第 26 讲：反馈应用实例：倒立摆
本讲内容：开环系统分析及反馈动态特性的选择；用二阶线性常系数微分方程描述系统行为；根轨迹分析；结合比例反馈与微分反馈实现倒立摆稳定；轨道上倒立摆的演示，以及改变动态特性和阻尼特性产生的影响。
授课教师：艾伦·V. 奥本海姆教授
英文字幕稿
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[MUSIC PLAYING]
PROFESSOR: Last time, we began a discussion of feedback.
And I briefly introduced a number of applications.
For example, the use of feedback in compensating for
non-ideal elements in a system.
And also, as another example, the use of feedback in
stabilizing unstable systems.
In this final lecture, what I'd like to do is focus in
more detail on the use of feedback to
stabilize unstable systems.
And the specific context in which I'd like to do that is
in the context of the inverted pendulum.
Now, I've referred to the inverted pendulum in several
past lectures.
And, basically, as everyone realizes, a pendulum is
essentially a rod with a weight on the bottom.
And, naturally, an inverted pendulum is just
that, upside down.
And so it's essentially a rod that's top-heavy.
And the idea with the inverted pendulum is, although by
itself it's unstable around a pivot point at the bottom, the
idea is to apply an external input, an acceleration,
essentially to keep it balanced.
And we've all probably done this type of thing at some
point in our lives, either childhood or not.
And so that's the system that I'd like to analyze and then
illustrate in this lecture.
Now, the context in which we'll do that is not with my
son's horse, but actually with an inverted pendulum which is
mounted on a cart.
And first I'd like to go through an analysis of it, and
then we'll actually see how the system works.
So basically, then, the system that we're talking about is
going to be a system which is a movable cart.
Mounted on the cart is a rod with a weight on the top.
And this then becomes the inverted pendulum.
And the cart has an external acceleration applied to it,
which is the input that we can apply to the system.
And then, in general, we can expect some disturbances.
And I'm going to represent the disturbances in terms of an
angular acceleration, which shows up around the weight at
the top of the rod.
So if we look at the system, then, as I've kind of
indicated it here, basically we have two inputs.
We have an input, which are the external disturbances,
which we can assume that we have no control over.
And then there is the acceleration that's apply to
the cart externally, or in the case of balancing my son's
horse, it's the movement of my hand.
And then, through the system dynamics, that ends up
influencing the angle of the rod.
And we'll think of the angle of the rod
as the system output.
And basically the idea, then, is to try to keep that angle
at 0 by applying the appropriate acceleration.
Now, as I've mentioned several times previously, if we know
exactly what the system dynamics are, and if we know
exactly what the external disturbances are, then
theoretically we can choose an acceleration for the cart
which will exactly balance the system, or balance the rod.
However, since the system is inherently unstable, any
deviation from that model-- in particular, any unanticipated
external disturbances--
will excite the instability.
And what that means is that the rod will fall.
So the idea, then, is, as we'll see, to stabilize the
system by placing feedback around it.
In particular through a measurement of the output
angle, processed through some appropriate feedback dynamics,
using that to control the acceleration of the cart.
And if we choose the feedback dynamics correctly, then, in
fact, we can end up with a stable system, even though the
open-loop system is unstable.
Well, as a first step, let's analyze the system in its
open-loop form.
In particular, let's analyze the system without feedback,
demonstrate that, indeed, it is unstable, and then see how
we can stabilize it using feedback.
So the system, then, is--
as we've indicated, the variables involved are the
measured angle, which represents the output, the
angular acceleration due to external disturbances, and
then there is the applied external
acceleration on the cart.
So these represent the variables, with L representing
the length of the rod, and s of t I indicate here, which we
won't really be paying attention to, as the position
of the cart.
And without going into the details specifically, what
we'll do is set up the equation, or write the
equation, in terms of balancing the acceleration.
And if you go through that process, then the basic
equation that you end up with is the equation that I
indicate here.
And this equation, then, tells us how the basic forces, or
accelerations, are balanced, where this is the second
derivative of the angle.
And then we have the acceleration due to gravity,
reflected through the angular acceleration times the sine of
the angle, the angular acceleration due to the
external disturbances, and finally, the angular
acceleration due to the motion of the cart.
Now, this equation, as it's written here, is a nonlinear
equation in the angle theta of t.
And what we'd like to do is linearize the equation.
And we'll linearize the equation by making the
assumption that the angle is very small, close to 0.
In other words, what we'll assume is that we're able to
keep the rod relatively vertical.
And our analysis, because we're linearizing the
equations, will obviously depend on that.
So making the assumption that the angle is small, we then
assume that the sine of the angle is approximately equal
to the angle, and the cosine of the angle is approximately
equal to 1.
That will then linearize this equation.
And the resulting equation, then, in its linearized form,
is the one that I indicate here.
So we have an equation of motion, which linearizes the
balance of the accelerations.
And what we see on the right-hand side of the
equation is the combined inputs due to the angular
acceleration of the rod and the acceleration of the cart.
And on the left-hand side of the equation, we have the
other forces due to the change in the angle.
So this is, then, the differential equation
associated with the open-loop system, the basic dynamics of
the system.
And if we apply the Laplace transform to this equation and
solve for the system function, then the basic equation that
we're left with expresses the Laplace transform of the
angle, equal to the system function, the open-loop system
function, times the Laplace transform of
the combined inputs.
And I remind you again that our assumption is that this is
an input that we have no control over.
This is the input that we can control.
And the two together, of course, form
the combined input.
Now, we can, of course, solve for the poles and 0's.
There are no 0's.
And there are two poles, since this is a second-order
denominator.
And looking at the poles in the s-plane then, we see that
we have a pair of poles, one at minus the square root of g
over L, and one at plus the square root of g over L.
And the important observation, then, is that while this pole
represents a stable pole, the right half-plane pole
represents an unstable pole.
And so this system, in fact, is an unstable system.
It's unstable because what we have is a pole in the right
half-plane for the open-loop system.
Now, before we see how to stabilize the system, let's,
in fact, look at the mechanical setup, which we
won't turn on for now, and just see essentially what this
instability means and what the physical orientation of the
equipment is.
So what we have, which you just saw a caricature of in
the view graph, is an inverted pendulum.
This represents the pendulum.
And it's mounted with the pivot point at the bottom.
It's mounted on a cart.
And here we have the cart.
And this is a pivot point.
And, as you can see, the cart can move back
and forth on a track.
And the external acceleration, which is applied to the cart,
is applied through this cable.
And that cable is controlled by a motor.
And we have the motor at this end of the track.
And so as the motor turns, then the cart will
move back and forth.
OK.
Now, since the system is not turned on
and there's no feedback--
in fact, the system, as we just saw in the transparency,
is an unstable system.
And what that instability means is that, for example, if
I set the angle to 0 and then let it go, then as soon as
there's a slight external disturbance, it
will start to fall.
Now, you can imagine that not only is the system unstable as
it is, but certainly can't accommodate changes in the
system dynamics.
For example, if we change, let's say, the
weight of the pendulum.
So if we thought, for example, of changing the weight by,
let's say, putting something like a glass with something in
it on top, and I think about trying to balance it by
itself, obviously that's hard.
In fact, I would say, without turning the system on,
guaranteed to be impossible.
So the basic system, as we've just analyzed it, is
inherently an unstable system.
The instability reflecting in the fact that the pendulum,
following its own natural forces, will tend to fall.
And now what we want to look at is how, through the use of
feedback, we can, in fact, stabilize the system.
So let's first, once again, look at the open-loop system.
And the open-loop system function that we saw was a
system function of this form, 1 over Ls squared minus g, and
that represents the system function associated with the
two inputs, one being the external disturbances, the
other being the externally applied acceleration.
And the system function here
represents the system dynamics.
And the resulting output is the angle.
Now, the feedback, the basic strategy behind the feedback,
is to in some way use the measured angle, make a
measurement of the angle, and use that through appropriate
feedback dynamics to stabilize the system.
And so with feedback applied around the system then, there
would be some measurement of the angle through some
appropriately chosen feedback dynamics.
And that then would determine for us the applied external
acceleration corresponding to what the motor does by pulling
the cable back and forth.
And we would like to choose G of s so that the overall
system is stable.
Now, as you recall from the lecture last time, with the
feedback around the system and expressed here in terms of
negative feedback, the overall transfer function then is the
transfer function associated with the basic feedback loop,
where H of s is the open-loop system and G of s corresponds
to the feedback dynamics.
Now, that's the basic strategy.
We haven't decided yet how to choose the feedback dynamics.
And that becomes the next step.
And we want to choose the dynamics in such a way that
the system ends up being stabilized.
Well, I think the thing to do is just simply begin with what
would seem to be the most obvious, which is a simple
measurement of the angle.
Let's take the angular measurements, feed that back,
let's say through a potentiometer and perhaps an
amplifier, so that there's some gain, and see if we can
stabilize the system simply through feedback which is
proportional to a measurement of the angle, what.
Is typically referred to as "proportional feedback." Well,
let's analyze the results of doing that.
Once again we have the basic feedback equation, where the
open-loop transfer function is what we had developed
previously, given by the second-order expression.
Now, using proportional feedback, we choose the
acceleration a of t to be directly proportional through
a gain constant or attenuation constant K1, directly
proportional to the measured angle theta of t.
And consequently the system function in the feedback path
is just simply a gain or attenuate K1.
So this, then, is the system function for
the feedback path.
Well, substituting this into the closed-loop expression,
then the overall expression for the Laplace transform of
the output angle is, as I indicate here, namely that
theta of s is proportional through this system function.
So the Laplace transform of the input X of t-- and let me
remind you that X of t, the external disturbances, now
represent the only input, since the other input
corresponding to the applied acceleration to the cart is
now controlled only through the feedback loop.
So we have this system function, then, for the
closed-loop system.
We recognize this once again as a second-order system.
And the poles of this second-order system are then
given by plus and minus the square root of g minus K1,
divided by L. So K1, the feedback constant, clearly
influences the position of the poles.
And let's look, in fact, at where the
poles are in the s-plane.
Well, first of all, with K1 equal to 0, which, of course,
is no feedback and corresponds to the open-loop system, we
have the poles where they were previously.
And if we now choose K1, let's say less than 0, then what
will happen as K1 becomes more and more negative, so that
this term is more and more positive, is that the left
half-plane pole will move further into the left
half-plane, the right half-plane pole will move
further into the right half-plane.
So clearly with K1 negative, this pole, which represents an
instability, becomes even more unstable.
Well, instead of K1 negative, let's try K1 positive and see
what happens.
And with K1 positive, what happens in this case is that
the left half-plane pole moves closer to the origin, the
right half-plane pole moves closer to the origin.
What one would hope is that they both end up in the left
half-plane at some point.
But, in fact, they don't.
What happens is that eventually they both reach the
origin, split at that point, and travel along the
j-omega-axis.
Now, this movement of the poles, as we vary K1 either
positive or negative, what we see is that with the open-loop
system, as we introduce feedback, those basic poles
move in the s-plane, either this way, as K1 becomes more
negative, or for this particular case, if K1 is
positive, they move in together, split, and move
along the j-omega-axis.
And that locus of the poles, in fact, is referred to in
feedback terminology as the "root locus." And as is
discussed in much more detail in the text, there are lots of
ways of determining the root locus for feedback systems
without explicitly solving for the roots.
For this particular example, in fact, we can determine the
root locus in the most straightforward way simply by
solving for the roots of the denominator
of the system function.
Now, notice that in this case, these poles, with K1 positive,
have moved together.
They move along the j-omega-axis.
And before they come together, the
system is clearly unstable.
Even when they come together and split, the system is
marginally stable, because, in fact, the system
would tend to oscillate.
And what that oscillation means is that with the
measurement of the angles, essentially if the poles are
operating on the j-omega-axis, what will happen is that
things will oscillate back and forth.
Perhaps with the cart moving back and forth trying to
compensate, and the rod sort of moving in
the opposite direction.
In any case, what's happened is, with just proportional
feedback, we apparently are unable to
stabilize the system.
Well, you could think that the reason, perhaps, is that we're
not responding fast enough.
For example, if the angle starts to change, perhaps, in
fact, we should make the feedback proportional to the
rate of change of angles, rather than
to the angle itself.
And so we could examine the possibility of using what's
referred to as "derivative feedback."
In derivative feedback, what we would do is to choose, for
the feedback equation, or for the feedback system function,
something reflecting a measurement of the derivative
of the angle.
And so here again, once again, we have the open-loop system
function with derivative feedback.
We'll attempt to--
or we will use this feedback.
And acceleration, which instead of being proportional
to the angle, as it was in the previous case, an acceleration
which is proportional to the derivative of the angle.
And so the basic feedback dynamics, or feedback system
function, is then G of s is equal to K2 times s.
The multiplication by s reflecting the fact that in
the time domain it's measuring the derivative of the angle.
The associated system function is then indicated here.
And if we solve again this second-order equation for its
roots, that tells us that the location of the poles are
given by this equation.
And so now what we would want to look at is how these poles
move as we vary the derivative feedback constant K2.
So let's look at the root locus for that case.
And first, once again, we have the basic open-loop poles.
And the open-loop poles consist of a pole on the left
half-plane and a pole on the right half-plane,
corresponding in this equation to K2 equal to 0.
That is, no feedback in the system.
If K2 is negative, then what you can see is that this real
part will become more negative.
Since K2 is squared here, this is still a positive quantity.
And, in fact, then with K2 less than 0, the root locus
that we get is indicated by this.
Now, notice, then, that this right half-plane pole is
becoming more unstable.
And the left half-plane pole likewise is
becoming more unstable.
And this point, by the way, corresponds to where this pole
ends up, or where the root locus ends up, when K2
eventually becomes infinite.
So clearly K2 negative is not going to stabilize the system.
Let's try K2 positive.
And the root locus dictated by this equation, then, is what I
indicate here.
The left half-plane pole moves further into the left
half-plane.
That's good.
That's getting more stable.
The right half-plane pole is moving closer to the left
half-plane, but unfortunately never gets there.
And, in fact, it's when K2 eventually becomes infinite
that this pole just gets to the point where the system
becomes marginally stable.
So what we found is that with proportional feedback, we
can't stabilize the system, with the derivative feedback,
we can't stabilize the system, by themselves.
And a logical next choice is to see if we can stabilize the
system by both measuring the angle and at the same time
being careful to be responsive to how fast that angle is
changing, so that if it's changing too fast, we can move
the cart or our hand under the inverted
pendulum more quickly.
Well, now then what we want to examine is the use of
proportional plus derivative feedback.
And in that case, we then have a choice for the acceleration,
which is proportional with one constant to the angle and
proportional with another constant to the
derivative of the angle.
And so the basic system function, then, with
proportional plus derivative feedback, is a system function
which is K1 plus K2 times s.
We then have an overall closed-loop system function,
theta of s, which is given by this equation.
And so the roots of this equation then represent the
poles of the closed-loop system.
And those poles involve two parameters.
They involve the parameter K2, which is proportional to the
derivative of the angle, and the constant K1, which is
proportional to the angle.
And we'll, first of all, examine this just with K2
positive, because what we can see is that as we vary K1, if
K2 were negative, that would, more or less immediately, put
poles into the right half-plane.
And the more negative K2 got, the larger this
term is going to get.
So, in fact, as it will turn out, we can stabilize the
system, provided that we choose K2 to be
greater than 0.
All right.
Now, with K2 greater than 0, what happens in the location
of the poles is that if K2 is greater than 0, and we choose
K1 greater than 0--
I'm sorry.
We choose K1 equal to 0, then in effect the influence of K2
is to shift the poles of the open-loop system slightly.
And so with K2 greater than 0 and K1 equal to 0, we have a
set of poles, which are indicated here, and so this is
just a shift, slight shift, to the left, depending on the
value of K2, a shift to the left of the open-loop poles.
Now, as we vary K1, and in particular we're going to
choose K1 greater than 0, what happens is that the poles will
begin to move together, as they did previously when we
looked at the variation of K1.
The poles will move together, reach a point where we have a
second-order pole, and then those poles will split and
move parallel to the j-omega-axis.
So what's indicated here is the root locus.
And what this represents, then, as long as we make K1
large enough so that this poles moves into the left
half-plane, is that it represents
now a stable system.
OK.
So what we've seen is that proportional feedback by
itself or derivative feedback by itself won't stabilize the
system, whereas with the right choice of feedback constants,
proportional plus derivative feedback will.
And we saw that basically by examining the root locus in
the s-plane.
All right.
Well, let's actually watch the system in action.
And I described it to you previously.
Basically, an inverted pendulum on a cart.
And so I still have it off.
And, of course, we have the pendulum.
And as I indicated, it's pivoted at the base.
And the angle is measured by a potentiometer that we have
attached to the pivot point.
And the measurement of the angle is fed back through this
wire to a motor that we have at the other end of the table.
And then that motor basically is used to provide the
acceleration to drive the cart.
OK.
Well, let's turn it on.
And when we turn it on, it'll take just
an instant to stabilize.
And fortunately we have the constants set right, and there
we have now the stabilization of an unstable system.
Remember that with the feedback off, the system is
unstable because the pendulum will fall, whereas now it's
stabilized.
Now, also, as you can see, not only have we stabilized it,
but we're able to compensate through the feedback to
changes in the external disturbances.
For example, by tapping it, because of the feedback and
the measurement of the angle, it will more or less
automatically stabilize.
Now, in addition to being stable in the presence of
external disturbances, it also remains stable and remains
balanced even if we were to change the system dynamics.
And let me just illustrate that with the glass that we've
talked about before.
Let's first not be too bold, and we'll take the liquid out
of the glass.
And presumably if it can adjust to changes in the
system dynamics, then if I put the glass on, in fact, it will
remain balanced.
And indeed it does.
And let me point out, by the way, that I don't have to be
very careful about exactly where I position the glass.
And furthermore, I can change the overall system even
further by, let's say for example, pouring a liquid in.
And now let me also comment that I've changed the physics
of it a little bit.
Because the liquid can slosh around a little bit, it
becomes a little more complicated a system.
But as you can see, it still remains balanced.
Now, if we really don't want to be too conservative at all,
we could wonder whether, with the feedback constants we
have, we could, in fact, balance the
pitcher on the top.
And, well, I guess we may as well give that a try.
And so now we're changing the mass at the top of the
pendulum by a considerable amount.
And, again, the system basically can respond to it.
Now, this is a fairly complicated system.
The liquid is sloshing around.
We, in fact, as you can see, have an instability right now,
although it's controlled.
And that's because the physics of the dynamics has changed.
And we can put a little bit more mass into the system, and
maybe or maybe not that will cut down on the instability.
OK.
Well, in fact, what happened there is that we increased the
mass at the top of the pendulum slightly.
And that provided just enough damping to
stabilize the system.
OK.
Well, with this lecture and this demonstration, this
concludes this entire set of lectures.
It concludes it especially if the pitcher happens to fall.
But, seriously, this concludes the set of lectures as we have
put them together.
And let me just comment that, as a professor of mine once
said, and which I've never forgotten, the purpose of a
set of lectures, or of a course, or, for that matter
anything that you study, is not really to cover a subject,
but to uncover the subject.
And I hope that, at least to some degree, we were able to
uncover the topic of signals and systems through this
series of lectures.
There are a lot of topics that we got only a very brief
glimpse into.
And I hope that at least what we've been able to do is get
you interested enough in them so that you'll pursue some of
these on your own.
And so I'd like to conclude by thanking you, both for your
patience and your interest.
And I hope that you have enough interest to pursue some
of these topics further.
Thank you.
FILMING DIRECTOR: Okay.
That's a wrap.