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8/4/2019 Optical Fiber Communication - Solution Manual-3
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Problem Solutions for Chapter 2
2-1.
E = 100cos 2108 t + 30( )ex + 20 cos 2108t 50( )ey+ 40cos 2108 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
2 sin2 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
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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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
E y
E0 y Ex
E 0xcos
2
= 1 Ex
E0x
2
sin2
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
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
n glass= 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
12cm, 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 + E z
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
q 21
rr
H
z
+ r
Ez
r
Hz
r
rEz
= 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
E z 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 (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)
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
u 2B
ja
J (ua) + A1uJ' (ua)
+1
w2 B
ja
J (ua) + A2wK' (wa)J (ua)
K(wa)
With k12 = 21 and k22 = 22 rewrite this as
B =ja
1
1
u2
+ 1w
2
k1
2J + k2
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
n22
n12
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V =2 (25m)
0.82 m (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( ) = (1.480)2 (1.478) 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 NA2 = (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)0.82m
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 =1.3 106 m
101
m= 1.310-5
For Lp = 2 m ny - nx =1.3 106 m
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
=1086at820nm
432at1300 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
9mm
2
= 1.39 cm/min
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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
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 = (2.510-3 cm)2 (105 cm) = 1.96 cm3
The mass M equals the density times the volume V:
M = V = (2.6 gm/cm3)(1.96 cm3) = 5.1 gm
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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/ m m3 / 2 )2
(70MN / 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
Yp
2
are the initial crack depth and the crack depth after proof testing, respectively, yields
b / 2 i
p
d = AYb
b 0
t p
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
AYb 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
t s
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 pbt p / B + pb2( )
m
b 2
0
m
= 1 - exp
L
L0
0
m
p
btp
/ B + p
b 2[ ]m
b 2
p
btp
+ s
bts
B
+ s
b 2
p
bt
p
/ B + p
b 2
m
b 2
1
= 1 - exp LNp
1 +
s
b tsp
btp+
s
p
b
B
s2t p
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 / m2 )2 s
0.3(350MN/ m2)[ ]210s= 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
L1
L0
= 1 - exp
2c0
m
L2
L0
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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Problem Solutions for Chapter 3
3-1.
dB/ km( )= 10
z
log P(0)
P(z)
=
10
z
log e pz
( )=10 p 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:
P1300(20 km) = - 8.2 dBm (0.6 dB/km)(20 km) = - 20.2 dBm = 9.55 W
P1550(20 km) = - 10.0 dBm (0.3 dB/km)(20 km) = - 16.0 dBm = 25.1 W
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) n2
2
2n2
(0)
we have
(r) = 1 + (2 - 1)n 2 (0) n2(r)
n2(0) n2
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 Kr 2( ) r3dr0
exp Kr 2( ) rdr0
=
1
2 K2 e xxdx
0
1
2K ex dx
0
=
1
K1!
0! =
1
K
Thus gi = 1 +(2 1)
Ka2
(b) p(a) = 0.1 P0 = P0e Ka 2
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 -u2a2
u 2a2 + w2a2=
w2
u 2 + w2
=2 k2n2
2
k2
n12 2 + 2 k2 n2
2=
2/ k2 n 22
n12 n2
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+ n2n
1+ n
2
=n
1(1 ) + n 2n
1+ n
2
= 1 -n
1
n1+ n
2
Letting n2 = n1(1 - ) then yields
/ k+ n2n
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 n2 = n1 1 ( ), we can rewrite the equation as
mod
L=
n1
c1
V
where the first tern is Equation (3-30). The difference is then given by the factor
1 V
= 1 2a
1
n1
2 n22( )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 er modal
=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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7
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 er modal = 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(16
5) + 2
6(1 65
) + 2
1/ 2
3
10 6
n1
2 L
20 3c
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8
3-20. We want to plot Eq. (3-30) as a function of , where int er modal 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 er modal
does not vary with
, we have
int er modal
L=
N1
2c
+ 2
+ 23 + 2
1/ 2 4C2( + 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/ kmat850nm1.026 102 ns/ kmat1300 nm
3-21. Using the same parameter values as in Prob. 3-18, except with = 0.001, we havefrom Eq. (3-41) int er modal /L = 7 ps/km, and from Eq. (3.45)
int ra modalL
=0.098ns /kmat850 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 + 2k2n1dn
1
dk
- 2 + 2
m
a2
+ 2 2
+ 2n1
2k
2( )2
+ 21
2k2n1 dn1
dk+ 2kn1
2 + n12k2
ddk
=L
c
kn1
N1
4 + 2
+ 2
m
a2
1
n12k
2
+2
N1 +n1k
2ddk
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9
=LN
1
c
kn1
1
4 + 2
m
M
+2
1 +4
with N1 = n1 + kdn1dk
and 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
a2k2n12
(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 + 2k2n1dn1
d+ 2kn1
2dk
d
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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
+ b2m
M
2
dm
0
M
= Lc
2
2
a2 2ab + 1 +b
2
2 + 1
=L
c
2
2
2 d2n1
d2
2
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11
- 2 2d2n
1
d2
N1C1
+ 1
+ (N1C1)2 42
( + 2)(3 + 2)
3-25. Plot of Eq. (3-57).
3-26. (a) D = 0
( )S0
= 50(0.07) = 3.5ps/(nm km)
(b) D =1500(0.09)
4 1
1310
1500
4
= 14.1ps/(nm km)
3-27. (a) From Eq. (3-48)
step
L =
n12 3c =
1.49(0.01)
2 3(3 108 ) =14.4
ns/ km
(b) From Eq. (3-47)
opt
L=
n12
20 3c=
1.49(0.01)2
20 3 (3 108 )=14.3ps/km
(c) 3.5 ps/km
3-28. (a) From Eq. (3-29)
mod = Tmax Tmin =
n1Lc
=(1.49)(0.01)(5 103 m)
3 108 m / s= 248ns
(b) From Eq. (3-48)
step
=n
1L
2 3c=
248
2 3= 71.7ns
(c) BT =0.2
step
= 2.8Mb / s
(d) BT L = (2.8MHz)(5km) =13.9 MHz km
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12
3-29. For = 0.95opt , we have
int er ( opt ) int er ( = opt )
=( opt)( + 2)
= 0.05
(0.015)(1.95)= 170%
For =1.05opt , we have
int er ( opt ) int er ( = opt )
=( opt )( + 2)
= +0.05
(0.015)(2.05)= +163%
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1
Problem Solutions for Chapter 4
4-1. From Eq. (4-1), ni = 2
2kBT
h2
3/2(memh)
3/4exp
-Eg
2kBT
= 22(1.38 1023 J / K)
(6.63 10 34 J.s) 2
3/ 2
T3 / 2
(.068)(.56)(9.11 10 31 kg)2[ ]
exp (1.55 4.3 104 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
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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 =1.240
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
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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(1310nm) = 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
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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
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5
.85 .75.22 103
=.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)
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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
R1R2
= 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
J s( )2
8 5nm( )2 16.19 1032 kg + 15.10 1031kg
=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
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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= nth
= 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
dnn B
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.
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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 + b2
2cos22t)
+ a3(b13cos31t + 3b1
2b2 cos21t cos 2t + 3b1b2
2cos 1t cos22t+ b2
3cos32t)
+a4(b14cos41t + 4b1
3b2 cos31t cos 2t + 6b1
2 b2
2cos21t cos22t
+ 4b1b23cos 1t cos32t + b2
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)]
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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
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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
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11
s
= 1.1110-5 exp{0.63/[(8.62510-5)(293)]} = 7.45105 hrs
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1
Problem Solutions for Chapter 5
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 s g = 3.6001.3053.600+1.305
2
= 0.219
Similarly, the relfectivity at the gel-to-fiber interface is
R g f =1.465 1.3051.465 +1.305
2
= 3.34 103
The total reflectivity then is R = Rs gR gf = 7.30 10
4
The power loss in decibels is (see Example 5-3)
L = 10 log (1 R) = 10 log (0.999)= 3.17 103 dB5-6. Substituting B() = B0 cosm into Eq. (5-3) for B(,), we have
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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
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3
Rg f =
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 a f =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)+ ...
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4
Therefore, ford
2a
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5
= -10 log
NA2R(0)
NA2E(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
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), I p = qhc
P0 = 0.593A
Then Q2 = 2qIpB = 2 1.6 10
19 C( )(0.593A)(150 106 Hz) = 2.84 1017 A2
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3
(b) DB
2 = 2 qID
B = 2 1.6 1019 C( )(1.0nA)(150 106 Hz) = 4.81 1020 A 2
(c) T2 = 4kBTRL
B =4 1.38 1023 J / K( )(293K)
500 150 10
6 Hz( )= 4.85 1015A 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
2R0 P0 m( )2 M 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 1016M2
2.176 1023 M 5/ 2 +1.656 1019
The value of M for maximum S/N is found from Eq. (6-19), with x = 0.5:
Moptimum = 62.1.
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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 I2pM
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
pnx =
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 - 2s L2p
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
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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 1012 F /m)(11.7)(5 108 m2 )2 105m
= 2.59 ns
(b) From Eq. (6-27), the carrier drift time is
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7
td =w
vd=
20 106 m4.4 104 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
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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.
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1
Problem Solutions for Chapter 7
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
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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( t x) 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
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3
eq 2 / 22
dq
V / 2 V
= 22 e x 2 dx
V / 2 2 2
where x = q/ 22
Then
Pe =1
2 22
2 2 e x 2 dx
V / 2 2 2
+ e x 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
p2 x 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 107errors/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
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4
P1(vth) =1
22e
( vV) 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)(6.6256 1034 Js)(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 = vout - vout
vN2
= vout vout[ ]2
= vout2
-2 vout
2
+ vout
2
= vout
2- vout
7-10. (a) Letting = fTb and using Eq. (7-40), Eq. (7-30) becomes
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5
Bbae =H p (0)
Hout (0)
2
Tb2 Hout' ()
Hp'
()
2
0
dTb =I 2
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 + 42f2 R2C 2( )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 = < >v2
s +< >v2
R +< >v2
I +< >v2
E
= 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( )vth 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):
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6
Pe = -1
exp y2( )
vth +b on2 on
dy =1
exp y2( )
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
bonI 2 + W
1/ 2
+ M 2+x
h
bonI2 (1
+ W
1/ 2
+Q(h/)M( h/)
1
2(2 + x)M1+ xb
onI
2
M 2+ x
h
bonI 2 + W
1/ 2 +
1
2(2 + x)M1+x b
onI
2(1 )
M 2+x
h
bonI 2 (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
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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
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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
1B
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
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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
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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)
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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
x2e162 x 2 cos4 x0
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:
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12
lim
1K =
lim
11+ 1+16
1+ xx2
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 )
1.6 1019C+
4(1.381023 J / K)(300K)(1.6 1019C)2105
0.543
+1
B
4(1.381023 J / K)(300 K)(.7)(1.6 1019 C)2 (.005 S)(10 5 ) 2
0.543
+2(10pF)
1.6 1019 C
2 4(1.38 1023 J / K)(300K)(.7)(.005S) 0.073 B
or
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14
WJFET3.511012
B+ 0.026B
and from Eq. (7-92) WBP =3.391013
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 +
1.7104
M
The receiver sensitivity in dBm is found from
Pr = 10 log bon(50 106 b / 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
(AR) 2 (AR) 2
1+ (2RC)2 f2
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+x
optB + 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 )2 P
r2
2(5 106/s) 3[1.6 1019 C(0.5Pr+108 )A]2 / 3
104 / J1.6561020
1 / 3
=1.530 1012 P
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
Problem Solutions for Chapter 8
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
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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
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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
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1
Problem Solutions for Chapter 9
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
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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(15 106 A) 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 6 A)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.
The carrier power is
C = 12
(mR0MP) 2 = 12
(15 105 A) 2
The quantum noise is
< >i2Q = 2q(R0P + ID)M2F(M)B = 2q(R0P + ID)M2.7B = 4.7610-10A2
Thus the carrier-to-noise ratio is
C
N=
12
(15 10 5 A)2
5.69 1013 A 2 + 4.76 1011 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
The carrier power is
C =1
2(mR0P) 2 =
1
2(15 106 A) 2
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4
The source noise is, with RIN = -135 dB/Hz = 3.16210-14 /Hz,
< >i2source = RIN (R0P)2 B = 5.6910-13A2
The quantum noise is
< >i2Q = 2q(R0P + ID)B = 9.510-14A2
The thermal noise is
< >i2T =4kBT
ReqFe = 8.2510-13A2
Thus the carrier-to-noise ratio is
C
N=
1
2(15 10 6 A)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
The carrier power is
C = 12
(mR0P) 2 =1
2(1.19 105 A) 2 = 7.0610-11A2
The source noise is, with RIN = -135 dB/Hz = 3.16210-14 /Hz,
< >i2source = RIN (R0P)2 B = 1.4210-13A2
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5
The quantum noise is
< >i2Q = 2q(R0P + ID)B = 4.810-14A2
The thermal noise is the same as in Part (a).
Thus the carrier-to-noise ratio is
C
N=
7.06 10 11A 2
1.42 1013 A 2 + 4.8 10 14A 2 + 8.25 1013 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(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
Problem Solutions for Chapter 10
10-1. In terms of wavelength, at a central wavelength of 1546 nm a 500-GHz channel
spacing is
= 2
cf = 1546 nm( )2
3 108 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
= 0.58dB
(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
A
in
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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= sin2 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
a j =Pi
in
Pj out = 10L Ij /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
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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
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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( )+ j sin kL / 2( )
M
12= M
21= j sin 2d( ) cos kL / 2( )
M 22 = cos 2d( ) cos kL / 2( ) j sin kL / 2( )The output powers are then given by
Pout ,1
= cos2 2d( ) cos2 kL / 2( )+ sin2 kL / 2( )[ ]Pin,1
+ sin2 2d( ) cos2 kL / 2( )[ ]Pin,2
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Pout ,2
= sin2 2d( ) cos2 kL / 2( )[ ]Pin,1
+ cos2 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)= 0.4mm and
L3 =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
2neff (4)= 0.75mm
In the second stage
L2 = c
2neff (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' .
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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
=244nm
2 sin(13.5) =244
2(0.2334)nm = 523nm
(b) From Eq. (10-47), Bragg
= 2neff
= 2(523nm)1.48 =1547 nm(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
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(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
= 126.4m
(b) From Eq. (10-57),
= x
L f n scd
m2 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 = (1.554 106 m)2
3 108 m / s 100.5GHz = 0.81nm
(c) From Eq. (10-60),
FSR =c
n gL=
3 108 m / s1.475(126.4m) = 1609GHz
Then
=
2
cFSR = (1.554 10
6 m)23 108 m / s 1609GHz =12.95nm
(d) Using the conditions
sin i i = 2(25m)9380m = 5.33 10
3 radians
and
sino o = 21.3 103 radians
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then from Eq. (10-59),
FSR c
ng [L + d i