240 likes | 337 Views
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
E N D
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
xr A – Ar e u b A
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
xr A – Ar - bgrT e 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 — 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 we want this to work for any xr that is a member of the set defined by Ar
c 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 Friedland’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, 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 control/ disturbance
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 — 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