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Problem Solutions for Chapter 2
2-1.
E = 100cos 2!108
t + 30 ex + 20 cos 2!108t" 50 ey
+ 40cos 2!108 t + 210( )ez
2-2. The general form is:
y = (amplitude) cos(#t - kz) = A cos [2!($t - z/%)]. Therefore
(a) amplitude = 8 m
(b) wavelength: 1/%= 0.8 m-1 so that %= 1.25 m
(c) #= 2!$= 2!(2) = 4!
(d) At t = 0 and z = 4 m we have
y = 8 cos [2!(-0.8 m-1)(4 m)]
= 8 cos [2!(-3.2)] = 2.472
2-3. For E in electron volts and %in m we have E =1.240
%
(a) At 0.82 m, E = 1.240/0.82 = 1.512 eV
At 1.32 m, E = 1.240/1.32 = 0.939 eV
At 1.55 m, E = 1.240/1.55 = 0.800 eV
(b) At 0.82 m, k = 2!/%= 7.662 m-1
At 1.32 m, k = 2!/%= 4.760 m-1
At 1.55 m, k = 2!/%= 4.054 m-1
2-4. x1= a1cos (#t - &1) and x2= a2cos (#t - &2)
Adding x1and x2yields
x1+ x2= a1[cos #t cos &1+ sin #t sin &1]
+ a2[cos #t cos &2+ sin #t sin &2]
= [a1cos &1+ a2cos &2] cos #t + [a1sin &1+ a2sin &2] sin #t
Since the a's and the &'s are constants, we can set
a1cos &1+ a2cos &2= A cos ' (1)
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a1sin &1+ a2sin &2= A sin ' (2)
provided that constant values of A and 'exist which satisfy these equations. To
verify this, first square both sides and add:
A2(sin2 '+ cos2 ') = a12
sin2
&1 +cos2
&1( )
+ a2
2 sin2 &2
+cos2 &2( )+ 2a1a2(sin &1sin &2+ cos &1cos &2)
or
A2= a1
2 +a2
2 + 2a1a2cos (&1- &2)
Dividing (2) by (1) gives
tan '=a
1sin&
1+a
2sin&
2
a1
cos&1
+a2cos&
2
Thus we can write
x = x1+ x2= A cos 'cos #t + A sin 'sin #t = A cos(#t - ')
2-5. First expand Eq. (2-3) as
Ey
E0 y= cos (#t - kz) cos &- sin (#t - kz) sin & (2.5-1)
Subtract from this the expression
E x
E0 xcos &= cos (#t - kz) cos &
to yield
Ey
E0 y
-ExE
0x
cos &= - sin (#t - kz) sin & (2.5-2)
Using the relation cos2(+ sin2(= 1, we use Eq. (2-2) to write
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3
sin2(#t - kz) = [1 - cos2(#t - kz)] = 1 " E
x
E0x
)
*+ ,
-.
2/
01
2
34 (2.5-3)
Squaring both sides of Eq. (2.5-2) and substituting it into Eq. (2.5-3) yields
E y
E0 y"
Ex
E 0xcos&
/
01
2
34
2
= 1 " Ex
E0x
)
*+ ,
-.
2/
01
2
34 sin
2&
Expanding the left-hand side and rearranging terms yields
Ex
E 0x
)
*+ ,
-.
2
+
Ey
E0y
)
*+
,
-.
2
- 2
Ex
E0x
)
*+ ,
-.
Ey
E0y
)
*+
,
-.
cos &= sin2
&
2-6. Plot of Eq. (2-7).
2-7. Linearly polarized wave.
2-8.
33 33
90 Glass
Air: n = 1.0
(a) Apply Snell's law
n1cos 51= n2cos 52
where n1= 1, 51= 33, and 52 = 90- 33= 57
6n2=cos 33
cos 57= 1.540
(b) The critical angle is found from
nglasssin 'glass = nairsin 'air
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with 'air= 90and nair= 1.0
6'critical= arcsin1
n glass= arcsin
1
1.540= 40.5
2-9
Air
Water
12 cm
r
5
Find 5cfrom Snell's law n1sin 51= n2sin 5c= 1
When n2= 1.33, then 5c= 48.75
Find r from tan 5c=r
12cm, which yields r = 13.7 cm.
2-10.
45
Using Snell's law nglasssin 5c = nalcoholsin 90
where 5c = 45we have
nglass=1.45
sin 45= 2.05
2-11. (a) Use either NA = n1
2 " n2
2( )1/ 2
= 0.242
or
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NA 7n1 28 = n12(n
1"n
2)
n1
= 0.243
(b) 50,max
= arcsin (NA/n) = arcsin0.242
1.0
)
*
,
-= 14
2-13. NA = n1
2 "n2
2( )1/ 2
= n1
2 "n1
2(1" 8)2[ ]1/ 2
= n1 28 " 82( )
1 / 2
Since 8
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jEr=1
9#
1
r
;Hz;'
+ jr:H')
*+ ,
-Substituting into Eq. (2-33b) we have
:9#
1r
;Hz;'
+ jr:H')
*+ ,
-+ ;E z
;r= j# H'
Solve for H'and let q2= #29- :2 to obtain Eq. (2-35d).
(d) Solve Eq. (2-34b) for jE'
jE'= -1
9# j:Hr +
;Hz;r
)*
,-
Substituting into Eq. (2-33a) we have
1
r
;Ez
;'-
:
9# j:Hr +
;Hz;r
)*
,-
= -j# Hr
Solve for Hrto obtain Eq. (2-35c).
(e) Substitute Eqs. (2-35c) and (2-35d) into Eq. (2-34c)
- j
q 21
r
;
;r:
;Hz;'
+9#r;Ez
;r
)
*+ ,
-"
;
;' :
;Hz;r
"9#
r
;Ez;'
)
*+ ,
-/
012
34= j9#Ez
Upon differentiating and multiplying by jq2/9#we obtain Eq. (2-36).
(f) Substitute Eqs. (2-35a) and (2-35b) into Eq. (2-33c)
-
j
q 2
1
r
;
;r :
;E z
;' " # r
;Hz
;r
)
*+ ,
-"
;
;' :
;Ez
;r +
#
r
;Hz
;'
)
*+ ,
-
/
01
2
34= -j#Hz
Upon differentiating and multiplying by jq2/9#we obtain Eq. (2-37).
2-15. For $= 0, from Eqs. (2-42) and (2-43) we have
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Ez= AJ0(ur) ej( #t " :z)
and Hz= BJ0(ur) ej( #t " :z)
We want to find the coefficients A and B. From Eqs. (2-47) and (2-51),
respectively, we have
C =J$ (ua)
K $ (wa)A and D =
J$ (ua)
K $ (wa)B
Substitute these into Eq. (2-50) to find B in terms of A:
Aj:$
a
)* ,
-
1
u2 +
1
w2
)* ,-= B#
J'$
(ua)
uJ $ (ua)+
K'$
(wa)
wK$(wa)
/
012
34
For $= 0, the right-hand side must be zero. Also for $= 0, either Eq. (2-55a) or (2-56a)
holds. Suppose Eq. (2-56a) holds, so that the term in square brackets on the right-hand
side in the above equation is not zero. Then we must have that B = 0, which from Eq. (2-
43) means that Hz= 0. Thus Eq. (2-56) corresponds to TM0mmodes.
For the other case, substitute Eqs. (2-47) and (2-51) into Eq. (2-52):
0 =1
u2
Bj:$
aJ $ (ua) +A#91uJ'$ (ua)
/
0
2
3
+1
w2 B
j:$
aJ$ (ua) +A#92w
K' $ (wa)J$ (ua)
K $(wa)
/
01 34
With k12 = #291and k22 = #292 rewrite this as
B$=ja
:#
1
1
u2
+ 1
w2
/
0
1
1
2
3
4
4k
1
2J$ + k2
2K $[ ]A
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V =2! (25m)
0.82 m (1.48)2 "(1.46)2[ ]
1/ 2
= 46.5
Using Eq. (2-61) M 7V2/2 =1081 at 820 nm.
Similarly, M = 417 at 1320 nm and M = 303 at 1550 nm. From Eq. (2-72)
Pclad
P
)*
,-
total
74
3M-1/2 =
4= 100%
3 1080= 4.1%
at 820 nm. Similarly, (Pclad/P)total = 6.6% at 1320 nm and 7.8% at 1550 nm.
2-20 (a) At 1320 nm we have from Eqs. (2-23) and (2-57) that V = 25 and M = 312.
(b) From Eq. (2-72) the power flow in the cladding is 7.5%.
2-21. (a) For single-mode operation, we need V
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2-23. For small values of 8we can write V 72!a
%n1 28
For a = 5 m we have 870.002, so that at 0.82 m
V 72! (5m)
0.82m1.45 2(0.002) = 3.514
Thus the fiber is no longer single-mode. From Figs. 2-18 and 2-19 we see that the LP01
and the LP11modes exist in the fiber at 0.82 m.
2-24.
2-25. From Eq. (2-77) Lp=2!
: =%
ny" nx
For Lp= 10 cm ny- nx=1.3 = 10"6 m
10"1
m= 1.3=10-5
For Lp= 2 m ny- nx=1.3 = 10"6 m
2 m= 6.5=10-7
Thus
6.5=10-7
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At %= 820 nm, M = 543 and at %= 1300 nm, M = 216.
For a step index fiber we can use Eq. (2-61)
Mstep7V 2
2
=1
2
2!a
%
)
*
,
-
2
n1
2 " n2
2( )
At %= 820 nm, Mstep= 1078 and at %= 1300 nm, Mstep= 429.
Alternatively, we can let (= >in Eq. (2-81):
Mstep=2!an1
%
)*
,-
2
8=1086at820nm
432at1300 nm
?@A
2-28. Using Eq. (2-23) we have
(a) NA = n12
"n22
( )1/ 2
= (1.60)2
" (1.49)2
[ ]1/ 2
= 0.58
(b) NA = (1.458)2 " (1.405)2[ ]
1/ 2
= 0.39
2-29. (a) From the Principle of the Conservation of Mass, the volume of a preform rod
section of length Lpreformand cross-sectional area A must equal the volume of the fiber
drawn from this section. The preform section of length Lpreformis drawn into a fiber of
length Lfiberin a time t. If S is the preform feed speed, then Lpreform= St. Similarly, if s is the
fiber drawing speed, then Lfiber= st. Thus, if D and d are the preform and fiber diameters,
respectively, then
Preform volume = Lpreform(D/2)2= St (D/2)2
and Fiber volume = Lfiber(d/2)2= st (d/2)2
Equating these yields
StD
2
)* ,-
2
= std
2
)* ,-
2
or s = SD
d
)* ,-
2
(b) S = sd
D
)*
,-
2
= 1.2 m/s0.125mm
9mm
)*
,-
2
= 1.39 cm/min
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2-30. Consider the following geometries of the preform and its corresponding fiber:
PREFORMFIBER
4 mm
3 mm
R
25 m
62.5 m
We want to find the thickness of the deposited layer (3 mm- R). This can be done by
comparing the ratios of the preform core-to-cladding cross-sectional areas and the fiber
core-to-cladding cross-sectional areas:
A preform core
Apreform clad=
Afibercore
Afiberclad
or
!(32 " R2 )
!(42 " 32 )=
!(25)2
! (62.5)2 "(25)
2[ ]
from which we have
R = 9" 7(25)
2
(62.5)2 "(25)
2
/
012
34
1/ 2
= 2.77 mm
Thus, thickness = 3 mm - 2.77 mm = 0.23 mm.
2-31. (a) The volume of a 1-km-long 50-m diameter fiber core is
V = !r2L = !(2.5=10-3cm)2(105 cm) = 1.96 cm3
The mass M equals the density Btimes the volume V:
M = BV = (2.6 gm/cm3)(1.96 cm3) = 5.1 gm
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(b) If R is the deposition rate, then the deposition time t is
t =M
R=
5.1gm
0.5gm / min= 10.2 min
2-32. Solving Eq. (2-82) for Cyields
C=K
YD
)*
,-
2
where Y = ! for surface flaws.
Thus
C=(20N / m m3 / 2 )2
(70MN/ m2 )2 != 2.60=10-4mm = 0.26 m
2-33. (a) To find the time to failure, we substitute Eq. (2-82) into Eq. (2-86) and
integrate (assuming that Dis independent of time):
C" b / 2 Ci
C f
E dC = AYbDb0
t
E dt
which yields
1
1 "b
2
Cf1" b / 2 " C i
1" b/ 2[ ]= AYbDbt
or
t =2
(b" 2)A(YD)b C i
(2 " b)/ 2 " Cf( 2" b)/ 2[ ]
(b) Rewriting the above expression in terms of K instead of Cyields
t =2
(b" 2)A(YD)b
Ki
YD
)*
,-
2" b
" Kf
YD
)*
,-
2 "b/
012
34
72Ki
2" b
(b" 2)A(YD)b if K i
b" 2>Kf
2" b
2-34. Substituting Eq. (2-82) into Eq. (2-86) gives
dC
dt= AKb= AYbCb/2Db
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Integrating this from Cito C p where
Ci=
K
YD i
)
*
+ ,
-
.2
and Cp=
K
YDp
)
*
+,
-
.
2
are the initial crack depth and the crack depth after proof testing, respectively, yields
C "b / 2 C i
C p
E dC = AYb
Db
0
t p
E dt
or
1
1 "b
2
Cp1" b / 2
" Ci1"b / 2[ ]= AYb D pb tp
for a constant stress Dp. Substituting for C i and C p gives
2
b" 2
)*
,-
K
Y
)*
,-
2"b
Di
b" 2 " Dp
b" 2[ ]= AYb D pb tpor
2
b" 2
)*
,-
K
Y
)*
,-
2"b1
AYb D i
b" 2" Dp
b" 2[ ]= B D ib"2 " Dpb"2[ ]= D pb
tp
which is Eq. (2-87).
When a static stress Dsis applied after proof testing, the time to failure is found from Eq.(2-86):
C "b / 2 C p
Cs
E dC = AYb Dsb
0
t s
E dt
where C s is the crack depth at the fiber failure point. Integrating (as above) we get Eq. (2-
89):
B Dpb"2 " Dsb"2[ ]= Dsb ts
Adding Eqs. (2-87) and (2-89) yields Eq. (2-90).
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2-36. The failure probability is given by Eq. (2-85). For equal failure probabilities of the
two fiber samples, F1= F2, or
1 - exp " D1c
D0
)
*+ ,
-.
m
L1
L0
/
01
34= 1 - exp "
D2cD0
)
*+ ,
-.
m
L2
L0
/
01
34
which implies that
D1c
D0
)
*+ ,
-.
m
L1
L0
=D
2c
D0
)
*+ ,
-.
m
L2
L0
or
D1cD
2c
=L
2
L1
)
*+ ,
-.
1 / m
If L1= 20 m, then D1c= 4.8 GN/m2
If L2= 1 km, then D2c= 3.9 GN/m2
Thus
4.83.9
)* ,-
m
= 100020
= 50
gives
m =log50
log(4.8/ 3.9)= 18.8
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!!(m) ""uv ""IR0.5 20.3 --
0.7 1.44 --
0.9 0.33 --
1.2 0.09 2.2)10-6
1.5 0.04 0.00722.0 0.02 23.2
3.0 0.009 7.5)104
3-6. From Eq. (3-4a) we have
!scat=8*3
3(4 (n
2+ 1)
2kBTf,T
=
8*3
3(0.63m)4 (1.46)2
+ 1[ ]2
(1.38)10-16dyne-cm/K)(1400 K)
)(6.8)10-12cm2/dyne)
= 0.883 km-1
To change to dB/km, multiply by 10 log e = 4.343: !scat= 3.8 dB/km
From Eq. (3-4b):
!scat= 8*3
3(4
n8p2kBTf,T = 1.16 km-1
= 5.0 dB/km
3-8. Plot of Eq. (3-7).
3-9. Plot of Eq. (3-9).
3-10. From Fig. 2-22, we make the estimates given in this table:
##m Pclad/P ""##m= ""11+ (""22 - ""11)Pclad/P 5 + 103Pclad/P01 0.02 3.0 + 0.02 5 + 20 = 25
11 0.05 3.0 + 0.05 5 + 50 = 55
21 0.10 3.0 + 0.10 5 + 100 = 105
02 0.16 3.0 + 0.16 5 + 160 = 165
31 0.19 3.0 + 0.19 5 + 190 = 195
12 0.31 3.0 + 0.31 5 + 310 = 315
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3
3-11. (a) We want to solve Eq. (3-12) for !gi. With != 2 in Eq. (2-78) and letting
-=n
2(0)+n 2
2
2n2
(0)
we have
!(r) = !1+ (!2- !1)n 2 (0)+n 2(r)
n2(0) +n 2
2 = !1+ (!2- !1)r 2
a2
Thus
!gi=!(r)p(r)rdr
0
.
/
p(r) rdr
0
.
/= !1+
(!2+ !
1)
a2
exp+ Kr2( )r 3dr0
.
/
exp +Kr2( )rdr0
.
/
To evaluate the integrals, let x = Kr2, so that dx = 2Krdr. Then
exp+Kr 2( )r3dr0
.
/
exp+Kr 2( )rdr0
.
/=
1
2 K2 e+ xxdx
0
.
/
1
2K e+x dx
0
.
/=
1
K1!
0! =
1
K
Thus !gi = !1+(!2+ ! 1)
Ka2
(b) p(a) = 0.1 P0= P0e+ Ka2
yields eKa 2
= 10.
From this we have Ka2= ln 10 = 2.3. Thus
!gi = !1+ (!2+ ! 1)2.3
= 0.57!1+ 0.43!2
3-12. With (in units of micrometers, we have
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n = 1 + 196.98
(13.4)2 + (1.24 / ()2012
345
1/ 2
To compare this with Fig. 3-12, calculate three representative points, for example,
(= 0.2, 0.6, and 1.0 m. Thus we have the following:
Wavelength !! Calculated n n from Fig. 3-12
0.2 m 1.548 1.550
0.6 m 1.457 1.458
1.0 m 1.451 1.450
3-13. (a) From Fig. 3-13,d6
d(780 ps/(nm-km) at 850 nm. Therefore, for the LED we
have from Eq. (3-20)
8mat
L=
d6
d(8(= [80 ps/(nm-km)](45 nm) = 3.6 ns/km
For a laser diode,
8mat
L= [80 ps/(nm-km)](2 nm) = 0.16 ns/km
(b) From Fig. 3-13,d6mat
d(
= 22 ps/(nm-km)
Therefore, Dmat(() = [22 ps/(nm-km)](75 nm) = 1.65 ns/km
3-14. (a) Using Eqs. (2-48), (2-49), and (2-57), Eq. (3-21) becomes
b = 1 -ua
V
9:
;
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5
, / k +n 2n
1+n
2
=n
1(1 + =) +n 2n
1+n
2
= 1 -n
1
n1+n
2
=
Letting n2= n1(1 - -) then yields
, / k +n 2n
1+n
2
= 1 - =2+ -
71 since =2+ -
=x
s=
n 2
n1=
n 1(1 + -)
n1= (1 - -)
The travel time of the highest order ray is thus
Tmax =n
1
c s
L
x
9:
;
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Therefore
Tmin
- Tmax
=Ln
1
c
1
1 + -+ 1
9:
;
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C2=3! + 2
2(! + 2)=
3 2 1 +6
5-
9:
;
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8
3-20. We want to plot Eq. (3-30) as a function of 8 ( , where 8 int er modal and
8 int ra modal are given by Eqs. (3-41) and (3-45). For A= 0 and != 2, we have C1=
0 and C2= 1/2. Since 8int er modal
does not vary with 8(
, we have
8int er modal
L=
N1-
2c
!
! +2
9:
;
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9
=LN
1
c
kn1
,1+
4-
! +2
m
M
9:
;
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10
Ignoringd-
d(and
dn1
d(terms yields
dM
d(=
2!
! +2a2k2n
1
2 - +1
(
9
:
;
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11
- 2 (2d 2n
1
d(29
:C ;
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3-29. For ! =0.95!opt , we have
8 int er (! H !opt )
8 int er (! = !opt )=
(! + ! opt)
-(! +2)=+
0.05
(0.015)(1.95)=+170%
For ! =1.05!opt , we have
8 int er (! H !opt )
8 int er (! = !opt )=
(! + ! opt )
-(! +2)= +
0.05
(0.015)(2.05)= +163%
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1
Problem Solutions for Chapter 4
4-1. From Eq. (4-1), ni= 2!"#
$%&2'kBT
h2
3/2(memh)
3/4 exp
!"#
$%&
-Eg
2kBT
= 22'(1.38(10)23 J / K)
(6.63(10)34 J.s) 2*
+,-
./
3/ 2
T3 / 2
(.068)(.56)(9.11(10)31
kg)2[ ]
(exp )(1.55) 4.3(10)4 T)eV
2(8.62(10)5 eV/K)T
*
+,-
./
= 4.15(1014T3/2
exp ) 1.55
2(8.62(10)5
)T
*
+,-
./exp
4.3(10)4
2(8.62(10)5
)
*
+,-
./
= 5.03(1015T3/2
exp!"#
$%&
-8991
T
4-2. The electron concentration in a p-type semiconductor is nP = ni= pi
Since both impurity and intrinsic atoms generate conduction holes, the total
conduction-hole concentration pP is
pP = NA+ ni= NA+ nP
From Eq. (4-2) we have that nP = n
2
i /pP . Then
pP = NA+ nP = NA+ n
2i /p
P or p
2
P - NApP - n2i = 0
so that
pP =
NA
2
!""#
$%%&
1 +4n
2i
N2A
+ 1
If ni
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2
x2+ 4.759x - 0.436 = 0. Solving this quadratic equation yields (taking the plus
sign only)
x =1
2[ ]- 4.759 + (4.759)2+ 4(.436) = 0.090
The emission wavelength is 1=1.240
1.540 = 805 nm.
(b) Eg= 1.424 + 1.266(0.15) + 0.266(0.15)2= 1.620 eV, so that
1=1.240
1.620 = 766 nm
4-4. (a) The lattice spacings are as follows:
a(BC) = a(GaAs) = 5.6536 Ao
a(BD) = a(GaP) = 5.4512 Ao
a(AC) = a(InAs) = 6.0590 Ao
a(AD) = a(InP) = 5.8696 Ao
a(x,y) = xy 5.6536 + x(1-y) 5.4512 + (1-x)y 6.0590 + (1-x)(1-y)5.8696
= 0.1894y - 0.4184x + 0.0130xy + 5.8696
(b) Substituting a(xy) = a(InP) = 5.8696 Ao
into the expression for a(xy) in (a),
we have
y =0.4184x
0.1894 - 0.0130x 0
0.4184x
0.1894 = 2.20x
(c) With x = 0.26 and y = 0.56, we have
Eg= 1.35 + 0.668(.26) - 1.17(.56) + 0.758(.26)2+ 0.18(.56)2
- .069(.26)(.56) - .322(.26)2(.56) + 0.03(.26)(.56)2= 0.956 eV
4-5. Differentiating the expression for E, we have
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3
2E = hc
1221 or 21 =
12
hc2E
For the same energy difference 2E, the spectral width 21is proportional to the
wavelength squared. Thus, for example,
211550
211310
= 1550
1310
#!
&$
2
=1.40
4-6. (a) From Eq. (4-10), the internal quantum efficiency is
3int
= 1
1 +25/ 90=0.783, and from Eq. (4-13) the internal power level is
Pint
=(0.783) hc(35mA)q(1310
nm)=26
mW
(b) From Eq. (4-16),
P = 1
3.5 3.5 + 1( )226 mW = 0.37mW
4-7. Plot of Eq. (4-18). Some representative values of P/P0are given in the table:
f in MHz P/P01 0.999
10 0.954
20 0.847
40 0.623
60 0.469
80 0.370
100 0.303
4-8. The 3-dB optical bandwidth is found from Eq. (4-21). It is the frequency f atwhich the expression is equal to -3; that is,
10 log 1
1 + 2'f4( )2[ ]
1/ 2
#
!"
&
$%=)3
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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 5= 1
gth=1
0.05
cmln!"#
$%&1
0.32
2 + 10 cm-1= 55.6 cm-1
(b) With R1= 0.9 and R2= 0.32,
gth=1
0.05 cm ln
+,*
./-1
0.9(0.32) + 10 cm-1= 34.9 cm-1
(c) From Eq. (4-37) 3ext= 3i(gth- 6)/gth;
thus for case (a): 3ext= 0.65(55.6 - 10)/55.6 = 0.53
For case (b): 3ext= 0.65(34.9 - 10)/34.9 = 0.46
4-10. Using Eq. (4-4) to find Egand Eq. (4-3) to find 1, we have for x = 0.03,
1=1.24
Eg =
1.24
1.424 + 1.266(0.3) + 0.266(0.3)2 = 1.462 m
From Eq. (4-38)
3ext= 0.8065 1(m)dP(mW)
dI(mA)
Taking dI/dP = 0.5 mW/mA, we have 3ext= 0.8065 (1.462)(0.5) = 0.590
4-11. (a) From the given values, D = 0.74, so that 5T= 0.216
Then neff
2= 10.75 and W = 3.45, yielding 5L= 0.856
(b) The total confinement factor then is 5= 0.185
4-12. From Eq. (4-46) the mode spacing is
21=12
2Ln =
(0.80 m)2
2(400 m)(3.6) = 0.22 nm
Therefore the number of modes in the range 0.75-to-0.85 m is
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5
.85) .75
.22(10)3 =
.1
.22 (103= 455 modes
4-13. (a) From Eq. (4-44) we have g(1) = (50 cm-1) exp+,
*
./
--
(1- 850 nm)2
2(32 nm)2
= (50 cm-1) exp+,*
./-
-(1- 850)2
2048
(b) On the plot of g(1) versus 1, drawing a horizontal line at g(1) = 6t
= 32.2 cm-1shows that lasing occurs in the region 820 nm < 1< 880 nm.
(c) From Eq. (4-47) the mode spacing is
21=12
2Ln =
(850)2
2(3.6)(400 m) = 0.25 nm
Therefore the number of modes in the range 820-to-880 nm is
N =880 - 820
0.25 = 240 modes
4-14. (a) Let Nm= n/1= m2L be the wave number (reciprocal wavelength) of mode m.
The difference 2N between adjacent modes is then
2N = Nm- Nm-1=1
2L (a-1)
We now want to relate 2N to the change 21in the free-space wavelength. Firstdifferentiate N with respect to 1:
dN
d1 =
d
d1!"#
$%&n
1 =
1
1dn
d1 -
n
12 = -
1
12!"#
$%&
n - 1dn
d1
Thus for an incremental change in wavenumber 2N, we have, in absolute values,
2N =1
12!"#
$%&
n - 1dn
d1 21 (a-2)
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6
Equating (a-1) and (a-2) then yields 21=12
2L!"#
$%&
n - 1dn
d1
(b) The mode spacing is 21=
(.85 m)2
2(4.5)(400 m) = 0.20 nm
4-15. (a) The reflectivity at the GaAs-air interface is
R1= R2= !"#
$%&n-1
n+1
2 =
!"#
$%&3.6-1
3.6+1
2 = 0.32
Then Jth=1
7 6+
1
2Lln
1
R1R2
*
+, ./= 2.65(103A/cm2
Therefore
Ith= Jth(l (w = (2.65(103A/cm2)(250(10-4cm)(100(10-4cm) = 663 mA
(b) Ith= (2.65(103A/cm2)(250(10-4cm)(10(10-4cm) = 66.3 mA
4-16. From the given equation
2E11 =1.43
eV + 6.6256( 10
)34
J 8 s( )
2
8 5nm( )2 1
6.19( 10)32 kg+ 1
5.10(10)31kg#!" &
$
=1.43 eV + 0.25eV =1.68eV
Thus the emission wavelength is 1= hc/E = 1.240/1.68 = 739 nm.
4-17. Plots of the external quantum efficiency and power output of a MQW laser.
4-18. From Eq. (4-48a) the effective refractive index is
ne=m1B29
=2(1570 nm)
2(460 nm) = 3.4
Then, from Eq. (4-48b), for m = 0
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7
1= 1B1
2
B
2neL!"#$%&1
2 = 1570 nm
(1.57 m)(1570 nm)
4(3.4)(300 m) = 1570 nm 1.20 nm
Therefore for m = 1, 1= 1B3(1.20 nm) = 1570 nm 3.60 nm
For m = 2, 1= 1B5(1.20 nm) = 1570 nm 6.0 nm
4-19. (a) Integrate the carrier-pair-density versus time equation from time 0 to td(time
for onset of stimulated emission). In this time the injected carrier pair density
changes from 0 to nth.
td
= dt0
t d
: = 1
J
qd)
n
4
0
n th
: dn =)4J
qd)
n
4
#
!" &
$n= 0
n= nth
=4
ln
J
J ) Jth
#
!" &
$%
where J = Ip/A and Jth= Ith/A. Therefore td= 4ln!"#
$%&Ip
Ip- Ith
(b) At time t = 0 we have n = nB, and at t = tdwe have n = nth. Therefore,
td= ;P(t) = P0[1 + 0.2m] = 1 mW
The minimum value is
P(t) = P0[1 - 2.36m] =0 which implies m >1
2.36 = 0.42
Therefore for the average value we have < >P(t) = P0[1 + 0.2(0.42)] >1 mW,
which implies
P0= 11.084 = 0.92 mW so thatP(t) = 0.92[1 + 0.42x(t)] mW and
i(t) = 10 P(t) = 9.2[1 + 0.42x(t)] mA
4-23. Substitute x(t) into y(t):
y(t) = a1b1cos ?1t + a1b2cos ?2t
+ a2(b12cos2?1t + 2b1b2cos ?1t cos ?2t + b2
2cos2?2t)
+ a3(b13cos3?1t + 3b1
2b2cos2?1t cos ?2t + 3b1b2
2cos ?1t cos2?2t+ b2
3cos3?2t)
+a4(b14cos4?1t + 4b1
3b2cos3?1t cos ?2t + 6b1
2 b2
2cos2?1t cos2?2t
+ 4b1b23cos ?1t cos3?2t + b2
4cos4?2t)
Use the following trigonometric relationships:
i) cos2x =1
2(1 + cos 2x)
ii) cos3x =1
4(cos 3x + 3cos x)
iii) cos4x =1
8(cos 4x + 4cos 2x + 3)
iv) 2cos x cos y = cos (x+y) + cos (x-y)
v) cos2x cos y =1
4 [cos (2x+y) + 2cos y + cos (2x-y)]
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9
vi) cos2x cos2y =1
4 [1 + cos 2x+ cos 2y +1
2 cos(2x+2y) +1
2 cos(2x-2y)]
vii) cos3x cos y =1
8 [cos (3x+y) + cos (3x-y) + 3cos (x+y) + 3cos (x-y)]
then
y(t) = 12+,*
./-a2b21+ a2b22+ 34a4b
41+ 3a4b
21b
22+ 34
a4b42 constantterms
+3
4+*
.-a3b
3
1+ 2a3b1b2
2 cos ?1t +3
4 a3+*
.-b
3
2+ 2b2
1b2 cos ?2tfundamental
terms
+b
2
1
2+*
.-a2 + a4b
2
1+ 3a4b2
2 cos 2?1t +b
2
2
2+*
.-a2 + a4b
2
2+ 3a4b2
1 cos 2?2t
2nd-order harmonic terms
+1
4 a3b
3
1 cos 3?1t +1
4 a3b
3
2 cos 3?2t 3rd-order harmonic terms
+1
8 a4b
4
1 cos 4?1t +1
8 a4b
4
2 cos 4?2t 4th-order harmonic terms
+ +*
.-a2b1b2+
3
2a4b3
1b2+3
2a4b1b3
2 [ ]cos (?1+?2)t + cos (?1-?2)t
2nd-order intermodulation terms
+3
4 a3b2
1 b2[ ]cos(2?1+?2)t + cos(2?1-?2)t +3
4 a3b1b2
2[ ]cos(2?2+?1)t + cos(2?2-?1)t
3rd-order intermodulation terms
+1
2 a4b3
1 b2[ ]cos(3?1+?2)t + cos(3?1-?2)t
+3
4 a4b2
1 b2
2[ ]cos(2?1+2?2)t + cos(2?1-2?2)t
+1
2 a4b1b3
2[ ]cos(3?2+?1)t + cos(3?2-?1)t 4th-order intermodulation
terms
This output is of the form
y(t) = A0+ A1(?1) cos ?1t + A2(?1) cos 2?1t + A3(?1) cos 3?1t
+ A4(?1) cos 4?1t + A1(?2) cos ?2t + A2(?2) cos 2?2t
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10
+ A3(?2) cos 3?2t + A4(?2) cos 4?2t + @m
@n
Bmncos(m?1+n?2)t
where An(?j) is the coefficient for the cos(n?j)t term.
4-24. From Eq. (4-58) P = P0 e-t/4m
where P0= 1 mW and4m= 2(5(104hrs) = 105
hrs.
(a) 1 month = 720 hours. Therefore:
P(1 month) = (1 mW) exp(-720/105) = 0.99 mW
(b) 1 year = 8760 hours. Therefore
P(1 year) = (1 mW) exp(-8760/105) = 0.92 mW
(c) 5 years = 5(8760 hours = 43,800 hours. Therefore
P(5 years) = (1 mW) exp(-43800/105) = 0.65 mW
4-25. From Eq. (4-60) 4s= K eEA/kBT
or ln4s= ln K + EA/kBT
where kB= 1.38(10-23J/K = 8.625(10-5eV/K
At T = 60C = 333K, we have
ln 4(104= ln K + EA/[(8.625(10-5eV)(333)]
or 10.60 = ln K + 34.82 EA (1)
At T = 90C = 363K, we have
ln 6500 = ln K + EA/[(8.625(10-5eV)(363)]
or 8.78 = ln K + 31.94 EA (2)
Solving (1) and (2) for EAand K yields
EA= 0.63 eV and k = 1.11(10-5hrs
Thus at T = 20C = 293K
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11
4s = 1.11(10-5exp{0.63/[(8.625(10-5)(293)]} = 7.45(105hrs
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1
Problem Solutions for Chapter 5
5-3. (a) cosL30 = 0.5
cos 30 = (0.5)1/L= 0.8660
L = log 0.5/log 0.8660 = 4.82
(b) cosT15 = 0.5
cos 15 = (0.5)1/T= 0.9659
T = log 0.5/log 0.9659 = 20.0
5-4. The source radius is less than the fiber radius, so Eq. (5-5) holds:
PLED-step= !2r2s B0(NA)
2= !2(2"10-3cm)2(100 W/cm2)(.22)2= 191 W
From Eq. (5-9)
PLED-graded= 2!2(2"10-3cm)2(100 W/cm2)(1.48)2(.01)#$%
&'(
1 -1
2)*+,2
5
2 = 159 W
5-5. Using Eq. (5-10), we have that the reflectivity at the source-to-gel interface is
R s- g = 3.600-1.3053.600+1.305*) ,+
2
= 0.219
Similarly, the relfectivity at the gel-to-fiber interface is
R g- f = 1.465 -1.305
1.465 +1.305
*)
,+
2
= 3.34"10-3
The total reflectivity then is R =Rs- g
Rg-f = 7.30"10
-4
The power loss in decibels is (see Example 5-3)
L =-10 log (1-R) =-10 log (0.999)=3.17 "10-3
dB
5-6. Substituting B(.) = B0cosm. into Eq. (5-3) for B(.,/), we have
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2
P =
01
12
0
rm
012
0
2!
#$%
&'(
2! 02
0
.0-maxcos3.sin .d. d.sr dr
Using
02
0
.0cos3.sin .d. = 02
0
.0
( )1 - sin2. sin .d(sin .) = 02
0
sin .0
( )x - x3 dx
we have
P = 2!
0112
0
rm
012
0
2!
#$%
&'(sin2.0-max
2-
sin4.0-max4
d.sr dr
= !
0112
0
rm
012
0
2!
#$% &'(NA2- 12NA
4 d.sr dr
=!2
[ ]2NA2- NA4 02
0
rm
r dr 02
0
2! d.s
5-7. (a) Let a = 25 m and NA = 0.16. For rs3a(NA) = 4 m, Eq. (5-17) holds. For
rs44 m, 5= 1.
(b) With a = 50 m and NA = 0.20, Eq. (5-17) holds for rs310 m. Otherwise, 5
= 1.
5-8. Using Eq. (5-10), the relfectivity at the gel-to-fiber interface is
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3
Rg- f =
1.485 -1.305
1.485 +1.305
*)
,+
2
= 4.16"10-3
The power loss is (see Example 5-3)
L =-10
log
(1-R) =-10
log
(0.9958) =0.018
dB
When there is no index-matching gel, the joint loss is
R a- f = 1.485 -1.000
1.485 +1.000
*)
,+
2
=0.038
The power loss is L =-10 log (1-R) =-10 log (0.962) = 0.17dB
5-9. Shaded area = (circle segment area) - (area of triangle) =1
2 sa -
1
2 cy
s = a.= a [2 arccos (y/a)] = 2a arccos)6*
+7,d
2a
c = 2#$%
&'(
a2-)6*+7,d
22 1/2
Therefore
Acommon
= 2(shaded area) = sa cy = 2a2arccos)6
*
+7
,d
2a - d
#$
%
&'
(a2-
)6
*
+7
,d2
4
1/2
5-10.
Coupling loss (dB) for
Given axial misalignments (m)
Core/cladding diameters
(m)
1 3 5 10
50/125 0.112 0.385 0.590 1.266
62.5/125 0.089 0.274 0.465 0.985
100/140 0.056 0.169 0.286 0.590
5-11. arccos x =!2
- arcsin x
For small values of x, arcsin x = x +x3
2(3) +
x5
2(4)(5) + ...
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4
Therefore, ford
2a
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5
= -10 log
#$$%
&''(NA
2
R(0)
NA2E(0)
5-15. For fibers with different 9values, where 9R
< 9E
LF= -10 log 5F = -10 log
k2NA2(0))66*
+77,9
R
29R
+ 4a2
k2NA2(0))66*
+77,9
E
29E
+ 4a2
= -10 log#$$%
&''(9
R(9
E+ 2)
9E
(9R
+ 2)
5-16. The splice losses are found from the sum of Eqs. (5-35) through (5-37). First find
NA(0) from Eq. (2-80b).
For fiber 1: NA1(0) =n
1 2: = 1.46 2 0.01( ) =0.206
For fiber 2: NA2
(0) = n1
2: = 1.48 2 0.015( ) = 0.256
(a)
The only loss is that from index-profile differences. From Eq. (5-37)
L1; 2 9( ) =-10 log
1.80(2.00 + 2)
2.00(1.80 + 2)= 0.24
dB
(b)
The losses result from core-size differences and NA differences.
L2;1
(a) =-20
log
50
62.5
*)
,+=
1.94
dB
L2;1
(NA) =-20
log
.206
.256
%#
(&= 1.89
dB
5-17. Plots of connector losses using Eq. (5-43).
5-18. When there are no losses due to extrinsic factors, Eq. (5-43) reduces to
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6
LSM;ff= -10 log
#$%
&'(4
)6*
+7,W1
W2+
W2W1
2
For W1= 0.9W2, we then have LSM;ff= -10 log #$%
&'(44.0446 = - 0.0482 dB
5-19. Plot of Eq. (5-44).
5-20. Plot of the throughput loss.
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1
Problem Solutions for Chapter 6
6-1. From Eqs. (6-4) and (6-5) with Rf= 0, != 1 - exp(-"sw)
To assist in making the plots, from Fig. P6-1, we have the following representativevalues of the absorption coefficient:
!!(m) ""s(cm-1)
.60 4.4#103
.65 2.9#103
.70 2.0#103
.75 1.4#103
.80 0.97#103
.85 630
.90 370
.95 190
1.00 70
6-2. Ip= qA $%
0
w
G(x) dx = qA &0"s$%
0
w
e-"sx
dx
= qA &0'( )*1 - e-"sw = qA P0(1 - Rf)h+A'( )*1 - e-"sw
6-3. From Eq. (6-6), R =!qh+
=!q,hc
= 0.8044 !, ( in m)
Plot R as a function of wavelength.
6-4. (a) Using the fact that Va-VB, rewrite the denominator as
1 - ./
0
12
3Va- IMRMVB
n
= 1 - ./
0
12
3VB- VB+ Va- IMRMVB
n
= 1 -./0
123
1 -VB- Va+ IMRM
VB
n
SinceVB- Va+ IMRM
VB
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2
1 -./0
123
1 -VB- Va+ IMRM
VB
n -1 -
'4(
)5*
1 -n( )VB- Va+ IMRM
VB
=
n( )VB- Va+ IMRM
VB -nIMRM
VB
Therefore, M0=IMIp
-VB
n( )VB- Va+ IMRM -
VBnIMRM
(b) M0=IMIp
=VB
nIMRM implies I
2M =
IpVBnRM
, so that M0=./0
123VB
nIpRM
1/2
6-5. < >i2
s(t) =
1
T$%0
T
i
2
s(t)dt =
627 $%
0
27/6
R
2
0P2(t) dt (where T = 27/6),
=627
R2
0 P2
0 $%
0
27/6
(1 + 2m cos 6t + m2cos26t) dt
Using
cos 6tdt =0
7 6
8 1
6
sin 6t t = 0t = 27 / 6
=0
and $%
0
27/6
cos26t dt =1
6
$9%
0
27
./0
1231
2+
1
2cos 2x dx =
76
we have < >i2s(t) = R2
0 P2
0./0
123
1 +m2
2
6-6. Same problem as Example 6-6: compare Eqs. (6-13), (6-14), and (6-17).
(a)First from Eq. (6-6), I p =!q,
hcP0 =0.593A
Then :Q2= 2qI pB =2 1.6# 10
;19C( )(0.593A)(150# 106 Hz) = 2.84# 10;17 A2
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6-9. 0 =d
dM./0
123S
N =
d
dM'44(
)55*1
2I2
pM2
2q(Ip+ ID)M2+x+ 2qIL+ 4kBT/RL
0 = I2p M -
(2+x)M1+x 2q(Ip+ ID)1
2I2pM
2
2q(Ip+ ID)M2+x+ 2qIL+ 4kBT/RL
Solving for M: M2+x
opt =2qIL+ 4kBT/RL
xq(Ip+ ID)
6-10. (a) Differentiating pn, we have
=pn
=x =
1
Lp.
/0
1
23
pn0+ Be
-"sw
e
(w-x)/Lp
- "sBe
-"sx
= 2pn=x
2= -
1
L2
p
./0
123
pn0+ Be-"sw
e(w-x)/Lp
+ "2
s Be-"sx
Substituting pnand= 2p
n
=x2into the left side of Eq. (6-23):
-Dp
L2
p
./0
123
pn0+ Be-"sw
e(w-x)/Lp
+ Dp"2
s Be-"sx
+1
>p
./0
123
pn0+ Be-"sw
e(w-x)/Lp
-B
>p
e-"sx
+ &0"se-"sx
=
'4(
)5*
B
./0
123
Dp"2
s -1
>p
+ &0"s e-"sx
where the first and third terms cancelled because L2
p = Dp>p .
Substituting in for B:
Left side =
'44(
)55*&0
Dp
"sL2p
1 - "2
s L2
p
./0
123
Dp"2
s -1
>p
+ &0"s e-"sx
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5
=&0Dp
'44(
)55*"sL
2
p
1 - "2s L
2p
.//0
1223"
2
s Lp- 1
>p
+ Dp"s e-"sx
=
&0Dp( )-Dp"s+ Dp"s e
-"sx
= 0 Thus Eq. (6-23) is satisfied.
(b) Jdiff= qDp=pn=x
0.
31
x =w
= qDp'4(
)5*1
Lp./0
123
pn0+ Be-"sw
- "sBe-"sw
= qDpB
.
/0
1
231
Lp- "s e
-"sw + qpn0
DpLp
= q&0
.//0
1223"sL
2p
1 - "2
s L2
p
./0
1231 - "sLp
Lp e
-"sw + qpn0
DpLp
= q&0"sLp
1 + "sLp e
-"sw + qpn0
DpLp
c) Adding Eqs. (6-21) and (6-25), we have
Jtotal= Jdrift+ Jdiffusion = q&0'4(
)5*
.0
13
1 - e-"sw
+./0
123"sLp
1 + "sLpe
-"sw + qpn0
DpLp
= q&0.//0
1223
1 -e-"sw
1 + "sLp e
-"sw + qpn0
DpLp
6-11. (a) To find the amplitude, consider
.0 13JtotJ*totsc
1/2 = q&0( )S S* 1/2 where S = 1 - e
-j6td
j6td
We want to find the value of 6tdat which ( )S S*1/2
=1
2 .
Evaluating ( )S S*1/2
, we have
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7
td=w
vd =
20 # 10;6 m
4.4# 104 m / s= 0.45 ns
c)1
"s = 10-3cm = 10 m =
1
2 w
Thus since most carriers are absorbed in the depletion region, the carrier diffusion
time is not important here. The detector response time is dominated by the RC
time constant.
6-13. (a) With k1-k2and keffdefined in Eq. (6-10), we have
(1) 1 -k1(1 - k1)
1 - k2 = 1 -
k1- k2
1
1 - k2 -1 -
k2- k2
1
1 - k2 = 1 keff
(2)(1 - k1)2
1 - k2 =
1 - 2k1+ k2
1
1 - k2 -
1 - 2k2+ k2
1
1 - k2
=1 - k21 - k2
-k2- k
2
1
1 - k2 = 1 - keff
Therefore Eq. (6-34) becomes Eq. (6-38):
Fe= keffMe+ 2(1 - keff) -1
Me(1 - keff) = keffMe+ ./0 1232 -
1
Me (1 - keff)
(b) With k1-k2and k'eff
defined in Eq. (6-40), we have
(1)k2(1 - k1)
k2
1(1 - k2) -
k2- k2
1
k2
1(1 - k2) = k
'eff
(2)(1 - k1)2k2
k2
1(1 - k2) =
k2- 2k1k2+ k2k2
1
k2
1(1 - k2)
-.0
13k2- k
2
1 - .0
13k
2
1- k2k2
1
k2
1(1 - k2) = k
'eff
- 1
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8
Therefore Eq. (6-35) becomes Eq. (6-39): Fh= k'eff
Mh-./0
123
2 -1
Mh(k
'eff
- 1)
6-14. (a) If only electrons cause ionization, then @= 0, so that from Eqs. (6-36) and (6-
37), k1= k2= 0 and keff= 0. Then from Eq. (6-38)
Fe= 2 -1
Me - 2 for large Me
(b) If "= @, then from Eqs. (6-36) and (6-37), k1= k2= 1 so that
keff= 1. Then, from Eq. (6-38), we have Fe= Me.
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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)ej2(ft
df = Rej2(ft
1+j2(fRC
df!&
&
'
Using the integral solution from Appendix B3:
ejpx
() +jx)n dx
!&
&
' =2((p) n !1e! )p
*(n)for p > 0, we have
hB(t) =1
2(C
ej 2(ft
1
2(RC+ jf
"#
$%
df!&
&
' =1
Ce-t/RC
7-3. Part (a):
hp!&
&
' (t) dt =1
+Tb ,-
-+Tb/2
+Tb/2
dt =1
+Tb#."
%/$+Tb
2+
+Tb2
= 1
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2
Part (b):
hp
!&
&
' (t) dt =1
2(
1
+Tb exp !
t2
2(+Tb )2
0
123
45!&
&
' dt
= 12(
1
+Tb
( +Tb 2 = 1 (see Appendix B3 for integral solution)
Part (c):
hp
!&
&
' (t) dt =1
+Tb
exp ! t
+Tb
0
123
45dt
0
&
' = - e!& ! e
!0[ ]= 1
7-4. The Fourier transform is
F[p(t)*q(t)] = p(t)* q(t)e!j2(ft
dt!&
&
' = q(x) p(t!x)e!j2(ft
dtdx!&
&
'!&
&
'
= q(x)e!j2(fx
p(t!x)e!j2(f( t ! x)
dtdx!&
&
'!&
&
'
= q(x)
e!j2(fx
dx
p(y)
e!j 2(fy
dy!&
&
'!&
&
' where y = t - x
= F[q(t)] F[p(t)] = F[p(t)] F[q(t)] = P(f) Q(f)
7-5. From Eq. (7-18) the probability for unbiased data (a = b = 0) is
Pe=1
2P0 (v th ) + P1(v th )[ ].
Substituting Eq. (7-20) and (7-22) for P0and P1, respectively, we have
Pe=1
2
1
2(6 2 e
!v 2 / 26 2dv
V / 2
&
' + e! (v !V ) 2 / 26 2
dv!&
V / 2
'0
12
45
In the first integral, let x = v/ 262 so that dv = 262 dx.
In the second integral, let q = v-V, so that dv = dq. The second integral thenbecomes
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3
e!q 2 / 26 2
dq!&
V / 2 !V
' = 262
e! x2
dx!&
!V / 2 26 2
' where x = q/ 2 62
Then
Pe=1
2
262
2(6 2 e
! x 2dx
V / 2 262
&
' + e! x 2
dx!&
! V / 2 26 2
'0
12
3
45
=1
2 ( e
!x 2
dx!&
&
' ! 2 e!x 2
dx0
V / 2 26 2
'0
12
3
45
Using the following relationships from Appendix B,
e
! p2 x 2
dx!&
&
' =(
p and2
( e!x 2
dx0
t
' = erf(t), we have
Pe=1
21! erf
V
26 2
"#
$%
0
123
45
7-6. (a) V = 1 volt and 6= 0.2 volts, so thatV
26= 2.5. From Fig. 7-6 for
V
26= 2.5,
we find Pe77810-3errors/bit. Thus there are (28105bits/second)(7810-3
errors/bit) = 1400 errors/second, so that
1
1400 errors/second = 7810-4seconds/error
(b) If V is doubled, thenV
26= 5 for which Pe73810-7errors/bit from Fig. 7-6.
Thus
1
(28 105
bits / sec ond)(3 8 10!7
errors/bit)= 16.7 seconds/error
7-7. (a) From Eqs. (7-20) and (7-22) we have
P0(vth) =1
2(62
e! v 2 / 26 2
dvV / 2
&
' =1
21! erf
V
26 2
"#
$%
0
12 45
and
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5
Bbae=H p (0)
Hout (0)
2
Tb2
Hout' (v2N = < >v2
s +< >v2
R +< >v2
I +< >v2
E
= 2q < >i0 < >m2 BbaeR2A2+4kBT
Rb BbaeR2A2+ SIBbaeR2A2+ SEBeA2
=#." %/$2q < >i0 M2+x+
4kB
T
Rb+ SI BbaeR2A2+ SEBeA2
=R2A2I2
Tb#."
%/$
2q < >i0 M2+x+4kBT
Rb+ SI+
SE
R2 +
(2(RCA)2
T3
b
SEI3
7-11. First let x = v! boff( )/ 2 6 off with dx = dv / 2 6off in the first part of Eq. (7-
49):
Pe=26off
2(6off
exp! x2( )vth ! b off
26off
&
' dx =1
( exp !x2( )
Q / 2
&
' dx
Similarly, let y = !v +bon
( )/ 2
6on( )so that
dy = -dv/ 26on( )in the second part of Eq. (7-49):
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6
Pe= -1
( exp !y
2( )&
! vth +b on26 on
' dy =1
( exp!y
2( )Q / 2
&
' dy
7-12. (a) Let x = V2 2 6
= K2 2
For K = 10, x = 3.536. Thus
Pe=e-x
2
2 (x = 2.97810-7errors/bit
(b) Given that Pe= 10-5=e-x
2
2 (x then e-x
2 = 2 ( 10-5x.
This holds for x 73, so that K = 2 2 x = 8.49.
7-13. Differentiating Eq. (7-54) with respect to M and setting dbon/dM = 0, we have
dbondM
= 0
= -Q(h: /9)
=2 M
2+ x 9
h:
"#
$% bonI 2 +W
0
123
45
1/ 2
+ M2+x 9
h:
"#
$% bonI2 (1 ! >
0
123
45+W
?@A
BCD
1/ 2
+Q(h: /9)
M( h:/9)
1
2(2 + x)M
1+ xb
onI
2
M 2+ x 9
h:
"#
$%
bon
I2 +W
0
123
45
1/ 2 +
1
2(2 +x)M
1+xb
onI
2(1! >)
M 2+x 9
h:
"#
$%
bon
I2
(1! > ) +W0
123
45
1/ 2
?
@E
AE
B
CE
DE
1 / 2
Letting G = M2+x#."
%/$9
h: bonI2for simplicity, yields
A@?
DCB
( )G + W1/2
+ [ ]G(1->) + W1/2
=G
2(2 + x)
A
@?
D
CB1
(G + W)
1/2+(1->)
[ ]G(1->) + W
1/2
Multiply by (G + W)1/2
[ ]G(1->) + W1/2
and rearrange terms to get
(G + W)1/2
120
453
W -Gx
2(1->) = [ ]G(1->) + W
1/2#."
%/$Gx
2- W
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7
Squaring both sides and collecting terms in powers of G, we obtain the quadratic
equation
G2120
453x2>
4(1->) + G
120
453x2>
4W(2->) - >W2(1+x) = 0
Solving this equation for G yields
G =
-x2
4W(2->)
AE@E?
DECEBx4
16W2(2->)2+ x2(1->)W2(1+x)
1
2
x2
2(1->)
=
W(2->)2(1->) AE
@E?
DECEB
1 - 120
453
1 + 16 #."
%/$1+x
x2
1->
(2->)2
1
2
where we have chosen the "+" sign. Equation (7-55) results by letting
G = M2+x
opt bonI2 #."
%/$9
h:
7-14. Substituting Eq. (7-55) for M2+xbon into the square root expressions in Eq. (7-
54) and solving Eq. (7-55) for M, Eq. (7-54) becomes
bon=
Q#."
%/$h:
9b
1/(2+x)
on
120
453h:
9W(2->)2I2(1->)
K1/(2+x)AE
@E?
DECEB
120
453W(2->)
2(1->)K + W
1
2+
120
453W
2(2->)K + W
1
2
Factoring out terms:
b(1+x)/(2+x)
on = Q#."
%/$h:
9
(1+x)/(2+x)
Wx/2(2+x)
I1/(2+x)
2
8AE@E?
DECEB
120
453(2->)
2(1->)K + 1
12+
120
4531
2(2->)K + 1
12
120
453(2->)K
2(1->)
1/(2+x)
or bon= Q(2+x)/(1+x)
#."
%/$h:
9 W
x/2(1+x)
I1/(1+x)
2 L
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8
7-15. In Eq. (7-59) we want to evaluate
lim
> F 1
(2! > )K
2(1! > )L
0
123
45
1+x
2+x
=lim
> F 1
(2 ! > )K
2(1! > )
0
123
45
1+x
2+x lim
> F 1
1
L
"#
$%
1+x
2+x
Consider first
lim
> F 1
(2 ! >)K
2(1! >)
0
12 45=
lim
> F1
(2 ! >)
2(1! > )!1+ 1 +B
(1! > )
(2! > )2
0
123
45
1
2?
@E
AE
B
CE
DE
where B = 16(1+x)/x2. Since >F1, we can expand the square root term in a
binomial series, so that
lim> F 1
(2 ! >)K2(1! >)
012
345= lim
> F1
(2 ! >)2(1! > )
!1+ 1 +12
B (1! > )(2 ! > )2
! Order(1 ! > )2012
345
?@A
BCD
=lim
> F 1
B
4(2 ! > )=
B
4 = 4
1+x
x2
Thuslim
> F1
(2 ! > )K
2(1! >)
0
123
45
1+x
2+x
=#."
%/$
41+x
x2
1+x
2+x
Next consider, using Eq. (7-58)
lim
> F 1
1
L
"#
$%
1+ x
2+x
=lim
> F1
(2 ! > )K
2(1! >)
0
123
45
1/(2+ x)
(2! >)
2(1! >)K +1
0
123
45
1
2
+ 1
2(2 ! > )K +10
134
1
2?
@E
AE
B
CE
D
From the above result, the first square root term is
120
453(2->)
2(1->)K + 1
1
2 =
120
453
41+x
x2+ 1
1
2 =
#."
%/$x2+ 4x + 4
x2
1
2 =
x + 2
x
From the expression for K in Eq. (7-55), we have thatlim
> F 1K = 0, so that
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9
lim
> F 1
1
2(2 ! > )K +10
134
1
2
= 1 Thus
lim> F 1
(2 ! > )2(1! > )
K +1012
345
1
2+ 1
2(2 ! >)K +10
134
1
2
?
@E
AE
B
C
DE
= x + 2x
+ 1 = 2(1 + x)x
Combining the above results yields
lim
> F 1
(2! > )K
2(1! > )L
0
123
45
1+x
2+x
=#."
%/$
41+x
x2
1+x
2+x
#."
%/$
41+x
x2
1
2+x
2(1 + x)
x =
2
x
so that
lim
> F1 M1+x
opt =W1/2
QI2 2
x
7-16. Using H'p(f) = 1 from Eq. (7-69) for the impulse input and Eq. (7-66) for the
raised cosine output, Eq. (7-41) yields
I2= Hout'
(
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10
Use Eq. (7-42) to find I3:
I3= Hout'
(
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12
lim
> F 1K =
lim
> F1 !1+ 1+16
1+x
x2"#
$%
1! >
(2 ! >) 20
123
45
1
2?
@E
AE
B
CE
DE
= -1 + 1 = 0
Also
lim
> F 1(1! >) = 0. Therefore from Eq. (7-58)
lim
> F 1L =
lim
> F1
2(1! >)
(2 ! > )K
0
123
45
1/(1+ x)
AE@E?
DECEB
120
453(2->)
2(1->)K + 1
1
2+ 1
2+x
1+x
Expanding the square root term in K yields
lim
> F 1
2 ! >
1! >K =
lim
> F1
2! >
1 ! >
"
#. $
%/ !1+ 1+
16
2
1 + x
x 2"#
$%
1! >
(2 ! > ) 2+order(1! >) 2
0
123
45
?@
A
BC
D
=lim
> F 1
8(1 + x)
x2
1
2 ! >
0
123
45=
8(1+x)
x2
Therefore
lim
> F 1
L =
1
20
4
532x2
8(1+x)
1/(1+x)
AE
@E?
DE
CEB
1
20
4
534(1+x)
x2 + 1
1
2+ 1
2+x
1+x
=120
4532x2
8(1+x)
1/(1+x)
AE@E?
DECEBx+2
x+ 1
2+x
1+x
= (1+x)#."
%/$2
x
x
1+x
7-21. (a) First we need to find L and L'. With x = 0.5 and >= 0.9, Eq. (7-56) yields K =
0.7824, so that from Eq. (7-58) we have L = 2.89. With H= 0.1, we have >' = >(1 -
H) = 0.9
>= 0.81. Thus L' = 3.166 from Eq. (7-80). Substituting these values into
Eq. (7-83) yields
y(H) = (1 + H)#."
%/$1
1 - H
2+x
1+x
L'
L = 1.1
#."
%/$1
.9
5/3
3.166
2.89 = 1.437
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13
Then 10 log y(H) = 10 log 1.437 = 1.57 dB
(b) Similarly, for x = 1.0, >= 0.9, and H= 0.1, we have L = 3.15 and L' = 3.35, so
that
y(H) = 1.1#."
%/$1
.9
3/2
3.35
3.15 = 1.37
Then 10 log y(H) = 10 log 1.37 = 1.37 dB
7-22. (a) First we need to find L and L'. With x = 0.5 and >= 0.9, Eq. (7-56) yields K =
0.7824, so that from Eq. (7-58) we have L = 2.89. With H= 0.1, we have >' = >(1 -
H) = 0.9>= 0.81. Thus L' = 3.166 from Eq. (7-80). Substituting these values into
Eq. (7-83) yields
y(H) = (1 + H)#."
%/$1
1 - H
2+x
1+x
L'
L = 1.1
#."
%/$1
.9
5/3
3.166
2.89 = 1.437
Then 10 log y(H) = 10 log 1.437 = 1.57 dB
(b) Similarly, for x = 1.0, >= 0.9, and H= 0.1, we have L = 3.15 and L' = 3.35, so
that
y(H) = 1.1#."
%/$1
.9
3/2
3.35
3.15 = 1.37
Then 10 log y(H) = 10 log 1.37 = 1.37 dB
7-23. Consider using a Si JFET with Igate= 0.01 nA. From Fig. 7-14 we have that +=
0.3 for >= 0.9. At += 0.3, Fig. 7-13 gives I2= 0.543 and I3= 0.073. Thus from
Eq. (7-86)
WJFET=1
B
2(.01nA )
1.6 8 10!19C+
4(1.388 10 !23 J / K)(300K)
(1.6 8 10!19C)2105 I
0
123
450.543
+1
B
4(1.38 8 10!23 J/ K)(300 K)(.7)
(1.6 8 10!19 C)2 (.005 S)(10 5 I) 20
123
450.543
+2((10pF)
1.6 8 10!19 C01
34
2 4(1.38 8 10!23 J / K)(300K)(.7)
(.005S)0.073 B
or
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14
WJFET73.518 1012
B+ 0.026B
and from Eq. (7-92) WBP=3.39 8 1013
B+ 0.0049B
7-24. We need to find bonfrom Eq. (7-57). From Fig. 7-9 we have Q = 6 for a 10-9
BER. To evaluate Eq. (7-57) we also need the values of W and L. With >= 0.9,
Fig. 7-14 gives += 0.3, so that Fig. 7-13 gives I2= 0.543 and I3= 0.073. Thus
from Eq. (7-86)
W =3.518 1012
B+ 0.026B = 3.518105+ 2.68105= 6.18105
Using Eq. (7-58) to find L yields L = 2.871 at >= 0.9 and x = 0.5. Substitutingthese values into eq. (7-57) we have
bon= (6)5/3(1.6810-19/0.7) (6.18105).5/3(0.543)1/1.52.871 = 7.97810-17J
Thus Pr= bonB = (7.97810-17J)(107b/s) = 7.97810-10W
or
Pr(dBm) = 10 log 7.97810-10= -61.0 dBm
7-25. From Eq. (7-96) the difference in the two amplifier designs is given by
JW =1
Bq22kBT
Rf I2= 3.528106for I2= 0.543 and >= 0.9.
From Eq. (7-57), the change in sensitivity is found from
10 log120
453WHZ+ JW
WHZ
x
2(1+x)
= 10 log120
4531.0 + 3.52
1.0
.5
3
= 10 log 1.29 = 1.09 dB
7-26. (a) For simplicity, let
D = M2+x#."
%/$9
h: I2 and F =
Q
M#."
%/$9
h:
so that Eq. (7-54) becomes, for >= 1,
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15
b = F[ ](Db + W)1/2+ W1/2
Squaring both sides and rearranging terms gives
b2
F2 - Db - 2W = 2W1/2
(Db + W)1/2
Squaring again and factoring out a "b2" term yields
b2- (2DF2)b + (F4D2- 4WF2) = 0
Solving this quadratic equation in b yields
b =1
2[ ]2DF2 4F4D2- 4F4D2+ 16WF2 = DF2+ 2F W
(where we chose the "+" sign) =h:9
#."
%/$
MxQ2I2+2Q
MW1/2
(b) With the given parameter values, we have
bon= 2.286810-19 39.1M0.5
+1.78 104
M
"
#. $
%
The receiver sensitivity in dBm is found from
Pr= 10 log bon(508106 b / s )
Representative values of Prfor several values of M are listed in the table below:
M Pr(dBm) M Pr(dBm)
30 - 50.49 80 -51.92
40 -51.14 90 -51.94
50 -51.52 100 -51.93
60 -51.74 110 -51.90
70 -51.86 120 -51.86
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7-27. Using Eq. (E-10) and the relationship
1
1+ x
a
"#
$%
2 dx0
&
' =(a2
from App. B, we have from Eq. (7-97)
BHZ=1
(AR) 2
(AR)2
1+ (2(RC)2 f2
0
&
' df=(2
1
2(RC =
1
4RC
where H(0) = AR. Similarly, from Eq. (7-98)
BTZ
=1
1
1
1+ 2(RCA
"#
$%
2
f2
0
&
' df=
(
2
A
2(RC =
A
4RC
7-28. To find the optimum value of M for a maximum S/N, differentiate Eq. (7-105)
with respect to M and set the result equal to zero:
d(S/N)
dM =
(Ipm)2M
2q(Ip+ID)M2+x B +4kBTB
ReqFT
-q(Ip+ID) (2+x) M1+x B (Ipm)2M2
120
453
2q(Ip+ID)M2+x B +4kBTB
ReqFT
2 = 0
Solving for M,
M2+x
opt =4kBTBFT/Req
q(Ip+ID)x
7-29. (a) For computational simplicity, let K = 4kBTBFT/Req; substituting Moptfrom
Problem 7-28 into Eq. (7-105) gives
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17
S
N =
1
2(Ipm)2M
2
opt
2q(Ip+ID)M2+x
optB + KB
=
1
2(Ipm)2
120
453K
q(Ip+ID)x
2
2+x
2q(Ip+ID)K
q(Ip+ID)xB + KB
=xm2I
2
p
2B(2+x) [ ]q(Ip+ID)x2/(2+x)#
."
%/$Req
4kBTFT
x/(2+x)
(b) If Ip>> ID, then
S
N =
xm2I2p
2B(2+x) ( )qx2/(2+x)
I2/(2+x)
p
#."
%/$Req
4kBTFT
x/(2+x)
=m2
2Bx(2+x)1220
4553
( )xIp
2(1+x)
q2(4kBTFT/Req)x
1/(2+x)
7-30. Substituting Ip= R0Printo the S/N expression in Prob. 7-29a,
S
N =
xm2
2B(2+x)
( )R0Pr
2
[ ]q(R0Pr+ID)x2/(2+x)#
."
%/$Req
4kBTFT
x/(2+x)
=(0.8)2 (0.5A / W )2 P
r2
2(5 8106 / s) 3[1.6 810!19 C(0.5Pr+10
!8 )A]2 / 3104 I / J
1.656 8 10!20"
#. $
%
1 / 3
=1.5308 1012 P
r2
0.5Pr+ 10
!8( )2 / 3 where Pris in watts.
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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)
2810-9 - 57 1.237 0.92
4810-9 - 54 4.669 6.69
1810-8 - 50 25.15 14.01
4810-8 - 44 253.5 24.04
1810-7 - 40 998.0 29.99
1810-6 - 30 2.48104 43.80
1810-5 - 20 5.28105 57.18
1810-4 - 10 1.138107 70.52
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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 thepin photodiode,
with 11 joints
PT= PS- PR= 11(lc) + !fL + system margin
= 0 dBm - (-45 dBm) = 11(2 dB) + (4 dB/km)L + 6 dB
which gives L = 4.25 km. The transmission distance cannot be met with these
components.
(b) For the APD
0 dBm - (-56 dBm) = 11(2 dB) + (4 dB/km)L + 6 dB
which gives L = 7.0 km. The transmission distance can be met with these
components.
8-3. From g(t) = ( )1 - e-2"Bt u(t) we have
#$
%&
1 - e-2"Bt10
= 0.1 and #$
%&
1 - e-2"Bt90
= 0.9
so that
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2
e-2"Bt10
= 0.9 and e-2"Bt90
= 0.1
Then
e2"Btr
= e2"B(t90-t10)
=.9
.1
= 9
It follows that
2"Btr= ln 9 or tr=ln 9
2"B =
0.35
B
8-4. (a) From Eq. (8-11) we have
1
2"' exp#(
($
%))&
-
t2
1/2
2'2 =1
21
2"' which yields t1/2= (2 ln 2)1/2'
(b) From Eq. (8-10), the 3-dB frequency is the point at which
G(*) =1
2 G(0), or exp
+,-
./0
-(2"f3dB)2'2
2 =
1
2
Using 'as defined in Eq. (8-13), we have
f3dB=(2 ln 2)1/2
2"' =
2 ln 2
"tFWHM =
0.44
tFWHM
8-5. From Eq. (8-9), the temporal response of the optical output from the fiber is
g(t) =1
2"'
exp
#
($
%
)&
-t2
2'2
If1eis the time required for g(t) to drop to g(0)/e, then
g(1e) =1
2"' exp
#($
%)&
-1e2
2'2 =
g(0)
e =
1
2"'e
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3
from which we have that 1e= 2 '. Since teis the full width of the pulse at the
1/e points, then te = 21e= 2 2 '.
From Eq. (8-10), the 3-dB frequency is the point at which
G(f3dB) = 12 G(0). Therefore with '= te/(2 2 )
G(f3dB) =1
2" exp
+,-
./0
-1
2(2"f3dB')2 =
1
2
1
2"
Solving for f3dB:
f3dB=2 ln 2
2"'
=2 ln 2
2"
2 2
te
=0.53
te
8-6. (a) We want to evaluate Eq. (8-17) for tsys.
Using Dmat= 0.07 ns/(nm-km), we have
tsys= 234
567
(2)2+ (0.07)2(1)2(7)2++,-
./0440(7)0.7
800
2
++,-
./0350
90
2 1/2
= 4.90 ns
The data pulse width is Tb=
1
B =
1
90 Mb/s = 11.1 ns
Thus 0.7Tb= 7.8 ns > tsys, so that the rise time meets the NRZ data requirements.
(b) For q = 1.0,
tsys= 234
567
(2)2+ (0.49)2++,-
./0440(7)
800
2
++,-
./0350
90
2 1/2
= 5.85 ns
8-7. We want to plot the following 4 curves of L vs B =1
Tb :
(a) Attenuation limit
PS- PR= 2(lc) + !fL + 6 dB, where PR= 9 log B - 68.5
so that L = (PS 9 log B + 62.5 - 2lc)/!f
(b) Material dispersion
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4
tmat= Dmat'8L = 0.7Tb or
L =0.7Tb
Dmat'8=
0.7
BDmat'8=
104
B (with B in Mb/s)
(c) Modal dispersion (one curve for q = 0.5 and one for q = 1)
tmod =0.440L
q
800 =
0.7
B or L =
+,-
./0
#($
%)&800
0.440.7
B
1/q
With B in Mb/s, L = 1273/B for q =1, and L = (1273/B)2for q = .5.
8-8. We want to plot the following 3 curves of L vs B =1
Tb :
(a) Attenuation limit
PS- PR= 2(lc) + !fL + 6 dB, where PR= 11.5 log B - 60.5, PS= -13 dBm, !f=
1.5 dB/km, and lc= 1 dB,
so that L = (39.5 - 11.5 log B)/1.5 with B in Mb/s.
(b) Modal dispersion (one curve for q = 0.5 and one for q = 1)
tmod =0.440Lq
800 =
0.7
B or L =
+,-
./0
#($
%)&800
0.440.7
B
1/q
With B in Mb/s, L = 1273/B for q =1, and L = (1273/B)2for q = .5.
8-9. The margin can be found from
PS- PR= lc+ 49(lsp) + 50!f+ noise penalty + system margin
-13 - (-39) = 0.5 + 49(.1) + 50(.35) + 1.5 + system margin
from which we have
system margin = 1.6 dB
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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
1 0 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 10-9BER,
Pmpn= -7.94 log (1 - 18k2"4h4) where for simplicity h = BZD'8
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 = 1.4910-4b/ps [to give D in ps/(nm.km)]:
0.5 = -7.94 log {1 - 18(0.3)2["(1.4910-4)(100)(3.5)]4D4}
or
0.5 = -7.94 log {1 - 9.097910-4D4} from which D = 2 ps/(nm.km)
B (Mb/s) D [ps/(nm.km)]
140 2
280 1
560 0.5
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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*10
+6A) 2
The source noise is, with RIN = -135 dB/Hz = 3.162*10-14/Hz,
< >i2
source
= RIN (R0P)2 B = 5.69*10-13A2
The quantum noise is
< >i2Q = 2q(R0P + ID)B = 9.5*10-14A2
The thermal noise is
< >i2
T =
4kBT
Req Fe= 8.25*10-13
A2
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3
Thus the carrier-to-noise ratio is
C
N =
1
2(15 *10+6 A)2
5.69*10+13
A2+ 9.5 *10
+14A
2+8.25 *10
+13A
2 = 75.6
or, in dB, C/N = 10 log 75.6 = 18.8 dB.
9-5. When an APD is used, the carrier power and the quantum noise change.
The carrier power is
C = 12(mR0MP) 2= 1
2(15 *10+5 A) 2
The quantum noise is
< >i2Q = 2q(R0P + ID)M2F(M)B = 2q(R0P + ID)M2.7B = 4.76*10-10A2
Thus the carrier-to-noise ratio is
C
N =
12
(15 *10 +5 A)2
5.69*10+13 A 2 + 4.76* 10+11 A2 +8.25 *10+13A2= 236.3
or, in dB,C
N = 23.7 dB
9-6. (a) The modulation index is m =
"
##$
%
&&'
(
i=1
32
(.044)21/2
= 0.25
The received power is P = -10 dBm = 100W
The carrier power is
C =1
2(mR0P) 2=
1
2(15*10+6 A) 2
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4
The source noise is, with RIN = -135 dB/Hz = 3.162*10-14/Hz,
< >i2source = RIN (R0P)2 B = 5.69*10-13A2
The quantum noise is
< >i2Q = 2q(R0P + ID)B = 9.5*10-14A2
The thermal noise is
< >i2T =4kBT
Req Fe= 8.25*10-13A2
Thus the carrier-to-noise ratio is
C
N =
1
2(15 *10+6 A)2
5.69*10+13
A2+ 9.5 *10
+14A
2+8.25 *10
+13A
2 = 75.6
or, in dB, C/N = 10 log 75.6 = 18.8 dB.
(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*10
+5A) 2= 7.06*10-11A2
The source noise is, with RIN = -135 dB/Hz = 3.162*10-14/Hz,
< >i2source = RIN (R0P)2 B = 1.42*10-13A2
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The quantum noise is
< >i2Q = 2q(R0P + ID)B = 4.8*10-14A2
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* 10+13 A 2 +4.8 *10+14A 2 +8.25 *10+13 A2= 69.6
or, in dB, C/N = 10 log 69.6 = 18.4 dB.
9-8. Using the expression from Prob. 9-7 with !,-= 0.05, f-= 0.05, and !,= f = 10
MHz, yields
RIN(f) =4R1R2
.
!,f2+ !,2
"$
%'
1 + e-4.!,-
- 2 e-2.!,-
cos(2.f-)
=4R1R2
. 1
20 MHz(.1442)
Taking the log and letting the result be less than -140 dB/Hz gives
-80.3 dB/Hz + 10 log R1R2< -140 dB/Hz
If R1= R2then 10 log R1R2= 20 log R1 < -60 dB
or 10 log R1= 10 log R2< -30 dB
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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
c
!f = 1546
nm( )2
3 # 108m / s
500 # 109s
$1 =4nm
The number of wavelength channels fitting into the 1536-to-1556 spectral band
then is
N = (1556 1536 nm)/4 nm = 5
10-2. (a) We first find P1by using Eq. (10-6):
10 log 200
W
P1
%
&' (
)*=2.7dB yields P1 = 10
(log
200$0.27 )=107.4W
Similarly, P2
=10(log 200$0.47) =67.8W
(b) From Eq. (10-5): Excess loss = 10
log
200
107.4 + 67.8
%&
()=
0.58
dB
(c) P1P
1+ P
2
=107.4175.2
=61% and P2P
1+P
2
= 67.8175.2
=39%
10-3. The following coupling percents are are realized when the pull length is stopped at
the designated points:
Coupling percents from input fiber to output 2
Points A B C D E F
1310 nm 25 50 75 90 100 0
1540 nm 50 88 100 90 50 100
10-4. From A out =s11Ain +s12Bin and Bout = s21Ain + s22 Bin = 0 , we have
Bin =$s21
s22A in and
Aout
= s11$
s12
s21
s22
+
,-.
/0A
in
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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=sin
20.4z( )exp$0.06z( ) =0.5
One can either plot both curves and find the intersection point, or solve the
equation numerically to yield z = 2.15 mm.
10-6. Since 1z2n , then for n
A> n
Bwe have 3
A< 3
B. Thus, since we need to have
3ALA= 3BLB, we need to have LA> LB.
10-7. From Eq. (10-6), the insertion loss LIjfor output port j is
LIj
= 10 log P
i $in
Pj $out
%
&'
(
)*
Let
a j = Pi $in
Pj$ out=10
L Ij /10
, where the values of LIjare given in Table P10-7.
Exit port no. 1 2 3 4 5 6 7
Value of aj 8.57 6.71 5.66 8.00 9.18 7.31 8.02
Then from Eq. (10-25) the excess loss is
10
log
Pin
Pj4
%
&'(
)*=10 log
Pin
Pin
1
a1
+ 1
a2
+... + 1
an
%
&' (
)*
%
&
'
''
(
)
*
**
= 10
log
1
0.95
%& () = 0.22
dB
10-8. (a) The coupling loss is found from the area mismatch between the fiber-core
endface areas and the coupling-rod cross-sectional area. If a is the fiber-core radius
and R is the coupling-rod radius, then the coupling loss is
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3
Lcoupling= 10 logPoutPin
= 10 log75a2
5R2 = 10 log
7(25)2
(150)2 = -7.11 dB
(b) Similarly, for the linear-plate coupler
Lcoupling= 10 log75a2
l6w = 10 log
75(25)2
800(50) = -4.64 dB
10-9. (a) The diameter of the circular coupling rod must be 1000 m, as shown in the
figure below. The coupling loss is
Lcoupling= 10 log
75a2
5R2 = 10 log7(100)2
(500)2 = -5.53 dB
(b) The size of the plate coupler must be 200 m by 2600 m.
The coupling loss is 10 log
75(100)2
200(2600) = -3.74 dB
10-10. The excess loss for a 2-by-2 coupler is given by Eq. (10-5), where P1= P2for a 3-
dB coupler. Thus,
200 m
400
m
Coupling rod
diameter
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4
Excess loss = 10 log P
0
P1
+P2
%
&' (
)*= 10 log
P0
2P1
%
&' (
)* = 0.1dB
This yields
P1
= P0
2%&
()
10 0.01 =0.977P02
%&
()
Thus the fractional power traversing the 3-dB coupler is FT= 0.977.
Then, from Eq. (10-27),
Total loss = $10 log log F
T
log 2$1
%
&' (
)log N = $10 log
log 0.977
log 2$ 1
%
&' (
)log 2
n 730
Solving for n yields
n7 $3
log 2log 0.977
log 2$ 1
%
&' (
)
=9.64
Thus, n = 9 and N =2n =2
9 = 512
10-11. For details, see Verbeek et al., Ref. 34, p. 1012
For the general case, from Eq. (10-29) we find
M
11= cos
23d( ) 8 cos
k!L / 2( )+ j
sin
k!L / 2( )
M
12=M
21=j sin 23d( )8 cos k!L / 2( )
M 22 =cos 23d( )8 cos k!L / 2( )$j sin k!L / 2( )
The output powers are then given by
Pout ,1
= cos2 23d( )8 cos2 k!L / 2( )+ sin2 k!L / 2( )[ ]Pin,1
+sin 2 23d( )8 cos2 k!L / 2( )[ ]Pin,2
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Pout ,2
= sin2 23d( ) 8cos2 k!L / 2( )[ ]Pin,1
+ cos 2 23d( )8 cos2 k!L / 2( ) + sin2 k!L / 2( )[ ]Pin,2
10-12. (a) The condition !9= 125 GHz is equivalent to having !"= 1 nm. Thus the
other three wavelengths are 1549, 1550, and 1551 nm.
(b) From Eqs. (10-42) and (10-43), we have
!L1 = c
2neff
(2!9)=0.4 mm and
!L3 = c
2neff
!9=0.8mm
10-13. An 8-to-1 multiplexer consists of three stages of 2# 2MZI multiplexers. The first
stage has four 2# 2MZIs, the second stage has two, and the final stage has one
2# 2MZI. Analogous to Fig. 10-14, the inputs to the first stage are (from top to
bottom) 9, 9+ 4!9, 9+ 2!9, 9+ 6!9, 9+ !9, 9+ 5!9, 9+ 3!9, 9+ 7!9.
In the first stage
!L1 = c
2neff(4!9)= 0.75mm
In the second stage
!L2 = c2neff(2!9)
=1.5
mm
In the third stage
!L 3 = c
2neff
(!9)=3.0
mm
10-14. (a) For a fixed input angle :, we differentiate both sides of the grating equation toget
cos ;d;=k
n'< d" or
d;d"
=k
n'
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Solving this fork
n'< and substituting into the
d;d"
equation yields
d;
d" =
2 sin ;
"cos ; =
2 tan ;
"
(b) For S = 0.01,
tan ;=,-+
/0.S"
2!"(1+m)
1/2
=0.01(1350)
2(26)(1+3)
+
,-.
/0
1/ 2
= 0.2548
or ;= 14.3
10-15. For 93% reflectivity
R =tanh2(3L) = 0.93 yields 3L = 2.0, so that L = 2.7 mm for 3= 0.75 mm-1.
10-16. See Bennion et al., Ref. 42, Fig. 2a.
10-17. Derivation of Eq. (10-49).
10-18. (a) From Eq. (10-45), the grating period is
< = "uv
2 sin;
2
= 244nm
2 sin(13.5)=
244
2(0.2334)nm =523nm
(b) From Eq. (10-47), "Bragg
= 2 =1$1/ 2 =0.827 , we have from Eq. (10-51),
3 =5?n>
"Bragg
=5 2.5 # 10 $4( )(0.827)
1.547 # 10$4 cm=4.2cm
$1
(d) From Eq. (10-49), !" = 1.547m( )
2
5(1.48)500 m
(2.1)2
+52[ ]1 / 2
= 3.9 nm
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(e) From Eq. (10-48), R max = tanh2(3L) =tanh
2(2.1) =(0.97)
2=94%
10-19. Derivation of Eq. (10-55).
10-20. (a) From Eq. (10-54),
!L =m" 0n c
=1181.554m
1.451= 126.4m
(b) From Eq. (10-57),
!9 = x
L f
nscd
m"2
nc
ng
= 25m
9.36# 103 m
1.453(3 # 10 8 m/s)(25 # 10 $6 m)
118(1.554 # 10 $6 m)2
1.451
1.475=100.5GHz
!" ="2
c!9 =
(1.554 # 10$6 m)2
3 # 108 m / s100.5GHz =0.81nm
(c) From Eq. (10-60),
!9FSR = c
n g!L=
3 # 108 m / s
1.475(126.4m)= 1609GHz
Then
!" =
"2
c!9FSR =
(1.554 # 10$6 m)2
3 # 108m / s
1609GHz =12.95nm
(d) Using the conditions
sin;i= ;
i= 2(25m)
9380m=5.33 # 10 $3 radians
and
sin;o = ;o =21.3 # 10$3
radians
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then from Eq. (10-59),
!9FSR = c
ng [!L + d; i +;o( )]
= 3 # 108 m / s
1.475 (126.4 # 10$6 m) + (25 # 10$6 m) 5.33 +21.3( ) # 10$3[ ]=1601GHz
10-21. The source spectral width is
!"signal
="2 9
c=
1550
nm( )2 (1.25 # 10 9
s$1)
(3 # 10 8
m/s) 109
nm / m( )=1 # 10$2
nm
Then from Eq. (10-61)
!"tune ="!neff
neff= 1550nm( ) 0.5%( ) =7.75nm
Thus, from Eq. (10-63)
N = !" tune10 " signal
= 7.75nm
10 0.01nm( )= 77
10-22. (a) From Eq. (10-64), the grating period is
< ="Bragg
2neff=
1550 nm
2 3.2( )= 242.2nm
(b) Again, from the grating equation,
!< = !"
2neff
=2.0nm
2 3.2( )= 0.3nm
10-23. (a) From Eq. (10-43)
!L = c
2neff!9=
"2
!"
1
2 neff=4.0mm
(b) !Leff =!n effL implies that !neff = 4mm
100mm=0.04 =4%
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10-24. For example, see C. R. Pollock, Fundamentals of Optoelectronics, Irwin, 1995,
Fig. 15.11, p. 439.
10-25. (a) The driving frequencies are found from
fa =9ov a !n
c=
va !n
"
Thus we have
Wavelength (nm) 1300 1546 1550 1554
Acoustic
frequency (MHz)
56.69 47.67 47.55 47.43
(b) The sensitivity is (4 nm)/(0.12 MHz) = 0.033 nm/kHz
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Problem Solutions for Chapter 11
11-1. (a) From Eq. (11-2), the pumping rate is
Rp=
I
qwdL=
100mA
(1.6! 10"19 C)(5m)(0.5 m)(200 m)
= 1.25 ! 1027 (electrons / cm 3) / s
(b) From Eq. (11-8), the maximum zero-signal gain is
g0= 0.3(1! 10 "20 m2 )(1ns) 1.25 ! 1033 (electrons/ m3) / s "
1.0 ! 1024 / m3
1ns
#
$%&
'(
= 750 m"1 =7.5cm"1
(c) From Eq. (11-7), the saturation photon density is
Nph;sat
= 1
0.3 (1! 10"20 m2 ) 2! 108 m / s( )(1ns)= 1.67 ! 1015 photons/ cm3
(d) From Eq. (11-4), the photon density is
N ph = Pin
)v
ghc wd( )
= 1.32 ! 1010 photons / cm3
11-2. Carrying out the integrals in Eq. (11-14) yields
g0L =lnP(L)
P(0)+
P(L) " P(0)
Pamp,sat
Then with P(0) = Pin, P(L) = Pout, G = Pout/Pin, and G0 =exp g 0L( )from Eq. (11-
10), we have
ln G 0 =g 0L =ln G + GP
in
Pamp,sat
" P
in
Pamp,sat
=ln G + 1 " G( ) P
in
Pamp,sat
Rearranging terms in the leftmost and rightmost parts then yields Eq. (11-15).
11-3. Plots of amplifier gains.
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11-4. Let G = G0/2 and Pin =Pout/ G = 2Pout,sat / G 0 . Then Eq. (11-15) yields
G0
2
=1+G
0P
amp.sat
2Pout.satln 2
Solving for Pout,satand with G0>> 1, we have
Pout .sat
=G
0ln 2
G0"2( )
Pamp.sat *(ln 2)Pamp.sat =