# Lecture Optical fibers

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```					                  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
16.711 Lecture 3 Optical fibers

Fiber modes --- single mode and multi-mode fibers
V-number
2a                                 2a
V         (n12  n2 )1/ 2 , Vcutoff 
2
(n12  n2 )1/ 2  2.41,
2

                                  c
Number of modes when V>>2.41
V2
M     ,
2
Normalized propagation constant
neff  n2
2      2

b               ,      b  (1.1428  0.996 / V ) 2 , for V between 1.5 – 2.5.
n12  n2
2

Mode field diameter (MFD)
1
2w  2a(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:

2a                                     V2
V             (n  n )
2     2 1/ 2
 91.44,   M     4181 ,

1     2
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:
2a
V            (n12  n2 )1/ 2  2.4, a < 2.1m
2


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
        ,   C~ 0.8dB/km m4
4
0.14dB loss @ 1.55m

• Bending loss and Bending radius
a
 exp(  R / Rc ), Rc             ,
n12  n2
3
16.711 Lecture 3 Optical fibers

Dispersions in single mode fiber
• Material dispersion
d             L       g   d   d 2n
vg       | 0 ,  g  ,         ( )  ( 2 ) ,
d             vg      L     d  c d

 d 2n
Dm  ( 2 ),          g  Dm L ,
c d

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  44 ps,      for the Laser
16.711 Lecture 3 Optical fibers

Dispersions in single mode fiber
• Waveguide dispersion
d             L         g    d     n2 (n1  n2 ) d 2 (Vb )
vg     | 0 ,  g  ,            ( )                  V          ,
d             vg         L     d            c        dV   2

 g   d    1.984N g 2            n2 (n1  n2 ) d 2 (Vb)
 ( )                , Dw                V         ,              g  Dm L ,
L     d   (2a) 2 2cn22
c        dV 2

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
 ( Dm  Dw ) ,
L

• 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
1
B              ,
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

1
B              ,   f 3dB  0.7 B,
2  1 / 2
• Bandwidth length product
0.25
BL         ,
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:
0.25
 1/ 2 / L  D  16 pskm1 ,   BL          36.9Gbs1km,
D
f 3dB  0.7 B  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 ],           ~                             2T02          0 
1
A(0,  )  A0 (2T0 ) exp( 
2 1/ 2
),                  ,
2 T0                                               2                     T0
dk                 1 d 2k
k ( )  k 0     | (   0 )         | ...,
d 0               2 d 2 0

k ( )        d                   d 2                        1
 ( )          0     | 0 (   0 )       | 0 ...   0  1   2 ( ) 2  ...,
c          d                   d 2                        2
~             ~             i
A( z,  )  A(0,  ) exp(  2 z ),
2
1  ~               i               A0               1
2 
A( z , t )         A(0,  0 ) exp(  2 z )d         exp[  2          ],
2               Q( z )       2T0 Q ( z )

i 2 z                    z
Q( z )  1           , T ( z )  [1  ( 22 ) 2 ]1/ 2 T0 ,
T02                     T0
16.711 Lecture 3 Optical fibers

Dispersion Management
• Pre compensation schemes
1.     Prechirp
Prechirped Gaussian Pulse:
~               2T02 1/ 2        2T02                                        (1  iC ) t 2
A(0,  )  A0 (        ) exp(             ),          A(0, t )  A0 exp[              ( ) ],
1  iC          2(1  iC )                                         2     T0
~            ~            i                 2T02 1/ 2        2T02          2 iCT0   2
2
A( z ,  )  A(0,  ) exp(  2 z )  A0 (
2
) exp[                              ],
2                 1  iC          2(1  C 2 )         (1  C 2 )
1
 0  (1  C 2 )1/ 2        ,
T0

1     ~              i               A0               1
A( z , t ) 
2    A(0,  0 ) exp(  2 z )d 
2               Q( z )
exp[  2
2T0 Q ( z )
],

(C  i )  2 z                  C 2 z     z
Q( z )  1                    , T ( z )  [(1  2 ) 2  ( 22 ) 2 ]1/ 2 T0 ,
T02                           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 ,    LD  T02 /  2 ,
1 C 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
TFWHM         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c
D   2  2 ,  2  24 ps / km,
2


C 2 z 2  2 z 2 1 / 2
T ( z )  [(1       2
)  ( 2 ) ] T0  T0 ,
T0        T0

z  12km,
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

n1[1  (  / a) ];   a,
n(  )  
 n1(1  )  n2 ;      a,

d 2  1 dn
      ,
dz  2
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
n1
 L           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:
n1
 L           2  0.026 ns,    BL 
0.25L
 9.6Gbs 1km,
20 3c                           

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