PROFESSOR: Last time we
introduced poles.
And in particular, we introduced
how to move from
the manipulation of feed-forward
and feedback
systems and the geometric
sequence that fell out into
using the base of that geometric
sequence to attempt
to predict the long-term
behavior of the system.
When we're solving for poles and
we're only interested in
long term behavior, one of the
easiest ways to do so to solve
for the roots of z, where z is a
substitution for 1 over R in
the denominator of the
system function.
Once we've done that, we
have a list of poles.
From that list of poles, we
would like to select, the
dominant pole, or the pole with
the greatest magnitude,
and then based on the magnitude
and period of that
pole we can determine what the
long term behavior of our
system looks like.
Today I'd like to mention
to you some notable
things about poles.
If you are interested in this
information or feedback and
controls in the general sense,
I highly recommend 6.003.
But here's some information you
should at least be aware
of as a consequence of 6.01.
The other thing I would like to
do is just walk through a
couple of pole problems to
familiarize you, or get you
more comfortable with the idea
of solving for the poles of a
system function, or looking at
the unit sample response of a
system function and then
graphing the poles.
The first thing that I want
to mention is pole-zero
cancellation.
And what do I mean
when I say that?
I mean that if both the
numerator and the denominator
have a degree of R in them, then
you're going to have both
a zero and a pole.
If the zero and the pole have
the same value associated with
them, you may be tempted
to cancel them out.
Unless both the zero and the
pole are equal to 0 --
don't do it.
The reason why is that when you
get to a implementation of
a real system, it is highly
unlikely that both the zero
and the pole will be implemented
to a degree of
accuracy that you will actually
see those two things
cancel out.
The only exception to this is
when both the pole and the
zero are equal to 0,
or in this case.
This you should feel free
to convert to this.
In almost any other situation,
don't factor.
The other thing I want to talk
about is repeated roots.
If you have a repeated root,
you'll have repeated poles.
This does get tricky when you're
talking about how to
add the unit response
of those poles.
But the long term behavior
of your system is
going to look the same.
So if both of these poles are
the dominant pole, then the
characteristics of both, which
are the same, are going to
determine what your long term
behavior looks like.
If they're not, then the
dominant pole is going to
determine what your long term
behavior looks like.
The last thing I want to mention
is superposition.
So far we've only talked about
the unit sample response of a
system function and how we use
poles to determine what the
long term behavior of our
system's going to be.
We can look at the response to
more complicated inputs than
the unit sample response,
or the delta.
In fact, one of the things we'
probably end up looking at
some point is the
step function.
The thing that you need to know
to go from talking about
unit sample response to any
other sort of response, is
that we're still working
with an LTI system.
What that means is if you take
the summation of your inputs,
and apply the system function
to that summation, it is the
same as the output that would
result from inputting all
those values at once.
The best way I would like to
explain it is by referring
again to if your function was
a system function, the same
property applies.
Now let's walk through
a pole problem.
Here I have a second order
system set up.
We've got two degrees of R. I
have feedback, and I can solve
for an expression of
y in terms of x.
In fact, let's do
that right now.
y is the result of the
summation or a linear
combination of x plus a delayed
signal of y scaled by
1.6 in a linear combination with
a delayed value of the
delayed value of y scaled
by negative 0.63.
There's my first degree.
For consistency's sake, there's
my second degree.
Let's first solve for
the system function.
If you're confused, I recommend
doing the algebra
from here to this expression.
You should get this
fraction out.
Our second step is to solve
for the roots of z.
Remember that z is equal to 1
over R in the denominator of
the system function.
In this case, we'll
be working with--
all I've done there is taken
every degree of R, substitute
it in for 1/z.
and then multiply it out, so
that I'm not working with z in
the denominator anymore.
I'm actually just working with
everything in the numerator.
If I follow this back out,
I get this expression.
And my poles are going
to be 0.7 and 0.9.
All right, based on my poles,
what are the properties of the
unit sample response
in the long term?
First thing I'm going to do is
look for the dominant pole
among the poles that I found.
In this case, I don't even have
to worry about finding
the length of the distance from
the origin for poles in
the complex plane.
All I have to worry about
is the magnitude of
poles on the real axis.
0.9 is my dominant pole, because
it's the largest pole.
0.9 is less than 1.
So I'm going to end up
with convergence.
Eventually, my system is
going to converge, or
tend towards 0.
The other interesting property
of my system is what is its
period, how does that relate
to what my function's
going to look like.
In this case, we're only working
on the positive real
axis, so the angle associated
with graphing this pole on the
complex plane is 0, so there is
no period for our system.
This means that our system
is going to converge
monotonically.
Now let's walk through some unit
sample responses and then
graph the poles that generated
those unit sample responses on
the unit circle, where this
is the complex plane.
Let's look at this
graph first.
The first thing that I notice
about this graph is that, like
in the previous example, we have
monotonic convergence.
We're tending towards 0, and
we're not alternating or
oscillating about the x-axis.
So I know I'm going to be
working somewhere along this
line before the edge
of the unit circle.
Because at the edge of the unit
circle, the distance from
the origin is equal to 1.
If you made me guess, then I
would look at the distance
here and compare it
to the distance at
the next time step.
I realize this is
a blackboard.
It's not entirely to scale.
But for the purposes of this
demonstration, I'd like to say
that the signal at this time
step is 0.5 the signal from
the previous times.
Likewise at the next time step,
I would like to say that
this signal is 0.5 the signal
from the previous time step,
and so on and so forth.
Therefore, I'm going to graph
my pole right here.
Let's take a look
at this graph.
I've drawn these squiggles to
indicate that the unit sample
response exceeds the bounds
of the space that I
gave for this graph.
So just assume that these values
are much larger than
I've drawn them.
The first thing that I notice
about this unit sample
response graph is the fact that
not only am I increasing
in a way that does not seem
to change in any way--
we're going to end
up diverging--
is that I'm actually alternating
about the x-axis.
And that particularly, that
if I were to call this an
oscillation, then I
would say it's an
oscillation with period 2.
This means that I'm working
with a negative real pole.
The fact that I'm diverging
means I'm working with a
negative real pole that has
magnitude greater than 1.
If you had to make me guess, I
would look at the distance
associated with this time
step, compare it to the
distance associated with
this time step.
And if you had to ask me, I
would say this is about 1.3
the value at the
previous step.
Likewise, if I were to look at
the next time step, I would
say that this increase
is about 30% of
the previous value.
I'm not even going
to try that one.
But what I'm trying to get
at is that you can use
comparisons of previous and
future time steps in order to
attempt to determine the
magnitude of the pole if
you're working with the
first order system.
If you're working with the
second order system, then it's
possible that you'll see
some really interesting
initialization effects.
And you should probably ask
one of us what's up.
But for this example,
we're going to put
our pole over here.
Here's the last graph I
want to talk about.
The first thing that I notice is
that it doesn't seem to be
diverging, but it doesn't
really seem to
be converging either.
If this is the case, then
I'm going to put
it on the unit circle.
The second thing that I notice
is that it's not monotonic and
it's not alternating.
This is oscillating.
So in order to determine what
angle I'm going to sign to my
unit simple response, I'm going
to count out the time
steps that it takes to cycle
through an entire period and
then from there figure out what
the angle would have to
be in order to determine a
period of that length.
So I start here.
I'm just going to count 1,
2, 3, 4, 5, 6, 7, 8 --
to complete one full
oscillation.
This means that my
period is 8.
If I have to divide 2pi by a
particular angle in order to
get out 8, I want to
divide by pi/4.
So at this point, I'm working
with a magnitude of about 1,
and I want this angle
to be about pi/4.
This concludes my tutorial
on solving poles.
Next time, we'll end up talking
about circuits.
