Lecture 25. Introduction to Control. in which we enlarge upon the simple intuitive control we’ve seen. We generally want a system to be in some “equilibrium” state. If the equilibrium is not stable, then we need a control to stabilize the state.
Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.
Lecture 25. Introduction to Control
in which we enlarge upon the simple intuitive control we’ve seen
We generally want a system to be in some “equilibrium” state
If the equilibrium is not stable, then we need a control to stabilize the state
I will talk a little bit about this in the abstract, but first,
let me repeat some of what we did Tuesday evening
I want to start on control at this point
We have open loop control and closed loop control
Open loop control is simply:
guess what input u we need to control the system and apply that
Closed loop control sits on top of open loop control in a sense we will shortly see.
In closed loop control we measure the error between what we want and what we have
and adjust the input to reduce the error: feedback control
About the simplest feedback control system we see in everyday life is cruise control
We want to go at a constant speed
If the wind doesn’t change and the road is smooth and level
we can do this with an open loop system
Otherwise we need a closed loop system
Recall the diagram from Lecture 1, and modify it to describe a cruise control
the open loop part
desired speed
INVERSE
PLANT
GOAL: SPEED
nominal fuel flow
Actual speed
+
+
PLANT:
DRIVE TRAIN
Input: fuel flow
disturbance


the closed loop part
error
Feedback: fuel flow
CONTROL
We have some open loop control — a guess as to the fuel flow
We have some closed loop control — correct the fuel flow if the speed is wrong
It happens that this is not good enough, but let’s just start naively
S flowimple first order model of a car driving along: drive force, air drag and disturbance
Nonlinear, with an open loop control
If s = 0, and v = vdwe’ll have an “equilibrium” that determines the open loop f
Split the force and the velocity into two parts (on the way to linearizing)
Substitute into the original ode
Expand flowv2, and cancel the common parts
Linearize by crossing out the square of the departure speed, v’
and the goal is to make v’ go to zero
Let’s say a little about possible disturbances flow
hills are probably the easiest to deal with analytically
mgsinf
f
I’ll say more as we go on
I’m not in a position to simply ask flowf’to cancel the disturbance
(because I don’t know what it is!)
I need some feedback mechanism to give me
more fuel when I am going too slow
and less fuel when I am going too fast
The linearized equation (still open loop)
Negative feed back from the velocity error
We do not need this whole apparatus to get a sense of how this works
Consider a hill, for which s(t) is constant, call it s0
We can find the particular solution by inspection
The homogeneous solution decays, leaving the particular solution,
and we see that we have a permanent error in the speed
The bigger K, the smaller the error, but we can’t make it go away
(and K will be limited by physical considerations in any case)
What we’ve done so far is called proportional (P) control this works
We can fix this problem by adding integral (I) control.
There is also derivative (D) control (and we’ll see that in another example)
PID control incorporates all three types, and you’ll hear the term
Add a variable and its ode this works
Let the force depend on both variables
Then
rearrange
define k
Convert to state space this works
We remember that x denotes the error
so the initial condition for this problem is y’= 0 = v’: x(0) = {0 0}T
The homogeneous this workssolution (closed loop without the disturbance)
and we see that it will decay as long as K2 > 0
What happens now when we go up a hill this works?
This means a nonzero disturbance, and it requires a particular solution
We can now let the displacement take care of the particular solution
Wait a minute here! this works
What’s going on!?
Have I pulled a fast one?
Not at all. Let’s think a little bit here.
What did we just do? this works
What can we say in general?
We did some linearization,
and we changed a one dimensional state into a two dimensional state
We also dealt with disturbances, which is actually an advanced topic
We can look at all of this in Mathematica if we have the time and inclination at the end of this lecture. For now, let’s look at some results for the second order control. I will scale to make the system nondimensional and of general interest.
The this worksscaled response to a constant hill is
The control still works very well, and tracks nicely once it is in place.
Let’s look at a much more complicated roadway
The scaled response looks like the scaled forcing is in place.
and the velocity error is minuscule — this really works
What did we do here? is in place.
We started with a one dimensional system
— and tried to find a force to cancel and exterior force
That didn’t work
We added a variable to the mix
found a new feedback
got the velocity to be controlled at the expense of its integral
about which we don’t care very much
Now let’s review this in a more abstract and general sense is in place.
(without worrying about disturbances for now)
Consider a basic single input linear system is in place.
We need a steady goal, for which
Typically ud = 0 and xd = 0, and I’ll assume that to be the case here
(xd might be the state corresponding to an inverted pendulum pointing up)
We can look at departures from the desired state (errors) is in place.
where remember that we want x’ —> 0
If u’ = 0, then the system is governed by
and its behavior depends on the eigenvalues of A
if they all have negative real parts, then x’ —> 0
We call that a stable system
If we have a stable system, we don’t need to control it is in place.
although we might want to add a control to make it more stable
If it is not stable, then we must add feedback control to make it stable
The behavior of this closed loop system depends on the eigenvalues of
We can get the eigenvalues of is in place.
by forming the determinant of
The terms in this equation will depend on the values of the gains
which are the components of the gain vector g
You might imagine that, since there will be as many of these as there are roots
that we can make the eigenvalues be anything we please.
This is often, but not always true. We’ll learn more about this in Lecture 31.
Let’s look at a simple example to see how this works is in place.
Last time we looked at an electric motor is in place.
and set it up as a second order system without being too specific
Make a vector equation out of this
or
Suppose we want the angle to be fixed at π/3 is in place.
The desired state satisfies the differential equations with no input voltage
We can write the differential equations for the primed quantities
We want the perturbations to go to zero
Is the homogeneous solution stable?
What are the eigenvalues of A?
There is one stable root and one marginally stable root ( quantitiess = 0).
The homogeneous solution in terms of the eigenvectors is
If the initial value of quantitiesq’ is not equal to its desired value (here 0)
then it will not ever get there.
The problem as posed is satisfied for everything equal to zero
but that’s not good enough
We need control.
If q’ is too big we want to make it smaller and vice versa
Let’s look at this in block diagram mode
We can close the loop by equationsfeeding the q’ signal back to the input
closed loop


w’
q’
feedback loop
So we have new equations governing the closed loop system equations
There’s no disturbance, so the equations are homogeneous
We’ve gone from an inhomogeneous set of equations to a homogeneous set
This what closing the loop does; there’s no more undetermined external input.
We want q and w to go to zero,
and that will depend on the eigenvalues of the new system
This will converge to zero for any positive homogeneous setg
Let’s put in some numbers: K = 0.429, R = 2.71, Ix = 0.061261 (10 cm steel disk)
We are overdamped for small g and underdamped for large g
We can get at the behavior by applying what we know about homogeneous problems
The eigenvectors homogeneous set
I will select the gain g = 0.2379 (to make some things come out nicely)
This leads to the eigenvaluess = 0.544 ± 0.544j
The homogeneous solution homogeneous set for the closed loop system is
With the numbers we have
At homogeneous sett = 0, we have
A little algebra
Now we have the complete solution in terms of the initial conditions
Let’s plot this and see what happens for homogeneous setq’0 = π/3 and w’0 = 0
I did this in something of an ad hoc fashion that did not really illustrate the general principle
Let me go back and repeat it more formally
I can clean up the algebra a bit with a definition
Let me define a gain vector really illustrate the general principle
If the output is the angle, then g1 is a proportional gain and g2 a derivative gain
The matrix
and the closed loop characteristic polynomial is the determinant of
which I can expand really illustrate the general principle
Denote the roots of this by s1 and s2
from which
I can choose any values of really illustrate the general principles1 and s2 and find g1 and g2 so that they will be the roots.
That’s what we needed to do today really illustrate the general principle
It was, maybe, a lot.
Let it sink in and we’ll deal with questions as we go along