Observer design by pole placement

ELEC 3035, Lecture 7: Observer design
Ivan Markovsky
• Observers
• Observer design by pole placement
• Duality between observer and controller design
• Pole placement by output feedback
ELEC 3035 (Part I, Lecture 7) Observer design 1 / 12

General observer design problem
Given dynamical system B ext with two types of external variables:
• observed variables w
• to-be-estimated variables z
B ext
w
z
find system (called observer) accepting w and producing z
w
Observer
z
We will consider the case: B = B i/s/o (A,B,C,D), w = (u,y), z = x.
Lecture 6: nonrecursive, feedforward observer for the initial state x(0)
Now our goal is recursive feedback observer for the current state x(t)
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Output feedback control
separation and certainty principles
We can extend a given state-feedback controller
to output feedback controller
u = Kx
u = K̂x
by using the observer state estimate ̂x in place of x.
u
Plant
y
K
̂x
Observer
Output feedback controller
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Internal model and feedback principles
The observer design is based on the following principles:
1. Internal model: the model run by u, gives an estimate ̂x for x
2. Feedback: correct the estimate ̂x, so that the error
x(t) −̂x(t) =: e(t) → 0 as t → ∞
Let the feedback be a linear function of the output error
feedback correction = L(y − ŷ)
Then the observer for the model B i/s/o (A,B,C,D) is
σ̂x = Âx + Bu − L(y − ŷ)
ŷ = Ĉx + Du
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Error dynamics
Our goal is to choose L, so that the state error e(t) → 0 as t → ∞.
The dynamics of e is
σe = σ(x −̂x)
= Ax + Bu − Âx − Bu + L(y − ŷ)
= A(x −̂x)+LC(x −̂x)
= (A+LC) e
} {{ }
A o
i.e., e ∈ B ss (A o ) — an autonomous LTI system.
Therefore, e(t) → 0 as t → ∞ is equivalent to stability of B ss (A o ).
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Comparison with state-feedback stabilization
In the state feedback stabilization problem we have
σx = Ax + Bu and u = Kx
which gives an autonomous LTI closed loop system
σx = (A+BK) x
} {{ }
A c
and the aim is to choose K , so that B ss (A c ) is stable.
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Observer design by pole placement
The condition e(t) → 0 as t → ∞ is a minimum requirement.
In fact we want e(t) → 0 fast
(possibly in a finite (small) number of steps deadbeat observer)
The error dynamics is governed by the poles of the matrix
A o := A+LC
so for desired error dynamics we can
select desired pole locations of A o and choose L to achieve them.
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Duality of the observer PP and controller PP problems
Observer PP problem: Choose L, so that
det ( zI −(A+LC) ) = p des (z)
Controller PP problem: Choose K , so that
det ( zI −(A+BK) ) = p des (z)
Observer PP is not a new problem:
det ( zI −(A+LC) ) ( (zI ) ) ⊤
= det −(A+LC)
(
)
= det zI −(A ⊤ + C ⊤ L ⊤ )
( )
= det zI −(Ã+ ˜B ˜K)
=⇒ observer PP is controller PP for the dual system.
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The results for state feedback PP can be restated for observer PP:
Theorem:
The eigenvalues of A+LC can be assigned choosing L
to any locations in C if and only if A,C is observable.
Observer canonical form ↔ Controller canonical form
Lemma:
• Let A,c and A ′ ,c ′ be two observable pairs and
• assume that A and A ′ have the same char. polynomials.
Then there is a unique similarity transformation given by the matrix
T := ( O(A ′ ,c ′ ) ) −1 O(A,c)
such that
T −1 AT = A ′ and cT = c ′ .
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Closed-loop system with output feedback controller
Consider the closed loop system
u
Plant
y
K
̂x
Observer
Output feedback controller
where
Plant: σx = Ax + Bu, y = Cx + Du
Observer:
State feedback controller:
σ̂x = Âx + Bu − L(y − Ĉx − Du)
u = K̂x
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Feedback controller:
σ̂x = (A+LC)̂x +(B + LD)u − Ly,
= (A+LC + BK + LDK)̂x − Ly
u = K̂x
Note: the feedback controller is a dynamical system
Closed-loop system:
[ [ ]
x̂x] A BK
σ =
−LC A+LC + BK][ x̂x
Note: closed-loop system order = plant order + controller order
Error equation:
[ [ ]
x A+BK −BK x
σ =
e]
0 A+LC][
e
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Example: output feedback deadbeat control
−2
−3
−4
1 x 105 0
t
1
−1
0
y
−1
−2
5 10 15 20
−3
2 x 105 t
u
5 10 15 20
15th order single-input open-loop system, 30 order closed-loop system
(The same system as the one used in the example of Lecture 1)
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