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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.
