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
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
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.
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 p
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 , he
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
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
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
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
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 rapid
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
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
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 r
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 e
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
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
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 disper
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 1