Lecture 11:

Controllability and
Observability
2019 Autumn

1
Observable/Unobservable Decomposition
 Theorem 6.O6: Consider an n-dimensional state
space equation with (A,B,C,D) and suppose
C
(O) = rank CA = n2 < n
⋮
CAn−1
i.e., the system is unobservable. Let the nn matrix of
p1
⋮
change of coordinates P be defined as P = pn2
⋮
pn
2
Observable/Unobservable Decomposition
where the first n2 rows of P are any n2 independent rows
in O, and the remaining are arbitrarily chosen so that P
is nonsingular.
 Then the equivalence transformation x=Px transforms
(A, B, C, D) to

3
Observable/Unobservable Decomposition
 The states in the new coordinates are decomposed
into decoupled from
xO : n2 observable states the output y
xO: n  n2 unobservable states

4
Observable/Unobservable Decomposition
 The reduced order state space equation of the
observable states
xO(t) = AOxO(t) + BOu(t)
y(t) = COxO(t) + Du(t)
is observable and has the same transfer function as the
original state equation (A, B, C, D).

5
 Kalman Decomposition
 The Kalman decomposition combines
the controllable/uncontrollable and
observable/unobservable
decompositions.

6
Kalman Decomposition
 Every state-space equation can be transformed, by
equivalence transformation, into a canonical form that
splits the states into
 Controllable and observable states
 Controllable but unobservable states
 Uncontrollable but observable states
 Uncontrollable and unobservable states

7
Kalman Decomposition
 The Kalman decomposition brings the system to the
form

8
Minimal realisation
 A minimal realisation of the system is obtained by
using only the controllable and observable states from
the Kalman decomposition.
xCO(t) = ACOxCO(t) + BCOu(t)
y(t) = CCOxCO(t) + Du(t)

9
Example: Kalman Decomposition
Consider the system in modal canonical form

 The first 1 is controllable and observable.

 2 is not controllable, although observable.
 3 is controllable and observable.

10
Example: Kalman Decomposition
 Thus a minimal realisation of this system is given by

with transfer function

11
Example: Kalman Decomposition

12
Controllability/Observability for
Jordan blocks
 In Jordan blocks, refer to
 The entries of B corresponding to the last row of

each Jordan block.

 The entries of C corresponding to the first column

of each Jordan block.

 They must be linearly independent to be controllable

or observable, respectively.

13
Example 6.10: Jordan blocks
 λ1 1 0 0 0 0 0 0 0 0
0 λ1 0 0 0 0 0  1 0 0 
 
0 0 λ1 0 0 0 0 0 1 0
   
x=  0 0 0 λ1 0 0 0  x  1 1 1 u
0 0 0 0 λ2 1 0 1 2 3
   
0 0 0 0 0 λ2 1 0 1 0
0 λ 2  1 1 
 0 0 0 0 0  1

1 1 2 0 0 2 1 
y= 1 0 1 2 0 1 1  x
1 0 2 3 0 2 0 

14
Example 6.10: Jordan blocks
 λ1 1 0 0 0 0 0 0 0 0
0 λ1 0 0 0 0 0  1 0 0 
 
0 0 λ1 0 0 0 0 0 1 0  L.I.
   
x=  0 0 0 λ1 0 0 0  x  1 1 1 u
0 0 0 0 λ2 1 0 1 2 3
   
0 0 0 0 0 λ2 1 0 1 0
0 λ 2  1 1 1  L.I.
 0 0 0 0 0 

1 1 2 0 0 2 1 
y= 1 0 1 2 0 1 1  x
1 0 2 3 0 2 0 

L.I. L.D. (=0)

15
Example 6.11
0 10 0  10 
0 01 0   9 
x=  x+ u
0 00 0  0
   
0 00 2   1 
y= 1 0 0 2 x

 The system is not controllable but observable.

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Discrete-time state equations

Controllability after sampling

17
Discrete-Time Systems
 For controllability and observability of a discrete-
time equation
x[k + 1] = Ax[k] + Bu[k]
y[k] = Cx[k] + Du[k]

 We can use the same controllability and observability

matrices rank tests that we have for continuous-time
systems.
(ℂ) = rank [B AB A2B ... An-1B] = n
 Controllable

18
Discrete-Time Systems
C
(O) = rank CA = n  Observable
⋮
CAn−1

 The procedures of canonical decompositions are

analogous.

19
Controllability after sampling
 Most control systems are implemented digitally, for
which we need a discrete-time model of the system.

 Using the Parameter Variation Formula, we have seen

how to obtain a discrete-time state space model that is
exact at the sampling instants.

20
Controllability after sampling

 If the continuous-time system is controllable, would

the discretised system be always controllable?

21
Controllability after sampling
 The controllability of the discretised system depends
on the sampling period T and the eigenvalues of the
continuous-time plant.

 Controllability can be lost after sampling.

22
Check controllability after sampling
 Theorem 6.9: If the pair (A, B) is controllable, then
the discretised pair (Ad, Bd) is controllable with
sampling time T if

whenever Re[i  j] = 0.

 This theorem gives a sufficient condition that
preserves controllability after sampling.

23
Check controllability after sampling
In other words …
 If A has only real eigenvalues, then the discretised
equation with any sampling period T > 0 is always
controllable.
 Suppose A has complex conjugate eigenvalues
𝛼 ± 𝑗𝛽; if T is not equal to any integer multiple of
𝜋/𝛽, then the discretised state equation is controllable.
 If T is equal to any integer multiple of 𝜋/𝛽, then the
discretised equation may not be controllable.

24
Example: controllability after sampling
 3 7 5 1 
x=  1 0 0  x+ 0  u
 0 1 0  0 
y=  0 1 2 x

 Since it is in the controllable canonical form, the

system is controllable.
 The pole are −1, −1 ± 𝑗2, hence Re[i  j] = 0.
 The differences in imaginary parts are 2 and 4.

25
Example: controllability after sampling
 Thus the discretised state space equation is
controllable if and only if
2𝜋𝑚 2𝜋𝑚
T≠ = 𝜋𝑚 and T ≠ = 0.5𝜋𝑚
2 4
for 𝑚 = 1,2, …
 The second condition includes the first.
 Hence, the discretised equation is controllable if and
2𝜋𝑚
only if T ≠ = 0.5𝜋𝑚 for any positive integer m.
4

26
Remarks
 A couple of final remarks concerning controllability
and sampling:
1. The sampling condition in Theorem 6.9 only applies
to systems with complex eigenvalues; a discretised
system with only real eigenvalues is controllable for
all T > 0 if its continuous-time counterpart is.
2. This is also applicable to the observability part.

27
Summary
We have learned
 Controllability
 Controllability matrix ℂ
 Observability
 Observability matrix O
 Controllability and Observability Gramian
 Controllability and Observability Decomposition
 Kalman Decomposition
 Discrete-time state equations
 Controllability after sampling.

28
Lecture W11
State Feedback and State
Estimators
2018 Autumn

29
 Contents
 Introduction
 State Feedback
 Regulation and Tracking
 State Estimator
 Feedback from Estimated States

30
Full State Feedback

31
Feedback Control via State Space
 Linear state space control theory involves modifying
the behaviour of an m-input/p-output/n-state system
(assuming D=0)
x(t) = Ax(t) + Bu(t)
y(t) = Cx(t)
which we call a plant/process, or (open-loop) state
equation, by an application of a control law of the form
u(t) = Nr(t)  Kx(t)
where r(t) is the new (reference) input signal.

32
Feedback Control via State Space

 The matrix K is the state feedback gain and N the

feedforward gain.
 Substituting u(t) into the state space equation gives
the closed-loop state equation
x(t) = (ABK)x(t) + BNr(t)
y(t) = Cx(t)

33
Static state feedback control
 State feedback with feedforward pre-compensation

 This type of control is said to be static, because u(t)

only depends on the present values of the state x(t)
and the reference r(t).
 Note that it requires that all states of the system be
measured.
34
State observer
 When not all the states of the system are measurable,
we rely on their estimation by means of an observer,
or state estimator, which reconstructs (or estimates)
the states from measurements of the inputs u(t) and
outputs y(t).
 The combination of state feedback and state
estimation yields a dynamic output feedback
controller.

35
State observer
 Output feedback by estimated state feedback

36
State observer

• Static control

• Dynamic
control

37
Objectives of control
 The basic objective of control

To learn how to design a linear control system by

dynamic output feedback (state feedback + state
observer) to satisfy the desired closed-loop system
specifications in stability and performance.

38
Objectives of control
We will learn:
 techniques for the design of the state feedback gain K
to achieve
 regulation and robust tracking

 pole placement (stabilisation)

 techniques for the design of observers

Considered systems:
 SISO systems

39
State Feedback Design for SISO
 We start by considering SISO systems, and the state
feedback control scheme with feedforward pre-
compensation: u(t) = Nr(t)  Kx(t)
 For simplicity, let us assume temporarily that N = 1.
u(t) = r(t)  Kx(t)

40
Invariance of controllability/observability

 An essential system property in state feedback is

controllability.

 Theorem (Invariance of controllability with state

feedback): For any KR1n, the pair (A–BK, B) is
controllable if and only if the pair (A, B) is
controllable.

 On the other hand, it is interesting to note that

observability is not invariant with respect to feedback.

41
Example
 Example: loss of observability after feedback

 The system is both controllable and observable.

42
Example
 The state feedback control
u(t) = r(t) – Kx(t), where K = [3 1],
yields the closed-loop state equations

0 2
 ℂ=  (ABK, B) is still controllable.
1 0

43
Example
1 1
 O=
2 2
  Observability matrix O is singular, and thus the
closed-loop system with this state feedback is not
observable.

 Observability is not invariant with respect to

feedback.

44
Eigenvalue assignment
 Eigenvalue assignment allows us to place the poles of
the state feedback system where we want them to be.
 Example: eigenvalue assignment by state feedback

1 3 1
x t = x t + u(t)
3 1 0

 (s) = (s  1)2  9 = s2  2s  8 = (s  4)(s + 2)

 Eigenvalues: s = 4, s = 2  The system is unstable.

45
Eigenvalue assignment
 The state feedback u = r  [k1 k2]x gives the closed-
loop system
x  (A  BK)x  Br
 1 3  k1 k 2   1 
     x   r
 3 1   0 0   0
1  k1 3  k 2  1 
    r
 3 1  0 

46
Eigenvalue assignment
 The new system matrix AK = A – BK has the
characteristic polynomial
K(s) = (s – 1 + k1)(s – 1) – 3(3 – k2)
= s2 + (k1 – 2)s + (3k2 – k1 – 8)
 Clearly, the roots of K(s), or equivalently, the
eigenvalues of the closed-loop system can be
arbitrarily assigned by a suitable choice of k1 and k2.

47
Eigenvalue assignment
 For example, for eigenvalues to be placed at 1j2,
the desired characteristic polynomial is
(s + 1 – j2)(s + 1 + j2) = s2 + 2s + 5
= s2 + (k1 – 2)s + (3k2 – k1 – 8).

 By equating k1 – 2 = 2 and 3k2 – k1 – 8 = 5, we obtain

k1 = 4 and k2 = 17/3.

 Thus, the feedback gain K = [4 17/3] will shift the

eigenvalues of the system from 4, 2 to 1j2.

48
Discussion
The previous example shows that
 State feedback allows us to place the eigenvalues of
the state feedback closed-loop system at any position
and that the state feedback gain can be computed by
direct substitution.

However,
 The method of the example is not practical for
systems of higher dimensions.
 It is not clear what role controllability did play in this
eigenvalue assignment.
49
State Feedback Design
 To formulate a general result we need to use the
controller canonical form (CCF) discussed in a
previous lecture.

 Recall that if ℂ = [ B AB ... An-1B ] is full row rank,

then we can always make a change of coordinates
x = Px
under which the state matrices have the form

50
State Feedback Design
 -α1 -α 2 -α n-1 -α n 
 1 0 0 0 

A=PAP -1 =  0 1 0 0 
 
 
 0 0 
 0 1
C=CP -1 =[β1 β2 β n-1 βn ]
1 
0 
 
B=PB=  0 
 
 
0 
 

51
State Feedback Design
 These matrices arise from the change of coordinates
x = Px where
C   B AB A n-1B

P -1  CC 1 with
C   B AB A n-1B


52
Theorem 8.3
 Theorem 8.3 (Eigenvalue Assignment by state
feedback): If the state equation
x(t) = Ax(t) + Bu(t)
y(t) = Cx(t)
is controllable, then the state feedback control law
u(t) = r(t) – Kx(t) with KR1n assigns the eigenvalues of
the closed-loop state equation
x(t) = (A–BK)x(t) + Br(t)
y(t) = Cx(t)
to any desired, arbitrary locations, provided that complex
eigenvalues are assigned in conjugate pairs.

53
Theorem 8.3
 Proof: If the system is controllable, we can take it to
its CCF by the change of coordinates x = Px, which
yields A =PAP-1 and B = PB.
 It is not difficult to verify that
C

and thus 𝑃−1 = ℂℂ−1 .

 On substituting x = Px in the state feedback law, we
have
u = r – Kx = r – KP–1x = r – Kx (K=KP-1)
54
Theorem 8.3
 Since A – BK = P(A – BK)P-1, we see that A – BK
and A – BK are similar, and thus have the same
eigenvalues.
 Now, say that 1, 2, ..., n are the desired closed-loop
eigenvalue locations.
 We can then generate the desired characteristic
polynomial

55
Theorem 8.3
 If we choose

the closed-loop state equation becomes (in the x

coordinates).

56
Theorem 8.3
 Then the state feedback equation becomes

+ Br(t)

57
58
59
Theorem 8.3
 Because the closed-loop evolution matrix A – BK is
still in companion form, we see from the last
expression that its characteristic polynomial is the
desired one K(s).

 Finally, from K = KP-1, we get that K = KP.

QED

60
Matrix ℂ
 The matrix ℂ can be easily built from the coefficients
1, 2, ..., n of the characteristic polynomial of A or
A as 1
1 α1 α 2 αn 2 α n 1 
0 1 α α n 3 αn 2 
 1 
0 0 1 αn 4 α n 3 
C 
 
0 0 0 1 α1 
 
0 0 0 0 1 

(contents in Ch. 7) P-1  CC 1

61
Effects on poles and zeros
 In closed-loop, once the eigenvalue assignment is
performed, the system transfer function from r to y is
given by

The transfer function has poles at the new, desired

locations, but the zeros of the system are the same as
the open-loop system.
 no effect on zeros
62
Procedure for pole placement by state
feedback
1. Obtain the coefficients of the open loop
characteristic polynomial
(s) = sn + 1sn-1 + 2sn-2 + ... + n
2. Form the controllability matrix
ℂ = [B AB ... An-1B ] and ℂ−1 such that
1
1 α1 α 2 αn 2 α n 1 
0 1 α α n 3 αn 2 
 1 
0 0 1 αn 4 α n 3 
C 
 
0 0 0 1 α1 
 
0 0 0 0 1 

63
Procedure for pole placement by
state feedback
3. Select the coefficients of the desired closed-loop
characteristic polynomial
K(s) = sn + α1 sn-1 + α2 sn-2 + ... + αn
and build the state feedback gain in x coordinates
K = [ α1 –1 α2 –2 ... αn –n]

4. Compute the state-feedback gain in the original x

coordinates
K =KP = Kℂℂ-1

64
State Feedback Stabilisation
 We have seen that if a state equation is controllable,
then we can assign its eigenvalues arbitrarily by state
feedback.

 But what happens when the state equation is not

controllable?

 We know that we can take any state equation to the

controllable/uncontrollable canonical form.

65
State Feedback Stabilisation

 Because the evolution matrix A is block-triangular, its

eigenvalues are the union of the eigenvalues of the
diagonal blocks: AC and AC .

66
State Feedback Stabilisation Law
 The state feedback law

yields the closed-loop system

67
State Feedback Stabilisation Law
 We see that the eigenvalues of AC are not affected by
the state feedback, so they remain unchanged.

 The value of K C is irrelevant.

 The uncontrollable states (associated with AC ) cannot
be affected.

68
State Feedback Stabilisation Law

 We conclude that the condition of controllability is

not only sufficient, but also necessary to place all
eigenvalues of A – BK in desired locations.
69
Stabilisability
 A notion of interest in control that is weaker than that
of controllability is that of Stabilisability.
 Definition (Stabilisability): The system
x(t) = Ax(t) + Bu(t)
y(t) = Cx(t)
is said to be stabilisable if AC is stable and if (AC , 𝐵C ) is
controllable.

70
Summary on (Full) State Feedback
 We have presented an overview of the process of
control design via state space methods.

 It involves the design of

 A state feedback gain K

 A state estimator (observer)

 We have shown that if the system is controllable, it is

possible to arbitrarily assign the eigenvalues of the
closed-loop state equation by a suitable choice of K.

71
Summary on (Full) State Feedback
 If the system is not controllable, then we can apply
state feedback of the controllable states only;
uncontrollable states cannot be affected.

 A system is thus stabilisable if those states that are

not controllable are already stable.

72
Assignment 2/3
 The assignment that has been uploaded on ABEEK is
due by 23:59, 2019-11-24.
 It will account for 7% of the final grade.
 There will be 1 more assignment, accounting for 7%
of the final grade; that is, three assignments will acco
unt for 20% for the final grade.
 Late homework will not be accepted.
 Note that the copier and copyee will each receive a
zero.

73