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After having understood the propagation characteristics
of optical fiber, now we would look into how
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we can make components using optical fibers.
The question is why do we want to make components
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using optical fiber and what kind of components
do we require.
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We are going to use this optical fiber in
telecommunication system or we can also use
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it in a sensing system. So, when we use the
fiber in a system we require several components.
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And if we have the components which are all
optical then the data rate is not compromised
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also the complexity of the system decreases.
So, we would like to have as many components
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as possible in optical domain itself and preferably
in optical fiber, because when we make components
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out of optical fiber then it is easy to splice
them with the transmission fiber or sensing
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fiber, the insertion loss is low compatibility
is better.
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So, in the next few lectures we are going
to study some components and devices made
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in optical fiber itself or using optical fiber.
What kind of components or devices we want
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to make or we want to study here are switches,
and then we have power splitter, wavelength
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filter.
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We may require multiplexer, de-multiplexers,
polarization controllers, fiber gratings,
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fiber amplifiers and dispersion compensating
fiber. So, we are going to study these components
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and devices in the next few lectures.
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A very basic components and very important
component using optical fiber is directional
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coupler. What is the directional coupler?
It is essentially a 4 port device. So, we
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have a fiber here a fiber coming out here,
a fiber coming out here, and fiber going inside
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here. What we can have using this well if
we launch power P in port 1, depending upon
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the parameters of this device or this component
the entire power may come out of port 3, then
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it is basically a switch. You are switching
power from port 1 to port 3. It may also happen
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that if you launch power in port 1 then 50
percent power comes out of port 2 and 50 percent
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out of port 3, then it is a power splitter.
Or you can have a desired ratio of power here.
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You can also have if you launch 2 wavelengths
lambda 1 and lambda 2 in port 1, then it may
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happen that lambda 1 comes out of port 2 and
lambda 2 comes out of port 3. Then such a
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component is demultiplexer. You can also input
lambda 1 from here and lambda 2 from here
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and both the wavelengths may come out of port
1 then it is wavelength multiplexer or WDM
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coupler. And all these components can be made
out of directional coupler.
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So, what is a directional coupler if we have
2 cores which are put together very close
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to each other and we launch light in core
1.
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Then as this light propagates through optical
fiber this composite system then, then there
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is a redistribution of power between the 2
cores and after a certain length all the power
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will come out of the other core. So, there
is a switching of power from core 1 to core
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2. So, if I look it look at it as a 4 port
device then this is port 1, port 2, port 3,
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port 4, and if I launch light in core 1 then
light may come out of core 2 in the output
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end.
How does it work? If I have 2 well separated
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course of course, there is a cladding.
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So, if I have 2 fibers which are very well
separated and these are single mode fibers,
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then this fiber supports this mode this fiber
supports this mode. And propagation of light
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in this fiber is not affected by the propagation
of light in this fiber, because they are very
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well separated. But if I bring them closer
then what may happen that now this is core
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1.
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And this is core 2 then this fiber has mode
which is like this, this fiber also has a
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mode which is like this.
So, these 2 model fields may combine like
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this, if the separation between the 2 course
is not very large. So, there are there is
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possibility combining these 2 fields in 2
ways, one way is they combine like this, another
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way is that the field in one core is like
this and in another core it is like this.
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So, these 2 fields may combine like this or
like this. We can see that this is symmetric
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field this is symmetric about this point the
middle point. And this is antisymmetric. Then
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what happens is that when you bring 2 course
together then this is not the mode of the
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composite system. The light in this core is
affected by this core and the light in this
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core is affected this core.
So, these are no more the modes of the system
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that is if I launch this pattern it will not
propagate as it is it will not retain it is
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shape. Similarly, if I propagate this distribution
in this course it will also not retain it
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is shape. So, these are no longer the modes
of the system, but now the modes of the system
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are this and this. So, if I launch this distribution
then it will retain it is shape. It will propagate
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along the composite structure without any
change in it is shape. And it will propagate
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of course, at certain phase velocity. Similarly,
this distribution will also propagate without
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any change in it is shape. So now, the modes
of the composite system are this and this,
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and these are known as normal modes of the
system or super modes of the system.
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So now if I launch this pattern into this
core then this can be understood as the super
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position of these 2 modes. When I launch this
it means I am exciting this mode and this
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mode equally, because you can see if I add
them up then this plus this will give you
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the power in this core while they will cancel
out. So, there would not be any power in this
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core.
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So, exciting core 1 with an appropriate field
pattern is equivalent to exciting symmetric
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and antisymmetric modes equally.
Now, since these are the modes of the system
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they will retain it is shape. So, I can understand
the propagation of this in the composite system
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in terms of propagation of symmetric and antisymmetric
modes because they retain their shape, but
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they have different velocities. So, as they
propagate, there is a phase shift accumulated
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between them and if it any intermediate distance
I want to know what is the field distribution
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then I will have to superpose these 2 modes
with their accumulated phase shift. And that
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will give me the resultant field pattern.
Now, if after a certain length the phase shift
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accumulated between them is pi then effectively
this mode flips with respect to his mode.
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So, if now I am interested in finding out
the resultant then the super position of these
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2 will give you cancellation of fields here
and addition of field here. So, the resultant
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field would be only in this core and no field
would be in this core. So, this is how there
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is switching of power from one core to another
core.
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So, if you launch light into this, then after
a certain distance which is known as Lc the
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coupling length. The minimum length at which
the entire field couples from core 1 to core
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2 is known as coupling length, then at this
length entire power will get transferred into
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the second core. I am assuming that these
2 corers are identical. If I let it propagate
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further then it will switch back to core 1
and then again to core 2; so there would be
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periodic switching of power between the 2
cores.
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I can understand this with the help of a very
simple example of a couple pendulum. I can
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take an analogy of couple pendulum. So, this
is the refractive index profile of the composite
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system, where I have 2 cores and this is the
symmetric state and I can also have antisymmetric
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state.
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So, I have symmetric mode and antisymmetric
mode of the system symmetric normal mode and
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antisymmetric normal mode. This is equivalent
to the modes of a couple pendulum; if I have
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a pendulum like this which I have coupled
with the help of this small string. Then now
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if in this kind of pendulum I push these 2
bobs in one directions simultaneously and
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then release them then the oscillations would
be like this. And this will keep on oscillating
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in this state for infinite time if there is
no air resistance.
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So, this is the mode of the system. And since
both of them are going in the same direction
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simultaneously they are moving they are moving
simultaneously in one direction and moving
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together, and then it is symmetric state equivalent
to symmetric mode of the directional coupler.
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I can also have another possibility that I
push one bob like this and pull another like
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this, and then release them then they will
do like this. And they will keep on doing
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like this forever if there is no air resistance,
then this is nothing but antisymmetric state
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or which is equivalent to antisymmetric super
mode of or antisymmetric normal mode of the
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system.
Now, what happens if I displace only one of
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them. If I displace only one of them and then
I release then what happens, we will see.
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Then this is equivalent to this is equivalent
to putting power into one core of the directional
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coupler. So, if you look at the actual picture,
then this is the symmetric mode.
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This color is changing it indicates that the
power is going along the length of fiber along
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the length of the directional coupler. So,
this is the field distribution at different
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lengths of the fiber as the color changes
it means the length is changing. So, I am
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changing the length, but it retains it is
shape. So, the symmetric mode it retains it
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is shape, and this is equivalent to the couple
pendulum when these 2 bobs are moving together.
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And they keep on doing like this.
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So, this is symmetric state of the couple
pendulum this is symmetric normal mode of
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the directional coupler. I can have antisymmetric
when I displace one on this direction another
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in this direction. So, they will keep on doing
like this. And this is how this field is also
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retaining it is shape when it propagates along
the length of the fiber.
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So, this is antisymmetric mode of the system
this is anti-symmetric state of couple pendulum.
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If I now displace only one, then what I see
that after certain time the rear one is stopped
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and the other one is in full swing. Then the
front one is stopped and the rear one is in
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full swing.
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So, there is periodic exchange of energy between
the 2 bobs. Similarly, here if I have all
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the powers here at z is equal to 0, then as
the power propagates along the length of the
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system then the power periodically transfers
from one core to another core. So, there is
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periodic coupling power couples back and forth
between the 2 cores. So, this is how by just
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a simple analogy of a couple pendulum we can
understand the phenomenon of coupling of power
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from one core to another core.
So, how the power evolves I can write the
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down the power as a function of z. z is the
length of the propagation, for a generalized
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case for a generalized case where I have 2
cores brought together separated by a low
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index medium which is cladding.
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And I take a generalized case where these
2 cores are not identical. So, the core 1
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supports a mode which has propagation constant
beta 1 and core 2 supports a mode which has
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a propagation constant beta 2. Then at z is
equal to 0 if I have entire power in core
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1 that is P1(0) is the power in core 1. Then
as the light propagates along the structure
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the power in core 1 varies according to this
and power in core 2 varies according to this.
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So, what we have? P1(z) is equal to P1(0)
1 minus kappa square over gamma square sin
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square gamma z. P2(z) is P1(0) kappa square
over gamma square sin square gamma z. This
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can be obtained by using what is known as
couple mode theory, we are not going into
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the details of that here in this course. What
is kappa and gamma? Well, kappa square over
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gamma square is given by 1 over 1 plus delta
beta square over 4 kappa square.
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So, gamma square is kappa square plus delta
beta square over 4 where delta beta is beta
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1 minus beta 2, which is also known as phase
miss match because beta 1 represents the propagation
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constant of the mode of core 1. And beta 2
represents the propagation constants of the
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mode of core 2. So, of course, if beta 1 and
beta 2 are different so there is a phase miss
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match between them. And kappa is called coupling
coefficient and it represents strength of
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interaction between the 2 cores.
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It depends on the fiber parameters, the core
separation and the wavelength. So, if I have
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2 identical fibers then beta 1 is equal to
beta 2 delta beta is equal to 0. And if delta
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beta is equal to 0 then gamma is equal to
kappa. If you go back to this expression let
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me retain this expression in on the board
that that P1(0) sorry P1(z) is equal to P1(0)
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times 1 minus kappa square over gamma square
kappa square over gamma square sin square
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gamma z. And P2(z) is equal to P1(0). P1(0)
and this is kappa square over gamma square
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sin square gamma z.
So, now if delta beta is equal to 0 then again
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if I go back here if delta beta is equal to
0 then gamma is equal to kappa. If gamma is
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equal to kappa then I will simply have P1(z)
as P1(0) cosine square kappa z. And P2(z)
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is equal to P1(0) sin square kappa z. If I
plot them with respect to z then I can see
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that P1(z) is cosine square kappa z varies
like this, and Pz(z) is sin square kappa z
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varies like this.
So, these are the lots of P1(z) over P1(0)
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this is P2(z) over P1(0). So, it is P1(z)
normalized with respect to the input power
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in P2 normalized with respect to input power.
What do I see here that at kappa z is equal
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to m pi if I have kappa z is equal to m pi
where m has integer values, then P1 is equal
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to P1(0) and P2 is equal to 0, which means
that at these points where kappa z is equal
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to m pi I have entire power in core 1 and
0 power 0 power in core 2.
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On the otherhand at kappa z is equal to m
plus half pi, P1 is equal to 0 and P2 is equal
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to P1(0), that is at these points at these
points where kappa z is equal to m plus half
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pi entire power is in core 2, and the power
in core 1 is 0. So, so at this length entire
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power switches from core 1 to core 2. The
power in core 1 drops down to 0, the power
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in core 2 increases to 1, normalized power.
So, the minimum distance this is the minimum
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distance at which the complete coupling takes
place then it is known as coupling length,
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which corresponds to which corresponds to
m is equal to 0 here. So, kappa z is equal
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to pi by 2 and z is equal to pi over 2 kappa.
So, the coupling length is pi over 2 kappa.
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So, it depends upon what is the coupling coefficient.
If the coupling coefficient is large then
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the switching of power from one core to another
core would happen at a very short distance.
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Let us have a feel of some numbers. If I have
a typical single mode fiber and I am working
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at a wavelength 1300 nanometer then typical
value of kappa is about 0.2 millimeter inverse.
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So, l c would be about 7.85 millimeter close
to one centimeter. So, in this length the
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entire power will get coupled from core 1
to core 2. I can also make a power splitter
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out of it if you work at the point where you
have the power in both the course equal than
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it is 50:50 coupler. So, you can work at this
length if the interaction length is this which
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is Lc by 2, then 50 percent power will be
in core 1 and 50 percent power would be in
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core 2 than it is power splitter. You can
have any ratio depending upon depending upon
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the length you choose you can have a ratio.
For example, if you choose this length then
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75 percent power in core 1 and 25 percent
power in core 2. So, you can choose any ratio
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for a splitting of power depending upon the
coupling length.
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So, when z is equal to Lc by 2 or equal to
pi over 4 kappa I see P1 is equal to P1(0)
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by 2 and P2 is equal to P1(0) by 2 then there
is a splitting equal spitting of power between
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the 2 cores then it is known as 50:50 coupler
or 3 dB power splitter and this will happen
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at certain wavelength because kappa is wavelength
dependent. So, if it is 50:50 power splitter
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at lambda not is equal to 1300 nanometer.
Then if I operate the same coupler at 1550
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nanometer the ratio of splitting would be
of course, different because kappa is different.
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So, that was the phase matched case where
delta beta is equal to 0 in general if I have
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2 non identical wave guides then delta beta
is not equal to 0, and then the fraction of
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power that appears in core 1 can be given
by P1(z) over P1(0), Which is 1 minus kappa
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square over gamma square sin square gamma
z.
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And the fraction of power that appears in
core 2 is zeta which is kappa square over
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gamma square sin square gamma z.
Now, I just look at it, if you look at this
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expression then I have this kappa square over
gamma square sitting here which means that,
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which means that I cannot have the maximum
value of zeta as 1. Of course, the maximum
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value of sin square gamma z is 1, but because
of this factor I cannot have maximum value
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of zeta as 1. It depends upon what is the
value of kappa square over gamma square. So,
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zeta max is kappa square over gamma square,
which is equal to 1 over 1 plus delta beta
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square over 4 kappa. Square since delta beta
over tau kappa whole square is always positive.
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And here it is 1 plus this thing. So, this
kappa square over gamma square will always
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be less than 1. So, I cannot have 100 percent
coupling from one core to another core, let
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us see how much we can have.
So, if I have delta beta over 2 kappa is equal
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to 0.2, I put it here then zeta max comes
out to be 0.96 that is 96 percent power can
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be coupled from core 1 to core 2 this is the
variation of eta and this is the variation
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of zeta. So, you can see that only 96 percent
power can be coupled from core 1 to core 2,
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but this is very small, delta beta is very
small here. If I have delta beta over 2 kappa
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is equal to 1, then this zeta max is just
0.5. It means that it means that the maximum
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power that can be coupled from core 1 to core
2 is 50 percent. So, it the power in core
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1 will oscillate like this, and power in core
2 will oscillate like this it will never cross
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50 percent mark here.
If I further increase delta beta, and it becomes
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4 kappa delta beta is equal to 4 kappa, then
zeta max is only 20 percent. So, if there
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is a larger phase mismatch between the 2 cores
between beta 1 and beta 2, then the coupling
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of power is not complete. Another thing that
I see here is that the gamma, what is the
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value of gamma? Since kappa square over gamma
square is this.
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And if I have delta beta over 2 kappa is equal
to 1 and this is how the power varies, then
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gamma is equal to square root kappa. And this
gamma basically tells you what is the spatial
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frequency of oscillation. If I increase delta
beta over 2 kappa from 1 to 2 then gamma becomes
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square root of 5 kappa then of course, the
frequency of oscillations increases here.
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So, the oscillations in power become more
rapid as the phase match as the phase mismatch
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increases.
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So, this is another thing that I see here.
This is how a 3D plot shows the 2 cases when
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delta beta is equal to 0, I launch power in
core 1 and then there is a complete transfer
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of power in core 2 at certain length. You
can see a dip which is which goes to 0 here
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and you can see a peak which comes to 1 here.
If I look it from the top and draw the contour
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plot then you launch power in core 1 and that
z is equal to Lc there is complete power in
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complete transfer of power in core 2 and here
the power in core 1 becomes 0.
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While when delta beta is not equal to 0, then
you can see that it is not maximized and there
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is a still some power left here. So, there
is no complete transfer of power, this is
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much more clear from this contour plot. So,
you launch power here, there is a transfer
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of power here at this distance the maximum
transfer, but this is not this is not 1, there
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is no complete transfer of power.
So, in the next lecture, after having understood
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the mechanism of light coupling between the
two cores in a directional coupler; in the
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next lecture we would look into some more
components which we can make out of this basic
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building block directional coupler.
Thank you.