自动控制原理 第十四讲 可控性和可观测性

0
B → CT Wc → Wo
c
K. J. Åström August, 2001
2
System Structure
The coordinates can be chosen so that a linear system has the following structure
d dt
xc xc ¯
Duality
Controllability of
dx = Ax + Bu dt is the same as observability for
dx = AT x dt y = BTx
Canonical Forms
Controllable canonical form
− a1 − a2 . . . − an 1 1 0 0 0 dz 0 1 0 = z+ u . . . dt . . . 0 0 0 0 y = b1 b2 . . . bn z + Du
Theme: A closer look at the controllability and observability and the structure of linear systems.
Introduction
• The concepts of controllability and observability introduced as conditions to solve problems of state feedback and observers • More insight • Kalmans decomposition • System structures • Cancellation of poles and zeros
xo x0 ¯
y = ( C1
where the state vector has been partitioned as x T
co ¯ x co x=
where the states xo are observable and xo ¯ not observable (quiet)
dx ˜ + B ( Lxm + lr r) = ( A − B L) x + B x dt ˜ dx ˜ = ( A − K C) x dt
The transfer function is given by the subsystem Sco
Observer error not controllable from r. Makes a lot of sense because we do not want reference signals to generate observer errors!
Controllability to observability through the transformation
A→A
T
Observable canonical form
− a1 1 − a 0 2 dz . . = . dt a − n−1 0 − an 0 y = 1 0 0... ... 0 b1 1 0 b2 . . z + u . 0 1 0 0 bn 0 z + Du
xc ¯o xc ¯o ¯
Kalmans Decomposition
Partitioning of state space • Sco controllable and observable • Sc o ¯ controllable not observable • Sc ¯o not controllable observable • Sc ¯o ¯ not controllable not observable
-Soc u Soc
System with State Feedback and Observers
Σ
Soc
y
dx = Ax + Bu dt y = Cx ˆ ) + lr r u = L( x m − x ˆ dx ˆ + Bu + K ( y − C x ˆ) = Ax dt
Soc
ˆ by x ˜ =x−x ˆ Replace x
dx1 = − x1 + u dt dx2 = − x2 + u dt y = x1 + x2
S
u
S Σ S
y
The linear combination x1 + x2 is controllable
S
The linear combination x1 − x2 is not observable
→ y(t) = u(t) + constant
The system can be written as
d dt x v
Take Laplace transforms (assuming all initial values zero)
A 0 B 0 0) x v x v
=
+
B 0
u
sY (s) = sU (s), → Y (s) = U (s)
c K. J. Åström August, 2001 3
Disturbance Observer
The following system has been used to model constant load disturbances
dx = Ax + B (u + v) dt dv =0 dt y = Cx
The Concepts
dx = Ax dt y = Cx
Algebraic Criteria
The system
dx = Ax + Bu, dt is controllable if the matrix y = Cx
Controllability: Assume that the system is at the origin initially. Can we find a control signal so that the state reaches a given position at a fixed time? Notice we do not require that it stays there! Observability: Can the state x be determined from observations of the output y over some time interval.
Kalmans Decomposition
A linear system can be transformed to the form
A dx 21 = 0 dt
0 y = ( C1 0 A11 0 A22 0 0 C2 A13 A23 A33 A43 0) x 0


=
A11 0
A12 A22
xc xc ¯
+
B1 0
u
B A24 2 x+ u 0 0
A44 0
B1

where the states xc are controllable and xc ¯ are non-controllable.
d dt xo xo ¯
=
A11 A21
0 A22 xo 0) xo ¯
Prototype of Non-controllable System
Two identical systems driven by the same input. Intuitively: no way to make the systems move in opposite ways. A simple example
u
s+a s+b
υ
s+b s+a
y
The system has the transfer function G (s) = 1. Natural questions: • Is the system equivalent to the system y = u? • What happens with the modes that are cancelled?
dx1 = − x1 + u dt dx2 = − x2 + u dt
Prototype of Non-observable System
Two identical systems whose outputs are added. Intuitively: no way to find out which system generated the output. A simple example
Wc = ( B AB A2 B . . . An−1 B )
has full rank. The system observable if the matrix C CA CA2 Wo = . . .
CAn−1
has full rank.
c K. J. Åström August, 2001 1
Example of Cancellation
Consider the system
Example of Cancellation ...
Introduce the state representation
dx1 = −ax1 + ( b − a)( x2 + u) dt dx2 = −bx2 + ( a − b)u dt y = x1 + v = x1 + x2 + u v = x2 + u
Lecture 14 - Controllability and Observability
K. J. Åström 1. Introduction 2. The Concepts 3. Structure of Linear Systems 4. Cancellation of Poles and Zeros 5. Summary
Cancellation of Poles and Zeros
Formal calculations with Laplace transforms sometimes leads to cancellation of poles and zeros. Consider the system
dy du = dt dt
→ y(t) = u(t)
y = (C
Notice that the state v which models the load disturbance is not controllable from u.
There are also design methods where it is deliberately attempted to cancel poles and zeros. It is important to understand what happens when this is done. The decomposition of a linear system gives good insight into what happens when working with transfer functions.
Change coordinates to
1 (Leabharlann Baidux1 + x2 ) 2 1 z2 = ( x1 − x2 ) 2 z1 = x1 = z1 + z2 x2 = z1 − z2
c
K. J. Åström August, 2001
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Example of Cancellation ...
u
Example of Cancellation ...