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[MUSIC PLAYING]

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ALAN OPPENHEIM:
In This lecture, I

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would like to demonstrate
the effects of sampling

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and aliasing, and also, some of
the properties of discrete time

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linear systems.

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What we'll be using
to demonstrate

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this is a programmable digital
filter, which is contained

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in this box,
programmable in the sense

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that many of the parameters
of the filter-- for example,

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the filter coefficients, the
coefficient and arithmetic word

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length, the sampling rate
etc, are easily changed.

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In other words, programmable.

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So this is the basic
digital filter.

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And then, of course, we have
some associated equipment

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to help us with
the demonstration.

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Well, we'll be returning to this
filter in a few minutes, when,

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together with my colleagues,
Mike Portnoff and Dave Harris,

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I'll be demonstrating
several of the ideas

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that we're about to talk about.

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00:01:43,030 --> 00:01:49,170
But first of all, let me
explain what the basic setup is.

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The programmable digital
filter consists, essentially

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of a system which is a sampler,
a continuous time, or C

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to D converter, which converts
an impulse train to a sequence,

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a digital filter to
obtain a filtered output

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sequence, a "discrete
time to time" converter

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to convert the sequence
back to an impulse train,

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and finally a D-sampling or
smoothing low-pass filter.

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00:02:24,780 --> 00:02:28,770
So at this point, we have
a continuous time input.

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At this point, we have a
continuous time impulse train,

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at this point, a sequence,
then a sequence here,

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an impulse train, and back
to a smooth, continuous time

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function.

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For the first part
of the demonstration,

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what I would like to
focus on is just simply

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the effects of
sampling and aliasing.

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And so for the first
part, I'll just

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simply choose this
digital filter

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to be an identity system.

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In other words, the impulse
response of this system

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is just simply an impulse.

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00:03:01,440 --> 00:03:03,900
In that case, this
overall system

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00:03:03,900 --> 00:03:07,080
collapses to a somewhat
simpler system,

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as I have on this
next viewgraph where

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we convert from a continuous
time input to a sequence

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and then back to the
continuous time input.

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And, in fact, we could really
collapse the A to D or C

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to D converter and
D to A converter

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together since we're simply
converting from an impulse

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train to a sequence and then
back to the same impulse train.

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00:03:36,420 --> 00:03:38,550
To see what the
effect of this system

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is in both the time domain
and frequency domain,

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we can look at the associated
time waveforms and spectra.

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On the left-hand side, we have
the associated time wave forms

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and sequences.

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On the right-hand side, the
associated Fourier transforms.

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So we can think of an input
continuous time function, which

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is of some general form, with
a band-limited spectrum--

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band-limited from minus
omega c to plus omega c.

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When we sample this
to obtain the impulse

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train with a sampling
period of capital T

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the associated spectrum is
then a periodic replication

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of this band-limited spectrum.

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So we have the Fourier
transform of x of a of t.

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Then the same thing
reproduce that multiples

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of 2 pi over capital T

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When we then convert
this to a sequence, that

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implies a frequency
normalization,

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a normalization of the frequency
axis, so that this periodicity

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gets converted to a
periodicity with a period of 2

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pi in the digital frequency
variable small omega.

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Otherwise, the general shape
of the Fourier transform

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stays the same.

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We then go back through
the system, converting.

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Back to an impulse
train and then finally,

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by low-pass
filtering, we extract

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just the one replication of
the original Fourier transform.

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And what we would recover
is x of a of j omega.

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We would recover
this exactly provided

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that the bandwidth is
small enough compared

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with the sampling frequency.

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If, on the other hand, omega
sub c is too large in relation

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to the sampling frequency,
then what we end up with

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is an interaction between
these two pieces of the Fourier

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transform.

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And that interaction is what's
referred to as aliasing.

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The effect aliasing
is most easily

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understood in terms of
a simple example, namely

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a sinusoidal input.

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00:05:58,870 --> 00:06:01,560
So let's consider,
specifically, what

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happens with the spectra in
the case of a sinusoidal input.

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Here, we have an
input cosine omega 0t,

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00:06:11,680 --> 00:06:16,470
an assumed sampling rate of
2 pi over capital T. This

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00:06:16,470 --> 00:06:22,620
then is the Fourier transform
of this sinusoid or cosine.

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00:06:22,620 --> 00:06:26,670
After sampling, that is just
simply periodically repeated

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with a period equal to
the sampling frequency.

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And we see that if omega 0, the
input frequency is low enough,

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then the original spectrum,
or Fourier transform,

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falls within the passband
of the low-pass filter.

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This dashed line corresponds
to the frequency response,

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an ideal frequency
response, associated

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with the D-sampling
low-pass filter.

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00:06:59,320 --> 00:07:04,650
And, of course, if the spectral
fall, as I've shown in here,

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00:07:04,650 --> 00:07:06,440
then there is no aliasing.

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00:07:06,440 --> 00:07:08,340
In other words, what we
recover at the output

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00:07:08,340 --> 00:07:10,860
of the low-pass
filter is just simply

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the original Fourier
transform, or equivalently,

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the original signal.

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Now, let's consider,
on the other hand,

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the effect of increasing
omega 0, the input frequency,

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and we can think in
particular of what

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the effect is on each one of
these impulses in the Fourier

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transform.

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As omega 0 increases, this
impulse moves to the right,

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this impulse moves to the left.

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00:07:43,000 --> 00:07:46,230
And likewise, in the
periodic replications,

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this impulse moves
down in frequency.

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00:07:48,750 --> 00:07:51,960
This impulse moves
up in frequency, etc.

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00:07:51,960 --> 00:07:58,740
Now, if omega sub s minus omega
0 is greater than omega 0,

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then these two impulses
haven't crossed.

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00:08:01,530 --> 00:08:08,190
However, if omega sub s minus
omega 0 is less than omega 0,

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00:08:08,190 --> 00:08:10,500
then the situation
that we have is

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00:08:10,500 --> 00:08:15,480
what I've illustrated here,
where now, the impulses that

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00:08:15,480 --> 00:08:18,120
lie in the passband
of the filter

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are at the frequency
omega sub s minus omega 0

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rather than at the
frequency omega 0.

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So as we think of increasing
the input frequency, what

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00:08:29,310 --> 00:08:32,280
happens for a while, is
that the output frequency

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00:08:32,280 --> 00:08:34,409
will correspondingly increase.

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00:08:34,409 --> 00:08:38,490
But after we've increased
omega 0 past this point,

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then the output of
the low-pass filter

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will decrease in
frequency because it's

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taken on the alias of a new
frequency or a new sinusoid.

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00:08:49,560 --> 00:08:51,570
And so the output,
in, that case is

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00:08:51,570 --> 00:08:54,510
cosine omega sub
s minus omega 0.

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And there's a
little t over here.

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There is-- I want to
demonstrate this effect.

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00:09:00,600 --> 00:09:02,760
There is another
effect that we'll

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00:09:02,760 --> 00:09:06,150
observe in the process
of demonstrating

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00:09:06,150 --> 00:09:10,500
this, because of the fact that
in any real-world situation,

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in fact, this
low-pass filter is not

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going to be an ideal low-pass
filter, as I've shown here

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but in fact, is going
to have some transition

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with associated with it.

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And so as omega 0 and
omega sub s minus omega 0

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get very close in frequency,
in essence, both of them

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are having some influence on
the output because of the fact

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that there isn't infinite
attenuation of one

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of the impulses an exact
replication of the other.

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So what we'll see
in that case, as we

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get an input frequency which
is close to half the sampling

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frequency, will see, in addition
to the effective aliasing

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that we want to
demonstrate, we'll

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see an effect
which, essentially,

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is a beating phenomenon.

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So let's move over
to the digital filter

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and demonstrate these effects,
where I remind you now

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that this filter, the
digital filter aspect of it,

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is just simply an identity
system so that it corresponds

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to sampling, and then
simply sampling and low-pass

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filtering.

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What we have as an input, if
we look at the oscilloscope,

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is on the upper trace, the input
sinusoid, on the lower trace,

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the output sinusoid.

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And as we see it right
now, the input sinusoid

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has chosen to be low enough
frequency so that, in fact,

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there is no aliasing.

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Well, let's now increase
the input frequency.

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And what we'll observe is
that the output frequency

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00:10:50,010 --> 00:10:52,380
increases likewise.

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00:10:52,380 --> 00:10:56,370
The output frequency is still
equal to the input frequency.

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00:10:56,370 --> 00:10:58,890
We're now getting
into the vicinity

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of half the sampling
frequency so

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00:11:00,900 --> 00:11:03,660
that what we're beginning
to see now in the output

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00:11:03,660 --> 00:11:06,090
is not just a sinusoidal output.

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But in fact, what we see
are the two components.

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In other words, we see the
beating effect due to the fact

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00:11:13,530 --> 00:11:18,490
that the low-pass filter is
not an ideal low-pass filter.

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00:11:18,490 --> 00:11:24,270
Now what we want to observe as
we sweep past half the sampling

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frequency is the aliasing
effect-- in other words,

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00:11:28,090 --> 00:11:30,240
the fact that the
output sinusoid

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will decrease in frequency.

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00:11:32,070 --> 00:11:35,850
Let's first sweep
back down to DC.

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00:11:35,850 --> 00:11:39,810
So the output sinusoid
follows the input sinusoid.

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00:11:39,810 --> 00:11:43,900
And then we'll sweep
automatically from 0

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00:11:43,900 --> 00:11:46,660
up to the sampling frequency.

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00:11:46,660 --> 00:11:48,740
And let's see that.

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So on the bottom trace
is the output sinusoid.

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00:11:51,390 --> 00:11:53,450
The top trace is
the input sinusoid.

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00:11:53,450 --> 00:11:56,190
Were now in the vicinity of
half the sampling frequency.

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00:11:56,190 --> 00:11:58,800
We're now past half
the sampling frequency.

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00:11:58,800 --> 00:12:01,830
And you see that the output
is decreasing in frequency

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00:12:01,830 --> 00:12:04,390
while the input was increasing.

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00:12:04,390 --> 00:12:07,330
Let's finally look
at that again.

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00:12:07,330 --> 00:12:11,430
But this time, let's also
listen to the output sinusoid.

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00:12:11,430 --> 00:12:14,010
And what you'll hear, in
addition to observing this

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00:12:14,010 --> 00:12:15,870
on the bottom
trace of the scope,

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00:12:15,870 --> 00:12:19,500
is the fact that the
output frequency first

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00:12:19,500 --> 00:12:23,670
increases, and then decreases,
even though the input frequency

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00:12:23,670 --> 00:12:25,472
is continuing to increase.

207
00:12:25,472 --> 00:12:26,430
So let's do that again.

208
00:12:26,430 --> 00:12:28,650
But now, in this case,
let's listen to the output.

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00:12:28,650 --> 00:12:31,638
[SOUND WAVES RISE AND LOWER]

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00:12:41,610 --> 00:12:45,690
OK, now what we would
now like to consider

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00:12:45,690 --> 00:12:49,200
is the effect of
actually carrying out

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00:12:49,200 --> 00:12:51,360
some digital
filtering in-between

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00:12:51,360 --> 00:12:54,430
the sampling and D-sampling.

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00:12:54,430 --> 00:13:00,510
And so let me return
to the basic system

215
00:13:00,510 --> 00:13:11,750
again where we had previously
removed this digital filter.

216
00:13:11,750 --> 00:13:13,400
And now, what we
want to consider

217
00:13:13,400 --> 00:13:18,080
is the effect of the overall
system, when we, in fact,

218
00:13:18,080 --> 00:13:20,660
insert an interesting,
or a more interesting,

219
00:13:20,660 --> 00:13:22,610
digital filter in the middle.

220
00:13:22,610 --> 00:13:26,390
The digital filter that
we're going to insert

221
00:13:26,390 --> 00:13:30,680
is a low-pass filter.

222
00:13:30,680 --> 00:13:33,740
And the impulse response
of the low-pass filter,

223
00:13:33,740 --> 00:13:36,290
or the unit sample response
of the low-pass filter,

224
00:13:36,290 --> 00:13:38,660
is, as I've shown up here.

225
00:13:38,660 --> 00:13:42,380
And it's a symmetric
unit sample response.

226
00:13:42,380 --> 00:13:46,730
And consequently, it corresponds
to a linear phase filter.

227
00:13:46,730 --> 00:13:48,740
The associated
frequency response

228
00:13:48,740 --> 00:13:53,420
I show down here, where this
is now the filter passband.

229
00:13:53,420 --> 00:13:55,010
This is the filter stop band.

230
00:13:55,010 --> 00:13:57,170
And, of course,
there is some ripple.

231
00:13:57,170 --> 00:14:00,320
There is an infinite
attenuation in the stopband.

232
00:14:00,320 --> 00:14:03,290
And I remind you of the
fact that, of course,

233
00:14:03,290 --> 00:14:07,370
the digital filter frequency
response must, by necessity, be

234
00:14:07,370 --> 00:14:11,720
periodic with a period of 2 pi.

235
00:14:11,720 --> 00:14:15,800
The cutoff frequency associated
with the particular filter

236
00:14:15,800 --> 00:14:19,610
that we want to
demonstrate is pi over 5,

237
00:14:19,610 --> 00:14:22,650
or one tenth of 2 pi.

238
00:14:22,650 --> 00:14:25,100
And the factor one
tenth is a factor

239
00:14:25,100 --> 00:14:29,010
that I'll want to
refer to again shortly.

240
00:14:29,010 --> 00:14:32,550
Now, the overall
system, of course,

241
00:14:32,550 --> 00:14:34,590
is a continuous time system.

242
00:14:34,590 --> 00:14:36,920
In other words, we have
a continuous time input.

243
00:14:36,920 --> 00:14:39,540
We have a continuous
time output.

244
00:14:39,540 --> 00:14:42,320
And the question
then is, what is

245
00:14:42,320 --> 00:14:46,490
the equivalent continuous
time system in relation

246
00:14:46,490 --> 00:14:49,820
to the digital filter
frequency response

247
00:14:49,820 --> 00:14:51,240
that we have illustrated here?

248
00:14:51,240 --> 00:14:54,470
In other words, what is the
equivalent frequency response

249
00:14:54,470 --> 00:14:59,450
of the corresponding
continuous time system?

250
00:14:59,450 --> 00:15:03,320
We can answer that
by simply referring

251
00:15:03,320 --> 00:15:07,040
to the basic definition
of frequency response

252
00:15:07,040 --> 00:15:09,710
for the continuous time
case and frequency response

253
00:15:09,710 --> 00:15:13,070
for the discrete time case.

254
00:15:13,070 --> 00:15:18,290
In the continuous time case, for
a linear time invariant filter,

255
00:15:18,290 --> 00:15:21,170
the frequency
response is defined

256
00:15:21,170 --> 00:15:26,210
as the gain change applied
to a complex exponential.

257
00:15:26,210 --> 00:15:30,300
So if we consider a complex
exponential as the input,

258
00:15:30,300 --> 00:15:33,980
then the output of the system
is a complex exponential

259
00:15:33,980 --> 00:15:39,000
at the same complex frequency,
but with an amplitude,

260
00:15:39,000 --> 00:15:42,440
which is equal to the frequency
response of the system

261
00:15:42,440 --> 00:15:44,960
at that frequency.

262
00:15:44,960 --> 00:15:48,350
Likewise, for a
discrete time system,

263
00:15:48,350 --> 00:15:51,170
we can consider a
complex exponential input

264
00:15:51,170 --> 00:15:53,600
at a frequency small omega.

265
00:15:53,600 --> 00:15:57,080
And the output is a
complex exponential

266
00:15:57,080 --> 00:16:01,220
at the same complex frequency,
with an amplitude change,

267
00:16:01,220 --> 00:16:05,480
which is the frequency
response of the digital filter,

268
00:16:05,480 --> 00:16:10,620
or discrete time filter, again
evaluated at that frequency.

269
00:16:10,620 --> 00:16:13,340
Well, simply from
these definitions,

270
00:16:13,340 --> 00:16:15,500
we can trace our way
through the system

271
00:16:15,500 --> 00:16:20,810
and see fairly easily what
the equivalent analog,

272
00:16:20,810 --> 00:16:25,910
or continuous time filter
frequency, response is.

273
00:16:25,910 --> 00:16:29,240
Let's consider the
overall system.

274
00:16:29,240 --> 00:16:33,590
And let's choose an input,
which is a complex exponential.

275
00:16:33,590 --> 00:16:36,800
And we'll choose the complex
exponential carefully

276
00:16:36,800 --> 00:16:39,620
to avoid aliasing.

277
00:16:39,620 --> 00:16:42,770
We then sample this
complex exponential

278
00:16:42,770 --> 00:16:44,870
and convert that to a sequence.

279
00:16:44,870 --> 00:16:46,810
And the sequence
values are there

280
00:16:46,810 --> 00:16:50,630
for e to the j omega
and capital T. Well,

281
00:16:50,630 --> 00:16:52,730
this is just the
discrete time complex

282
00:16:52,730 --> 00:16:58,190
exponential with a frequency of
capital omega times capital T.

283
00:16:58,190 --> 00:17:01,400
So the output of
the digital filter

284
00:17:01,400 --> 00:17:04,880
is a complex exponential with
the same complex frequency,

285
00:17:04,880 --> 00:17:08,910
capital omega times capital
T, with an amplitude,

286
00:17:08,910 --> 00:17:13,400
which is the frequency response
of this filter evaluated

287
00:17:13,400 --> 00:17:15,780
at the frequency of the input--

288
00:17:15,780 --> 00:17:20,450
In other words, evaluated at
capital omega times capital T.

289
00:17:20,450 --> 00:17:23,300
Then we convert that
back to an impulse train,

290
00:17:23,300 --> 00:17:25,550
and finally, low-pass filter.

291
00:17:25,550 --> 00:17:27,990
And the output of
the low-pass filter

292
00:17:27,990 --> 00:17:35,000
then has the same amplitude,
but now multiplying

293
00:17:35,000 --> 00:17:37,820
a continuous time
complex exponential

294
00:17:37,820 --> 00:17:39,890
at the original input frequency.

295
00:17:39,890 --> 00:17:43,160
So simply from the
definition comparing this

296
00:17:43,160 --> 00:17:46,940
to the input from the definition
of the continuous time

297
00:17:46,940 --> 00:17:50,570
frequency response, the
continuous time frequency

298
00:17:50,570 --> 00:17:53,180
response is equal to this term.

299
00:17:53,180 --> 00:17:56,060
In other words, it's
the frequency response

300
00:17:56,060 --> 00:18:01,040
of the digital filter, but with
a rescaling of the frequency

301
00:18:01,040 --> 00:18:02,010
axis--

302
00:18:02,010 --> 00:18:04,490
in other words, with
the digital frequency

303
00:18:04,490 --> 00:18:10,920
variable small omega replaced by
capital omega times capital T.

304
00:18:10,920 --> 00:18:14,880
The consequence of that for the
particular digital filter that

305
00:18:14,880 --> 00:18:18,720
we're talking about-- or, in
fact, for any digital filter--

306
00:18:18,720 --> 00:18:22,320
is that the
continuous time filter

307
00:18:22,320 --> 00:18:26,640
frequency response has the same
shape as the digital filter

308
00:18:26,640 --> 00:18:31,980
frequency response, but has
a rescaled frequency axis--

309
00:18:31,980 --> 00:18:35,550
rescaled according
to this scaling.

310
00:18:35,550 --> 00:18:38,190
And in essence, what the
rescaling corresponds

311
00:18:38,190 --> 00:18:42,060
to-- and I think you can
verify this on your own--

312
00:18:42,060 --> 00:18:50,220
is to reconvert or rescale the
frequency 2 pi in small omega

313
00:18:50,220 --> 00:18:54,510
to the sampling
frequency in large omega.

314
00:18:54,510 --> 00:18:57,630
And the upshot of
all of this, is

315
00:18:57,630 --> 00:19:00,540
that this cutoff frequency,
which in a digital filter

316
00:19:00,540 --> 00:19:07,320
is a pi over 5, is now rescaled
to pi over 5 capital T.

317
00:19:07,320 --> 00:19:10,710
And what we would observe
is that as capital T,

318
00:19:10,710 --> 00:19:13,410
the sampling period
changes, then

319
00:19:13,410 --> 00:19:16,110
the bandwidth, or
the cutoff frequency,

320
00:19:16,110 --> 00:19:19,830
of the continuous time
filter changes also.

321
00:19:19,830 --> 00:19:24,780
So let me remind you of the
fact that the digital filter had

322
00:19:24,780 --> 00:19:30,060
a cutoff frequency
which was 1/10 of 2 pi.

323
00:19:30,060 --> 00:19:34,860
2 pi gets rescaled to
the sampling frequency

324
00:19:34,860 --> 00:19:37,680
in the continuous time domain.

325
00:19:37,680 --> 00:19:40,530
And the filter cutoff
frequency will then

326
00:19:40,530 --> 00:19:46,960
get rescaled to one tenth of
the filter sampling frequency.

327
00:19:46,960 --> 00:19:49,120
Let's illustrate
some of these effects

328
00:19:49,120 --> 00:19:51,730
with the digital filter.

329
00:19:51,730 --> 00:19:56,590
What we have, as I said was,
is a low pass filter impulse

330
00:19:56,590 --> 00:19:58,840
response and frequency response.

331
00:19:58,840 --> 00:20:01,690
First let's look at
the impulse response

332
00:20:01,690 --> 00:20:05,180
of the filter of the overall
system-- in other words,

333
00:20:05,180 --> 00:20:10,480
after the D-sampling
low-pass filter.

334
00:20:10,480 --> 00:20:16,510
What we see here is the impulse
response of the overall system.

335
00:20:16,510 --> 00:20:20,440
And we observe, for one thing,
that it's a symmetrical impulse

336
00:20:20,440 --> 00:20:23,620
response-- in other words,
corresponds to a linear phase

337
00:20:23,620 --> 00:20:24,710
filter.

338
00:20:24,710 --> 00:20:26,860
We can also look at
the impulse response

339
00:20:26,860 --> 00:20:29,340
before the D-sampling
low-pass filter.

340
00:20:29,340 --> 00:20:31,270
Lets take out that the
D-sampling low-pass

341
00:20:31,270 --> 00:20:33,520
filter slowly.

342
00:20:33,520 --> 00:20:37,540
And what we observe is
basically the output

343
00:20:37,540 --> 00:20:40,840
of the digital-to-analog
converter, which,

344
00:20:40,840 --> 00:20:44,470
of course, is a staircase or
boxcar function, not an impulse

345
00:20:44,470 --> 00:20:45,280
train.

346
00:20:45,280 --> 00:20:48,130
In the real world, the output
of a [? D-day ?] converter

347
00:20:48,130 --> 00:20:51,070
generally is a boxcar
type of function.

348
00:20:51,070 --> 00:20:54,010
We can put the D sampling
filter back in now.

349
00:20:54,010 --> 00:20:57,760
And notice that the effect
of the D-sampling filter is

350
00:20:57,760 --> 00:21:03,670
basically to smooth out the
rough edges in the boxcar

351
00:21:03,670 --> 00:21:06,526
output from the [?
D-day ?] converter.

352
00:21:06,526 --> 00:21:09,010
All right, now what
we'd like to demonstrate

353
00:21:09,010 --> 00:21:14,140
is the actual frequency response
of the overall continuous time

354
00:21:14,140 --> 00:21:15,130
filter.

355
00:21:15,130 --> 00:21:19,570
And we can do that by
sweeping the system

356
00:21:19,570 --> 00:21:21,430
with a sinusoidal input.

357
00:21:21,430 --> 00:21:23,150
And what we expect
to see, of course,

358
00:21:23,150 --> 00:21:26,080
is as the sinusoidal
input frequency gets

359
00:21:26,080 --> 00:21:29,440
past the effective
cutoff frequency,

360
00:21:29,440 --> 00:21:34,150
then the output sinusoid
is greatly attenuated.

361
00:21:34,150 --> 00:21:36,970
Let's now sweep the
filter frequency response.

362
00:21:42,140 --> 00:21:46,160
And there is the filter
cutoff frequency.

363
00:21:51,250 --> 00:21:55,270
We can demonstrate the
filter characteristics

364
00:21:55,270 --> 00:21:57,320
in several other ways.

365
00:21:57,320 --> 00:22:02,890
One way is to choose, as a
display, instead of the output

366
00:22:02,890 --> 00:22:06,760
as a function of time, we can
display the output sinusoid

367
00:22:06,760 --> 00:22:08,690
as a function of frequency.

368
00:22:08,690 --> 00:22:12,040
And so we'll observe that
on the left-hand scope,

369
00:22:12,040 --> 00:22:13,540
while on the
right-hand scope, we'll

370
00:22:13,540 --> 00:22:16,540
have the same trace
that we just saw, namely

371
00:22:16,540 --> 00:22:19,750
two traces the upper trace
as the input sinusoid,

372
00:22:19,750 --> 00:22:22,420
the lower trace as
the output sinusoid.

373
00:22:22,420 --> 00:22:26,050
And in addition to observing
the frequency response,

374
00:22:26,050 --> 00:22:29,200
let's also listen to
the output sinusoid

375
00:22:29,200 --> 00:22:32,710
and observe the
attenuation in the output

376
00:22:32,710 --> 00:22:35,050
as we go from the filter
passband to the filter

377
00:22:35,050 --> 00:22:36,100
stopband.

378
00:22:36,100 --> 00:22:39,160
Again, a 20-kilohertz
sampling rate and a sweep

379
00:22:39,160 --> 00:22:40,900
range from 0 to 10 kilohertz.

380
00:22:40,900 --> 00:22:43,184
[SOUND WAVES RISE]

381
00:22:48,957 --> 00:22:50,790
Now, of course we're
in the filter stopband.

382
00:22:53,780 --> 00:22:59,930
Now, if we increase the sweep
range from 10 kilohertz to 20

383
00:22:59,930 --> 00:23:03,420
kilohertz so that the sweep
range is equal to the sampling

384
00:23:03,420 --> 00:23:06,210
frequency, in essence,
that corresponds

385
00:23:06,210 --> 00:23:10,710
to sweeping out the digital
filter from 0 to 2 pi.

386
00:23:10,710 --> 00:23:12,930
And in that case,
we'll begin to see

387
00:23:12,930 --> 00:23:15,840
some of the periodicity in
the digital filter frequency

388
00:23:15,840 --> 00:23:16,930
response.

389
00:23:16,930 --> 00:23:20,490
So let's do that now with
a 20-kilohertz sampling

390
00:23:20,490 --> 00:23:23,445
rate and a sweep range
of 0 to 20 kilohertz.

391
00:23:23,445 --> 00:23:25,676
[SOUND WAVES RISE AND FALL]

392
00:23:28,460 --> 00:23:36,770
Now we come near 2 pi, we
get back into the passband,

393
00:23:36,770 --> 00:23:41,060
and finally, back to a
0- to 10-kilohertz sweep

394
00:23:41,060 --> 00:23:45,500
so they were again, sweeping
only from 0 to pi with regard

395
00:23:45,500 --> 00:23:47,863
to the digital filter.

396
00:23:47,863 --> 00:23:50,809
[SOUND WAVES RISE]

397
00:23:58,670 --> 00:24:01,560
OK, now, what we would
like to demonstrate

398
00:24:01,560 --> 00:24:04,840
is the effect of changing
the sampling frequency.

399
00:24:04,840 --> 00:24:08,790
And we know that the sampling--
that the effective filter

400
00:24:08,790 --> 00:24:13,440
cutoff frequency is tied
to the sampling frequency,

401
00:24:13,440 --> 00:24:16,050
and for this particular
filter, corresponds

402
00:24:16,050 --> 00:24:18,990
to a tenth of the
sampling frequency.

403
00:24:18,990 --> 00:24:21,570
Consequently, if we double
the sampling frequency,

404
00:24:21,570 --> 00:24:25,770
we should double the effective
filter passband width

405
00:24:25,770 --> 00:24:29,010
or double the filter
cutoff frequency.

406
00:24:29,010 --> 00:24:31,380
And so let's do that now.

407
00:24:31,380 --> 00:24:34,140
Again, a 0 to 10
kilohertz sweep range,

408
00:24:34,140 --> 00:24:38,307
but a 40-kilohertz
sampling frequency.

409
00:24:38,307 --> 00:24:40,602
[SOUND WAVES RISE]

410
00:24:40,602 --> 00:24:45,040
And we should observe that the
filter cutoff frequency has now

411
00:24:45,040 --> 00:24:47,200
doubled out to 4 kilohertz.

412
00:24:49,850 --> 00:24:54,260
Now let's begin to decrease
the filter sampling frequency.

413
00:24:54,260 --> 00:24:56,890
So from 40, let's change
the sampling frequency

414
00:24:56,890 --> 00:24:58,620
to 20 kilohertz.

415
00:24:58,620 --> 00:25:03,354
And we should see the cutoff
frequency cut in half.

416
00:25:03,354 --> 00:25:05,844
[SOUND WAVES RISE]

417
00:25:12,330 --> 00:25:13,970
Now we can go even further.

418
00:25:13,970 --> 00:25:16,910
We can cut the sampling
frequency down to 10 kilohertz.

419
00:25:16,910 --> 00:25:20,160
And remember that the sweep
range is 0 to 10 kilohertz.

420
00:25:20,160 --> 00:25:24,506
So now we'll be
sweeping from 0 to 2 pi.

421
00:25:24,506 --> 00:25:26,876
[SOUND WAVES RISE]

422
00:25:30,200 --> 00:25:33,490
So as we get close to 2 pi,
we'll see the passband again.

423
00:25:33,490 --> 00:25:35,560
[SOUND WAVES FALL]

424
00:25:35,560 --> 00:25:40,200
And now, let's cut down the
sampling frequency even further

425
00:25:40,200 --> 00:25:43,965
to 5 kilohertz.

426
00:25:43,965 --> 00:25:46,310
[SOUND WAVES RISE]

427
00:25:47,250 --> 00:25:48,928
Here we are at 2 pi.

428
00:25:48,928 --> 00:25:51,320
[SOUND WAVES RISE]

429
00:25:51,320 --> 00:25:53,000
And then at 4 pi.

430
00:25:53,000 --> 00:25:54,590
[SOUND WAVES FALL]

431
00:25:54,590 --> 00:25:58,640
Now finally, let's
demonstrate this effect

432
00:25:58,640 --> 00:26:02,570
of changing the
effective filter cutoff

433
00:26:02,570 --> 00:26:06,200
frequency by changing the
sampling rate by carrying out

434
00:26:06,200 --> 00:26:09,380
some low-pass filtering
on some live audio.

435
00:26:09,380 --> 00:26:13,520
And we'll demonstrate this
by listening to the audio,

436
00:26:13,520 --> 00:26:20,750
and also, observing the audio
on on a single trace, namely,

437
00:26:20,750 --> 00:26:22,430
the time waveform.

438
00:26:22,430 --> 00:26:26,680
And we'll begin it with a
sampling frequency of 40

439
00:26:26,680 --> 00:26:31,550
kilohertz, change that then to
20 kilohertz, 10 kilohertz, 5,

440
00:26:31,550 --> 00:26:34,580
and then 2 and 1/2,
corresponding to a filter

441
00:26:34,580 --> 00:26:38,570
cutoff frequency then of 4
kilohertz, then 2 kilohertz,

442
00:26:38,570 --> 00:26:43,520
then 1 kilohertz,
then 500, , 250 etc.

443
00:26:43,520 --> 00:26:46,250
So let's begin 40 kilohertz.

444
00:26:46,250 --> 00:26:49,916
And then we'll
work our way down.

445
00:26:49,916 --> 00:26:52,286
[MUSIC PLAYING]

446
00:26:54,660 --> 00:26:58,470
Now, let's reduce that to
a 20-kilohertz frequency

447
00:26:58,470 --> 00:27:00,220
or a 2-kilohertz filter.

448
00:27:00,220 --> 00:27:02,720
[MUSIC PLAYING]

449
00:27:05,550 --> 00:27:07,760
And 10 kilohertz
sampling frequency.

450
00:27:07,760 --> 00:27:13,130
[MUSIC PLAYING]

451
00:27:13,130 --> 00:27:15,850
And finally, a 5-kilohertz
sampling frequency

452
00:27:15,850 --> 00:27:20,480
corresponding to make 500-cycle
equivalent analog filter.

453
00:27:20,480 --> 00:27:22,815
[MUSIC PLAYING]

454
00:27:25,620 --> 00:27:27,370
All right, now let's
finally conclude

455
00:27:27,370 --> 00:27:31,020
by returning to a little
higher quality ragtime

456
00:27:31,020 --> 00:27:34,580
by changing the sampling
rates back to 40 kilohertz.

457
00:27:34,580 --> 00:27:37,330
[MUSIC PLAYING]