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[MUSIC PLAYING]
ALAN OPPENHEIM: In the last several lectures,
we've considered a number of procedures
for designing digital filters for the class
of infinite impulse response filters.
And the two primary design techniques
that we considered, in fact, the two useful design
techniques that we developed, were the techniques
which we refer to as impulse invariance, and also
the bilinear transformation.
You recall, hopefully, that both of these techniques
are techniques that correspond to mapping a continuous time
or analog filter to a digital filter.
What I'd like to do in this lecture
is consider an example of the design
of a digital low-pass filter using each of these techniques,
using impulse invariance and the bilinear transformation.
And hopefully this will give you a better feeling for how
both of these techniques work.
The class of analog filters that I would like to use
are the class of so-called analog Butterworth filters,
which are a useful class of lowpass analog
and digital filters, and also a relatively simple class.
To first of all, define the class of analog filters
that we're considering, they are the class
of analog Butterworth filters.
And this class of filters is defined by the analog system
function H sub a of j omega magnitude squared of the form
1 over 1 plus j omega divided by j omega sub c to the 2N.
Now, there are two parameters in this expression.
One is what's referred to often as the cutoff frequency, omega
sub c.
And the second is the order of the filter,
namely N. Let me remind you that we're
looking at here an expression for the squared magnitude
function.
And so we have a 2N here.
But in fact, N is what corresponds
to the order of the filter.
Well, if we look at this frequency response
characteristic, we observe that at omega equal to 0,
the gain or the magnitude squared is equal to unity.
At omega equal to omega sub c, the j's of course cancel out.
At omega equal to omega sub c, this term then is equal to 1,
and we have for the magnitude squared function, a half.
Or for the magnitude function, which
is what we're looking at here, 1 divided
by the square root of 2.
So at omega equal to omega sub c,
an analog Butterworth filter has a magnitude which is down by 1
over the square root of 2.
And the other important characteristic
is that the frequency response is a monotonic function
of capital Omega.
So that as capital Omega increases from 0 to infinity,
the magnitude or the frequency response
is a monotonic function.
Well, we observed that the cutoff frequency, the parameter
omega sub c, adjusts in a sense where we might divide
the frequency characteristic into passband, transition band,
and stopband.
We have another parameter, which is the parameter
N, the order of the filter.
And basically the effect that the parameter N has
is to affect the shape of this frequency response in the sense
that if capital N is larger, the frequency response tends
to be flatter longer, so that for a larger N than what I've
drawn with the black curve, the frequency response might
look more like this and then drop off sharper.
Whereas if capital N is smaller, then it
will be less flat in the passband
and drop off more slowly.
So in fact, the way that I've drawn it here,
this is the behavior of the frequency response curve
as the parameter capital N increases.
So, the higher the order of the filter,
the sharper the drop from what we
could consider as a passband region into a stopband region.
And obviously then, the sharper the filter we want,
the higher order filter we would design.
Now, to look at the pole 0 pattern
or the 0 pattern really of the Butterworth filter,
we can convert this expression into an expression relating
the magnitude squared function expressed
in terms of the Laplace transform variable s.
And so this expression corresponds to the product
H sub a of s times H sub a of minus s, evaluated at s
equals j omega, in which case the j omega
is what corresponds to s, and so H sub a of s times H sub
a of minus s is of the form 1 over 1 plus
s divided by j omega sub c raised to the 2N power.
Well, what we recognize is that this expression has poles,
if we solve this equation.
This expression has poles at of course the roots
of the denominator polynomial or at s sub p equal to minus 1
to the over 2N times j omega sub c.
That's simply solving the equation with this denominator
polynomial equal to 0.
Well, we can see more specifically
where these poles lie in the s plane
by recognizing that these are the 2N roots of minus 1.
And then we have a j, which corresponds
to e to the j pi over 2.
So we can rewrite this expression for the poles
as s sub p equal to e to the j pi over 2N.
That's one of the roots of minus 1, plus k times 2 pi
where k is an integer.
This generates the roots of minus 1.
We need to multiply by j, and we can do that
by multiplying by e to the j pi over 2, which
is equal to j times omega sub c, which is the cutoff frequency
parameter for the Butterworth filter.
So this then specifies where the pole locations
are for the Butterworth filter.
For example, if we consider the case,
let's say with n equal to 3, k let's say equal to 0.
That says then that we have a pole at e
to the j pi divided by 6, e to the j pi over 6, times
e to the j pi over 2.
In other words, we have a pole in the s plane
at an angle of pi over 6 plus pi over 2,
which corresponds to this location.
Then we'll have poles distributed in angle
around a circle in the s plane, distributed
in an angle where the angular spacing is pi over 3.
The pi over 3 arising from a consideration of 2 pi k
divided by 2N, where in this case, N is equal to 6.
So these poles are distributed equally spaced in angle
on a circle in the s plane.
For N equal to 3, the angular spacing is pi over 3.
For k equal to 0 that first pole corresponds to this one,
and then we continue around.
Now these are poles distributed on a circle, which
tends to look like the z plane.
But I stress that what we're talking about here
is the s plane.
In particular, we're talking about the analog frequency
response design as the class of analog Butterworth filters,
which we're then going to map to digital filters.
So this is a circle in the s plane.
It's often referred to as the Butterworth circle.
And it has a radius which is equal to omega sub c.
In designing an analog Butterworth filter then,
the procedure is to appropriately choose
the parameters omega sub c and capital
N. Those are the parameters of the analog Butterworth filter.
And then from the poles distributed on the Butterworth
circle, this gives us the pole locations for the product,
H sub a of s times H sub a of minus s.
We can then simply associate with the transfer function
H sub a of s.
The poles in the left half plane,
to generate a stable filter, and then the ones which
are contributed by the factor H sub a of minus s
are the corresponding image poles in the right half plane.
The procedure then is to design a squared magnitude function.
Talking about a Butterworth filter,
we choose these parameters omega sub c
and N. That tells us where the poles are located
on the Butterworth circle.
From the pole locations for the squared magnitude function,
then we can factor that into H sub a of s times H sub
a of minus s, by choosing the poles that
fall in the left half plane, to give us
a causal, stable analog filter.
All right.
Well, that is an analog filter.
What we would like to talk about is
the design of a digital filter.
And the procedure that we're going to go through
is to choose a set of specs for the digital filter,
and then decide on what the corresponding specifications
should be for the analog filter, depending on whether we're
using impulse invariance or the bilinear transformation,
design the analog filter, and then map
the result to a digital filter using
one of those two techniques.
So let's choose a set of digital filter specifications
for a digital lowpass filter with a passband which
occurs for frequencies now in the digital domain,
for frequencies below 0.2 pi, and a stopband region that
occurs for frequencies greater than 0.3 pi,
and the transition band then, of course, is the region
in between those.
So, we would like to design a digital filter, which
stays within some limits in this passband region,
drops below some limit, delta sub s in the stopband region.
And we'll choose as the limits in the passband,
unity for the upper limit and 1 minus delta sub p for the lower
limit.
Where delta 1 minus delta sub p might not
and in fact it won't turn out to be 1 over the square root of 2,
in other words, this cutoff frequency
will not necessarily correspond to the parameter omega sub
c for the Butterworth filter.
All right.
Well, so let's choose a set of specifications.
The ones that I've chosen are the requirement
that the frequency response characteristic be within 1db
of unity in the passband.
In other words, the requirement that 1 minus deltas sub
p be greater than or equal to minus 1db.
So this deviation must be less than 1db.
And then the requirement that in the stopband
the frequency response characteristic
is below minus 15db.
In other words, the stopband attenuation
for frequencies greater than 0.3 pi is greater than minus 15db.
It's further down than minus 15db.
So delta sub s is less than or equal to minus 15db.
The one other specification that we'll impose,
or in fact, it's really a consequence that
will result from using a Butterworth filter,
is the fact that the frequency characteristic is monotonic.
So we have the passband edge, the stopband edge,
the passband ripple or deviation,
and the stopband ripple or deviation.
And we can, of course, rewrite these specifications
in terms of requirements on the transfer function or frequency
response, the magnitude of the frequency response.
The requirement that the deviation be less than 1db
in the passband is the requirement then
that 20 log to the base 10 of the magnitude of the frequency
response--
that's the deviation db--
be greater than or equal to minus 1, or that the magnitude
be greater than or equal to 10 to the minus 0.05.
And of course, we wanted also to be less than unity,
and that's an implied condition.
So it must be less than unity, and it
must be greater than or equal to 10 to the minus 0.05
for frequencies from 0 to 0.2 pi.
So in particular, since it's monotonic,
we'll impose that condition at the frequency point 0.2 pi,
corresponding to the end of the stopband
and the beginning of the transition band.
The second condition that we impose
is the condition that by the time
we get to omega equal to 0.3 pi, we
would like the frequency response to be down at least 15
db, and more if possible.
But the condition is that the 20 log
to the base 10 of the magnitude at omega equal to 0.3 pi
must be less than or equal to minus 15.
So this says we must be further down than 15 db,
or we can translate that into a condition on the magnitude,
so that the magnitude of that frequency is less than
or equal to 10 to the minus 0.75.
Well, let's consider two--
we want to consider two design procedures.
Let's first consider the impulse invariant design procedure,
in which case we would like to convert these specifications
on the digital filter to corresponding specifications
on the analog filter.
Now, we know that from discussions
that we've had previously that if we have an analog frequency
response and we apply impulse invariance to the impulse
response, the digital frequency response consists
of the analog frequency response scaled in frequency so
that capital Omega is equal to little omega divided
by capital T. And then that scaled replica of H
sub a added to itself delayed by little omega equal to 2 pi, 4
pi, et cetera.
That is multiples of 2 pi.
So, that means then that the digital filter is scaled
repeated replicas of the analog filter,
and so we would choose the analog filter
to have similar specifications to the digital filter,
but with the corresponding frequency parameters dictated
by this transformation, capital Omega equal to little omega,
divided by capital T. Well, that of course neglects aliasing,
and in fact as we go through the impulse invariant design,
we'll initially neglect aliasing.
And a rationale for doing this or a reason why, in fact, it's
not unreasonable to do that, will emerge
as we go through the example.
So for the time being, in fact, we're
going to neglect aliasing, so that we'll
think of the analog and digital frequency responses
as simply related by this substitution of frequency
variable, capital Omega equal to little omega
divided by capital T.
Well, that leads us then to the analog frequency response
which I've depicted here.
We had a factor of 1 over capital T in the amplitude
of this expression.
So for the digital filter to result
with an amplitude of unity at dc,
we would scale the analog frequency response by capital
T. So the important values then in capital Omega
are the ones corresponding to little omega equal to 0.2 pi,
or since capital Omega is equal to little omega divided
by capital T. We have a passband edge occurring at capital
Omega equal to 0.2 pi over capital T,
and a stopband edge equal to 0.3 pi over capital T.
At this passband edge, according to the specifications
on the digital filter that we want,
we would like the frequency response for frequencies
below this--
In other words, in the passband--
to be between capital T and capital T times 10
to the minus 0.05.
That's the guarantee that we have less than 1 db
ripple in the passband.
And for frequencies greater than 0.3 pi,
we would like to be below capital T times 10
to the minus 0.75.
That's the guarantee that we're down 15 db or more
in the stopband.
The way I've drawn it here, I've implied that I will exactly
meet the specifications at the passband edge,
and exactly meet the specifications
at the stopband edge.
You can imagine, of course, that if I did that and then
converted to a digital filter using impulse invariance
that there would be an effect due to aliasing, which
in fact would result in my not meeting
exactly the specifications at the stopband edge.
And that's a point that we'll deal with in a few minutes.
All right, well now we have our specifications.
Let me incidentally point out that there's
this parameter capital T that we're carrying around.
And as I'll justify partly on the basis of this example,
the parameter capital T is an irrelevant parameter.
It's a parameter that will disappear.
We'll carry it through with us for this one example.
But for other examples and problems
that you work through the study guide,
you should have absorbed the fact
that capital T is irrelevant, and in fact capital
T can be chosen equal to unity.
But let's carry it through in this example,
and just see how it disappears.
All right.
Well, so we want the analog Butterworth filter
meaning these specifications.
And we want it at omega equal to 0
to have a gain magnitude squared function to be T squared,
so that the magnitude is T. So here we
have the form then of the analog Butterworth filter.
And in order to meet these specifications
at this frequency and this one, we then have two equations.
One equation says that at capital Omega
equal to 0.2 pi over capital T, we
would like the value of the magnitude function
to be T times 10 to the minus 0.05,
which implies that this denominator should
be equal to 10 to the 0.1.
Second of all, at the stopband frequency,
0.3 pi divided by capital T, the fact
that we want the magnitude to be T times 10 to the minus 0.75
implies the magnitude squared is of course the square of that.
And we want then the denominator to be equal to 10 to the 1.5.
So we have two points that we're trying to nail down.
We have two parameters, omega sub c and N.
From these two points that we're trying to nail down,
we get two equations.
And we can now solve these equations for our parameters,
omega sub c and capital N.
Well, if we do that, and it's a little tedious to do it,
let me point out incidentally since we're
trying to keep track of this capital T,
that capital T drops down into the numerator
and so, in fact, we can solve for capital N and omega sub
c times capital T. In other words,
we can just move this down here algebraically.
And so we have then instead of the parameter omega sub c,
we have the parameter omega sub c times capital T.
If you grind through these two equations,
what results is a value for N equal to 5.8858 and probably
some other digits.
And omega sub c capital T equal to 0.70474.
And we could imagine that at this point
we're done, except that the parameter N
corresponding to the order of the filter, of course,
has to be an integer.
So, what that means is that we can't use the value of N
as I've indicated here.
We need an integer value.
Also if we were to round this down or truncate
it down to a choice of N equal to 5,
then since the order is lower you
can imagine that the filter characteristics
won't be as good.
In other words, we're not going to meet these specifications.
So the other choice then is to use instead of this value of N,
use a value of N equal to 6.
In other words, round up to the next integer value.
And if we do that then these two equations
will no longer be satisfied.
We're now using a higher order filter
than these equations would dictate.
Of course, we're using an integer order,
because that's what we need.
We're using a sixth order filter.
And what that means is that if we choose N equal to 6,
we can satisfy this equation, or we can satisfy this equation,
but we can't satisfy both of them.
Well, which one should we try to meet
and which one should we exceed?
Well, we know that with impulse invariance
there is the issue of aliasing.
And so in fact, in impulse invariant design
it will often happen that although we design exactly
to meet the stopband specifications,
the stopband specifications won't
be met because of aliasing.
Consequently, given the flexibility that we have,
due to the fact that we can actually implement a higher
order filter than these equations
dictate, to compensate for aliasing
we can exceed the stopband specifications
and meet exactly the passband specifications.
So if we substitute for N, N equals 6
into this equation that then requires
that omega sub c T be such that we exactly
meet the passband specifications, in which case
this equation won't be met.
In fact, what will happen at the stopband
edge is that the stopband specifications
will be exceeded.
If we do this, then we have for the equation
that we solve for omega sub c T from,
the equation which specified the passband edge, which is where
we're going to meet the specs.
And what results then is for omega sub c T, this value,
rather than the previous value that had been obtained
from the two equations.
Well, from these parameters then,
we have that omega sub c is 0.7032 divided by capital T.
We can substitute these parameters into the form
for the squared magnitude function, H sub a of s times H
sub a of minus s, and actually it's
this raised to the 2N, which is the 2 times
sixth power, which is the 12th power.
So this then is the form for the analog squared magnitude
function.
Now notice that the parameter capital T still
hasn't disappeared.
We still have it.
We have it there, and we have it here.
But notice that when we substitute in for omega sub c,
this number divided by capital T if you
track through that algebra, capital T
reappears in this expression as a multiplier on s.
That is we had 0.7032 divided by capital T.
That gets reflected back up into this numerator,
so that we have s times capital T. Well,
let's see what that means.
Suppose that capital T were equal to 1.
And we got an analog impulse response, H sub a of T,
or the system function H sub a of s.
If instead we reinserted capital T
and considered H sub a of s times
capital T multiplied by capital T
that impulse response would be related to this one
by simply a time scaling.
In other words, this one would be
replaced by H sub a of t divided by capital T. Well,
let's look at this and ask what happens when we sample it
at samples spaced by capital T.
So, we look at this impulse response
with T replaced by n times capital
T, as we've specified here.
And we observe that no matter what we pick capital
T to be that parameter disappears.
Because little t equal to n times
capital T gives us samples H sub a
of n, which are exactly the same samples that we would
get no matter how we chose the parameter capital T,
or equivalently whether or not we chose the parameter
capital T equal to unity.
I think that it requires a little reflection
to absorb this point.
It's basically tied to the fact that the frequency response
is specified for the digital filter,
not for the analog filter.
And specifying the frequency response
for the digital filter, in effect,
specifies capital Omega times capital T. Well,
I leave it to you to think about this a little.
But the important point is that, in fact this parameter
T, if we specify the filter in terms
of digital specifications, this parameter capital T
is really irrelevant.
All right.
Well, so here is our analog design.
And we can obtain H sub a of s from this,
according to the method that we talked
about previously taking the left half plane
poles of this expression.
Converting those to a digital filter
by using the transformation that we discussed
in the previous two lectures, expanding,
for example, in terms of residues and poles,
and then applying the impulse invariant transformation.
And if we do that, what results is a digital lowpass filter,
which was the lowpass filter that we set out to design.
And let me just show how those specifications,
or how that design comes out.
So this, for the example that we've been talking about,
is a graph of the frequency response characteristics.
Here is the magnitude as a function of digital frequency
omega.
And here is the same frequency response
plotted in terms of db.
And we see that we had in the specifications
a passband edge at 0.2 pi, a stopband edge at 0.3 pi.
At 0.2 pi we asked that the gain be down by 1db.
And of course, it's hard to tell exactly that we're down 1db,
but in fact we are.
At omega equal to 0.3 pi, our specification
was that we wanted to be down by 15 db.
In fact, we overcompensated, because of the fact
that we had to round the order of the filter up to an integer
value.
And you can see, in fact, that if you look at this carefully,
we exceed the specifications of 0.3 pi by a little bit.
In other words, we're down by a little more than 15 db.
In fact, for this particular filter
you can imagine that the effective aliasing is, in fact,
rather minimal.
Because this frequency response, by the time we get out to pi
is essentially 0.
Here we're down to minus 75 db.
And that's, of course, the amount
that would get aliased back in.
So this is the resulting filter frequency characteristics
for the impulse invariant design where
we exceed slightly the stopband edge specifications.
Now, the next thing that I'd like to go onto
is the corresponding design, the design
or the corresponding Butterworth filter,
now using the bilinear transformation
rather than using impulse invariance.
I remind you that the basic idea with the bilinear
transformation, as we've discussed in the last lectures,
is that we basically choose the analog specifications
by pre-warping the critical frequencies
for the digital filter through the nonlinear distortion
curve corresponding to the bilinear transformation.
So, we would start here for the digital filter.
Of course, here we're talking about a monotonic filter rather
than one that ripples.
But we have a passband edge at 0.2 pi,
a stopband edge at 0.3 pi.
We reflect those critical frequencies
through the arctangent curve to obtain
the corresponding critical frequencies
for the analog Butterworth filter.
When the analog Butterworth filter is designed,
we then obtain the digital Butterworth filter
by transforming this filter back through the bilinear
transformation.
So, let's look at how this design works then.
We have, as we've just reflected on that curve,
we begin with the digital frequency specifications,
reflect those through the bilinear warping curve
to the analog specifications, and consequently
the critical frequencies--
and let me emphasize first of all that
just as with impulse invariance, you
can show that the parameter capital
T is an irrelevant parameter.
We could, in fact, carry it all the way through.
And what would happen is that we would find that no matter
what we chose capital T to be, the digital filter design
that resulted would be exactly the same.
So in fact, for this example I'll work through it simply
with capital T equal to 1.
All right.
With capital T equal to 1, we reflect
these critical frequencies into this passband frequency
edge and this stopband edge.
And consequently, the analog filter--
this is the analog filter--
the analog filter that we design is one that
meets these specifications.
And consequently we have, again, the two equations
that say that at the passband edge,
we want the magnitude to be within 1db of unity.
And at the stopband edge, we want the magnitude
to be less than, down by more than 15 db.
And the two equations that result
are these, exactly the same as we had in the impulse invariant
case, except that the frequencies are different.
So now substituting in for the form of the analog Butterworth
filter into these expressions, and setting this up
for equality in these expressions,
what results then are the two equations which
are identical to the two equations
that we had for the impulse invariant design,
except that the frequencies at which the passband and stopband
edges of the analog Butterworth filter fall are different.
They're different because of the warping of the bilinear curve,
or the pre-warping that we have to apply.
So, these are the two equations.
And if we solve these two equations for capital N,
the resulting value for capital N is 5.3 and some digits.
Notice incidentally that it's less
than it was for the impulse invariant design.
And in general that's a characteristic
of the bilinear transformation.
But so it's less, but it is not an integer.
So the first thing we can ask again
is, could we just simply truncate this down or round it
down to the nearest integer value, in other words,
N equals 5.
But of course if we did that, then
we wouldn't be able to pick omega sub c,
so that these equations were satisfied or so
that the specifications were met or exceeded.
So in fact, we need to choose, we have to choose N equal to 6.
We have to choose the next highest integer value.
Choosing N equal to 6, we then have the choice
of exactly satisfying this equation,
or exactly satisfying this equation, or in fact
not satisfying either one of them,
but playing back and forth between the two.
Well let's do it in a particular way.
We can, for example, exactly meet
the stopband specifications.
In other words, we can use N equals 6 in this equation,
and solve for omega sub c, in which case
the passband specifications will be exceeded.
Recall that in the impulse invariant design
we did it the other way around.
We met exactly the passband specifications, in which case
the stopband specs were exceeded.
But there we had a good reason for doing that.
The reason that we did that is because we knew in advance
that in any case the passband specs wouldn't be met exactly,
because of the effect of aliasing.
For the bilinear transformation, we
don't have a problem with aliasing.
So, in fact, we're free to play with this flexibility in a more
general way.
All right.
Well, in any case then if we meet the stopband specs,
substituting N equals 6 in this equation,
and solving for omega sub c; we end up
with a value for omega sub c, which is 0.76622.
And so this then gives us the parameters omega sub c
and N for the analog Butterworth filter.
We can then use those specifications,
again to generate the pole pattern
for the squared magnitude function H sub a
of s times H sub a of minus s.
Factor this by choosing the left half plane poles in s plane,
and then mapping this function to a digital filter using
the bilinear transformation.
And let's now look at how this filter comes out as compared
with the digital Butterworth filter that we obtained
by using impulse invariance.
Well again, I have a display of the magnitude of the frequency
response and also the gain in db.
From what you may recall from the curve
that I showed previously, this looks roughly the same.
It's a monotonic filter and drops off.
It's down by some amount at 0.2 pi,
and the stopband is at 0.3 pi.
In particular, according to the specifications,
we wanted this to be down by no more than 1db at 0.2 pi.
Well, it's hard again to say that this is 1db or less.
But if we look at 0.3 pi, we can see
that this is in fact exactly down by 15 db.
And that's the way that we designed it.
We designed it to exactly meet the stopband cutoff
and exceed the passband cutoff.
And then the frequency response continues on from there.
It's interesting, in fact, to compare the design
that we got from the bilinear transformation with the design
that we got previously with impulse invariance.
And let me do that by superimposing the two of them.
So that the one that I'm putting on now
is the impulse invariant design, and let
me just lay them on top of each other.
OK.
You can see incidentally that looking at the frequency
response on a magnitude plot there, in fact,
are some differences.
This line has gotten a little thicker.
That is, the two don't exactly superimpose.
But you can't really see the differences.
Whereas you can when we look at this on a logarithmic plot.
In particular, this is the frequency response
that corresponded to the bilinear transformation.
This is the frequency response that corresponded
to impulse invariance.
And this is, in general, a characteristic
of the use of the bilinear transformation.
It's a result essentially of the frequency distortion,
because we've taken the frequency
axis and the continuous time frequency axis.
As you run along it at a constant rate,
it gets compressed more and more along the digital frequency
axis.
That wasn't well said.
But what I really mean to say is as we walk along here
at a constant rate, the farther we walk, the faster we're
running along the j omega axis.
And that's why, in fact, the use of the bilinear transformation
on a lowpass filter will give better attenuation
at the higher frequencies than impulse invariance will.
And I don't know if you can really
see the fact that the passband specs were exceeded
for the bilinear design.
In fact, when you look at these view graphs in the study guide,
I think you'll see that for impulse invariance
the curve is slightly lower than it is for the bilinear
transformation
OK well, look.
This is a simple example that hopefully at least illustrates
for you the basic procedure involved
in carrying through a design of an infinite impulse response
filter using impulse invariance and the bilinear
transformation.
There are other examples that are considered in the text.
There is a discussion of digital Chebyshev filters,
in other words digital designs beginning
from an analog Chebyshev filter.
And you'll also have an opportunity
to carry through some designs in the study guide.
A point that I would like to mention briefly
is that most of the discussion, in fact
all of the discussion that we've gone through,
has focused entirely on lowpass filters.
In fact, the discussion generalizes
in a fairly straightforward way to bandpass, bandstop,
highpass, et cetera filters.
In particular, as is discussed in some detail in the text,
there are a class of frequency transformations
that allow lowpass designs to get
converted to bandstop, bandpass, highpass, et cetera designs.
So this concludes our discussion of design procedures
for recursive or infinite impulse response
digital filters.
In the next lecture I'll discuss the design of finite impulse
response filters, and in particular
introduce a number of designs, one of which
corresponds to a class of algorithmic designs.
Thank you.
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