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PROFESSOR: Today I want to talk a little bit about
designing control systems.
This will finish up our discussions
on signals and systems.
Let me then just briefly review where we are.
Hopefully this might help you also for perspective with
regard to thinking about the exam tonight.
We've looked at a bunch of different kinds of
representations for discrete time systems.
The easiest, most concise method we looked at was the
representation using a difference equation.
That's mathematically as concise as you can get.
But it doesn't tell you important things like who's
the input and who's the output and what are all the different
ways that you can get through the system
from input to output?
So for that question block diagrams are nice.
Block diagrams are graphical.
It makes it very easy to see when there is, for example, a
cyclic path through the network.
But they're graphic.
They're not nearly so concise as difference equations.
So then we went on to operators.
Operators are just as concise as difference equations but
they contain additional information because the
operators have an implicit argument.
So there's an input, which is the argument to the operator,
and there's an output, which is the output of the operator.
So you can tell who is the input and who is the output.
So that's good.
That sort of combines the strengths of difference
equations and block diagrams.
You end up with a concise representation that has
complete information about the signal flow paths.
Furthermore, you can analyze the operators by using
polynomial mathematics and that gives rise to the notion
of a system functional.
And that's a very nice closure because that represents an
abstraction that lets us think about a whole system as though
it were just one part, one thing, one operator.
So we use that structure then, all of those representations,
to try to learn about feedback.
And first off, in the block diagram it's very easy to see
that any time you have feedback -- feedback so
enormously powerful that we want to use it in design --
but you can see immediately from the structure of the
block diagram that if you have feedback then you have cycles.
Why is that interesting?
Well, that's interesting because if you have cycles
then even transient inputs can generate persistent outputs.
So that's a kind of behavior that we would like to
understand.
From the nature of feedback it generates cycles.
From the nature of cycles it generates persistent responses
even if there's no input.
And we saw that we could characterize those by thinking
about those responses for one part at a time.
And those parts we thought of as poles and the responses to
a single pole we called modes.
So we thought through a way of decomposing the response of a
complicated system in terms of a number of additive
components that are based on poles.
Poles are just the base of a geometric sequence.
So today then what I want to do is use that framework to
think about design.
How would you optimize the design of a controller?
Looking back to where we've been, way back in Lab 4,
ancient history, we looked at how you could program the
robot to approach a wall.
And we saw that depending on how you set up that system you
could get very different performances.
And what we'd like to do is have a way that we can design
for performance without actually building it.
The kind of thing that we built in the lab, Lab 4, was
so easy that building it to determine its behaviors was
not a bad problem.
But In general if you were building a 777, there's more
than one pole.
And you wouldn't necessarily want to test drive all of the
bad configurations.
So we'd like to be able to understand that kind of a
problem analytically.
We'd like to be able to analyze it.
So using the different representations you can
generate a very concise representation just thinking
in terms of difference equations.
You all did this in Lab 4.
And you get a single difference equation that tells
you in principle everything there is that you could know,
but not in a form that's very easy to analyze.
It's a bit better if you translate the difference
equation into a block diagram because now you can see that
this system of equations has in fact two
feedback loops in it.
Two cycles.
Two things that might potentially generate
persistent responses to transient signals, which could
then degrade performance.
If the transient signal lasts ten years it might be a bad
controller.
If the 777 hits turbulence and never stabilizes
that would be bad.
If small disturbances got bigger with time that might be
bad, right?
So we can see that this simple controller described by these
difference equations has the potential to do
that sort of stuff.
And we'd like to understand, when does it?
The easiest way to think about analyzing this is to focus
first on the inner loop and ask the question, what's the
functional representation for that box which we would call
an accumulator?
This box, this thing, accumulates at its output, the
sum of all the things that ever came in so we call it an
accumulator.
So what's the functional representation of an
accumulator?
Well, we just do polynomial math.
Easy so we can recognize from the block diagram that the
signal Y could be constructed by applying R to W.
But we can also see in the block diagram that W is the
sum of X and Y. And then if we take the left hand side and
the right hand side of this double equation we get
something that involves just X and Y, which we can solve for
the ratio Y over X. Which then says that the ratio is R over
(1 minus R).
That's a functional representation for the effect
of the accumulation.
That's also something that comes up so frequently in the
design of control systems that we give it a name.
We call this Black's Equation.
And it's especially useful to avoid these little trivial
steps in algebra, to just jump to the answer.
So let's see that everybody's following me.
The equation for this box is the thing that we will call
Black's Equation.
It's not mysterious.
It's something that you could derive, so derive it.
Figure out the functional form for the system that goes from
X to Y and figure out which of these forms is correct.
(1) through (4), or (5) if none of the above applies.
So take 30 seconds, figure out the answer.
I'm going to ask you to raise your hand.
You're free to talk to your neighbors.
OK.
So everybody who raised your hands tell me what the
right answer is.
OK, wonderful.
It's about 95%.
No.
It's about 100% correct out of about 95% participation.
So the answer you can form just like we did before, no
particular tricks, using simple algebra.
Simple algebra you get F over (1 minus FG).
The thing I want you to recognize is kind of a
graphical way of thinking about that.
And you can just memorize F over (1 minus
FG) and that's fine.
But there's some interesting things that the designer
thinks about.
This functional form is F, that's the forward gain.
That's the gain through the system starting at the input
and going directly to the output.
So this form says that the closed loop gain, the system
function that results when the loop is closed, is just the
forward gain, F, divided by (1 minus the loop gain).
The loop gain is the product around the loop once.
So that's just a convenient way of thinking about it.
Any time you have a feedback system of this type you can
think about the closed loop response as being the forward
path F divided by (1 minus loop gain FG).
So the answer was 1.
And generally we'll see two different ways of representing
the system.
Sometimes we'll represent it with a plus as we did with the
accumulator.
Many times we'll represent it with a minus.
That's the way we think about control problems.
We put in the minus because we'd like to think about the
controller as trying to make something go to 0.
So when you take the difference, that gives us an
error signal.
And then we can think about the controller being the thing
that drives that error to 0.
But however you think about it, there's two forms that are
very closely related.
They really differ by just a minus sign which you could
think of as just multiplying G. So it's sort of like the
right hand side is just a minus G plugged into
the left hand side.
So then the way you use this idea, you think about the
block diagram and you say, OK I can replace this thing with
an equivalent system which is R over (1
minus R) and then repeat.
So the R over (1 minus R) means that, if you think about
Black's Equation now for this loop, you should think about
the forward gain, K minus T, R over (1 minus R), that's this.
Divided by (1 plus the loop gain).
So 1 plus, and the loop gain, well, the wire down here just
has a gain of 1.
So the loop gain and the forward gain
are the same thing.
So you get this kind of expression which
simplifies to this.
There's two things I want you to see about this.
First off, I want you to see that even though the
simple-minded way of plugging in said that we should have
got a quotient of quotients N over D over N over D, it's
simplified to a single ratio.
If you design a system out of just adders, gains and delays,
that will always be true.
There's a closure.
It will always be the case that the functional that
represents a system of that form will always have the
property that it's a polynomial in R divided by
another polynomial in R, that's just the way
polynomials work.
The other thing is that you can now start to interpret
what the behavior of this could represent in a simpler
form than thinking about this.
So this kind of a representation that leads to
intuition about what the behavior should be is very
helpful when you're thinking about design.
And in particular this particular thing says that if
we think about a simpler system that could generate
that same response we can generate some intuition for
how we would like to set the parameter k.
So in particular this system functional which we generated
for this system could equally apply to that system.
Is that clear?
So there's a numerator, which I've represented here.
There's a numerator, which has an R in it and has
a minus kT in it.
I'm going to, for reasons that you'll see in a minute, I'm
going to call something P0 because it's the pole.
So the numerator I've represented here and this
denominator has this form.
And I wrote it that way because this is the canonical
form for the way of thinking about a pole.
So what I showed is that even though this was a more
complicated system you can think about it as the cascade
of a delay, a gain, and a pole.
The pole can be calculated from the gain, the pole is the
multiplier for R, so the pole is (1 plus kT).
So if I were to choose kT to be minus 0.2 then the pole
would be at 0.8.
If the pole is at 0.8 then the mode, the natural response to
the pole, would have the form P to the n.
They always have the form geometric P to the n, so it
would look like 0.8 to the n.
Because of the pre-multiplier of 1 minus P0 the whole thing
gets multiplied by 0.2 and because of the delay the whole
thing gets shifted to the right.
The important thing is that by thinking about manipulating
this as an operator we can recognize and simplify the
form of the behavior.
That gives us an intuitive grasp over how
to best choose kT.
It's all clear?
Now, the behaviors that we're interested in are not always
unit-sample responses.
We do unit-sample responses because they're the easiest
possible thing we could think of, right?
A unit-sample is the simplest non-zero signal.
A unit-sample is the signal that is different from 0 in
exactly one place --
the easiest possible place, 0.
And it has its easy-as-possible
non-zero value --
1.
So we focus on the unit-sample signal because it's the
easiest possible signal we could think about.
But when we're thinking about behaviors we're often thinking
about other things.
Often we'll think about the step response.
Here I have illustrated the way you would
measure a step response.
A step response is what would happen if the output were
initially 0, if we were at rest, and suddenly we turned
on a signal that was constant at 1.
That would happen in the robot case if we started the robot
close to the wall, at rest, near 0 --
so that the output signal is close to 0, but the desired
input was a meter behind.
Then that would be an input signal that started at time
equals 0 equal to 1 and persisted forever at 1.
And the result then would be what we
call the step response.
The step response is typically easier to measure in the lab
than is the unit-sample response.
So we use the unit-sample response when we're thinking
analytically, when we're doing calculations, and we use the
step response when we're in the lab
trying to measure something.
And the whole theory wouldn't be very useful if there
weren't a close relationship between those two things.
This diagram illustrates the relationship.
If we think about a system H for which we would like to
find the step response, the step response to that system
is what would the system do if you put the
unit-step into the system.
I've represented the unit-step here as u[n].
u[n], the signal that is 0 for n less than 0, and 1 for n
bigger than or equal to 0 --
is just the accumulation of the delta function, the
unit-sample.
So this system, the cascade of an accumulator with H, would
measure the step response of H if it were driven with the
sample signal.
Because of the properties of polynomials and because block
diagrams follow the rules for polynomials, we can flip these
whenever the systems both start at rest, and if we flip
those we get a different interpretation.
What this says is that if you were to take H and stimulate
it with the unit-sample you would get h, little h, which
we would call the unit-sample response because it's the
response to the system when the input is the unit-sample.
So if you measured h with the unit-sample rather then with
the unit-step you would get the unit-sample response from
which you could generate the step response by running it
through an accumulator.
So what that says is there's a close association, there's a
close relationship, between the unit-sample response and
the unit-step response.
One is the accumulation of the other.
The unit-step response is the accumulated response to the
unit-sample response.
So that means that in that previous example where we saw
that setting kT equal to minus 0.2 resulted in this
unit-sample response, that would correspond to this
unit-step response.
All you do is for every sample you calculate, for this
response you calculate the sum of say, n equals 0, you would
take the sum of all of the previous answers in H[n].
It's the accumulation.
So it starts at 0 since the sum of all those numbers is 0.
Then at time 1 it becomes the sum from here back so it
becomes 0.2.
Then here it's the sum from here back.
And if you add these all up it becomes a number that
approaches 1.
Not surprisingly, right?
If you've got a feedback system and if you started the
robot up against the wall and the desired position was one
meter behind you would monotonically approach 1, OK?
And what you can see is that if you change the value of the
pole, here I've changed the kT from minus 0.2 which is what
the previous answer was, to minus 0.8, I've changed the
value of the pole, the unit-sample
response got faster.
And the unit-step response also got faster.
The point is there are different kinds of performance
metrics that we might want to use, unit-sample response,
unit-step response, but the responses of all of them, you
can tell something about the response to all of them from
the response of the unit-sample signal.
That's why we focus so much on the unit-sample signal.
It's not because it's the most popular thing
to use in the lab.
It's because it's the easiest thing to calculate with and it
gives us insight into things that we would like to measure
in the lab.
So for this very simple system what you can show is that
there's only a few possible behaviors, a few
categories of behaviors.
If you were to choose kT to be between 0 and 1, then the
pole, which is (1 plus kT) would also be between 0 and 1.
Since the system has a single pole you can say a lot about
the response from the numerical value of the pole.
If the pole is between 0 and 1 then the response is going to
be monotonic and converging.
That results because the unit-sample response was
positive only and decayed towards 0.
Because it's positive only it means monotonic --
goes to 0 and makes it converge.
So you can infer properties about the unit-step's response
from properties of the pole just like we could infer
properties of the unit-sample response.
If you changed kT to be between minus 2 and 1 you get
a P0 that's between minus 1 and 0.
Again that's just that equation.
That says that the response will be alternating.
So the sign of the unit-sample response goes
positive then negative.
It still converges in the sense that the unit sample
response approaches 0.
And what that means for the unit step-response is that the
unit-step response will converge toward 1.
The sign doesn't alternate around 0, the sign
alternates around 1.
So again, you can infer the properties of the unit-step
response from the properties of the unit-sample response.
And if you have kT that is less than minus 2 then you get
a P0 that's less than minus 1 and
that's a divergent response.
So the point is you can infer properties about the control
system by thinking about the poles of the system where here
I've illustrated it for a simple system that
only has one pole.
OK.
I told you a bunch of facts.
Now you figure out something.
How would I choose k for this system to get the quote, "best
performance?"
So which value of kT would give the fastest convergence
for the unit-sample signal?
OK, participation is down but the hit rate is still good.
Virtually everybody who volunteered to answer got the
right answer.
The most popular answer was (2).
Why is the answer (2)?
What's the range of possibilities
that we could get?
If we choose k, or kT, we could choose kT to be--
what's the range of kT that we could use?
Minus 2 to 0?
[INAUDIBLE] we could use kT, any real number, right?
We couldn't use imaginary numbers because that doesn't
sort of make sense for a real system.
But we could choose any real number.
The real numbers map, according to this chart, the
real numbers map to a different real number.
If you choose kT you can figure out where is the pole
by that mapping.
Where would you put the pole to get the fastest response?
If you have your choice of putting the pole anywhere on
this red line, that red line or that red line, where would
you put it and why?
AUDIENCE: Just inside the [INAUDIBLE]
PROFESSOR: Putting it inside the unit circle would probably
be a good idea because?
AUDIENCE: [UNINTELLIGIBLE]
PROFESSOR: Yeah.
If you didn't put it inside the unit
circle it wouldn't converge.
That's right.
So you like [UNINTELLIGIBLE] the inside.
Given the choice of anywhere here and anywhere here which
would you choose?
How would you choose it?
Yeah.
AUDIENCE: Derive.
PROFESSOR: Derive.
And how do you get that?
AUDIENCE: [UNINTELLIGIBLE]
PROFESSOR: What would happen if it was close to
[UNINTELLIGIBLE]?
It converges quite quickly, right?
Poles always converge geometrically.
The base of the geometric is the pole value.
So you'd like the pole to be as small as possible to get
the convergence as fast as possible.
That make sense to everybody?
So in particular for this example if you chose kT to be
minus 1 in that limit then this entire factor goes away.
So the entire response degenerates to R and R is not
instantaneous but it's pretty fast.
What that says is that you get to the final
value in one step.
So if the input consisted of a unit sample, which has
non-zero value only at 0, the output would have non-zero
value only at 1, right?
Thinking about the way that works in practice, think about
the robot and think about we're trying to
drive toward the wall.
If we made kT be minus one, and just for the sake of being
concrete let me say that T is about 1/10.
That's what the sampling period is for the robots we
use in the lab.
If T were 1/10 then the best k would be minus 10.
And what that says is that if we were 1 meter away from
where we want to be we would set the velocity to 10 meters
per second.
What that says is that if we started 1 meter away from
where we want to be, so this is intended to represent
position on the same axis that this has showed, so if we
started here and we wanted to be here, time is plotted down.
If we use the rule that we just specified then we would
set the velocity given this condition which is 1 meter
away from where we want to be.
We would set the velocity to be 10.
If we set the velocity to be 10 then after 1 unit of time,
after 1/10 of a second, we are 1 meter to the right, which
just happens to be exactly where we want to be.
Had we chosen k to be bigger we would have overshot.
Had we chosen k to be smaller we would have undershot.
k equals 10 gave us precisely the right answer so that we
get there in one fell swoop.
Then on the very next step we would compute a velocity of 0
because we are at where we want to be so
we would stay there.
And that condition would persist forever.
The idea would be this simple system provides a way that we
could set the gain so we could get to where we want
to be in one step.
It's hard to beat that.
The problem that results and the reason you didn't see that
good behavior in the lab was that the sensors in the robot
don't work instantaneously.
They introduce delay.
And as an idealization of that delay I want to think through
the same problem.
But now let's say that the sensor delays the input to the
sensor which is the output of the system.
Let's say that the sensor introduces a delay of 1, so
now instead of reporting d sensed, which was d0[n], it
reports d0[n minus 1].
So now what would happen?
Now with the delay, if I started here and if by some
mysterious process I was here at time n equals 1, then I
would calculate my new velocity.
What would be my new velocity here?
I'm right where I want to be.
What would be my new velocity if I assume that the sensor
has a delay of [UNINTELLIGIBLE]?
AUDIENCE: 10.
PROFESSOR: 10.
Because of the delay the sensor is reporting that I'm a
meter away from where I want to be.
So the controller calculates, oh, I need to
go forward a meter.
I'll set the velocity to 10.
So having set the velocity to 10 and then one step goes by,
now we're completely on the wrong side.
That's what happens when you put delay into the system.
So because we're basing this decision on where we were last
time we go to the wrong place.
So now we're here.
What will the controller say next?
AUDIENCE: [UNINTELLIGIBLE]
PROFESSOR: Stay here.
You're in a great place.
I'm really 1 meter too close.
In fact I banged into the wall.
But the sensor is telling me I'm exactly where I want to be
so stay here.
Say I didn't kill myself, what will the velocity next be?
Minus 10.
So now I tell myself to go back.
That's probably a good move.
But now I still think I'm too close to the wall so I tell
myself to continue to back up.
The idea is that I get poor performance.
The delay had a devastating effect on the way that the
controller worked.
Even though it's a tiny change to the way the system works it
has a devastating effect on behavior.
We'd like to be able to predict that without having to
measure it.
Here's the same equations except that I put a delay in
the sensor.
Here is the same block diagram but I've represented a delay
in the sensor path.
So now the question is, what's the new functional
representation for that control system?
So what's the answer?
Can you rate the functional form for this system as one of
(1), (2), (3), or (4), or is it none of the above?
About 1/3 participation and about 100% correct.
The answer's four.
You get to use Black's Equation or however you'd like
to think about that.
You can think about reducing the inner loop the same as we
did before and then think about this as forward over (1
plus loop gain) but now the loop gain has R squared in it
instead of R. We get this form.
How does this form differ from when the R wasn't here?
What's the difference between R not there and R is there?
What's the answer?
Anyone?
AUDIENCE: [UNINTELLIGIBLE]
PROFESSOR: The squared term.
So this term in the previous form was just an R and in this
form is a square and so what's that do?
What's the importance of the fact that
there's a square there?
Two poles, right?
We now have a polynomial in the denominator that is
quadratic in R.
And what that's going to do is it's going to give us two
poles instead of one.
The importance of that is that now we're going to have to
think through-- we previously categorized what were all the
behaviors you could get from one pole.
The behaviors you can get from one pole were monotonic
divergence, non-monotonic alternating divergence,
monotonic convergence, alternating convergence.
So there were four behaviors that were
possible with one pole.
Now we have to think through what are all the possible
behaviors that we could get with two poles.
Different problem.
Hopefully they're related.
So here, the way we would find out what the poles are is take
this expression, substitute for every R, 1 over z, turn
the ratio of polynomials in R into a ratio of
polynomials in z.
To do that in this case I had to multiply numerator and
denominator by z squared.
Having done that I get a second order polynomial in z
in the bottom so there's two poles which are the roots of
that polynomial.
And that's just a quadratic equation.
The interesting thing now is to map out what are all the
possible behaviors that that system can give us.
It's important to realize that's a simple generalization
of what we saw before.
It will be the case that any system that we construct out
of adders, gains and delays will have the property that we
can write the system functional as a ratio of
polynomials in R. By the factor theorem we will always
be able to factor the denominator.
And by the notion of partial fractions we'll always be able
to write some complicated expression like that in terms
of a sum of parts.
Each part being first order.
The intuition we get from this is that what we ought to do is
factor the denominator, find the poles, and associate a
behavior with each of those poles.
Here's what the problem looks like for the two pole problem.
If we have the general form given here and if we start by
thinking about kT having a small magnitude, if kT has a
small magnitude then we have 1/2 plus or minus the square
root of 1/2 squared.
So that's 1/2 plus or minus 1/2.
That's 0 or 1.
So the poles for this system, if you make k be very small,
the poles are at 0, near 0 and near 1.
Is that a good system response or a bad system response?
Bad.
Why?
Well, we're trying to think through the behavior of the
second order system by thinking about the separate
behaviors of each of the poles.
Is this a good pole or a bad pole?
Why?
The response is always pole [UNINTELLIGIBLE].
The mode associated with the response at a pole near one is
something near one to the end.
That never converges.
If you start with some error the error persists forever.
Well, that's not good.
If wind turbulence knocks you into a decline in your
airplane and it persists forever, that's not good.
You would like those things to damp out.
So this pole is bad.
How about that pole?
That one has a response that decays quickly.
But the problem is that when you add the two pieces
together, that was the reason I showed you this
decomposition, you can think about the polynomial being
factored and being broken into a number of parts.
The part that's associated with the pole near one has a
response that goes for a long time.
So that will asymptotically dominate your response.
So we refer to this as a dominant pole.
This pole dominates the response.
That's a way of inferring the behavior of two poles from the
sum of single poles.
In this particular case there's one pole that matters
more than the other one.
So we call that pole the dominant pole.
If you were to make kT more negative, So here's the
general form.
If you make kT negative, you can make the thing under the
radical sign go towards 0.
If you made the thing under the radical sign go to 0 then
you would get two poles at 1/2.
So here we would see that if kT were minus 1/4, if kT were
minus 1/4 we would have 1/2 squared, which is plus 1/4.
Minus 1/4 would give us 0 under the radical.
So we would get two poles at 1/2.
Is that good or bad?
Well, it's better than the previous example, right?
Because each of those poles is associated with the response
where the error gets half what it used to be on every step.
So it converges.
If going from 0 to minus 1/4 is good, then going to minus
1/2 might be better, right?
If you continue that trend, say you make kT be minus 1, if
kT is minus 1 then you get 1/2 squared minus one.
So 1/2 squared is 1/4, minus 1 would be minus 3/4.
That gives us a complex pole here.
So we get two poles that are right on the unit circle.
What's that mean?
That means oscillations.
Oscillations is something you can't get with one pole with a
real system.
Oscillations result from a poll that has
an imaginary component.
If the system is real you could only get
such poles in pairs.
So it's this pair that makes sense for
a real valued system.
And that gives rise to oscillations and that's
exactly what we saw here.
So we can associate the oscillations that we saw in
the simulated lab experiment with poles that have imaginary
components.
So what would be the period of the oscillation in the system
given by 1/2 plus j root 3 over 2?
AUDIENCE: [INAUDIBLE]
PROFESSOR: Excuse me?
Excuse me?
AUDIENCE: Where did the root three come from?
PROFESSOR: The previous page.
If you substitute minus 1.
So what's the period of the oscillation?
So the period's represented by 5 converged into 6.
So how do you get 6?
The easiest way to think about that is to think about the
poles being expressed in polar notation.
The poles we previously said were 1/2 plus or minus
j root 3 over 2.
That's the same as e plus or minus j pi over 3.
It's easier to use that form because if you take that form,
so if you think about e to the j, what was it, 2 pi over 3?
Pi over 3.
So if you think about that form, that's the pole, we can
write that that way.
Then the inside has a magnitude of 1.
So we can think about that just being a magnitude of 1
and an angle of pi over 3.
So when you raise that to the n, the magnitude to the n, one
to the n is always 1, and the angle raised to the n, it just
increases linearly with n.
So the angle goes from pi over 3 to 2pi over 3 to pi to 4pi
over 3, et cetera.
So you can think about this going from pi over 3, 2pi over
3, pi, 4, 5, 6.
It takes n equals 6 to get around to where it started so
the period is 6.
If you were to further change the game, if you were to make
it even more negative, the poles would go outside the
unit circle.
And then what would happen?
AUDIENCE: [UNINTELLIGIBLE]
PROFESSOR: Right.
So that's completely unacceptable.
The point is that by changing the gain you can get any
behavior on this figure which is called the root locus.
So root meaning the root of a polynomial.
Locus meaning the acceptable values of points.
So the root locus shows you all the possible behaviors
they could result from this system.
So given that root locus, how would you choose k to make
your system response as fast as you could?
So what value of kT would you want?
Everyone raise your hands.
That's very good.
So the most popular answer is number (2).
So why would the answer be number (2)?
What do you look at?
Yeah?
AUDIENCE: [INAUDIBLE]
PROFESSOR: Remember what we're trying to do.
We're trying to infer properties of the behavior of
this second order system from the pole locations.
We know that there's an expansion that lets us expand
the system in terms of the sum of two first order responses.
The slowest of the first order responses will dominate
eventually.
So what we need to look at is the
slowest of the two responses.
We would like to know of the two poles
which one is the slowest?
The slowest is the one that's closest to the unit circle.
So we would get the fastest response when the slowest one
is as fast as possible.
As the poles initially go toward each other from 0 to 1
this one is getting faster, this one is getting slower.
So the slowest one is this one.
So the slowest one is fastest when they meet.
And then when they diverge does the slowest one get
faster or slower?
It's already slower because it gets
closer to the unit circle.
So you get the fastest response whenever you get the
two poles both colliding at 1/2 and that was the case that
happened when kT was minus 1/4 from two or three slides ago.
So the idea then is to try to infer what would be the
behavior of this higher order system by thinking about the
behaviors of the individual components, here the poles.
And what we saw was something that's in fact a very
important general trend.
What we saw was that we first analyzed the wall finder
system assuming there was no delay in the sensor.
And we found that that system was characterized by a single
pole and we had the design freedom of putting that pole
anywhere we wanted to on the real axis.
And that allowed us to choose the pole to be at 0 which gave
terrific performance.
The interesting thing that happened when you add just one
more pole by putting a delay in the sensor, you make the
system more complicated and now you can't possibly get
nearly so good behavior.
The behavior is a lot worse than it was before.
And in fact, if you were to do the same kind of analysis by
putting yet another delay in the sensor you would find even
worse behavior.
The idea then, the generalization of the way the
behaviors is working, generally speaking adding
delays inside a feedback loop is a destabilizing thing.
Generally as the number of delays increases you end up
having to back off on the maximum gain that you can use
because the system becomes less stable.
So the overall moral is that delays are bad generally.
I mean, you could concoct some kind of a weird scheme where
that wouldn't be true.
But it's actually hard to concoct such a weird scheme.
In general, and in virtually every physical system that
you'll run into, adding delays makes the
system harder to stabilize.
And that's the big message.
And the system that we looked at in the lab, the wall finder
was actually quite hard because the number
of delays was large.
If you try to track where delays can enter the robot
system they get in at very many different places.
In the physical sensor, in the microprocessor, in the
conversion from analog to digital, there's a number of
delays in that system.
And that's why it becomes hard to stabilize.
OK.
So that's the main content for today.
What I want to do is give you one more practice question.
The big problem that I want you to think about from today
is how do you characterize performance?
When we had a single pole performance was easy to talk
about because performance was diverging monotonically,
diverging alternating, converging monotonically,
converging alternating.
There were four kinds of behaviors.
When we went to second order we saw some new behaviors.
It could become oscillatory.
What I'd like you to do now is think not just about those
properties but many other properties.
So here's some questions.
Think about the system on the top and I'd like you to infer
properties about that system.
In particular, does this system have three poles?
Is the unit sample response, is there a way to write that
as the sum of three geometric sequences?
What's the unit sample response?
And is one of the poles that z equals 1?
So think about the system, think about five ways of
characterizing it and tell me how many of those five
characterizations is correct.
So how many of the properties are true?
AUDIENCE: [UNINTELLIGIBLE]
Probably 2/3 correct?
PROFESSOR: How many poles?
How do you get three?
Where are the poles?
AUDIENCE: [UNINTELLIGIBLE]
PROFESSOR: How do I find poles?
What do I do?
Yes?
AUDIENCE: You use Black's Equation [UNINTELLIGIBLE]
the top you can express [UNINTELLIGIBLE]
so use Black's Equation to express the system function as
R cubed over (1 minus R-cubed) then--
PROFESSOR: That's right.
AUDIENCE: --the denominator as an order of 3 and
you combine the 3s.
PROFESSOR: So a little more formally, we would take this
thing and we would rewrite that with R goes to 1 over z.
So we get 1 over z cubed, 1 minus (1 over z cubed), which
is then, clearing the z cubes we would get 1 over
(z cubed minus 1).
How many poles?
Three.
What are the poles of z cubed minus 1?
Three poles in what?
AUDIENCE: 0 [UNINTELLIGIBLE]
PROFESSOR: And so let's vote.
Let's take a vote.
There's two poles at z equals 1.
Yes?
No?
AUDIENCE: [UNINTELLIGIBLE]
PROFESSOR: Why not?
AUDIENCE: [UNINTELLIGIBLE]
PROFESSOR: So there's two poles.
I made a [UNINTELLIGIBLE] plane.
Where's the poles?
Well, you could factor it, right?
If you factored it you'd find that there is
a pole at 1, right?
But then there's two more poles like that.
So the poles are the three roots of 1, which can be
written like 1 e to the j, 2pi over 3, and e to the j minus
2pi over 3.
Which pole was the dominant pole?
AUDIENCE: [UNINTELLIGIBLE]
PROFESSOR: OK, bad question.
What's a better question?
AUDIENCE: How many dominant poles are there?
PROFESSOR: How many dominant poles are there?
That's a much better question, yes.
There's sort of three poles that are
equally dominant, right?
They all have the same magnitude.
Why do we talk about dominant poles?
What are dominant poles good for?
If I told you that I had a pole at 3 and a pole at minus
1, which one's the dominant pole?
AUDIENCE: 3.
PROFESSOR: Why?
AUDIENCE: Greater magnitude.
PROFESSOR: Greater magnitude.
Why do we care?
We don't care, right?
What's good about the dominant pole?
Well, we can write this response as something that
looks like three to the n plus minus 1 to the n.
If you let n get big enough the only one that
matters is 3 to the n.
So if all you care about is exactly how the plane was
flying the instant before it hit the ground then you would
only need to worry about long time.
And if you only worry about long time you only need to
worry about the pole that's worst behaved.
That's where the concept comes from.
So none of these poles are particularly worse behaved
than the others.
What's the unit-sample response
associated with that pole?
We have a name for that [INAUDIBLE] right?
AUDIENCE: It's huge.
PROFESSOR: It's [UNINTELLIGIBLE].
What's the unit-sample response
associated with this pole?
Well, it's got a complex value right?
So the unit-sample response associated with that pole is e
to the j 2 pi over 3 n.
That's a complex number.
That's 1 at time 0 and e to the j 2 pi over 3n at time 1
and e to the j 4 pi over 3 at time two.
So it goes from here at 0 to here at 1 to here at 2,3, 4,
5, 6, 7, 8.
What's the period of this pole?
What's the period of the unit-sample responses
associated with that pole?
3.
Because it takes 3 to get around to
where you started again.
What's the period of this pole?
3.
You just spin around backward.
What's the period of the response
associated with that pole?
Bad question.
All right, what's a better question?
Is there a period associated with it?
You could say that period 1 [UNINTELLIGIBLE]
definition of period.
What's the period of this pole?
Dumb question, right?
Period implies repeat.
If the response repeats itself after some time then we would
say the response is periodic.
Neither of those poles, well, the minus 1 is.
Is the minus 1 pole periodic?
Yes.
What's the difference between periodic and alternation?
Does a [UNINTELLIGIBLE]
alternate?
AUDIENCE: Yes.
PROFESSOR: Does it oscillate?
AUDIENCE: No.
PROFESSOR: Bad question.
Alternate is a word that we invented for one pole.
Because the response alternated in sine.
The unit-sample response in one pole, where the pole is a
negative number, alternated in sine.
So we gave that a name.
Alternation is not necessarily something that we would like
to associate with a higher order system.
Periodic is perfectly reasonable to talk about for a
higher order system.
Periodic merely means that if y of n were a periodic signal
that I could express y of n plus n as y of n.
That would be periodic.
If the thing repeats itself we would say it's periodic.
So the point of going over this stuff is just to give you
some exercise in thinking about how to think about
properties of systems.
We develop properties initially
thinking about one pole.
Those properties were easy.
Converging, diverging, monotonic, alternating.
When we try to think about corresponding properties of
higher order systems we can't simply map the simple
properties of first order systems into the other.
We have to think about more complicated things.
Then we think about things like dominant.
If one of the poles has a bigger magnitude than the
other then for large times we can ignore the smaller one.
What happens for short time?
Does this response monotonically increase with
time for all time?
No.
Since the response associated with minus 1 alternates in
sine, for short times, for times with n close to 0, that
can be just as important as this one.
So the dominant pole idea tells you how things work when
you have large times.
It doesn't necessarily tell you how things work when you
have small times.
How about, the unit sample response is the sum of three
geometrics.
Yes or no?
What are the three geometrics?
And the answer to that's yes.
That's very important.
The three geometrics over here are this pole to the n plus
this pole to the n plus something that goes with this
other pole to the n.
Now it's a weighted sum but the weights are not
necessarily 0.
The slide that I showed you for the partial fraction
decomposition, you can always write a higher order system as
a sum of first order factors.
That's the partial fraction expansion.
That doesn't mean the weights are all unity.
Number (2), can you write the unit-sample response as the
sum of three geometric signals?
Yes.
There it is.
And if you're really good at complex math you could find
out a, b and c.
And that would tell you the unit-sample response and that
would tell you the answer to (3) and (4).
Is the unit-sample response 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0,
1, 0, 0, 0, 1?
Or 1, 0, 0, 0, 1, 1 --
whatever.
Is it one of those two or something different?
And how do you figure it out?
How do you figure out the unit?
Is the unit-sample response--
Is number (2) correct?
Is the unit-sample response 0, 0, 0, 1, 0, 0,
0, 1, 0, 0, 0, 1--
Yes?
How do you know that?
You could solve that equation.
Is there an easier way?
Yeah?
AUDIENCE: I wrote it as a difference equation.
PROFESSOR: Just write the difference equation.
Exactly.
So even though I bad-mouth difference equations a lot,
here it's easy.
In fact, you can see it in the network.
Thinking about the difference equation would be easy.
Thinking about the network would be easy.
If we think about the unit-sample response of this
thing started at rest, rest means this is 0, this is 0,
and this is 0 initially.
Unit-sample response means this becomes 1 at time 0.
At time 0 this is 1, this is 0, this is 0, this is 0.
So if I think about what's the time response look like and I
did plus, this is 0, 1, 0, 0, 0.
The first answer is 0.
Clock ticks.
What happens?
This is 1.
This becomes 1.
This doesn't change.
That doesn't change.
This goes to 0.
0 comes around here.
That goes to 0.
So that's the answer at time 1.
What happens at time 2?
Just keep working it.
The clock ticks, this goes to 1, this goes to 0, these stay
0, this stays 0.
That's the answer at time equals 2.
Now the clock ticks.
Now this comes over to here, that means it comes back here.
This is still 0.
That comes to 1.
That's the answer for time 3.
Now the clock ticks.
And you can see the whole thing would just
repeat itself now.
Is the response periodic?
Yes.
The response is periodic.
What's the period?
3.
And I can see [INAUDIBLE] if the [INAUDIBLE] there it's
going to have be related to the period over here.
These periods are not same.
This period is 3, this period is 3, this period is 1, if you
want to call that a period.
But they are related.
OK.
The point of this exercise is to illustrate two things.
We inferred properties of first order systems by looking
at a single pole which for a real system could only behave
in one of four different ways.
Second order system introduced new behaviors.
Now we can oscillate, which we couldn't do before.
Having got to oscillation, oscillation came about because
of complex numbers.
If you go to higher order systems nothing
new happens in algebra.
There's no such thing as meta-complex numbers, right?
Complex is as bad as it gets.
So you can have complex numbers.
The higher order behaviors can still have complex numbers but
you have to think when we ask you, what's the property of a
higher order system.
You can think about it in terms of the individual parts
but it requires some thinking.
OK.
Good luck tonight.
See you then.