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So, the last class, we were discussing what
happens during the sampling phase and in
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that connection we were looking at the noise
module of a resistor. So, it turns out that
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if
you have a resistor in thermal equilibrium
4
00:00:25,950 --> 00:00:31,740
with surroundings at an absolute temperature
T. Then there is a noise associated with it
5
00:00:31,740 --> 00:00:41,840
this is due to random motion of charge carriers
inside the resistor. And the spectral density
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00:00:41,840 --> 00:00:54,600
of this noise turns out to be 4 k T R on
Whole Square per Hertz. We will not get into
7
00:00:54,600 --> 00:01:01,449
the details of the math behind spectral
density and such. But what I would like to
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00:01:01,449 --> 00:01:07,670
mention here is that it is important get a
physical field for what this means? And what
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00:01:07,670 --> 00:01:17,620
this means is the following I took a resistor
and monitored the voltage across it I would
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00:01:17,620 --> 00:01:25,620
see some random waveform.
Now, if I took that random waveform and filtered
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00:01:25,620 --> 00:01:33,750
it through an ideal band pass filter with
a bandwidth of 1 Hertz and a center frequency
12
00:01:33,750 --> 00:01:41,900
of f naught. There will be some waveform
at the output of this band pass filter is
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00:01:41,900 --> 00:01:44,630
not it, because you had taken a band pass
filter you
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the input is some signal. The output is bound
to be some noise like waveform, please
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00:01:52,140 --> 00:01:57,540
note that a band pass filter has 2 attributes;
one is the center frequency which in this
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00:01:57,540 --> 00:01:58,540
case
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00:01:58,540 --> 00:02:05,170
we have denoted by f naught and the other
one is the bandwidth which is fixed at 1
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00:02:05,170 --> 00:02:12,090
Hertz. Now, it turns out that if you do this
experiment and measure the mean square
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00:02:12,090 --> 00:02:17,590
value of the waveform at the output of the
band pass filter that mean square value will
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00:02:17,590 --> 00:02:26,470
be
4 k T times R on volt square. The input is
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00:02:26,470 --> 00:02:29,410
dimensions of volts the output of the filter
also
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00:02:29,410 --> 00:02:37,430
has dimensions of volts. So, the mean square
value has dimensions of volt square couple
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00:02:37,430 --> 00:02:48,551
of things that one should notice what is.
So, striking about this relationship the mean
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00:02:48,551 --> 00:02:57,290
square voltage is 4 k T R on volt square and
the key point to note is that this is completely
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00:02:57,290 --> 00:03:06,250
independent of. The center frequency of
the band pass filter you understand this means
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00:03:06,250 --> 00:03:14,950
that regardless of what f naught is as long
as the bandwidth of the band pass filter is
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00:03:14,950 --> 00:03:24,790
1 Hertz. The mean square value at the output
of the band pass filter is simply 4 k T times
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00:03:24,790 --> 00:03:33,100
R on is this clear. So, you can think of this
as
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00:03:33,100 --> 00:03:45,990
you know a sinusoid with roughly this power
at I mean you can think of spectral density.
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Therefore, in a very loose and you know non
mathematical way as a sinusoid at that
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00:03:56,450 --> 00:04:11,680
frequency with this with a power given by
4 k T R on. So, since this basically means
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00:04:11,680 --> 00:04:15,430
that
regardless of f naught if the mean square
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00:04:15,430 --> 00:04:21,000
value is 4 k T times R on volt square. It
means
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00:04:21,000 --> 00:04:30,419
that as far as the input signal here is concerned
it means that it must have energy or
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00:04:30,419 --> 00:04:38,430
power at the same concentration regardless
of frequency. I mean finally, what is the
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00:04:38,430 --> 00:04:40,900
band
pass filter doing? It is taking a signal and
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00:04:40,900 --> 00:04:48,300
only selecting those frequencies around f
naught in a narrow bandwidth of 1 Hertz. Now,
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00:04:48,300 --> 00:04:53,319
regardless of what I make f naught if I
am getting the same power at the output. It
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00:04:53,319 --> 00:05:01,710
means that the input process which is driving
the band pass filter has a strength which
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00:05:01,710 --> 00:05:11,910
is independent of frequency correct such a
process is what is called a white process.
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00:05:11,910 --> 00:05:27,089
So, this is white noise and the reason for
calling it white is that it consist of all
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00:05:27,089 --> 00:05:39,289
frequencies. All frequencies just like white
light which consist of all frequencies. Now,
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00:05:39,289 --> 00:06:00,430
let us see what happens when I take the track
and hold it is in the track phase and the
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00:06:00,430 --> 00:06:18,849
input is 0. So, track and hold in the track
phase when the input is 0 v n represents the
45
00:06:18,849 --> 00:06:39,669
what does v n represents noise of the resistor
correct now during the track phase
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00:06:39,669 --> 00:06:56,680
therefore, the output voltage even though
the input is 0 the output is not 0 why? Basically
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00:06:56,680 --> 00:07:04,229
as far as the noise source is concerned it
sees a R c network and the output voltage
48
00:07:04,229 --> 00:07:09,550
is the
voltage across the capacitor. So, as far as
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00:07:09,550 --> 00:07:13,180
the output voltage is concerned it is nothing
but
50
00:07:13,180 --> 00:07:26,699
a low pass filtered version of whatever noise
there is in the resistor, you understand.
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00:07:26,699 --> 00:07:29,300
And
this noise was often also called thermal noise,
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00:07:29,300 --> 00:07:36,089
because of because the origin of this effect
is because of random motion of electrons which
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00:07:36,089 --> 00:07:47,259
depends on temperature.
So, physically speaking this voltage is now
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00:07:47,259 --> 00:07:56,479
no longer 0 even though the input is 0 there
will be some voltage fluctuation there. And
55
00:07:56,479 --> 00:08:01,849
that voltage fluctuation is nothing but the
noise waveform v n filtered by a first order
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00:08:01,849 --> 00:08:11,370
low pass filter found by the R c network.
Before we get into the math of what happens
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00:08:11,370 --> 00:08:16,249
with the output. Physically let us try and
understand what happens when we open the switch.
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00:08:16,249 --> 00:08:29,039
So, what do you think will happen
during the hold phase? So, the switched was
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00:08:29,039 --> 00:08:39,790
closed it is in the track phase and the noise
waveform on the resistor is something. And
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00:08:39,790 --> 00:08:51,060
correspondingly the output is also doing
something given the, if we know this waveform
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00:08:51,060 --> 00:08:58,560
we can compute the waveform at the
output. Now, suddenly what are you doing?
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00:08:58,560 --> 00:09:01,560
Opening the.
You are opening the switch. So, what do you
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think happens at the output?
Disturbance noise by the hold.
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00:09:07,620 --> 00:09:12,800
So, if you suddenly open the switch which
is what will happen when you go from the
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00:09:12,800 --> 00:09:17,940
track phase to the hold phase. Whatever voltage
was there on the capacitor just before
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00:09:17,940 --> 00:09:21,690
you opened the switch will tend to.
Retain that way.
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00:09:21,690 --> 00:09:27,589
Stay that way correct and you think that voltage
will be 0.
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00:09:27,589 --> 00:09:31,720
No.
No because that depends on the noise waveform.
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00:09:31,720 --> 00:09:37,730
So, what is this telling us that even if
there was a no signal, because of the noise
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00:09:37,730 --> 00:09:43,380
of the resistor the output voltage is not
0 it is
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00:09:43,380 --> 00:09:50,589
some random quantity because it is a function
of the noise waveform you understand. So,
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what I want to impress on you is the fact
that even if there was no input the held voltage
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on the capacitor has got a noise component,
does it make sense. And given that the
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00:10:11,339 --> 00:10:17,220
average of the input is 0, can you comment
on the average of the noise component which
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00:10:17,220 --> 00:10:32,519
is stored on the capacitor. The average value
of the resistive noise is 0 correct first
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00:10:32,519 --> 00:10:45,250
of all
why does this make physical sense or does
77
00:10:45,250 --> 00:10:48,130
it sounds strange that the average value of
v n
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00:10:48,130 --> 00:10:52,970
is 0. Or it is not strange at all and this
is. In fact, what you must expect.
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00:10:52,970 --> 00:10:56,520
Sir randomly noise is.
Pardon.
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00:10:56,520 --> 00:11:02,600
Noise is coming due to the motion of the electrons
and volts.
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00:11:02,600 --> 00:11:05,550
All that is fine coming because of you know
motion of electrons. Let us say it moves one
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side if it has to come back otherwise normally.
Is there a more tongue itching argument
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00:11:11,759 --> 00:11:14,480
you can give.
We are not giving any input sir.
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00:11:14,480 --> 00:11:20,959
So, what I mean. So, so as to the battery
you are not giving any input still the output
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00:11:20,959 --> 00:11:26,020
of
the. At T is equal to minus infinity plus
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00:11:26,020 --> 00:11:37,399
infinity it is non 0. No I did not understand.
Since the magnitude is not tending to 0 at
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00:11:37,399 --> 00:11:41,889
any point.
No, my question is the following if I plot
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00:11:41,889 --> 00:11:46,529
if I record this waveform it turns out that
the
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00:11:46,529 --> 00:11:47,529
average is.
0.
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00:11:47,529 --> 00:11:54,459
0; now, the question is does it seem reasonable
that the average is 0 or does it seem
91
00:11:54,459 --> 00:11:58,530
unreasonable? Of course, it is a scientific
fact. So, this we cannot dispute it , but
92
00:11:58,530 --> 00:12:04,589
does it
make sense or does it not make sense it makes
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00:12:04,589 --> 00:12:09,660
sense. Because if it was not then there is
no need for a battery if the output average
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00:12:09,660 --> 00:12:16,730
value is not 0, I will hook up hundred resistors
in series. And the average values across the
95
00:12:16,730 --> 00:12:23,269
resistance will be some finite number from
which I can generate.
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00:12:23,269 --> 00:12:27,970
Power .
Power. So, since; obviously, that is not possible
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00:12:27,970 --> 00:12:33,630
it must mean that the average value must
be 0 notwithstanding whether electrons are
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00:12:33,630 --> 00:12:50,600
moving one way or the other inside you
understand. Now, given that the average value
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00:12:50,600 --> 00:12:56,769
of v n is 0 can you now comment on the
average value of the noise quantity that is
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00:12:56,769 --> 00:13:01,930
sampled on the capacitor. Please note that
v n
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00:13:01,930 --> 00:13:08,000
appears in series with the input correct.
Just like the input is sampled when you open
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00:13:08,000 --> 00:13:10,589
the
switch the post switch does not know what
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00:13:10,589 --> 00:13:16,160
is input and what is noise when you open it
samples. Whatever is there on the capacitor
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is simply the voltage across it at that instant
in this particular example it happens to be
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simply v n, because v n happens to be 0
correct. So, given that the average value
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of v n is 0 and the output voltage is nothing
but
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00:13:32,959 --> 00:13:37,720
a filtered version of v n. Can you comment
on the average value of this noise voltage
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00:13:37,720 --> 00:13:43,899
which is sampled on the capacitor?
It has to be 0.
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00:13:43,899 --> 00:13:56,610
0 is that clear? So, in other words what we
have seen now is that the held voltage has
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00:13:56,610 --> 00:14:04,139
a
noise component
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00:14:04,139 --> 00:14:17,870
whose value is mean is 0 correct. So, if the
mean is 0 I mean the other
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00:14:17,870 --> 00:14:24,129
thing we would like to be in interested in
is what happens to the mean square value.
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00:14:24,129 --> 00:14:26,870
Just
like, we are interested in the mean square
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00:14:26,870 --> 00:14:32,300
value of v n. We are also likely we are also
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00:14:32,300 --> 00:14:42,810
going to be interested in the mean square
value of the noise component which is sampled
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00:14:42,810 --> 00:14:56,029
on the capacitor you understand. So, in other
words the average value of the noise
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00:14:56,029 --> 00:15:03,149
voltage sampled on this capacitor here is
got to be the same as the average value of
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00:15:03,149 --> 00:15:06,209
the
waveform. That exists across the capacitor
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00:15:06,209 --> 00:15:12,160
when the switch is closed correct, because
the
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00:15:12,160 --> 00:15:18,410
switch is closed, you suddenly open the switch
whatever existed at that node before you
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00:15:18,410 --> 00:15:22,810
opened the switch is what is stored in the
capacitor. So, the mean square value of the
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00:15:22,810 --> 00:15:30,880
voltage sampled on the capacitor due to the
noise is simply equal to the mean square
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00:15:30,880 --> 00:15:38,019
value of the voltage waveform. That exist
across the capacitor when the switch is closed.
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00:15:38,019 --> 00:15:46,930
So, now how do you figured out what the mean
square value of this waveform is.
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00:15:46,930 --> 00:15:52,749
Transfer function.
Pardon.
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00:15:52,749 --> 00:16:05,190
4 k T R into.
Transfer function whole square.
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00:16:05,190 --> 00:16:11,350
Transfer function whole square and what is
the intuition behind that.
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00:16:11,350 --> 00:16:23,350
So, what is happening? The answer is indeed.
So, we have a noise source with some
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00:16:23,350 --> 00:16:34,779
spectral density S v of f which happens to
be 4 k T R on hold square per hertz. Now,
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00:16:34,779 --> 00:16:45,370
all I
have done is taken this noise source
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00:16:45,370 --> 00:16:48,990
and passed this through a filter which has
a transfer
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00:16:48,990 --> 00:17:02,980
function of H of f. And I am interested in
finding the mean square value of this
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00:17:02,980 --> 00:17:13,559
waveform clearly this is a filter a linear
filter being driven by a random waveform.
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00:17:13,559 --> 00:17:16,760
So,
the output is also a random waveform which
135
00:17:16,760 --> 00:17:20,410
depends on the input as well as the impulse
response of the filter or the transfer function
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00:17:20,410 --> 00:17:29,930
of the filter. So, if we now go back to the
intuition, we developed with regard to what
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00:17:29,930 --> 00:17:35,900
this S v of f means that will help us get
an
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00:17:35,900 --> 00:17:42,660
understanding of how to compute the mean square
value of the output voltage. So, what
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00:17:42,660 --> 00:17:51,880
is the meaning of S v of f? What did we just
discuss?
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00:17:51,880 --> 00:17:59,890
What S v of f means is that if you take this
noise voltage conceptually pass it through
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00:17:59,890 --> 00:18:03,300
a
band pass filter centered at some frequency
142
00:18:03,300 --> 00:18:08,710
f naught with a bandwidth of 1 Hertz. The
mean square value at the output of this filter
143
00:18:08,710 --> 00:18:21,240
is going to be 4 k T R on. Now, if I took
a
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00:18:21,240 --> 00:18:28,260
sinusoid. So, basically if I look at this
voltage and pass it through a band pass filter
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00:18:28,260 --> 00:18:33,420
this is
going to be the variance the mean square value.
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00:18:33,420 --> 00:18:43,930
Now, if I took this a sinusoid with this
power and pass this through a filter with
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00:18:43,930 --> 00:18:48,940
a frequency response H of f. What is the
meaning of H of f? It means that if I take
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00:18:48,940 --> 00:18:55,350
a sinusoid and pass it at a frequency f, the
output sinusoid will be at an amplitude.
149
00:18:55,350 --> 00:19:00,130
Modulus of H of f.
Modulus of H of f times the input.
150
00:19:00,130 --> 00:19:03,490
Sinusoid.
Sinusoid or the power at the output of the
151
00:19:03,490 --> 00:19:06,920
sinusoid at the output of the filter will
be.
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00:19:06,920 --> 00:19:11,740
Square.
Mod H of f the whole square times the
153
00:19:11,740 --> 00:19:14,240
Input power.
Input power correct and what is the input
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00:19:14,240 --> 00:19:20,900
power in a narrow frequency range around f
naught around some frequency f?
155
00:19:20,900 --> 00:19:23,960
4 k T into
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00:19:23,960 --> 00:19:33,430
In general it will be S v of f is the power
in 1 Hertz around a frequency f correct. Now,
157
00:19:33,430 --> 00:19:38,310
if
I took this and pass this through an amplifier
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00:19:38,310 --> 00:19:41,900
or a transfer function with the gain H of
f
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00:19:41,900 --> 00:19:49,590
the power at the output will therefore, be
S v of f which is the input power multiplied
160
00:19:49,590 --> 00:19:51,510
by.
H of f .
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00:19:51,510 --> 00:20:04,410
Mod H of f the whole square correct. So, if
you look at the output power spectral density.
162
00:20:04,410 --> 00:20:11,050
And this is simply the power of the output
waveform in a 1 hertz bandwidth around.
163
00:20:11,050 --> 00:20:21,890
Center frequency.
A frequency f. Now, clearly the input the
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00:20:21,890 --> 00:20:27,680
S v of f that is the noise process is white
which
165
00:20:27,680 --> 00:20:31,200
means that S v of f is independent of frequency.
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00:20:31,200 --> 00:20:41,080
And that turned out to be 4 k T R on volt
square per hertz and what is mod H of f whole
167
00:20:41,080 --> 00:20:50,031
square for s?
1 by 1 plus omega square.
168
00:20:50,031 --> 00:20:55,200
Omega square R square c square and please
note this is H of f. So, what should it be
169
00:20:55,200 --> 00:21:07,050
4 pi
square F square R square R on square c square.
170
00:21:07,050 --> 00:21:21,190
So, in other words S v out of f is this and
let us not lose sight of the units volts square
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00:21:21,190 --> 00:21:32,920
per hertz, does it make sense to you people.
The input noise source had a uniform power
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00:21:32,920 --> 00:21:38,150
as a function of frequency you pass that
through a filter whose gain is varying as
173
00:21:38,150 --> 00:21:41,061
a function of frequency. So, it follows that
the
174
00:21:41,061 --> 00:21:51,110
output power is the input power multiplied
by square of the gain at that frequency. So,
175
00:21:51,110 --> 00:21:59,000
the output spectral density will be the input
spectral density multiplied by mod H of f
176
00:21:59,000 --> 00:22:07,070
the
whole square correct. So, now what do you
177
00:22:07,070 --> 00:22:22,940
think would be the mean square value of the
output waveform. See what this means is that
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00:22:22,940 --> 00:22:29,340
if I took the, what does it mean? Let us
recollect again if I took the output waveform
179
00:22:29,340 --> 00:22:34,980
pass this through a band pass filter at a
frequency f and a bandwidth of 1 Hertz. The
180
00:22:34,980 --> 00:22:42,950
mean square value I get will be this quantity
let us plot it to get some more understanding.
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00:22:42,950 --> 00:22:47,970
And see how will this look like at DC what
will it be?
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00:22:47,970 --> 00:23:02,560
4 k T R on.
4 k T R on at will decrease at 1 by 2 pi R
183
00:23:02,560 --> 00:23:07,700
on times c what will happen to the power?
It become half.
184
00:23:07,700 --> 00:23:23,960
It become half I will do this. So, why does
this shape make intuitive sense? How is it
185
00:23:23,960 --> 00:23:27,670
that
even though the input has a frequency content
186
00:23:27,670 --> 00:23:32,890
which is constant with frequency? Why is
it that the output frequency content seems
187
00:23:32,890 --> 00:23:36,000
to be decreasing?
Sir.
188
00:23:36,000 --> 00:23:41,840
Pardon.
Filter in there.
189
00:23:41,840 --> 00:23:46,750
There is a filter which is selectively letting
only low frequency components get through.
190
00:23:46,750 --> 00:23:51,920
So, it makes sense that only the low frequency
components of the noise are passed
191
00:23:51,920 --> 00:24:03,621
through with a gain of 1. As frequency increases
the gain is reducing correct now we can
192
00:24:03,621 --> 00:24:15,350
think of. So, you can think of this as many
sinusoids closely spaced together each at
193
00:24:15,350 --> 00:24:19,380
1
Hertz spacing and each sinusoid has got a
194
00:24:19,380 --> 00:24:28,570
power which is 4 k t R on. R on times 1 by
1
195
00:24:28,570 --> 00:24:37,770
plus 4 pi square f square does it make sense.
So, you can think of this as a whole bunch
196
00:24:37,770 --> 00:24:47,170
of sine waves all separated by 1 hertz where
each sine waves mean square value or
197
00:24:47,170 --> 00:24:55,800
power for a sine wave means square value is
same as the I mean as the power. So, the
198
00:24:55,800 --> 00:25:00,840
mean square value of each sinusoid is this
quantity here.
199
00:25:00,840 --> 00:25:07,470
So, now, when you add all these sinusoids
all of these are at different frequencies
200
00:25:07,470 --> 00:25:12,080
mind,
you when you add many sinusoids have different
201
00:25:12,080 --> 00:25:15,060
frequencies. So, what can you conclude
202
00:25:15,060 --> 00:25:19,690
about the power of the composite waveform
or the mean square value of the composite
203
00:25:19,690 --> 00:25:32,550
waveform. You understand question I am asking
you have many sinusoids all at different
204
00:25:32,550 --> 00:25:40,091
frequencies. Each one of these sinusoids has
some power when you add when you when
205
00:25:40,091 --> 00:25:44,200
you construct a waveform which is the sum
of all these sinusoids. Can you comment on
206
00:25:44,200 --> 00:25:54,330
the power of this composite waveform?
They will add up.
207
00:25:54,330 --> 00:26:05,380
They will simply add up why what happens if
all the sinusoids were of the same
208
00:26:05,380 --> 00:26:31,021
frequency can you add up powers? Sinusoids
are orthogonal to each square root of
209
00:26:31,021 --> 00:26:33,819
square no simply divided, we have to add the
roots and squares. No if they are no if they
210
00:26:33,819 --> 00:26:34,819
are same frequency and same phase then the
amplitudes will add up. So, why is it that
211
00:26:34,819 --> 00:26:44,610
now you say that I can simply add the powers.
Otherwise if it is not and simply a Fourier
212
00:26:44,610 --> 00:26:49,440
series frequency multiply in the orthogonal.
I mean see you have a waveform v 1 v 2 v 3
213
00:26:49,440 --> 00:26:56,910
and so on . So, all I am saying is mean square
I mean v 1 square plus average value of v
214
00:26:56,910 --> 00:27:04,250
1 square plus v 2 square plus v 3 square is
not the same. As in general is not the same
215
00:27:04,250 --> 00:27:08,970
as
average value of v 1 plus v 2 plus v 3 the
216
00:27:08,970 --> 00:27:16,400
whole square when will it be the same.
When all the cross terms become 0
217
00:27:16,400 --> 00:27:22,390
They are 0 and if you have sinusoids when
will all the cross terms becomes 0. If its
218
00:27:22,390 --> 00:27:25,820
multiplied when there is. When there 2 of
when the 2 waveforms are of different
219
00:27:25,820 --> 00:27:30,390
frequencies. The cross terms all the average
value of all the cross terms becomes 0 you
220
00:27:30,390 --> 00:27:36,560
understand which is why you can add the powers
in general. If we just add many
221
00:27:36,560 --> 00:27:42,280
waveforms the power of the combined waveform
is simply not the same as the sum of
222
00:27:42,280 --> 00:27:54,020
the individual powers is this clear. No let
me just what I was trying to say.
223
00:27:54,020 --> 00:28:05,200
So, let us say I had a waveform a cos omega
1 T another waveform a 2 cos omega 2 T
224
00:28:05,200 --> 00:28:12,820
the mean square value of this chap is a 1
square by 2 the mean square value of this
225
00:28:12,820 --> 00:28:17,520
guy is
a 2 square by 2. Now, let us say I add these
226
00:28:17,520 --> 00:28:23,680
2 together then I will get a 1 cos omega 1
T
227
00:28:23,680 --> 00:28:36,100
plus a 2 cos omega 2 T, what is the mean square
value of this character A 1 square by 2
228
00:28:36,100 --> 00:28:46,070
plus a 2 square by 2? Only if Omega 1 is not
equal, omega 1 is not the same as omega 2
229
00:28:46,070 --> 00:28:54,850
if omega 1 equals omega 2, what is the mean
square value? It will be 2 a square 2 a 1
230
00:28:54,850 --> 00:28:56,310
I
mean or a 1 plus a 2.
231
00:28:56,310 --> 00:29:03,260
Whole square by 2.
Whole square by 2 you understand you are clear.
232
00:29:03,260 --> 00:29:09,380
So, even though a lot of you said
simply you can add up the powers, that is
233
00:29:09,380 --> 00:29:13,060
only possible because all these different
tones
234
00:29:13,060 --> 00:29:21,760
are at at different frequencies. Now, therefore,
the mean square value of the waveform at
235
00:29:21,760 --> 00:29:33,530
the output of the filter can simply be got
in by adding the powers in individual each
236
00:29:33,530 --> 00:29:35,280
infinite signal bandwidth.
237
00:29:35,280 --> 00:29:49,820
So, v out square average is simply integral
0 to infinity of the output power spectral
238
00:29:49,820 --> 00:30:02,200
density integrated over all frequencies. Of
course, you must bear in mind that none of
239
00:30:02,200 --> 00:30:08,890
this is I mean none of what I have told you
is really rigorous from a random processes
240
00:30:08,890 --> 00:30:15,460
slash communication point of view. But in
this course; obviously, it is not possible
241
00:30:15,460 --> 00:30:19,720
for
us to get into mathematical details of spectral
242
00:30:19,720 --> 00:30:21,490
density and so on. All the time interested
in
243
00:30:21,490 --> 00:30:30,900
giving you is some feel for why those formulae
make sense you understand this. Now,
244
00:30:30,900 --> 00:30:38,930
why the mean square value is simply the integral
of the power spectral density? Now,
245
00:30:38,930 --> 00:30:48,250
what is the power spectral density? It is
nothing but 4 k T R on which is the input
246
00:30:48,250 --> 00:30:55,181
power
spectral density times 1 by 1 plus 4 pi square
247
00:30:55,181 --> 00:31:22,340
f square R on square c square d f 0 to
infinity and we use a change of variable.
248
00:31:22,340 --> 00:31:32,310
So, choose for example, 4 pi sorry if I choose
2
249
00:31:32,310 --> 00:31:51,070
pi f R on c as u. Then this integral simply
becomes the mean square value of the output
250
00:31:51,070 --> 00:32:05,840
is
4 k T R on times d f becomes d u by 2 pi R
251
00:32:05,840 --> 00:32:22,240
on times c times 1 plus u square. And this
integral transforms 0 to infinity is this
252
00:32:22,240 --> 00:32:23,660
clear.
253
00:32:23,660 --> 00:32:44,810
And magically we see that this is 4 k T by
2 pi into C, because the R on goes away. And
254
00:32:44,810 --> 00:32:58,890
this is integral 0 to infinity of d u by 1
plus u square which is 4 k T by 2 pi c tan
255
00:32:58,890 --> 00:33:08,090
inverse u
evaluated from 0 to infinity. Now, what is
256
00:33:08,090 --> 00:33:16,240
this pi by 2? So, the grand result is that
v out
257
00:33:16,240 --> 00:33:26,620
square is k T by.
258
00:33:26,620 --> 00:33:40,100
So, in English what this results means is
that if I took a track and hold with 0 input
259
00:33:40,100 --> 00:33:54,100
and a
capacitor the switch can have an arbitrary
260
00:33:54,100 --> 00:34:07,900
resistance. And because of the noise associated
with the resistor of the switch when I open
261
00:34:07,900 --> 00:34:12,040
the switch when I transition from the track
262
00:34:12,040 --> 00:34:22,970
phase to the hold phase the voltage stored
on the capacitor is not 0 is not 0 which is
263
00:34:22,970 --> 00:34:25,630
what
1 would expect for a noiseless situation.
264
00:34:25,630 --> 00:34:30,149
There is some random voltage on the switch
that
265
00:34:30,149 --> 00:34:36,940
random voltage is because of noise. And the
mean square value of that random voltage
266
00:34:36,940 --> 00:34:49,520
which is stored on the switch on the capacitor
is k T by c or else spend a couple of
267
00:34:49,520 --> 00:34:56,950
minutes staring at this. And see why this
might make sense one thing is very intriguing
268
00:34:56,950 --> 00:35:02,950
is
that the mean square noise sampled on the
269
00:35:02,950 --> 00:35:06,380
capacitor is independent of.
R on
270
00:35:06,380 --> 00:35:11,420
R on, but what does the mechanism that is
producing noise.
271
00:35:11,420 --> 00:35:15,940
R.
It is R. So, at first sight it seems somewhat
272
00:35:15,940 --> 00:35:28,700
strange that the resistor is the 1 which is
producing noise. But magically the mean square
273
00:35:28,700 --> 00:35:33,650
value sampled on the capacitor is
independent of the resistor it only depends
274
00:35:33,650 --> 00:35:51,480
on the capacitance. So, why do you think this
makes sense? Because that noise source is
275
00:35:51,480 --> 00:36:06,470
dependent on temperature. But clear the final
expression for noise has temperature in it.
276
00:36:06,470 --> 00:36:12,600
So, that is.
Sir this we have taken when the switch is
277
00:36:12,600 --> 00:36:20,130
open. So, any noise not coming
Good argument, but no. So, please note that
278
00:36:20,130 --> 00:36:27,310
the value of we are sampling which is held
on the capacitor has something to do with
279
00:36:27,310 --> 00:36:34,120
the resistor is not it. So, why do you think
this
280
00:36:34,120 --> 00:36:45,500
makes sense?
R is constant with frequency.
281
00:36:45,500 --> 00:36:55,800
So, what? So, in other words the mean square
value is simply the integral of the noise
282
00:36:55,800 --> 00:37:02,620
spectral density across all frequencies which
is which means that this is simply the area
283
00:37:02,620 --> 00:37:07,580
under this curve. So, if I reduce resistance
what is happening?
284
00:37:07,580 --> 00:37:10,470
Bandwidth.
Two things are happening one.
285
00:37:10,470 --> 00:37:22,530
Bandwidth peak.
The peak spectral density is reducing; however,
286
00:37:22,530 --> 00:37:28,900
what other effect is the bandwidth is.
Increasing.
287
00:37:28,900 --> 00:37:37,710
Increasing; so, the bandwidth is now become
twice as it was before. So, if I reduce R
288
00:37:37,710 --> 00:37:52,450
by
a factor of 2 this is what happens. And on
289
00:37:52,450 --> 00:37:57,990
the other hand if I increased R, what would
happen? The bandwidth will come down the spectral
290
00:37:57,990 --> 00:38:04,220
density at DC will increase , but in
all 3 cases the area under the curve remains
291
00:38:04,220 --> 00:38:08,680
the same. And why it makes sense is that if
I
292
00:38:08,680 --> 00:38:16,610
increase resistance it is true that the noise
spectral density of the resistor is is higher
293
00:38:16,610 --> 00:38:17,820
or
lower.
294
00:38:17,820 --> 00:38:18,820
Higher.
295
00:38:18,820 --> 00:38:24,930
It is higher, because the noise spectral density
of the of the corresponding to the resistor
296
00:38:24,930 --> 00:38:27,240
is 4 k t.
R.
297
00:38:27,240 --> 00:38:32,920
R is directly dependent on the resistance.
However, the voltage across the capacitor
298
00:38:32,920 --> 00:38:37,770
is
not simply the noise of the resistor it is
299
00:38:37,770 --> 00:38:42,761
that which is being filtered by an R c filter.
So, if
300
00:38:42,761 --> 00:38:48,900
you increase the resistance while it is true
that the spectral density of the noise source
301
00:38:48,900 --> 00:38:51,290
of
the resistor increases simultaneously the
302
00:38:51,290 --> 00:38:53,150
bandwidth is..
Reducing.
303
00:38:53,150 --> 00:39:02,860
Reducing the R c filterâ€™s bandwidth is proportional
to 1 by R c. So, if the R increases it is
304
00:39:02,860 --> 00:39:11,540
accompanied by reduction in the bandwidth
of the track and hold. And these 2 are effects
305
00:39:11,540 --> 00:39:20,200
which are working in the opposite direction.
And it seems I mean it does not seem
306
00:39:20,200 --> 00:39:31,750
surprising that they cancel out or rather
you know while being it being an absolutely
307
00:39:31,750 --> 00:39:42,520
constant is perhaps surprising. You can at
least see that these are 2 opposing effects
308
00:39:42,520 --> 00:39:47,720
and
luckily they cancel out making the mean square
309
00:39:47,720 --> 00:40:00,100
value of the noise voltage held on the
capacitor independent of the resistance of
310
00:40:00,100 --> 00:40:03,560
the switch it only depends on the capacitor.
311
00:40:03,560 --> 00:40:10,040
Now, what implications does this have on the
track and hold design? What this means is
312
00:40:10,040 --> 00:40:22,810
couple of things? So, now, if there was an
input voltage in addition to the noise of
313
00:40:22,810 --> 00:40:29,240
the
switch which I? So, let us denote model the
314
00:40:29,240 --> 00:40:33,370
switch as the noise source, it is on resistance
315
00:40:33,370 --> 00:40:46,250
and an ideal switch. So, this is a real switch
and I have here the sampling capacitor. So,
316
00:40:46,250 --> 00:40:59,380
the voltage sampled on the capacitor is v
n of k T plus. What is the voltage sample
317
00:40:59,380 --> 00:41:00,380
of the
capacitor?
318
00:41:00,380 --> 00:41:19,030
Is small very small
Plus some v n out of whatever k T where the
319
00:41:19,030 --> 00:41:39,990
mean square value of v n out is independent
of the resistance. So, in other words the
320
00:41:39,990 --> 00:41:46,030
very act of sampling has already, what is
this?
321
00:41:46,030 --> 00:41:53,370
What I mean? Do you think this a good thing
or a bad thing ideally what you want to
322
00:41:53,370 --> 00:42:00,590
store on the capacitor V in of k t. You want
v in of k T I mean please do not mistake this
323
00:42:00,590 --> 00:42:11,490
k T for Boltzmann constant and temperature
this case the index in time. And T is the
324
00:42:11,490 --> 00:42:16,740
sampling rate I mean your sampling period.
So, perhaps oh I think this is a bad idea
325
00:42:16,740 --> 00:42:24,530
to I
will call this m times yeah I mean n times
326
00:42:24,530 --> 00:42:26,570
T sub S T sub S.
Sir.
327
00:42:26,570 --> 00:42:30,690
Yes.
This R on cancelation it is happening, because
328
00:42:30,690 --> 00:42:36,369
it is a first order system R on s.
Yes.
329
00:42:36,369 --> 00:42:42,690
Suppose it is becoming second order 10 R v
1.
330
00:42:42,690 --> 00:42:51,520
Well. So, it turns out that it does even if
it is I mean what you are saying is if you
331
00:42:51,520 --> 00:42:57,690
had a
higher order system you know does this cancel
332
00:42:57,690 --> 00:43:03,000
out it turns out that it does you can prove
it you know. In fact, it is not very difficult
333
00:43:03,000 --> 00:43:07,280
to prove, but it can be from what I have
discussed today you should be able to prove
334
00:43:07,280 --> 00:43:10,869
that that is indeed the case. So, if I take
a
335
00:43:10,869 --> 00:43:18,740
general R c network and find the mean square
value across the output of some capacitor.
336
00:43:18,740 --> 00:43:24,880
If I increase all the resistors by the same
factor it will turn out that if you measure
337
00:43:24,880 --> 00:43:29,430
the
mean square noise everything will the increase
338
00:43:29,430 --> 00:43:34,840
in the spectral density will be
compensated by the decrease in the bandwidth
339
00:43:34,840 --> 00:43:41,980
and you will get the same mean square
value.
340
00:43:41,980 --> 00:43:46,960
So, what I wish to point out therefore, is
that the very act of sampling the input on
341
00:43:46,960 --> 00:43:50,060
the
switch I mean using the switch. And the capacitor
342
00:43:50,060 --> 00:43:59,610
has already corrupted the input you
understand and the degree to which it is corrupted
343
00:43:59,610 --> 00:44:05,740
is dependent on is only dependent on
344
00:44:05,740 --> 00:44:18,110
k T of course, and c. So, if you have to reduce
the noise due to the sampling action of the
345
00:44:18,110 --> 00:44:26,330
switch and the capacitor. There are I mean
there is only 1 choice in practice I mean
346
00:44:26,330 --> 00:44:30,410
you
are often not do not have freedom to change
347
00:44:30,410 --> 00:44:35,920
the temperature at which the system
operates and due to cannot change Boltzmann
348
00:44:35,920 --> 00:44:42,360
constant. So, the only thing left is to
increase c. And this makes intuitive sense
349
00:44:42,360 --> 00:44:47,180
as we just discussed before we went into this
discussion. We said that if the sampling capacitance
350
00:44:47,180 --> 00:44:55,910
is small you are now signing up for
getting easily disturbed by noise external
351
00:44:55,910 --> 00:45:00,810
disturbance.
You know extraneous disturbances and therefore,
352
00:45:00,810 --> 00:45:06,710
this does not seem very surprising at
And what this is telling you is that if you
353
00:45:06,710 --> 00:45:14,220
make the capacitance bigger the noise
associated with sampling will be will be small.
354
00:45:14,220 --> 00:45:23,250
So, this makes intuitive sense. So, now,
can we answer the question? There are any
355
00:45:23,250 --> 00:45:30,160
number of choices of R and c which will give
us the same tracking bandwidth correct. So,
356
00:45:30,160 --> 00:45:35,630
is there. So, now, can we revisit that
question we had earlier we said that they
357
00:45:35,630 --> 00:45:37,570
may be we can choose a very bad switch with
a
358
00:45:37,570 --> 00:45:46,290
large lot of resistance and a very small capacitance.
So, what comment do you have now
359
00:45:46,290 --> 00:45:53,190
given that we have had this discussion?
Some reasonable value is.
360
00:45:53,190 --> 00:45:58,690
So, if you get too greedy and choose a use
a very bad switch and a very small capacitor
361
00:45:58,690 --> 00:46:05,600
claiming that any way the R c times constant
is what matters for a given tracking
362
00:46:05,600 --> 00:46:09,610
bandwidth. What you will end up with is that?
Noise component.
363
00:46:09,610 --> 00:46:15,980
The noise component which is held on the capacitor
after opening the switch will be
364
00:46:15,980 --> 00:46:32,330
large since you are now added the since you
are now added noise to the input signal. It
365
00:46:32,330 --> 00:46:38,150
now makes sense to talk of the degree to which
you corrupted the signal by adding this
366
00:46:38,150 --> 00:46:44,640
noise and a number. I mean what kind of measure
do you is commonly used? I mean
367
00:46:44,640 --> 00:46:49,960
how, what do you think you can characterize
signal to noise ratio. So, you can use a
368
00:46:49,960 --> 00:46:54,710
signal to noise ratio as a number to characterize
the degree to which the sampling
369
00:46:54,710 --> 00:47:11,360
operation has corrupted the input. Before
I get there 1 minor point that I want to add
370
00:47:11,360 --> 00:47:16,230
is
that let us say you are sampling; you close
371
00:47:16,230 --> 00:47:22,100
the switch; you opened it once; you open the
switch. There is some noise voltage on the
372
00:47:22,100 --> 00:47:27,970
capacitor in addition to the signal. The next
time again you close the switch to get into
373
00:47:27,970 --> 00:47:39,280
the track mode and you open the switch again.
Now you will have another value for the noise,
374
00:47:39,280 --> 00:47:40,990
can you comment on the correlation
375
00:47:40,990 --> 00:47:48,330
between the noise you had earlier in the previous
cycle with the noise you had in this
376
00:47:48,330 --> 00:47:49,780
cycle.
Uncorrelated.
377
00:47:49,780 --> 00:47:56,040
They are uncorrelated why?
By random 1 twice it is.
378
00:47:56,040 --> 00:48:02,180
So, what in auto correlation function you
will have an impulse set. Yes I mean I agree
379
00:48:02,180 --> 00:48:14,760
that this noise source successive samples
of v n are uncorrelated that is absolutely
380
00:48:14,760 --> 00:48:21,070
correct. What I am asking for is not that
what I am saying is let me draw the diagram
381
00:48:21,070 --> 00:48:26,560
when there is no input v n R on.
382
00:48:26,560 --> 00:48:41,470
So, at time instant n equal to 0 this was
the state of the network and I open the switch.
383
00:48:41,470 --> 00:49:11,530
So, some voltage v n out of 0 is held on the
capacitor then what do I do for n equal to
384
00:49:11,530 --> 00:49:20,410
1
again I close the switch and I open it again.
385
00:49:20,410 --> 00:49:34,080
And I get v n out of 1 can I comment on the
correlation between this voltage and this
386
00:49:34,080 --> 00:49:45,150
voltage. In other words can we say anything
specific about v n out? In the next sample
387
00:49:45,150 --> 00:49:49,380
given that we know what it was in this
sample? Yes, no.
388
00:49:49,380 --> 00:49:50,450
No.
No why?
389
00:49:50,450 --> 00:50:30,740
Capacitor voltage is it is like sampling the
output of an R c filter whose input is a thermal
390
00:50:30,740 --> 00:50:31,740
noise.
Correct.
391
00:50:31,740 --> 00:50:35,630
So, if you are sampling then there is if the
correlation has to be there between T and
392
00:50:35,630 --> 00:50:36,630
T
plus t.
393
00:50:36,630 --> 00:50:38,350
Correct.
That is that would be the successive samples
394
00:50:38,350 --> 00:50:39,580
here.
Correct.
395
00:50:39,580 --> 00:50:44,930
Because the thermal noise is quite and its
auto correlation function is a delta function.
396
00:50:44,930 --> 00:50:48,760
Yes.
X of T into x of T plus T that correlation
397
00:50:48,760 --> 00:50:49,990
will be 0 that is for the input.
Input.
398
00:50:49,990 --> 00:50:51,850
Correct.
No, but actually the time separated between
399
00:50:51,850 --> 00:51:08,420
T and T equal to 0 and T equal to T they are
the successive samples here. So, not quite
400
00:51:08,420 --> 00:51:14,380
please note. So, n equal to 0 we open the
switch a voltage the noise trapped on this
401
00:51:14,380 --> 00:51:23,280
capacitor is v n out of 0. Then what is
happening? We close the switch then again
402
00:51:23,280 --> 00:51:32,490
what happens the noise I mean the input
noise waveform is connected to the capacitor
403
00:51:32,490 --> 00:51:38,050
what happens on the. So, what is? So, v n
out of 0 now behaves like the initial charge
404
00:51:38,050 --> 00:51:47,440
on the capacitor correct. And so, it is
basically like having an initial charge on
405
00:51:47,440 --> 00:51:51,551
the capacitor and it is being driven by a
noise
406
00:51:51,551 --> 00:52:02,160
source And as he pointed out between T and
2 T or between 0 and T the waveform v n of
407
00:52:02,160 --> 00:52:06,650
T is is completely.
Uncorrelated.
408
00:52:06,650 --> 00:52:09,100
Uncorrelated and independent of what happened
in the previous clock? Previous clock
409
00:52:09,100 --> 00:52:16,520
cycle correct, but what happens to v n out
of 0 which was stuck on the capacitor when
410
00:52:16,520 --> 00:52:22,500
you began it might discharge at point. It
might or will might.
411
00:52:22,500 --> 00:52:26,160
Sir depends upon R on.
It depends on R on correct.
412
00:52:26,160 --> 00:52:30,330
It might discharge.
Might.
413
00:52:30,330 --> 00:52:36,450
It can charge or discharge they charge or
discharge. See you can think of this R c circuit
414
00:52:36,450 --> 00:52:41,770
as having 2 components there was if you have
an R c network. And there was some
415
00:52:41,770 --> 00:52:50,090
voltage v 1 on this capacitor and there was
some input. You can deal with the initial
416
00:52:50,090 --> 00:52:57,110
conditions and the input as using superposition
correct this is you are all in agreement
417
00:52:57,110 --> 00:53:01,920
with. So, let us see since you are only interested
in what happens to be 1. I am going to
418
00:53:01,920 --> 00:53:06,210
say I am going to not worry about the input
for the time being. So, if you have an R c
419
00:53:06,210 --> 00:53:11,830
network like this what you think will happen
to v 1.
420
00:53:11,830 --> 00:53:14,110
Discharge.
It will discharge.
421
00:53:14,110 --> 00:53:19,500
Definitely discharge it is not it might discharge,
it will discharge and what is the time
422
00:53:19,500 --> 00:53:22,030
constant associated of the discharge?
R c.
423
00:53:22,030 --> 00:53:27,590
R c correct; now can you comment on R c versus
a clock period. R c is very very less
424
00:53:27,590 --> 00:53:34,950
than T R c is very very less than t. T v 1
it should be given. Pardon.
425
00:53:34,950 --> 00:53:41,890
Clock period should be much much higher.
Very high or so, the clock period must be
426
00:53:41,890 --> 00:53:44,750
much much large than.
R c R c time constant.
427
00:53:44,750 --> 00:53:48,170
The R c time constant and why is that chosen
that way?
428
00:53:48,170 --> 00:53:54,050
It discharge.
What happens if a, if I mean why should the
429
00:53:54,050 --> 00:53:55,690
clock period be much larger than
430
00:53:55,690 --> 00:53:58,381
Discharge otherwise it would not discharge
.
431
00:53:58,381 --> 00:54:02,290
Please note that we want we are interested
in having a high tracking bandwidth. The
432
00:54:02,290 --> 00:54:08,480
tracking bandwidth I mean if you if you are
interested in sampling at a signal at a rate
433
00:54:08,480 --> 00:54:14,680
f
S. Then the signal bandwidth is the maximum
434
00:54:14,680 --> 00:54:20,359
signal bandwidth you would be interested
in is f S by 2 correct. So, in other words
435
00:54:20,359 --> 00:54:23,680
if you want to have to be able to track a
signal
436
00:54:23,680 --> 00:54:29,609
with the frequency as large as f S by 2. It
must follow that the tracking bandwidth must
437
00:54:29,609 --> 00:54:37,540
be much higher than f S by 2 which is equivalent
to saying that the R c time constant
438
00:54:37,540 --> 00:54:41,460
must be much smaller than.
Clock period.
439
00:54:41,460 --> 00:54:46,530
The clock period; because the clock period
is related to the maximum signal period and
440
00:54:46,530 --> 00:54:54,960
if you want to track a high frequency corresponding
to f S by 2 the R c time constant
441
00:54:54,960 --> 00:55:04,560
must be much smaller than. I mean give or
take a constant must be much smaller than
442
00:55:04,560 --> 00:55:07,760
the
clock period. Now, if the R c time constant
443
00:55:07,760 --> 00:55:12,750
is very very small compared to the clock
period, what do you think will happen to v
444
00:55:12,750 --> 00:55:24,250
1 between? It will weakly discharge weakly
discharge. Weakly discharge which means that
445
00:55:24,250 --> 00:55:31,430
at the end of the next cycle v 1 is
completely discharge. And the new voltage
446
00:55:31,430 --> 00:55:38,640
held on c 1 on the sampling capacitor is
largely a function of only the input waveform
447
00:55:38,640 --> 00:55:46,990
during that next cycle, which as we all
understand is completely uncorrelated. Because
448
00:55:46,990 --> 00:55:52,910
it is a white noise it is completely
uncorrelated from period to period which is
449
00:55:52,910 --> 00:56:01,980
why there is no correlation between or
virtually there is almost no correlation between
450
00:56:01,980 --> 00:56:10,119
successive samples of the output. Output
is this clear please note that will not be
451
00:56:10,119 --> 00:56:16,210
the case if the R c time constant is large
larger.
452
00:56:16,210 --> 00:56:21,660
Is too large if I deliberately choose the
R c time constant to be very large then what
453
00:56:21,660 --> 00:56:29,280
happens? There are 2 things happening 1 v
1 is decaying and the input noise voltage
454
00:56:29,280 --> 00:56:33,900
is
trying to charge up the capacitor correct.
455
00:56:33,900 --> 00:56:39,119
So, if the time constant is too large then
that
456
00:56:39,119 --> 00:56:44,099
decaying will not happen which means that
at the end of the second clock cycle. The
457
00:56:44,099 --> 00:56:53,420
voltage on the capacitor has got some part
of v 1 plus something else which basically
458
00:56:53,420 --> 00:57:01,240
means that the previous held noise. And this
noise are noise voltage are related which
459
00:57:01,240 --> 00:57:07,359
means they are I mean there is correlation
does it make sense? But in practice, we do
460
00:57:07,359 --> 00:57:09,680
not
have to worry about it, because by design
461
00:57:09,680 --> 00:57:17,070
we would always make R c times the R c time
constant to be much smaller than a.
462
00:57:17,070 --> 00:57:18,070
Clock period.
463
00:57:18,070 --> 00:57:27,980
Clock period; so, in the next class, we will
evaluate signal to noise ratios. And then
464
00:57:27,980 --> 00:57:31,369
get to
implementation of a practical switch which
465
00:57:31,369 --> 00:57:34,410
uses devices which you know very well the
mosfet.
466
00:57:34,410 --> 00:57:35,839
Thank you.