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
   • The Graded index fibers
                              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

The Graded index 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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