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8/11/2019 Modal Analysis of Step-Index Fibers
1/25
ECE 4006/5166 Guided Wave Optics
Robert R. McLeod, University of Colorado 65
Step-index silica fiber
Material and fabrication
Types and naming of modes
Derivation and solution of the WE
Solution of the WE
TE/TM modes
Hybrid modes LP modes
Modal analysis of step-index fibersIntroduction
8/11/2019 Modal Analysis of Step-Index Fibers
2/25
ECE 4006/5166 Guided Wave Optics
Robert R. McLeod, University of Colorado 66
Optical properties
SiO2 & SiO2/GeO2
0.6 0.8 1 1.2 1.4 1.6 1.8 2
l @mmD
1.47
1.48
1.49
1.5
1.51
ng
0.6 0.8 1 1.2 1.4 1.6 1.8 2
l @mmD
-600
-500
-400
-300
-200
-100
0
D
@
spH
mn
mk
LD
Loss[dB/km]
Modal analysis of step-index fibersMaterials and fabrication
Absorption:
Large at small wavelength due to Rayleigh scattering off of inhomogeneities
in the glass
Large at long wavelength due to molecular vibrational resonances
(absorption) farther out in the IR.
Is remarkably low loss around 1.5 microns
Dispersion:
Pure and doped silica have nearly identical dispersion
Zero dispersion around 1.3 microns
0.6 0.8 1 1.2 1.4 1.6 1.8 2
l@mmD
1.44
1.45
1.46
1.47
1.48
n
8/11/2019 Modal Analysis of Step-Index Fibers
3/25
ECE 4006/5166 Guided Wave Optics
Robert R. McLeod, University of Colorado 67
Fabrication of the preform
Modal analysis of step-index fibersMaterials and fabrication
8/11/2019 Modal Analysis of Step-Index Fibers
4/25
ECE 4006/5166 Guided Wave Optics
Robert R. McLeod, University of Colorado 68
Drawing fiber
4000o F
10-20m/sec
Several km of fiber are typical on a single reel.
Modal analysis of step-index fibersMaterials and fabrication
8/11/2019 Modal Analysis of Step-Index Fibers
5/25
ECE 4006/5166 Guided Wave Optics
Robert R. McLeod, University of Colorado 69
Types of modes
Exact solution
Meridonal rays (=0): TE & TM
Skew rays (0): HE & EH
Weakly guiding approximation
LP modes
Can be expressed as sum of TE,
TM, HE, EH that become
degenerate for smalln
Modal analysis of step-index fibersTypes of modes
8/11/2019 Modal Analysis of Step-Index Fibers
6/25
ECE 4006/5166 Guided Wave Optics
Robert R. McLeod, University of Colorado 70
Wave equation in cylindrical
for Ezand Hz
Monochromatic vector WE
zEyExEE zyx 2222
++=
r Cartesian vector
Laplacian
0
2
0
2
=
=+ EEkE
rrr
Scalar simplification
( ) ( ) ( ) ( )zzrEzrErzrEzrE zr ,,,,,,,, ++=r
Write E in cylindrical coordinates:
E radial
E azimuthal
E radial Ez
Ez does not couple to Er and E fields:
Modal analysis of step-index fibersDerivation of the wave equation
011
0
2
02
2
2
2
2
2
0
2
=+
+
+
=+
zzzz
zz
Ekz
EE
rr
Er
rr
EkE
Scalar WE for Ez
2 in cylindrical coord.
Ez
Note that Hzobeys the same W.E.
8/11/2019 Modal Analysis of Step-Index Fibers
7/25
ECE 4006/5166 Guided Wave Optics
Robert R. McLeod, University of Colorado 71
Solution of the WESeparation of variables
Modal analysis of step-index fibersSolution of the wave equation
( ) ( ) ( ) ( )zZrRzrEz =,, Separation of variables
Plug into wave equation. Note now ordinary differential eq.
011 202
2
2
2
22
2
=+++
+ ZRk
dzZdR
ddRZ
rdrdR
rdrRdZ
Multiply by r2/RZ
0111 22
02
22
2
2
2
22 =++
+
+ rk
dz
Zd
Zr
d
d
dr
dRr
dr
Rdr
R
Assume sinusoidal dependence in z: Z(z) = exp[-j z]
( ) 011 222022
2
22 =+
+
+ rk
d
d
dr
dRr
dr
Rdr
R
Depends on r Depends on rDepends on
( ) 22
2222
02
22 11
=
=+
+
d
drk
dr
dRr
dr
Rdr
R
8/11/2019 Modal Analysis of Step-Index Fibers
8/25
ECE 4006/5166 Guided Wave Optics
Robert R. McLeod, University of Colorado 72
Solution of the WERadial and azimuthal functions
( )
jeA
d
d ==
2
2
2
Solution of azimuthal equation. Since must be periodic,=integer.
02
222
0
2
2
22 =
+
+ R
rkr
dr
dRr
dr
Rdr
Solution of radial equation.
Note similarity to d2/dr2 + (k2 kz2-(/r)2) = 0. Solutions:
Modal analysis of step-index fibersSolution of the wave equation
0 2 4 6 8 10
z
-0.4
-0.2
0
0.2
0.4
0.6
0.8
1
0 1 2 3 4
z
0.5
1
1.5
2
2.5
3
3.5
4( )0
222
=+
+ fzfzfz
( ) ( )42
2
largecos
zzJ
zz( ) z
zz ezK
21
large
( )0
222
=+
+ fzfzfz
zero
1 2.4048 3.8317 5.1356 6.3802 7.5883 8.7715
2 5.5201 7.0156 8.4172 9.7610 11.0647 12.3386
3 8.6537 10.1735 11.6198 13.0152 14.3725 15.7002
4 11.7915 13.3237 14.7960 16.2235 17.6160 18.9801
5 14.9309 16.4706 17.9598 19.4094 20.8269 22.2178
Zeros of J:
Core: nco > N Core: ncl < N
8/11/2019 Modal Analysis of Step-Index Fibers
9/25
ECE 4006/5166 Guided Wave Optics
Robert R. McLeod, University of Colorado 73
Form of fields
( ) ( )
( )
( ) ( )
( )
>
0 has two possible angularphases and all have two possible polarizations
Modal analysis of step-index fibersLP modes
8/11/2019 Modal Analysis of Step-Index Fibers
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ECE 4006/5166 Guided Wave Optics
Summary The solution of the wave-equation in cylindrical
coordinates yields J in the core (like cos) and K in thecladding (like e-x).
Matching these functions across the boundary with theEM boundary conditions yields a complexcharacteristic equation for which has different
solutions m for each angular mode number. If there is no angular variation (=0), the solutions are
TE0m and TM0m.
Otherwise the solutions involve all 6 field componentsand are labeled HEm and HMm.
The HE11 mode is ~linearly polarized and never cutsoff.
The number of modes in terms of Vis roughly equal tothe number in a slab waveguide of the sameparameters, squared.
TE/TM modes occur for V> 2.405.
Increases in the decrease the maximum m by ~1/2. In the limit of weak guiding, sets of these modes have
degenerate effective indices and can thus be linearly
combined. The resulting LP modes are linearly polarized, have
small axial fields and have two degeneratepolarizations and (for modes with angular variation)two angular phases.
Modal analysis of step-index fibersSummary