Download FOC-lecture3

MODAL PROPAGATION INSIDE AN OPTICAL
FIBER (WAVE MODEL)
1. Basics
1. The ray model does not predict whether, after total internal reflection there will be
some field in the cladding. Also it does not predict that rays can be launched at only
discreet angles in an optical fiber. So there are certain limitations to ray model.
2. For a complete description of light propagation inside an optical fiber we have to go in
for a more rigorous model, called the WAVE MODEL. Here we shall discuss the
propagation of light, treating light as an electromagnetic wave.
3. Energy gets guided inside an optical fiber i.e. the energy is confined inside the core
and the field exponentially dies down in the cladding from the core-cladding
interface.
4. In a practical fiber the cladding surrounded by a protecting layer. Generally, by
time the field reaches that layer, the field dies down significantly so that
protecting layer does not affect the propagation of the wave significantly.
In other words, we can take the cladding to be of infinite size in our analysis.
So whole propagation of light is governed by the core cladding interface and
interface between the cladding and other protecting layers does not affect
propagation.
2. Cylindrical Geometry and Maxwell’s Equations :
Figure (2.1)
the
the
the
the
1. As shown above in Figure 1, the core of the optical fiber is a cylinder of radius a , and
of refractive index n1 . The light energy propagates along the axis of the fiber. The
refractive index of cladding is n2 and the cladding is of infinite radius.
2.
The appropriate coordinate system to analyze this problem is the cylindrical
coordinate system, ( r ,  , z ) . The wave propagates in the z direction and it has some
distribution in the cross sectional plane defined by r ,  . Any radial direction from the
center of fiber is denoted by r and the azimuthal angle measured from a reference
axis in the cross-sectional plane is denoted by  .
3. For any electromagnetic problem we start with the Maxwell’s equations. Here we
investigate the propagation of light in the fiber without worrying about the origin of
the light inside the fiber. In other words we assume that the Maxwell’s equations
which govern the electromagnetic radiations inside the fiber are source free.
4. Maxwell’s equations for a source-free medium (i.e., the charge density and the
conduction current densities in the medium are zero):
(a) .D  0
(2-1)
(b) .B  0
(2-2)
B
(c)  E  
(2-3)
t
D
(d)  H  
(2-4)
t
where D is the electric displacement vector, B is the magnetic flux density, E
is the electric field, and H is the magnetic field intensity.
We have two more equations, called the constitutive relations as
B  H
& D E
Where  is the permeability of the medium and  is the permittivity of the
medium.
Now we have to decoupled the equations (3) & (4),
For this take curl of equation (3)
   E   
B
t
Substituting for B   H and interchanging the space and time derivatives, we get

  ( H )
t

   (  H ) (Since we assume the homogeneous medium  is
t
not a function of space).
Now substituting from eqn.(4) we get
 E  
  D 


t  t 
 
     E
t t
2
 E
    E    2
t
From vector algebra we have the identity
    E  
    E     E   2 E
so in the above equation can be simplified to
2 E
    E    2 E    2
t
For a homogeneous medium  is not a function of space so eqn (1) gives
E  0
Therefore we get
2 E
 E   2
t
2
(2-5)
This equation is called the Wave Equation.
If we do the similar analysis for the magnetic field we get the same wave equation for
the magnetic field
 2 H  
2 H
t 2
(2-6)
To analyse the propagation of light inside an optical fiber, we have to solve the wave
equation for appropriate boundary conditions
5. Since we have chosen the cylindrical coordinate system  r ,  , z  , we
write the wave equation for the electric field in the cylindrical coordinates as:
 2 E 1 E 1  2 E  2 E
2 E
(2-7)





r 2 r r r 2  2 z 2
t 2
So this is the generalized wave equation in a cylindrical coordinate system.
6. The electric field, E is a vector quantity having three components and the
magnetic filed, H is a vector quantity also having three components. So we have
total 6 field components. However all these components are not independent of
each other because E & H are related through the Maxwell’s equations. We can
take two components as independent components and express the remaining four
components in terms of the independent components. Since in this case the wave
propagates along the axis of the fiber i.e., in the z direction, generally the two
components, Ez & H z are taken as independent components ( also called the
longitudinal components) and we express the other four transverse field
components i.e., Er , E , H r , H in terms of these two components.
Wave equation is solved for these two components and from the solution we get, we
go to the Maxwell’s equations and get the other four components.
7. (a) The transverse components are related to the longitudinal components as
Er 
 j  Ez  H z 



q 2  r
r  
E 
H z 
 j   Ez
 

2 
q  r 
r 
Hr 
 j  H z  Ez 



q 2  r
r  
H 
E 
 j   H z
  z 
2 
q  r 
r 
where q 2   2    2 (  is the propagation constant of the wave along the
axis of the fiber.)
b)
The wave equation is to be solved for Ez and H z which are scalar quantities. So in
general the wave equation is solved for a scalar function  .
That is, Ez , H z 
Writing the above wave equation in terms of scalar function  we get
 2 1  1  2  2
 2





r 2 r r r 2  2 z 2
t 2
________ (2-8)
Let us assume that the fields are time harmonic fields, that is, all electric and
magnetic field components have time variation e j t .

e jt

 j
t
2
  2
t 2
So equation (8) becomes
 2 1  1  2  2

 2
 2   2 
2
2
r
r r r 
z
c) Energy has to propagate along the axis of fiber, so that the solution is a traveling
wave type solution along the axis that is, the z direction. If wave travels in z
direction then its z-variation should be e  j  z . That is to say that
 e j  z ,

2
  2
2
z
d) From Figure1, if we move only in the  direction i.e., in the azimuthal direction in a
cross sectional plane, after one complete rotation we reach to the same location. i.e.
the function  is periodic in  over 2 . In  direction, the function is a harmonic
function that is,
 e j
where  is an integer. This functional form represents a field which will repeat itself
after one rotation or when  changes by 2 .
8. For a general solution of the wave equation we apply separation of variables. We
assume the solution as
  R  r     Z  z  T t 
We have already defined the functions ( ), Z ( z ), T (t ) . Substituting for these functions
in the wave equation only unknown function remains to be evaluated is R ( r ) .
The wave equation therefore becomes

d 2 R 1 dR   2

  2   2   2   R  0
2
dr
r dr  r

________ (2-9)
Since q 2   2    2 as was defined earlier, the final wave equation will be
d 2 R 1 dR  2  2 

q  2 R  0
dr 2 r dr 
r 
_________ (2-10)
This equation is the Bessel’s equation and solutions of this equation are called the
Bessel functions.
Field distribution in the transverse plane in the radial direction will follow the Bessel’s
equation and the solution of these equation would be Bessel functions.
So we find that the variation of field in transverse direction is nothing but the Bessel
functions.
9. We have a variety of solutions of Bessel’s equations depending upon the parameters
 and q .  is an integer and a positive quantity.
Depending upon the choice of q i.e., a) real , b) imaginary, c) complex,
we get different solutions to the Bessel’s equations. So to choose the proper Bessel
function we must have the physical understanding of which model is most appropriate
for our configuration.
Figure (2.2)
a)
We have seen from the ray model that the rays can be launched at discreet angles
inside an optical fiber. For a particular angle there in no single ray. All rays which
lie on the surface of a cone are equip-probable rays. Basically the ring of rays is
simultaneously launched inside the optical fiber (as shown in the figure 2). Each ray
has a wave front associated with it. If we now take any cross sectional plane, these
sets of wave fronts which essentially represent rays of the same angle will give
interference. Somewhere the interference is destructive and somewhere it is
constructive. So when the wave fronts move along the fiber inside the core they
exhibit field distribution which will have maxima and minima.
b) When total internal reflection takes place, then the field must die down away from
the core cladding boundary. If the field does not decay, then the energy is not
guided and it moves in some other directions or the energy is lost. Here since we
are interested in the guided fields, we see only those field distribution which die
down away from the core-cladding interface.
c) So now from the physical understanding of the problem we seek a solution which
has an oscillatory behavior inside the core and decaying behavior inside the
cladding. Any other solution is not acceptable because these solutions are not
consistent with the physical understanding of the modal propagation.
10. There are three different types of Bessel functions depending upon the nature of q .
(a)
If q is real then the solutions are called
 J  qr  Bessel functions
 N (qr ) Neumann functions
The quantity  is called the order of the function and ( qr ) is called the argument of
the function. Plots of the two functions as a function of their arguments are shown in the
Figure 3-4.
Figure (2.3)
Figure (2.4)
As we see in figure 3, except J 0 all the other Bessel functions start from zero as the
argument goes to zero. Only J o starts from 1 as its argument is zero. All of them
have oscillatory behavior and their amplitude slowly decreases as the argument
increases.
Figure 4 shows the behavior of the Neumann function as a function of its argument,
qr . The important thing to note is, the Bessel functions J are finite for all
values of the argument , whereas the Neumann functions finite at all
arguments except zero. When the argument tends to zero, the Neumann
functions tend to  .
(b)
If q is imaginary, we get solutions of the Bessel’s equation as
I  qr / j 
K (qr / j )
These are called the Modified Bessel functions of first and second kind respectively.
Note: Since q is imaginary, (qr / j ) is a real quantity. So the argument of the modified
Bessel functions is real.
Figure (2.5)
Figure(2.6)
The modified Bessel functions are shown in the figures 5 and 6. The I functions
are monotonically increasing functions of  qr / j  , and K functions are
monotonically decreasing functions of (qr / j ) .
(c) If q is complex
Then the solutions are called the Hankel functions
H(1)  qr  1st kind
H(2)  qr  2nd kind
In our analysis, q will be either real or imaginary. Therefore we have to deal with
only Bessel, Neumann and Modified Bessel functions only. The Hankel functions are
needed for analyzing propagation in a lossy medium.
(d)
Now q 2   2    2 , where  is the propagation constant of the wave along the
z direction. If we assume the situation is lossless i.e. when the wave travels
in z direction, its amplitude does not change as a function of z , then  , which is
defined as phase change per unit length, should be a real quantity.
If  becomes imaginary, the function e j  z becomes an exponentially decaying
function, and there is no wave propagation.
(e)
For wave
propagation inside an optical fiber we assume that the materials are
lossless materials.
Then the dielectric constant  is also a real quantity.
Now  2   real quantity
 2  real quantity so
q 2   2    2 is also a real quantity albeit it can be positive or negative. In
other words q can be real or imaginary depending upon whether  2  is greater or
lesser than  2 . Hence for a lossless situation, solution is nt Henckel function.
(If we consider the losses, then the solution is Henckel functions of 1st &2nd
kind.)
So for a lossless case, we have a solutions either Bessel and Neumann functions
or modified Bessel functions of first and second kind.
(f) As far as guided wave propagation is concerned, there is oscillatory behavior
inside the core and in cladding the field must die down monotonically. Therefore
inside the core Modified Bessel functions is not the proper solution. Only Bessel
function or Neumann function could be solutions inside the core.
Let us take dielectric constant  r1 ( n12 ) for the core. Then we have
q 2   2  0 r1   2   2  0 n12   2
For oscillatory kind of behavior of the solution inside the core, q 2 is positive
because q is a real number. Therefore for a guided mode we must have
 2 1   2 i.e.,     0 n1  0 n1
Where  0 is the wave number in free space and  0 n1 ( 1 ) is nothing but an wave
number in an unbound medium filled with a material of refractive index n1 .
(g)
So q 2  12   2 , this quantity should be positive i.e. propagation constant of guided
mode should be less than the propagation constant of a wave traveling in unbound
medium having a refractive index n1 . Let us write inside the core
q 2  u 2  12   2
(h)
Let us now consider the two functions, Bessel function (fig.3) and Neumann
function (fig.4).
Bessel’s function  All these function J  qr  will always be finite for every
value of r .
Neumann Function  All these functions N (qr ) start from  at r  0 and then
show oscillatory behavior. If we take the physical situation for modal propagation
function,  represents the field amplitude inside the fiber.
For the core r  0 represents the axis of fiber. Therefore if Neumann function is
chosen as the solution, the field strength would be  at the axis of the fiber which is
inconsistent with the physical conditions. The fields should be finite all over the
cross section. So Neumann functions cannot be the solution of the problem if r  0
point included in the region.
Therefore we conclude that only J  qr  is the appropriate solution for the
modal fields inside the core of an optical fiber.
11. Field distribution inside the cladding has be of monotonically decaying kind. We
therefore must have q imaginary in the cladding. We hence should have
 2  2   2 where  2 is the refractive index of the cladding.
(a) Since q 2   2  2   2 is negative, let we define a real quantity w ( q / j ) such that
w2  q 2   2  2 .
Let us now look at the modified Bessel’s functions, as shown in figures 5 & 6. For
modified Bessel’s function of the 1st kind, as r increases that is, as we move away
from the axis of the fiber the field will monotonically increase and when r   field
would be infinite. Since the energy source is inside the core, the fields cannot grow
indefinitely away from the core. The only acceptable situation is that the field dies
down when we go away from the core or for larger values of r . This behavior is
coorectly given by the Modified Bessel function of second kind.
So we conclude that the 1st kind of modified Bessel’s function is not appropriate
solution in the cladding. The correct solution would be only Modified Bessel
function of 2nd kind, K (wr ) .
(b) The fields inside the core are given by J (ur ) and in the cladding are given by
K (wr ) .
(c) For a guided mode, the propagation constant lies between two limits 1 and  2 .
If 1    2 then we generate a field distribution which will have an oscillatory
behavior in the core and an exponentially decaying behavior in the cladding. The
energy then is propagated along fiber without any loss.
(d) Field distribution: From the solution of the wave equation we get the longitudinal
fields inside the core and the cladding.
Inside Core  r  a 
Electric field:
Ez1  AJ (ur )e j  j  z  jt
Magnetic field:
H z1  BJ  ur  e j  j z  jt
____________ (2-11a)
____________ (2-11b)
In Cladding  r  a 
Electric field:
Ez 2  CK  wr  e j  j z  jt
Magnetic field:
H z 2  DK  wr  e j  j z  jt
_____________ (2-11c)
_____________ (2-11d)
Where A, B, C , D are arbitrary constants which are to be evaluated from the boundary
conditions.
(e) Once we get the longitudinal components of the electric and magnetic fields, then
from the above relation we find the transverse field components. Now we write down
the entire field component inside the core and outside the core i.e. in the cladding.
Applying the boundary conditions i.e., the tangential component of electric field and
the tangential component of magnetic field should be continuous along the core
cladding boundary, we get what is called the characteristic equation of the mode.
(f)
Core:
Er1 , E 1 , Ez1
Cladding: Er 2 , E 2 , Ez 2
H r 1 , H  1 , H z1
H r 2 , H 2 , H z 2
The tangential components at the core cladding boundary are the 
z components.
The boundary conditions give that at r  a :
1.
E 1  E 2
2.
3.
Ez1  Ez 2
H1  H 2
and the
4.
H z1  H z 2
Now these equations are in terms of A, B, C , D arbitrary constants and the phase
constant  .
(g) We find the propagation constant of the wave,  by eliminating the arbitrary
constants. Elimination of the arbitrary constants gives the characteristic equation of
mode inside an optical fiber as
Characteristic Equation (eigen Value Equation)
 J '  ua  K '  wa  

 2 J '  ua 
  2
K '  wa  


  22 

  1

wK  wa  
a
 uJ  ua  wK  wa  

 uJ  ua 


2
1 
1
 2  2  _____ (2-12a)
w 
u
This equation contains three unknowns, u , w,  . However, we two more equations
u 2   2  1   2
______ (2-12b)
w2   2   2  2
______ (2-12c)
So using the equations (12a) ,(12b) & (12c) we can find the modal propagation constant
.
(h) We have to use numerical technique to solve the characteristic equation for a given
value of  . all We get multiple solutions to the problem and each solution gives one
mode for a given value of  . So depending upon the fiber radius, different number of
modes will propagate in an optical fiber.
3. Modes of optical fiber
1. A mode can be characterized by two parameters,  and the solution number.
2 (a) If we take H z  0 , all field components are expressed in terms of Ez and whatever
field we get, will not have any magnetic field component in the direction of
propagation. We call this mode as Transverse Magnetic mode.
Similarly if Ez  0 , then the mode is called the Transfer Electric mode.
(b) If both the longitudinal components of the fileds are present then we call the
mode a Hybrid mode. This is a combination of both TE and TM modes.
(c)
1.
2.
3.
So inside an optical fiber we have three types of modes
TE modes
TM modes
Hybrid modes
(c) In a hybrid mode, if we calculate the contribution by Ez & H z to the transfer fields,
one of them i.e. Ez or H z would dominate. Depending upon which of them
contributes more, we can sub-classify the Hybrid modes.
 Ez Dominates: EH mode
 H z Dominates: HE mode
4. Each of the above three modes are characterized by two indices,  m . The mode are
therefore designated as
TE m , TM m & EH m , HE m
5. What do these two indices physically mean?
In a rectangular wave guide, they represent the number of half cycle variations and
the number of zero crossings in the x direction and in y directions. The same thing in
applicable in this case also.
(a) If   0 , the function e j does not change when we move along  , so essentially
the field distribution is circularly symmetric.
(b) If   1 we get the one cycle variation and if   2 we get two cycle variations and so
on in the azimuth.
The first index of the mode therefore gives us the zero crossings in the azimuth of the
cross section of the optical fiber. Second index tells us how many zero crossing we
get if we move in radial direction. So if we fix all other parameters and just move
radially outwards, how many zero crossings the field variation would see is
essentially given by the second index. If we don’t hit any zero crossing then we have
m  1, if we have one zero crossing then m  2 , and so on. The indices are quite
similar to that of the rectangular wave guide.
6. For   0 the characteristic equation (12a) becomes

K '  wa  
 J '  ua  K '  wa  

 2 J '  ua 


  22 

  1
  0 ______ (2-13a)
wK  wa  
 uJ  ua  wK  wa  

 uJ  ua 


So either of the brackets is equal to zero. So if we take circularly symmetric mode for
which   0 , the characteristic equation split into two parts. Both separately show a
transcendental equation or characteristic equations.
(a)
If we take first bracket as equal to zero, the equation denotes the characteristic
equation of transverse electric mode. Similarly, the second bracket equal to zero,
denotes the characteristic equation of the transverse magnetic mode.
(b) Since   0 represents transfer electric and transfer magnetic modes, these modes
have a field distribution which is essentially circularly symmetric .If   0 then
we always get a field distribution which is hybrid. So the transverse electric and
transverse magnetic fields have only radial variation, and they not have any
variation in the  direction.
(c)
If we take the first bracket as equal to zero, we get
 J '  ua  K '  wa  


  0 ________ (2-13b)
 uJ  ua  wK  wa  
This is the characteristic equation for TE0m mode.
We get multiple solutions from this equation because J function is an oscillatory
Function .The first index of this function is essentially zero since   0 . Then
only the equation is split into two parts , so the equation (13b) becomes
J 0 '  ua  K 0 '  wa 

 0 The prime indicates derivatives of Bessel’s function
uJ 0  ua  wK 0  wa 
and Hanckel functions with respect to total argument. Equation (13b) becomes
J1  ua 
uJ 0  ua 

K1  wa 
wK 0  wa 
0