1 = 1.69 0
x
1 = 0
2 = 1.44 0
2 = 0
z
in = 80o
80
ki
Ei
Hi
(a) n1 = r1 = 1.69 = 1.3 and n2 = r 2 = 1.44 = 1.2 .
n2
o
c = 67.38 . Since angle of incidence is larger than the critical
n1
Critical angle is c = sin 1
angle, there is total internal ref
ECE 135
Homework 5 solution
1. (a)
Spontaneous emission rate is A 108 s 1 .
(b)
Stimulate emission rate is
dN 2
AN 2 .
dt
Rstim B f f df
0
If input radiation is white f is constant so it can be taken out of the integral. So
Rstim B f f df B f .
0
Relatio
ECE 135
Midterm solution
1.
3
1 j
j
j
j 628.318 z
1 j j 628.318 z
4
3.6e 4 a x 14.4
E 0.3
V/m
a x 1.2e a y e
a y e
2
2
(a)
This is a plane wave since constant phase surfaces are planes. It has two counter-traveling components
so it is a stan
ECE 135
Spring 2016
Homework 2 solution
1. We can form a propagating slab mode as the interference pattern formed by the incident and
reflected plane waves shown below. In order for the interference pattern to be a mode it should
satisfy the symmetry prop
ECE 135
Homework 7 solution
1. Before lasing starts there is only spontaneous emission. So we can neglect the stimulated emission
term in the carrier rate equation. Hence
dN I N
dt q c
dN
qN
0 , hence DC steady state solution is I
.
dt
c
In DC steady
ECE 135
Homework 3 solution
1.
Material index variation as a function of wavelength is shown below.
n
0
(a)
Dispersion is the variation of group delay as a function of wavelength. In other words
D
d
d L
g
.
d
d vg
For dispersion specification we consi
ECE 135
Homework 4 solution
1.
We are given an optical communications system with the following parameters:
o 1.55 m
B 5 Gbps
TFWHM 100 ps
C 6
The dispersion limited fiber length, with chirp is given by
2 20
ps 2
km
1
, where is the rms width of
4B
the
ECE 135
HW 6 solution
1.
Heterojunctionpnjunction,calculationofthebuiltinpotential
E0
2
1
Ec 2
Ec
Ec1
Ef 2
qV0
Eg1
Eg 2
Ef1
Ev
Ev1
Ev 2
n N d 2 Nc 2e
Ec 2 E f 2
Ec 2 E f 2 kT ln
kT
Nc 2
Nd 2
p N a1 N v1e
E f 1 Ev1
E f 1 Ev1 kT ln
qV0 Ec Eg1 kT ln
kT
N v1
ECE 135
Homework 1 solution
3 1
x z
2
2
j 2
1. Ei a x 2a y 3a z e
x
1 0
1 0
2 3 0
2 0
z
i
ki
(a)
j k xk z
Ei E0i e x z
3 1
k x 2
2 m
1 1
k z 2
2 m
k 2 k x2 k z2
2
3 1 2
1
23
2
k 2
2 2 4
4 4
2 2
2
0 , 0
x
z
i
ki
i k xi
k zi
3
k x k
UNIVERSITY OF CALIFORNIA
Santa Barbara
Electrical and Computer Engineering Department
ECE 135
Spring 2010
Optical Fiber Communications
This is a course covering the technologically important and rapidly developing fiber optic
communications. In this cours
UNIVERSITY OF CALIFORNIA
Santa Barbara
Department of Electrical and Computer Engineering
ECE 135
MIDTERM
Spring 2010
During this exam you can refer to one 8.5"x11" sheet of notes only. In your answers make your
reasoning very clear and show your work expl
ECE 135
MIDTERM SOLUTION
1.
See class notes.
2.
(a)
Broadening factor of a Gaussian pulse of initial rms width 0 is given as
2
2
2
L
L
2 C 2 L
= 1 +
+ (1 + V2 ) 2 2 + (1 + C 2 + V2 ) 3 3 , where V = 2 0 .
2
2
4 2
2 0
0
2 0
0
is the rms spectra
A plane wave of free space wavelength 1.5 m is incident on a dielectric interface as shown
below. Since medium 1 has a higher index of refraction than medium 2, it is possible to have
total internal reflection at this interface.
1 = 1.69 0
1 = 0
x
2 = 1
The figure below shows a symmetrical slab waveguide.
0 = 1.5 m
Upper
cladding
n2
x=
Core
a
Lower
cladding
n1
x=0
a
2
x
z
a
x=
2
n2
The electric field of the fundamental mode of this waveguide has an electric is given as
a
j14.37 z
for x
a y E0 cos(1.06
The dispersion slope and zero dispersion wavelength values for two different fibers are given as
ps
ps
S1 = 0.09
, 01 = 1.312 m , S 2 = 0.3
, 02 = 1.312 m . Remember that
2
km-nm
km-nm 2
dispersion, D ( ) and dispersion slope S are related as D ( ) = S (
When source spectral width is negligible and 2 = 0 , for Gaussian input pulses of rms pulse
(a)
width 0 the rms output pulse width after a transmission length of L is given as:
2
L
= + (1 + C ) 3 2 .
4 2
0
2
2
0
2
The limitation to the bit rate can b
An optical communication system is operating with linearly chirped Gaussian input pulses of
chirp parameter C. Assume that 2 = 0 and source linewidth is very narrow. Consider only the
effect of dispersion slope due to finite 3 .
(a)
Find an expression for
ECE 135
Homework 8 solution
1. (a)
n+
p
w
p+
x
Netcharge(cm3)
1018
n+
wn wA
1014
4x1015
wp
w
x
p
p+
1018
(V/cm)
p
2
x
Note that in each region:
d q
ND N A
dx
q
1.6 1019
1.5 107 V / cm
8.85 1014 12
d q 18
V
10 1.5 1011
dx
cm 2
d
q
V
In p :
1018 1.5