Let us continue our discussion on the modes
of a planar symmetric waveguide. In the last
lecture we had obtained the modal fields and
propagation constants of symmetric and antisymmetric
modes of asymmetric dielectric planar waveguide.
Let us now look more carefully at these modes
and understand what do they actually represent.
What are the modal fields actually and what
do the propagation constants represent.
So, for that let me write down the field of
for example, TE0 mode or any symmetric mode.
So, it is given in the core or in the guiding
film it is given by Ey of x is equal to a
cosine kappa x. If I write down the complete
solution that is the z part and t part also,
then it would be a cosine kappa x e to the
power i omega t minus beta z.
Let me expand this cosine kappa x it is in
the same way as I had done for planar mirror
waveguide. So, it is E to the power I kappa
x plus E to the power minus I kappa x divided
by 2, times E to the power i omega t minus
beta z. So, let me arrange the terms in a
particular passion. And then what I see that
this term has E to the power i omega t minus
beta z minus kappa x and this has E to the
power i omega t minus beta z plus kappa x.
So, this is nothing but a plane wave propagating
in plus x-z direction, making certain angle
with z axis and this is another plane wave
which is propagating in minus x-z direction
making certain angle from z axis.
So, I have this field comes out to be the
superposition of 2 plane waves one going in
plus x-z direction another going in minus
x-z direction. So, these are the 2 constituent
plane waves.
And what happens is if I do this kappa square
plus beta square it comes out to be k naught
n1 and k naught, n1 is nothing but the propagation
constant with which a plane wave propagates
in an infinitely extended medium of refractive
index n1.
So, kappa is nothing but, kappa is nothing
but the component of this along x. So, if
I resolve this in x and z direction then k
naught n1 sin theta will give me kappa. So,
it is the component of k naught n1 along x.
And beta is the component of this k naught
n1 along z. So, in x direction I will get
from here and here I will get 2 counter propagating
waves, and these 2 counter propagating waves
gives me standing wave in x direction. So,
the energy stands in x direction it does not
flow out.
However this standing wave pattern flows in
z direction with propagation constant beta,
exactly in the same way as planar mirror waveguide.
So, these modal fields these modal fields
are the standing wave patterns in x direction,
because I have Ey of x Ey of x and this is
standing wave pattern.
And this Ey of x is propagating with propagation
constant beta. So, if you look at it again
this E (x, z, t) is equal to Ey of x E to
the power i omega t minus beta z. Beta is
k naught n effective. Which basically tells
with what velocity this particular field this
is standing wave will propagate. If you remember
that in one of the previous lectures I had
said that intuitively I can I can understand
the effective index of the mode as effective
refractive index felt by a particular field
distribution because the field is now distributed
in n1 and n2 regions. So, the effective refractive
index would be somewhere between n2 and n1.
And that was intuitively correct and that
was not rigorously correct. And I see that
rigorously, rigorously n effective is nothing
but n effective is nothing but the velocity
with which it will it will define the velocity
with which a particular field pattern will
travel ok.
Rigorously I cannot get I cannot get from
the distribution of power in n1 and n2 regions.
So, this n effective is nothing but n1 cos
theta. So, it is the component of the plane
wave k naught n1 in z direction. And this
distribution the effective refractive index
as the distribution of field into 2 regions
is also not rigorously correct in the sense
that that I cannot get the effective index
from that field distribution in that way.
And also if you look at planar mirror waveguide
then you will say that the field is always
in n1 region then why it should have different
effective indices for different modes and
why it should be different from n1 or refractive
index n.
Similarly, if you talk about radiation mode
in radiation modes the energy is distributed
in n2 region it goes up to infinity, and the
effective index for radiation modes as we
will see later it is it is less than n2. So,
rigorously effective index of the mode is
nothing but the component of this along z
direction. And it gives me the velocity with
which a particular standing wave pattern will
travel in z direction. And since different
modes are different in standing wave patterns
and constitute different plane waves which
are at different angles from z axis. So, n
effective would be different.
If I look at the condition for guided mode
then it is beta over k naught lies between
n2 and n1. And beta over k naught is equal
to n2 gives me the cutoff of a particular
guided mode. From here what I get beta over
k naught is nothing but n1 cos theta. So,
if I replace this beta over k naught by n1
cos theta here. And if I look at it this cos
theta is nothing but sin phi if phi is the
angle which the plane wave bakes with the
normal to the interface of n1 and n2 regions
than this is n2 less than n1 sin phi is less
than n1.
Let me divide the whole thing by n1. So, this
gives me n2 over n1 is smaller than sin phi
is smaller than 1 or sin phi is greater than
n2 over n1. This is nothing but the condition
for total internal reflection. So, I automatically
get that that the cutoff condition translates
to the condition for total internal reflection.
If this angle phi is smaller than if this
angle phi is smaller than sin inverse of n2
over n1, then this wave will not undergo total
internal reflection at this interface and
it would be refracted. And the corresponding
mode would be radiated out it would not be
guided anymore. So, this is how I can understand
the mode of a waveguide.
Let us work out again some examples. Let me
consider a waveguide with n1 is equal to 1.5
n2 is equal to 1.46 d is equal to 2 micrometer.
And it supports 2 modes TE0 and TE1 mode at
lambda is equal to 1 micrometer. And if I
find out the effective indices of these 2
modes they come out to be 1.4905 and 1.4665
respectively.
Now, let me calculate the penetration depths
of these modes. And the angles of constituent
plane waves from the waveguide axis. I know
that penetration depth is given by 1 over
gamma, where gamma is equal to beta square
minus k naught square n2 square. Or k naught
times square root of n effective square minus
n2 square. So, k naught is 2 pi over lambda
naught lambda naught is given n effective
and n2 are given. So, if I now calculate these
gamma for both the modes, then for TE0 mode
gamma is equal to 1.884 micrometer inverse
and correspondingly penetration depth is 0.53
micrometer.
For TE1 mode gamma comes out to be 0.867 micrometer
inverse. And correspondingly the penetration
depth comes out to be 1.15 micrometer. What
are the angles of constituent plane waves
from the waveguide axis? Well I know that
these angles are given by n effective is equal
to n1 cos theta. So, for these values of n
effective I can immediately find out the values
of theta, because I know the value of n1 also.
So, for TE0 mode this theta would be cos inverse
n effective over n1. So, it would be 6.45
degrees and for TE1 it would be 12.13 degrees.
So, TE0 mode is nothing but is nothing but
the superposition of 2 plane waves, making
angles making angles plus minus 6.45 degrees
from the waveguide axis. And TE1 mode is the
super position of 2 plane waves which make
angles plus minus 12.13 degrees from the waveguide
axis.
So, after doing the analysis of the waveguide
for TE modes, now in the same manner we can
do the analysis for TM modes transfers magnetic
modes. If I go back and see that for a waveguide
which has confinement in x direction, that
is refractive index variation in x direction
and propagation in z direction, then the non-vanishing
components of E and H are Hy, Ex and Ez.
And these 3 components are related by these
3 relations. The procedure to obtain the modes
and modal fields for TM modes is exactly the
same as we had done in the case of TE polarization.
So, what I need to do? I need to find out
a differential equation for example, in Hy
and solve it.
So, to find out the differential equation
in Hy, I substitute for Ex and Ez from these
2 equations into this equation, and rearrange
the terms. Then I get a differential equation
in Hy as d2Hy over dx square minus 1 over
n square dn square over dx. dHy over dx plus
k naught square n square minus beta square
Hy is equal to 0. You may see that you may
notice that this equation is different from
the wave equation that we had got for the
TE case. In the TE case this term was not
there, excuse me in the TE case this term
was not there, but here I have this term.
So, let us see how do we take care of this
now we need to solve this equation for given
n square of x to find the modes. So, this
is the waveguide which we are considering
again symmetric dielectric planar waveguide
where n of x is given like this.
Procedure is again the same that I need to
write this equation in this region and in
this region. So, I do this for mode x less
than d by 2 I get this equation. Why? Although
I have this term in general for a variation
n square of x, but when I write this equation
in the individual regions in this kind of
a step index waveguide the refractive index
is uniform and when the effective index as
uniform in this region. So, this term goes
off. So, I have got this equation for mode
x less than d by 2, and this equation for
mode x greater than d by 2 and you can see
these equations are exactly the same as we
had obtained in the case of TE polarization.
So, again I define this is kappa square and
this is gamma square, and since I am interested
in guided modes. So, kappa square and gamma
square are positive.
So, these are the equations and the solutions
are again the same exactly the same, Hy is
equal to a cosine kappa x plus b sin kappa
x for mode x less than d by 2 and the decaying
solutions for mode x greater than d by 2.
And again A B C and D can be determined with
the help of boundary conditions.
In this case also I can make use of the symmetry
of the problem and divide these solutions
into symmetric and antisymmetric. So, for
symmetric modes Hy would be a cosine kappa
x in the guiding film, and Ce to the power
minus gamma mode x for mode x greater than
d by 2 that is in these regions. So, what
next? What I should do? Now I should apply
the boundary conditions, to this kind of modal
field. And what are the boundary conditions
the boundary conditions are again the same
the tangential components of E and H should
be continuous at x is equal to plus minus
d by 2.
What are the tangential components here? Well
if I look at TM modes, the non-vanishing field
components are Hy, Ex and Ez. So, the components
which are tangential to x is equal to plus
minus d by 2 planes are Hy and Ez. So, Hy
and Ez should be continuous. How Ez is related
to Hy? We will look at this. So, Ez is related
to Hy with some constant times 1 over n square
dHy over dx. So, this because this n square
depends upon depends on x. So, I should take
it to that side. So now, now the boundary
conditions give me Hy and 1 over n square
dHy over dx should be continuous. So, this
is the difference with the TE case.
In TE case I had the boundary conditions the
boundary condition led to Ey and dEy over
dx should be continuous at the boundaries,
but here I have Hy and 1 over n square dHy
over dx should be continuous at the boundary.
So, I apply these so first I apply that the
field should be continuous Hy is continuous
at x is equal to let us say plus d by 2. So,
a cosine kappa d by 2 is equal to C e to the
power minus gamma d by 2, and then I apply
this 1 over n square dHy over dx should be
continuous at x is equal to plus d by 2, then
it is on this side I have 1 over n square
minus a sin kappa d by 2 is equal to minus
1 over n2 square gamma C e to the power minus
gamma d by 2.
So, this gives me the transcendental equation,
if I divide this by this. Kappa tan kappa
d by 2 is equal to n1 square over n2 square
times gamma or by multiplying with d by 2
on both the sides I get kappa d by 2 tan kappa
d by 2 is equal to n1 square over n2 square
times gamma d by 2. So, this is the transcendental
equation the difference is this factor. I
have this factor n1 square over n2 square
extra in the case of TM polarization.
So, for symmetric modes, if I now convert
this equation into psi and V then I know psi
is nothing but kappa d by 2, then in exactly
the same way as I had done for TE case for
symmetric modes I will have the equation as
psi tan psi is equal to n1 square over n2
square V square minus psi square and this
is the field. Similarly for antisymmetric
modes the equation would be transformed to
minus psi cot psi is equal to n1 square over
n2 square, square root of V square minus psi
square and this is the field.
So, the transcendental equation is this. So,
if I use the graphical solution to solve this.
So, I should equate this on the right hand
side and left hand side both to eta in this
case as well as in this case. So, for symmetric
mode I will have eta is equal to psi tan psi
and for antisymmetric is as well as eta is
equal to minus psi cot psi and from the right
hand side I will get psi square over V square
plus eta square over V square times n1 square
over n2 square whole square is equal to 1.
In the case of TE modes this factor was not
there and I had the circles here psi square
plus eta square is equal to V square. But
now, but now I no more have circles, but this
is the equation of an ellipse. So, I have
ellipse here, and what is the major axis of
the ellipse it is along eta. So, what I will
have to do now? To have graphical solutions
I will have to plot these 2 together for symmetric
modes and these 2 together for antisymmetric
modes and look for points of intersections.
So, I do this. So, again these red curves
show eta is equal to n1 square over n2 square
psi tan psi.
And so, this is for symmetric modes this is
for antisymmetric modes. So, I this is eta
is equal to psi tan psi it is eta is equal
to minus psi cot psi. Now the solid lines
the solid lines represent the circles, psi
square plus eta square is equal to V square.
And correspond to TE modes while these ellipses
which are represented by dotted lines they
are the ellipse corresponding to the TM modes.
So, these are the ellipse corresponding to
TM modes.
What I see here if I take a particular value
of V, if I take a particular value of V and
see and see the points of intersection of
these TE and TM modes, then it should not
be V naught it should be V. So, for example,
for V is equal to 6, for V is equal to 6 the
point of intersection corresponding to TE
mode is somewhere here. And for TM mode it
is at slightly larger value of psi. So, what
I have for a given for a given mode whether
it is m is equal to 0 or m is equal to 1 or
m is equal to 2, psi for TM mode is always
greater than psi for TE mode. And since psi
is equal to kappa d by 2 which is d by 2 square
root of k naught square n1 square minus beta
square, then beta for TM is always less than
beta of TE.
In terms of normalized parameters in the same
way as I had done for TE case I can also write
down the transcendental equation for TM modes,
symmetric and antisymmetric.
So, equations are similar, only thing is this
extra factor of n1 square over n2 square in
both the cases symmetric and antisymmetric
cases.
Cut off conditions. What are the cut off conditions?
Well beta over k naught should be equal to
n2, which means b should be equal to 0, if
I put b is equal to 0 here and here I will
get the cut off conditions for symmetric and
antisymmetric modes. And you can see that
since b has to be 0 if I put b is equal to
0 then this factor is absorbed here in 0.
So, it would not have any effect it would
not have any effect on the cut offs.
So, for symmetric modes the cut off condition
remains Vc tan Vc is equal to 0, and for antisymmetric
mode it remains Vc cot Vc is equal to 0. So,
in general I have the cut offs for TMm modes
as Vcm is equal to m pi by 2, which is the
same as in the case of TE modes ok.
Let us now plot b-V curves. So, so that I
will have to plot for a given value of n1
over n2, let me see how does it look like
schematically for n1 over n2 is equal to 1.5.
So, the first thing is the cut offs cut offs
are the same. So, the plots will start from
the same value for TE and TM. So, if I plot
it for TE0 mode then for TE0 mode it both
will start from 0 because they have the same
cut off. So, TE0 mode goes like this and TM0
will also start from here, but whether this
curve would be below this or above this. And
I see that since the propagation constants
of TM modes are smaller than the propagation
constants of TE modes. So, the curve corresponding
to TM mode would lie below the curve corresponding
to TE modes. So, it will go like this, for
m is equal to 1, So this would be TE1 and
this is TM1 similarly TE2 and TM2.
Again these curves are universal for given
value of n1 over n2 now. So, here I should
also take care what is this ratio n1 over
n2. Modal fields, if you look at modal fields
then again in the same way the modal field
for TM0 would look like this. This is for
low index contrast waveguide and for TM1 mode
it looks like this. What I should take care
here is and why it I should pay attention
to is that in case of TE polarization Ey and
dEy over dx was continuous.
So, the field as well as it is slope was continuous
at the boundaries x is equal to plus minus
d by 2, but you remember that in TM polarization
Hy and 1 over n square dHy over dx are continuous
at the boundaries.
So, the slope is not continuous. Although
here I see that the slope is also continuous
at the boundary, but rigorously it is not
continuous. It is the discontinuity in the
slope is. So, small that it does not show
up because if you look at n1 and n2. n1 and
n2 are very close. So, if it is weakly guiding
if the waveguide is weekly guiding the n1
the values of n1 and n2 are very close to
each other than that would not show up here.
If on the other hand if I plot this for high
index contrast waveguide you now look at n
the values of n1 and n2, the factor n1 over
n2 is now 3.5.
Now, if I plot the modal fields the modal
fields for TE0 mode are like this there is
no discontinuity in the slope while here this
discontinuity shows up, because the factor
n1 over n2 is very large.
So, this is the first thing I will see, similarly
in the case of TM1 mode there would be huge
discontinuity. Again just as in the case of
TE modes in TM modes also mth TM mode will
have m number of 0s, and since the propagation
constant of TM mode this is wrong it should
be it should be beta TM is less than beta
TE please make correction. So, the penetration
depth of TM modes are larger, you can see
the penetration depth of TM modes are larger
than the penetration depths of TE modes.
I can look at some examples if I consider
this waveguide with n1 is equal to 1.5 n2
is equal to 1.46 d is equal to 2 micrometer.
At lambda is equal to 1 micrometer n effective
of TE0 mode comes out to be this and of TM0
mode comes out to be this. So, you can see
if the index difference is very small, TE0
and TM0 modes have nearly the same propagation
constants, very close to each other. And if
I now calculate the penetration depths for
the 2 modes the penetration depths are also
very close to each other 0.5305, 0.5329 differences
only about 0.45 percent, but if I take high
index contrast waveguide where n1 is 3.5 n2
is 1. Then n effective of TE0 mode is 3.1916
and that of TM0 mode is 2.92. So, which is
very different there is huge difference between
the propagation constants. Now, if I calculate
the penetration depths then the difference
is about 10 and half percent.
With this I have done the analysis of dielectric
planar waveguide for symmetric modes as well
as antisymmetric modes. I have calculated
the propagation constants, I have found out
the modal fields, their behaviors, penetration
depth. Now only thing that remains is how
much energy do they carry, how much energy
how much power these modes carry. We will
go into that in the next lecture.
Thank you.
