Lecture 36 Alternate (Preferred) Control Method
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Lecture 36 Alternate (Preferred) Control Method. I did the magnetic suspension tracking problem by finding an auxiliary input. This is hard to do in general, and it is limited to a single problem. We can do something else, which I will introduce this evening

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Lecture 36 Alternate (Preferred) Control Method

I did the magnetic suspension tracking problem by finding an auxiliary input

This is hard to do in general, and it is limited to a single problem

We can do something else, which I will introduce this evening

and use for the rest of the semester.

If we have time, I want to look at PS 10

looks like a simple problem with some extra forcing terms

terms that we in some sense know, because we’ve picked xr

Instead of finding aur. find some Arsuch that

with that choice we can write

and we can draw a block diagram of this


A – Ar





What about Ar?

Clearly the class of constant reference states has Ar = 0

Suppose we have a pair of second order equations with the state

If we are to look for harmonic reference states

What about the magnetic suspension problem?

The desired state is

and we can write

Back to the development

We want e to go to zero

We know how to make e go to zero in the absence of tracking

We simply check for controllability, and then find relevant gains

for a full state feedback control u =- gTe

We can extend this idea by choosing u to depend on e and xr

We can draw a block diagram of this closed loop system


A – Ar - bgrT


A –bgT

We have to do the g part first

Then we can move on to the gr part

As it happens we cannot always make this give us a complete vector zero for e

We settle for:

The closed loop system must be asymptotically stable

Some components of e can be held at zero

The eigenvalues have negative real parts/lie in the left half plane.

That closed loop system means the system without xr

The following argument is not the be all and end all

— sometimes we can do better, but let’s look at it anyway

We can ask that the derivative of the error goes to zero, from which

is stable, hence invertible and we can write

If xr is as big as the space, with k components, then e has kcomponents

There are k components of gr but that turns out not to mean that I can fix this

We’ll see this best in terms of examples.

We can make some of the error vanish — from which

the output, as defined by c

We get the nicest result when the output has as many elements as the input

In our case we want to look at single input-single output (SISO) systems

put in the matrix dimensions to help us understand

The left hand side is a scalar — that part of e that we make disappear

rearrange from which

we want this to work for any xr that is a member of the set defined by Ar

c from which is not a square matrix, so it doesn’t have an inverse;

we have to do the whole thing at once

is a scalar for a SISO system, so its inverse is very simple

Let’s look at example from from whichFriedland’s book

(It’s an abbreviated missile control system . . .)

We are going to have a desired (reference) normal acceleration

which we will compare to the actual normal acceleration in terms of an error

We can write the normal acceleration in term of the angle a

He assumes that the command signal is slowly varying, acceleration

so that its time derivative can be neglected

He keeps all the rest of this, and the algebra gets very intense.

He eventually winds up with a third order system for the state vector

with the single input u, and the equations, after some algebra are



Eigenvalues state vector of A: -100, -1.664 ± 16.6599j

The first thing he needs to do is see about the undisturbed system

He moves the poles to add damping and keep a quick response time

He does this using the Bass-Gura procedure,

which is equivalent to what we do, though we haven’t looked at it

I will do it our way, and get slightly different results because I don’t quite

match what he is doing. We can discuss this another time.

The next thing he does is to apply the new stuff — system

Find ag0to cancel errors. If we follow him we’ll get the same result

We can now go to Mathematica and look at this problem