1) exp

-

( - 850 nm)2

2(32 nm)2

= (50 cm-1) exp

-( - 850)2

2048

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

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

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

=2

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/ = m2L be the wave number (reciprocal wavelength) of mode m.

The difference N between adjacent modes is then

N = Nm - Nm-1 =1

2L(a-1)

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

dN

d=

d

d

n

=

1

dn

d-

n

2= -

1

2

n -

dn

d

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

N =1

2

n - dn

d (a-2)


34/116

6

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

2L

n - dn

d

(b) The mode spacing is =(.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

+

1

2Lln

1

R 1R 2

= 2.65103 A/cm2

Therefore

Ith = Jth l w = (2.65103 A/cm2)(25010-4 cm)(10010-4 cm) = 663 mA

(b) Ith = (2.65103 A/cm2)(25010-4 cm)(1010-4 cm) = 66.3 mA

4-16. From the given equation

E11 =1.43 eV + 6.6256 1034

Js( )2

8 5nm( )2 16.19 1032 kg + 15.10 1031 kg

=1.43 eV + 0.25eV =1.68eVThus the emission wavelength is = 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 =mB2 =

2(1570 nm)

2(460 nm)= 3.4

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


35/116

7

= B

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, = B 3(1.20 nm) = 1570 nm 3.60 nm

For m = 2, = B 5(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 = dt

0

t d

= 1Jqd

n

0

n th

dn = Jqd n

n= 0

n= n th

= ln JJ J

th

where J = Ip/A and Jth = Ith/A. Therefore td = ln

Ip

Ip - Ith

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

td =

0

td

dt =1

J

qd

n

dnnB

n th

= lnJ

qd nB

J

qd

n th

In the steady state before a pulse is applied, nB = JB/qd. When a pulse is applied,the current density becomes I/A = J = JB + Jp = (IB + Ip)/A

Therefore, td = ln

I - IB

I - Ith

= ln

Ip

Ip + IB - Ith

4-20. A common-emitter transistor configuration:

4-21. Laser transmitter design.


36/116

8

4-22. Since the dc component of x(t) is 0.2, its range is -2.36 < x(t) < 2.76. The power

has the form P(t) = P0[1 + mx(t)] where we need to find m and P0. The average

value is

< >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) = a1b1 cos 1t + a1b2 cos 2t

+ a2(b12cos21t + 2b1b2 cos 1t cos 2t + b 2

2cos22t)

+ a3(b13cos31t + 3b1

2b2 cos21t cos 2t + 3b1b 2

2cos 1t cos22t+ b 2

3cos32t)

+a4(b14cos41t + 4b1

3b2 cos31t cos 2t + 6b1

2b

2

2cos21t cos22t

+ 4b1b 23cos 1t cos32t + b 2

4cos42t)

Use the following trigonometric relationships:

i) cos2 x =1

2(1 + cos 2x)

ii) cos3 x =1

4 (cos 3x + 3cos x)

iii) cos4 x =1

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

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

v) cos2 x cos y =1

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


37/116

9

vi) cos2 x cos2 y =1

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

2 cos(2x+2y) +1

2 cos(2x-2y)]

vii) cos3 x cos y =1

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

then

y(t) = 12 a2b

21 + a2b

22 + 34

a4b41 + 3a4b21b

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 21t +b

2

2

2

a2 + a4b

2

2 + 3a4b2

1 cos 22t

2nd-order harmonic terms

+1

4a3b

3

1 cos 31t +1

4a3b

3

2 cos 32t 3rd-order harmonic terms

+1

8a4b

41 cos 41t +

1

8a4b

42 cos 42t 4th-order harmonic terms

+

a2b1b2 +

3

2 a4b3

1b2 +3

2 a4b1b3

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

2nd-order intermodulation terms

+3

4 a3b2

1 b2[ ]cos(21+2)t + cos(21-2)t +3

4 a3b1b2

2[ ]cos(22+1)t + cos(22-1)t

3rd-order intermodulation terms

+1

2 a4b3

1 b2[ ]cos(31+2)t + cos(31-2)t

+3

4 a4b2

1 b2

2[ ]cos(21+22)t + cos(21-22)t

+1

2 a4b1b32[ ]cos(32+1)t + cos(32-1)t 4th-order intermodulation

terms

This output is of the form

y(t) = A0 + A1(1) cos 1t + A2(1) cos 21t + A3(1) cos 31t

+ A4(1) cos 41t + A1(2) cos 2t + A2(2) cos 22t


38/116

10

+ A3(2) cos 32t + A4(2) cos 42t + m

n

Bmn cos(m1+n2)t

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

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

where P0 = 1 mW and m = 2(5104 hrs) = 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 = 58760 hours = 43,800 hours. Therefore

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

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

or ln s = ln K + EA/kBT

where kB = 1.3810-23 J/K = 8.62510-5 eV/K

At T = 60C = 333K, we have

ln 4104 = ln K + EA/[(8.62510-5 eV)(333)]

or 10.60 = ln K + 34.82 EA (1)

At T = 90C = 363K, we have

ln 6500 = ln K + EA/[(8.62510-5 eV)(363)]

or 8.78 = ln K + 31.94 EA (2)

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

EA = 0.63 eV and k = 1.1110-5 hrs

Thus at T = 20C = 293K


39/116

11

s

= 1.1110-5 exp{0.63/[(8.62510-5)(293)]} = 7.45105 hrs


40/116

1


5-3. (a) cosL 30 = 0.5

cos 30 = (0.5)1/L = 0.8660

L = log 0.5/log 0.8660 = 4.82

(b) cosT 15 = 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 = 2r2

s B0(NA)2 = 2(210-3 cm)2(100 W/cm2)(.22)2 = 191 W

From Eq. (5-9)

PLED-graded = 22(210-3 cm)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 sg = 3.600 1.3053.600 +1.305

2

= 0.219

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

R gf =1.465 1.3051.465 +1.305

2

= 3.34 103

The total reflectivity then is R = RsgR gf = 7.30 10

4

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

L = 10 log (1 R ) = 10 log ( 0.999) = 3.17103 dB5-6. Substituting B() = B0 cosm into Eq. (5-3) for B(,), we have


41/116

2

P =

0

rm

0

2

2 0

0-maxcos3 sin d ds r dr

Using

0

0cos3 sin d =

0

0( )1 - sin2 sin d(sin ) =

0

sin 0( )x - x3 dx

we have

P = 2

0

rm

0

2

sin20-max

2-

sin40-max4

ds r dr

=

0

rm

0

2

NA2 - 12 NA4 ds r dr

=2

[ ]2NA2 - NA4 0

rm

r dr 0

2ds

5-7. (a) Let a = 25 m and NA = 0.16. For rs a(NA) = 4 m, Eq. (5-17) holds. Forrs 4 m, = 1.(b) With a = 50 m and NA = 0.20, Eq. (5-17) holds for rs 10 m. Otherwise,

= 1.

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


42/116

3

Rgf =

1.485 1.3051.485 +1.305

2

= 4.16103

The power loss is (see Example 5-3)

L = 10 log (1 R ) = 10 log ( 0. 9958) = 0.018dBWhen there is no index-matching gel, the joint loss is

R af =1.485 1.0001.485 +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

2sa -

1

2cy

s = a = a [2 arccos (y/a)] = 2a arccos

d

2a

c = 2

a2 -

d

22 1/2

Therefore

Acommon

= 2(shaded area) = sa cy = 2a2 arccos

d2a

- d

a2 -

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

NA2R(0)

NA2

E(0)

5-15. For fibers with different

values, where

Ri2

s(t) =

1

T0

T

i

2

s(t)dt =

2

0

2/

R

2

0 P2(t) dt (where T = 2/),

=2 R

2

0 P2

0 0

2/(1 + 2m cos t + m2 cos2t) dt

Using

cos tdt =0

2 /

1

sin t t = 0t = 2/ = 0

and 0

2/cos2t dt =

1

0

2

1

2+

1

2cos 2x dx =

we have < >i2s(t) = R2

0 P2

0

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), Ip = qhc

P0 = 0.593A

Then Q2 = 2qIpB = 2 1.610

19 C( )(0. 593A )(150 106 Hz) = 2.84 1017 A 2


48/116

3

(b) DB

2 = 2 qID

B = 2 1.6 1019 C( )(1.0nA )(150 10 6Hz) = 4.81 10 20 A 2

(c) T2 = 4kBTR

L

B =4 1.38 10 23 J/K( )(293K)

500 150 106 Hz( )= 4.85 10 15 A 2

6-7. Using R 0 =qhc

= 0.58 A/W, we have from Eqs. (6-4), (6-11b), (6-15), and (6-

17)

S

NQ

=

1

2R0P0 m( )2M 2

2qIpBM1/ 2

M2 =

R0P0m2

4qBM1/2= 6.5651012 P0

S

NDB

=( )R0P0m 2

4qIDBM1/2= 3.7981022 P

2

0

S

NDS

=( )R0P0m

2M2

4qILB= 3.7981026 P20

S

NT

=

1

2( )R0P0m2M2

4kBTB/RL= 7.3331022 P

2

0

where P0 is given in watts. To convert P0 = 10-n W to dBm, use 10 log

P0

10-3=

10(3-n) dBm

6-8. Using Eq. (6-18) we have

S

N=

1

2( )R0P0m2

M2

2qB(R0P0 + ID)M5/2 + 2qILB + 4kBTB/RL

=1.215 10 16M2

2.176 1023M 5/ 2 +1.656 10 19

The value of M for maximum S/N is found from Eq. (6-19), with x = 0.5:

Moptimum = 62.1.


49/116

4

6-9. 0 =d

dM

S

N=

d

dM

1

2 I2

pM2

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

0 = I2

p M -

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

2 I2

pM2

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

p nx =

1

Lp

pn0 + Be

-sw

e

(w-x)/Lp

- sBe

-sx

2pnx2

= -1

L2

p

pn0 + Be-sw

e(w-x)/Lp

+ 2s Be-sx

Substituting pn and2pnx2

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

-Dp

L2

p

pn0 + Be-sw

e(w-x)/Lp

+ Dp2

s Be-sx

+1

p

pn0 + Be-sw

e(w-x)/Lp

-B

p

e-sx

+ 0s e-sx

=

B

Dp2

s -1

p

+ 0s e-sx

where the first and third terms cancelled because L2p = Dpp .

Substituting in for B:

Left side =

0

Dp

sL2

p

1 - 2

s L2

p

Dp2

s -1

p

+ 0s e-sx


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5

=0Dp

sL

2

p

1 - 2

s L2

p

2s Lp - 1

p

+ Dps e-sx

=

0Dp( )-Dps + Dps e

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

(b) Jdiff= qDppnx

x =w

= qDp

1

Lp

pn0 + Be-sw

- sBe-sw

= qDp B

1

Lp- s e

-sw+ qpn0

DpLp

= q0

sL

2

p

1 - 2

s L2

p

1 - sLp

Lpe-sw

+ qpn0DpLp

= q0sLp

1 + sLpe-sw

+ qpn0DpLp

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

Jtotal = Jdrift + Jdiffusion = q0

1 - e-sw

+

sLp

1 + sLpe-sw

+ qpn0DpLp

= q0

1 -e-sw

1 + sLpe-sw

+ qpn0DpLp

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

Jtot J*totsc

1/2 = q0( )S S* 1/2 where S = 1 - e-jtd

jtd

We want to find the value oftd at which ( )S S*1/2

=1

2.

Evaluating ( )S S*1/2

, we have


51/116

6

( )S S*1/2

=

1 - e

-jtd

jtd

1 - e

+jtd

-jtd

1/2

= 1 - e+

jtd + e-jtd + 1

1/2

td=

( )2 - 2 cos td 1/2

td

=[ ]( )1 - cos td /2

1/2

td/2=

sin

td

2

td2

= sinc

td

2

We want to find values oftd where ( )S S*1/2

=1

2

.

x sinc x x sinc x

0.0 1.000 0.5 0.637

0.1 0.984 0.6 0.505

0.2 0.935 0.7 0.368

0.3 0.858 0.8 0.234

0.4 0.757 0.9 0.109

By extrapolation, we find sinc x = 0.707 at x = 0.442.

Thus td2 = 0.442 which implies td = 0.884

(b) From Eq. (6-27) we have td =w

vd=

1

svd. Then

td = 2f3-dB td = 2f3-dB1

svd= 0.884 or

f3-dB = 0.884 svd/2

6-12. (a) The RC time constant is

RC =R

0KsA

w=

(104 )(8.85 10 12F /m )(11. 7)(5 10 8 m 2 )2 10 5m

= 2.59 ns

(b) From Eq. (6-27), the carrier drift time is


52/116

7

td =w

vd=

20 10 6m4. 4 10 4 m /s

= 0.45 ns

c)1

s= 10-3 cm = 10 m =

1

2w

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 k2 and 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 + 2 -

1

Me (1 - keff)

(b) With k1 k2 and 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)

k2 - k

2

1 - k

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 -

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)ej2ftdf = R ej2ft

1 + j2fRCdf

Using the integral solution from Appendix B3:

ejpx

( + jx) ndx

=2(p) n1ep

(n )for p > 0, we have

hB(t) =1

2C ej 2ft

1

2RC+ jf

df

=1

Ce-t/RC

7-3. Part (a):

hp

(t) dt = 1Tb -Tb/2

Tb/2

dt =1

Tb

Tb

2+

Tb2

= 1


55/116

2

Part (b):

hp

(t) dt = 12

1Tb

exp t22(Tb )

2

dt

= 12

1

Tb

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

Part (c):

hp

(t) dt = 1Tb

exp tT

b

dt

0

= - e e0[ ]= 1

7-4. The Fourier transform is

F[p(t)*q(t)] = p( t) * q (t)ej2 ft dt

= q(x) p( t x )ej2ftdtdx

= q(x) ej2fx p( t x )ej2f ( tx )dtdx

= q(x) ej2fx dx p( y)ej 2fydy

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 P0 and P1, respectively, we have

Pe =1

2 1

2 2 ev 2/ 2 2 dv

V / 2

+ e(v V ) 2 /2 2 dv

V /2

In the first integral, let x = v/ 22 so that dv = 22 dx.

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


56/116

3

eq 2 / 22

dq

V / 2 V

= 22 ex 2 dx

V / 2 2 2

where x = q/ 22

Then

Pe =1

2 22

2 2 ex 2 dx

V / 2 2 2

+ ex 2 dx

V / 2 22

=1

2 ex 2 dx

2 ex 2 dx0

V / 2 22

Using the following relationships from Appendix B,

e

p2x 2 dx

=

p and2

ex 2 dx

0

t

= erf(t), we have

Pe =1

21 erf

V

2 2

7-6. (a) V = 1 volt and = 0.2 volts, so that V2

= 2.5. From Fig. 7-6 forV

2= 2.5,

we find Pe 710-3 errors/bit. Thus there are (2105 bits/second)(710-3

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

1

1400 errors/second= 710-4 seconds/error

(b) If V is doubled, thenV

2= 5 for which Pe 310-7 errors/bit from Fig. 7-6.

Thus

1

(2 105 bits /sec ond)(3 10 7errors /bit )= 16.7 seconds/error

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

P0(vth) =1

22 e v 2 / 2 2 dv

V / 2

=1

21 erf

V

2 2

and


57/116

4

P1(vth) =1

22e

( v V) 2 /2 2dv

V / 2

=1

21 erf

V

2 2

Then for V = V1 and = 0.20V1

P0(vth) =1

2

1 - erf

1

2(.2) 2=

1

2

1 - erf

2

0.8

=1

2[ ]1 - erf( )1.768 =

1

2( )1 - 0.987 = 0.0065

Likewise, for V = V1 and = 0.24V1

P1(vth) =1

2

1 - erf

1

2(.24) 2=

1

2

1 - erf

2

0.96

=1

2[ ]1 - erf( )1.473 =

1

2( )1 - 0.963 = 0.0185

(b) Pe = 0.65(0.0185) + 0.35(0.0065) = 0.0143

(c) Pe = 0.5(0.0185) + 0.5(0.0065) = 0.0125

7-8. From Eq. (7-1), the average number of electron-hole pairs generated in a time t is

N = Eh = Pthc/ = 0.65(25 1010

W)(1 109

s)(1.3 106

m )(6. 625610 34Js)(3 108 m /s) = 10.6

Then, from Eq. (7-2)

P(n) = Nne-N

n!= (10.6)5

e-10.6

5!=

133822

120e-10.6 = 0.05 = 5%

7-9. v N = v out - v out

vN2

= vout vout[ ]2

= vout2

-2 vout

2

+ vout

2

= vout

2- vout

2

7-10. (a) Letting = fTb and using Eq. (7-40), Eq. (7-30) becomes


58/116

5

Bbae =Hp(0)

Hout(0)

2

Tb2 Hout'

()Hp

'()

2

0

dTb =I2

Tb

since Hp(0) = 1 and Hout(0) = Tb. Similarly, Eq. (7-33) becomes

Be =H

p(0)

Hout

( 0)

2

Hout( f)H

p(f )

1 + j2fRC( )2

0

df

=1

Tb2 Hout( f )

Hp(f )

2

0

1 + 42f2R2C2( )df

=1

Tb2 Hout( f )

Hp(f )

2

0

df+ 2RC( )

2

Tb2 Hout(f )

Hp(f )

2

0

f2 df= I2

Tb+

(2RC)2

T3b

I3

(b) From Eqs. (7-29), (7-31), (7-32), and (7-34), Eq. (7-28) becomes

< >v2N = < >v2s +< >v

2R +< >v

2I +< >v

2E

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

RbBbaeR2A2 + SIBbaeR2A2 + SEBeA2

= 2q < >i0 M2+x +4k

BT

Rb+ SI BbaeR2A2 + SEBeA2

=R2A2I2

Tb

2q < >i0 M2+x +4kBT

Rb+ SI +

SE

R2+

(2RCA)2

T3

b

SEI3

7-11. First let x = v boff( )/ 2 off( )with dx = dv / 2 off( ) in the first part of Eq. (7-49):

Pe =2off

2off

exp x2( )v th b off

2off

dx = 1 exp x2( )

Q / 2

dx

Similarly, let y = v + bon

( )/ 2 on( ) so thatdy = -dv/ 2

on( )in the second part of Eq. (7-49):


59/116

6

Pe = -1

exp y 2( )

v th +b on2 on

dy =1

exp y 2( )

Q / 2

dy

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

= K2 2

For K = 10, x = 3.536. Thus

Pe =e-x

2

2 x= 2.9710-7 errors/bit

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

2

2 xthen e-x

2= 2 10-5 x.

This holds for x 3, 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/)

2M

2+x h

b onI2 + W

1/ 2

+ M 2+x

h

bonI2(1

+ W

1/ 2

+Q(h/)

M( h/)

1

2(2 +x)M1+x b

onI

2

M2+x

h

b onI2 + W

1/ 2 +

1

2(2 + x)M1+x b

onI

2(1 )

M2+x

h

bonI2(1 ) + W

1/ 2

1 / 2

Letting G = M2+x

h bonI2 for simplicity, yields

( )G + W

1/2+ [ ]G(1-) + W

1/2=

G

2(2 + x)

1

(G + W)

1/2 +(1-)

[ ]G(1-) + W1/2

Multiply by (G + W)1/2

[ ]G(1-) + W1/2

and rearrange terms to get

(G + W)1/2

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

G2

x2

4(1-) + G

x2

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

Solving this equation for G yields

G =

-x2

4W(2-)

x4

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

1

2

x2

2(1-)

=

W(2-)2(1-)

1 -

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

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 b

1/(2+x)

on

h

W(2-)2I2(1-)

K1/(2+x)

W(2-)

2(1-) K + W

1

2+

W

2(2-)K + W

1

2

Factoring out terms:

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

on = Q

h

(1+x)/(2+x)

Wx/2(2+x)

I1/(2+x)

2

(2-)

2(1-) K + 112

+

1

2(2-)K + 1

12

(2-)K

2(1-)1/(2+x)

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

h

Wx/2(1+x)

I1/(1+x)

2 L


61/116

8

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

lim

1(2 )K2(1 )L

1+x2+x

=lim

1(2 )K2(1 )

1+x2+x lim

11

L

1+x2+x

Consider first

lim

1(2 )K2(1 )

=

lim

1 (2 )

2(1 )1 + 1 + B

(1 )(2 )2

1

2

where B = 16(1+x)/x2 . Since 1, we can expand the square root term in abinomial series, so that

lim 1

(2 )K2(1 )

= lim

1 (2 )

2(1 )1 + 1 + 1

2B (1 )

( 2 )2 Order(1 )2

=lim

1 B

4(2 )=

B

4= 4

1+x

x2

Thuslim

1(2 )K2(1 )

1+x2+x

=

41+x

x2

1+x

2+x

Next consider, using Eq. (7-58)

lim

1 1

L

1+x2+x

=lim

1(2 )K2(1 )

1/(2+x )

(2 )2(1 )

K +1

1

2

+1

2(2 )K +1

1

2

From the above result, the first square root term is

(2-)

2(1-) K + 1

1

2=

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

1K = 0, so that


62/116

9

lim

11

2(2 )K +1

1

2

= 1 Thus

lim 1

(2 )2(1 )

K +1

1

2+ 1

2(2 )K +1

1

2

= x + 2x

+ 1 = 2(1 + x)x

Combining the above results yields

lim

1(2 )K2(1 )L

1+x2+x

=

41+x

x2

1+x

2+x

41+x

x2

1

2+x

2(1 + x)

x=

2

x

so that

lim

1 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'

()2 d

0

=

1

2 Hout' () 2 d

=1

2

-1-2

1-2

d +

1-2

1+2

1

8

1-sin

-2

2d +

-1+

2

-

1-2

1

8

1-sin

-2

2d

Letting y = -

2 we have

I2 = 12(1 - ) +

4

-2

2

[ ]1 - 2sin y + sin2y dy

=1

2(1 - ) +

4

- 0 +

2

=1

2

1 -4


63/116

10

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

I3 = Hout'

()2

2 0

d =1

2 Hout' () 2 2

d

=1

2

-1-2

1-2

2 d +

1-2

1+2

1

8

1-sin

-2

22d +

-1+

2

-1-2

1

8

1-sin

-2

22d

Letting y = -

2

I3 =1

3

1-

2

3+

4

-2

2

[ ]1 - 2sin y + sin2y

2y2

2 +y +

1

4dy

=1

3

1-

2

3+

4

-2

2

2y2

2 +1

4 ( )1 + sin2y dy -

2

-2

2

y sin y dy

where only even terms in "y" are nonzero. Using the relationships

0

2

sin2y dy =4

;

y sin ydy = -y cos y + sin y

and

x2 sin x2 dx = x3

6-

x2

4- 1

8sin 2x - x cos 2x

4

we have

I3 =1

3

1-

2

3+

2

2

2

1

3

2

3+

1

6

2

3+

8

+1

4

2+

4

-2 (1)


64/116

11

=316

1

2 -1

6- 2

1

2 -1

8-

32

+1

24

7-17. Substituting Eq. (7-64) and (7-66) into Eq. (7-41), with s2 = 422 and = 1, wehave

I2 =

0

H'out()

H'p()

2d =

1

4

0

1

es22

1-sin

-

2

2d

=

0

1

es22 1 + cos

22d =

0

1

es22cos4 2 d

Letting x =2

yields I2 =2

0

2

e162x2

cos4 x dx

Similarly, using Eqs. (7-64) and (7-66), Eq. (7-42) becomes

I3 =

0

1

es22

cos4

2 2d =

2

3

x2e162x 2 cos4x0

2

dx

7-18. Plot of I2 versus for a gaussian input pulse:

7-19. Plot of I3 versus for a gaussian input pulse:

7-20. Consider firstlim

1K:


65/116

12

lim

1K =

lim

11+ 1+16

1+ xx

2

1 ( 2 ) 2

1

2

= -1 + 1 = 0

Also

lim

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

lim

1L =

lim

12(1 )

(2 )K

1/(1+x )

(2-)

2(1-) K + 1

1

2+ 1

2+x

1+x

Expanding the square root term in K yields

lim

1 2

1 K =

lim

12 1

1+ 1+

16

2

1 +xx

2

1 ( 2 ) 2

+ order(1 ) 2

=lim

18(1+x)

x2

1

2

=

8(1+x)

x2

Therefore

lim

1L =

2x2

8(1+x)

1/(1+x)

4(1+x)

x2 + 1

1

2+ 1

2+x

1+x

=

2x2

8(1+x)

1/(1+x)

x+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 = 0.1, we have ' = (1 -

) = 0.9

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

Eq. (7-83) yields

y() = (1 + )

1

1 -

2+x

1+x

L'

L= 1.1

1

.9

5/3

3.166

2.89= 1.437


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13

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

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

that

y() = 1.1

1

.9

3/2

3.35

3.15= 1.37

Then 10 log y() = 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 = 0.1, we have ' = (1 -) = 0.9= 0.81. Thus L' = 3.166 from Eq. (7-80). Substituting these values intoEq. (7-83) yields

y() = (1 + )

1

1 -

2+x

1+x

L'

L= 1.1

1

.9

5/3

3.166

2.89= 1.437

Then 10 log y() = 10 log 1.437 = 1.57 dB(b) Similarly, for x = 1.0, = 0.9, and = 0.1, we have L = 3.15 and L' = 3.35, sothat

y() = 1.1

1

.9

3/2

3.35

3.15= 1.37

Then 10 log y() = 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. 61019 C+

4(1.3810 23J/K)(300 K)(1.61019 C)210 5

0.543

+1

B

4(1.381023 J/K)(300 K)(.7)(1.61019 C)2(. 005 S)(10 5 ) 2

0.543

+2(10pF)

1.61019 C

24(1.38 1023 J/K)(300 K)(. 7)

(. 005S) 0.073 Bor


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14

WJFET3.511012

B+ 0.026B

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

B+ 0.0049B

7-24. We need to find bon from 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 =3.511012

B+ 0.026B = 3.51105 + 2.6105 = 6.1105

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 (1.610-19/0.7) (6.1105).5/3 (0.543)1/1.5 2.871 = 7.9710-17 J

Thus Pr = bonB = (7.9710-17 J)(107 b/s) = 7.9710-10 W

or

Pr(dBm) = 10 log 7.9710-10 = -61.0 dBm

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

W =1

Bq22kBT

RfI2 = 3.52106 for I2 = 0.543 and = 0.9.

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

10 log

WHZ + W

WHZ

x

2(1+x)

= 10 log

1.0 + 3.52

1.0

.5

3

= 10 log 1.29 = 1.09 dB

7-26. (a) For simplicity, let

D = M2+x

h I2 and F =Q

M

h

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


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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

MxQ2I2 +2Q

MW1/2

(b) With the given parameter values, we have

bon = 2.28610-19 39.1M0.5 +

1. 710 4

M

The receiver sensitivity in dBm is found from

Pr = 10 log b on(50 10 6b/s)[ ]Representative values of Pr for 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


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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

(A R )2 ( A R )

2

1+( 2RC)2f2

0

df =2

1

2RC =1

4RC

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

BTZ

=1

1 1

1+ 2RCA

2

f2

0

df =

2

A

2RC=

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

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 Mopt from

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


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17

S

N=

1

2(Ipm)2M

2

opt

2q(Ip+ID)M2+xopt

B + KB=

1

2(Ipm)2

K

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

Req4kBTFT

x/(2+x)

=m2

2Bx(2+x)

( )xIp

2(1+x)

q2(4kBTFT/Req)x

1/(2+x)

7-30. Substituting Ip = R0Pr into 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 )2P

r2

2(5 106/s) 3[1. 6 1019 C(0.5Pr+108 )A ]2 / 3

10 4 /J1.6561020

1 / 3

=1.530 1012P

r 2

0.5Pr+ 108( )2 / 3

where Pr is in watts.


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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)

210-9 - 57 1.237 0.92

410-9 - 54 4.669 6.69

110-8 - 50 25.15 14.01

410-8 - 44 253.5 24.04

110-7 - 40 998.0 29.99

110-6 - 30 2.4104 43.80

110-5 - 20 5.2105 57.18

110-4 - 10 1.13107 70.52


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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 the pin 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-2Bt u(t) we have

1 - e-2Bt10

= 0.1 and

1 - e-2Bt90

= 0.9

so that


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2

e-2Bt10

= 0.9 and e-2Bt90

= 0.1

Then

e2Btr

= e2B(t90-t10)

=.9

.1

= 9

It follows that

2Btr = ln 9 or tr =ln 9

2B =0.35

B

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

1

2 exp

-

t2

1/2

22 =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

2G(0), or exp

-(2f3dB)22

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

22

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

g(e) =1

2exp

-e2

22 =g(0)

e=

1

2e


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3

from which we have that e = 2 . Since te is the full width of the pulse at the

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

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

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

G(f3dB) =1

2exp

-1

2(2f3dB)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 =

(2)2 + (0.07)2(1)2(7)2 +

440(7)0.7

800

2

+

350

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 =

(2)2 + (0.49)2 +

440(7)

800

2

+

350

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


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4

tmat = Dmat L = 0.7Tb or

L =0.7Tb

Dmat=

0.7

BDmat=

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

Bor L =

800

0.440.7

B

1/q

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

Bor L =

800

0.440.7

B

1/q

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

8-9. The margin can be found from

PS - PR = lc + 49(lsp) + 50f+ 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


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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

10 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 10-9 BER,

Pmpn = -7.94 log (1 - 18k24h4) where for simplicity h = BZD

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 = 1.410-4 b/ps [to give D in ps/(nm.km)]:

0.5 = -7.94 log {1 - 18(0.3)2[(1.410-4)(100)(3.5)]4D4}or

0.5 = -7.94 log {1 - 9.09710-4D4} 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 106A ) 2

The source noise is, with RIN = -135 dB/Hz = 3.16210-14 /Hz,

< >i2

source

= RIN (R0P)2 B = 5.6910-13A2

The quantum noise is

< >i2Q = 2q(R0P + ID)B = 9.510-14A2

The thermal noise is

< >i2

T =

4kBT

Req Fe = 8.2510-13

A2


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3

Thus the carrier-to-noise ratio is

C

N=

1

2(15 10 6A )2

5.69 1013

A2 + 9. 5 10

14A

2 + 8.25 1013

A2 = 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 = 12

(15 105A )2


< >i2Q = 2q(R0P + ID)M2F(M)B = 2q(R0P + ID)M2.7B = 4.7610-10A2


C

N=

12

(15 10 5A )2

5.69 1013A 2 + 4.76 1011A 2 + 8.25 10 13A 2= 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 106A ) 2


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4


< >i2source = RIN (R0P)2 B = 5.6910-13A2


< >i2Q = 2q(R0P + ID)B = 9.510-14A2

The thermal noise is

< >i2T =4kBT

ReqFe = 8.2510-13A2


C

N=

1

2(15 10 6A )2

5.69 1013

A2 + 9. 5 10

14A

2 + 8.25 1013

A2 = 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 = 12

(1. 19 10 5A ) 2 = 7.0610-11A2


< >i2source = RIN (R0P)2 B = 1.4210-13A2


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5


< >i2Q = 2q(R0P + ID)B = 4.810-14A2

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


C

N=

7.0610 11A 2

1.42 1013A 2 + 4. 8 10 14A 2 + 8.25 10 13A 2= 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(2f)

=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 = R2 then 10 log R1R2 = 20 log R1 < -60 dB

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


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1


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

spacing is

= 2

cf = 1546nm( )2

3 10 8 m /s 500 109 s1 = 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 P1 by using Eq. (10-6):

10 log 200WP

1

= 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 200107.4 + 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 = s11A in + s12B in and B out =s21A in +s22B in = 0 , we have

B in = s21

s22A in and

Aout = s11

s12

s21

s22

A

in


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2

Then

T =A out

A in

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 2 0.4z( )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

z n , then for n

A> n

Bwe have

A LB.

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

LIj = 10 log

Pi in

Pj out

Let

aj =Pi

in

Pj out = 10LIj /10

, where the values of LIj are 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 P

in

Pj

= 10 log P

in

Pin

1

a1

+1

a2

+ . . . +1

an

=10 log

1

0.95

= 0.22dB

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 log7a2R2 = 10 log

7(25)2

(150)2= -7.11 dB

(b) Similarly, for the linear-plate coupler

Lcoupling = 10 log7a2lw = 10 log

7(25)2800(50)

= -4.64 dB

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

Lcoupling = 10 log

7a2

R2 = 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

7(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 = P2 for a 3-

dB coupler. Thus,

200 m

400

m

Coupling rod

diameter


87/116

4

Excess loss = 10 log P0P

1+ P

2

= 10 log P0

2P1

= 0.1dB

This yields

P1

= P02

10

0.01 = 0.977 P02

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

Then, from Eq. (10-27),

Total loss = 10 log log FTlog 2 1

log N = 10 log log 0.977

log 2 1

log 2 n 30

Solving for n yields

n 3

log 2 log 0.977log 2 1

= 9.64

Thus, n = 9 and N = 2n = 29 = 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 2d( ) cos kL /2( )+ jsin kL / 2( )

M

12= M

21= jsin 2d( ) cos kL /2( )

M 22 = cos 2d( ) cos kL /2( ) jsin kL / 2( )The output powers are then given by

Pout,1

= cos2 2d( ) cos2 kL /2( )+sin2 kL /2( )[ ]Pin,1

+ sin 2 2d( ) cos2 kL /2( )[ ]Pin,2


88/116

5

Pout ,2

= sin 2 2d( ) cos 2 kL / 2( )[ ]Pin,1

+ cos 2 2d( ) cos2 kL /2( ) +sin2 kL /2( )[ ]Pin,2

10-12. (a) The condition = 125 GHz is equivalent to having = 1 nm. Thus theother three wavelengths are 1549, 1550, and 1551 nm.

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

L1 =c

2neff

(2)= 0. 4 mm and

L 3 =c

2neff

= 0. 8mm

10-13. An 8-to-1 multiplexer consists of three stages of2 2 MZI multiplexers. The first

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

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

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

In the first stage

L1 =c

2n eff (4)= 0.75mm

In the second stage

L 2 = c

2 neff (2 )= 1. 5mm

In the third stage

L 3 =c

2neff

()= 3. 0mm

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

cos d =k

n' d ordd =

k

n' cos

If, then the grating equation becomes 2 sin =kn' .


89/116

6

Solving this fork

n' and substituting into thedd equation yields

d

d=

2 sin

cos =

2 tan

(b) For S = 0.01,

tan =

S

2 (1+m)

1/2

=0.01(1350)

2(26)(1 + 3)

1/ 2

= 0.2548

or = 14.3

10-15. For 93% reflectivity

R = tanh 2(L ) = 0.93 yields L = 2.0, so that L = 2.7 mm for = 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

=244 nm

2 sin(13.5) =244

2(0.2334)nm = 523nm

(b) From Eq. (10-47), Bragg = 2n eff = 2(523nm )1.48 =1547nm(c) Using = 1 1/ 2 = 0.827, we have from Eq. (10-51),

=n

Bragg

= 2. 5 10 4( )(0. 827)

1.547 104 cm = 4. 2cm1

(d) From Eq. (10-49), = 1.547m( )2

(1. 48)500 m (2. 1)2 + 2[ ]1 / 2 = 3. 9 nm


90/116

7

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

= 118 1.554m1.451

= 126.4 m

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

= x

Lf nscd

m2 n c

ng

=25m

9.36 103 m 1.453(3 10 8 m /s)(25 10 6m )

118(1. 554 10 6m )2 1.451

1.475= 100.5GHz

=2

c = (1.554 10 6m) 2

3 108 m /s 100.5GHz = 0.8

88758088 Solution Manual Optical Fiber Communication Gerd Keiser 3rd Ed 1

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Transcript of 88758088 Solution Manual Optical Fiber Communication Gerd Keiser 3rd Ed 1