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.