EE 4392 Introduction To Optical Systems Instructor: Dr Mohammed Zamshed Ali Light Wave Fundamentals

1
WELCOME
EE 4392Introduction To Optical
SystemsInstructor Dr Mohammed Zamshed
AliLight Wave Fundamentals Fall
2007Department of Electrical EngineeringUniversi
ty of Texas at Dallas
2
Overview of Light Wave Technology

• Electromagnetic Wave
• Dispersion, Pulse Distortion, Information Rate
• Polarization
• Resonant Cavities
• Reflection at a Plane Boundary
• Critical Angle Reflection

3
Electromagnetic Wave

• Wave Properties
• Velocity, Power, Polarization, Interference,
  Refraction

The following graph shows a wave traveling in the
z direction with t1 lt t2 lt t3
4
Electromagnetic Wave
Waves traveling in the z direction are
represented by the uniform plane wave equation
(3.1)
This is a solution of the wave equation.
(3.2)
5
Electromagnetic Wave
The frequency is f v/l The radian frequency
is w 2pf The wave amplitude is E0 and the wave
phase is f wt-kz
6
Electromagnetic Wave
Relationships for the Propagation Factor k v
c/n k w/v w/(c/n) wn/c In free space, n
1, so that k ko w/c In general, then k kon
7
Electromagnetic Wave
(3.6)
l wavelength in the medium lo wavelength in
free space Then,
(3.7)
8
Power
Electromagnetic Wave - Power
Power in a resistor Pr V2/R The circuit power
is proportional to the voltage (V) squared. In an
optical beam, define Intensity I E2 Since the
power in a wave is proportional to the square of
the electric field, the optical power is
proportional to the intensity I.
9
Electromagnetic Wave - Power
Next define Irradiance S S Power Density
(watts/m2) For a plane wave, the irradiance and
intensity are given by
10
Electromagnetic Wave - Power
Now
We conclude that the intensity is proportional to
the irradiance.
11
Electromagnetic Wave

• Recall the plane wave given by

(3.1)
This expression represents a wave traveling with
zero loss.

• If loss occurs, the field is represented by

(3.8)
? Is the attenuation coefficient The frequency
and phase do not vary with loss, only the
amplitude of the wave Eoe-az changes with loss.
12
Electromagnetic Wave
The intensity of the beam is proportional to the
square of the electric field. For a path length L
in a lossy medium, the power diminishes by a
factor of
The corresponding power reduction in decibels is
This will be a negative number for propagation
through a lossy medium. From this equation we can
compute the dB loss per kilometer (g) in terms of
the attenuation coefficient.
13
Electromagnetic Wave
Wave Traveling in a Lossy Medium
t1
t2
Electric Field
Distance (z)
14
Dispersion, Pulse Distortion, Information Rate
When we write E Eosin (wt kz), we imply a
single frequency source.
Normalized Power
Frequency
Radio oscillators approximate this pretty well.
However, optical sources do not produce single
frequencies. They produce a range of frequencies
(range of wavelengths).
15
Spectrum of an optical source
Example Emission Spectrum of an Optical Source
1
Normalized Power
.5
f1
f2
Frequency
f
?f
?f source bandwidth (range of frequencies
emitted by the source).
16
Spectrum of an optical source
Alternatively, we can plot the wavelength
emission spectrum as follows
1
Normalized Power
.5
?
Wavelength
?1
?2
??
?? linewidth or spectral width
17
Spectrum of an optical source
Example If ? 0.82 ?m, ?? 30 nm
so we have 3.7 bandwidth.
The conversion between wavelength and frequency
is
(3.9)
18
Spectrum of an optical source
Proof
19
Spectrum of an optical source
Define the mean wavelength as
Then
The mean frequency is f c/l
20
Spectrum of an optical source
Now, we have
Thus, completing the proof of (3.9).
21
Spectrum of an optical source
Spectral Widths for Typical Light Sources
Source Spectral Width ?? (nm) LED 20-100
Laser Diode 1-5 NdYAG-Laser 0.1
HeNe Laser 0.002
22
Spectrum of an optical source
If ?? 0, (?f 0), the source is perfectly
coherent. It is monochromatic. Laser diodes
are more coherent than LEDs, but are not
perfectly coherent. We will see how source
bandwidth limits the information capacity of
fiber transmission lines.
23
Material Dispersion and Pulse Distortion
Recall that v c/n. For glass, n varies with
wavelength. Thus, waves of different wavelengths
(frequencies) travel at different
speeds. Dispersion Property of velocity
variation with wavelength
24
Material Dispersion and Pulse Distortion
Material Dispersion Dispersion caused by the
material. Waveguide Dispersion Dispersion
caused by the structure of the waveguide.
25
Material Dispersion and Pulse Distortion
Consider a pulse of light emitted by a source
which contains a range of wavelengths (say ?1,
?2, ?3)
Input Power
Output Power
T
T ??
t
t
Fastest wavelength
Arrives first
?1
?1
t
t
?2
?2
t
t
Arrives last
?3
?3
Slowest wavelength
t
t
26
Material Dispersion and Pulse Distortion
Because of dispersion, the components of the
input pulse at ?1, ?2, and ?3 travel at
different speeds and thus arrive at the receiver
at different times. The output is widened by an
amount we label as ??.
27
Material Dispersion and Pulse Distortion

• We conclude the following
• The longer the path the greater the pulse
  spread.
• The greater the source spectral width, the
  greater the pulse spread.

Dispersion also distorts an analog signal
waveform, as illustrated on the next slide.
28
Material Dispersion and Pulse Distortion
Input Power
Output Power
Pac,in
Pac,out
?1
?1
?2
?2
t
t
Slower wavelength
Pac,out lt Pac,in Information is contained in
the amplitude variation.
29
DISPERSION IN SiO2 (SILICON DIOXIDE) GLASS
Refractive Index Variation for SiO2
Inflection Point
n
1.45
0
?o
First Derivative
?o
0
dn/dl
n
30
DISPERSION IN SiO2 (SILICON DIOXIDE) GLASS
Second Derivative
n
dn2/dl2
0
?o
Inflection point for SiO2 glass occurs near
wavelength
31
Material Dispersion and Pulse Distortion
Find the amount of pulse spread due to material
dispersion. Let ? time of travel of a pulse
over path length L.
t/L
(t/L)2
t/L
(t/L)1
??
?
?
?1
?2
With No Dispersion
With Dispersion Present
32
Material Dispersion and Pulse Distortion
Referring to the previous slide, the source
linewidth is taken to be (with ?2 gt ?1) ?? ?2
- ?1 The pulse spread per unit length is then
(3.10)
where ?1 is the fastest and ?2 is the slowest
wavelength.
33
Material Dispersion and Pulse Distortion
Referring to the preceding plot of t/L vs.
l D(t/L)/Dl d(t/L)/dl so that the pulse
spread per unit length can be calculated
from D(t/L) d(t/L)/dl Dl (3.12) The
actual spread would then be
34
Material Dispersion and Pulse Distortion
Repeating the result from the preceding slide
D(t/L) d(t/L)/dl Dl (t/L) Dl
(3.12) There are two distinct terms here which
determine the pulse spread. The first is the
slope of the t/L curve and the second is the
linewidth of the source. The linewidth will be
available from manufacturer's data or must be
measured. The slope will be calculated in the
next lecture.
35
Material Dispersion and Pulse Distortion
Previously, we showed that
(3.12)
Further analysis shows that
(3.13)
The prime and double prime denote first and
second derivatives, respectively.
36
Material Dispersion and Pulse Distortion
Proof Pulses travel at a speed called the group
velocity u. The group velocity is given by
37
Material Dispersion and Pulse Distortion
The pulse travel time is thus
This is the pulse travel time per unit of path
length.
38
Material Dispersion and Pulse Distortion
We need the following to prove (3.13)
(? is the free space value) If n ? ?(?), then
(?/L) 0 and there is no dispersion and no
pulse spread.
39
Material Dispersion and Pulse Distortion
Consider the term dn/df
40
Material Dispersion and Pulse Distortion
(3.13)
Completing the proof.
41
Material Dispersion and Pulse Distortion
Define material dispersion M
Combining (3.12) and (3.13)
(3.14)
M is in picoseconds of pulse spread per nanometer
of source spectral width and per kilometer of
fiber length. The graph of M appears on the
next slide.
42
Material Dispersion and Pulse Distortion
110
M (ps/nm.km)
1.3
1.55
?(?m)
0.82
-20
43
Material Dispersion and Pulse Distortion
For M gt 0 (wavelengths less than about 1.3 mm)
Wavelength ?2 arrives before wavelength ?1
Energy at wavelength ?2 travels faster than
energy at wavelength ?1. (Recall that ?2 is
greater than ?1)
44
Material Dispersion and Pulse Distortion
For M lt 0 (wavelengths greater than about 1.3 mm)
So that ? 1 travels faster than wavelength ?2.
At ? ?1.3 ?m, M 0 , and there is no material
dispersive pulse spreading!
45
Material Dispersion and Pulse Distortion
Example Consider an LED at ? 0.82 ?m, L 10
km, and ?? 20 nm. Find ?(?/L). From the
graph, at 0.82 mm, M 110 ps/nmkm.
46
Material Dispersion and Pulse Distortion
Change the wavelength to ? 1.5 ?m, ?? 50 nm.
At 1.5 mm, M -15 ps/nmkm. Then
47
Material Dispersion and Pulse Distortion
Example ? 0.82 ?m, ?? 1 nm. M 110 ps/nmkm
Example ? 1.5 ?m, ?? 1 nm. M -15 ps/nmkm
48
Material Dispersion and Pulse Distortion
Between 1200 nm and 1600 nm, M is given by
Mo -0.095 ps/(nm2km) and ?o is the zero
dispersion wavelength (? 1300 nm). The
wavelengths are in nm.
49
Optical Solitons
SOLITON A soliton is a pulse that travel without
spreading. The refractive index of glass depends
upon the pulse intensity. This fiber
nonlinearity is used to counter the effects of
dispersion. The leading edge of the pulse can be
slowed down, and the trailing edge speeded up, to
reduce spreading. Thus, the pulse must be
properly shaped.
50
Optical Solitons
The nonlinearity is such that solitons are only
produced at wavelengths longer than the
zero-dispersion wavelength in glass fibers.
Compensation to overcome pulse broadening is only
possible in the longer wavelength region range
1300 to 1600 nm.
51
Optical Solitons
Solitons overcome the bandwidth limitations of
the fiber, but not the attenuation. Optical
amplifiers are needed along the transmission path
to maintain the pulse energy above the minimum
required for soliton production.
Amplifier
Fiber
52
Information Rate

• Information Rate is limited by pulse spreading
  due to Dispersion

• Consider sinusoidal modulation of the light
  source with modulation frequency f. The
  transmitted optical power can be written as
• PT 21cos(2?ft) (1)
• Modulation period T
• Then
• T 1/f

53
Information Rate
MODULATED OPTICAL WAVE
Power
Time
54
Information Rate

• The limit on the allowable pulse spread will be
  taken to be

(2)

• If the spread just equals half the modulation
  period, then the power variation becomes zero.

55
Information Rate
Blue l1 Red l2
consder
Power
T/2
Dt
Time
This spread puts peaks over nulls and reduces the
total power variation to zero. Then PR 2.
56
Information Rate
Analytically, we find
57
Information Rate

• From (2) we have the requirement that
• R 1/T lt 1/(2Dt) (3)

so that the modulation frequency has the limits
(4)

• The maximum modulation frequency is then

58
Information Rate
This modulation frequency turns out to be the
3-dB bandwidth. The signal is actually reduced by
half (3-dB) at this modulation frequency. Thus,
we can write
(5)
59
Information Rate

• The total signal loss due to spreading has two
  parts and can be expressed by the equation

(6)
La Loss due to absorption and scattering. This
is the fixed loss. Lf Modulation (message)
frequency dependent loss due to pulse spreading.
60
Information Rate
The modulation frequency dependent loss is given
by
(7)
Example Suppose f f3-dB. Compute the loss.
61
Information Rate
Example Suppose f ltlt f3-dB. Compute the loss.
As expected, the equation predicts no modulation
frequency loss for modulation frequencies well
below the 3-dB frequency.
62
Information Rate
Example Suppose f 0.1 f3-dB. Compute the loss.
63
Information Rate

• The maximum frequency length product is
  calculated from (5) as follows

(5)
(8)
(3.16)
Optical 3-dB Bandwidth Length Product
64
Information Rate

• Frequency at which Lf 1.5 dB.
• Use (7) for Lf

(7)
Solving for the frequency at which the loss is
1.5 dB, we obtain
65
Information Rate

• A loss of 1.5 dB in optical power yields a loss
  of 3 dB in the corresponding electrical power.

• Thus, the electrical 3-dB bandwidth occurs when
  the optical loss is 1.5 dB.
• We found that the modulation frequency at which
  the optical loss is 1.5 dB was

66
Information Rate
Thus, the electrical 3-dB bandwidth length
product is
(3.19)
67
Information Rate

• Information Rate for Return-to-Zero (RZ) digital
  signal.

Power
1 1 1 0 1
tp
t
0 T 2T 3T 4T 5T
6T 7T
tp T/2 pulse width R 1/T data rate, b/s
68
Information Rate
Spectrum of the RZ Signal
Power Spectral Density (Watts/Hz)
0
Frequency
The required transmission bandwidth is
69
Information Rate
From equation (3.19)
The RZ rate length product is then
(3.20)
70
Information Rate
We obtain the same result by allowing a pulse
spread of 70 of the initial pulse duration.
As on the preceding slide.
71
Information Rate

• Information Rate for Non-Return-to-Zero (NRZ)
  digital signal.

Power
1 1 1 0 1
tp
t
T 2T 3T 4T 5T
6T 7T
0
72
Information Rate
Spectrum of the (NRZ) Signal
Power Spectral Density (Watts/Hz)
Frequency
0
The required transmission bandwidth is
73
Information Rate
The allowed data rate is then
Use the electrical 3-dB bandwidth
74
Information Rate
The NRZ rate length product is then
(3.21)
Comparing the results for the RZ and NRZ data
rates, we see that
75
BANDWIDTH DATA RATE SUMMARY
76
POLARIZATION
Picture of a wave, linearly polarized in x
direction and traveling in the z direction.
E
x
v
z
y
77
POLARIZATION
Picture of a wave, linearly polarized in y
direction and traveling in the z direction.
v
x
E
z
y
An electric field that points in just one
direction is said to be linearly polarized
because it always points along a single line.
78
POLARIZATION
The two orthogonal linear polarizations are the
plane wave modes of an unbounded media. They can
exist simultaneously. The actual polarization is
determined by the polarization of the light
source and by other polarization sensitive
components in the optical system.
79
POLARIZATION
If the direction of electric field E varies
randomly (as shown) the wave is
unpolarized. Most fibers depolarize the
input light. Only special fibers maintain the
light polarization.
y
E
x
80
Resonant Cavities

• Resonant Cavities are formed by two mirrors, one
  at each end of the fiber cable
• Used for maintaining zero power variation

L
E
z
Mirror
Mirror
81
Resonant Cavities
L is the Length of the cavity
(3.22)
The cavity must be an integral number of half
wavelengths long to support a wave.
82
Resonant Cavities
The resonant wavelengths are
The corresponding resonant frequencies are
83
Resonant Cavities
Resonance Spectrum
Frequency
The resonant frequency spacing is
(3.25)
84
Resonant Cavities
The free space wavelength spread corresponding to
?fc is ??c calculated from
(3.26)
This equation refers to the free space
wavelengths.
85
Resonant Cavities
Example Consider an AlGaAs laser cavity. L
0.3 mm 300 ?m n 3.6 ?o 0.82 ?m. Find
the cavity resonant wavelength spacing ??c
86
Resonant Cavities
Example Suppose the AlGaAs, LD has a spectral
width of 2 nm. Determine the number of
longitudinal modes in the output.
87
Resonant Cavities
Gain (AlGaAs)
0.82 ?m
?
2 nm
Cavity Resonances
??c
?
0.82 ?m
88
Resonant Cavities
Laser Output
2 nm
??c
?
0.82 ?m
The laser emits 6 longitudinal modes.
A laser emitting only one longitudinal mode is a
single-mode laser.
89
REFLECTION AT A PLANE BOUNDARY

• Light reflected at the input and connector gap
  should be low as this reduces optical power to
  be transmitted
• Internal Reflection at the core boundary should
  be high to keep the light inside the fiber

n1
n2
Incident Wave
Transmitted Wave
Reflected Wave
Boundary
90
REFLECTION AT A PLANE BOUNDARY
Reflection Coefficient
91
REFLECTION AT A PLANE BOUNDARY
Transmission coefficient as
92
REFLECTION AT A PLANE BOUNDARY
Defining Reflectance as
This result is valid for normal incidence.
93
REFLECTION AT A PLANE BOUNDARY
For air-to-glass, compute the transmitted power.
4 power reflected. 96 power transmitted. In
dB, the transmitted power is 10log(0.96)
-0.177 dB Typically we round this off to 0.2 dB
(omitting the minus sign). This is called the
Fresnel loss.
94
REFLECTION AT A PLANE BOUNDARY
Consider arbitrary incidence
Et
?
n1 n2
Er
?
?t
?r
?i
?
Ei
Perpendicular Polarization (s)
95
REFLECTION AT A PLANE BOUNDARY
Consider arbitrary incidence
Et
n1 n2
Er
?t
?r
?i
Ei
Parallel Polarization (p)
96
REFLECTION AT A PLANE BOUNDARY
Plane of Incidence Defined by the normal to the
boundary and the ray direction of the incident
beam.
x
z
Incident
Boundary
The xz plane is the plane of incidence in this
example.
97
REFLECTION AT A PLANE BOUNDARY
Fresnels Law of Reflection
For parallel polarization the result is
(3.29)
For perpendicular polarization, the result is
(3.30)
Note that ? may be complex.
98
REFLECTION AT A PLANE BOUNDARY
Plots of ?p and ?s for n1 1 (air), n2 1.48
(glass)
Perpendicular (rs)
rp
rs
Parallel (rp)
Angle (q)
99
REFLECTION AT A PLANE BOUNDARY
From equation (3.29) for parallel polarization,
we can get total transmission (no reflection)
if
The angle satisfying this equation is the
Brewster angle ?B. The solution is
100
REFLECTION AT A PLANE BOUNDARY
Compute ?B for air-to-glass and glass-to-air
For n1 1, n2 1.5 (air-to-glass)
For n1 1.5, n2 1 (glass-to-air)
For perpendicular polarization there is no
Brewster angle.
101
REFLECTION AT A PLANE BOUNDARY
Antireflection Coatings We have just seen that we
can transmit a beam from one material to another
without reflection under Brewster angle
conditions. We can also transmit with no (or very
little) reflection by placing a coating between
the two materials as indicated on the figure on
the next slide.
102
REFLECTION AT A PLANE BOUNDARY
n1
n2
n3
l/4
The thickness of the coating is a quarter
wavelength. The reflectance for this
configuration is
103
REFLECTION AT A PLANE BOUNDARY
Clearly, the reflectance becomes zero if
A coating that reduces the reflectance is called
an antireflection (AR) coating. Example Compute
the reflectance when a quarter wavelength of
magnesium fluoride (n 1.38) is coated onto a
piece of glass (n 1.5).
104
REFLECTION AT A PLANE BOUNDARY
Solution The reflectance is
Without the coating, the air-to-glass interface
would have a reflectance of 4. The AR coating
has reduced the reflectance to 1.4.
105
Critical Angle Reflection

• Critical Angle is the minimum incident angle to
  maintain total reflection

Review Fresnels Law of Reflection
For parallel polarization
(3.29)
For perpendicular polarization
(3.30)
106
Critical Angle Reflection
From equations (3.29) and (3.30), we find that
n1 sin qi n2
107
Critical Angle Reflection
The incident angle satisfying this equation is
the angle whose sine is given by
(3.34)
Call the solution ?c, the critical angle.
?c only exists if n1 gt n2. That is, travel from
a high index to a low index material. This
result is valid for both polarizations.
108
Critical Angle Reflection
If
then
is purely imaginary.
109
Critical Angle Reflection
Under this condition, equations (3.29) and (3.30)
can be written in the form
where A, B, C, and D are real and j is the
imaginary term
110
Critical Angle Reflection
Then
We conclude that there is complete reflection
(called critical angle reflection) for all rays
which satisfy the condition
111
Critical Angle Reflection
Consider waves undergoing critical angle
reflections
n1 n2
In region n1 we have a standing wave due to the
interference of the incident and reflected
waves.
112
Critical Angle Reflection
In region n2 the electric field is not zero. The
boundary conditions require the electric field
to be continuous at the boundary. The field in
n2 is a decaying exponentially. Its total
z-dependence is
where the attenuation coefficient is given by
113
Critical Angle Reflection
Consider a wave where
n1 n2
Envelope
Evanescent Wave
e-?z
E
z
Standing Wave
The decaying wave carries no power in the
z-direction.
114
Critical Angle Reflection
At the critical angle,
In this case, there is no decay. The wave
penetrates deeply into the second medium as
indicated on the next page.
115
Critical Angle Reflection
As ?i increases, ? increases and the decay
becomes greater.
? 0, ? ?c
E
e-?z
?i gt ?c
z
0
As ?i increases from qc towards 90o, a increases
and the evanescent field penetrates less and less
into the second medium.
qi