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00:00:09,990 --> 00:00:17,279
So, in the last class, we were reviewing some
basics of sampling, so let us quickly recap
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what we did in the last class.
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So, let us say we had a continuous time signal
x of t and we sample it, at a uniform rate
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every T seconds, then we get a discrete time
signal where time is quantized, but
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amplitude is continuous. And we were trying
to relate the spectrum of the continuous
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time signal to the spectrum of the discrete
time signal. And we said that, if the
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continuous time signal had a spectrum x of
a, and we denote the discrete time Fourier
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transform of the sequence x of K T by x of
e to the j omega, which is sigma overall k
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00:01:12,700 --> 00:01:22,560
x
of K T e to the minus j omega k.
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And to remind ourselves that, this signal
here corresponds to continuous time, I will
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label
this chap with a subscript x c of f, and to
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remind ourselves that this is a discrete time
spectrum, I will call that x d of e to the
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j omega. And yesterday we saw that x d of
e to
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the j omega, can be simply obtained by forming
the sum over all k of x c of f minus k by
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T times 1 over T and then, we replace 2 pi
f times T with omega.
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Refer Slide Time: 02:47)
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And therefore, x d of e to the j omega
is simply sigma overall k x c of
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omega by T times
2 pi minus k by T. And instead of writing
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1 by T all over the place, if I denote 1 by
T by
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the symbol f s stands for sampling frequency,
then x d of e to the j omega can be written
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as 1 over T times sigma over all k x c of
f s by 2 pi times omega minus 2 pi k. And
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one
must remember, that x d is periodic with 2
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pi, and when a continuous time sinusoid of
a
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frequency f 1 is sampled at a rate f s. What
kind of sinusoidal will that result in, in
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the
discrete time domain, let us say you had a
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continuous time sign wave with the frequency
f 1.
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When you sample it at a rate f s, it will
become a discrete time signal, what do you
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think
the frequency of that discrete time signal
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would be.
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It will be f 1 by f s times 2 pi, so in particular
D C, will transform to D C I am just
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putting that down here again D C will transform
to D C f s by 4 will transform to f s
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corresponds to 2 pi. So, f s by will correspond
to 2 pi by 4, which is pi by 2 f s by 2 will
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00:05:53,180 --> 00:06:09,169
correspond to pi and f s corresponds to 2
pi, and a very easy way of drawing the
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spectrum is to do the following. If the continuous
time signal had this spectrum, how do
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you draw the discrete time spectrum, the first
thing would be to make copies of this with
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the period, let us say this bandwidth was
B Hertz, I will make copies of this at f s.
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So, let me just do that here, the next step
is to multiply all of this with 1 over T and
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then,
replace
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Scale
Scale the x axis where B becomes B by F s
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times 2 pi, and f s becomes 2 pi 2 f s
becomes 4 pi and so on, and as we all agreed
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00:07:55,919 --> 00:08:02,259
yesterday, the discrete time spectrum is
periodic with a period 2 pi. So, it does not
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really matter which particular period you
choose, and it is common to choose the range
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from...
Minus pi
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Minus pi to 2 pi, is that clear.
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Now, couple of things I would like to bring
to your attention, the first thing is that
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00:08:31,389 --> 00:08:36,990
if you
had a continuous time signal at a frequency
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f 1, it will transform to as we discussed
just 1
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2 pi times f 1 plus f 1 by f s. Now, what
if you had a continuous time signal which
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was f
1 plus f s, the frequency of the sinusoid
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was f 1 plus f s, if you sample it at f s
what do
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you think it will look like in the discrete
time domain.
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Please notice that it, may say is 2 pi times
f 1 plus f s
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By f s right which is 2 pi into f 1 by f s
plus 2 pi, but we know that the discrete time
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spectrum is periodic, so everything is modulo
2 pi, so you can remove the 2 pi. So, what
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00:09:46,410 --> 00:09:57,130
this is telling you is that, a frequency f
1 and a frequency f 1 plus f s when sampled
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will
look like the same frequency in the discrete
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time domain this is nothing but, aliasing.
And this makes a sense also, because from
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the Nyquist theorem we know that, if the
input frequency is greater than f s by 2,
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then it will result in aliasing.
This is any illustration of that, it just
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says that if you have a frequency f 1 plus
f s, it will
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look after sampling just like f 1. Now, by
the same token f 1 plus k times f s, after
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sampling will also look like 2 pi f 1 by f
s, when this is assuming that f 1 lies between
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0
and f s by 2. So, in other words when you
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sample a continuous time signal, there are
many frequencies that can masquerade like
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the, mean there are many signals which kind
of map down to the same discrete time sinusoid.
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So, if you are not careful before sampling,
then you can be thoroughly mistaken as to
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what the continuous time signal is, please
note that the idea in the whole the idea behind
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sampling is to be able to eventually reconstruct
the input signal in some fashion. If you
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have lost information while sampling or you
made errors during sampling, it is very
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difficult to recover.
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So, which is why as we were discussing yesterday
we said that, while it is true that the
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desired signal may be of only a small bandwidth,
there is always accompanying noise,
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whose bandwidth could be much wider than the,
the bandwidth of the desired signal.
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Now, we should not get into the mistaken notion
that, the desired signal is only got a
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bandwidth B, so I will only sample at 2 B.
So, if we do not do anything, then noise
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components which are much broader band than
to be, as we just saw will all alias to in
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band, there by degrading the in band signal
to noise ratio.
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So, the way around this problem is do not
sample the signal directly, but to put a filter,
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call the anti alias filter which make sure
that the bandwidth of the signal
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is what you
think, it should be. And then the output of
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the anti alias filter is sampled and thanks
to
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the anti alias filter that, the signal bandwidth
does not exceed B, which means that I am
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safe if I sample at 2 B this is what you have
seen in the communication text books.
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00:13:26,889 --> 00:13:33,629
And when we are dealing with mathematical
abstraction, it is often very convenient to
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assume brick-wall type filters and so on.
So, in your communication classes very likely
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that you have seen a block diagram, where
you have an ideal low pass filter, whose
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bandwidth is B, in other words it allows everything
to pass through below B, and cuts off
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everything beyond a bandwidth B. And by now
you should know that this is not possible,
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any practical filter that you build with a
finite order will only have a, only thing
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you can
do is make the transition band narrower and
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narrower.
You cannot make the pass band absolutely flat,
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you cannot make the stop band at
attenuation infinite, which is what one would
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tend to believe; if we saw the ideal block
diagram of a anti alias filter, it would look
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like a brick-wall filter with one say ideal
anti
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alias filter.
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You would expect the frequency response to
look like this, where this is 0 and this is
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1
this is the bandwidth B, a practical filter
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can only approximate this in some sense. So,
you will find that a the pass band is not
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as flat as you would want, the transition
band is
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not as sharp as you would want and the stop
band attenuation is not as large as you
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would want. Now, given this information can
you comment on a other any
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considerations you think for the sampling
rate, in other words we know that if a signal
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has a bandwidth B, we need to at least sample
at 2 B.
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And we also know that we must put an anti
alias filter, now the question I am asking
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00:15:32,550 --> 00:15:38,580
you
is, do you think it makes sense to sample
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at not at 2 B, but at 4 B, 2 B satisfies an
Nyquist criterion, 4 B satisfies the Nyquist
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criterion, 40 B satisfies the Nyquist criterion.
It is natural to wonder whether should I just
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sample at Nyquist, after all Nyquist is telling
me that, if I sample it at least the twice
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the bandwidth, I will be able to reconstruct
no
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problem.
The question is should I sample only at 2
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B is, 2 B good enough or I am doing better,
if I
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sample at 4 B or am I doing even better if
I sample at 8 B, do you have any comments,
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is
there anything we gain, it seems like it is
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more difficult to sample at a higher rate.
So,
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question is do we gain anything at all in
this bargain.
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Very good, so the suggestion is the following,
so let me draw spectrum let us say this is
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0
I will only draw the positive half, this is
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f I will draw the characteristics of the anti
alias
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filter in red here . And let me draw this
on a log scale, so this is log
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magnitude of the anti alias filter response
and because, the filter has got a finite order
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as
you move away from the band edge, this frequency
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below B is the, so called pass band of
the filter.
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The portion beyond B is the stop band of the
filter, because the filter is a practical
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1, it is
a real 1, the order is finite which means
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that you can never have infinite sub band
rejection, over a band of frequencies. You
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can have it at one frequency or two
frequencies, but not over a over a band, and
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if you have an all pole filter and all pole
filter is 1 where, the numerator is 1, H of
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s is of form 1 by D of s. So, if you have
an n-th
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order filter for large frequencies, how does
the how do you think the attenuation will
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go.
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For frequencies far away from the band edge,
it must go down as 1 by omega to the
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power n which is 20 N D B per decay correct.
So, if this is a linear scale on the x axis
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and a log scale on the y axis it will look
like this, if I plot a log log plot it will
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be a
straight line with a slope of minus 20 N D
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B per decay. Now, let us consider two
situations, one the signal frequency is the
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desired signal is here, let us consider two
sampling rates, one where the sampling rate
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is f s, and other one where the sampling rate
is let me call this f s 1.
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And the other one where the sampling rate
is f s 2, so now, can you comment on the
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00:20:09,360 --> 00:20:14,730
consequences of these choices f s 1 versus
f s 2, what do you think becomes more
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00:20:14,730 --> 00:20:21,600
simplified, if you chose f s 2 versus f s
1.
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noise component will the higher frequency
component,
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the aliasing of the i
So, the first thing we need to understand
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is that, if there was no noise at all we would
not
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have to worry about the anti alias filter,
because the sampling rate evidently is much
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higher than twice the bandwidth. The problem
or the reason why we need to have an anti
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00:20:48,950 --> 00:20:55,640
alias filter in the first place is to filter
of noise, and where do we want to specifically
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get
rid of noise, what is the job of the anti
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alias filter.
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Within the bandwidth of the same thing
See you cannot
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Where outside the required band, the question
is it important to get rid of noise all
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outside, I mean completely outside the signal
band or at their specific locations where
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you really want to get rid of noise.
B and f s
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Between
B and f s
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No, it will be a little more specific B and
f s of course, it is good but
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Where do we want to get rid of noise, we want
to get rid of noise at all those frequencies
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which can potentially alias to the in band
frequencies, which is that in band frequency
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0
to B. So, now we need to figure out which
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all frequencies will alias to the range 0
to B,
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let us start with 0, which all frequencies
were alias to 0.
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Obviously 0 will translate the 0, then
f s 1
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Fs1
F s 2
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00:22:10,490 --> 00:22:14,299
Twice f s 1
3 f s 1
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All my integer multiples of f s 1 will all
alias to
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D 0
So, we need to definitely get rid of noise
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at f s 1, now what frequencies will alias
to B
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Fs1
Plus B
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Plus B 2 f s 1 plus B and so on, must remember
also that there is minus B, so what will
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alias to minus B
Studded:
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00:22:41,580 --> 00:22:48,840
Minus B plus f s 1 minus b plus 2 f s 1 and
so on, so now we know what all frequencies
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alias to 0, what all frequencies alias to
B. So, we know I mean now all the range between
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also will alias to something between 0 and
B, so f s 1 plus B, this is f s 1 plus B and
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f s 1
minus B is this range here. So, this is a
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band of frequencies where signal will alias
to
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minus B to B you understand, while it is true
that it will be great, if we had an anti alias
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filter which would just cut off everything
beyond B, we know that in practice it is not
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possible.
So, the next question is where do we really
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want to cut off noise, and we really want
to
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cut off noise at frequencies which can alias
down to the range.
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Minus B to
Minus B to B and that will be in the if we
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chose f s 1 as the sampling rate it, will
be f s 1
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00:24:00,000 --> 00:24:10,409
minus B to f s 1 plus B, now if instead we
had chosen f s 2 as the sampling rate, what
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00:24:10,409 --> 00:24:12,890
do
we see which all frequencies will now alias
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to base band...
F s 2 minus
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F s 2 minus B to f s 2 plus B, now can you
comment on which of these is a better choice
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00:24:27,360 --> 00:24:35,029
of sampling frequency and why?
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F s 2, why is f s 2 is a better choice
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So, if we see that if we chose f s 2, then
the rejection, for a given anti alias filter,
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the
rejection of the anti alias filter
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Is higher
Will be higher for a higher choice of sampling
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frequency, so even though it appears that
it is harder work to do to sample at a higher
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rate, while Nyquist just dictates it twice
of
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the bandwidth is enough. We see that choosing
a sampling rate, which is much higher
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than what was dictated by Nyquist is advantageous,
in terms of the design of the anti
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00:25:26,769 --> 00:25:39,309
alias filter. Now, let us take this argument
the other way, if I insisted that I want to
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sample at Nyquist, in other words my sampling
rate is 2 B, what do you think the
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characteristics of the anti alias filter must
be.
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00:25:51,650 --> 00:25:55,809
It has to be
Sharper
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Lot sharper than what we have now does it
make sense.
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So, choice to sampling frequency relative
the Nyquist frequency, what is the Nyquist
200
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frequency?
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The Nyquist frequency is the minimum sampling
rate you require, to sample a signal of
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bandwidth B, so the Nyquist frequency in this
particular example is 2 B. So, the choice
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of the sampling frequency related to the Nyquist
frequency, has implications on
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the
design of the anti alias filter. So, to compare
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different sampling rates, it makes sense to
only compare it with respect to the Nyquist
206
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rate, so 2 B is the minimum sampling rate
required to be able to reconstruct a signal
207
00:27:37,649 --> 00:27:42,700
with bandwidth B.
If you are sampling at a rate over and above
208
00:27:42,700 --> 00:27:47,429
2 B, it means that your, because sampling
at
209
00:27:47,429 --> 00:27:52,670
Nyquist your sampling, if you are sampling
at a rate higher than Nyquist your
210
00:27:52,670 --> 00:28:11,769
oversampling over. So, the ratio f s by 2
B is called the oversampling ratio or the
211
00:28:11,769 --> 00:28:18,919
abbreviated as the OSR, so we will repeatedly
keep using these things, this abbreviation
212
00:28:18,919 --> 00:28:30,190
later in the course, you understand why this
definition makes sense. So, now to rephrase
213
00:28:30,190 --> 00:28:37,350
all that we have discussed with this new jargon,
if my oversampling ratio is high, what
214
00:28:37,350 --> 00:28:41,420
does it mean is my anti aliasing filter design
easier or more difficult.
215
00:28:41,420 --> 00:28:44,779
Easier
It is easier, and other thing we need to bear
216
00:28:44,779 --> 00:28:49,830
in mind is that, it is only in the text books
that you have an ideal filter, it is only
217
00:28:49,830 --> 00:28:54,269
in the text books where you say you have a
filter
218
00:28:54,269 --> 00:28:59,230
with the bandwidth B, and the bandwidth is
actually B. In practice whenever you say I
219
00:28:59,230 --> 00:29:07,480
am going to have a filter B, you must be prepared
to take variations in the bandwidth.
220
00:29:07,480 --> 00:29:13,990
Because, no real system will have a bandwidth
which is absolutely fixed, it is bandwidth
221
00:29:13,990 --> 00:29:19,399
will vary perhaps, because of temperature
variation, because of manufacturing tolerances
222
00:29:19,399 --> 00:29:27,799
and so on. So, in practice it is never possible
to ensure a bandwidth which is exactly what
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00:29:27,799 --> 00:29:30,039
you want.
224
00:29:30,039 --> 00:29:53,409
So, one high OSR means that the anti alias
filter, need not be very sharp why, because
225
00:29:53,409 --> 00:29:58,760
if
you want to reject the alias band to a certain
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00:29:58,760 --> 00:30:06,169
level, if you increase the oversampling ratio,
which is the ratio of the sampling rate to
227
00:30:06,169 --> 00:30:14,570
twice the signal bandwidth. Then, your anti
alias filter must satisfy, in the pass band
228
00:30:14,570 --> 00:30:16,990
you want your anti aliasing filter to have
a gain
229
00:30:16,990 --> 00:30:26,210
of 1, in the alias band you want it to have
a gain of 0. Of course, we do not really expect
230
00:30:26,210 --> 00:30:32,220
that gain to be 0, we want it to be some small
number, so if we say we want the gain in
231
00:30:32,220 --> 00:30:41,760
the alias band to be at least smaller than
in at most some number like this.
232
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Then the filter design problem becomes a lot
simpler, because the filter needs to have
233
00:30:47,190 --> 00:30:52,940
a
transition band which is doing this. On the
234
00:30:52,940 --> 00:30:59,770
other hand, if my oversampling ratio is small,
then what happens I need to have the same
235
00:30:59,770 --> 00:31:06,460
rejection in the stop band, and I need to
have
236
00:31:06,460 --> 00:31:10,450
the same transmission in the pass band. So,
this means that my transition band from pass
237
00:31:10,450 --> 00:31:19,259
band to stop band must be very sharp, and
designing sharp filters is more difficult
238
00:31:19,259 --> 00:31:23,220
than
designing filters which roll off gently, without
239
00:31:23,220 --> 00:31:29,100
good getting into the theory, I mean while
this seems at least intuitively satisfying.
240
00:31:29,100 --> 00:31:43,059
Not only that can you comment on the effect
of variation of this filter corner, as you
241
00:31:43,059 --> 00:31:52,201
change OSR, in other words do you think I
can tolerate a bigger variation of the band
242
00:31:52,201 --> 00:31:58,320
edge, the motivation being that no practical
filter will have a band edge, which is fixed
243
00:31:58,320 --> 00:32:03,520
in
frequency, it will vary. The question now
244
00:32:03,520 --> 00:32:09,049
is, does it make any difference if I increase
the
245
00:32:09,049 --> 00:32:13,460
OSR, in other words can I tolerate the larger
variation of band edge frequency with the
246
00:32:13,460 --> 00:32:23,009
higher OSR.
Yes
247
00:32:23,009 --> 00:32:41,690
Any do you understand the question no, see
we know that the anti alias filter bandwidth
248
00:32:41,690 --> 00:32:46,649
cannot be fixed, there will be some tolerance
it will move. And you want to make sure in
249
00:32:46,649 --> 00:32:53,230
spite of this the band edge moving, you want
to make sure that it rejects the alias band
250
00:32:53,230 --> 00:33:00,899
to
some degree. Now, the question is will you
251
00:33:00,899 --> 00:33:03,360
be able to tolerate a larger variation of
this
252
00:33:03,360 --> 00:33:11,399
band edge, when the oversampling ratio is
small or when the oversampling ratio is large,
253
00:33:11,399 --> 00:33:16,140
so it is a its very straight forward.
So, if the oversampling ratio is large, not
254
00:33:16,140 --> 00:33:18,679
only is the filter design easier from a point
of
255
00:33:18,679 --> 00:33:28,370
view of the width of the transition band,
it will be lot more tolerant to variations
256
00:33:28,370 --> 00:33:32,650
in the
band edge frequency to see this, imagine what
257
00:33:32,650 --> 00:33:44,039
happens if you have Nyquist sampling,
the filter must be really really sharp. And
258
00:33:44,039 --> 00:33:48,720
now if the bandwidth even moves a little bit,
if
259
00:33:48,720 --> 00:33:53,599
the bandwidth reduces what happens, it will
get rid of the alias for sure, but it will
260
00:33:53,599 --> 00:33:56,470
also
cut off some of the desired signal.
261
00:33:56,470 --> 00:34:05,650
On the other hand, if the bandwidth increases,
then what will you see some of that alias
262
00:34:05,650 --> 00:34:13,429
band is not properly rejected, so while it
is true that Nyquist sampling will is all
263
00:34:13,429 --> 00:34:15,820
that is
necessary to be able to reconstruct the signal
264
00:34:15,820 --> 00:34:22,780
properly, there are some very very practical
reasons, why you would want to actually over
265
00:34:22,780 --> 00:34:31,940
sample the input signal. So, and typically
you would never have a system where the signal
266
00:34:31,940 --> 00:34:37,110
bandwidth is say B Hertz and you
sample exactly at 2 B.
267
00:34:37,110 --> 00:34:43,380
This would make the job of the filter designer
very very difficult of course, you can say
268
00:34:43,380 --> 00:34:48,129
some of my problem is somebody else problem,
but you could I mean, in the next project
269
00:34:48,129 --> 00:34:54,149
you could be that somebody else you understand.
So, it is very common to have little bit
270
00:34:54,149 --> 00:35:03,920
of oversampling, so that the job of the filter
designer is made easy. Of course, pushing
271
00:35:03,920 --> 00:35:11,920
the sampling rate high is also not at all
trivial effect, that basically means that
272
00:35:11,920 --> 00:35:15,390
your
circuits have to work that much faster.
273
00:35:15,390 --> 00:35:21,569
But, system design is trade off between these
possibilities, if you try to make your job
274
00:35:21,569 --> 00:35:27,050
easy somebody else’s job becomes a lot more
difficult to do, if you want to make that
275
00:35:27,050 --> 00:35:34,100
guys job easy then your job becomes very difficult
to do. So, both of you sit together and
276
00:35:34,100 --> 00:35:42,780
figure what works best for both of you.
277
00:35:42,780 --> 00:35:45,860
Meaning
On the transition band what we are talking
278
00:35:45,860 --> 00:35:48,630
about, suppose we are having the
low sampling rate
279
00:35:48,630 --> 00:35:49,870
Correct.
And transition has
280
00:35:49,870 --> 00:35:54,760
Will be very sharp
So, whether any chance of that stability criteria
281
00:35:54,760 --> 00:35:59,930
Well, it is true that if you have the comment
we made was, that if your oversampling
282
00:35:59,930 --> 00:36:05,850
ratio is very small then the filter has to
be extremely sharp. And the comment he made
283
00:36:05,850 --> 00:36:11,359
was that, if the filter has to be extremely
sharp, then that response must only be possible
284
00:36:11,359 --> 00:36:19,569
by poles whose quality factors are extremely
high, only then you can get a sharp roll off.
285
00:36:19,569 --> 00:36:25,069
And yes, that is indeed a challenge, once
you are very high cube poles it turns out
286
00:36:25,069 --> 00:36:27,500
that
the sensitivity of the circuit to component
287
00:36:27,500 --> 00:36:32,380
variations also becomes high.
As you might have seen in perhaps your digital
288
00:36:32,380 --> 00:36:37,200
filter design class, the same thing also
holds for analogue filters. So, whenever you
289
00:36:37,200 --> 00:36:42,240
want to make something very rapidly in the
frequency domain, sensitivity to component
290
00:36:42,240 --> 00:36:47,950
tolerances, noise etcetera at those
frequencies becomes large. And therefore,
291
00:36:47,950 --> 00:36:52,170
you would like to try and avoid very sharp
filters if possible.
292
00:36:52,170 --> 00:36:53,170
Sir
Yes
293
00:36:53,170 --> 00:36:57,520
Bandwidth criteria whatever we are talking
tolerance, how much percentage we
294
00:36:57,520 --> 00:37:04,200
will take in this, suppose we are having the
say for your audio signal 20 kilohertz
295
00:37:04,200 --> 00:37:11,589
suppose we are talking, how much we can go
for the sampling ratio.
296
00:37:11,589 --> 00:37:17,480
The question he asked was, if you have a signal
with the certain bandwidth how will you
297
00:37:17,480 --> 00:37:23,800
choose your oversampling ration, so will you
choose a very small number or will you
298
00:37:23,800 --> 00:37:28,160
chose a very large number, this is very situation
specific. When you are dealing with low
299
00:37:28,160 --> 00:37:33,220
signal bandwidths for example, audio it is
very easy to sample at a higher rate given
300
00:37:33,220 --> 00:37:40,740
today’s technology constraints. For example,
24 kilohertz audio signal. a common
301
00:37:40,740 --> 00:37:44,800
sampling rate to use is say 6 Megahertz. 6.144
Megahertz.
302
00:37:44,800 --> 00:37:55,040
So, if let me just take an example since he
has brought it up, so the signal bandwidth
303
00:37:55,040 --> 00:38:02,359
is
24 kilohertz, the Nyquist bandwidth is what
304
00:38:02,359 --> 00:38:16,640
Nyquist rate is 48 kilohertz and if the
sampling rate is 6.144 Megahertz, what is
305
00:38:16,640 --> 00:38:39,480
the oversampling ratio.
Around 120
306
00:38:39,480 --> 00:38:51,430
It is 128, and this turns out that is a fairly
common thing to do, as you can see the alias
307
00:38:51,430 --> 00:38:59,580
requirements given that the oversampling ratio
is 128, the anti alias filter is to have can
308
00:38:59,580 --> 00:39:06,050
be actually very very gentle. Because, it
needs to pass it needs to have a flat gain
309
00:39:06,050 --> 00:39:10,300
up to 24
kilohertz and it needs to have an attenuation
310
00:39:10,300 --> 00:39:16,800
at 6 point something Megahertz. So, very
often a simple R C or a combination of RC,
311
00:39:16,800 --> 00:39:25,510
a passive RC filters all that is needed and
one might also ask given that, I am doing
312
00:39:25,510 --> 00:39:27,619
all this extra work I am, I need to sample
only
313
00:39:27,619 --> 00:39:30,859
at 48 kilohertz, but now I am sampling at
much higher rate.
314
00:39:30,859 --> 00:39:43,079
Can I exploit this to improve circuit properties,
can I exploit this to a larger degree than
315
00:39:43,079 --> 00:39:48,540
simply saying anti alias filtering becomes
easy, it turns out that this is the subject
316
00:39:48,540 --> 00:39:55,200
or what
is called the oversampling form in the family
317
00:39:55,200 --> 00:39:58,280
of weighted e convertors, called
oversampling analogue digital convertors.
318
00:39:58,280 --> 00:40:05,130
Where we exploit the fact that, there is your
oversampling significantly, oversampling significantly
319
00:40:05,130 --> 00:40:09,150
means that you are sampling at a
much higher rate than is necessary which means,
320
00:40:09,150 --> 00:40:15,750
there is a lot of correlation between
successive samples of the signal.
321
00:40:15,750 --> 00:40:26,720
In other words, if you are watching a movie
it is like watching slow motion there is or
322
00:40:26,720 --> 00:40:31,540
if
you are one of those T v soap box, it is like
323
00:40:31,540 --> 00:40:35,630
[FL] you watch today and you watch
tomorrow and it will looks like the same thing.
324
00:40:35,630 --> 00:40:40,640
There is no difference between
successive samples, which is basically telling
325
00:40:40,640 --> 00:40:44,411
us Nyquist state is very low you come back
2 years, later and watch and you will know
326
00:40:44,411 --> 00:40:53,160
perfectly well what is happening.
So, later on this course, we will see how
327
00:40:53,160 --> 00:41:00,440
oversampling can be exploited, not only to
simplify the bandwidth of, I mean simplify
328
00:41:00,440 --> 00:41:02,319
the requirements of the anti alias filter,
you
329
00:41:02,319 --> 00:41:11,200
can also use it to like to improve the performance
of the A to D converter. So, of course,
330
00:41:11,200 --> 00:41:16,550
now if the signal bandwidth becomes very high,
it may become impractical to be able to
331
00:41:16,550 --> 00:41:24,980
sample it at in an oversampling ratio this
large. In which case you have to settle for
332
00:41:24,980 --> 00:41:27,730
more
modest values of oversampling ratio and then,
333
00:41:27,730 --> 00:41:33,450
simply because you are not able to build
circuits, which can sample this fast.
334
00:41:33,450 --> 00:41:44,349
Now, the next thing I wanted to talk about
is related to what I just said, sometimes
335
00:41:44,349 --> 00:41:49,920
you
want to sample a signal at a very high rate,
336
00:41:49,920 --> 00:41:55,030
that could be A, because you want to
oversample or B, simply because the signal
337
00:41:55,030 --> 00:42:04,460
bandwidth is extremely high. A case in point
being front ends of oscilloscopes, today to
338
00:42:04,460 --> 00:42:08,220
test your high speed circuits you need an
oscilloscope, which is much higher in speed.
339
00:42:08,220 --> 00:42:12,250
Only then, you will be able to test
something which is high speed to begin with,
340
00:42:12,250 --> 00:42:17,450
so oscilloscope for front ends have been a
big application area for requiring, higher
341
00:42:17,450 --> 00:42:27,440
and higher and higher sampling speeds.
Unfortunately device technology is may not
342
00:42:27,440 --> 00:42:31,990
be advancing at the rate you require. let
us
343
00:42:31,990 --> 00:42:37,080
say you want to build a sample and hold which
samples at say 40 Gigahertz, because you
344
00:42:37,080 --> 00:42:43,710
want to test something. Then, it may not be
possible first of all to be able to build
345
00:42:43,710 --> 00:42:48,869
sample
and hold, which can operate at such high speed.
346
00:42:48,869 --> 00:42:52,319
So, one way around that what do you
347
00:42:52,319 --> 00:42:57,130
think you can do, if one fellow cannot do
the job quickly enough, what do you think
348
00:42:57,130 --> 00:43:04,960
you
will do, you put two guys on the job, so it
349
00:43:04,960 --> 00:43:09,710
is the same thing here.
So, you can have many sample and holds which
350
00:43:09,710 --> 00:43:16,540
are working at a lower rate, and put
them together and make it look like a single
351
00:43:16,540 --> 00:43:19,960
sample, and hold working at a higher rate.
352
00:43:19,960 --> 00:43:33,570
So, let us take and this technique is called
time interleaving, where you have many
353
00:43:33,570 --> 00:43:41,380
sample and holds working parallelly at lower
speeds making, a single unit which appears
354
00:43:41,380 --> 00:43:47,809
as if it is working fast. So, now you say
then, I do not know how to design a really
355
00:43:47,809 --> 00:43:49,410
high
speed sample and hold, however I know how
356
00:43:49,410 --> 00:43:51,490
to design a low speed sample and hold, if
I
357
00:43:51,490 --> 00:43:56,280
put many of the sample and holds together
in some fashion. I will be able to hopefully
358
00:43:56,280 --> 00:44:03,270
combine the outputs, in such a way as to make
a sample and hold which looks like a high
359
00:44:03,270 --> 00:44:14,760
speed, it seems like a reasonable idea.
So, this
360
00:44:14,760 --> 00:44:18,770
is the principle of what is called time interleaving
and the basic idea is like this,
361
00:44:18,770 --> 00:44:26,150
let us say we had a signal here something
like this, and you want to sample this at
362
00:44:26,150 --> 00:44:28,210
some
rate. However, you are not able to build a
363
00:44:28,210 --> 00:44:30,920
sample and hold which travels at that high
rate,
364
00:44:30,920 --> 00:44:39,461
so the simplest case I am going to use two
sample and holds, one which samples all the
365
00:44:39,461 --> 00:45:04,430
even samples. And one which samples the odd
one, so the circles and the crosses
366
00:45:04,430 --> 00:45:19,510
represent, outputs coming from in different
sample and holds.
367
00:45:19,510 --> 00:45:28,350
So, please note that, even though the effective
sampling rate is f s each individual
368
00:45:28,350 --> 00:45:36,040
sample, and hold is working at the rate in
this particular case, f s by 2 this can be
369
00:45:36,040 --> 00:45:44,890
extended to n sample and holds sampling systems
operating in parallel. So, let us try and
370
00:45:44,890 --> 00:45:56,200
first analyse this also serves as a good way
of seeing, if we understand this continuous
371
00:45:56,200 --> 00:46:00,660
time, to discrete time conversion and spectrum
and all that properly.
372
00:46:00,660 --> 00:46:14,160
So, if I had a continuous time signal, one
equivalent way of representing this system
373
00:46:14,160 --> 00:46:25,859
mathematically, is to say I have two samplers
operating at a sampling rate of f s by 2
374
00:46:25,859 --> 00:46:37,589
which means that, they are sampling at 2 T.
But, mathematically I can get the samples
375
00:46:37,589 --> 00:46:54,650
if
I take the signal advance it by T and sample
376
00:46:54,650 --> 00:47:03,780
it at 2 T, so both the sample and holds of
these sampling systems the switches are being
377
00:47:03,780 --> 00:47:11,980
closed simultaneously. It is equivalent to
it you can either skew the input signal, or
378
00:47:11,980 --> 00:47:17,200
you can skew the stamping clocks, I have
chosen simply for mathematical convenience.
379
00:47:17,200 --> 00:47:24,670
I have chosen to advance one of the signals,
it does not I mean have a advanced version
380
00:47:24,670 --> 00:47:35,109
of the signal, it does not as long as mathematically
equivalent it does not matter. So, now
381
00:47:35,109 --> 00:47:44,570
here I have a discrete time signal, which
is the samples of the continuous time signal
382
00:47:44,570 --> 00:47:57,970
taken at even instance of time. And here what
do I have, I have again a discrete time
383
00:47:57,970 --> 00:48:10,630
signal, where the samples are taken at odd
instances of time. Now, how do I reconstruct
384
00:48:10,630 --> 00:48:20,650
when I want to use these two signals, to make
the output look like a sample and hold
385
00:48:20,650 --> 00:48:23,290
which was operating at f s.
386
00:48:23,290 --> 00:48:28,320
So, what do you think I should do to these
two output discrete time signals, no please
387
00:48:28,320 --> 00:48:31,750
if
you if you simply add these two signals what
388
00:48:31,750 --> 00:48:46,190
will happen, you need to switch before I
what is the rate of these samples f s by 2,
389
00:48:46,190 --> 00:48:49,200
if I simply add the two sequences what will
be
390
00:48:49,200 --> 00:48:51,720
the rate be
391
00:48:51,720 --> 00:48:53,559
It will still be
F s by 2
392
00:48:53,559 --> 00:48:57,349
F s by 2, but I need to eventually get to
a rate of
393
00:48:57,349 --> 00:49:00,559
F s
F s, so what do you think I should do
394
00:49:00,559 --> 00:49:14,900
Multiply, during one time period and during
the second, set at the next second
395
00:49:14,900 --> 00:49:20,770
signal of some sample
See these samples are
396
00:49:20,770 --> 00:49:32,220
No, once you sampled only I have list of samples
there is no more time, so only sequence
397
00:49:32,220 --> 00:49:43,999
of numbers how will you generate the
( interpolate it, one signal interpolate
398
00:49:43,999 --> 00:49:50,890
I am understand what you are saying, but what
technically needs to be done is you must
399
00:49:50,890 --> 00:49:54,160
first insert you must increase the sampling
rates of
400
00:49:54,160 --> 00:49:58,190
Individually
Individual strips that means, you must
401
00:49:58,190 --> 00:50:10,299
Inside 0
Up sample each of these sequences, so now
402
00:50:10,299 --> 00:50:16,359
up sampling means you insert zeros, every
your up sample were factor of 2, so you insert
403
00:50:16,359 --> 00:50:21,430
zeroes and then, what should you do, you
404
00:50:21,430 --> 00:50:38,549
must put a delay here and now what should
you do, you simply add these two. So, now,
405
00:50:38,549 --> 00:50:46,930
if I put all of this in a big box this should
look exactly like
406
00:50:46,930 --> 00:50:51,680
Sample
A sample and hold
407
00:50:51,680 --> 00:50:57,000
F s
Operating at f s, so obviously, the first
408
00:50:57,000 --> 00:51:02,440
question that would come to you is apart from
here , why would you want to do this, and
409
00:51:02,440 --> 00:51:07,010
the answer is that the
sampling operation which I claim to be difficult
410
00:51:07,010 --> 00:51:22,990
to implement is now happening at half
the rate. And so let us try and figure out
411
00:51:22,990 --> 00:51:30,000
the spectra at various places in the signal
chain,
412
00:51:30,000 --> 00:51:47,819
this is also good exercise to do to see if
we understand, I mean you already know the
413
00:51:47,819 --> 00:51:52,049
final answer, we know what the spectrum must
be at h.
414
00:51:52,049 --> 00:52:05,710
Let us remind ourselves with that the x d
of e to the j omega at h must be 1 by T, it
415
00:52:05,710 --> 00:52:07,790
looks
mind you like a sample and hold which is running
416
00:52:07,790 --> 00:52:25,089
at f s, so it must look like 1 over T
sigma k x c of what did we see just now f
417
00:52:25,089 --> 00:52:42,559
s by 2 pi times omega minus 2 pi k. This is
what we must get at h, so the spectrum at
418
00:52:42,559 --> 00:52:48,900
a is simply, the signal at a discrete time
or
419
00:52:48,900 --> 00:52:49,900
continuous time?
Continuous time
420
00:52:49,900 --> 00:53:00,450
Continuous time, so this is x c of f at B
how do the signal look like, please note that
421
00:53:00,450 --> 00:53:06,740
the
signal at B is sampled at f s by 2, so all
422
00:53:06,740 --> 00:53:14,300
we need to do is replace T with 2 T. So, is
that at
423
00:53:14,300 --> 00:53:29,470
b is it a discrete time signal or a continuous
time signal, after sampling
424
00:53:29,470 --> 00:53:38,730
it is discrete, so
this is nothing but, sigma k x c of what should
425
00:53:38,730 --> 00:54:03,931
I do now, f s by 4 pi times omega minus 2
pi k, which I will write as f s by 2 pi. And
426
00:54:03,931 --> 00:54:36,890
pi times k, but actually I think I made a
mistake, I think it is. Now, what is the spectrum
427
00:54:36,890 --> 00:54:58,789
at C, C mind you is this signal here,
what do you think that is, it is up sampled
428
00:54:58,789 --> 00:55:06,420
so you are inserting 0, so what is.
429
00:55:06,420 --> 00:55:08,950
It is simply scaling of the
Axis
430
00:55:08,950 --> 00:55:10,510
Frequency axis
Frequency axis
431
00:55:10,510 --> 00:55:22,960
So, what should I do I replace,
432
00:55:22,960 --> 00:55:32,230
F s by 2 pi I must replace
433
00:55:32,230 --> 00:55:33,940
No
434
00:55:33,940 --> 00:55:38,880
Omega with
2 omega
435
00:55:38,880 --> 00:55:44,349
2 omega
So, what must this become
436
00:55:44,349 --> 00:55:48,619
(
Omega minus
437
00:55:48,619 --> 00:56:08,630
Pi k
Now, what about the spectrum at d, d is this
438
00:56:08,630 --> 00:56:11,339
signal here is it continuous time or discrete
time?
439
00:56:11,339 --> 00:56:14,140
Continuous time
It is continuous time, so what is the spectrum
440
00:56:14,140 --> 00:56:22,530
at d, x of f e to the x e of f e to the minus,
e
441
00:56:22,530 --> 00:56:37,109
to the j 2 pi f times T, because it is advancing
it is plus T, now what is the spectrum at
442
00:56:37,109 --> 00:56:47,420
e,
e is nothing but, the sampled version of the
443
00:56:47,420 --> 00:57:01,720
continuous time spectrum. So, it must be of
the form 1 by 2 T sum over all k x c of f
444
00:57:01,720 --> 00:57:14,660
s by 2 pi times omega by 2 minus pi k times
e to
445
00:57:14,660 --> 00:57:28,220
the j 2 pi. And what should be do, mean how
do we get the discrete time expression, you
446
00:57:28,220 --> 00:57:41,440
now you form the term f minus k times f s,
and replace f with please note that you must
447
00:57:41,440 --> 00:57:54,609
replace 2 pi f T with here, the sampling rate
is 2 T.
448
00:57:54,609 --> 00:58:14,569
And therefore, sampling rate period is 2 T,
so sampling rate is 1 by sampling rate is
449
00:58:14,569 --> 00:58:24,900
f s
by 2, the delay is T, and I have f and how
450
00:58:24,900 --> 00:58:28,650
do I what must I do here I must replace 2
pi f
451
00:58:28,650 --> 00:58:44,750
times 2 T with omega. So, how does this look
like f must be replaced with omega by 2 pi
452
00:58:44,750 --> 00:58:59,819
times f s by does it make sense, it is the
same expressions I am manipulating them with
453
00:58:59,819 --> 00:59:02,589
f
s by 2.
454
00:59:02,589 --> 00:59:48,470
Rather than, and which simplifies to recalling
that f s times T is 1, we simply see that
455
00:59:48,470 --> 00:59:51,640
this
expression is omega by 2 the first term minus
456
00:59:51,640 --> 01:00:09,339
k pi. So, that is the spectrum at e what
457
01:00:09,339 --> 01:00:27,599
would be the spectrum at f, at f it is simply
1 over 2 T sigma k x c of f s by 2 pi times
458
01:00:27,599 --> 01:00:52,030
omega minus pi k
times e to the j omega minus k pi. Now, what
459
01:00:52,030 --> 01:01:08,900
are we doing, we are
delaying this by one sample, so this is what
460
01:01:08,900 --> 01:01:20,720
how does that look like, g is simply a stuff
at
461
01:01:20,720 --> 01:01:49,829
f e to the minus j omega.
So, what goes away the plus j omega and the
462
01:01:49,829 --> 01:01:55,910
minus j omega go away that makes
intuitive sense, because we have advanced
463
01:01:55,910 --> 01:02:02,089
the signal here by 1 by a factor T. And we
are
464
01:02:02,089 --> 01:02:11,460
delaying it by 1 sample, which is also I mean
in time it is time T, I mean at c and f what
465
01:02:11,460 --> 01:02:20,770
is the sampling rate?
F s
466
01:02:20,770 --> 01:02:33,440
So, that goes away, so this simply becomes
e to the minus j k time times pi and what
467
01:02:33,440 --> 01:02:36,740
are
we doing finally, we are adding the stuff
468
01:02:36,740 --> 01:02:42,559
at c with the stuff at f, let me just copy
and
469
01:02:42,559 --> 01:03:04,970
paste this is what we had per things at c.
So, now if we add I mean what is so funny
470
01:03:04,970 --> 01:03:14,750
about e to the minus j k pi, it is minus 1
to the power n, so for even values of k it
471
01:03:14,750 --> 01:03:22,050
is 1 for
odd values of k it is minus 1. So, k equal
472
01:03:22,050 --> 01:03:43,329
to 0 how does this look 1 by T x c of f s
by 2 pi
473
01:03:43,329 --> 01:04:00,400
times omega, for k equal to 1 what happens?
((Refer Time: 64: 10))
474
01:04:00,400 --> 01:04:20,390
For k equal to 1, this argument is the same
as this argument, this is 1 where as this
475
01:04:20,390 --> 01:04:25,579
is
minus 1, so when you add the 2, the 2 go away
476
01:04:25,579 --> 01:04:41,630
and you get 0, so going the same way for
all odd k, these things simply vanish you
477
01:04:41,630 --> 01:04:53,069
understand. So, this is only valid for, these
things will be non zero only for, so that
478
01:04:53,069 --> 01:04:58,920
h it will be of the form 1 by T times sigma
over
479
01:04:58,920 --> 01:05:28,349
all k, all even k which I can simply say as
f s by 2 pi times omega minus 2 pi k, which
480
01:05:28,349 --> 01:05:34,099
is
the same as what we get for a sample and hold,
481
01:05:34,099 --> 01:05:41,680
operating at the pole rate.
Well this is where the maths stops from the
482
01:05:41,680 --> 01:05:47,590
communication point of view, or the signal
processing point of view, when you make a
483
01:05:47,590 --> 01:05:54,670
practical system like this, there are a whole
bunch of problems. For instance, there might
484
01:05:54,670 --> 01:06:00,150
be an offset here, which may not be there
in
485
01:06:00,150 --> 01:06:08,420
that path, in other words there is a D C offset
which is different in both parts. The next
486
01:06:08,420 --> 01:06:14,329
thing is that the gains of both the paths
may not be exactly the same, the third thing
487
01:06:14,329 --> 01:06:17,980
is
that, in practice you are not going to implement
488
01:06:17,980 --> 01:06:21,740
it like this, you are going to one switch
is
489
01:06:21,740 --> 01:06:25,069
going to be sampling at 2 T, 4 T and so on.
490
01:06:25,069 --> 01:06:31,840
The other switch is going to be sampling at
1 T, 3 T, 5 t and so on, but in practice what
491
01:06:31,840 --> 01:06:35,230
happens is that 2 independent switches, so
the exact sampling instance of both these
492
01:06:35,230 --> 01:06:42,490
switches may not be may not be exactly T apart.
So, you ideally want the first sample
493
01:06:42,490 --> 01:06:50,779
and hold to sample the second sample and hold
must exactly sample the time T later, in
494
01:06:50,779 --> 01:06:55,680
other words, both of these are sampling at
a sampling rate of f s by 2 that is every
495
01:06:55,680 --> 01:06:59,839
2 T,
but the offset between their sampling instance,
496
01:06:59,839 --> 01:07:05,849
must be exactly T.
Otherwise, it will be like this is a small
497
01:07:05,849 --> 01:07:12,510
skew, then it will be like one is sampling
here,
498
01:07:12,510 --> 01:07:17,779
the other one is sampling here, this guy is
sampling again here, the other one is sampling
499
01:07:17,779 --> 01:07:25,299
here and so on. So, this corresponds to some
kind of non uniform sampling of the input,
500
01:07:25,299 --> 01:07:31,330
all these will have will cause artifacts in
the discrete time spectrum which must be
501
01:07:31,330 --> 01:07:40,010
addressed, you understand. So, we will continue
with the effect of these artifacts which
502
01:07:40,010 --> 01:07:44,680
are fundamental things, which will happen
every time you implement time sample
503
01:07:44,680 --> 01:07:48,289
sampling system, we will see this in the next
class.