1
00:00:18,650 --> 00:00:28,930
Now after having understood, most of the things
that is the transmission medium which is an
2
00:00:28,930 --> 00:00:42,690
optical fiber the components devices in between
optical sources that is a transmitter end
3
00:00:42,690 --> 00:00:52,470
and optical detectors that is the receiver
end. So, we have now studied all the segments
4
00:00:52,470 --> 00:01:01,850
all the sections individually. Now it is time
to put them together and look into system
5
00:01:01,850 --> 00:01:03,200
design aspects.
6
00:01:03,200 --> 00:01:13,080
So, what we have the overall system that we
have electric signal, which comes out from
7
00:01:13,080 --> 00:01:21,080
data, data is in the form of electric signal
which drives light source generates optical
8
00:01:21,080 --> 00:01:33,170
pulses, then this optical energy is coupled
to transmission medium which is optical fiber
9
00:01:33,170 --> 00:01:39,690
using connector, then in between we can have
several optical fibers.
10
00:01:39,690 --> 00:01:47,420
We can have different components and devices
based on optical fiber itself, then at the
11
00:01:47,420 --> 00:01:56,010
receiver end we need to couple this optical
energy into the optical detector. So, again
12
00:01:56,010 --> 00:02:02,240
we will require a connector and ultimately
we will get electrical signal output from
13
00:02:02,240 --> 00:02:10,310
optical detector. So, this is the complete
system, now let us look at the design aspects
14
00:02:10,310 --> 00:02:19,890
of this. So, the very first thing is light
source for long haul telecom system one would
15
00:02:19,890 --> 00:02:29,889
prefer laser that laser diodes as light source,
with operating wavelength 1300 nanometer or
16
00:02:29,889 --> 00:02:37,849
1550 nanometer.
Line width should be one nanometer or less
17
00:02:37,849 --> 00:02:48,519
and modulation speed should be in excess of
several Gbps or the light source should be
18
00:02:48,519 --> 00:02:56,370
compatible to modulation speeds which are
in excess of several Gbps basically you can
19
00:02:56,370 --> 00:03:02,409
directly modulate the light source using the
injection current or you can use external
20
00:03:02,409 --> 00:03:14,810
modulator. Directly modulating the light source
has some consequences and it cannot go beyond
21
00:03:14,810 --> 00:03:22,860
few Gbps. So, for very high speed you will
require external modulators and your laser
22
00:03:22,860 --> 00:03:34,010
diode should be compatible to that.
Another thing is that when we design the system,
23
00:03:34,010 --> 00:03:39,739
then we would have to look into what input
power of light source would be needed taking
24
00:03:39,739 --> 00:03:47,650
into account all the components transmission
medium and sensitivity of the detector.
25
00:03:47,650 --> 00:03:54,280
Then we have connectors at several places
took up a light from light source to transmission
26
00:03:54,280 --> 00:04:01,499
fiber and from fiber to the optical detector,
and in between also we put several optical
27
00:04:01,499 --> 00:04:08,700
fiber based components. So, we will require
connectors. So, these connectors will cause
28
00:04:08,700 --> 00:04:14,239
some loss each connector will have certain
loss. So, we will have to take into account
29
00:04:14,239 --> 00:04:24,250
that.
Then in the transmission medium we will have
30
00:04:24,250 --> 00:04:31,960
we may have to join several fibers together
or we may have to join some specialty fibers
31
00:04:31,960 --> 00:04:40,750
for certain operations with the transmission
fiber. So, we would require splices. So, we
32
00:04:40,750 --> 00:04:51,190
will have to take into account the loss at
each splice then of course, the transmission
33
00:04:51,190 --> 00:04:54,360
loss of the fiber itself.
34
00:04:54,360 --> 00:05:00,300
So, whatever fiber we are using we will have
to take into account the loss of that fiber
35
00:05:00,300 --> 00:05:08,560
at the operating wavelength. So, this loss
depends upon the fiber used and of course,
36
00:05:08,560 --> 00:05:10,090
the wavelength of operation.
37
00:05:10,090 --> 00:05:16,310
If you look at certain fibers which is G.652.A
& B, and this is G.652.C & D.
38
00:05:16,310 --> 00:05:24,289
So, the loss coefficient as a function of
wavelength which is admitted is like this,
39
00:05:24,289 --> 00:05:35,110
which has been adapted from ITU-T manual on
optical fibers cables and systems. So, where
40
00:05:35,110 --> 00:05:43,789
I can see that if I look at 1550 nanometer.
So, the maximum loss that can be admitted
41
00:05:43,789 --> 00:05:57,410
is about 0.275 dB per kilometer and minimum
is around 0.2 dB per kilometer; however, if
42
00:05:57,410 --> 00:06:04,470
I look at this. So, at 1550 the loss limits
are similar, but if I compare this fiber with
43
00:06:04,470 --> 00:06:12,810
this fiber also operates near 1380 nanometer
wavelength where this fiber does not operate.
44
00:06:12,810 --> 00:06:20,500
So, here you have water peaks in this fiber.
So, this fiber cannot be used here; however,
45
00:06:20,500 --> 00:06:32,860
this fiber can be used in the entire wavelength
range from 1300 nanometer to 1600 nanometer.
46
00:06:32,860 --> 00:06:42,740
Then another very important aspect is to design
or to choose optical detector and look into
47
00:06:42,740 --> 00:06:52,229
various characteristics of optical detector.
The detector should of course, have high sensitivity
48
00:06:52,229 --> 00:07:00,300
should support high data rate and should have
high signal to noise ratio so that we can
49
00:07:00,300 --> 00:07:10,569
fish out signal from the noise. So, that the
signal is tends out quite well from the noise
50
00:07:10,569 --> 00:07:18,930
and it should have low bit error rate. So,
very important aspect of any detector is how
51
00:07:18,930 --> 00:07:27,010
well it can distinguish signal from the noise,
and that is particularly when your signal
52
00:07:27,010 --> 00:07:36,960
levels drop down to very low levels. So, the
noise should be as low as possible and in
53
00:07:36,960 --> 00:07:43,039
fact, the ratio between the signal and noise
should be as large as possible.
54
00:07:43,039 --> 00:07:51,889
So, it is defined in the form of a ratio which
is known as SNR signal to noise ratio which
55
00:07:51,889 --> 00:08:00,379
is nothing, but the ratio of signal power
which comes from photo current divided by
56
00:08:00,379 --> 00:08:08,159
photo detector noise power plus amplifier
noise power. So, the total noise signal power
57
00:08:08,159 --> 00:08:19,759
divided by total noise power. In a photo detector
you have various processes by which the noise
58
00:08:19,759 --> 00:08:27,539
is generated; one is short noise which is
due to statistical nature of generation and
59
00:08:27,539 --> 00:08:35,779
collection of electrons that constitute photo
current. So, photo current is because of the
60
00:08:35,779 --> 00:08:45,370
stream of electrons which is generated and
these electrons are generated at random times
61
00:08:45,370 --> 00:08:53,820
and collected at random times.
So, because of that there is a variance in
62
00:08:53,820 --> 00:09:03,560
the current and this leads to noise because
of the random nature of this. So, the noise
63
00:09:03,560 --> 00:09:11,020
variance is given by sigma s square is equal
to 2 times e times ip. ip is the photocurrent
64
00:09:11,020 --> 00:09:20,350
times delta f where e is the charge of an
electron delta f is the effective noise bandwidth.
65
00:09:20,350 --> 00:09:28,200
So, this is an ampere square. So, the square
root of this basically gives you the noise
66
00:09:28,200 --> 00:09:38,880
current corresponding to short noise then
another process is the dark current. So, even
67
00:09:38,880 --> 00:09:46,750
if there is no optical energy if you do not
incident any light on photo detector, then
68
00:09:46,750 --> 00:09:54,700
in the bias circuit you will still have certain
current bulk current and surface leakage currents
69
00:09:54,700 --> 00:10:00,160
and these kind of currents are known as dark
current because of this you will have noise
70
00:10:00,160 --> 00:10:05,060
variance.
Noise variance corresponding to bulk current
71
00:10:05,060 --> 00:10:17,630
is given by sigma DB square is equal to two
e iB delta f, where iB is the bulk dark current
72
00:10:17,630 --> 00:10:26,400
and because of surface dark current it is
given as sigma d s square two e iD delta f
73
00:10:26,400 --> 00:10:37,690
where iD is the surface dark current. Then
in the photo detector circuit you will have
74
00:10:37,690 --> 00:10:48,730
to put a load resistor and when you put a
load resistor then there is thermal noise
75
00:10:48,730 --> 00:10:58,390
which is generated and it is because of random
thermal motion of electrons in the load resistor.
76
00:10:58,390 --> 00:11:05,060
Because of this the noise variance is given
by sigma T square is equal to 4 kBT divided
77
00:11:05,060 --> 00:11:14,580
by RL times delta f where kB is the Boltzmann
constant and RL is the load resistor.
78
00:11:14,580 --> 00:11:24,950
So, you have all these processes by which
the noise is generated and your signal should
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00:11:24,950 --> 00:11:39,220
be very high as compared to these noises.
Let us work out an example and see what kind
80
00:11:39,220 --> 00:11:48,310
of noise currents or noise variance we get
in a typical semiconductor photo detector.
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00:11:48,310 --> 00:11:56,730
So, I take an example of indium gallium arsenide
Pin photodiode at lambda naught is equal to
82
00:11:56,730 --> 00:12:08,720
1300 nanometer, which has iD is equal to 4
nano amperes, quantum efficiency 90 percent,
83
00:12:08,720 --> 00:12:23,230
load resistor is 1 kilo om and Pin is minus
35 d B m and receiver bandwidth is 20 megahertz.
84
00:12:23,230 --> 00:12:32,350
So, I would like to find out various noise
terms and SNR at 300 k neglecting the surface
85
00:12:32,350 --> 00:12:39,030
leakage current.
So, this is iD is the bulk leakage currents
86
00:12:39,030 --> 00:12:53,000
and here I am neglecting the surface leakage
current. So, I know that this is Pin, Pin
87
00:12:53,000 --> 00:13:01,540
is minus 35 dBm which corresponds to 3.16
into 10 to the power minus 4 milli watts.
88
00:13:01,540 --> 00:13:12,850
So, this much Pin will generate a photo current
depending upon this value of eta. So, I can
89
00:13:12,850 --> 00:13:21,291
immediately find out what would be the photo
current generated with this p n. So, the photo
90
00:13:21,291 --> 00:13:29,620
current generated would be given by ip is
equal to eta e lambda over h c times Pin.
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00:13:29,620 --> 00:13:37,940
So, if I plug in all these numbers this comes
out to be 0.3 micro amperes.
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00:13:37,940 --> 00:13:47,760
Now, if I look at short noise then it is given
by two e times ip times delta f and ip is
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00:13:47,760 --> 00:13:56,260
this much delta f is 20 megahertz. So, this
gives me sigma s square is equal to 1.9 into
94
00:13:56,260 --> 00:14:06,450
10 to the power minus 18 amperes square. Dark
current noise variance is given by two e iD
95
00:14:06,450 --> 00:14:19,080
delta f and iD is 4 nano amperes. So, this
gives me dark current noise variance is 2.56
96
00:14:19,080 --> 00:14:28,190
into 10 to the power minus 20 amperes square.
Thermal noise variance is given by this if
97
00:14:28,190 --> 00:14:35,571
I plug in all the numbers from here into this
then it comes out to be 3.31 into 10 to the
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00:14:35,571 --> 00:14:38,940
power of minus 16 ampere square.
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00:14:38,940 --> 00:14:49,450
So, if I put all of them together then I have
short noise, the dark current noise and thermal
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00:14:49,450 --> 00:14:59,950
noise, and you can see that thermal noise
dominates over dark current and short noise.
101
00:14:59,950 --> 00:15:09,130
So, what would be the total noise? Total noise
would be the sum of all these and it comes
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00:15:09,130 --> 00:15:17,500
out to be 3.33 into 10 to the power minus
16. So, most of the contribution comes only
103
00:15:17,500 --> 00:15:28,980
from thermal noise here from here I can calculate
signal to noise ratio which is given by ip
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00:15:28,980 --> 00:15:40,810
square divided by sigma square because this
is equivalent to the photocurrent power and
105
00:15:40,810 --> 00:15:49,580
this is the noise power goes as I square.
So, when I calculate this then SNR comes out
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00:15:49,580 --> 00:16:01,340
to be 270. So, this is how for any given photo
detectors I can find out various noises and
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00:16:01,340 --> 00:16:10,690
signal to noise ratio. Then this kind of noise
can lead to errors in measurement there is
108
00:16:10,690 --> 00:16:17,840
I want to measure one a high bit comes a bit
one comes, then there is a finite probability
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00:16:17,840 --> 00:16:25,060
of detecting it as 0 and when a 0 comes then
there is a finite probability of detecting
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00:16:25,060 --> 00:16:37,130
it as one. So, because of this there is error
in the measurement, and this is defined as
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00:16:37,130 --> 00:16:45,110
bit error rate which is a way to measure the
rate of error occurrence in a digital data
112
00:16:45,110 --> 00:16:46,140
stream.
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00:16:46,140 --> 00:16:57,790
So, BER is defined as number of errors occurring
over time interval t divided by total number
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00:16:57,790 --> 00:17:06,600
of pulses 1s and 0s over time interval t.
If the number of error occurrence says is
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00:17:06,600 --> 00:17:15,620
Ne and the total number of pulses or total
number of bits in time interval t is Nt then
116
00:17:15,620 --> 00:17:24,410
BER is Ne over Nt and of course, this can
be written as Ne divided by Bt where B is
117
00:17:24,410 --> 00:17:38,670
the bit rate. You can also define it as BER
is equal to P 1 by 0 plus P 0 y 1 divided
118
00:17:38,670 --> 00:17:46,180
by 2 where P 10 is the probability of deciding
one when 0 has actually been received and
119
00:17:46,180 --> 00:17:50,840
this is the probability of deciding 0 when
one has actually been received.
120
00:17:50,840 --> 00:18:00,930
You can write in write BER in terms of SNR
as 1 over 2, 1 minus error function is square
121
00:18:00,930 --> 00:18:09,960
root of SNR divided by 2 root 2 or you can
approximate it to 2 divided by pi times SNR
122
00:18:09,960 --> 00:18:21,490
to the power half times e to the power minus
SNR by 8, and from here I can immediately
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00:18:21,490 --> 00:18:31,750
see that for BER is equal to 10 to the power
minus 9, SNR should be approximately equal
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00:18:31,750 --> 00:18:32,930
to 144.
125
00:18:32,930 --> 00:18:43,790
So, what is the sensitivity of a detector
for a given bit rate and SNR so that I can
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00:18:43,790 --> 00:18:54,340
distinguish my 1 from 0 very well. It is given
in terms of minimum received power what should
127
00:18:54,340 --> 00:19:05,180
be the minimum power that a receiver should
receive, and it is given by B divided by responsivity
128
00:19:05,180 --> 00:19:14,130
bit rate divided by responsivity times two
square root of 2 pi kBT times C times SNR.
129
00:19:14,130 --> 00:19:22,940
Where C is the capacitance of photo detector
T is the temperature and of course, SNR is
130
00:19:22,940 --> 00:19:36,110
the signal to noise ratio and here this particular
expression corresponds to NRZ format non return
131
00:19:36,110 --> 00:19:48,050
to 0 format where delta f is B by 2. Let us
see an example for a given photo detector
132
00:19:48,050 --> 00:20:00,400
the capacitance is one pic farad, responsivity
is let us say 0.5 milliamps per milli watt
133
00:20:00,400 --> 00:20:07,360
then what would be the sensitivity of the
detector or minimum received power for one
134
00:20:07,360 --> 00:20:16,770
Gbps link if I want to have a BER of 10 to
the power minus 9 which actually corresponds
135
00:20:16,770 --> 00:20:25,750
to SNR of 144.
So, if I plug in the numbers then Pmin can
136
00:20:25,750 --> 00:20:33,030
be given by this expression, and I put all
the numbers here then it will come out to
137
00:20:33,030 --> 00:20:50,210
be 3.87 micro watts or minus 24 dBm. What
is the ultimate detection sensitivity? And
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00:20:50,210 --> 00:21:01,590
this ultimate detection sensitivity is provided
by short noise limited system. If I take an
139
00:21:01,590 --> 00:21:11,250
ideal detector where there is no thermal noise
no dark current noise and the quantum efficiency
140
00:21:11,250 --> 00:21:19,900
is one then bit rate is given by e to the
power minus Np divided by 2.
141
00:21:19,900 --> 00:21:25,980
Where Np is the average number of photons
per bit for BER is equal to 10 to the power
142
00:21:25,980 --> 00:21:35,440
minus 9, if you put it here then Np will come
out to be about 20. So, this is the quantum
143
00:21:35,440 --> 00:21:43,900
limit and corresponds to the ultimate detection
sensitivity.
144
00:21:43,900 --> 00:21:55,860
What is the corresponding minimum power on
the detector? Well if I know the minimum number
145
00:21:55,860 --> 00:22:03,731
of photons per bit then I can find out the
corresponding power. So, these are the number
146
00:22:03,731 --> 00:22:12,090
of photons divided by the bit slot. So, number
of photons per second times energy of one
147
00:22:12,090 --> 00:22:24,210
photon. So, this would be the Pmin and since
T is equal to 1 over B for NRZ system then
148
00:22:24,210 --> 00:22:35,130
this can be written as Np times h nu time
B. Let us see how much it is for one Gbps
149
00:22:35,130 --> 00:22:44,790
system at 1550 nanometer wavelength, if I
work this out from here it comes out to be
150
00:22:44,790 --> 00:22:55,740
about 2.7 nano watts or minus 55.7 d B m which
is really very small.
151
00:22:55,740 --> 00:23:04,820
But most of the practical systems operate
at 20 dB higher level that is instead of Np
152
00:23:04,820 --> 00:23:14,210
is equal to 20 which is ultimate quantum limit,
we operate them at 20 dB higher level which
153
00:23:14,210 --> 00:23:27,350
is 100 times. So, Np is equal to 2000 typically,
now let us power budget our optical fiber
154
00:23:27,350 --> 00:23:28,760
link.
155
00:23:28,760 --> 00:23:39,750
So, as we have seen that we have the light
source then we have connector splices loss
156
00:23:39,750 --> 00:23:51,220
due to transmission medium, then sensitivity
of the detectors and so on. So, how much power
157
00:23:51,220 --> 00:23:58,140
my light source should have or what should
be the sensitivity of the detector if I take
158
00:23:58,140 --> 00:24:05,610
into account all the losses.
So, let us power budget it. Let me have the
159
00:24:05,610 --> 00:24:17,040
source power as Pi in dBm everything I will
I will write in dB units here for easy calculation,
160
00:24:17,040 --> 00:24:27,010
let us say there are Nc number of connectors
with l c loss each loss is again in dB, and
161
00:24:27,010 --> 00:24:32,559
the link has length l and transmission loss
coefficient is alpha. So, transmission loss
162
00:24:32,559 --> 00:24:44,610
is alpha times L, they are Ns number of splices
with l s dB loss each then one has to keep
163
00:24:44,610 --> 00:24:52,540
certain power margin of about 6 to 8 dB to
account for losses from a splices and components
164
00:24:52,540 --> 00:25:03,690
which might be added later on in the system,
and if the detector sensitivity is Pmin in
165
00:25:03,690 --> 00:25:16,880
dBm, then what should I have that the source
power minus losses due to connectors, minus
166
00:25:16,880 --> 00:25:22,080
losses due to splices minus transmission loss
minus the power margin.
167
00:25:22,080 --> 00:25:31,780
So, if I subtract all these losses from my
input power, then this is what the power is
168
00:25:31,780 --> 00:25:39,990
left at the output end, and it should be greater
than or equal to the sensitivity of the detector
169
00:25:39,990 --> 00:25:48,590
which is Pmin. So, I should be able to fulfill
this condition or I can say that the source
170
00:25:48,590 --> 00:25:56,070
power should be greater than this which is
one or the same thing, let us understand it
171
00:25:56,070 --> 00:26:00,440
with the help of an example.
172
00:26:00,440 --> 00:26:08,880
Let me have a forty kilometer link with 0.5
dB kilometer 0.5 dB per kilometer loss receiver
173
00:26:08,880 --> 00:26:16,800
sensitivity is minus 39 dBm.
I have four splices with 0.5 dB loss each
174
00:26:16,800 --> 00:26:25,910
two connectors with 1 dB loss each and let
me put a power margin of 6 dB, then what is
175
00:26:25,910 --> 00:26:34,240
the minimum source power which I would require.
Well the minimum source power would be Pmin
176
00:26:34,240 --> 00:26:42,559
plus Nclc plus Nsls plus alpha L plus Pm and
since everything is in dB unit. So, I just
177
00:26:42,559 --> 00:26:54,020
put them here and add them up. So, this gives
me Pi minimum equal to minus nine dBm which
178
00:26:54,020 --> 00:27:02,820
in linear units is nothing, but 0.13 milli
watts. So, this much power I should have from
179
00:27:02,820 --> 00:27:12,370
my source.
Now, let us look into dispersion limited system
180
00:27:12,370 --> 00:27:20,380
and attenuation limited system what is the
attenuation limited system? In attenuation
181
00:27:20,380 --> 00:27:25,880
limited system dispersion is not a major concern
and repeater less length of the link is limited
182
00:27:25,880 --> 00:27:27,060
by attenuation.
183
00:27:27,060 --> 00:27:39,380
So, how do I get this? Well to receive data
at the output end with certain accuracy, I
184
00:27:39,380 --> 00:27:48,920
need to have minimum number of photons per
bit information and if B is the bit rate and
185
00:27:48,920 --> 00:27:59,510
T is the bit duration, then the minimum average
received optical power should be Np h nu divided
186
00:27:59,510 --> 00:28:09,190
by 2 T which is Np h nu B divided by 2.
I will tell you why I have taken the factor
187
00:28:09,190 --> 00:28:22,210
two here as compared to in the previous slide,
because I have in my system 0s and 1s. So,
188
00:28:22,210 --> 00:28:32,330
I have n number of 0s n number of 1s and I
assume that all these 0s and 1s have equal
189
00:28:32,330 --> 00:28:41,970
probability of occurring occurrence. So, the
average power per bit would be the power here
190
00:28:41,970 --> 00:28:48,970
divided by 2, 50 percent of that. So, this
would be the minimum average received optical
191
00:28:48,970 --> 00:28:50,260
power.
192
00:28:50,260 --> 00:28:56,500
Now, if alpha is the loss coefficient dB per
kilometer, then alpha is equal to ten divided
193
00:28:56,500 --> 00:29:19,370
by L log Pi over Pout and. So, the Lmax is
equal to 10 divided by alpha log Pi divided
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00:29:19,370 --> 00:29:25,799
by Pr which is minimum average received power,
and if I put the expression for that from
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previous slide it comes out to be like this.
So, if I consider a 2.5 Gbps link at 1300
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nanometer wavelength with Pi is equal to one
milli watt and P is equal to 1000, minimum
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number of photons which are required and at
1300 nanometer wavelength alpha is 0.4 d B
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per kilometer then Lmax from here will come
out to be 93 kilometers. If I use the same
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link at 1550 nanometer wavelength then loss
goes down from 0.4 d B per kilometer to 0.2d
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B per kilometer and Lmax will become 190 kilometers.
Next is dispersion limited system where attenuation
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is not a major concern and repeater less length
of the link is decided by the dispersion in
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the system.
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And a commonly used criterion for maximum
allowed dispersion is that the broadening
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should be less than or equal to T by 4 where
T is the bit slot or bit duration and since
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B is equal to 1 over T. So, I can write this
as 4 times delta tau times B is less than
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or equal to 1.
Now, if I have a link which has a dispersion
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of dispersion coefficient D which is in pico
seconds per kilometer nanometers. So, delta
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tau would be D times L length of the link
times delta lambda 0 which is the spectral
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00:31:19,559 --> 00:31:28,900
width of the source. So, if I put this one
here then I get this is four times D times
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L times B times delta lambda naught should
be less than or equal to 1, this gives me
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what is the bandwidth length product for a
given dispersion. So, this bandwidth length
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product B times L should be smaller than or
equal to one over 4 D times delta lambda naught,
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D is in picoseconds per kilometer nanometers
delta lambda naught I put in nanometers.
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So, this comes out to be 1 Tbps kilo meter
divided by 4D times delta lambda naught where
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D is in picoseconds per kilometer nanometers
and delta lambda naught is in nanometers or
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00:32:15,890 --> 00:32:22,030
I can write it down because one Tbps over
4 is 250 Gbps. So, it is 250 Gbps kilometers
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divided by D delta lambda naught, let us work
out an example and conclude.
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So, if I have a system with B is equal to
1 picoseconds per kilometer nanometer at 1300
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nanometer wavelength with delta lambda naught
is equal to 1 nanometer which is the typical
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00:32:45,610 --> 00:32:51,030
system at 1300 nanometer wavelength you have
very small dispersion.
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00:32:51,030 --> 00:33:00,750
Then b times till B times L would be less
than or equal to 250 Gbps kilometer here because
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D is 1 delta lambda naught is one which means
that which means that if I want to send pulses
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at the rate 2.5 Gbps, I can have maximum repeater
less length of the link is100 kilometer.
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So, this is all in system design aspects.
In the next lecture I would look into certain
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measurement techniques in optical fibers.
Thank you.