Transcript Document

16.711 Lecture 3 Optical fibers
Last lecture
• Geometric optic view of waveguide, numeric aperture
• Symmetric planar dielectric Slab waveguide
• Modal and waveguide dispersion in palnar waveguide
• Rectangular waveguide, effective index method
16.711 Lecture 3 Optical fibers
Today
• Fiber modes
• Fiber Losses
• Dispersion in single-mode fibers
• Dispersion induced limitations
• Dispersion management
• The Graded index fibers
16.711 Lecture 3 Optical fibers
Fiber modes --- single mode and multi-mode fibers
V-number
V
2a

(n12  n22 )1/ 2 , Vcutoff 
2a
c
(n12  n22 )1/ 2  2.41,
Number of modes when V>>2.41
V2
M
,
2
Normalized propagation constant
b
2
neff
 n22
n12  n22
,
b  (1.1428 0.996/ V )2 , for V between 1.5 – 2.5.
Mode field diameter (MFD)
2 w  2a (1 
1
),
V
16.711 Lecture 3 Optical fibers
Examples --- single mode and multi-mode fibers
1. Calculate the number of allowed modes in a multimode step index fiber, a = 100 m, core
index of 1.468 and a cladding index of 1.447 at the wavelength of 850nm.
Solution:
V
2a

(n  n )
2
1
2 1/ 2
2
 91.44,
V2
M
 4181,
2
2. What should be the core radius of a single mode fiber that has the core index of 1.468 and the
cladding index of 1.447 at the wavelength of 1.3m.
Solution:
V
2a

(n12  n22 )1/ 2  2.4, a < 2.1m
3. Calculate the mode field diameter of a single mode fiber that has the core index of 1.458 and
the cladding index of 1.452 at the wavelength of 1.3m.
Solution:
2w0  2a(1  1 / V )  10.1m,
16.711 Lecture 3 Optical fibers
Fiber loss
• Material absorption
silica electron resonance <0.4m
OH vibrational resonance ~ 2.73 m
Harmonic and combination tones ~1.39 m
1.24 m, 0.95 m
• Rayleigh scattering
Local microscopic fluctuations in density

C
4
,
C~ 0.8dB/km m4
0.14dB loss @ 1.55m
• Bending loss and Bending radius
 exp(R / Rc ), Rc 
a
,
n12  n23
16.711 Lecture 3 Optical fibers
Dispersions in single mode fiber
• Material dispersion
vg 
d
L
| 0 ,  g  ,
d
vg
 d 2n
Dm  ( 2 ),
c d
 g
d
 d 2n
 ( )  ( 2 ) ,
L
d
c d
 g  Dm L,
Example --- material dispersion
Calculate the material dispersion effect for LED with line width of 100nm and a laser with a
line width of 2nm for a fiber with dispersion coefficient of Dm = 22pskm-1nm-1 at 1310nm.
Solution:
  DmL  2.2ns, for the LED
  Dm L  44ps,
for the Laser
16.711 Lecture 3 Optical fibers
Dispersions in single mode fiber
• Waveguide dispersion
d
L
vg 
| 0 ,  g  ,
d
vg
 g
d
n2 (n1  n2 ) d 2 (Vb )
 ( )  
V
,
2
L
d
c
dV
 g
1.984N g 2
n2 (n1  n2 ) d 2 (Vb )
d
V
,
 ( ) 
, Dw  
c
dV 2
L
d
(2a) 2 2cn22
 g  Dm L,
Example --- waveguide dispersion
n2 = 1.48, and delta n = 0.2 percent. Calculate Dw at 1310nm.
Solution:
b  (1.1428 0.996/ V )2 , for V between 1.5 – 2.5.
d 2 (Vb )
V
 0.26,
dV 2
n2 (n1  n2 ) d 2 (Vb )
Dw  
V
 1.9 ps /(nm  km),
c
dV 2
16.711 Lecture 3 Optical fibers
• chromatic dispersion (material plus waveduide dispersion)
 g
L
 ( Dm  Dw ) ,
• material dispersion is determined by
the material composition of a fiber.
• waveguide dispersion is determined
by the waveguide index profile of a
fiber
16.711 Lecture 3 Optical fibers
• Polarization mode dispersion
 g
 D p  ,
L
• fiber is not perfectly symmetric,
inhomogeneous.
• refractive index is not isotropic.
• dispersion flattened fibers:
Use waveguide geometry and
index profiles to compensate
the material dispersion
16.711 Lecture 3 Optical fibers
• Dispersion induced limitations
• For RZ bit With no intersymbol interference
B
1
,
2 1/ 2
• For NRZ bit With no intersymbol interference
1
B
,
 1/ 2
16.711 Lecture 3 Optical fibers
Dispersion induced limitations
• Optical and Electrical Bandwidth
B
1
,
2 1/ 2
f 3dB  0.7 B,
• Bandwidth length product
BL 
0.25
,
D
16.711 Lecture 3 Optical fibers
Dispersion induced limitations
Example --- bit rate and bandwidth
Calculate the bandwidth and length product for an optical fiber with chromatic dispersion
coefficient 8pskm-1nm-1 and optical bandwidth for 10km of this kind of fiber and linewidth of
2nm.
Solution:
1/ 2 / L  D  16 pskm1 ,
BL 
0.25
 36.9Gbs 1km ,
D
f3dB  0.7B  2.8GHz,
• Fiber limiting factor absorption or dispersion?
Loss  0.25dB 10km  2.5dB,
16.711 Lecture 3 Optical fibers
Dispersion Management
• Pre compensation schemes
1.
Prechirp
Gaussian Pulse:
1 t
A(0, t )  A0 exp[ ( ) 2 ],
2 T0
 2T02
~
2 1/ 2
A(0,  )  A0 (2T0 ) exp(
),
2
 0 
dk
1 d 2k
k ( )  k0 
| (   0 ) 
| ...,
d 0
2 d 2 0
k ( )
d
d 2
1
 ( ) 
 0 
|0 (   0 ) 
|
...







 2 ( ) 2  ...,
0
1
2 0
c
d
d
2
i
~
~
A( z,  )  A(0,  ) exp(  2 z ),
2
A0
1  ~
i
1
A( z, t ) 
A
(
0
,

)
exp(

z


)
d



exp[

],
0
2
2



2
2
2T0 Q( z )
Q( z )
Q( z )  1 
 2 z 2 1/ 2
i 2 z
,
T
(
z
)

[
1

(
) ] T0 ,
T02
T02
1
,
T0
16.711 Lecture 3 Optical fibers
Dispersion Management
• Pre compensation schemes
1.
Prechirp
Prechirped Gaussian Pulse:
2T02 1/ 2
 2T02
~
A(0,  )  A0 (
) exp(
),
1  iC
2(1  iC )
A(0, t )  A0 exp[
(1  iC ) t 2
( ) ],
2
T0
2
2T02 1/ 2
 2T02
i
~
~
2
2 iCT0   2
A( z,  )  A(0,  ) exp(  2 z )  A0 (
) exp[
 
],
2
1  iC
2(1  C 2 )
(1  C 2 )
1
 0  (1  C 2 )1/ 2 ,
T0
A( z, t ) 
1
2
Q( z )  1 

A0
i
1
~
A
(
0
,

)
exp(

z


)
d



exp[

],
0
2
2

2
2T0 Q( z )
Q( z )
(C  i)  2 z
C 2 z 2  2 z 2 1/ 2
,
T
(
z
)

[(
1

)  ( 2 ) ] T0 ,
2
2
T0
T0
T0
16.711 Lecture 3 Optical fibers
Dispersion Management
1.
Prechirp
With T1/T0 = sqrt(2), the transmission distance is:
C  1  2C 2
L
LD ,
1 C 2
LD  T02 /  2 ,
16.711 Lecture 3 Optical fibers
Dispersion Management
Examples:
1. What’s the dispersion limited transmission distance for a 1.55m light wave system making
use of direct modulation at 10Gb/s? D = 17ps(km-nm). Assume that frequency chirping
broadens the guassian-shape by a factor of 6 from its transform limited width.
Solution:
1
 5 10 11 ( s ), T0  TFWHM / 1.66  31011 s,
2B
1
 0  (1  C 2 )1/ 2 , C  5.9,
T0
2
2c
D   2  2 ,  2  24 ps / km,

TFWHM 
T ( z )  [(1 
z  12km,
C 2 z 2  2 z 2 1/ 2
)  ( 2 ) ] T0  T0 ,
2
T0
T0
16.711 Lecture 3 Optical fibers
Dispersion compensation fiber or dispersion shifted fiber
•
Why dispersion compensation fiber:
• for long haul fiber optic communication.
• All–optical solution
D1L1  D2 L2  0
•
Approaches
 d 2n
Dm  ( 2 ),
c d
• longer wavelength has
a larger index.
make the waveguide
weakly guided so that
longer wavelength has a
lower index.
16.711 Lecture 3 Optical fibers
The Graded index fibers
n1[1  (  / a) ];   a,
n(  )  
  a,
 n1(1  )  n2 ;
d 2  1 dn

,
2
dz
n d
D1L1  D2 L2  0
•
  0 cos( pz)  0 ' sin( pz),
Approaches
p  (2 / a 2 )1/ 2 , z  2 / p,
Only valid for paraxial approximation
General case Intermode dispersion
 L
n1
2 ,
20 3c
Calculate the BL product of a grade index filber of 50m core with refractive index of n1 =
1.480 and n2 = 1.460. At 1.3 m.
Solution:
 L
n1
2  0.026ns,
20 3c
BL 
0.25 L

 9.6Gbs 1km ,