Lecture 29: Preliminaries to Formal Control. Quick review of last time. State transformations I: diagonalization. Introduction to the ideas of controllability and observability. What did we do last time?. We looked at outputs and found transfer functions.
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Lecture 29: Preliminaries to Formal Control
Quick review of last time
State transformations I: diagonalization
Introduction to the ideas of controllability and observability
We looked at outputs and found transfer functions
The we took the transfer function and used It to find a new equivalent system
The examples we did gave a simpler equivalent system
typically an A matrix in companion form
Let me review some of the results from last time: suspension and pendulum
Put these numbers into our picture
z2
m2 = 2.479 lbsec2/in
m2
k3
c3
k3 = 143 lb/in
c3 = 15.06 lbsec/in
z1
m1 = 0.194 lbsec2/in
m1
k1 = 1198 lb/in
k1
zG
I can form the transfer function between the ground motion and the axle motion
The numerator is not constant, so we will want to use the method we discussed last time
This defines a state, but it is not the state we started with!
The left hand equation defines the dynamics in terms of a new state
We can make a block diagram of the system
and we can add the output
y
37515
356217
+
u

83.7
6970
37515
356217
Let me add a couple of comments
We can make a new dynamical system from the block diagram
I have not addressed the connection between the old state and the new state
I’ll deal with that systematically another day
For now, let’s look at the dynamics of the two systems
Let’s compare the A matrices for the two systems
The original matrix
The new matrix
They have the same eigenvalues
This will modify the matrix A and make the input u the voltage e
so that we have a modified A and b
Last time we looked at the angle as the output
Today I’d like to look at the cart position as my output
This leads me to
From which we get a new state block diagram — much simpler than the old one
+
u

1/25000
12005
249900
117649
We have discovered that we can solve the problem
In terms of a transfer matrix
or in the common case of a SISO system, a scalar transfer function
We only looked at the single inputsingle output (SISO) case
For which the transfer matrix is a scalar, the transfer function
If the numerator of H is not constant, we had to split the transfer function
This defined a new state z, and we didn’t know how to connect z and x
Today I want to look at transformations in general,
and at two very special transformations: diagonalization and companion form
I’m going to start with diagonalization, which is a close relative of modal analysis
We’ll eventually settle on companion form,
which is what we’ve been getting informally from our transfer function games
we’ll get a formal approach to that on Thursday
What would we like to do?
The eigenvectors are the state space equivalent of the modal vectors of modal analysis
Can we make the A matrix diagonal — uncoupling the eigenvectors?
The answer is yes, sometimes . . .
We need to transform the state x to a new state z that will separate the eigenvectors
A linear transformation between two vectors can be written
T is a constant matrix, and we want transformations such that it has an inverse
general transformation
The equation becomes
and we can simplify
We would be happy if TAT1 were diagonal
If A has distinct eigenvectors we can make that happen
we candiagonalizeA.
Let V be a matrix whose columns are the eigenvectors of A
then
Thus if we choose Tto be V1 we can diagonalizeA
In that case we’ll have
or, in terms of the individual equations
Uncoupled first order ordinary differential equations, which we know how to solve
Block diagrams on the next slide
So if I want the full state,
I can find z and then use T1 (here V) to obtain the physical state in term of z
All the left hand multipliers have to be nonzero for the input to reach all the zs
All the right hand multipliers have to be nonzero for all the zs to contribute to the output
What does all this have to say about the examples we’ve worked?
Two masses and some springs
The auto suspension
Artificial example 5A
The inverted pendulum on a motordriven cart
Put in numbers and generally simplify a bit: k1 = 1 = k2, k3 = k, m1 = 1 = m2
I can do this without loss of generality by appropriate scaling
we have one variable parameter — the strength of the coupling spring
The matrix of eigenvectors, V = T1 is given by
It is important to note that I can multiply the eigenvectors by a constant
without changing anything.
If you don’t like complex displacements, multiply by j
The two coupling matrices are given by
Both are complete, so the input affects all the components of z
and the output contains all the components of z
We can draw the block diagram
What is the interpretation of z?
The first two lines involve y1 and y2 moving together (sum of x1 and x2)
the second two lines involve them moving in the opposite directions
The coefficients reflect the eigenvalues: ±J and ±j√(1+2k)
The differential equations, supposing z positive up
Define a physically reasonable state
Block diagram of the original problem
red denotes the upper pair
x3
x1
u = zG
x4
x2
blue denotes the lower pair
Put numbers into the picture and the system
z2
m2 = 2.479 lbsec2/in
m2
k3
c3
k3 = 143 lb/in
c3 = 15.06 lbsec/in
z1
m1 = 0.194 lbsec2/in
m1
k1 = 1198 lb/in
k1
zG
Let me define an output. I’ll choose body position: z2 = x2
We looked at this using transfer functions to find a new state
y
The new state block diagram — much simpler than the old one
37515
356217
+
u

83.7
6970
37515
356217
How does this compare to diagonalization?
The eigenvector matrix V = T1
Check the transformed input and output matrices
They are complete and we are OK:
the input goes everywhere and the output sees everything
Let’s draw the block diagram (it looks a lot like the one we just saw)
Diagonalization gives us y in terms of u without knowing what z represents
If we want x in terms of z or vice versa, we need to do a little more
It’s numerically tedious; we can look in Mathematica if we want
Right now I don’t want
Friedland gives an example of order reduction — 5A
and I’d like to take a look at how that goes — what happens
It’s not a real physical system, but nonetheless instructive
We started on this last time and managed to reduce the 4th order system
to a first order system — a really weird result
We can use diagonalization to understand how this happened
The eigenvalues of A are 4, 3, 2, 1
They are independent and span the space. The inverse is
Let’s look at the input and output vectors for the z state
We see that the input does not reach every piece of z
and that the output does not see every part of z
We can draw a new block diagram for this transformed state
The only path from u to y is that shown in red below
It corresponds to the one dimensional transfer function
y
2
u
1
What do the solutions to this look like?
Note that each z responds at a single eigenfrequency — they are eigenfunctions
The input and the output are only connected through z4
which suggests an explanation of the first order system we got before
and found a reduced block diagram
y
+
u
the single eigenvalue
that contributes to the
output
1
What can we say about the relations between x and z?
We need z in terms of x to get the initial conditions for z
We need x in terms of z to convert the solution to a physical state
If we start from rest, z1 and z3 remain zero, and then we’ll have
u reaches all the components of x, but we are missing two of the eigenfunctions
Before we go on to the next example,
let’s say a little about controllability and observability
The goal of control is to use u to make x do what we want: u has to reach the whole state
If it cannot, then the system is uncontrollable
We are going to do this using feedback: we may have to see the whole state
If we cannot, then the system is unobservable
Let’s talk about this in the context of Example 5A
the system is not controllable
4
z1 and z2 do not appear in the output
the system is not observable
u
3
y
2
u
1
We will develop better criteria for controllability and observability going forward
These will involve the companion form
All I want to do for now is to indicate that not all systems can be controlled
I’ll do one last example for this afternoon
This will modify the matrix A and make the input u the voltage e
so that we have a modified A and b
(This is only to three significant figures; the two 953s are not actually equal)
It’s invertible and we can find the diagonalized problem
The transformed b and c matrices (vectors, here) are given by
The input vector is complete; the output vector is not.
u
u
u
u
y
3.17
3.16
0
0.47
z4 is simply the integral of u
The eigenvalues are unscaled; the other blocks are shown 1000 times bigger than
they actually are
In all the cases where we have done both modifications —
transfer function to new state and diagonalization —
the resulting new states have been different, with very different block diagrams
The different block diagrams correspond to different A and B matrices
Recall the suspension example
This happens to be a partial companion form —
full companion form also changes B, and we’ll get to this
y
37515
356217
+
u

83.7
6970
37515
356217
Let’s compare the A matrices for the two methods
The diagonalized matrix we just did
The companion form matrix