自动控制原理 第十二讲 状态反馈

n
we have
Wc = ( B , AB ) = dx = dt
0 0 0 1 1 0 1 0
Ak−1 B
k=1 0
t
, Controllable!
1 0
α k (s)u(s)ds = −a + be− At
n−1
x+
u
Left side a combination of vectors B , AB , . . . , A can be来自百度文库solved if the matrix
Try to do same thing for
dx = dt 0 0 1 0 x+ 0) x 1 0 u
Controllability
dx = Ax + Bu dt Find control signal that moves the system from x(0) = a to x(t) = b. Hence
Choosing
2 l1 = 2ζ ω 0, l2 = ω 0
,
B=
b11 b21
b12 b22
,
C=
c11 c21
y = Cx = ( 1
• Which matrices can be added? • Which matrices can be multiplied?
det A = a11 a22 − a12 a21 A−1 = 1 det A a22
Wc = ( B AB . . . An−1 B )
B . Equation
we have
Wc = ( B , AB ) =
1 0 0 0
Not Controllable!
can be inverted. Controllability!
A Special Case
The system −a1 −a2 . . . an−1 −an 1 1 0 0 0 0 dz 0 1 0 0 0 = z + u dt . . . . . .
0 0 0 0
The feedback
u = −l1 z1 − l2 z2 − . . . − ln z n + lr r l1 = p1 − a1 , l2 = p2 − a2 , . . . , ln = pn − an
gives the characteristic polynomial
sn + p1 sn−1 + p2 sn−2 + . . . + pn
Lecture 12 - State Feedback
K. J. Åström 1. Introduction 2. An Example 3. The General Case 4. Integral Action 5. Summary
Theme: Using the state for control.
Control law
u = −l1 x1 − l2 x2 + lr r
gives the characteristic polyno2 mial s2 + 2ζ ω 0 s + ω 0 . The closed loop system is
Y ( s) = r lr R( s ) 2 s2 + 2ζ ω 0 s + ω 0
The General Case
Problem solved for special case. Can we transform a given system to this case? Start with
dx = Ax + Bu dt
A system
The General Case
dx = Ax + Bu dt which is controllable can be transformed to
0 0 1 0 0
is controllable
1 −a1 a2 ... 1 − a2 0 1 −a1 ... 0 0 1 ... Wc = . . .
0 0 0
...
1

c
has the characteristic polynomial
State Feedback
Assume that the process is described by
dx = Ax + Bu dt y = Cx
The Mathematical Tool - Matrices
Please refresh your knowledge of matrices. • What is a matrix • Matrix algebra: addition, multiplication, notice AB = B A (a very useful property), transpose • Matlab • Linear equations and inverses Ax = B , x = A−1 B • Eigenvalues and eigenvectors
s − a11
Characteristic equation
s 2 + l1 s + l2 = 0 Y ( s) =
−a21
−a12 s − a22
2 ω0 R( s ) 2 s2 + 2ζ ω 0 s + ω 0
= s2 − (a11 + a22 )s + a11 a22 − a12 a21
Example
Simple Examples
Consider the following matrices and their transpose
A=
a11 a21 a12 a22
Example - The Car
Process model
c12 c22 c13 c23
dx = dt 0 0 1 0 x+ 0)x 0 1 u
Ae = λ e ( A − λ I)e = 0 det A − λ I = 0
The general linear controller is: u = − Lx + lr r The closed loop system then becomes
dx = Ax + Bu = Ax + B (− Lx + lr r) = ( A − B L) x + Blr r dt
0 0 1 0 0
A Special Case
−a1 −a2 . . . an−1 −an 1 1 0 0 0 0 dz 0 1 0 0 0 = z + u dt . . . . . .
0 0 1 0 0
The feedback u = −l1 z1 − l2 z2 − . . . − ln zn + lr r gives −a1 − l1 −a2 − l2 . . . −an − ln lr 1 0 0 0 dz 0 1 0 0 = z + r dt . . . . . .
Introduction
• The simple design method becomes very cumbersome for systems of high order • The PID controller predicts based on linear extrapolation • Can we do something better • Where to look for inspiration? • State a number of variables that summarizes the past that is useful for prediction • The future behavior can be predicted from the state • Can we find a general controller? • The state is an ideal basis for control • We will focus on the predictive part, reference values and integral action will be dealt with later
c K. J. Åström August, 2001 2
Controllability ...
t 0
Examples
dx = dt
0 0 1 0
e− As Bu(s)ds = −a + be− At
x+
0 1
u
It follows from Cayley-Hamiltons theorem (a matrix satisfies it own characteristic equation) that
0 0 1 0 0
The Characteristic Equation
The system −a1 −a2 . . . −an−1 −an 1 1 0 0 0 0 dz 0 1 0 0 0 = z + u dt . . . . . .
The closed loop system has the characteristic equation
det(sI − A + B L) = 0
Can we choose L so that this equation has specified poles?
c K. J. Åström August, 2001 1
t lr 0
y = Cx = ( 1
Control law u = −l1 x1 − l2 x2 + lr r. Closed loop system
dx = dt
− l1
0 1 − l2 0
x+
b = x(t) = e a +
At 0
eA(t−s) Bu(s)ds
r
Characteristic equation
s n + a1 s n − 1 + a2 s n − 2 + . . . + a n
K. J. Åström August, 2001
3
A Special Case which is Easy
−a1 −a2 . . . an−1 −an 1 1 0 0 0 0 dz 0 1 0 0 0 = z + u dt . . . . . .
Closed loop system
−a12 a11
−a21
dx = dt
0
1
− l1
− l2
x+
0 lr
2 Choosing lr = ω 0 gives the correct steady state value.
Characteristic polynomial det (sI − A)
det
s ( s + l1 ) = s 2 + l1 s = 0
This implies
t 0
e− As Bu(s)ds = −a + be− At
We cannot obtain a desired characteristic polynomial. Why? The control signal does not influence the second state! Some conditions are required!