Solution Manual optical fiber communication gerd keiser 3rd ed

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1
Problem Solutions for Chapter 2
2-1.
 
E = 100cos 2p108 t + 30°( ) ex + 20 cos 2p108t - 50°( ) ey
+ 40cos 2p108 t + 210°( ) ez
2-2. The general form is:
y = (amplitude) cos(wt - kz) = A cos [2p(nt - z/l)]. Therefore
(a) amplitude = 8 mm
(b) wavelength: 1/l = 0.8 mm-1 so that l = 1.25 mm
(c) w = 2pn = 2p(2) = 4p
(d) At t = 0 and z = 4 mm we have
y = 8 cos [2p(-0.8 mm-1)(4 mm)]
 = 8 cos [2p(-3.2)] = 2.472
2-3. For E in electron volts and l in mm we have E = 1.240
l
(a) At 0.82 mm, E = 1.240/0.82 = 1.512 eV
 At 1.32 mm, E = 1.240/1.32 = 0.939 eV
 At 1.55 mm, E = 1.240/1.55 = 0.800 eV
(b) At 0.82 mm, k = 2p/l = 7.662 mm-1
 At 1.32 mm, k = 2p/l = 4.760 mm-1
 At 1.55 mm, k = 2p/l = 4.054 mm-1
2-4. x1 = a1 cos (wt - d1) and x2 = a2 cos (wt - d2)
Adding x1 and x2 yields
x1 + x2 = a1 [cos wt cos d1 + sin wt sin d1]
+ a2 [cos wt cos d2 + sin wt sin d2]
= [a1 cos d1 + a2 cos d2] cos wt + [a1 sin d1 + a2 sin d2] sin wt
Since the a's and the d's are constants, we can set
a1 cos d1 + a2 cos d2 = A cos f (1)
2
a1 sin d1 + a2 sin d2 = A sin f (2)
provided that constant values of A and f exist which satisfy these equations. To
verify this, first square both sides and add:
A2 (sin2 f + cos2 f) = a1
2 sin2 d1 + cos
2 d1( )
+ a2
2 sin2 d2 + cos
2 d2( ) + 2a1a2 (sin d1 sin d2 + cos d1 cos d2)
or
A2 = a1
2 + a2
2 + 2a1a2 cos (d1 - d2)
Dividing (2) by (1) gives
tan f = 
a1 sind1 + a2 sind2
a1 cosd1 + a2 cosd2
Thus we can write
x = x1 + x2 = A cos f cos wt + A sin f sin wt = A cos(wt - f)
2-5. First expand Eq. (2-3) as
Ey
E0 y
= cos (wt - kz) cos d - sin (wt - kz) sin d (2.5-1)
Subtract from this the expression
E x
E0 x
cos d = cos (wt - kz) cos d
to yield
 
Ey
E0 y
-
Ex
E0x
 cos d = - sin (wt - kz) sin d (2.5-2)
Using the relation cos2 a + sin2 a = 1, we use Eq. (2-2) to write
3
sin2 (wt - kz) = [1 - cos2 (wt - kz)] = 1 -
Ex
E0x
æ 
è 
ç ö 
ø 
÷ 
2é 
ë 
ê 
ù 
û 
ú (2.5-3)
Squaring both sides of Eq. (2.5-2) and substituting it into Eq. (2.5-3) yields
Ey
E0 y
- Ex
E0x
cosd
é 
ë ê 
ù 
û ú 
2
= 1 -
Ex
E0x
æ 
è 
ç ö 
ø 
÷ 
2é 
ë 
ê 
ù 
û 
ú sin2 d
Expanding the left-hand side and rearranging terms yields
Ex
E0x
æ 
è 
ç ö 
ø 
÷ 
2
+ 
Ey
E0y
æ 
è 
ç ö 
ø 
÷ 
2
- 2 
Ex
E0x
æ 
è 
ç ö 
ø 
÷ Ey
E0y
æ 
è 
ç ö 
ø 
÷ cos d = sin2 d
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 q1 = n2 cos q2
where n1 = 1, q1 = 33°, and q2 = 90° - 33° = 57°
\ n2 = 
 
cos 33°
cos 57°
= 1.540
(b) The critical angle is found from
nglass sin fglass = nair sin fair
4
with fair = 90° and nair = 1.0
\ fcritical = arcsin 
1
n glass
= arcsin 
1
1.540
= 40.5°
2-9
Air
Water
12 cm
r
q
Find qc from Snell's law n1 sin q1 = n2 sin qc = 1
When n2 = 1.33, then qc = 48.75°
Find r from tan qc = 
 
r
12 cm
 , which yields r = 13.7 cm.
2-10.
45 °
Using Snell's law nglass sin qc = nalcohol sin 90°
where qc = 45° we have
nglass = 
 
1.45
sin 45°
= 2.05
2-11. (a) Use either NA = n1
2 - n2
2( )1/ 2 = 0.242
or
5
NA » n1 2D = n1
2(n1 - n2)
n1
= 0.243
(b) q0,max = arcsin (NA/n) = arcsin 
0.242
1.0
æ 
è 
ö 
ø = 14°
2-13. NA = n1
2 - n2
2( )1/ 2 = n12 - n12(1- D)2[ ]
1/ 2
= n1 2D - D
2( )1 / 2
Since D << 1, D2 << D; \ NA » n1 2D
2-14. (a) Solve Eq. (2-34a) for jHf:
jHf = j 
ew
b
 Er - 
1
br
¶Hz
¶f
Substituting into Eq. (2-33b) we have
 j b Er + 
¶Ez
¶r
= wm j
ew
b
Er -
1
br
¶Hz
¶f
é 
ë ê 
ù 
û ú 
 
Solve for Er and let q2 = w2em - b2 to obtain Eq. (2-35a).
(b) Solve Eq. (2-34b) for jHr:
jHr = -j 
ew
b
Ef - 
1
b
¶Hz
¶r
Substituting into Eq. (2-33a) we have
 j b Ef + 
1
r
¶Ez
¶f
= -wm - j
ew
b
Ef -
1
b
¶Hz
¶r
é 
ë ê 
ù 
û ú 
 
Solve for Ef and let q2 = w2em - b2 to obtain Eq. (2-35b).
(c) Solve Eq. (2-34a) for jEr:
6
jEr = 
1
ew
1
r
¶Hz
¶f
+ jrbHf
æ 
è 
ç ö 
ø 
Substituting into Eq. (2-33b) we have
 
b
ew
1
r
¶Hz
¶f
+ jrbHf
æ 
è 
ç ö 
ø 
+ 
¶Ez
¶r
= jwm Hf
Solve for Hf and let q2 = w2em - b2 to obtain Eq. (2-35d).
(d) Solve Eq. (2-34b) for jEf
jEf = - 
1
ew
jbHr +
¶Hz
¶r
æ 
è 
ö 
ø Substituting into Eq. (2-33a) we have
1
r
¶Ez
¶f
- 
b
ew
jbHr +
¶Hz
¶r
æ 
è 
ö 
ø = -jwm Hr
Solve for Hr to obtain Eq. (2-35c).
(e) Substitute Eqs. (2-35c) and (2-35d) into Eq. (2-34c)
-
j
q 2
1
r
¶
¶r
b
¶Hz
¶f
+ ewr
¶Ez
¶r
æ 
è 
ç ö 
ø 
-
¶
¶f
b
¶Hz
¶r
-
ew
r
¶Ez
¶f
æ 
è 
ç ö 
ø 
é 
ë ê 
ù 
û ú 
= jewEz
Upon differentiating and multiplying by jq2/ew we obtain Eq. (2-36).
(f) Substitute Eqs. (2-35a) and (2-35b) into Eq. (2-33c)
- j
q2
1
r
¶
¶r
b
¶Ez
¶f
- mw r
¶Hz
¶r
æ 
è 
ç ö 
ø 
-
¶
¶f
b
¶Ez
¶r
+
mw
r
¶Hz
¶f
æ 
è 
ç ö 
ø 
é 
ë ê 
ù 
û ú 
= -jmwHz
Upon differentiating and multiplying by jq2/ew we obtain Eq. (2-37).
2-15. For n = 0, from Eqs. (2-42) and (2-43) we have
7
Ez = AJ0(ur) e
j(wt - bz) and Hz = BJ0(ur) e
j(wt - bz)
We want to find the coefficients A and B. From Eqs. (2-47) and (2-51),
respectively, we have
C = 
Jn (ua)
K n (wa)
 A and D = 
Jn (ua)
K n (wa)
 B
Substitute these into Eq. (2-50) to find B in terms of A:
A 
jbn
a
æ 
è 
ö 
ø 
1
u2
+
1
w2
æ 
è 
ö 
ø = Bwm 
J' n (ua)
uJ n (ua)
+
K' n (wa)
wKn(wa)
é 
ë ê 
ù 
û ú 
For n = 0, the right-hand side must be zero. Also for n = 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 2
B
jbn
a
J n(ua) + Awe 1uJ'n (ua)
é 
ë 
ù 
û 
 + 
1
w2
B
jbn
a
Jn(ua) + Awe2w
K' n (wa)Jn(ua)
K n(wa)
é 
ë ê 
ù 
û ú 
With k1
2 = w2me1 and k2
2 = w2me2 rewrite this as
Bn = 
 
ja
bwm
 
1
1
u 2
+ 1
w2
é 
ë 
ê 
ê 
ù 
û 
ú 
ú 
 k1
2Jn + k2
2K n[ ] A
8
where Jn and Kn are defined in Eq. (2-54). If for n = 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
D = 
n 1
2 - n2
2
2n1
2 = 
1
2
1 -
n2
2
n1
2
æ 
è 
ç ö 
ø 
÷ 
D << 1 implies n1 » n2
Thus using Eq. (2-46), which states that n2k = k2 £ b £ k1 = n1k, we have
 n2
2k2 = k2
2 » n1
2k2 = k1
2 » b2
2-17.
2-18. (a) From Eqs. (2-59) and (2-61) we have
M »
2p2a2
l2
n1
2 - n2
2( )= 2p
2a2
l2
NA( )2
a =
M
2p
æ 
è 
ö 
ø 
1/ 2 l
NA
=
1000
2
æ 
è 
ö 
ø 
1/ 2 0.85mm
0.2p
= 30.25mm
Therefore, D = 2a =60.5 mm
(b) M =
2p2 30.25mm( )2
1.32mm( )2
0.2( )2 = 414
(c) At 1550 nm, M = 300
2-19. From Eq. (2-58),
9
V = 
 
2p (25 mm)
0.82 mm
 (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
æ 
è 
ö 
ø 
total
» 
4
3
M-1/2 = 
4 ´100%3 1080
= 4.1%
at 820 nm. Similarly, (Pclad/P)total = 6.6% at 1320 nm and 7.8% at 1550 nm.
2-20 (a) At 1320 nm we have from Eqs. (2-23) and (2-57) that V = 25 and M = 312.
(b) From Eq. (2-72) the power flow in the cladding is 7.5%.
2-21. (a) For single-mode operation, we need V £ 2.40.
Solving Eq. (2-58) for the core radius a
 a = 
Vl
2p
n1
2 - n2
2( )-1/ 2 = 2.40(1.32mm)
2p (1.480)2 - (1.478)2[ ]1/ 2
= 6.55 mm
(b) From Eq. (2-23)
NA = n1
2 - n2
2( )1/ 2 = (1.480)2 - (1.478)2[ ]1/ 2 = 0.077
(c) From Eq. (2-23), NA = n sin q0,max. When n = 1.0 then
q0,max = arcsin 
NA
n
æ 
è 
ö 
ø = arcsin 
0.077
1.0
æ 
è 
ö 
ø = 4.4°
2-22. n2 = n1
2 - NA2 = (1.458)2 - (0.3)2 = 1.427
a = lV
2pNA
= 
(1.30)(75)
2p(0.3)
= 52 mm
10
2-23. For small values of D we can write V » 2pa
l
 n1 2D
For a = 5 mm we have D » 0.002, so that at 0.82 mm
V » 
 
2p (5 mm)
0.82 mm
 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 mm.
2-24.
2-25. From Eq. (2-77) Lp = 
2p
b
= 
l
n y - 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 £ ny - nx £ 1.3´ 10-5
2-26. We want to plot n(r) from n2 to n1. From Eq. (2-78)
n(r) = n1 1- 2D(r / a)
a[ ]1 / 2 = 1.48 1- 0.02(r / 25)a[ ]1 / 2
n2 is found from Eq. (2-79): n2 = n1(1 - D) = 1.465
2-27. From Eq. (2-81)
 
M =
a
a + 2
 a2k2n1
2D =
a
a + 2
 
2pan1
l
æ 
è 
ö 
ø 
2
D
where
D = 
n1 - n2
n1
= 0.0135
11
At l = 820 nm, M = 543 and at l = 1300 nm, M = 216.
For a step index fiber we can use Eq. (2-61)
Mstep » 
V 2
2
 = 
1
2
2pa
l
æ 
è 
ö 
ø 
2
n1
2 - n2
2( )
At l = 820 nm, Mstep = 1078 and at l = 1300 nm, Mstep = 429.
Alternatively, we can let a = ¥ in Eq. (2-81):
Mstep = 
2pan1
l
æ 
è 
ö 
ø 
2
D = 
 
1086 at 820 nm
432 at 1300 nm
ì 
í 
î 
2-28. Using Eq. (2-23) we have
(a) NA = n1
2 - n2
2( )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
St 
D
2
æ 
è 
ö 
ø 
2
= st 
d
2
æ 
è 
ö 
ø 
2
or s = S 
D
d
æ 
è 
ö 
ø 
2
(b) S = s 
d
D
æ 
è 
ö 
ø 
2
= 1.2 m/s 
 
0.125 mm
9 mm
æ 
è 
ö 
ø 
2
= 1.39 cm/min
12
2-30. Consider the following geometries of the preform and its corresponding fiber:
PREFORM
FIBER
4 mm
3 mm
R
25 mm 
62.5 mm
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
= 
 
A fiber core
Afiber clad
or
p(32 - R2 )
p(42 - 32 )
= 
 
p (25)2
p (62.5)2 - (25)2[ ]
from which we have
R = 9 -
7(25)2
(62.5)2 - (25)2
é 
ë ê 
ù 
û ú 
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-mm diameter fiber core is
V = pr2L = p (2.5´10-3 cm)2 (105 cm) = 1.96 cm3
The mass M equals the density r times the volume V:
M = rV = (2.6 gm/cm3)(1.96 cm3) = 5.1 gm
13
(b) If R is the deposition rate, then the deposition time t is
t = 
M
R
= 
 
5.1 gm
0.5 gm /min
= 10.2 min
2-32. Solving Eq. (2-82) for c yields
c = 
K
Ys
æ 
è 
ö 
ø 
2
where Y = p for surface flaws.
Thus
c = 
 
(20 N / mm3 / 2 )2
(70 MN / m2 )2 p
= 2.60´10-4 mm = 0.26 mm
2-33. (a) To find the time to failure, we substitute Eq. (2-82) into Eq. (2-86) and
integrate (assuming that s is independent of time):
 
 c- b / 2 
ci
c f
ò dc = AYbsb
0
t
ò dt
which yields
1
1 - b
2
cf
1- b / 2 - c i
1- b/ 2[ ] = AYbsbt
or
t = 
2
(b - 2)A(Ys)b
c i
(2- b) / 2 - cf
(2- b) / 2[ ]
(b) Rewriting the above expression in terms of K instead of c yields
t = 2
(b - 2)A(Ys)b
Ki
Ys
æ 
è 
ö 
ø 
2- b
- Kf
Ys
æ 
è 
ö 
ø 
2 -bé 
ë ê 
ù 
û ú 
 » 
2Ki
2- b
(b - 2)A(Ys)b
if K i
b- 2 << K f
b- 2 or K i
2 -b >> Kf
2- b
2-34. Substituting Eq. (2-82) into Eq. (2-86) gives
dc
dt
= AKb = AYbcb/2sb
14
Integrating this from c i to c p where
c i = 
K
Ys i
æ 
è 
ç ö 
ø 
÷ 
2
and c p = 
K
Ysp
æ 
è 
ç ö 
ø 
÷ 
2
are the initial crack depth and the crack depth after proof testing, respectively, yields
 
 c -b / 2 
c i
c p
ò dc = AYb
 
 sb 
0
t p
ò dt
or
 
1
1 - b
2
 cp
1- b / 2 - ci
1-b / 2[ ]= AYb s pb tp
for a constant stress sp. Substituting for c i and c p gives
2
b - 2
æ 
è 
ö 
ø 
K
Y
æ 
è 
ö 
ø 
2-b
s i
b- 2 - sp
b- 2[ ]= AYb s pb tp
or
2
b - 2
æ 
è 
ö 
ø 
K
Y
æ 
è 
ö 
ø 
2-b 1
AYb
s i
b- 2 - sp
b- 2[ ]= B s ib-2 - spb-2[ ]= s pb tp
which is Eq. (2-87).
When a static stress ss is applied after proof testing, the time to failure is found from Eq.
(2-86):
 
 c -b / 2 
c p
cs
ò dc = AYb 
 
ss
b 
0
t s
ò dt
where c s is the crack depth at the fiber failure point. Integrating (as above) we get Eq. (2-
89):
B sp
b-2 - ss
b-2[ ]= ssb ts
Adding Eqs. (2-87) and (2-89) yields Eq. (2-90).
15
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 -
L
L0
sp
btp + ss
bts( )/ B + ssb-2[ ]
m
b-2
s0
m -
sp
bt p / B + sp
b-2( )
m
b- 2
s0
m
ì 
í 
ï 
î ï 
ü 
ý 
ï 
þ ï 
æ 
è 
ç 
ç 
ö 
ø 
÷ 
÷ 
= 1 - exp
-
L
L0s0
m sp
btp / B + sp
b- 2[ ]
m
b -2
sp
bt p + ss
bts
B
æ 
è 
ç ö 
ø 
÷ + ss
b -2
sp
btp / B + sp
b- 2
é 
ë 
ê 
ê 
ê 
ù 
û 
ú 
ú 
ú 
m
b- 2
ì 
í 
ï 
ï 
î 
ï 
ï 
ü 
ý 
ï 
ï 
þ 
ï 
ï 
- 1
æ 
è 
ç 
ç 
ç 
ç 
ö 
ø 
÷ 
÷ 
÷ 
÷ 
= 1 - exp -LN p
1 +
ss
b ts
sp
btp
+
ss
sp
æ 
è 
ç ö 
ø 
÷ 
b
B
ss
2t p
é 
ë 
ê 
ù 
û 
ú 
m
b -2
1 + B
sp
2t p
-1
ì 
í 
ï 
ï 
î 
ï 
ï 
ü 
ý 
ï 
ï 
þ 
ï 
ï 
æ 
è 
ç 
ç 
ç 
ç 
ç 
ö 
ø 
÷ 
÷ 
÷ 
÷ 
÷ 
» 1 - exp -LN p 1+
ss
bts
sp
b tp
æ 
è 
ç ö 
ø 
÷ 1
1+ B
sp
2 tp
é 
ë 
ê 
ê 
ê 
ù 
û 
ú 
ú 
ú 
m
b-2
- 1
ì 
í 
ï ï 
î 
ï 
ï 
ü 
ý 
ï ï 
þ 
ï 
ï 
æ 
è 
ç 
ç 
ç ç 
ö 
ø 
÷ 
÷ 
÷ 
(b) For the term given by Eq. (2-96) we have
s s
sp
æ 
è 
ç ö 
ø 
÷ 
b
B
ss
2tp
= (0.3)15 
 
0.5 (MN /m2 )2 s
0.3 (350 MN / m2)[ ]210 s
= 6.5´10-14
Thus this term can be neglected.
16
2-36. The failure probability is given by Eq. (2-85). For equal failure probabilities of the
two fiber samples, F1 = F2, or
1 - exp-
s1cs0
æ 
è 
ç ö 
ø 
÷ 
m
L1
L0
é 
ë 
ê 
ù 
û 
ú = 1 - exp-
s2c
s0
æ 
è 
ç ö 
ø 
÷ 
m
L2
L0
é 
ë 
ê 
ù 
û 
ú 
which implies that
s1c
s0
æ 
è 
ç ö 
ø 
÷ 
m
L1
L0
= 
s2c
s0
æ 
è 
ç ö 
ø 
÷ 
m
L2
L0
or
s1c
s2c
= 
L2
L1
æ 
è 
ç ö 
ø 
÷ 
1 / m
If L1 = 20 m, then s1c= 4.8 GN/m2
If L2 = 1 km, then s2c= 3.9 GN/m2
Thus
4.8
3.9
æ 
è 
ö 
ø 
m
= 
1000
20
= 50
gives
m = 
log 50
log(4.8/ 3.9)
= 18.8
1
Problem Solutions for Chapter 3
3-1.
 
a dB/ km( )= 10
z
 log 
P(0)
P(z)
é 
ë ê 
ù 
û ú 
=
10
z
 log ea p
z( )
=10a p log e = 4.343 a p(1/ km)
3-2. Since the attenuations are given in dB/km, first find the power levels in dBm for
100 mW and 150 mW. These are, respectively,
P(100 mW) = 10 log (100 mW/1.0 mW) = 10 log (0.10) = - 10.0 dBm
P(150 mW) = 10 log (150 mW/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 mW
P1550(8 km) = - 10.0 dBm – (0.3 dB/km)(8 km) = - 12.4 dBm = 57.5 mW
(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 mW
P1550(20 km) = - 10.0 dBm – (0.3 dB/km)(20 km) = - 16.0 dBm = 25.1 mW
3-3. From Eq. (3-1c) with Pout = 0.45 Pin
a = (10/3.5 km) log (1/0.45) = 1.0 dB/km
3-4. (a) Pin = Pout 10aL/10 = (0.3 mW) 101.5(12)/10 = 18.9 mW
(b) Pin = Pout 10aL/10 = (0.3 mW) 102.5(12)/10 = 300 mW
3-5. With l in Eqs. (3-2b) and (3-3) given in mm, we have the following representative
points for auv and aIR:
2
ll (mmm) aauv aaIR
0.5 20.3 --
0.7 1.44 --
0.9 0.33 --
1.2 0.09 2.2´ 10-6
1.5 0.04 0.0072
2.0 0.02 23.2
3.0 0.009 7.5´ 104
3-6. From Eq. (3-4a) we have
ascat = 
8p3
3l4
(n2 -1)2 kBTfbT
= 
 
8p3
3(0.63 mm)4
(1.46)2 - 1[ ]2 (1.38´10-16 dyne-cm/K)(1400 K)
´(6.8´ 10-12 cm2/dyne)
= 0.883 km-1
To change to dB/km, multiply by 10 log e = 4.343: ascat = 3.8 dB/km
From Eq. (3-4b):
ascat = 
8p3
3l4
n8p2 kBTfbT = 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:
nnm Pclad/P aannm = aa11 + (aa2 2 - aa11)Pclad/P 5 + 103Pclad/P
01 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
3
3-11. (a) We want to solve Eq. (3-12) for agi. With a = 2 in Eq. (2-78) and letting
D = 
n 2 (0) - n2
2
2n2 (0)
we have
a(r) = a1 + (a2 - a1) 
n2 (0) - n2(r)
n 2(0) - n2
2 = a1 + (a2 - a1) 
r2
a2
Thus
agi = 
 
 a(r) p(r) r dr
0
¥
ò
 p(r) r dr
0
¥
ò
= a1 + 
 
(a2 - a1)
a2
 
 exp -Kr2( ) r3dr
0
¥
ò
 exp -Kr2( ) r dr
0
¥
ò
To evaluate the integrals, let x = Kr2, so that dx = 2Krdr. Then
 
 exp -Kr 2( ) r3dr
0
¥
ò
 exp -Kr 2( ) r dr
0
¥
ò
= 
 
1
2K2
 e- x x dx
0
¥
ò
1
2K
 e-x dx
0
¥
ò
= 
 
1
K
 1!
0!
= 
1
K
Thus agi = a1 + 
(a2 - a1)
Ka2
(b) p(a) = 0.1 P0 = P0 e
- Ka 2 yields eKa
2
 = 10.
From this we have Ka2 = ln 10 = 2.3. Thus
agi = a1 + 
(a2 - a1)
2.3
= 0.57a1 + 0.43a2
3-12. With l in units of micrometers, we have
4
n = 1 +
196.98
(13.4)2 - (1.24 / l)2
ì 
í 
î 
ü 
ý 
þ 
1/ 2
To compare this with Fig. 3-12, calculate three representative points, for example,
l = 0.2, 0.6, and 1.0 mm. Thus we have the following:
Wavelength ll Calculated n n from Fig. 3-12
0.2 mm 1.548 1.550
0.6 mm 1.457 1.458
1.0 mm 1.451 1.450
 
3-13. (a) From Fig. 3-13, 
dt
dl
» 80 ps/(nm-km) at 850 nm. Therefore, for the LED we
have from Eq. (3-20)
smat
L
= 
dt
dl
 sl = [80 ps/(nm-km)](45 nm) = 3.6 ns/km
For a laser diode,
smat
L
 = [80 ps/(nm-km)](2 nm) = 0.16 ns/km
(b) From Fig. 3-13, 
dtmat
dl
= 22 ps/(nm-km)
Therefore, Dmat(l) = [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
æ 
è 
ö 
ø 
2
= 1 - 
u2a2
u2a2 + w2a2
= 
w2
u 2 + w2
= 
b2 - k2n2
2
k2 n1
2 - b2 + b2 - k2 n2
2 = 
b2 / k2 - n2
2
n1
2 - n2
2
(b) Expand b as b = 
(b / k + n 2 )(b / k - n2 )
(n1 + n 2 )(n1 - n 2 )
Since n2< b/k < n1 , let b/k = n1(1 - d) where 0 < d < D << 1. Thus,
5
b / k + n2
n1 + n2
= 
n1(1 -d) + n2
n1 + n2
= 1 - 
n1
n 1 + n2
d
Letting n2 = n1(1 - D) then yields
b / k + n2
n1 + n2
= 1 - 
d
2 - D
» 1 since 
d
2 - D
 << 1
Therefore, b » 
b / k - n2
n1 - n2
or b = k[bn1D + n2]
From n2 = n1(1 - D) we have
n1 = n2(1 - D)-1 = n2(1 + D + D2 + ...) » n2(1 + D)
Therefore, b = k[b n2(1 + D)D + n2] » k n2(bD + 1)
3-16. The time delay between the highest and lowest order modes can be found from the
travel time difference between the two rays shown here.
The travel time of each ray is given by
sin f = 
x
s
 = 
n 2
n1
 = 
n 1(1 - D)
n1
 = (1 - D)
The travel time of the highest order ray is thus
Tmax = 
 
n1
c
 s 
L
x
æ 
è 
ö 
ø 
é 
ë ê 
ù 
û ú 
= 
n1L
c
1
1 - D
For the axial ray the travel time is Tmin = 
Ln1
c
jq a
s
x
6
Therefore
Tmin - Tmax = 
Ln1
c
1
1 - D
- 1æ è 
ö 
ø = 
Ln1
c
D
1 - D
» 
Ln1D
c
3-17. Since n2 = n1 1 - D( ), we can rewrite the equation as
smod
L
=
n1D
c
1 -
p
V
æ 
è 
ö 
ø 
where the first tern is Equation (3-30). The difference is then given by the factor
1 -
p
V
= 1 -
pl
2a
1
n1
2 - n2
2( )1/ 2
» 1 -
pl
2a
1
n1 2D
At 1300 nm this factor is 1 -
p(1.3)
2 62.5( )
1
1.48 2(0.015)
=1 - 0.127 = 0.873
3-18. For e = 0 and in the limit of a ® ¥ we have
C1 = 1, C2 = 
3
2
, 
a
a + 1
= 1, 
a + 2
3a + 2
= 
1
3
, 
a +1
2a +1
= 1
2
,
and 
(a + 1)2
(5a + 2)(3a + 2)
=
1
15
Thus Eq. (3-41) becomes
s int er mod al = 
Ln1D
2 3c
1+ 3D +
12
5
D2æ è 
ö 
ø 
1/ 2
» 
Ln1D
2 3c
3-19. For e = 0 we have that a = 2(1 - 
6
5
D). Thus C1 and C2 in Eq. (3-42) become
(ignoring small terms such as D3, D4, ...)
C1 = 
a - 2
a + 2
 
=
2 1 -
6
5
Dæ è 
ö 
ø - 2
2 1 - 6
5
Dæ è 
ö 
ø + 2
=
- 3
5
D
1- 3
5
D
» - 3
5
D 1+
3
5
Dæ è 
ö 
ø 
7
C2 = 
3a - 2
2(a + 2)
=
3 2 1 -
6
5
Dæ è 
ö 
ø 
é 
ë ê 
ù 
û ú 
- 2
2 2 1- 6
5
Dæ è 
ö 
ø + 2
é 
ë ê 
ù 
û ú 
=
1 -
9
5
D
2 1 - 3
5
Dæ è 
ö 
ø 
Evaluating the factors in Eq. (3-41) yields:
(a) C1
2 » 
9
25
 D2
(b)
4C1C2 (a + 1)D
2a +1
= 
4D -
3
5
Dæ è 
ö 
ø 1 -
9
5
Dæ è 
ö 
ø 2 1-
6
5
Dæ è 
ö 
ø + 1
é 
ë ê 
ù 
û ú 
1 - 3
5
Dæ è 
ö 
ø 2 1-
3
5
Dæ è 
ö 
ø 4 1 -
6
5
Dæ è 
ö 
ø + 1
é 
ë ê 
ù 
û ú 
= 
-
18D
1 -
11
D +
18
25
æ 
è 
ö 
ø 
5 1
5
Dæ è 
ö 
ø 
2 24
25
Dæ ö ø 
18
25
 D2
(c) 
16D2C2
2 (a + 1)2
(5a + 2)(3a + 2)
= 
16D2 1-
9
5
Dæ è 
ö 
ø 
2
2(1 -
6
5
D) +1é 
ë 
ù 
û 
2
4 1 - 3
5
Dæ è 
ö 
ø 
2
10(1 - 6
5
D) + 2é 
ë 
ù 
û 
6(1 - 6
5
D) + 2é 
ë 
ù 
û 
= 
16D2 1-
9
5
Dæ è 
ö 
ø 
2
9 1 -
4
5
Dæ è 
ö 
ø 
2
96(1 - D)(1 - 9
10
D)4 1- 3
5
Dæ è 
ö 
ø 
2 » 
9
24
 D2
Therefore,
s int er mod al = 
Ln1D
2c
a
a +1
æ 
è 
ö 
ø 
a + 2
3a + 2
æ 
è 
ö 
ø 
1/ 2 925
D2 - 18
25
D2 + 9
24
D2æ è 
ö 
ø 
1/ 2
= 
Ln1D
2
2c
2(1 -
6
5
D)
2(1 - 6
5
D) +1é 
ë 
ù 
û 
2(1-
6
5
D) + 2
6(1- 6
5
D) + 2
é 
ë 
ê 
ê 
ù 
û 
ú 
ú 
1/ 2
3
10 6
» 
n1D
2 L
20 3c
8
3-20. We want to plot Eq. (3-30) as a function of sl , where s int er mod al and
s int ra mod al are given by Eqs. (3-41) and (3-45). For e = 0 and a = 2, we have C1 =
0 and C2 = 1/2. Since s int er mod al does not vary with sl , we have
s int er mod al
L
= 
N1D
2c
a
a + 2
æ 
è 
ö 
ø 
a + 2
3a + 2
æ 
è 
ö 
ø 
1/ 2 4DC2(a + 1)
(5a + 2)(3a + 2)
= 0.070 ns/km
With C1 = 0 we have from Eq. (3-45)
s int ra mod al = 
1
c
sl
l
-l2
d2n1
dl2
æ 
è 
ç ö 
ø 
= 
 
0.098sl ns / km at 850 nm
1.026 ´10-2 sl ns / km at 1300 nm
ì 
í 
î 
3-21. Using the same parameter values as in Prob. 3-18, except with D = 0.001, we have
from Eq. (3-41) s int er mod al /L = 7 ps/km, and from Eq. (3.45)
s int ra mod al
L
= 
 
0.098sl ns / km at 850 nm
0.0103sl ns / km at 1300 nm
ì 
í 
î 
The plot of 
s
L
= 
1
L
s int er
2 + s int ra
2( )1/ 2 vs sl :
3-22. Substituting Eq. (3-34) into Eq. (3-33)
tg = 
L
c
db
dk
= 
L
c
1
2b
2kn1
2 + 2k2n1
dn1
dk
ì í î 
- 2
a + 2
a
m
a 2
æ 
è 
ö 
ø 
a
a+ 2 2
a + 2
n1
2k2D( )
2
a+2
-1
´ D 2k2n1
dn 1
dk
+ 2kn1
2 +
n1
2k2
D
dD
dk
é 
ë ê 
ù 
û ú 
ü 
ý 
þ 
= 
L
c
kn1
b
N1 -
4D
a + 2
a + 2
a
m
a2
1
n1
2k2D
æ 
è 
ç ö 
ø 
÷ 
a
a +2
N1 +
n1k
2D
dD
dk
æ 
è 
ö 
ø 
é 
ë 
ê 
ê 
ù 
û 
ú 
ú 
9
= 
LN1
c
kn1
b
1-
4D
a + 2
m
M
æ 
è 
ö 
ø 
a
a +2
1 +
e
4
æ 
è 
ö 
ø 
é 
ë 
ê 
ù 
û 
ú 
with N1 = n1 + k 
dn1
dk
and where M is given by Eq. (2-97) and e is defined in Eq.
(3-36b).
3-23. From Eq. (3-39), ignoring terms of order D2,
l
dt
dl
 = 
L
c
dN1
dl
1 +
a - e - 2
a + 2
m
M
æ 
è 
ö 
ø 
a
a+ 2
é 
ë 
ê 
ù 
û 
ú 
+ 
LN1
c
a - e - 2
a + 2
d
dl
D
m
M
æ 
è 
ö 
ø 
a
a +2
é 
ë 
ê 
ù 
û 
ú 
where
N1 = n1 - l 
dn1
dl
and M = 
a
a + 2
a2k2n1
2 D
(a)
dN1
dl
= 
d
dl
n1 - l
dn1
dl
æ 
è 
ö 
ø = - l 
d2n1
dl2
Thus ignoring the term involving D 
d2n1
dl2
 , the first term in square brackets
becomes - 
L
c
l2 
d2n1
dl2
(b)
d
dl
D
m
M
æ 
è 
ö 
ø 
a
a+ 2é 
ë 
ê 
ù 
û 
ú = m
a
a+ 2 dD
dl
1
M
æ 
è 
ö 
ø 
a
a +2
+ D
-a
a + 2
dM
dl
1
M
æ 
è 
ö 
ø 
a
a +2
+1é 
ë 
ê 
ù 
û 
ú 
(c)
dM
dl
= 
a
a + 2
a2 
d
dl
k2n1
2 D( )
= 
a
a + 2
 a2 
dD
dl
k2n1
2 + 2k2Dn1
dn1
dl
+ 2kn1
2D
dk
dl
æ 
è 
ö 
ø 
10
Ignoring 
dD
dl
and 
dn1
dl
terms yields
dM
dl
= 
2a
a + 2
a2k2n1
2 D -
1
l
æ 
è 
ö 
ø = - 
2M
l
so that
d
dl
D
m
M
æ 
è 
ö 
ø 
a
a+ 2
é 
ë 
ê 
ù 
û 
ú = 
D
l
2a
a + 2
m
M
æ 
è 
ö 
ø 
a
a +2
. Therefore
l
dt
dl
 = - 
L
c
l2 
d2n1
dl2
+ 
LN1
c
a - e - 2
a + 2
2aD
a + 2
m
M
æ 
è 
ö 
ø 
a
a +2
3-24. Let a = l2 
d2n1
dl2
; b = N1C1D 
2a
a + 2
; g = 
a
a + 2
Then from Eqs. (3-32), (3-43), and (3-44) we have
s int ra mod al
2 = 
 
L2
sl
l
æ 
è 
ö 
ø 
2 1
M
 l 
dtg
dl
æ 
è 
ö 
ø 
m= 0
M
å
2
= 
 
L
c
æ 
è 
ö 
ø 
2 sl
l
æ 
è 
ö 
ø 
2
1
M
 -a + b 
m
M
æ 
è 
ö 
ø 
gé 
ë ê 
ù 
û ú m= 0
M
å
2
» 
L
c
æ 
è 
ö 
ø 
2 sl
l
æ 
è 
ö 
ø 
2
1
M
-a + b
m
M
æ 
è 
ö 
ø 
gé 
ë ê 
ù 
û ú 
2
dm
0
M
ò
= 
 
L
c
æ 
è 
ö 
ø 
2 sl
l
æ 
è 
ö 
ø 
2
 
1
M
 a 2 - 2ab 
m
M
æ 
è 
ö 
ø 
g
+ b2
m
M
æ 
è 
ö 
ø 
2gé 
ë ê 
ù 
û ú 
 dm
0
M
ò
= 
L
c
æ 
è 
ö 
ø 
2 sl
l
æ 
è 
ö 
ø 
2
a2 -
2ab
g + 1
+
b2
2g + 1
é 
ë ê 
ù 
û ú 
= 
L
c
æ 
è 
ö 
ø 
2 sl
l
æ 
è 
ö 
ø 
2
-l2 d
2n1
dl2
æ 
è 
ç ö 
ø 
2é 
ë 
ê 
11
- 2 l2
d2n1
dl2
æ 
è 
ç ö 
ø 
N1C1D 
a
a + 1
+ (N1C1D)
2 4a2
(a + 2)(3a + 2)
ù 
û ú 
3-25. Plot of Eq. (3-57).
3-26. (a) 
 
D = l - l0( )S0 = -50(0.07) = -3.5 ps /(nm - km)
(b) 
 
D =
1500(0.09)
4
 1 -
1310
1500
æ 
è 
ö 
ø 
4é 
ë ê 
ù 
û ú 
= 14.1 ps /(nm - km)
3-27. (a) From Eq. (3-48)
 
s step
L
= n1
D
2 3 c
= 1.49(0.01)
2 3 (3 ´ 108 )
=14.4 ns / km
(b) From Eq. (3-47)
 
sopt
L
=
n1D
2
20 3 c
=
1.49(0.01)2
20 3 (3 ´108 )
=14.3 ps / km
(c) 3.5 ps/km
3-28. (a) From Eq. (3-29)
 
smod = Tmax - Tmin =
n1DL
c
=
(1.49)(0.01)(5 ´103 m)
3 ´ 108 m / s
= 248 ns
(b) From Eq. (3-48)
 
sstep =
n1DL
2 3 c
=
248
2 3
= 71.7 ns
(c) 
 
BT =
0.2
sstep
= 2.8 Mb / s
(d) BT × L = (2.8 MHz)(5 km) =13.9 MHz × km
12
3-29. For a = 0.95aopt , we have
s int er (a ¹ aopt )
s int er (a = aopt )
=
(a - a opt)
D(a + 2)
= -
0.05
(0.015)(1.95)
= -170%
For a =1.05aopt , we have
s int er (a ¹ aopt )
s int er (a = aopt )
=
(a - a opt )
D(a + 2)
= +
0.05
(0.015)(2.05)
= +163%
1
Problem Solutions for Chapter 4
4-1. From Eq. (4-1), ni = 2 è
ç
æ
ø
÷
ö2pkBT
h2
3/2
 (memh)
3/4
 exp 
è
ç
æ
ø
÷
ö
- 
Eg
2kBT
 
= 2 
2p(1.38 ´10-23 J / K)
(6.63 ´10 -34 J.s)2
é 
ë ê 
ù 
û ú 
3/ 2
T 3 / 2 (.068)(.56)(9.11´10 -31 kg)2[ ]3/ 4
´ exp -
(1.55 - 4.3 ´ 10-4 T)eV
2(8.62 ´10-5 eV / K)T
é 
ë ê 
ù 
û ú 
= 4.15´1014 T
3/2
 exp -
1.55
2(8.62 ´10 -5 )T
é 
ë ê 
ù 
û ú 
exp 
4.3 ´10 -4
2(8.62 ´10-5 )
é 
ë ê 
ù 
û ú 
= 5.03´1015 T
3/2
 exp èç
æ
ø÷
ö- 
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
2
i /pP or p
2
P - NApP - n
2
i = 0
so that
pP = 
NA
2
èç
çæ
ø÷
÷ö1 + 
4n
2
i
N
2
A
 + 1 
If ni << NA , which is generally the case, then to a good approximation
pP » NA and nP = n
2
i /pP » n
2
i /NA 
4-3. (a) From Eq. (4-4) we have 1.540 = 1.424 + 1.266x + 0.266x2 or
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 l = 
1.240
1.540 = 805 nm.
(b) Eg = 1.424 + 1.266(0.15) + 0.266(0.15)2 = 1.620 eV, so that
l = 
1.240
1.620 = 766 nm
4-4. (a) The lattice spacings are as follows:
a(BC) = a(GaAs) = 5.6536 A
o
a(BD) = a(GaP) = 5.4512 A
o
a(AC) = a(InAs) = 6.0590 A
o
a(AD) = a(InP) = 5.8696 A
o
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 A
o
 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
3
DE =
hc
l2
Dl or Dl =
l2
hc
DE
For the same energy difference DE, the spectral width Dl is proportional to the
wavelength squared. Thus, for example,
Dl1550
Dl1310
= 1550
1310
æ 
è 
ö 
ø 
2
= 1.40
4-6. (a) From Eq. (4-10), the internal quantum efficiency is
hint =
1
1 + 25/ 90
= 0.783, and from Eq. (4-13) the internal power level is
 
Pint = (0.783)
hc(35 mA)
q(1310 nm)
= 26 mW
(b) From Eq. (4-16),
 
P =
1
3.5 3.5 + 1( )2
26 mW = 0.37 mW
4-7. Plot of Eq. (4-18). Some representative values of P/P0 are given in the table:
f in MHz P/P0
1 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 at
which the expression is equal to -3; that is,
10 log
1
1 + 2pft( )2[ ]1/ 2
æ 
è 
ç 
ö 
ø 
÷ = -3
4
With a 5-ns lifetime, we find 
 
f =
1
2p 5 ns( )
100.6 - 1( )= 9.5 MHz
4-9. (a) Using Eq. (4-28) with G = 1
gth = 
 
1
0.05 cm
ln èç
æ
ø÷
ö1
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 ëê
é
ûú
ù1
0.9(0.32) + 10 cm
-1 = 34.9 cm-1
(c) From Eq. (4-37) hext = hi (gth - a )/gth ;
thus for case (a): hext = 0.65(55.6 - 10)/55.6 = 0.53
For case (b): hext = 0.65(34.9 - 10)/34.9 = 0.46
4-10. Using Eq. (4-4) to find Eg and Eq. (4-3) to find l, we have for x = 0.03,
l = 
1.24
Eg
 = 
1.24
1.424 + 1.266(0.3) + 0.266(0.3)2
 = 1.462 mm
From Eq. (4-38)
hext = 0.8065 l(mm) 
dP(mW)
dI(mA) 
Taking dI/dP = 0.5 mW/mA, we have hext = 0.8065 (1.462)(0.5) = 0.590
4-11. (a) From the given values, D = 0.74, so that GT = 0.216
Then n eff
2 = 10.75 and W = 3.45, yielding GL = 0.856
(b) The total confinement factor then is G = 0.185
4-12. From Eq. (4-46) the mode spacing is
Dl = 
l2
2Ln = 
(0.80 mm)2
2(400 mm)(3.6) = 0.22 nm
Therefore the number of modes in the range 0.75-to-0.85 mm is
5
.85 - .75
.22 ´ 10-3
= 
.1
.22 ´ 10
3 = 455 modes
4-13. (a) From Eq. (4-44) we have g(l) = (50 cm-1) exp 
ë
ê
é
û
ú
ù
- 
(l - 850 nm)2
2(32 nm)2
 
= (50 cm-1) exp ëê
é
ûú
ù
- 
(l - 850)2
2048 
(b) On the plot of g(l) versus l, drawing a horizontal line at g(l) = at
= 32.2 cm-1 shows that lasing occurs in the region 820 nm < l < 880 nm.
(c) From Eq. (4-47) the mode spacing is
Dl = 
l2
2Ln = 
(850)2
2(3.6)(400 mm) = 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/l = 
m
2L be the wave number (reciprocal wavelength) of mode m.
The difference DN between adjacent modes is then
DN = Nm - Nm-1 = 
1
2L (a-1)
We now want to relate DN to the change Dl in the free-space wavelength. First
differentiate N with respect to l:
dN
dl = 
d
dl èç
æ
ø÷
ön
l = 
1
l 
dn
dl - 
n
l2 = - 
1
l2 èç
æ
ø÷
ön - l 
dn
dl 
Thus for an incremental change in wavenumber DN, we have, in absolute values,
DN = 
1
l2 èç
æ
ø÷
ön - l 
dn
dl Dl (a-2)
6
Equating (a-1) and (a-2) then yields Dl = 
l2
2L
èç
æ
ø÷
ön - l 
dn
dl
 
(b) The mode spacing isDl = 
(.85 mm)2
2(4.5)(400 mm) = 0.20 nm
4-15. (a) The reflectivity at the GaAs-air interface is
R1 = R2 = èç
æ
ø÷
ön-1
n+1
2
 = èç
æ
ø÷
ö3.6-1
3.6+1
2
 = 0.32
Then Jth = 
1
b
a +
1
2L
ln
1
R1R2
é 
ë ê 
ù 
û ú 
= 2.65´103 A/cm2
Therefore
Ith = Jth ´ l ´ w = (2.65´103 A/cm2)(250´10-4 cm)(100´10-4 cm) = 663 mA
(b) Ith = (2.65´103 A/cm2)(250´10-4 cm)(10´10-4 cm) = 66.3 mA
4-16. From the given equation
 
DE11 =1.43 eV +
6.6256 ´ 10-34 J × s( )2
8 5 nm( )2
1
6.19 ´ 10-32 kg
+ 1
5.10 ´10-31 kg
æ 
è 
ç ö 
ø 
=1.43 eV + 0.25 eV =1.68 eV
Thus the emission wavelength is l = 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 = 
mlB
2L = 
2(1570 nm)
2(460 nm) = 3.4
Then, from Eq. (4-48b), for m = 0
7
l = lB ± 
l
2
B
2neL
 èç
æ
ø÷
ö1
2 = 1570 nm ± 
(1.57 mm)(1570 nm)
4(3.4)(300 mm) = 1570 nm ± 1.20 nm
Therefore for m = 1, l = lB ± 3(1.20 nm) = 1570 nm ± 3.60 nm
For m = 2, l = lB ± 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.
 
t d = dt
0
t d
ò = 
1
J
qd
- n
t
 
0
n th
ò dn = -t 
J
qd
-
n
t
æ 
è 
ç ö 
ø 
 
n= 0
n= nth
= t ln 
J
J - J th
æ 
è 
ç ö 
ø 
÷ 
where J = Ip/A and Jth = Ith/A. Therefore td = t ln è
ç
æ
ø
÷
öIp
Ip - Ith
 
(b) At time t = 0 we have n = nB, and at t = td we have n = nth. Therefore,
td = õó
0
td
 dt = 
 
 
1
J
qd
- n
t
 dn
nB
n th
ò = t ln 
J
qd
- nB
t
J
qd
- n th
t
æ 
è 
ç 
ç 
ö 
ø 
÷ 
÷ 
In the steady state before a pulse is applied, nB = JBt/qd. When a pulse is applied,
the current density becomes I/A = J = JB + Jp = (IB + Ip)/A
Therefore, td = t ln 
è
ç
æ
ø
÷
öI - IB
I - I
th
 = t ln 
è
ç
æ
ø
÷
öIp
Ip + IB - Ith
 
4-20. A common-emitter transistor configuration:
4-21. Laser transmitter design.
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] ³ 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 = 
1
1.084 = 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 w1t + a1b2 cos w2t
+ a2(b1
2 cos2 w1t + 2b1b2 cos w1t cos w2t + b2
2 cos2 w2t)
+ a3(b1
3 cos3 w1t + 3b1
2 b2 cos2 w1t cos w2t + 3b1b2
2 cos w1t cos2 w2t+ b2
3 cos3 w2t)
+a4(b1
4 cos4 w1t + 4b1
3 b2 cos3 w1t cos w2t + 6b1
2 b2
2 cos2 w1t cos2 w2t
+ 4b1b2
3 cos w1t cos3 w2t + b2
4 cos4 w2t)
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)]
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) = 
1
2 ëê
é
ûú
ùa2b
2
1 + a2b
2
2 + 
3
4 a4b
4
1 + 3a4b
2
1b
2
2 + 
3
4 a4b
4
2 
constant
terms 
+ 
3
4ë
é
û
ùa3b
3
1 + 2a3b1b
2
2 cos w1t + 
3
4 a3ë
é
û
ùb
3
2 + 2b
2
1b2 cos w2t 
fundamental
terms 
+ 
b
2
1
2 ë
é
û
ùa2 + a4b
2
1 + 3a4b
2
2 cos 2w1t + 
b
2
2
2 ë
é
û
ùa2 + a4b
2
2 + 3a4b
2
1 cos 2w2t
2nd-order harmonic terms
+ 
1
4 a3b
3
1 cos 3w1t + 
1
4 a3b
3
2 cos 3w2t 3rd-order harmonic terms
+ 
1
8 a4b
4
1 cos 4w1t + 
1
8 a4b
4
2 cos 4w2t 4th-order harmonic terms
+ ë
é
û
ùa2b1b2 + 
3
2 a4b
3
1b2 + 
3
2 a4b1b
3
2 [ ]cos (w1+w2)t + cos (w1-w2)t 
2nd-order intermodulationterms
+ 
3
4 a3b
2
1 b2 [ ]cos(2w1+w2)t + cos(2w1-w2)t + 
3
4 a3b1b
2
2 [ ]cos(2w2+w1)t + cos(2w2-w1)t 
3rd-order intermodulation terms
+ 
1
2 a4b
3
1 b2 [ ]cos(3w1+w2)t + cos(3w1-w2)t 
+ 
3
4 a4b
2
1 b
2
2 [ ]cos(2w1+2w2)t + cos(2w1-2w2)t 
+ 
1
2 a4b1b
3
2 [ ]cos(3w2+w1)t + cos(3w2-w1)t 4th-order intermodulation
terms
This output is of the form
y(t) = A0 + A1(w1) cos w1t + A2(w1) cos 2w1t + A3(w1) cos 3w1t
+ A4(w1) cos 4w1t + A1(w2) cos w2t + A2(w2) cos 2w2t
10
+ A3(w2) cos 3w2t + A4(w2) cos 4w2t + å
m
 
 å
n
 
 Bmn cos(mw1+nw2)t 
where An(wj) is the coefficient for the cos(nwj)t term.
4-24. From Eq. (4-58) P = P0 e
-t/tm where P0 = 1 mW and tm = 2(5´104 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 = 5´8760 hours = 43,800 hours. Therefore
P(5 years) = (1 mW) exp(-43800/105) = 0.65 mW
4-25. From Eq. (4-60) ts = K e
EA/kBT or ln ts = ln K + EA/kBT
where kB = 1.38´10-23 J/°K = 8.625´10-5 eV/°K
At T = 60°C = 333°K, we have
ln 4´ 104 = ln K + EA/[(8.625´10-5 eV)(333)]
or 10.60 = ln K + 34.82 EA (1)
At T = 90°C = 363°K, we have
ln 6500 = ln K + EA/[(8.625´10-5 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 = 1.11´10-5 hrs
Thus at T = 20°C = 293°K
11
t
s
 = 1.11´10-5 exp{0.63/[(8.625´10-5)(293)]} = 7.45´105 hrs
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 = p2r
2
s B0(NA)2 = p2(2´ 10-3 cm)2(100 W/cm2)(.22)2 = 191 mW
From Eq. (5-9)
PLED-graded = 2p2(2´ 10-3 cm)2(100 W/cm2)(1.48)2(.01)ëê
é
ûú
ù
1 - 
1
2 è
æ
ø
ö2
5
2
 = 159 mW
5-5. Using Eq. (5-10), we have that the reflectivity at the source-to-gel interface is
R s- g =
3.600 -1.305
3.600 +1.305
æ 
è 
ö 
ø 
2
= 0.219
Similarly, the relfectivity at the gel-to-fiber interface is
R g- f =
1.465 -1.305
1.465 +1.305
æ 
è 
ö 
ø 
2
= 3.34 ´ 10-3
The total reflectivity then is R = R s- gR g-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(q) = B0 cosm q into Eq. (5-3) for B(q,f), we have
2
P = 
õô
ôó
0
rm
 
õ
ô
ó
0
2p
 
ë
ê
é
û
ú
ù
 2p õó
0
q0-max
 cos3 q sin q dq dqs r dr 
Using
õó
0
q0
 cos3 q sin q dq = õó
0
q0
 ( )1 - sin2 q sin q d(sin q) = õó
0
sin q0
 ( )x - x3 dx 
we have
P = 2p 
õô
ôó
0
rm
 
õ
ô
ó
0
2p
 ë
ê
é
û
ú
ùsin2 q0-max
2 - 
sin4 q0-max
4 dqs r dr 
= p 
õô
ôó
0
rm
 
õô
ó
0
2p
 ëê
é
ûú
ùNA2 - 
1
2 NA
4 dqs r dr 
= 
p
2 [ ]2NA
2 - NA4 õó
0
rm
 r drõó
0
2p
 dqs 
5-7. (a) Let a = 25 mm and NA = 0.16. For rs ³ a(NA) = 4 mm, Eq. (5-17) holds. For
rs £ 4 mm, h = 1.
(b) With a = 50 mm and NA = 0.20, Eq. (5-17) holds for rs ³ 10 mm. Otherwise, h
= 1.
5-8. Using Eq. (5-10), the relfectivity at the gel-to-fiber interface is
3
R g- f =
1.485 -1.305
1.485 +1.305
æ 
è 
ö 
ø 
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
æ 
è 
ö 
ø 
2
= 0.038
The power loss is L = -10 log (1 - R) = -10 log (0.962) = 0.17 dB
5-9. Shaded area = (circle segment area) - (area of triangle) = 
1
2 sa - 
1
2 cy
s = aq = a [2 arccos (y/a)] = 2a arccos èç
æ
ø÷
öd
2a 
c = 2 
ëê
é
ûú
ùa2 - èç
æ
ø÷
öd
2
2 1/2
 
Therefore
Acommon = 2(shaded area) = sa – cy = 2a2 arccos èç
æ
ø÷
öd
2a - d ëê
é
ûú
ù
a2 - èç
æ
ø÷
öd2
4
1/2
 
5-10.
Coupling loss (dB) for
Given axial misalignments (mmm)
Core/cladding diameters
(mmm)
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 = 
p
2 - arcsin x
For small values of x,arcsin x = x + 
x3
2(3) + 
x5
2(4)(5) + ...
4
Therefore, for 
d
2a << 1, we havearccos 
d
2a » 
p
2 - 
d
2a 
Thus Eq. (5-30) becomes PT = 
2
p Pèç
æ
ø÷
öp
2 - 
d
2a - 
5d
6a = P èç
æ
ø÷
ö1 - 
8d
3pa 
d/a PT/P (Eq.5-30) PT/P (Eq.5-31)
0.00 1.00 1.00
0.05 0.9576 0.9576
0.10 0.9152 0.9151
0.15 0.8729 0.8727
0.20 0.8309 0.8302
0.25 0.7890 0.7878
0.30 0.7475 0.7454
0.35 0.7063 0.7029
0.40 0.6656 0.6605
5-12. Plots of mechanical misalignment losses.
5-13. From Eq. (5-20) the coupling efficiency h
F
 is given by the ratio of the number of
modes in the receiving fiber to the number of modes in the emitting fiber, where
the number of modes M is found from Eq. (5-19). Therefore
h
F
 = 
MaR
MaE
 = 
k2NA2(0) 
èç
æ
ø÷
ö1
2 -
1
a+2a
2
R
k2NA2(0) 
èç
æ
ø÷
ö1
2 -
1
a+2a
2
E
 = 
a
2
R
a
2
E
 
Therefore from Eq. (5-21) the coupling loss for aR £ aE is LF = -10 log 
è
ç
ç
æ
ø
÷
÷
öa
2
R
a
2
E
 
5-14. For fibers with different NAs, where NAR < NAE
LF = -10 log hF = -10 log 
MR
ME
 = -10 log 
k2NA
2
R(0) è
ç
æ
ø
÷
öa
2a+4a
2
k2NA
2
E(0) è
ç
æ
ø
÷
öa
2a+4a
2
 
5
= -10 log 
ë
ê
ê
é
û
ú
ú
ùNA
2
R(0)
NA
2
E(0)
 
5-15. For fibers with different a values, where a
R
 < a
E
 
LF = -10 log hF = -10 log 
k2NA2(0) 
è
ç
ç
æ
ø
÷
÷
öaR
2a
R
 + 4 a
2
k2NA2(0) 
è
ç
ç
æ
ø
÷
÷
öaE
2a
E
 + 4 a
2
 = -10 log 
ë
ê
ê
é
û
ú
ú
ùaR(aE + 2)
a
E
(a
R
 + 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) = n1 2D = 1.46 2 0.01( ) = 0.206
For fiber 2: NA2 (0) = n1 2D = 1.48 2 0.015( ) = 0.256
(a) The only loss is that from index-profile differences. From Eq. (5-37)
 
L1® 2 a( ) = -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
æ 
è 
ö 
ø = 1.94 dB
 
L2®1(NA) = -20 log 
.206
.256
é 
ë 
ù 
û 
= 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
6
LSM;ff = -10 log 
ë
ê
é
û
ú
ù4
è
ç
æ
ø
÷
öW1
W2
 + 
W2
W1
2 
For W1 = 0.9W2 , we then have LSM;ff = -10 log ëê
é
ûú
ù4
4.0446 = - 0.0482 dB
5-19. Plot of Eq. (5-44).
5-20. Plot of the throughput loss.
1
Problem Solutions for Chapter 6
6-1. From Eqs. (6-4) and (6-5) with Rf = 0, h = 1 - exp(-asw)
To assist in making the plots, from Fig. P6-1, we have the following representative
values of the absorption coefficient:
ll (mmm) aas (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 õó
0
w
 G(x) dx = qA F0 as õ
ó
0
w
 e
-asx
 dx 
= qA F0 ë
é
û
ù
1 - e
-asw
 = qA 
P0(1 - Rf)
hnA ë
é
û
ù
1 - e
-asw
 
6-3. From Eq. (6-6), R = 
hq
hn = 
hql
hc = 0.8044 hl ( in mm)
Plot R as a function of wavelength.
6-4. (a) Usingthe fact that Va » VB, rewrite the denominator as
1 - 
è
ç
æ
ø
÷
öVa - IMRM
VB
n
 = 1 - 
è
ç
æ
ø
÷
öVB - VB + Va - IMRM
VB
n
 
= 1 - 
è
ç
æ
ø
÷
ö
1 - 
VB - Va + IMRM
VB
n
 
Since 
VB - Va + IMRM
VB
 << 1, we can expand the term in parenthesis:
2
1 - 
è
ç
æ
ø
÷
ö
1 - 
VB - Va + IMRM
VB
n
 » 1 - 
ë
ê
é
û
ú
ù
1 - 
n( )VB - Va + IMRM
VB
 
= 
n( )VB - Va + IMRM
VB
 » 
nIMRM
VB
 
Therefore, M0 = 
IM
Ip
 » 
VB
n( )VB - Va + IMRM
 » 
VB
nIMRM
 
(b) M0 = 
IM
Ip
 = 
VB
nIMRM
 implies I
2
M = 
IpVB
nRM
 , so that M0 = è
ç
æ
ø
÷
öVB
nIpRM
1/2
 
6-5. < > i2s(t) = 1T õó
0
T
 i
2
s(t)dt = 
w
2p õ
ó
0
2p/w
 R
2
0 P2(t) dt (where T = 2p/w),
= 
w
2p R
2
0 P
2
0 õó
0
2p/w
 (1 + 2m cos wt + m2 cos2 wt) dt 
Using
 
 cos wt dt =
0
2 p / w
ò
1
w
 sin wt t = 0
t = 2p / w = 0
and õó
0
2p/w
 cos2 wt dt = 
1
w õô
ó
0
2p
 èç
æ
ø÷
ö1
2 + 
1
2cos 2x dx = 
p
w 
we have < > i2s(t) = R
2
0 P
2
0 èç
æ
ø÷
ö
1 + 
m2
2 
6-6. Same problem as Example 6-6: compare Eqs. (6-13), (6-14), and (6-17).
(a) First from Eq. (6-6), 
 
Ip =
hql
hc
P0 = 0.593 mA
Then sQ
2 = 2qIpB = 2 1.6 ´10
-19
 C( )(0.593 mA)(150 ´ 106 Hz) = 2.84 ´ 10-17 A2
3
(b) sDB
2 = 2qIDB = 2 1.6 ´10
-19 C( )(1.0 nA)(150 ´ 106 Hz) = 4.81´ 10-20 A 2
(c) 
 
sT
2 = 4kBT
RL
B =
4 1.38 ´ 10-23 J / K( )(293 K)
500 W
150 ´106 Hz( )= 4.85 ´ 10-15 A 2
6-7. Using R0 = 
hql
hc = 0.58 A/W, we have from Eqs. (6-4), (6-11b), (6-15), and (6-
17)
èç
æ
ø÷
öS
N Q
 = 
1
2
R0 P0 m( )
2
M 2
2qIpBM
1/ 2M2
= 
R0P0m2
4qBM1/2
 = 6.565´1012 P0
èç
æ
ø÷
öS
N DB
 = 
( )R0P0m
2
4qIDBM1/2
 = 3.798´1022 P
2
0 
èç
æ
ø÷
öS
N DS
 = 
( )R0P0m
2
M2
4qILB
 = 3.798´1026 P
2
0 
èç
æ
ø÷
öS
N T
 = 
1
2 ( )R0P0m
2
M2
4kBTB/RL
 = 7.333´1022 P
2
0 
where P0 is given in watts. To convert P0 = 10-n W to dBm, use 10 log è
ç
æ
ø
÷
öP0
10-3
 =
10(3-n) dBm
6-8. Using Eq. (6-18) we have
S
N = 
1
2 ( )R0P0m
2
M2
2qB(R0P0 + ID)M5/2 + 2qILB + 4kBTB/RL
 
= 
1.215 ´10-16M2
2.176 ´ 10-23 M 5/ 2 +1.656 ´ 10-19
The value of M for maximum S/N is found from Eq. (6-19), with x = 0.5:
Moptimum = 62.1.
4
6-9. 0 = 
d
dM èç
æ
ø÷
öS
N = 
d
dM ë
ê
ê
é
û
ú
ú
ù1
2 I
2
pM2
2q(Ip + ID)M2+x + 2qIL + 4kBT/RL
 
0 = I
2
p M - 
(2+x)M1+x 2q(Ip + ID)
1
2 I
2
pM2
2q(Ip + ID)M2+x + 2qIL + 4kBT/RL
 
Solving for M:M
2+x
opt = 
2qIL + 4kBT/RL
xq(Ip + ID)
 
6-10. (a) Differentiating pn, we have
¶pn
¶x
= 
1
Lp
 èç
æ
ø÷
ö
pn0 + Be
-asw
 e
(w-x)/Lp
 - asBe
-asx
 
¶2pn
¶x2
= - 
1
L
2
p
 èç
æ
ø÷
ö
pn0 + Be
-asw
 e
(w-x)/Lp
 + a
2
s Be
-asx
 
Substituting pn and 
¶2pn
¶x2
 into the left side of Eq. (6-23):
- 
Dp
L
2
p
 èç
æ
ø÷
ö
pn0 + Be
-asw
 e
(w-x)/Lp
 + Dp a
2
s Be
-asx
 
+ 
1
t
p
 èç
æ
ø÷
ö
pn0 + Be
-asw
 e
(w-x)/Lp
 - 
B
t
p
 e
-asx
 + F0 as e
-asx
 
= 
ë
ê
é
û
ú
ùB
è
ç
æ
ø
÷
öDp a
2
s - 
1
t
p
 + F0 as e
-asx
 
where the first and third terms cancelled because L
2
p = Dptp .
Substituting in for B:
Left side = 
ë
ê
ê
é
û
ú
ú
ùF0
Dp
 
asL
2
p
1 - a
2
s L
2
p
 
è
ç
æ
ø
÷
öDp a
2
s - 
1
t
p
 + F0 as e
-asx
 
5
= 
F0
Dp
 
ë
ê
ê
é
û
ú
ú
ùasL
2
p
1 - a
2
s L
2
p
 
è
ç
ç
æ
ø
÷
÷
öa
2
s Lp - 1
t
p
 + Dpas e
-asx
 
= 
F0
Dp
 ( )-Dpas + Dpas e
-asx
 = 0 Thus Eq. (6-23) is satisfied.
(b) Jdiff = qDp 
¶pn
¶x
æ 
è 
ö 
ø 
x =w
= qDp ë
ê
é
û
ú
ù1
Lp
 èç
æ
ø÷
ö
pn0 + Be
-asw
 - asBe
-asw
 
= qDp Bèç
æ
ø÷
ö1
Lp
 - as e
-asw
 + qpn0 
Dp
Lp
 
= qF0 
è
ç
ç
æ
ø
÷
÷
öasL
2
p
1 - a
2
s L
2
p
 
è
ç
æ
ø
÷
ö1 - asLp
Lp
 e
-asw
 + qpn0 
Dp
Lp
 
= qF0 
asLp
1 + asLp
 e
-asw
 + qpn0 
Dp
Lp
 
c) Adding Eqs. (6-21) and (6-25), we have
Jtotal = Jdrift + Jdiffusion = qF0 
ë
ê
é
û
ú
ù
è
æ
ø
ö
1 - e
-asw
 + 
è
ç
æ
ø
÷
öasLp
1 + asLp
 e
-asw
 + qpn0 
Dp
Lp
 
= qF0 
è
ç
ç
æ
ø
÷
÷
ö
1 - 
e
-asw
1 + asLp
 e
-asw
 + qpn0 
Dp
Lp
 
6-11. (a) To find the amplitude, consider
è
æ
ø
öJ tot J
*
tot
sc
1/2
 = qF0 ( )S S*
1/2
 where S = 
1 - e
-jwtd
jwtd
 
We want to find the value of wtd at which ( )S S*
1/2
 = 
1
2
 .
Evaluating ( )S S*
1/2
 , we have
6
( )S S*
1/2
 = 
ë
ê
ê
é
û
ú
ú
ù
è
ç
ç
æ
ø
÷
÷
ö
1 - e
-jwtd
jwtd è
ç
ç
æ
ø
÷
÷
ö
1 - e
+jwtd
-jwtd
1/2
 
= 
ë
é
û
ù
1 - è
æ
ø
ö
e
+jwtd
 + e
-jwtd
 + 1
1/2
wtd
 = 
( )2 - 2 cos wtd
1/2
wtd
 
= 
[ ]( )1 - cos wtd /2
1/2
wtd/2
 = 
sin è
ç
æ
ø
÷
öwtd
2
wtd
2
 = sinc è
ç
æ
ø
÷
öwtd
2 
We want to find values of wtd 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 
wtd
2 = 0.442 which implies wtd = 0.884
(b) From Eq. (6-27) we have td = 
w
vd
 = 
1
asvd
 . Then
wtd = 2pf3-dB td = 2pf3-dB 
1
asvd
 = 0.884 or
f3-dB = 0.884 asvd/2p
6-12. (a) The RC time constant is
RC = 
Re
0
KsA
w = 
(104 W)(8.85 ´10-12F /m)(11.7)(5 ´ 10-8 m2 )
2 ´ 10-5m
= 2.59 ns
(b) From Eq. (6-27), the carrier drift time is
7
td = 
w
vd
 = 
20 ´10-6 m
4.4 ´ 104 m / s
= 0.45 ns
c) 
1
as
 = 10-3 cm = 10 mm = 
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 » k2 and keff defined in Eq. (6-10), we have
(1) 1 - 
k1(1 - k1)
1 - k2
 = 1 - 
k1 - k
2
1
1 - k2
 » 1 - 
k2 - k
2
1
1 - k2
 = 1 – keff
(2)
(1 - k1)2
1 - k2
 = 
1 - 2k1 + k
2
1
1 - k2
 » 
1 - 2k2 + k
2
1
1 - k2
 
= 
1 - k2
1 - 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 + èç
æ
ø÷
ö2 - 
1
Me
(1 - keff) 
(b) With k1 » k2 and k
'
eff defined in Eq. (6-40), we have
(1)
k2(1 - k1)
k
2
1(1 - k2)
 » 
k2 - k
2
1
k
2
1(1 - k2)
 = k'eff 
(2)
(1 - k1)2k2
k
2
1(1 - k2)
 = 
k2 - 2k1k2 + k2k
2
1
k
2
1(1 - k2)
 
» 
è
æ
ø
ök2 - k
2
1 - è
æ
ø
ök
2
1 - k2k
2
1
k
2
1(1 - k2)
 = k'eff - 1
8
Therefore Eq. (6-35) becomes Eq. (6-39): Fh = k
'
eff Mh - èç
æ
ø÷
ö2 - 
1
Mh
(k'eff - 1) 
6-14. (a) If only electrons cause ionization, then b = 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 a = b, then from Eqs. (6-36) and (6-37), k1 = k2 = 1 so that
keff = 1. Then, from Eq. (6-38), we have Fe = Me.
1
Problem Solutions forChapter 7
7-1. We want to compare F1 = kM + (1 - k)2 -
1
M
æ 
è 
ö 
ø 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 3.00
9 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
-¥
¥
ò (f )e j2pft df = R 
 
e j2pft
1 + j2pfRC
 df
-¥
¥
ò
Using the integral solution from Appendix B3:
 
e jpx
(b + jx)n
 dx
-¥
¥
ò = 
2p(p)n-1e- bp
G(n)
for p > 0, we have
hB(t) = 
 
1
2pC
 
e j 2pft
1
2pRC
+ jfæ è 
ö 
ø 
 df
-¥
¥
ò = 
1
C
e-t/RC
7-3. Part (a):
 
hp
-¥
¥
ò (t) dt = 
1
aTb
 õó
-aTb/2
aTb/2
 dt = 
1
aTbè
ç
æ
ø
÷
öaTb
2 + 
aTb
2 = 1
2
Part (b):
 
hp
-¥
¥
ò (t) dt = 
 
1
2p
 
1
aTb
 exp -
t2
2(aTb )
2
é 
ë ê 
ù 
û ú 
 
-¥
¥
ò dt
= 
1
2p
1
aTb
p aTb 2 = 1 (see Appendix B3 for integral solution)
Part (c):
 
hp
-¥
¥
ò (t) dt = 
 
1
aTb
 exp -
t
aTb
é 
ë ê 
ù 
û ú 
 dt
0
¥
ò = - e-¥ - e-0[ ]= 1
7-4. The Fourier transform is
F[p(t)*q(t)] = 
 
 p(t)* q(t)e- j2 pft dt
-¥
¥
ò = 
 
 q(x) p(t - x) e- j2pft dt dx
-¥
¥
ò
-¥
¥
ò
= 
 
 q(x) e- j2pfx p(t - x) e- j2pf( t - x) dt dx
-¥
¥
ò
-¥
¥
ò
= 
 
 q(x) e- j2pfx dx p(y) e- j 2pfy dy
-¥
¥
ò
-¥
¥
ò 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
2
P0 (v th ) + P1(v th )[ ].
Substituting Eq. (7-20) and (7-22) for P0 and P1, respectively, we have
Pe = 
 
1
2
 
1
2ps 2
 e-v
2 / 2s 2 dv
V / 2
¥
ò + e- (v-V )
2 / 2s 2 dv
-¥
V / 2
ò
é 
ë ê 
ù 
û ú 
In the first integral, let x = v/2s2 so that dv = 2s2 dx.
In the second integral, let q = v-V, so that dv = dq. The second integral then
becomes
3
e-q
2 / 2s2 dq
-¥
V / 2 -V
ò = 
 
2s2 e- x
2
 dx
-¥
-V / 2 2s 2
ò where x = q/2s2
Then
Pe = 
 
1
2
 
2s2
2ps 2
 e- x
2
 dx
V / 2 2s 2
¥
ò + e- x
2
 dx
-¥
- V / 2 2s2
ò
é 
ë 
ê 
ù 
û 
ú 
= 
 
1
2 p
 e-x
2
 dx
-¥
¥
ò - 2 e-x
2
 dx
0
V / 2 2s2
ò
é 
ë 
ê 
ù 
û 
ú 
Using the following relationships from Appendix B,
 
 e- p
2x 2
 dx
-¥
¥
ò = 
p
p
and 
 
2
p
 e-x
2
 dx
0
t
ò = erf(t), we have
Pe = 
1
2
1- erf
V
2s 2
æ 
è 
ö 
ø 
é 
ë ê 
ù 
û ú 
7-6. (a) V = 1 volt and s = 0.2 volts, so that V
2s
 = 2.5. From Fig. 7-6 for 
V
2s
 = 2.5,
we find Pe » 7´ 10-3 errors/bit. Thus there are (2´105 bits/second)(7´10-3
errors/bit) = 1400 errors/second, so that
1
1400 errors/second = 7´10
-4 seconds/error
(b) If V is doubled, then 
V
2s
 = 5 for which Pe » 3´ 10-7 errors/bit from Fig. 7-6.
Thus
1
(2 ´ 105 bits / sec ond)(3 ´10 -7errors / bit )
= 16.7 seconds/error
7-7. (a) From Eqs. (7-20) and (7-22) we have
P0(vth) = 
 
1
2ps2
 e- v
2 / 2s 2
 dv
V / 2
¥
ò = 
1
2
1- erf
V
2s 2
æ 
è 
ö 
ø 
é 
ë ê 
ù 
û ú 
and
4
P1(vth) = 
1
2ps2
e- (v- V)
2 / 2s 2 dv
-¥
V / 2
ò = 
1
2
1- erf
V
2s 2
æ 
è 
ö 
ø 
é 
ë ê 
ù 
û ú 
Then for V = V1 and s = 0.20V1
P0(vth) = 
1
2 ë
ê
é
û
ú
ù1 - erf
è
ç
æ
ø
÷
ö1
2(.2) 2
 = 
1
2 ë
ê
é
û
ú
ù
1 - erfè
ç
æ
ø
÷
ö2
0.8 
= 
1
2 [ ]1 - erf( )1.768 = 
1
2 ( )1 - 0.987 = 0.0065
Likewise, for V = V1 and s = 0.24V1
P1(vth) = 
1
2 ë
ê
é
û
ú
ù1 - erf
è
ç
æ
ø
÷
ö1
2(.24)2
 = 
1
2 ë
ê
é
û
ú
ù
1 - erfè
ç
æ
ø
÷
ö2
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 = 
hE
hn = 
hPt
hc/l = 
0.65(25 ´10-10W)(1 ´10-9 s)(1.3 ´ 10-6 m)
(6.6256 ´10 -34 Js)(3 ´108 m / s)
= 10.6
Then, from Eq. (7-2)
P(n) = Nn 
e-N
n! = (10.6)
5 
e-10.6
5! = 
133822
120 e
-10.6 = 0.05 = 5%
7-9. v N = vout - vout
vN
2 = vout - vout[ ]
2
= vout
2 -2 vout
2
+ vout
2
= vout
2 - vout
2
7-10. (a) Letting f = fTb and using Eq. (7-40), Eq. (7-30) becomes
5
Bbae = 
 
H p (0)
Hout (0)
2
 Tb
2
 
Hout
' (f)
Hp
' (f)
2
0
¥
ò 
df
Tb
= 
I2
Tb
since Hp(0) = 1 and Hout(0) = Tb. Similarly, Eq. (7-33) becomes
Be = 
 
H p (0)
Hout (0)
2
 
Hout (f)
Hp (f)
1 + j2pfRC( )
2
0
¥
ò df
= 
 
1
Tb
2 
Hout (f)
Hp (f)
2
0
¥
ò 1 + 4p2f2 R2C2( ) df
= 
 
1
Tb
2 
Hout (f)
Hp (f)
2
0
¥
ò df + 
 
2pRC( )2
Tb
2 
Hout (f)
Hp (f)
2
0
¥
ò f2 df = 
I2
Tb
 + 
(2pRC)2
T
3
b
 I3
(b) From Eqs. (7-29), (7-31), (7-32), and (7-34), Eq. (7-28) becomes
< > v2N = < > v2s +< > v2R +< > v2I +< > v2E 
= 2q < >i0 < >m2 BbaeR2A2 + 
4kBT
Rb
 BbaeR2A2 + SIBbaeR2A2 + SEBeA2
= 
è
ç
æ
ø
÷
ö
2q < >i0 M2+x + 
4kBT
Rb
 + SI BbaeR2A2 + SEBeA2
= 
R2A2I2
Tb
 
è
ç
æ
ø
÷
ö
2q < >i0 M2+x + 
4kBT
Rb
 + SI + 
SE
R2
 + 
(2pRCA)2
T
3
b
 SEI3
7-11. First let 
 
x = v - boff( )/ 2 soff( ) with dx = dv / 2 soff( ) in the first part of Eq. (7-
49):
Pe = 
 
2soff
2psoff
 exp -x2( )
vth - boff
2soff
¥
ò dx = 
 
1
p
 exp -x2( )
Q / 2
¥
ò dx
Similarly, let 
 
y = -v + bon( )/ 2 son( ) so that
dy = -dv/ 2son( )in the second part of Eq. (7-49):
6
Pe = - 
 
1
p
 exp -y2( ) 
¥
- vth +b on
2s on
ò dy = 
 
1
p
 exp -y2( ) 
Q / 2
¥
ò dy
7-12. (a) Let x = 
V
2 2 s
 = 
K
2 2
 For K = 10, x = 3.536. Thus
Pe = 
e-x2
2 p x
 = 2.97´10-7 errors/bit
(b) Given that Pe = 10-5 = 
e-x2
2 p x
 then e-x2 = 2 p 10-5 x.
This holds for x » 3, so that K = 22 x = 8.49.
7-13. Differentiating Eq. (7-54) with respect to M and setting dbon/dM = 0, we have
dbon
dM = 0
= - 
 
Q(hn / h)
M2
M2+ x
h
hn
æ 
è 
ö 
ø bonI2 + W
é 
ë ê 
ù 
û ú 
1/ 2
+ M 2+x
h
hn
æ 
è 
ö 
ø bonI2 (1 - g
é 
ë ê 
ù 
û ú 
+ W
ì 
í 
î 
ü 
ý 
þ 
1/ 2
+ 
 
Q(hn / h)
M(hn/ h)
1
2
(2 + x)M1+ x bonI2
M 2+ x
h
hn
æ 
è 
ö 
ø bonI2 + W
é 
ë ê 
ù 
û ú 
1/ 2 +
1
2
(2 + x)M1+x bonI2 (1- g)
M 2+x
h
hn
æ 
è 
ö 
ø bonI2 (1- g) + W
é 
ë ê 
ù 
û ú 
1/ 2
ì 
í 
ï 
î 
ï 
ü 
ý 
ï 
þ 
ï 
1 / 2
Letting G = M2+x
è
ç
æ
ø
÷
öh
hn
 bonI2 for simplicity, yields
î
í
ì
þ
ý
ü
( )G + W
1/2
 + [ ]G(1-g) + W
1/2
 = 
G
2 (2 + x)
î
í
ì
þ
ý
ü1
(G + W)
1/2 + 
(1-g)
[ ]G(1-g) + W
1/2 
Multiply by (G + W)
1/2
 [ ]G(1-g) + W
1/2
 and rearrange terms to get
(G + W)
1/2
 ëê
é
ûú
ùW - 
Gx
2 (1-g) = [ ]G(1-g) + W
1/2
 èç
æ
ø÷
öGx
2 - W 
7
Squaring both sides and collecting terms in powers of G, we obtain the quadratic
equation
G2ëê
é
ûúùx2g
4 (1-g) 
 + Gëê
é
ûú
ùx2g
4 W(2-g) - gW
2(1+x) = 0
Solving this equation for G yields
G = 
- 
x2
4 W(2-g) ± îï
í
ïì
þï
ý
ïüx4
16 W
2(2-g)2 + x2(1-g)W2(1+x)
1
2
x2
2 (1-g)
 
= 
W(2-g)
2(1-g) îï
í
ïì
þï
ý
ïü
1 - 
ë
ê
é
û
ú
ù
1 + 16 
èç
æ
ø÷
ö1+x
x2
 
1-g
(2-g)2
1
2
 
where we have chosen the "+" sign. Equation (7-55) results by letting
G = M
2+x
opt bonI2 è
ç
æ
ø
÷
öh
hn 
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 
è
ç
æ
ø
÷
öhn
h b
1/(2+x)
on
ë
ê
é
û
ú
ùhn
h 
W(2-g)
2I2(1-g)
 K
1/(2+x) îï
í
ïì
þï
ý
ïü
ë
ê
é
û
ú
ùW(2-g)
2(1-g) K + W
1
2
 + ëê
é
ûú
ùW
2 (2-g)K + W
1
2
 
Factoring out terms:
b
(1+x)/(2+x)
on = Qè
ç
æ
ø
÷
öhn
h
(1+x)/(2+x)
 W
x/2(2+x)
 I
1/(2+x)
2 
´ 
îï
í
ïì
þï
ý
ïü
ë
ê
é
û
ú
ù(2-g)
2(1-g) K + 1
1
2
 + ëê
é
ûú
ù1
2 (2-g)K + 1
1
2
 ¸ 
ë
ê
é
û
ú
ù(2-g)K
2(1-g)
1/(2+x)
 
or bon = Q
(2+x)/(1+x)
è
ç
æ
ø
÷
öhn
h W
x/2(1+x)
 I
1/(1+x)
2 L
8
7-15. In Eq. (7-59) we want to evaluate
lim
g ®1
(2 - g)K
2(1- g )L
é 
ë ê 
ù 
û ú 
1+x
2+x
= 
lim
g ®1
(2 - g)K
2(1- g)
é 
ë ê 
ù 
û ú 
1+x
2+x lim
g ®1
1
L
æ 
è 
ö 
ø 
1+x
2+x
Consider first
lim
g ®1
(2 - g)K
2(1- g)
é 
ë ê 
ù 
û ú 
= 
 
lim
g ®1
 
(2 - g)
2(1- g )
-1 + 1 + B
(1- g )
(2 - g)2
é 
ë ê 
ù 
û ú 
1
2
ì 
í 
ï 
î ï 
ü 
ý 
ï 
þ ï 
where B = 16(1+x)/x2 . Since g®1, we can expand the square root term in a
binomial series, so that
lim
g ®1
(2 - g)K
2(1- g)
é 
ë ê 
ù 
û ú 
= 
 
lim
g ®1
 
(2 - g)
2(1- g )
-1 + 1 +
1
2
B
(1- g)
(2 - g)2
- Order(1- g )2
é 
ë ê 
ù 
û ú 
ì 
í 
î 
ü 
ý 
þ 
= 
 
lim
g ®1 
B
4(2 - g )
= 
B
4 = 4 
1+x
x2
 
Thus
lim
g ®1
(2 - g)K
2(1- g)
é 
ë ê 
ù 
û ú 
1+x
2+x
= 
èç
æ
ø÷
ö4 
1+x
x2
1+x
2+x
 
Next consider, using Eq. (7-58)
 
lim
g ®1
 
1
L
æ 
è 
ö 
ø 
1+ x
2+x
= 
lim
g ®1
(2 - g)K
2(1- g)
é 
ë ê 
ù 
û ú 
1/(2+ x)
¸ (2 - g)
2(1- g)
K + 1
é 
ë ê 
ù 
û ú 
1
2
+
1
2
(2 - g )K +1é 
ë 
ù 
û 
1
2
ì 
í 
ï 
î ï 
ü 
ý 
ï 
þ ï 
 
From the above result, the first square root term is
ë
ê
é
û
ú
ù(2-g)
2(1-g) K + 1
1
2
 = 
ëê
é
ûú
ù4 
1+x
x2
 + 1
1
2
 = 
è
ç
æ
ø
÷
öx2 + 4x + 4
x2
1
2
 = 
x + 2
x 
From the expression for K in Eq. (7-55), we have that 
lim
g ®1
 K = 0, so that
9
lim
g ®1
1
2
(2 - g )K +1é 
ë 
ù 
û 
1
2
= 1 Thus
lim
g ®1
(2 - g )
2(1- g )
K + 1
é 
ë ê 
ù 
û ú 
1
2
+ 1
2
(2 - g)K +1é ë 
ù 
û 
1
2
ì 
í 
ï 
î ï 
ü 
ý 
ï 
þ ï 
= 
x + 2
x + 1 = 
2(1 + x)
x 
Combining the above results yields
lim
g ®1
(2 - g)K
2(1- g )L
é 
ë ê 
ù 
û ú 
1+x
2+x
= 
èç
æ
ø÷
ö4 
1+x
x2
1+x
2+x
 
èç
æ
ø÷
ö4 
1+x
x2
1
2+x
 
2(1 + x)
x = 
2
x 
so that
lim
g ®1
 M
1+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
' (f)
2
 df
0
¥
ò =
 
1
2
 Hout
' (f)
2
 df
-¥
¥
ò
= 
1
2 õó
-
1-b
2
1-b
2
 df + 
õ
ô
ó
1-b
2
1+b
2
 
1
8ë
ê
é
û
ú
ù
1-sin 
è
ç
æ
ø
÷
öpf
b - 
p
2b
2
df + 
õ
ô
ó
- 
1+b
2
- 
1-b
2
 
1
8ë
ê
é
û
ú
ù
1-sin 
è
ç
æ
ø
÷
öpf
b - 
p
2b
2
df 
Letting y = 
pf
b - 
p
2b we have
I2 = 
1
2 (1 - b) + 
b
4põ
ó
-
p
2
p
2
 [ ]1 - 2sin y + sin2y dy
= 
1
2 (1 - b) + 
b
4p ëê
é
ûú
ùp - 0 + 
p
2 = 
1
2 èç
æ
ø÷
ö
1 - 
b
4 
10
Use Eq. (7-42) to find I3:
I3 = 
 
 Hout
' (f)
2
f2 
0
¥
ò df =
1
2
 Hout
' (f)
2
f2
-¥
¥
ò df
= 
1
2 õó
-
1-b
2
1-b
2
 f2 df + 
õ
ô
ó
1-b
2
1+b
2
 
1
8ë
ê
é
û
ú
ù
1-sin 
è
ç
æ
ø
÷
öpf
b - 
p
2b
2
f2df + 
õ
ô
ó
- 
1+b
2
- 
1-b
2
 
1
8ë
ê
é
û
ú
ù
1-sin 
è
ç
æ
ø
÷
öpf
b - 
p
2b
2
f2df 
Letting y = 
pf
b - 
p
2b 
I3 = 
1
3 èç
æ
ø÷
ö1-b
2
3
 + 
b
4põ
ô
ó
-
p
2
p
2
 [ ]1 - 2sin y + sin2y
ë
ê
é
û
ú
ùb2y2
p2 + 
by
p + 
1
4 dy
= 
1
3 èç
æ
ø÷
ö1-b
2
3
 + 
b
4p
ë
ê
ê
é
û
ú
ú
ù
 
õ
ô
ó
-
p
2
p
2
 
è
ç
æ
ø
÷
öb2y2
p2 + 
1
4 ( )1 + sin
2y dy - 
2b
p õó
-
p
2
p
2
y sin y dy 
where only even terms in "y" are nonzero. Using the relationships
õó
0
p
2
 sin2y dy = 
p
4 ; õó
 
 
 y sin ydy = -y cos y + sin y
and õó
 
 
 x2 sin x2 dx = 
x3
6 - èç
æ
ø÷
öx2
4 - 
1
8 sin 2x - 
x cos 2x
4 
we have
I3 = 
1
3 èç
æ
ø÷
ö1-b
2
3
 + 
b
2p î
í
ì
þ
ý
üb2
p2 ë
ê
é
û
ú
ù1
3èç
æ
ø÷
öp
2
3
 + 
1
6èç
æ
ø÷
öp
2
3
 + 
p
8 + 
1
4èç
æ
ø÷
öp
2 + 
p
4 - 
2b
p (1) 
11
= 
b3
16 èç
æ
ø÷
ö1
p2 - 
1
6 - b
2 
èç
æ
ø÷
ö1
p2 - 
1
8 - 
b
32 + 
1
24 
7-17. Substituting Eq. (7-64) and (7-66) into Eq. (7-41), with s2 = 4p2a2 and b = 1, we
have
I2 = 
õ
ô
ó
0
•
 
ï
ï
ï
ï
ï
ïH'out(f)
H
'
p(f)
2
 df = 
1
4õ
ô
ó
0
1
e
s2f2
ëê
é
ûú
ù
1-sin èç
æ
ø÷
öpf - 
p
2
2
df 
= 
õ
ô
ó
0
1
e
s2f2
èç
æ
ø÷
ö1 + cos pf
2
2
df = 
õô
ó
0
1
e
s2f2
cos4èç
æ
ø÷
ö pf
2 df
Letting x = 
 pf
2 yields I2 = 
2
p õ
ó
0
 p
2
e
16a2x2
cos4 x dx
Similarly, using Eqs. (7-64) and (7-66), Eq. (7-42) becomes
I3 = õô
ó
0
1
e
s2f2
cos4 èç
æ
ø÷
ö pf
2 f
2df = 
 
2
p
æ 
è 
ö 
ø 
3
 x2e16a
2 x2
 cos4 x
0
p
2
ò dx
7-18. Plot of I2 versus a for a gaussian input pulse:
7-19. Plot of I3 versus a for a gaussian input pulse:
7-20. Consider first 
lim
g ®1
 K:
12
lim
g ®1
 K = 
lim
g ®1
-1+ 1+16
1+ x
x2
æ 
è 
ö 
ø 
1- g
(2 - g)2
é 
ë ê 
ù 
û ú 
1
2
ì 
í 
ï 
î ï 
ü 
ý 
ï 
þ ï 
= -1 + 1 = 0
Also 
lim
g ®1
(1- g) = 0. Therefore from Eq. (7-58)
lim
g ®1
 L = 
lim
g ®1
2(1- g)
(2 - g)K
é 
ë ê 
ù 
û ú 
1/(1+ x)
 
îï
í
ïì
þï
ý
ïü
ë
ê
é
û
ú
ù(2-g)
2(1-g) K + 1
1
2
 + 1
2+x
1+x
 
Expanding the square root term in K yields
 
lim
g ®1
 
2 - g
1- g
K = 
lim
g ®1
2 - g
1 - g
æ 
è 
ç ö 
ø 
÷ -1+ 1+
16
2
1 + x
x 2
æ 
è 
ö 
ø 
1- g
(2 - g)2
+ order(1- g)2
é 
ë ê 
ù 
û ú 
ì 
í 
î 
ü 
ý 
þ 
= 
lim
g ®1
8(1 + x)
x2
1
2 - g
é 
ë ê 
ù 
û ú 
= 
8(1+x)
x2
 
Therefore
lim
g ®1
 L = ëê
é
ûú
ù2x2
8(1+x)
1/(1+x)
 
îï
í
ïì
þï
ý
ïü
ëê
é
ûú
ù4(1+x)
x2
 + 1
1
2
 + 1
2+x
1+x
 
= ëê
é
ûú
ù2x2
8(1+x)
1/(1+x)
 
îï
í
ïì
þï
ý
ïüx+2
x +1
2+x
1+x
 = (1+x) èç
æ
ø÷
ö2
x
x
1+x
 
7-21. (a) First we need to find L and L'. With x = 0.5 and g = 0.9, Eq. (7-56) yields K =
0.7824, so that from Eq. (7-58) we have L = 2.89. With e = 0.1, we have g' = g(1 -
e) = 0.9g = 0.81. Thus L' = 3.166 from Eq. (7-80). Substituting these values into
Eq. (7-83) yields
y(e) = (1 + e)
èç
æ
ø÷
ö1
1 - e
2+x
1+x
 
L'
L = 1.1èç
æ
ø÷
ö1
.9
5/3
 
3.166
2.89 = 1.437
13
Then 10 log y(e) = 10 log 1.437 = 1.57 dB
(b) Similarly, for x = 1.0, g = 0.9, and e = 0.1, we have L = 3.15 and L' = 3.35, so
that
y(e) = 1.1èç
æ
ø÷
ö1
.9
3/2
 
3.35
3.15 = 1.37
Then 10 log y(e) = 10 log 1.37 = 1.37 dB
7-22. (a) First we need to find L and L'. With x = 0.5 and g = 0.9, Eq. (7-56) yields K =
0.7824, so that from Eq. (7-58) we have L = 2.89. With e = 0.1, we have g' = g(1 -
e) = 0.9g = 0.81. Thus L' = 3.166 from Eq. (7-80). Substituting these values into
Eq. (7-83) yields
y(e) = (1 + e)
èç
æ
ø÷
ö1
1 - e
2+x
1+x
 
L'
L = 1.1èç
æ
ø÷
ö1
.9
5/3
 
3.166
2.89 = 1.437
Then 10 log y(e) = 10 log 1.437 = 1.57 dB
(b) Similarly, for x = 1.0, g = 0.9, and e = 0.1, we have L = 3.15 and L' = 3.35, so
that
y(e) = 1.1èç
æ
ø÷
ö1
.9
3/2
 
3.35
3.15 = 1.37
Then 10 log y(e) = 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 a =
0.3 for g = 0.9. At a = 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 ´10-19C
+
4(1.38´10 -23 J / K)(300K)
(1.6 ´ 10-19C)2105 W
é 
ë ê 
ù 
û ú 
 0.543
+ 
 
1
B
4(1.38´10-23 J / K)(300 K)(.7)
(1.6 ´ 10-19 C)2 (.005 S)(105 W)2
é 
ë ê 
ù 
û ú 
 0.543
+ 
 
2p(10 pF)
1.6 ´10-19 C
é 
ë 
ù 
û 
2 4(1.38 ´10-23 J / K)(300 K)(.7)
(.005 S)
 0.073 B
or
14
WJFET » 
3.51´1012
B
+ 0.026B
and from Eq. (7-92) WBP = 
3.39 ´1013
B
+ 0.0049B
7-24. We need to find bon from Eq. (7-57). From Fig. 7-9 we have Q = 6 for a 10-9
BER. To evaluate Eq. (7-57) we also need the values of W and L. With g = 0.9,
Fig. 7-14 gives a = 0.3, so that Fig. 7-13 gives I2 = 0.543 and I3 = 0.073. Thus
from Eq. (7-86)
W = 
3.51´1012
B
+ 0.026B = 3.51´105 + 2.6´105 = 6.1´105
Using Eq. (7-58) to find L yields L = 2.871 at g = 0.9 and x = 0.5. Substituting
these values into eq. (7-57) we have
bon = (6)5/3 (1.6´10-19/0.7) (6.1´105).5/3 (0.543)1/1.5 2.871 = 7.97´10-17 J
Thus Pr = bonB = (7.97´10-17 J)(107 b/s) = 7.97´10-10 W
or
Pr(dBm) = 10 log 7.97´10-10 = -61.0 dBm
7-25. From Eq. (7-96) the difference in the two amplifier designs is given by
DW = 
1
Bq2
 
2kBT
Rf
 I2 = 3.52´106 for I2 = 0.543 and g = 0.9.
From Eq. (7-57), the change in sensitivity is found from
10 log 
ë
ê
é
û
ú
ùWHZ + DW
WHZ
x
2(1+x)
 = 10 log ëê
é
ûú
ù1.0 + 3.52
1.0
.5
3
 = 10 log 1.29 = 1.09 dB
7-26. (a) For simplicity, let
D = M2+x
è
ç
æ
ø
÷
öh
hn 
 I2 and F = 
Q
M è
ç
æ
ø
÷
öh
hn 
so that Eq. (7-54) becomes, for g = 1,
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 [ ]2DF
2 ± 4F4D2 - 4F4D2 + 16WF2 = DF2 + 2FW 
(where we chose the "+" sign)= 
hn
h èç
æ
ø÷
öMxQ2I2 + 
2Q
M W
1/2 
(b) With the given parameter values, we have
bon = 2.286´10-19 39.1M
0.5 +
1.7´ 104
M
æ 
è 
ç ö 
ø 
The receiver sensitivity in dBm is found from
Pr = 10 log bon(50 ´10
6 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
16
7-27. Using Eq. (E-10) and the relationship
 
 
1
1+
x
a
æ 
è 
ö 
ø 
2 dx0
¥
ò = 
pa
2 
from App. B, we have from Eq. (7-97)
BHZ = 
 
1
(AR)2
 
(AR)2
1+ (2pRC)2 f 2
 
0
¥
ò df = 
p
2 
1
2pRC = 
1
4RC 
where H(0) = AR. Similarly, from Eq. (7-98)
BTZ = 
 
1
1
 
1
1+
2pRC
A
æ 
è 
ö 
ø 
2
f2
 
0
¥
ò df = 
p
2 
A
2pRC = 
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
Req
 FT
 
- 
q(Ip+ID) (2+x) M1+x B (Ipm)2M2
ë
ê
é
û
ú
ù
2q(Ip+ID)M2+x B + 
4kBTB
Req
 FT
2 = 0
Solving for M,
M
2+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
17
S
N = 
1
2 (Ipm)
2M
2
opt
 2q(Ip+ID)M
2+x
opt
 B + KB
 = 
1
2 (Ipm)
2 
ëê
é
ûú
ùK
q(Ip+ID)x
2
2+x
2q(Ip+ID)K
q(Ip+ID)x
 B + KB
 
= 
xm2I
2
p
2B(2+x) [ ]q(Ip+ID)x
2/(2+x) è
ç
æ
ø
÷
öReq
4kBTFT
x/(2+x)
 
(b) If Ip >> ID, then
S
N = 
xm2I
2
p
2B(2+x) ( )qx
2/(2+x)
I
2/(2+x)
p
 
è
ç
æ
ø
÷
öReq
4kBTFT
x/(2+x)
 
= 
m2
2Bx(2+x) ë
ê
ê
é
û
ú
ú
ù
( )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)x
2/(2+x) è
ç
æ
ø
÷
öReq
4kBTFT
x/(2+x)
 
= 
 
(0.8)2 (0.5 A / W)2 Pr
2
2(5 ´ 106 / s) 3 [1.6 ´ 10-19 C(0.5Pr + 10
-8 ) A]2 / 3
104 W / J
1.656 ´ 10-20
æ 
è 
ç ö 
ø 
1 / 3
= 
1.530 ´1012 P
r 2
0.5Pr + 10
-8( )2 / 3
where Pr is in watts.
18
We want to plot 10 log (S/N) versus 10 log 
Pr
1 mW . Representative values are
shown in the following table:
Pr (W) Pr (dBm) S/N 10 log (S/N) (dB)
2´ 10-9 - 57 1.237 0.92
4´ 10-9 - 54 4.669 6.69
1´ 10-8 - 50 25.15 14.01
4´ 10-8 - 44 253.5 24.04
1´ 10-7 - 40 998.0 29.99
1´ 10-6 - 30 2.4´ 104 43.80
1´ 10-5 - 20 5.2´ 105 57.18
1´ 10-4 - 10 1.13´107 70.52
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) + afL + 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) + afL + 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-2pBt u(t) we have
è
æ
ø
ö
1 - e
-2pBt10
 = 0.1and è
æ
ø
ö
1 - e
-2pBt90
 = 0.9
so that
2
e
-2pBt10
 = 0.9 ande
-2pBt90
 = 0.1
Then
e
2pBtr
 = e
2pB(t90-t10)
 = 
.9
.1 = 9
It follows that
2pBtr = ln 9 or tr = 
ln 9
2pB = 
0.35
B 
8-4. (a) From Eq. (8-11) we have
1
2p s
 exp
èç
çæ
ø÷
÷ö
- 
t
2
1/2
2s2 = 
1
2 
1
2p s
 which yields t1/2 = (2 ln 2)1/2 s
(b) From Eq. (8-10), the 3-dB frequency