Topic 3 Dielectric Waveguides and Optical Fibers 2-1 Symmetric ...

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Topic 3
Dielectric Waveguides and
Optical Fibers
Kasap Chapter 2
2-1 Symmetric Planar Dielectric
Slab Waveguide
1
3
Contents (Kasap Chapter 2)
1 Symmetric Planar Dielectric Slab Waveguide
2 Modal and Waveguide Dispersion in the Planar Waveguide
3 Step Index Fiber
4 Numerical Aperture
5 Dispersion in Single Mode Fibers
6 Bit Rate, Dispersion, Electrical and Optical Bandwidth
7 The Graded Index (GRIN) Optical Fiber
8 Light Absorption and Scattering
9 Attenuation in Optical Fibers
10 Fiber Manufacture
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n2 cladding
Light
Light
n2 n1 > n core
2
A planar dielectric waveguide has a central rectangular region of
higher refractive index n1 than the surrounding region which has
a refractive index n2. It is assumed that the waveguide is
infinitely wide and the central region is of thickness 2a. It is
illuminated at one end by a monochromatic light source.
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Light
?1999 S.O. Kasap, Optoelectronics (Prentice Hall)
2
Ligh
4

TIR at B & C
k1 (AB+BC) + phase change due to TIR = m(2π)
κ
E
λ
θ
θ θ
?1999 S.O. Kasap, Optoelectronics (Prentice Hall)
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k 1
A
β
B
C
n 2
Light
n 1
n 2
d = 2a
A light ray travelling in the guide must interfere constructively with itself to
propagate successfully. Otherwise destructive interference will destroy the
wave.
1
E
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2
θ
A
k 1
θ
A′
C
n 2
π−2θ
n 1
n 2
B′
2θ−π/2
Two arbitrary waves 1 and 2 that are initially in phase must remain in phase
after reflections. Otherwise the two will interfere destructively and cancel each
other.
?1999 S.O. Kasap, Optoelectronics (Prentice Hall)
B
θ
2a
y
x
1
y
x
z
z
5
7
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Waveguide Condition
k = kn =
1
1
2 1
π n / λ
For constructive interference, the phase difference between A
and C must be a multiple of 2π
Δφ(
AC ) = k ( AB + BC ) − 2φ
= m(
2π
)
1
1
d
BC = AB = BC cos( 2θ
)
cosθ
2
AB + BC = BC cos( 2θ
) + BC = BC[(
2cos
θ −1)
+ 1]
= 2d
cosθ
[ 2d
cosθ
] − 2φ
m(
2π
)
→ k =
Dividing (2) by 2 we obtain the waveguide condition
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⎡
⎢
⎣
πn
( 2a)
⎤
θm
− φm
= mπ
λ ⎥ cos
⎦
2 1
(1)
(2)
(3)
Resolve the wavevector k 1 into two propagation constants,
β and κ, along and perpendicular to the the guide axis z
m
⎛ 2πn1
⎞
βm = k1 sinθ
m = ⎜ ⎟sinθ m
⎝ λ ⎠
⎛ 2πn1
⎞
κ m = k1 cosθ
m = ⎜ ⎟cosθ m
⎝ λ ⎠
Φ = k φ ( a − y)
cosθ
−φ
( 1AC
− m)
− k1A'
C = 2k1
Φ
m
= Φ
m
y
( y) = mπ
− ( mπ
+ φm
)
a
m
m
6
8

Φm = ( k1AC −φm
) − k1A'
C = 2k1(
a − y)
cosθ
m −φm
y
Φ m = Φ m(
y) = mπ
− ( mπ
+ φm
)
a
E y,
z,
t)
= E cos( ωt
− β z + κ y + Φ
1( 0
m m m
E2 m
( y,
z,
t)
= E0
cos( ωt − βm
z −κ
y)
Guide center
1
E
2
θ
A
k
θ
n2 C
A′
a y
Interference of waves such as 1 and 2 leads to a standing wave pattern along the ydirection
which propagates along z.
9
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?1999 S O K O l i (P ti H ll)
Field of guided wave
Field of evanescent wave
(exponential decay)
E(y)
m = 0
a−y
y
π−2θ
E y,
z,
t)
= 2E
1
1
cos( κ m y + Φ m)
cos( ωt
− βm
z + Φ
2
2
( y,
z,
t)
= 2E
( y)
cos( ωt − β z)
( 0 m
E m
m
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?1999 S O K Ot l t i (P ti H ll)
y
n 2
Light
n 1
n 2
x
)
E(y,z,t) = E(y)cos(ωt ? β 0 z)
The electric field pattern of the lowest mode traveling wave along the
guide. This mode has m = 0 and the lowest θ. It is often referred to as the
glazing incidence ray. It has the highest phase velocity along the guide. 11
z
)
E y,
z,
t)
= E cos( ωt
− β z + κ y + Φ
1( 0
m m m
E2( y,
z,
t)
= E0
cos( ωt − βm
z −κ
m y)
E y,
z,
t)
= 2E
1
1
cos( κ m y + Φ m)
cos( ωt
− βm
z + Φ
2
2
( 0 m
E m
m
( y,
z,
t)
= 2E
( y)
cos( ωt − β z)
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E(y)
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Standing wave along y
y
n 2
Cladding
)
Traveling wave along z
m = 0 m = 1 m = 2
Core
2a
n 1
n 2
Cladding
The electric field patterns of the first three modes (m = 0, 1, 2)
traveling wave along the guide. Notice different extents of field
penetration into the cladding.
?1999 S.O. Kasap, Optoelectronics (Prentice Hall)
10
12
)

• mode of propagation
• mode number
• high order mode
θm smaller, more bouncing, more penetration
• lowest order mode (m=0) : fundamental mode
θm ~90 deg., travels axially
Intensity
0
Light pulse
t
Cladding
Core
High order mode
Axial
Low order mode
Broadened
light pulse
Intensity
Spread, Δτ
Schematic illustration of light propagation in a slab dielectric waveguide. Light pulse
entering the waveguide breaks up into various modes which then propagate at different
group velocities down the guide. At the end of the guide, the modes combine to
constitute the output light pulse which is broader than the input light pulse.
13
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?1999 S O K Ot l t i (P ti H ll)
O
y
B //
B z
z
x (into paper)
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B y
(a) TE mode
E ⊥
θ
θ
E //
E z
(b) TM mode
E y
B ⊥
θ θ
Possible modes can be classified in terms of (a) transelectric field (TE)
and (b) transmagnetic field (TM). Plane of incidence is the paper.
?1999 S.O. Kasap, Optoelectronics (Prentice Hall)
15
t
TIR condition
From (3)
mode number
Single and Multimode Waveguides
sinθ > sinθ
m
≤ V
m 2
−φ
π
V< π/2, m=0 is the only possibility and only the fundamental mode (m=0)
Propagates along the dielectric slab waveguide, which is then termed
single mode planar waveguide.
14
At λc that leads to V= π/2: cutoff wavelength
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c
2 πa
2 2 /
V = ( n1
− n2
)
λ
V-number
= V-parameter
= normalized thickness
= normalized frequency
1 2
For given λ, V depends on the waveguide parameters a, n1 , n2 .
V that makes m=0 single mode
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⎡
⎢
⎣
2 1
TE and TM modes
TEm modes,
transverse electric field modes
E = E⊥
= Exxˆ
TMm modes,
transverse magnetic field modes
E = E = E yˆ + E zˆ
// y z
φ
πn
( 2a)
⎤
θm
− φm
= mπ
λ ⎥ cos
⎦
The phase change that accompanies TIR depends
on the polarization and is different for E and E
⊥
//
However for n1
− n2

Example 2.1.1: Waveguide modes
Consider a planar dielectric guide with a core thickness 20μm,
n 1 = 1.455, n 2 = 1.440, light wavelength of 900nm. Given the
waveguide condition in Eq(3) and the expression for φ in TIR
for the TE mode,
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⎡
⎢
⎣
πn
( 2a)
⎤
θm
− φm
= mπ
λ ⎥ cos
⎦
2 1
using a graphical solution, find angles θ m for all the modes.
What is your conclusion?
Example 2.1.2: V-number and the number of modes
Using Eq. (9), mode number =
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≤ V
m 2
−φ
estimate the number of modes that can be supported in a planar
dielectric waveguide that is 100μm wide and has n 1 = 1.490, n 2 =
1.470 at the free-space source wavelength λ = 1μm. Compare your
estimation with the formula:
M = Int(2V/π) + 1 Int(x): integer function
π
17
19
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tan(ak 1 cosθ m − mπ/2)
10
5
0
θ c
m = 1, odd
m = 0, even
82° 84° 86° 88° 90°
f(θ m)
89.17°
88.34°
87.52°
86.68°
Modes in a planar dielectric waveguide can be determined by
plotting the LHS and the RHS of eq. (11).
?1999 S.O. Kasap, Optoelectronics (Prentice Hall)
Example 2.1.3: Mode field width (MFD), 2w 0
The field distribution along y penetrates into the cladding as depicted
in the following figure
The extent of the electric field across the guide is therefore more than
2a. Within the core, the field distribution is harmonic whereas from
the boundary into the cladding, the field decays exponentially:
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E cladding (y’) = E cladding (0) exp (-α cladding y’)
Find the mode field width (MFD) 2w 0 = 2a + 2δ, where δ = 1/ α cladding
θ m
18
20

2-2 Modal and Waveguide
Dispersion in the Planar Waveguide
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Waveguide Dispersion Diagram
4/9/2009
• What is important is the group velocity along the
guide, the velocity at which the energy or information
is transported.
• The higher modes penetrate more into the cladding
where the refractive index is smaller and the waves
travel faster.
• Waveguide condition
⎡2πn
1(
2a)
⎤
⎢
θm
− φm
= mπ
λ ⎥ cos
⎣ ⎦
21
23
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ω cut-off
ω
Slope = c/n 2
TE 2
TE 1
TE 0
Slope = c/n 1
Schematic dispersion diagram, ω vs. β for the slab waveguide for various TEm. modes.
ωcut ff corresponds to V = π/2. The group velocity vg at any ω is the slope of the ω vs. β
curve at that frequency.
?1999 S.O. Kasap, Optoelectronics (Prentice Hall)
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Waveguide Dispersion Diagram
The ω vs. βm characteristics
The slope dω
at frequency ω is the group velocity vg dβ
m
The group velocity at one frequency changes from
one mode to another.
For a given mode the group velocity changes with the
frequency.
The cutoff frequency ωcut−off
corresponds to the cutoff
condition λ = λ when V
= π
c
2
β m
22
24

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ω cut-off
ω
Slope = c/n 2
TE 2
TE 1
TE 0
Slope = c/n 1
Schematic dispersion diagram, ω vs. β for the slab waveguide for various TEm. modes.
ωcut ff corresponds to V = π/2. The group velocity vg at any ω is the slope of the ω vs. β
curve at that frequency.
?1999 S.O. Kasap, Optoelectronics (Prentice Hall)
Fig. 10
E(y)
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Intramodal Dispersion
λ 1 > λ c
v g1
y
ω 1 < ω cut-off
Cladding
Core
Cladding
λ 2 > λ 1
ω 2 < ω 1
β m
y
v g2 > v g1
The electric field of TE0 mode extends more into the
cladding as the wavelength increases. As more of the field
is carried by the cladding, the group velocity increases.
?1999 S.O. Kasap, Optoelectronics (Prentice Hall)
25
27
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Intermodal Dispersion
Modal dispersion
L L
Δτ = −
v v
Δτ n
≈
L
g min g max
n
c
1 −
Example: n 1 = 1.48, n 2 = 1.46
τ1/2
2
V g,min = c/n 1
V g,max = c/n 2
Δτ
≈ 67ns
/ km
L
Δ Half intensity points that is smaller than the full width
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Intramodal Dispersion
Waveguide Dispersion
Material Dispersion: n = n(λ) or n(ω)
Dispersion in WG
- Intermodal dispersion: for multimode waveguide
- Intramodal dispersion:
waveguide dispersion
material dispersion
Intermodal dispersion >> Intramodal dispersion
26
28

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2-3 Step Index Fiber
1
1
Fiber axis
Along the fiber
Meridional ray
2
Fiber axis
3
Skew ray
Ray path along the fiber
2
4
5
3
5
1, 3
4
1
2
Ray path projected
on to a plane normal
to fiber axis
2
3
29
(a) A meridiona
ray always
crosses the fibe
axis.
(b) A skew ray
does not have
to cross the
fiber axis. It
zigzags around
the fiber axis.
Illustration of the difference between a meridional ray and a skew ray.
Numbers represent reflections of the ray.
?1999 S.O. Kasap, Optoelectronics (Prentice Hall)
HE and EH
Hybrid modes
31
y
n 2 n 1
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n
Cladding
Core
φ
r z Fiber axis
The step index optical fiber. The central region, the core, has greater refractive
index than the outer region, the cladding. The fiber has cylindrical symmetry. We
use the coordinates r, φ, z to represent any point in the fiber. Cladding is
normally much thicker than shown.
?1999 S.O. Kasap, Optoelectronics (Prentice Hall)
Linear polarized modes
Core
Cladding
(a) The electric field
of the fundamental
mode
E 01
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E
(b) The intensity in
the fundamental
mode LP 01
?1999 S.O. Kasap, Optoelectronics (Prentice Hall)
r
ELP = Elm(
r,
ϕ) exp j(
ωt
− βlmz)
The electric field distribution of the fundamental mod
in the transverse plane to the fiber axis z. The light
intensity is greatest at the center of the fiber. Intensity
patterns in LP 01, LP 11 and LP 21 modes.
y
(c) The intensity
in LP 11
30
(d) The intensity
in LP 21
32

Normalized index difference
Linear polarized modes
Normalized frequency or
V-number
Normalized index difference
Single mode cutoff frequency
Number of modes
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n1 − n
Δ =
n
1
2
ELP = Elm(
r,
ϕ) exp j(
ωt
− βlmz)
V
2πa
λ
2πa
λ
2 2 1/
2
= ( n1
− n2
) = ( 2n1
n
n1 − n n − n
Δ =
n
V
Normalized propagation constant
cut−off
Example 2.3.1: A multimode fiber
2
V
2 2
2 1 2 ≈ 2
1 2n1
2πa 2 2 1/
2
( n1
− n2
)
λc
=
M ≈
2
2
( / k)
− n
b =
n − n
β
2
1
typical Δ

Example 2.3.4: Group velocity and delay
Consider a single mode fiber with core and cladding indices of 1.448
and 1.440, core radius of 3μm, operating at 1.5μm. Given that
we can approximate the fundamental mode normalized
propagation constant b by
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b ≈ (1.1428 – 0.996 / V ) 2 1.5 < V < 2.5
(1) Calculate the propagation constant β.
(2) Change the operating wavelength to λ’ by a small amount,
0.01%, and then recalculate the new propagation constant β’.
(3) Then determine the group velocity v g of the fundamental mode at
1.5μm, and the group delay τ g over 1 km of fiber.
αα max
n 0
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n 2
n 1
θ < θ c
Fiber axis
Lost
B
θ > θ c
Cladding
Core
©1999 S.O. Kasap, Optoelectronics (Prentice Hall)
Propagates
A
37
Maximumacceptance angle
α max is that which just gives
totalinternalreflectionatthe
core-cladding interface, i.e.
when α=α maxthen θ=θ c.
Rays with α>α max (e.g. ray
B) become refracted and
penetrate the cladding and are
eventually lost.
39
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Snell’s law -->
Numerical Aperture
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2-4 Numerical Aperture
Maximum acceptance
angle α max
sinα
max n
=
sin( 90 −θ ) n
sinα
max
=
c
1
0
2 2 ( n − n )
1
n
0
( ) 2 / 1 2 2
n n
NA = −
sinα
max
1
=
0
2
NA
n
2πa
V = NA
λ
1/
2
2
38
40

Example 2.4.1: A multimode fiber and total acceptance angle
A step index fiber has a core diameter of 100μm and a refractive index
of 1.48. The cladding has a refractive index of 1.460. Calculate
(a) the numerical aperture of the fiber,
(b) acceptance angle from air,
(c) number of modes sustained
when the source wavelength is 850nm.
Example 2.4.2: A single mode fiber
A typical single mode optical fiber has a core of diameter 8μm and a
refractive index of 1.46. The normalized index difference is 0.3%. The
cladding diameter is 125μm.
(a) Calculate the numerical aperture of the fiber
(b) Calculate the acceptance angle of the fiber
2-5 Dispersion in Single Mode
Fibers
(c) What is the single mode cut-off wavelength λc of the fiber? 41
42
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Intensity Intensity Intensity
λ 1
Spectrum, λ
λ o
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Emitter
Very short
light pulse
λ 2
Input
λ
0
Cladding
vg (λ1 )
Core
vg (λ2 )
t
Output
τ
Spread, τ
All excitation sources are inherently non-monochromatic and emit within a
spectrum, λ, of wavelengths. Waves in the guide with different free space
wavelengths travel at different group velocities due to the wavelength dependence
of n 1. The waves arrive at the end of the fiber at different times and hence result in
a broadened output pulse.
?1999 S.O. Kasap, Optoelectronics (Prentice Hall)
43
t
Material Dispersion
Material Dispersion
Coefficient
Fundamental Mode
group delay time
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Material Dispersion
Δτ
=
L
Dm
Δλ
2
λ ⎛ d n ⎞
− ⎜
⎟
c ⎝ dλ
⎠
Dm ≈ 2
τ
g
=
1 dβ01
=
dω
v g
44

Dispersion coefficient (ps km
30
-1 nm-1 )
20
10
0
-10
-20
-30
1.1
901 37500 ?1999 光電導論 S.O. Kasap, Optoelectronics (Prentice Hall)
λ 0
Dm
Dm + Dw
D w
1.2 1.3 1.4 1.5 1.6
λ (μm)
Material dispersion coefficient (D m) for the core material (taken as
SiO 2), waveguide dispersion coefficient (D w) (a = 4.2 μm) and the
total or chromatic dispersion coefficient D ch (= D m + D w) as a
function of free space wavelength, λ.
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Chromatic Dispersion or Total Dispersion
Chromatic Dispersion
Chromatic Dispersion
Coefficient
Δτ
= | D
L
m
+ D | Δλ
w
D D D + =
m
Dispersion Shifted Fiber (DSF): to design the waveguide dispersion to
shift the zero (material) dispersion from 1330 nm to 1550 nm
by reducing the core radius and increasing the core doping
Dispersion flattened Fiber: to use multiple clad fiber to control the
total chromatic dispersion that is flattened between two wavelength
DFF
w
45
47
Waveguide Dispersion
Waveguide Dispersion
Coefficient
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20
10
-10
-20
-30
Waveguide Dispersion
Δτ
=
L
D
w
Dw
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?1999 S O K Ot l t i (P ti H ll)
Δλ
1.
984N
≈ − 2
( 2πa)
2cn
Dispersion coefficient (ps km -1 nm -1 )
30
0
λ 1
D w
D m
λ 2
1.1 1.2 1.3 1.4 1.5 1.6 1.7
λ (μm)
g 2
2
2
D ch = D m + D w
n
r
46
Thin layer of cladding
with a depressed index
Dispersion flattened fiber example. The material dispersion coefficient (Dm) for the
core material and waveguide dispersion coefficient (Dw) for the doubly clad fiber
result in a flattened small chromatic dispersion between λ1 and λ2. 48

Profile Dispersion: dispersion due to Δ=Δ(λ)
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Polarization Modal Dispersion (PMD)
Δτ
=
L
Dp
Δλ
(originates from material dispersion)
Polarization Dispersion: dispersion due to an-isotropy, n 1x-n 1y
Due to anisotropic composition, geometry, strain …
Different group delays even with monochromatic source
Example 2.5.1: Material dispersion
By convention , the width Δλ of the wavelength spectrum of the
source and the dispersion Δτ refer to half-power widths and not
widths from one extreme end to the other.
Δλ1/2: width of intensity vs. wavelength spectrum between the half
intensity points (i.e. linewidth)
Δτ1/2: width of the output light intensity vs. time signal between the
half-intensity points
(a) Estimate the material dispersion effect per km of silica fiber
operated from a light emitting diode (LED) emitting at 1.55μm
with a linewidth of 100nm.
(b) What is the material dispersion effect per km of silica fiber
operated from a laser diode emitting at the same wavelength with
a linewidth of 2nm?
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49
51
Polarization modal dispersion
n 1 x // x
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t
n 1 y // y
E
Input light pulse
E y
Output light pulse
z
Core
E x
E y
Intensity
E x
Δτ
Δτ = Pulse spread
Suppose that the core refractive index has different values along two orthogonal
directions corresponding to electric field oscillation direction (polarizations). We can
take x and y axes along these directions. An input light will travel along the fiber with E x
and E y polarizations having different group velocities and hence arrive at the output at
different times
?1999 S.O. Kasap, Optoelectronics (Prentice Hall)
Example 2.5.2: Material, waveguide, and chromatic dispersion
Consider a single mode optical fiber with a core of SiO2- 13.5%GeO2 for which the
material and waveguide dispersion coefficients are shown in the following figure.
Suppose the fiber is excited from a 1.5μm laser source with a linewidth Δλ1/2 of 2 nm.
(a) What is the dispersion per km of fiber if the core diameter 2a is 8 μm?
(b) What should be the core diameter for zero chromatic dispersion at λ = 1.5μm?
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Dispersion coefficient (ps km -1 nm -1 )
20
10
0
–10
SiO 2 -13.5%GeO 2
–20
1.2 1.3 1.4 1.5 1.6
λ (μm)
D w
D m
a (μm)
4.0
3.5
3.0
2.5
t
50
52

2-6 Bit Rate, Dispersion,
Electrical and Optical Bandwidth
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RZ
Information
Digital signal
t
Emitter
Input
Photodetector
Information
Output
Input Intensity
Output Intensity
² τ1/2
Very short
light pulses
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0
T
Fiber
t
0
~2Δτ1/2 An optical fiber link for transmitting digital information and the effect of
dispersion in the fiber on the output pulses.
?1999 S.O. Kasap, Optoelectronics (Prentice Hall)
53
55
t
Bit rate capacity B (bits/sec) is related to the dispersion characteristics.
The dispersion is measured by the pulse spread:
Full width at half power (FWHP) Δτ1/
2
Full width at half maximum (FWHM)
There is no inter-symbol interference
peak-to-peak separation 2 Δτ
1/
2
the maximum bit rate
Intuitive return-to-zero (RZ) bite rate (or data rate)
Non-return to zero (NRZ) bite rate 2B
901 37500 光電導論
NRZ
−1
/ 2
=
Output optical power
1
0.61
0.5
Bit Rate and Dispersion
e 0.
607 Δτ
2σ
2σ
rms =
Δτ1/2
T = 4σ
901 37500 ?1999 S.O. 光電導論 Kasap, Optoelectronics (Prentice Hall)
0.
5
B ≈
Δτ
A Gaussian output light pulse and some tolerable intersymbol
interference between two consecutive output light pulses (y-axis in
relative units). At time t = σ from the pulse center, the relative
magnitude is e-1/2 = 0.607 and full width root mean square (rms)
spread is Δτrms = 2σ.
1/
2
t
54
56

For a Gaussian pulse,
Root-mean-square deviation or
Root-mean-square (rms) dispersion
Full-width rms time spread: between the rms points of the pulse
Maximum RZ bit rate and rms dispersion
Maximum bit rate X distance
Total rms dispersion
901 37500 光電導論
901 37500 光電導論
σ
τ 2σ
rms
0.
25 0.
59
B ≈ =
σ Δτ1
2
= Δ
0.
25L
0.
25
BL ≈ =
σ | D |
2
2
1 −t
2
σ
ht = e 2
()
σ = 0. 425Δτ
1 2
2
intermodal
2 ( 2πσ
)
σ = σ + σ
f op
0.
19
≈ 0. 75B
≈
σ
12
ch
σ λ
2
intramodal
57
59
Sinusoidal signal
Emitter
t
f = Modulation frequency
P i = Input light power
0
Optical and Electrical Bandwidth
Optical bandwidth for Gaussian dispersion
t
f op
Optical
Input
0.
19
≈ 0. 75B
≈
σ
Fiber
0
Optical
Output
P o = Output light power
Photodetector
Sinusoidal electrical signal
An optical fiber link for transmitting analog signals and the effect of dispersion 58in
the
901 fiber 37500 on the 光電導論 bandwidth, fop. Example 2.6.1: Bit rate and dispersion
t
Electrical signal (photocurrent)
1
0.707
1 kHz 1 MHz 1 GHz
fel Po / Pi 0.1
0.05
1 kHz 1 MHz 1 GHz
fop Consider an optical fiber with a chromatic dispersion coefficient
8 ps km-1 nm-1 at an operation wavelength of 1.5μm. Calculate
(a) the bit rate × distance product (BL),
(b) the optical and electrical bandwidths
for a 10km fiber if a laser diode source with a FWHP linewidth Δλ 1/2
of 2 nm is used.
901 37500 光電導論
60
f
f

2-7 The Graded Index (GRIN)
Optical Fiber
901 37500 光電導論
B'
nc c We can visualize a graded index
fiber by imagining a stratified
B
θB' c/nb
θB' A
Ray 2
B''
nb b
medium with the layers of refractive
indices na > nb > nc ... Consider two
close rays 1 and 2 launched from O
O
2 θB
1 c/na θA M
Ray 1
na a
O'
at the same time but with slightly
different launching angles. Ray 1
just suffers total internal reflection.
Ray 2 becomes refracted at B and
reflected at B'.
?1999 S.O. Kasap, Optoelectronics (Prentice Hall)
901 37500 光電導論
61
63
O
O
901 37500 光電導論
O' O''
3
2
1
3
2
1
2
3
?1999 S.O. Kasap, Optoelectronics (Prentice Hall)
(a) TIR (b)
TIR
n decreases step by step from one layer
to next upper layer; very thin layers.
901 37500 光電導論
n 2
n 2
n 2
n 1
n 1
n
n
(a) Multimode step
index fiber. Ray paths
are different so that
rays arrive at different
times.
(b) Graded index fiber.
Ray paths are different
but so are the velocities
along the paths so that
all the rays arrive at the
same time.
Continuous decrease in n gives a ray
path changing continuously.
(a) A ray in thinly stratifed medium becomes refracted as it passes from one
layer to the next upper layer with lower n and eventually its angle satisfies TIR
(b) In a medium where n decreases continuously the path of the ray bends
continuously.
?1999 S.O. Kasap, Optoelectronics (Prentice Hall)
62
64

The profile index
The coefficient of index grating
The intermodal dispersion is
minimum when
Dispersion in graded-index fiber
901 37500 光電導論
γ ⎡ ⎛ r ⎞ ⎤
n = n1
⎢1−
2Δ⎜
⎟ ⎥
⎢⎣
⎝ a ⎠ ⎥⎦
n = n
2
4 + 2Δ
γ = ≈ 2 1
2 + 3Δ
σ
intermode
L
1/
2
( − Δ)
n1
≈ Δ
20 3c
2
; r < a
; r = a
Example 2.7.1: Dispersion in a graded-index fiber and bit rate
Consider a graded index fiber whose core has a diameter of 50μm and
a refractive index of n 1 = 1.480. The cladding has n 2 = 1.460. If this
fiber is used at 1.30μm with a laser diode that has a vary narrow
linewidth, what will the bit rate × distance produce be?
Evaluate the BL product if this were a multimode step index fiber
given that the light output is nearly rectangular so that σ≈0.29 Δτ in
which Δτ is the full spread.
901 37500 光電導論
65
67
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901 37500 光電導論
2-8 Light Absorption and
Scattering
66
68

E
901 37500 光電導論
Medium
?1999 S.O. Kasap, Optoelectronics (Prentice Hall)
901 37500 光電導論
Incident wave
k
z
Attenuation of light in the
direction of propagation.
A dielectric particle smaller than wavelength
Scattered waves
Through wave
Rayleigh scattering involves the polarization of a small dielectric
particle or a region that is much smaller than the light wavelength.
The field forces dipole oscillations in the particle (by polarizing it)
which leads to the emission of EM waves in "many" directions so
that a portion of the light energy is directed away from the incident
beam.
?1999 S.O. Kasap, Optoelectronics (Prentice Hall)
69
71
E x
901 37500 光電導論
A solid with ions
Light direction
k
Lattice absorption through a crystal. The field in the wave
oscillates the ions which consequently generate "mechanical"
waves in the crystal; energy is thereby transferred from the wave
to lattice vibrations.
?1999 S.O. Kasap, Optoelectronics (Prentice Hall)
Ionic polarization
usu. IR region
high absorption
Generates lattice waves
Others:
microwave oven
molecule vibration
2-9 Attenuation in Optical Fibers
901 37500 光電導論
z
70
72

901 37500 光電導論
10
5
1.0
0.5
0.1
0.05
(2.8/3 μm)
Rayleigh
scattering
OH - absorption peaks
(1/(2/2.8+1/9) μm) 1550 nm Si-O
1310 nm
Lattice
absorption
(9 μm)
0.6 0.8 1.0 1.2 1.4 1.6 1.8 2.0
Wavelength (痠) μm
Illustration of a typical attenuation vs. wavelength characteristics
of a silica based optical fiber. There are two communications
channels at 1310 nm and 1550 nm.
?1999 S.O. Kasap, Optoelectronics (Prentice Hall)
10
5
1.0
0.5
0.1
0.05
901 37500 光電導論
Rayleigh
scattering
(2.8/2 μm)
OH - absorption peaks
1310 nm
Lattice
absorption
0.6 0.8 1.0 1.2 1.4 1.6 1.8 2.0
Wavelength (痠) μm
Illustration of a typical attenuation vs. wavelength characteristics
of a silica based optical fiber. There are two communications
channels at 1310 nm and 1550 nm.
?1999 S.O. Kasap, Optoelectronics (Prentice Hall)
1550 nm
73
75
901 37500 光電導論
4/23/2009
901 37500 光電導論
Attenuation coefficient
Attenuation coefficient
in dB/length
Rayleigh scattering
in silica
Field distribution
θ θ
Pout in
1
α = −
P
dP
dx
⎛ P
=
⎜
L ⎝ P
ln
1
α
in
out
⎞
⎟
⎠
= P exp( −αL)
1 ⎛ P ⎞ in
α =
⎜
⎟
dB 10log
L ⎝ Pout
⎠
10
α dB = α = 4.
34α
ln( 10)
Cladding
Core
3
8π
α R ≈ 4
3λ
2 2
( n −1)
βT
kBT
f
θ
θ > θc θ′
R
Microbending
θ′ < θ
74
Escaping wave
Sharp bends change the local waveguide geometry that can lead to waves
escaping. The zigzagging ray suddenly finds itself with an incidence
angle θ′ that gives rise to either a transmitted wave, or to a greater
cladding penetration; the field reaches the outside medium and some light
energy is lost.
?1999 S.O. Kasap, Optoelectronics (Prentice Hall)
76

α B (m -1 ) for 10 cm of bend
10 2
10 −1
10 −2
10 −3
901 37500 光電導論 ?1999 S.O. Kasap, Optoelectronics (Prentice Hall)
901 37500 光電導論
10
1
λ = 633 nm
V ≈ 2.08
λ = 790 nm
V ≈ 1.67
0 2 4 6 8 10 12 14 16 18
Radius of curvature (mm)
Measured microbending loss for a 10 cm fiber bent by different amounts of radius of
curvature R. Single mode fiber with a core diameter of 3.9 μm, cladding radius 48 μm,
Δ = 0.004, NA = 0.11, V ≈ 1.67 and 2.08 (Data extracted and replotted with Δ correction
from, A.J. Harris and P.F. Castle, IEEE J. Light Wave Technology, Vol. LT14, pp. 34-
40, 1986; see original article for discussion of peaks in αB vs. R at 790 nm).
77
Example 2.9.1: Rayleigh scattering limit
What is the attenuation due to Rayleigh scattering at around the
λ = 1.55μm window given that pure silica (SiO 2 ) has the following
properties: T f = 1730°C (softening temperature);
β T = 7 x 10 -11 m 2 N -1 (at high temperatures);
n = 1.4446 at 1.5μm.
α
≈
3
=
R
8π
≈
3λ
3
4
2 ( n −1)
3
8π
2 2 −11
−23
( 1.
4446 −1)
( 7×
10 )( 1.
38×
10 )( 1.
730 + 273)
−6
4
( 1.
55×
10 )
3.
27×
10
−5
Attenuation
α
dB
= 4.
34α
m
R
−1
=
2
=
β
T
k
B
T
3.
27×
10
f
−2
km
−1
−2
−1
−1
( 4.
34)(
3.
27×
10 km ) = 0.
142dBkm
79
Example 2.9.1: Rayleigh scattering limit
What is the attenuation due to Rayleigh scattering at around the
λ = 1.55μm window given that pure silica (SiO 2) has the following
properties: T f = 1730°C (softening temperature);
β T = 7 x 10 -11 m 2 N -1 (at high temperatures);
n = 1.4446 at 1.5μm.
901 37500 光電導論
Example 2.9.2: Attenuation along an optical fiber
The optical power launched into a single-mode fiber from a laser
diode is approximately 1 mW. The photodetector at the output
required a minimum power of 10 nW to provide a clear signal (above
noise). The fiber operates at 1.3μm and has an attenuation coefficient
of 0.4 dB km -1 . What is the maximum length of fiber that can be used
without inserting a repeater (to regenerate the signal)?
α
dB
1
L =
α
901 37500 光電導論
1 ⎛ P
= 10log
L ⎜
⎝ P
dB
in
out
⎛ P
10log
⎜
⎝ P
in
out
⎞
⎟
⎠
−3
⎞ 1 ⎛10
⎞
⎟ = 10log
= 125km
8
0.
4 ⎜
10 ⎟ −
⎠ ⎝ ⎠
The signal has to be amplified after a distance of about 50 ~
100km, and eventually regenerated by using a repeater.
78
80

901 37500 光電導論
n
n 1
n 2
901 37500 光電導論
2-10 Fiber Manufacture
r
Buffer tube: d = 1mm
Protective polymerinc coating
Cladding: d = 125 - 150 μm
Core: d = 8 - 10 μm
The cross section of a typical single-mode fiber with a tight buffer
tube. (d = diameter)
?1999 S.O. Kasap, Optoelectronics (Prentice Hall)
81
83
901 37500 光電導論
Preform feed
Thickness
monitoring gauge
Fiber Drawing
Furnace 2000蚓
Polymer coater
Ultraviolet light or furnace
for curing
Take-up drum
Capstan
Schematic illustration of a fiber drawing tower.
?1999 S.O. Kasap, Optoelectronics (Prentice Hall)
Vapors: SiCl 4 + GeCl 4 + O 2
Fuel: H 2
Burner
Deposited Ge doped SiO 2
(a)
Outside Vapor Deposition (OVD)
Deposited soot
Target rod
Rotate mandrel
?1999 S.O. Kasap, Optoelectronics (Prentice Hall)
901 37500 光電導論
(b)
Drying gases
Porous soot
preform with hole
Clear solid
glass preform
Furnace
(c)
Preform
82
84
Furnace
Drawn fiber
Schematic illustration of OVD and the preform preparation for fiber drawing. (a)
Reaction of gases in the burner flame produces glass soot that deposits on to the outside
surface of the mandrel. (b) The mandrel is removed and the hollow porous soot preform
is consolidated; the soot particles are sintered, fused, together to form a clear glass rod.
(c) The consolidated glass rod is used as a preform in fiber drawing.

Example 2.10.1: Fiber drawing
In a certain fiber production process a preform of length 110cm and
diameter 20 mm is used to draw a fiber. Suppose that the fiber
drawing rate is 5 m/s. What is the maximum length of the fiber that
can be drawn from this preform if the last 10 cm of the preform is
not drawn and the fiber diameter is 125 μm? How long does it take
to draw the fiber?
L d
L
f
f
2
f
=
= L d
901 37500 光電導論
2
p p
−3
( 1.
1−
0.
1m)(
20×
10 m)
−6
2
( 125×
10 m)
rate = 5m
/ s
Length(
km)
Time(
hrs)
=
=
Rate(
km / hr)
2
= 25600m
= 25.
6km
25600m
( 5m
/ s)(
60×
60s
/ hr)
= 1.
4hrs.
Typical drawing rates are in the range 5 – 20m/s so that 1.4hrs would be on
the long-side.
85