Separation Principle. Controllers & Observers Plant Controller Observer Picture.
Transcript of Separation Principle. Controllers & Observers Plant Controller Observer Picture.
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Separation Principle
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Controllers & Observers
( ) ( ) ( )
ˆ
( )
ˆ( ) ( ) ( )
(
ˆ( ) ( ) ( ) ( )
)
) (
u t Kx t
x t A L
x t Ax t B
C x t Bu t
u t y t Cx
y t
t
L
r t
Plant
Controller
Observer
Picture
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Composite state model
( ) ( ) ( ) ( ) ( )
ˆ ˆ( ) ( ) ( ) (
ˆ( ) ( )
) ( )
ˆ( )ˆ
(
( )(
)
0(
))
)(
x t A LC x t Bu t Ly t
L
A B x t B
C A LC BK x tx t
x t Ax t Bu t y t Cx t
x
u t Kx t r t
Kr t
t
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Similarity Transformation
1
( )
( ) 0
ˆ ˆ( )
(ˆ( )ˆ( )
0 0
0
()
0
) Kr t
A B x t B
C B
x t I x xP
e t I I x x
IP P
I I
I A B I
I I C I I
A BK I
A LC A
L A LC BK x tx t
L A LC B
C
K
L I
K
I
x t
0
( ) ( )
( ) 0 ( ) 0
A BK BK
A LC
x t A BK BK x t Br
e t A LC e t
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Controllability of Composite System
2 2 1
( ) ( )
( ) 0 ( ) 0
0 0 0 0
nc c c
Comp
x t A BK BK x t Br
e t A LC e t
B A B A A
Composite system cannot be controllable. (rows of zeros)•Look at transfer function of composite system
•Poles and zeros
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Stability Discussion
• K stabilizes “control system”
• L stabilizes “observer”
• Values of K and L are largely independent of each other
• Observer must be “more stable” than controller
• Separation Principle
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Satellite Example
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More Stability
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Energy Storage & State
component Energy/Power
Resistor I2R=V2/R
Capacitor, v V2/(2C)
Inductor, i LI2/2
Friction element Bv2/2
Mass, v Mv2/2
Spring, dx K(dx)2/2
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Free Response and Energy
• Free response indicates how energy stored in the system “leaves” the system
( ) (0)Atx t e x• If x(t) goes to zero, then the energy initially
stored in the system (kinetic, potential, inductors, capacitors) is dissipated in friction elements, resistors, or radiates to the ambient world
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Forced Response and Energy
• The forced response shows how energy from the input function (and bias sources) enters the states
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Stability and Energy Dissipation
• The conceptual (not literal) connection between energy stored in the states and whether or not it dissipates to zero in the free response, leads to a powerful test of stability associated with the name of Lyapunov.
• See Chapter 8: pps. 193, 198, 200, 209
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Definitions
• The origin is stable if, for any given value of ε > 0, there is a value of δ(ε) > 0 such that if ||x(0)|| < δ then ||x(t)|| < ε for all t > 0 (see Fig. 8.1 pg. 195).
• The origin is asymptotically stable if it is stable, and if there exists δ’ > 0 such that whenever ||x(0)|| < δ’ then limt||x(t)|| = 0.
• Diagonal systems
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Energy Functions & Lyapunov stability
• Lyapunov functions energy functions• Consider a mechanical system with no
energy sources, consider any function built up from kinetic and potential energy of its components. This function does not increase as time passes.
• Consider an electrical system with no energy sources, consider any function built from the energy stored in the capacitors and inductors. This function does not increase as time passes.
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2nd Method of Lyapunov
Along every solution x(t) contained in Ω
1. If v(x(t)) is decreasing except at the origin, then the origin is asymptotically stable, and all initial conditions inside Ω produce trajectories that approach the origin;
2. If v(x(t)) is increasing except at the origin, then the origin is unstable.
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Energy Functions for LTI systems• v(x) = xTPx, P symmetric positive-definite.
( ) ( )
( )
( )
T T
T T
T T T
T T
T
T
x t Ax t
v x x Px x Px
Ax Px x PAx
x A Px x PAx
x A P PA x
x Qx
A P PA Q
• Choose real, symmetric, positive definite Q.
• Find, if possible, real, symmetric, positive definite P
• If successful, conclude stability.
• Approach generalizes to non-linear systems
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Positive/Negative (semi-) definite matrices, x != 0
• xTQx > 0 positive definite• xTQx < 0 negative definite• xTQx >= 0 positive semi-definite• xTQx <= negative semi-definite
• Symmetric QT=Q• Real every element of Q is real• Q is positive definite if, for some orthogonal S, STQS is
– Diagonal,– Has non-zero, positive elements on the diagonal
• Diagonal matrix examples
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Positive Definiteness
• A real symmetric matrix M is positive definite if and only if any one of the following conditions holds:
– All eigenvalues of M are positive
– There exists a nonsingular matrix N such that M = NTN
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Connection between Q and P• Q is any positive definite matrix and let
( ) ( ) ( ), ( ) (0)T Atv t x t Qx t x t e x • Total energy dissipated is
0
0
0 0
( ( )) ( ) ( ) (0) (0)
(0) (0) (0) (0)
T
Tt
A t At
t
t t tT T A t At
t t t
T T
v t dt x t Qx t dt x e Qe x
e Qe
dt
x x x xdt P
• Now consider ATP+PA
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ATP+PA=?
0
0 0
0 0
0
lim
0
T T
T T
T T
T
T
T T
t tT
t tA t At A t At
t t
A t At A t At A t A
A t At A t A
t t
t
t
t
t
t t
A t At
P P e Qe dt e Qe dt
e Qe e Qe
A A A A
dA e Qe
e Qe e Q
A dt dtdt
e Q e
Q Q
e
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Solving Lyapunov Eqns
• ATP+PA=-Q– Given Q, A, find P– “vectorize” to get Mp=q– Solve for p– Reconstruct P from p
– Solutions exist “if and only if” λi+ λj ≠0 for eigenvalues of A, including when i = j.
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Satellite Example
1,1 1,2 1,1 1,2 1,1 1,2
1,2 2,2 1,2 2,2 1,2 2,2
1,2 1
2
,1
2
,2
,2 1,2
1,2
2,2
0 1
0 0
0 1 0 0
0 0 1 0
2
0
T
A
AP PA Q
p p p p q q
p p p p q q
p q
p q
q
p q
Q cannot be totally arbitrary since q2,2 must be 0.Can Q and P be positive definite?
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Satellite Example
1,1 1,2 1,1 1,1
1,2 1,1 1,2
2 21,1 1,1 1,22 2 2
1,1 1,2 1,1 1,2 , /
21,1 1,2 1,1 , /
2,
0 1
0 0
/ 2
0 / 2
4det( ) ( )
2
det( ) ( )( ) / 4
det( )
T
Q
P
A
A AP PA Q
q q p qQ P
q q q
q q qsI Q s s q q s q s q
sI P s p s q q
sI A s
, , , ,0 0A A A A
Q cannot be positive definite since it has a negative EV.
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With Feedback
11 12 11 12 11 12
12 22 12 22 12 22
12 11
11 12 2
22 11 1
2 12 12 22 22
2 12
0 1
2 2
0 1 0 2
2 2 1 2
2
2 2 4 4
0 2 0
2
2 2
2 1
0 4 4
fb
Tfb fb
A A BK
A P PA Q
p p p p q q
p p p p q q
p q
p p p
p p p q
q p p q
11 111
12 12
22 22
11 11 12 22
12 11
22 11 22
6 4 11
4 0 084 0 2
6 4 /8
/ 2
2 / 4
p q
p q MP Q M
p q
p q q q
p q
p q q
Any Q produces a P.Does pos-def Q produce pos-def P?
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With Feedback
11 12 11 12 11 12
12 22 12 22 12 22
11
12
22
211 22 12
0 1
2 2
0 1 0 2 1 0
2 2 1 2 0 2
1
1/ 2
1
det( ) ( )( ) ( 1)( 2)
det(
Tfb fb fbA A BK A P PA Q
p p p p q q
p p p p q q
p
p
p
sI Q s q s q q s s
s
2 2 2
11 22 11 22 12) ( ) 2 3/ 4
( 1.5)( .5)
I P s p p s p p p s s
s s
Any Q produces a P.Does pos-def Q produce pos-def P?
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With Feedback and Observer
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Uncancelled pole-zero pairs
• CCF– Observable?
• OCF– Controllable?
• Subspaces– CO– Not C, O– Not O, C– Not O, Not C
• Controllable if input affects all state components directly or indirectly
• Observable if all state components affects output directly or indirectly
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Example
23 2
0
1( )
3 3 10 1 0 0
0 0 1 0 1 1 0
1 3 3 1
0
( )
( )
0 1 0
1 2 1 1 0
1
2
??
1
10 1
0
0
0
c
c
o
sH s
s s s
x x u y x
rank
rank
z z u y z
z Px
s
P
s
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Kalman Canonical form (1)
0 0
00
c c cc c cc c
c c c
o c o oo
o oo c o o
x A A x Bu y C C x
x A x
x A x Bu y C x
x A A x B
Uncontrollable states can affect controllable states.Controllable states cannot affect uncontrollable states.
Observable states can affect unobservable states.Unobservable states cannot affect observable states.
Picture.
Compute Transfer functions.
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Kalman Canonical Form (2)
?0
? ?
0 0 0 0
0 ? 0
0 0
0
?
0co co co co
co co co co
co co co
co co co
co
coco co
co
co
x A x B
x A x Bdu
x A xdt
x A x
x
xy C C
x
x
Compute Transfer Function.
Depends only on Observable and controllable portion.
Hence lower degree polynomials.
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Design via Similarity Transformations -- Picture
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Signal Energy
• Example 7, pg. 202
• Define signal energy
• Compute energy in free response due to x(0)
• P is solution to Lyapunov Eqn.
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Optimal Control• See Example 8, Pg 203
• Introduce SVFB, need to choose K
• Define quadratic cost functional in terms of – Positive definite R– Positive semidefinite Q
• Substitute y=Cx, u=-Kx+u’
• Substitute free response, with initial state x(0)
• Compare integrand with signal energy integrand, solve resulting Lyapunov eq.