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In the last lecture we had seen that when
optical pulses propagate through optical fiber
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then they attenuate.
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I had also mentioned that they broaden in
this lecture we are going to see how the pulses
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are broadened in an optical fiber and in particular
in a multimode fiber.
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So, you are going to look at broadening of
pulses; why do they broaden, mechanism of
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broadening, in this lecture intermodal dispersion,
then what is the intermodal dispersion coefficient,
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then we are going to look at bit rate length
product.
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You will see that for a given bit rate I cannot
go for a very long length or if I want to
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keep a particular length of the fiber then
bit rate I will have to compromise with.
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We will see in this lecture what is the need
for cladded fiber, and how can I minimize
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intermodal dispersion.
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So, again I look at this slide, go back to
this slide and particularly focus on the broadening
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of pulses.
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So, the input pulse train gets broaden.
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And let us see what happens because of this
broadening.
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So, this is the input pulse train; let us
say this is the sensitivity of the detector
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this green line represents the sensitivity
of the detector.
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So, anything which is below this would be
recorded as 0 and anything which is much above
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this is 1.
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So, when I send this pulse train through L
length of optical fiber then the pulses become
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like this.
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And in this particular scenario what I see
that if this is the sensing level then this
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is the high pulse this is low, so high and
low regions are well distinguished.
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So, these pulses are resolvable.
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If here I have 1, here I have 0 then I will
receive it as 1 and 0 and my information is
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intact.
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While, if the broadening is much larger than
this then the pulses will overlap and everywhere
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I will see 1, I will lose my 0s here and I
will lose the information which I want to
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retrieve at the output end.
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So, question is what causes the broadening
of these pulses so that we can take care of
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this.
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For that let us first look at the shape of
pulses what kind of pulses we are talking
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about and the most popular shape of the pulse
is Gaussian.
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What is the Gaussian?
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Gaussian is this particular type of variation
which is expressed as P0 e to the power minus
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2 t minus t0 square over tau naught square.
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Where, this pulse is centered around t is
equal to t0.
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So, this is called the position of pulse or
the center of pulse and 2 tau 0 is the width
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of the pulse where the power has dropped down
to 1 over e square of its peak value.
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So, this 2 tau naught is full width and we
usually call tau naught as width of the pulse.
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If this pulse is travelling, if I let it travel
then a distance Z the pulse becomes this its
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amplitude changes and also its width changes
and position also.
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Now, I have a pulse which becomes something
like this, which is now centered at t is equal
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to tz whose power has gone down to Pz and
the width has become now tau.
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Where tau can be given by tau naught square
plus delta tau square, your delta tau is known
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as broadening of the pulse and that is what
we want to find out how much is the broadening.
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You may notice here that in case of Gaussian
pulse after travelling the Gaussian pulse
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remains Gaussian.
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However, its amplitude and width changes,
but the function remains the same.
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So, why it is broadened?
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What I see that if I look at a multimode fiber
then in a multimode fiber I can have various
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ray paths: if I launch a pulse of light into
this fiber then some of the energy can go
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into this ray path, some of the energy can
go into this blue one, some of the energy
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can go into red one.
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And what happens is I can immediately see
that if the energy is going into this path
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then it travels the least distance to traverse
a certain length of the fiber.
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However this red one it takes longer path.
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So, the light is coupled into all these ray
paths then these ray past take different times
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to reach the output end and because of this
there is broadening.
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So, I send the pulse like this and at output
and I receive the pulse like this.
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And this kind of broadening is known as intermodal
dispersion, because different ray paths can
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be called different modes of propagation of
the system.
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So, this is known as intermodal dispersion.
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Let us now find out how much broadening happens
due to this intermodal dispersion.
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So, I have again three rays here white one
which is axial ray, then there is the green
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one and the red one.
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So, there can be several rays various rays
like this.
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In order to find out the expression for broadening
let me consider a ray which makes an angle
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theta from the fiber axis and correspondingly
it makes an angle phi from the normal to the
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core cladding interface and this ray goes
like this.
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When it reaches path B when it reaches point
B it repeats itself after this.
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So, I can see these two points are kind of
identical.
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So, if I can find out the time taken by this
ray in traversing this part AB then I can
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find out what would be the time taken by traversing
L length of the fiber, because that L length
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can be comprising these several AB’s.
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So, what would be the time taken in reaching
from A to B?
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If it follows this path red path well it is
very simple: AC plus CB divided by the velocity
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of light in the refractive index region of
n 1, which is C by n1.
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So, t is equal to AC plus CB divided by c
over n1.
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Now from here I can see AC cos theta is equal
to AB by 2 and similarly CB cos theta is equal
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to AB by 2.
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I can immediately see that this angle theta
would be equal to this angle theta, because
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this theta is equal to this theta and then
it should be equal to this.
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Now, if I find out AC plus CB then AC plus
CB from here is AB over cos theta.
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And in this way I will have t is equal to
n1 AB over c cos theta.
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And therefore, for L length of the fiber it
would be n1 L over c cos theta.
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So, it should come somewhere here.
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Since time taken for traversing AB length
of the fiber is n1 AB over c cos theta.
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So, the time taken in traversing L length
of the fiber would be n1 L over c cos theta.
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So, this is what I have.
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Now, what is the range of theta for light
guidance by total internal reflection?
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You can have an axial ray which corresponds
to theta is equal to 0 and then you can have
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a ray which corresponds to theta is equal
to theta C, which in turn corresponds to the
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critical angle.
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So, your minimum time would be for axial ray
and it is corresponding to theta is equal
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to 0 given by n 1 L over C. While the maximum
time would be taken by the ray which makes
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an angle theta C from the axis, so, t max
is equal to n1 L over c cos theta C. And because
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of the time difference between this and this
I will have broadening.
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From here I can immediately have cos theta
C is equal to sine phi C which is n2 over
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n1.
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So, t max is n1 square L over c n2, ok.
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Now, if all the rays are excited simultaneously
at the input end then the time interval occupied
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by the rays at the output end now would be
tmax minus tmin which are represent by delta
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tau broadening, and it would be simply n1
L by c times n1 over n2 minus 1.
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Let us work out an example: how much broadening
I am going to have for a typical fiber.
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If I have a typical multimode fiber and I
send a very narrow pulse or impulse through
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that fiber then what would be the broadening.
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For a typical multimode fiber n1 is equal
to 1.5 and relative index difference between
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the core and the cladding is typically 0.01
or 1 percent.
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So, now for these values if I calculate delta
tau which is given by n1 L by c n1 by n2 minus
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1 and let me do it for 1 kilometer length
of the fiber L is equal to 1 kilometer then
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I find out the delta tau comes out to be 50
nanoseconds: 50 nanoseconds for 1 kilometer;
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which means that if I send an impulse then
it will become a pulse of 50 nanoseconds duration
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after traveling through 1 kilometer length
of the fiber.
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So, let me send this very narrow pulse through
this optical fiber and as it propagates through
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optical fiber that it broadens.
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So, you can see that it broadens.
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So, this pulse at the input end will come
out to be like this at output and after 1
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kilometer length of the fiber.
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Let me see what happens in the case of pulse
strain.
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If I now send two pulses like this which are
separated in time by 50 nanoseconds and let
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me send it to 1 kilometer length of an optical
fiber, then what happens is that as the pulses
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propagate through fiber they broaden, you
can see the broadening.
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Now you see they have started overlapping.
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And around this point you can see that if
the sensitivity of detector is somewhere here
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then you cannot make out where is your 0 where
is your 1 they have completely overlapped,
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you cannot distinguish between the two pulses
and here at 1 kilometer length you cannot
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say that there were two pulses initially,
ok.
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So, they are unresolved you cannot resolve
these pulses.
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So, what you will have to do?
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If you want to keep the length of the fiber
as 1 kilometer and you still want to resolve
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them at output end then you will have to increase
the separation between the pulses.
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So, that they are resolvable at the output
end.
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So, let me now increase the separation between
the pulses and let me make it 100 nanosecond.
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Now if I propagate these through this fiber.
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Now, I can see the pulses are broadened, but
they are resolvable, they are still resolvable
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ok.
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So, for a given broadening this was the case
for delta tau corresponding to 50 nanoseconds.
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So, for a given broadening or for a given
fiber if I want to keep the length constant
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1 kilometer then in order to retrieve information
at output end I will have to have large separation
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between the pulses; large separation means
that you are now able to send less number
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of pulses per second, because separation between
the pulses has been increased, which means
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that your data rate is now smaller, ok.
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In fact, we can relate this delta tau to this
data rate or the information carrying capacity
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the data rate or bit rate I can relate this
delta tau to this bit rate B, ok.
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A precise relation between B and tau depends
upon several things; like pulse shape or coding
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scheme.
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But intuitively it is clear that this delta
tau should be less than TB; what is TB, TB
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is the allocated bit slot which is nothing
but 1 over B if B is the bit rate then TB
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is equal to 1 over B is the allocated bit
slot; which means that this B times delta
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tau should be less than 1.
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And delta tau I have from previous slide n1
L over c n1 minus n2 minus 1.
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So, B times this should be less than 1.
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Or B times L should be less than n to c over
n1 times n1 minus n2.
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So, for a given fiber where n1 and n2 are
fixed your B times L is constant.
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So, for a given fiber B times L is constant
that is bit rate length product is always
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constant.
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What is the implication of this?
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Let us consider two cases.
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Case one: it is of un-cladded fiber, where
I have core of glass of refractive index 1.5
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and (Refer Time: 18:12) works as cladding.
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So, n1 is equal to 1.
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In this case let me find out this B times
L, and B times L if I plug in these values
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here a small c is the speed of light 3 into
10 to the power 8 meters per second.
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Then it comes out to be 0.4 Gbps times meter,
which means that you can send pulses at the
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rate of 0.4 GB per second or 100 MB per second
only over 1 meter length of the fiber.
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So, if you send pulses at this rate the information
can be retrieved just within 1 meter, otherwise
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you will lose information.
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of course, you would not like to send information
only for a meter.
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If I talk about kilometer then 1 kilometer
length of the fiber can sustain a bit rate
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only up to 0.4 megabits per second.
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And if you remember that if I am trying to
send a video then video requires a data rate
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of about 100 mbps.
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So, you cannot use it.
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Now like we take another case, we are which
is of cladded fiber where n1 is equal to 1.5
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and delta the relative index difference is
1 percent then what is B times L. Then B times
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L comes out to be 20 Gbps over 1 meter or
1 mbps over 20 kilometers So, you can send
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data at the rate of 1 megabits per second
over 20 kilometer length of the fiber or if
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you want to keep it within campus typically
2 kilometer length then the data rate can
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be 10 mbps.
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So, if still it is not sufficient; 10 mbps
is nothing actually.
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So, we will have to take care of this.
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So, what I see here that is smaller is the
index difference; if smaller is the index
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difference between the core and cladding higher
would be the data.
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That is why I cannot use unclad fiber, I cannot
use unclad fiber and I need a cladding to
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have a telecom fiber.
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So, this clearly shows the importance of the
cladding in the fiber.
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Question is how can I minimize this intermodal
dispersion.
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So, for that let me look at the origin of
this intermodal dispersion.
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And the origin is there different rays have
different transit times through particular
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length of the fiber.
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So, if somehow I can minimize the difference
in transit times corresponding to different
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ray paths then I can minimize intermodal dispersion.
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For that I can use a graded index fiber.
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What happens in a graded index fiber?
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For an axial rate the path is like this.
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So, the path is shortest, but here I have
highest refractive index.
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So, the velocity of light is the smallest.
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So, ray 1 has shortest path, but lowest velocity
also, it is slowest.
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While this ray it has longer path, but it
traverses in medium from here to here the
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refractive index changes and in fact refractive
index decreases.
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So, the velocity increases.
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And for this ray this ray spends most of its
time in a smaller refractive index region.
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So, it will have largest velocity.
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So, in this way the difference between the
transit times of different rays can be reduced.
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If I look at typical graded index fiber defined
by power low profile then this is the refractive
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index variation in the core.
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And if I find out the value of delta tau the
expression for delta tau for a fiber which
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corresponds to q is equal to 2 and called
parabolic index fiber then delta tau comes
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out to be like this.
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Now, if I take a fiber with n2 is equal to
1.45 and delta is equal to 1 percent then
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00:23:53,861 --> 00:24:00,849
delta tau comes out to be about 0.25 nano
seconds per kilometer.
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We call that for step index fiber; for a step
index fiber delta tau was something like 50
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nanoseconds per kilometer.
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So, I can bring it down from 50 nanoseconds
per kilometer to 0.25 nano seconds per kilometer.
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I can also find out what is the optimum profile,
what is the optimal value of q for which my
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intermodal dispersion would be minimum.
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And it can be shown that this optimum profile
of course, depends upon the value of n1 and
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n2 and it can be given y2 times n2 over n1
or approximated by 2 minus 2 delta.
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So, now if I take n1 is equal to 1.46 and
delta is equal to 1 percent then for step
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index fiber which corresponds to q is equal
to infinity the intermodal dispersion is 50
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nanoseconds per kilometer, for parabolic index
fiber which corresponds to q is equal to 2
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this dispersion is 0.25 nano seconds per kilometer.
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And the optimum profile across at around 1.98,
q is equal to 1.98.
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And intermodal dispersion can be brought down
to 0.0625 nano seconds per kilometer.
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So, you can tremendously bring down the intermodal
dispersion if you are using a graded index
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fiber.
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Of course, if you want to completely get rid
of intermodal dispersion then you should use
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a single mode fiber.
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The question comes: why cannot we completely
get rid of intermodal dispersion and you single
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mode fiber instead of doing all this.
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The answer is that this fiber has core diameter
of about 50 micro meter, and single mode fiber
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has a core diameter of about 10 micrometer.
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So, if you go from multimode to single mode
then it becomes very expensive, because your
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components will become more precise they have
to be compactified.
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So, cost increases.
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That is why if you do not want to have very
high data rate you want to use it within a
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campus; in short distance communication, in
local area networks, it is referable that
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you use a multimode fiber with optimum profile.
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So, let us now look at ray path and pulse
propagation in step index fiber.
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So, what we have done here?
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We have launched these pulses, I have launched
this pulse and the red one corresponds to
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the energy coupled into the axial ray and
blue one is the energy coupled into the ray
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which is launched at an angle.
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00:27:22,710 --> 00:27:30,789
And let us see when the light propagates along
these two paths what happens to this.
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00:27:30,789 --> 00:27:38,710
So, you can see that in the time when this
axial ray reaches here this comes somewhere
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here.
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00:27:39,730 --> 00:27:52,919
And you can see that the axial rate reaches
faster while this ray at an angle reaches
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slower, it reaches later.
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So, by the time this blue one reaches here
the red one will go much ahead.
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So, there would be broadening because of this.
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If I now see the same thing for optimum index
fiber then by the time the red one reaches
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00:28:18,960 --> 00:28:22,480
here the blue one reaches here.
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00:28:22,480 --> 00:28:28,549
So, they traverse the same length of the fiber;
it reaches red one reaches here blue also
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00:28:28,549 --> 00:28:29,609
reaches here.
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00:28:29,609 --> 00:28:36,729
So, they go along and at the output end both
of them will reach almost at the same time.
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So, the intermodal dispersion is minimized.
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So, what we have learnt in this lecture.
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00:28:46,219 --> 00:28:57,779
In a multimode fiber the pulse spreads during
propagation through the fiber and the broadening
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of this pulse is attributed to different transit
times corresponding to various possible ray
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paths.
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00:29:06,389 --> 00:29:09,719
And this is known as intermodal dispersion.
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00:29:09,719 --> 00:29:18,619
Typical broadening in my step index multimode
fiber is 50 nanoseconds over 1 kilometer.
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00:29:18,619 --> 00:29:24,659
The bit rate length product in a fiber is
a constant and it depends only on the refractive
265
00:29:24,659 --> 00:29:28,970
index profile of the fiber.
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00:29:28,970 --> 00:29:36,710
In a multimode fiber the intermodal dispersion
can be minimized by using a near parabolic
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00:29:36,710 --> 00:29:39,820
profile in the core.
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00:29:39,820 --> 00:29:43,239
And such fibers are useful in local area networks.
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Thank you.