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This is the third part of the module transmission
characteristics of an optical fiber. In the
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first and second modules we had seen that
the attenuation and intermodal dispersion
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they limit the repeater less length of the
link. In this lecture we will look into what
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would be the implication of the wavelength
content of the source, the flow of the lecture
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is something like this we will talk about
chromatic dispersion, then what is an optical
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pulse, what our faith and group velocities,
then material dispersion then since the fiber
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is telecom fiber is made of pure silica glass.
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So, we would look into the material dispersion
of pure silica glass, and then what is the
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total dispersion in a multimode fiber.
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So, I come back to this slide again, where
we had seen that the repeater less length
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of the link for a given data rate, depends
upon attenuation and broadening of the pulse.
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Broadening of the pulse we had already seen
in the last lecture, the material dispersion
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where light is coupled into various ray paths,
and these ray paths take different times to
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reach at the output end of the fiber. So,
light coupled into these various ray paths
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takes different times to reach at the output
end of the fiber and these causes, what is
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known as the intermodal dispersion.
Today we would see if we have a source and
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which has certain wavelength content, what
would be the implication of this on the broadening
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of pulses.
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I know that if I launch a beam of white light
into a prism then I see dispersion, I see
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various colors appearing at various angles,
and this is purely due to the refractive index
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wavelength dependence of refractive index
of the material of the prism and the geometry
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of the prism enables these different colors
coming out at different angles, this is known
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as chromatic dispersion.
Now, if I have a fiber then made of glass,
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then fiber material also has this characteristic,
the refractive index of the material at different
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wavelengths is different, what would be the
implication of that. So, for that let me first
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look at how do I transmit data in an optical
fiber, I transmit data in the form of pulses
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and when I switch on a laser and switch it
off I generate a pulse.
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If I look into this pulse, then as I will
see a little later that this pulse is essentially
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a superposition of large number of harmonic
waves of different wavelengths or slightly
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different frequencies, and when I use a light
source like LED or a laser diode, for creating
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these pulses to send data into an optical
fiber, then these light sources have finite
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line width.
For example, an LED has a typical line width
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of 20 to 30 nanometers, while a laser diode
although it is highly monochromatic it also
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has a certain line width. Its line width can
be of the order of a nanometer and if it is
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very highly chromatic and very good laser
source then it can be 0.1 nanometers or so,
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but it has certain line width. So, all the
wavelengths components which fall in the line
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width of these light sources, would now experience
different refractive index of the material
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and they will travel with different velocities
and that should give rise to dispersion.
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To understand that and this kind of dispersion
is known as material dispersion. To understand
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that let us first find out it with what velocity
this pulse travels in an optical fiber or
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in a material. Even if we forget about optical
fiber even if it is infinitely extended materials
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with what velocity this pulse travels in the
fiber; does it travel in the same way as a
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continuous waves a harmonic wave let us look
into it.
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So, if I first consider a single harmonic
wave of angular frequency omega which is propagating
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in z direction, then I can write the displacement
of this wave as y is equal to a cosine omega
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t minus kz where a is the amplitude, omega
is angular frequency, k is the wave vector.
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And if I plot it at certain z if I observe
the amplitude of this the displacement of
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this at certain z for all the times, then
it will vary with time something like this.
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Now, the velocity of phase fronts I can find
out from here, since what is the velocity
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of phase fronts what are phase fronts, phase
fronts are the surfaces of constant phase,
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and here the phase is omega t minus kz.
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So, surface of constant phase would be given
by omega t minus kz is equal to constant,
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now I can find out the velocity of phase fronts
from here I just differentiate it, then omega
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dt minus kdz is equal to 0 that is what I
have there, and this will give me dz over
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dt is equal to omega over k, and this dz over
dt is nothing, but the velocity of the phase
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front and this is known as the phase velocity.
So, when a single harmonic wave travels in
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a medium, then it goes with this velocity.
For light waves in free space this is nothing,
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but c and can be given by omega divided by
k naught where k naught is the wave vector
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in free space now let us consider 2 harmonic
waves a very close angular frequencies omega
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1 and omega 2 and very close wave vectors
k1 and k2.
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So, I can write them as y1 is equal to a cosine
omega 1 t minus k1z, and y2 is equal to a
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cosine omega 2 t minus k2z, for simplicity
I have taken the same amplitudes of these
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waves.
Now, I superpose them when I superpose them
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then y becomes y1 plus y2 given by this and
if I simplify this what do I get? I get y
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is equal to 2 a cosine, omega 2 minus omega
1 by 2t, minus k2 minus k1 by 2 z times cosine
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of omega 1 plus omega 2 by 2t, minus k1 plus
k2 by 2 z. Now since omega 1 is very close
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to omega 2, and k1 is very close to k2 let
me write them down as omega and k, and also
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assume that omega 2 minus omega 1 is equal
to delta omega, and k2 minus k1 is equal to
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delta k, then I can write this down as 2 a
cosine delta omega t by 2 minus delta kz by
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2 times cosine omega t minus kz.
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Now, let us examine this if I examine this.
This is nothing, but effect a term which contains
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a frequency omega while this term contains
a very small frequency delta omega. So, this
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is nothing, but high frequency constituent
waves and this is the envelope which is low
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frequency. If I plot them if I plot this y
now as a function of t for any given z what
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do I see? The amplitude both something like
this. So, I have this red one is the envelope
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low frequency envelope, and blue one is the
high frequency constituent waves. And if I
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plot the intensity of this which is proportional
to y2, then it comes out like this.
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So, what I have observed that if I superpose
2 waves I make groups a train of group, I
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generate a group of waves. And these groups
of waves are travelling with certain velocity,
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which is the velocity of the envelope from
here and if I find out the velocity of the
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envelope from here it comes out to be delta
omega over delta k. So, this is group velocity.
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So, this group of wave travels with velocity
vg which is delta omega over delta k. However,
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the constituent waves individual constituent
waves they travel with velocity omega over
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k, which is the phase velocity of the constituent
waves.
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So, this is a series of group, but if I have
a pulse and isolated pulse then it is a group
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of very large number of such harmonic waves
with continuous frequency variation in omega
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and k. So, instead of taking the superposition
of 2 ways, I take the superposition of 100
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ways and then 1000 waves then I can find out
that these groups are separated ok.
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And I can create an individual group or wave
packet is also known as wave packet. If I
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have this very large number of harmonic waves,
with continuous variation in omega and k,
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and in such a situation delta omega over delta
k would now become d omega over dk, and the
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group velocity will be given by d omega over
dk.
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So, now if I send this pulse through a material
that material can be the material of optical
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fiber, then it is a group of waves.
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And now I want to find out how the different
frequency components travel in this material
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and what is the implication of that. I know
the constituent waves of this wave packet
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or the pulse, will have propagation constant
given by k naught times n omega, where k naught
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is omega by c and n of omega is frequency
dependent refractive index of the medium.
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Now, I can find out the velocity from here
group velocity, because the pulse will travel
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with the group velocity vg. I know vg is equal
to d omega over dk from the previous slide.
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I have vg is equal to d omega over dk, but
remember that ultimately what I want to do
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is, I want to find out the transit time to
L length of the fiber and that would be given
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by Ly vg. So, instead of working out the expression
for vg let me work out the expression for
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1 over v g because that is how it will appear
ultimately.
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So, I find out L over vg from here, L over
vg would be dk over d omega. So, I take differential.
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So, I take differential of k with respect
to omega, I take derivative of k. So, it comes
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out to be L by cn of omega plus omega times
dn over d omega, and since I work with wavelength
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instead of frequencies in practical situations
I always work with wavelength. So, let me
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obtain this vg in terms of wavelength lambda
naught, lambda naught is given by 2 pi c over
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omega, and I have vg in terms of omega like
this.
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So, now let me find out this omega times dn
over d omega in terms of lambda. So, omega
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over dn over d omega would be 2 pi c over
lambda naught, times dn over d lambda naught,
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times d lambda naught over d omega and lambda
naught is given by this. So, it would be simply
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2 pi c over lambda naught dn over d lambda
naught, minus 2 pi c over omega square, and
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omega again I can put as 2 pi c over lambda
naught, and if I do that I get omega times
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dn over d omega is equal to minus lambda naught
dn over d lambda naught.
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So, I put it back into this expression, and
I get the expression of 1 over vg in terms
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of wavelength. So, L over vg is now L over
c, n lambda naught minus lambda naught times
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dn over d lambda naught.
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Now look at it, if I have a single wave having
wavelengths lambda naught and there is there
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are no other wavelengths components, then
this dn over d lambda naught is 0 and I simply
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get L over vg is equal to n by c or vg is
equal to c by n which is nothing, but the
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phase velocity. But when I have a group of
wave then this extra term comes into picture.
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From here you might think that in this way
the refractive index which is felt by the
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group of waves has decreased. You can see
that now the refractive index from individual
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wave to group of waves, now changes as n lambda
minus lambda naught times dn over d lambda
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naught.
So, the group of waves will feel this refractive
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index of the medium, has it decreased? If
I compare it with the individual wave no because
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dn over d lambda naught is negative. So, the
refractive index which is also refractive
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index of the group of waves which is also
known as group index has basically increased.
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So, group of index has increased and this
is how the group velocity has decreased now.
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So, individual waves they travel with faster
velocity, but the group moved with a slower
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velocity group moves bit slower much lower,
others pull others pull them down. One individual
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wave tries to move fast, but others is no
you cannot go that fast and they pull, they
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pull them down. So, group index is always
larger than the index of the individual wave.
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Now, let us find out what is the time taken
in traversing length L of the fiber. So, this
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would be given by L by vg, and simply L by
c times group index and this group index is
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of course, a function of lambda. Now let us
look at the spectral characteristics of the
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source, what wavelength what values of lambda
naught I have. See if I take the LED then
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this is the line width, and if I take a laser
diode this is the line width, and all the
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wavelengths components which fall into these
line width would now contribute here they
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have different transit times. What would be
the implication of this on the broadening
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how much would be the broadening. So, it is
very simple, that if the source has the spectral
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width delta lambda naught, they the temporal
broadening of the pulse would be simply given
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by delta tau is equal to d tau over d lambda
naught, times delta lambda naught while considering
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only first order term.
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And since tau is equal to L by c, n lambda
naught minus lambda naught times dn over d
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lambda naught then this would be simply now
I just take d tau over d lambda naught from
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here. So, dn over d lambda naught minus dn
over d lambda naught minus lambda naught times
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d2n over d lambda naught square, times delta
lambda naught and this would simply be minus
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L by c times lambda naught, d2n over d lambda
naught square times delta lambda naught. And
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usually what we do we define this broadening
in terms of dispersion coefficient, and here
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it would be called material dispersion coefficient.
Since the fiber length is measured in kilometers
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and the line width of the source is measured
in nanometers. So, I find the broadening of
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the pulse per kilo meter length of the fiber,
and per nanometer spectral width of the source
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and. So, the dispersion coefficient is given
as delta tau divided by per kilometer length
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of the fiber per nanometer spectral width
of the source, and we usually represent this
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delta tau in picoseconds per kilometer nanometers.
So, this would now be simply minus lambda
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naught by c, d2n over d lambda naught square.
Let us work out with the dimensions here.
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So, if you go back it is lambda naught it
is d2n over d lambda naught square. So, what
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I do now I multiply numerator and denominator
by lambda naught.
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So, I can write this down like this. The motivation
for writing this down is that now I can get
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this as a dimensionless quantity. So, now,
the dimensions are coming from here. So, let
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me express lambda naught in micrometer and
c in kilometers per second, then the dimensions
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would be seconds per kilometer times micrometer,
and let me convert it into picoseconds per
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kilometer nanometers. So, this would be 10
to the power 12 picoseconds per kilo meters
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times thousand nanometers. So, it would simply
be this much times 10 to the power 9 picoseconds
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per kilometer nanometer.
So, you can use this expression to calculate
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material dispersion coefficient in picoseconds
per kilometer nanometer, provided that you
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put the velocity of light c in kilometers
per second and wavelength of light in micrometer.
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So, that is what I have written here. So,
now, let us look. So, if I again go back and
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see that this material dispersion coefficient
depends upon how the refractive index of the
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material changes with wavelength.
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So, I need to know what is the second derivative
of n with respect to lambda, and this is the
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characteristic of a material for different
materials it would be different. So, different
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materials will exhibit different material
dispersion characteristics.
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Let me do this for pure silica glass, it is
fused silica and the variation of effective
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index with wavelength or pure silica glass
is given by this, where various constants
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have these values and lambda naught is measured
in micrometer. What is done is basically you
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measure the refractive index of the material
at different wavelengths, and then fit this
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kind of relationship to find out the values
of these constants. So, these values are also
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experiment dependent, but these are quite
accepted values in the literature.
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So, now what I do? I find out the variation
of n with respect to lambda. I plot the variation
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of n with respect to lambda or fused silica
glass, and it looks like this. Then dn over
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d lambda naught in the units of micro meter
inverse it goes like this, and what I find
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that it changes its slope from here to here,
and it has got a maximum somewhere here and
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it tells me that and remember that in dispersion
coefficient I need to have d2n over d lambda
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naught square, which means that it should
cross 0 somewhere here. So, now, I plot d2n
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over d lambda naught square in micrometer
to the power minus 2, and I see the variation
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looks like this, and it crosses 0 somewhere
here. And if I find out d2n over d lambda
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naught squares at certain wavelengths say
800 nanometer or 0.8 micrometer, which was
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the wavelength, used earlier for optical fiber
communication around 800 or 850 nanometer.
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So, I find that is d2n over d lambda naught
square is about 0.04 micrometer to the power
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00:26:34,740 --> 00:26:49,850
minus 2, and here if I plot this material
dispersion coefficient using this in picoseconds
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00:26:49,850 --> 00:26:55,909
per kilometer nanometer, then corresponding
D mat goes something like this and I find
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00:26:55,909 --> 00:27:01,830
that since there is zero crossing around this
wavelength. So, material dispersion is zero
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00:27:01,830 --> 00:27:12,559
at this wavelength, and this wavelength is
around 1.27 micrometer. Now let me work out
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00:27:12,559 --> 00:27:18,979
some numbers for a typical led source, if
I take lambda naught 0.8 micrometer, delta
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00:27:18,979 --> 00:27:25,480
lambda naught 25 nanometer and from here if
I find out d mat I have already seen that
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00:27:25,480 --> 00:27:30,610
d2n over d lambda naught square is 0.04 micrometer
to the power minus 2.
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00:27:30,610 --> 00:27:40,169
So, if I put these numbers here I find that
material dispersion coefficient at this wavelength
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00:27:40,169 --> 00:27:47,599
comes out to be about 106 picoseconds per
kilo meter nanometer. Corresponding broadening
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00:27:47,599 --> 00:27:57,659
will come out to be minus 2.7 nano seconds
per kilometer well broadening would be always
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00:27:57,659 --> 00:28:03,440
I have to take the magnitude of that. This
dispersion coefficient is negative, what is
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00:28:03,440 --> 00:28:08,639
the meaning of negative and positive we will
learn as we go along.
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00:28:08,639 --> 00:28:17,380
But there would be a broadening of about p
nanosecond every kilometer. If I take a laser
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00:28:17,380 --> 00:28:28,609
diode at 1.55 micrometer and delta lambda
naught about 2 nanometers, at this wavelength
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00:28:28,609 --> 00:28:34,940
d2n over d lambda naught square comes out
to be minus 0.004 micrometer to the power
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00:28:34,940 --> 00:28:40,999
minus 2, and it gives me material dispersion
coefficient as 20 picoseconds per kilometer
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00:28:40,999 --> 00:28:45,669
nanometer and now the broadening is only 40
picoseconds.
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00:28:45,669 --> 00:28:56,299
So, you can see when I go from an LED at 800
nanometer wavelength to a laser diode at 1550
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00:28:56,299 --> 00:29:03,070
nanometer wavelength, the material dispersion
comes down drastically from 3 nano seconds
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00:29:03,070 --> 00:29:10,570
per kilometer if you go back, 3 nano seconds
per kilometer to 40 picoseconds per kilometer
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00:29:10,570 --> 00:29:19,350
and if I use a wavelength around 1.27 micrometer
then it is almost 0. So, this wavelength is
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00:29:19,350 --> 00:29:28,729
also known as 0 material dispersion wavelength
and that is why the earlier fibers the fiber
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00:29:28,729 --> 00:29:38,350
which is already laid on seabed, most of that
fiber is optimized around 1.27 or 1.3 micrometer
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00:29:38,350 --> 00:29:39,350
wavelength.
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00:29:39,350 --> 00:29:45,049
Now, let us look at pulse propagation at different
wavelengths this is the LED at 800 nanometer
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00:29:45,049 --> 00:29:52,730
with line width 25 nanometer and broadening
of 2.7 nano seconds per kilometer. So, if
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00:29:52,730 --> 00:30:01,919
I now send this pulse I see that it broadens
very quickly, when I use a laser diode at
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00:30:01,919 --> 00:30:06,669
1550 nanometer wavelength which gives me a
broadening of 40 picoseconds per kilometer
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00:30:06,669 --> 00:30:12,730
nanometer, then it also broadens the pulse
there, but the broadening is not that much.
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00:30:12,730 --> 00:30:20,419
But if I use 1270 nanometer laser diode then
the broadening is very small around one picoseconds
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00:30:20,419 --> 00:30:29,369
per kilometer length of the fiber and at this
wavelength the pulses is retain their shape
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00:30:29,369 --> 00:30:30,990
over very long distances.
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00:30:30,990 --> 00:30:43,580
Then there is total dispersion in multimode
fiber, total dispersion will comprise both
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00:30:43,580 --> 00:30:52,340
the intermodal and material. So, I can then
find out the total dispersion by using the
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00:30:52,340 --> 00:30:59,279
information of intermodal as well as material
dispersion. Then the maximum bit rate I can
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00:30:59,279 --> 00:31:08,710
find out if I know the total dispersion by
0.7 over delta tau, and I know that beam max
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00:31:08,710 --> 00:31:17,729
times delta tau should be less than one, but
I have taken this vector 0.7 here it is corresponding
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00:31:17,729 --> 00:31:26,450
to non-return to zero coding. Now I take 2
examples for a step index, multimode fiber
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00:31:26,450 --> 00:31:36,289
where n2 is equal to 1.45 delta is equal to
0.01 and I have wavelengths which is 850 nanometer
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00:31:36,289 --> 00:31:45,799
please LED of 25 nanometer line width.
So, here delta tau m material dispersion is
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00:31:45,799 --> 00:31:50,989
about 290 seconds per kilometer, while intermodal
dispersion is 50 nanoseconds per kilometer.
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00:31:50,989 --> 00:31:58,249
So, total dispersion comes out to be 50 nanoseconds
per kilometer this dominates, and if I find
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00:31:58,249 --> 00:32:05,919
out Bmax from here it is about 40 Mbps over
a kilometer. If I take a graded index fiber
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00:32:05,919 --> 00:32:11,960
graded index multimode fiber with q is equal
to 2 which is which has got parabolic index
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00:32:11,960 --> 00:32:18,849
variation, then I find that intermodal dispersion
comes down to 0.25 nano seconds per kilometer,
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00:32:18,849 --> 00:32:26,639
while material dispersion is 1.7 nano seconds
per kilometer. So, this dominates total dispersion
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00:32:26,639 --> 00:32:36,799
is about 1.72 and Bmax can be 400 Mbps over
a kilometer 400 Mbps over a kilometer. So,
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00:32:36,799 --> 00:32:42,820
it increases from 14 Mbps to 400 mbps.
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00:32:42,820 --> 00:32:51,550
So, what we have learnt in this lecture that,
an optical pulse comprises a group of large
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00:32:51,550 --> 00:32:57,999
number of harmonic waves of different frequencies
or wavelengths, these constituent waves of
244
00:32:57,999 --> 00:33:04,470
pulse have different phase velocities, the
envelope of the pulse moves with a velocity
245
00:33:04,470 --> 00:33:09,330
called group velocity, different velocities
of constituent waves give rise to what is
246
00:33:09,330 --> 00:33:15,000
known as material dispersion and material
dispersion coefficient we usually represent
247
00:33:15,000 --> 00:33:21,700
in terms of picoseconds per kilometer nanometer.
Pure silica glass has 0 dispersion around
248
00:33:21,700 --> 00:33:30,229
1.27 micrometer wavelength, and total dispersion
will contain now both intermodal as well as
249
00:33:30,229 --> 00:33:37,909
material dispersion, in a multimode step index
fiber the intermodal dispersion will dominate
250
00:33:37,909 --> 00:33:42,559
while in a graded index multimode fiber material
dispersion dominates.
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00:33:42,559 --> 00:33:51,139
So, this is all about some basic transmission
characteristics of an optical fiber. In the
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00:33:51,139 --> 00:33:59,539
next few modules we will do some rigorous
analysis of light propagation in optical fibers,
253
00:33:59,539 --> 00:34:05,960
before that we will understand how light propagates
in infinitely extended medium, then a very
254
00:34:05,960 --> 00:34:13,200
simple planar waveguide. So, we will do in
the next few lectures now.
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00:34:13,200 --> 00:34:13,669
Thank you.