Recursive Estimation
Raffaello D’Andrea
Spring 2011
Problem Set:
Observers and the Separation Principle
Last updated: July 22, 2011
Notes:
• Please report any errors found in this problem set to the teaching assistants
([email protected] or [email protected]).
Problem Set
Problem 1
For the LTI system
x(k) = A x(k−1)
z(k) = H x(k),
show that the following equivalence statement holds:
(HA, A) detectable
⇔
(H, A) detectable.
Problem 2
In this problem you will design linear state observers for two different LTI systems using pole
placement.
a)
Design a linear, static gain state observer for the discrete-time LTI system
0 −6
1
x(k) =
x(k−1) +
u(k)
1 5
1
z(k) = 0 1 x(k).
The poles of the observer dynamics should be placed at p1 = 0.8 and p2 = 0.7. You can
use the MATLAB command place to obtain the observer gain K.
Derive a state space representation of the closed loop system consisting of the plant and
the state observer. Simulate in MATLAB the observer error for the initial conditions
x(0) = (1, 1) and x̂(0) = (0, 0) and the input u(k) = 0 for all k.
b)
Design a linear, static gain state observer for the discrete-time LTI system

6
5.25
x(k) = 
 −5
−1
1 0
z(k) =
0 0



0.5
0.5
1
6.5


−0.5 1.25 6.5 
 x(k−1) +  0.5  u(k)


0.5 
0.5
0
−5.5
−0.5
−0.5
0
−0.5
1 2
x(k)
1 1
The given system is detectable, but not observable.
Since the system is unobservable, you cannot arbitrarily place all poles of the observer
dynamics. In order to design the observer, you first have to decompose the system into
an observable and unobservable subsystem using an appropriate state transformation (you
can do this using the MATLAB command obsvf). Then, place the poles of the observable
subsystem at p1 = 0.8 and p2 = 0.7 (using MATLAB command place). Compute the
observer gain matrix K in the original state coordinates and the poles of the observer
dynamics (four poles in total).
Derive a state space representation of the closed loop system consisting of the plant and
the state observer. Simulate in MATLAB the observer error for the initial conditions
x(0) = (1, 1, 1, 1) and x̂(0) = (0, 0, 0, 0) and the input u(k) = 0 for all k.
2
Problem 3
Derive a state space model of the closed loop system consisting of the plant
x(k) = A x(k−1) + B u(k) + v(k)
z(k) = H x(k) + w(k)
with process noise v(k) and measurement noise w(k); the linear observer
x̂(k) = A x̂(k−1) + B u(k) + K z(k) − ẑ(k)
ẑ(k) = H A x̂(k−1) + B u(k) ;
and the static feedback controller
u(k) = F x̂(k−1).
Use (x(k), e(k)) as the state vector, where e(k) := x(k) − x̂(k).
Remark: The same analysis was done in the lecture, except that, in the lecture, the case without
noise (v(k) = 0 and w(k) = 0 for all k) was considered.
Problem 4
Come up with an example, where the separation principle does not hold: that is, where the
combination of a state observer with stable dynamics and a state-feedback controller with stable
dynamics leads to an unstable closed loop system.
3
Sample solutions
Problem 1
We first show the implication
(HA, A) detectable
⇒
(H, A) detectable
by contraposition (that is, by showing that (H, A) not being detectable implies that (HA, A) is
not detectable). Assume (H, A) is not detectable. Then, there exist λ with |λ| ≥ 1 and v 6= 0
such that
Av = λv
Av = λv
Av = λv
⇒
⇒
,
Hv = 0
Hλv = 0
HAv = 0
which implies (HA, A) is not detectable.
To show the opposite direction,
(H, A) detectable
⇒
(HA, A) detectable,
assume (HA, A) is not detectable. Then, there exist λ with |λ| ≥ 1 and v 6= 0 such that
Av = λv
Av = λv
Av = λv
⇒
⇒
,
HAv = 0
Hλv = 0
Hv = 0
which implies (H, A) is not detectable.
Problem 2
The MATLAB code for both problem parts is available on the class website.
For the simulation of the observer response to an initial deviation, we derive a state space
representation of the plant together with the observer. For the observer state, we have
x̂(k) = Ax̂(k−1) + Bu(k) + K z(k) − ẑ(k)
= Ax̂(k−1) + Bu(k) + KH Ax(k−1) + Bu(k) − Ax̂(k−1) − Bu(k)
= (I − KH)Ax̂(k−1) + KHAx(k−1) + Bu(k).
Hence, the closed-loop system consisting of plant and observer reads
B
x(k−1)
A
0
x(k)
u(k).
+
=
B
KHA (I − KH)A x̂(k−1)
x̂(k)
Notice from the simulations that the estimation error e(k) := x(k) − x̂(k) converges to 0, even
though the system states diverge.
Problem 3
First, consider the error dynamics of the state observer:
e(k) = x(k) − x̂(k)
= Ax(k−1) + Bu(k) + v(k)
− Ax̂(k−1) − Bu(k) − KH Ax(k−1) + Bu(k) + v(k) − Kw(k)
+ KHAx̂(k−1) + KHBu(k)
= (I − KH)Ae(k−1) + (I − KH)v(k) − Kw(k).
4
And the system state with state feedback:
x(k) = Ax(k−1) + BF x̂(k−1) + v(k)
= Ax(k−1) + BF x̂(k−1) + x(k−1) − x(k−1) + v(k)
= (A + BF )x(k−1) − BF e(k−1) + v(k).
The complete closed loop system can therefore be written as
x(k)
A + BF
−BF
x(k−1)
I
0
=
+
v(k) +
w(k).
e(k)
0
(I − KH)A e(k−1)
I − KH
−K
Notice that the stability of the linear system (the eigenvalues of the system matrix) is not affected
by including the noise in the analysis (compare with result from lecture).
Problem 4
We summarize a counterexample for the separation principle, which has been presented in [1].
Therein, it is shown that, for a nonlinear discrete-time system, the combination of an exponentially stable state-feedback controller and a state observer with exponentially stable error
dynamics may result in an unstable output feedback control system.
Consider the nonlinear, discrete-time system
x(k) = x(k−1) + x3 (k−1) + u(k).
(1)
The state feedback controller
u(k) = −x(k−1) − x3 (k−1)
(2)
makes the closed loop system exponentially stable; in fact, x(k) = 0 for all k > 0.
Now assume that instead of the true state x(k), an estimate x̂(k) of the state is used for feedback.
The estimate is obtained from a state observer with exponentially decaying estimation error
e(k) := x(k) − x̂(k); that is,
e(k) = α e(k−1) ,
0<α<1
(3)
with some decay rate α. An example of such an observer1 with access to the measurement
z(k) = x(k) is given by
x̂(k) = α x̂(k−1) + (1−α) z(k−1) + z 3 (k−1) + u(k).
(4)
Using the state estimate feedback controller,
3
u(k) = −x̂(k−1) − x̂3 (k−1) = − x(k−1) − e(k−1) − x(k−1) − e(k−1) ,
the closed loop system is governed by (combining equations (1), (3), and (5))
x(k) = 3 x(k−1) e(k−1) x(k−1) − e(k−1) + e(k−1) + e3 (k−1)
e(k) = α e(k−1).
1
(5)
(6)
(7)
We note that the problem of an observer design for a scalar system in a deterministic scenario (that is, without
sensor noise) is an artificial one, since with a measurement of the only state, z(k) = x(k), there is really no need
for an observer. Nonetheless, we want to show only the existence of a counterexample for the separation principle
here and consider a scalar system for the sake of simplicity.
5
In [1], it is proven that the set
1
2
2
Ma := (x, e) ∈ R : x > 0, 0 < e ≤ a, xe ≥ a +
3α
is an invariant set for solutions of (6), (7); that is, if (x(k0 ), e(k0 )) is in Ma for some k0 , then
(x(k), e(k)) remains in Ma for all k ≥ k0 . For the details of this proof, refer to [1].
Now assume (x(0), e(0)) ∈ Ma .2 Then, for all k,
x(k) e(k) ≥ a2 +
1
> a2 .
3α
(8)
Because of (3), we have e(k) → 0 as k → ∞. Therefore, (8) implies that x(k) → ∞ as k → ∞.
That is, the overall closed loop system is not (globally) stable even though the state-feedback
controller and the state observer are stable individually.
References
[1] V. Sundarapandian, “A counterexample for the global separation principle for discrete-time
nonlinear systems,” Applied Mathematics Letters, vol. 18, no. 7, pp. 765–768, 2005.
2
Notice that such initial values exist. For example, for e(0) > 0, we can choose a = e(0) and let x(0) ≥
e(0) + 3α 1e(0) , then (x(0), e(0)) ∈ Me(0) .
6