# Dispersion in Single-Mode Fibers

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```					                      Dispersion in Single-Mode Fibers
The impact of dispersion on the input signal can be summarized as follows:

 0t   0 z
E ( x, y , z , t )   ( x, y )  e  j ( )                              s (t , z )    (1)
                                                  
Transverse field            Carrier        Pulse Envelope
distributi on
~
s (t , z )  F 1[ A ()  H ()]                                                              (2)
~
A ()  s (t ,0)
where, “F” stands for the Fourier transform.

In this section, we mainly focus on the pulse envelope s(t , z ) . Let us assume that the
transverse field distribution of the laser output is same as that of the fiber so that we don’t
have to worry about it. Let the pulse envelope of the laser output be si (t ) , i.e.
~
si (t )  s(t ,0) and A()  F[si (t )]  Si ()                       (3)
The fiber can be imagined as a linear system with the transfer function H () . Let the
fiber be of length L, inverse group speed 1 and dispersion coefficient  2 . Let the pulse
envelope of the output end of the fiber be so (t ) , i.e.
s(t , L)  so (t )
(4)
F[ so (t )]  S o ()  H ()  S i ()

si (t )                           so (t )
H ()
Si ()                            S o (  )  H ( ) S i (  )

Step 1: Input pulse envelope si (t ) is given.
Take its Fourier transform to obtain Si () .
Step 2: Multiply Si () by H () to get the output spectrum, So () .
Step 3: Take the inverse Fourier transform of So () to obtain the output pulse
envelope so (t ) .

Total field distribution at the output is obtained by
E ( x, y, L, t )   ( x, y )  e  j ( 0t   0 L )  s 0 (t )                         (5)
Example: (Prob.2 of tutorial 1)
Screen
Free space

Laser                     L
Z=0                      Z=L
Optical fiber               Screen

Laser                     L

Fig. a

The laser shown in the figure operates at 375 THz. It is turned on for 100ps and then
turned off. Sketch the pulse envelope at the screen if the medium is (a) free space (b)
fiber.
From the tutorial 1, we know that the laser output (ignoring the transverse field
distribution) can be written as
envelope 

                 carrier

F (t ,0)  rect (t / T0 )  A  cos(2f 0t )
f 0  375 THz ,          T0  100 ps                         (6)
Envelope
A
Carrier

t

(a) In tutorial 1, we have seen that the electric field intensity at the screen (z=L) is
given by
envelope
   
t  T1                
carrier

F (t , L)  rect (       )  A  cos[2f 0 (t  T1 )]
T0
T1  L / c, c  velocity of light in the free space             (7)
Let the pulse envelope by s0 (t ) , i.e.
t  T1
s0 (t )  rect (        )
T0                                            (8)
si (t )                           so (t )

t
100 ps                             100 ps

L/c  T1

At the laser                      At the screen

We see that the pulse envelope is delayed by T1  L / c which is the propagation
delay.
(b) In the case of an optical fiber, first let us consider the case  2  0 ,
si (t )  rect (t / T0 )
Step 1 :        S i ( f )  T0  sin c( f T0 )
H ( f )  e j 2f1L                                                 (9)

So ( f )  Si ( f )H ( f )

Step 2:                      T0  sin c( fT0 )  e j 2f1L                           (10)
Recall the Fourier transform property
g (t  T0 )  G ( f )  e j 2fT             0
(11)
Step 3: Using (10), (11), the output pulse envelope can be written as:
so (t )  si (t  1L)
t  1L
 rect (             )
T0                                                (12)

si (t )                            so (t )

t
100 ps                            100 ps

1 L
At the laser                       At the screen

When  2  0 , step 2 can be written as
So ( f )  T0  sin c( fT0 )  e j 2f1L j ( 2f ) 2 L / 2
2

(13)
It is not possible to find the inverse Fourier transform of Eq.(13) and one needs to do it
numerically using FFT (Fast Fourier Transform). Please look at the matlab code posted
on the web. The effect of dispersion on the input pulse is to broaden the pulse. The output
pulse might look like this:

si (t )                              so (t )

t
100 ps                               100 ps

1 L
At the laser                         At the screen

Relation between field and power
The power/unit area (= optical intensity) of an electromagnetic wave is given by
E2
I x
2                                                                 (1)
where E x is the electric field intensity and  is the impedance of a medium. The above
relation follows from the Poynting vector. The power flowing through any area A normal
to the direction of propagation is
E x2 A
P                                               (2)
2
2
2                      E
When complex notation is used, E x should be replaced by x . The electric field
intensity in a fiber is given by

j (
E x ( x, y, z, t )  ( x, y )  e  0t    s(t , z )
0z)
                     
Transverse field         Carrier       Pulse Envelope
distributi on                                                      (3)
Using Eqs. (2) and (3), we have

| s (t , z | 2  2 ( x, y ) A
or P  K | s (t , z ) |
2
P
2

In this course, we set the constant K to unity to obtain

P | s (t , z ) | 2

Fiber Design Considerations
Cut-off Wavelength
For high bit-rate (>2.5GHz) and long-haul (>100Km) applications, it is absolutely
essential that the fiber is single-moded. The single mode condition is given by
2a 2
V        (n1  n2 )1 / 2  2.405
2


For examples, if   1.55m , a  4m and (n1  n2 )  0.1 , V  1.62 . Therefore, this
2   2 1/ 2

fiber is single-moded at this wavelength. However, if   0.7 m corresponding to the
optical communication in the visible spectrum, V  3.59 and the fiber is not single-
moded at this wavelength. For the given fiber parameters, we can define the cut-off
wavelength as
2a (n12  n2 )1 / 2
2
C 
2.405
Cut-off wavelength is the frequency above which the optical fiber is single-moded.

Fiber Loss
Fiber
Pin                                      Pout

L
In an optical fiber, the power attenuation is governed by
dP
 P
dz                                                                       (5)
where  is the attenuation coefficient and P is the optical power. Solving Eq.(5),
dP
 dz
P
or
ln P  z  k
P  k  e z
If Pin is the input power, k can be calculated as
P(0)  k  Pin
If the fiber length is L, the output power can be written as
Pout  Pin exp( L)
The loss power in dB unit through the fiber of length L is given by
P
loss(dB)  10 log 10 out  10 (L) log 10 e  4.343  L
Pin
The loss per unit length (dB/Km)  4.343 .  
Here,  is the attenuation coefficient in the unit of Km .
1

Example
1
A fiber of length 40Km has a loss coefficient of 0.046Km . Find the total loss.
The loss per unit length (dB/Km) = 4.343 0.046  0.2dB/ Km .
Total loss = 0.2  40  8dB

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 views: 13 posted: 11/13/2011 language: English pages: 6