Transcript Optical Fiber Communications

Optical Fiber Communications
Lecture 10
Topics
• Single Mode Fiber
– Mode Field Diameter
– Propagation Modes in Single Mode Fiber
• Graded Index Fiber
Single Mode Fiber
• Single Mode fiber are constructed by
– letting dimensions of core diameter be a few
wavelengths and
– by having small index difference between core and
cladding
• In practice core cladding index difference
varies between 0.1 and 1.0 percent
• Typical single mode fiber may have a core
radius of 3 micron and NA =0.1 at wavelength
= 0.8 micron
Mode Field Diameter
• Geometric distribution of light in propagation
mode is important when predicting
performance characteristics.
• Mode Field Diameter (MFD) is analogues to
core diameter in multimode fibers except not
all light that propagates is carried in the core
• Many models for characterizing MFD with
main consideration of how to approximate
electric field distribution
MFD
• Assume electric field distribution to be
Gaussian
E(r )  E0 exp(r 2 / W02 )
• Where r is radius, Eo is field at zero radius, and
Wo is the width of the electric field
distribution
• Take width 2Wo of MFD to be 2e-1radius of
optical electric field (e-2 of optical power)
MFD
• The MFD width 2Wo can be defined for LP01
mode as


3 2
2
r
E
(
r
)
dr
 


2Wo  2 0


2
  rE (r )dr 
 0


1/ 2
Propagation Modes in Single Mode
Fiber
• In single mode fiber there are two
independent degenerate propagation modes
• These modes are similar but their polarization
planes are orthogonal
• Electric field of light propagating along the
fiber is linear superposition of these two
polarization mode
Fig. 2-24: Polarizations of fundamental mode
contd
• Choose one of the modes to have its
transverse electric field along x direction and
the other in y direction
• In ideal fiber, two modes are degenerate with
equal propagation constants and will maintain
polarization state injected into the fiber
• In actual fiber there are imperfections, such as
asymmetrical lateral stress, noncircular cores
and variations in refractive index profiles
• These imperfections break circular symmetry
contd
• The modes propagate with different phase
velocities and the difference between their
effective refractive indices is called
birefringence
• Or
B f  ny  nx

2

(n y  nx )
• If light is injected into the fiber so that both
modes are excited, then one will be delayed in
phase
contd
• When this phase difference is an integral
m*pi, the two mode will beat at this point and
input polarization state will be reproduced.
• The length over which this beating occurs is
the fiber beat length
Lp  2 / 
Example
• A single mode optical fiber has a beat length
of 8 cm at 1300nm, the modal birefringence

1.3x106
5
B f  ny  nx 


1
.
63
x
10
Lp
8x102
• Or

2
2

 78.5m1
Lp 0.08
Graded Index Fiber Structure
• In graded index fiber, core refractive index
decreases continuously with increasing radial
distance r from center of fiber and constant in
cladding
r  1/ 2


n
[
1

2

(
) ]
for 0  r  a
 1

n(r )  
a

1
/
2
n (1  2)  n (1  )  n for r  a
1
2
 1

• Alpha defines the shape of the index profile
• As Alpha goes to infinity, above reduces to
step index
n n
n n
• The index difference is


2
1
2n12
2
2
1
2
n1
contd
• NA is more complex that step index fiber since
it is function of position across the core
• Geometrical optics considerations show that
light incident on fiber core at position r will
propagate only if it within NA(r)
• Local numerical aperture is defined as

[n 2 (r )  n22 ]1/ 2  NA(0) 1  (r / a)
NA(r )  

0 for r  0
• And
NA(0)  [n (0)  n ]
2
2 1/ 2
2

for r  a 





2 1/ 2
2
 n n
2
1
 n1 2
contd
• Number of bound modes
M

 2
a k n 
2
2
2
1
Examples
If a = 9.5 micron, find n2 in order to design a
single mode fiber, if n1=1.465.
Solution,
V  2.405  (2a /  ) n12  n22  (2 4.25 /  ) 1.4652  n22
  820nm, n2  1.463
  1300nm, n2  1.46
  1550nm, n2  1.458
The longer the wavelength, the larger refractive
index difference is needed to maintain single
mode condition, for a given fiber
Examples
• Compute the number of modes for a fiber
whose core diameter is 50 micron. Assume
that n1=1.48 and n2=1.46. Wavelength = 0.8
micron.
• Solution
V0.82 m  (2a /  ) n12  n22  (2 25 / 0.82) 1.48 2  1.46 2  46.45
For large V, the total number of modes
supported can be estimated as
M  V 2 / 2  46.452 / 2  1079
Example
• What is the maximum core radius allowed for
a glass fiber having n1=1.465 and n2=1.46 if
the fiber is to support only one mode at
wavelength of 1250nm.
• Solution
Vcritical  2.405  (2a /  ) n12  n22


a  Vcritical /(2 n12  n22 )  2.405x1.25 / 2 1.4652  1.462  3.956m
HW
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Due 2/11/08
2-2
2-8
2-11
2-18
2-19
2-21
2-24
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