Page : 1 EE406 Control Systems Lecture 18 : State Space Design
UCSI University Faculty of Engineering
Kuala Lumpur, Malaysia Department of Mechatronics
Lecture 18
State Space Design
Mohd Sulhi bin Azman Lecturer
Department of Mechatronics UCSI University
1 August 2011
Contents
• Open loop and closed loop system representation
• Controller and observer design via
pole-placement method
Page : 3 EE406 Control Systems Lecture 18 : State Space Design
State Space Representation
• Open-loop control system representation:• Closed-loop control system representation with feedback:
Pole Placement Design Method
• The pole-placement design method is one of the
many method used to design the controller and
observer for a system.
• The problem of placing the poles at the desired
location (on the s-plane) is called the
pole-placement design method.
Page : 5 EE406 Control Systems Lecture 18 : State Space Design
Controller Design via Pole Placement Method
• To design a controller with pole-placement
method, the system considered must be
completely controllable.
• Furthermore, all state variables are assumed
measureable and available for feedback, and
then only the poles of the closed-loop system
maybe placed at any desired locations by means
of state feedback through an appropriate state
feedback gain.
Controller Design via Pole Placement Method
• Consider the following state equations:
• The open-loop state space representation of the plant is given by:
A Bu y C
= +
=
x x
x
Page : 7 EE406 Control Systems Lecture 18 : State Space Design
Controller Design via Pole Placement Method
• The closed-loop representation is given as:
• In pole placement method, another feedback is
introduced to feed the state back to the
control (input) signal.
Controller Design via Pole Placement Method
• Now, the control signal is:
• Substituting gives:
• The solution of the above equation is:
u
= −
K
x
(
)
(
)
(
)
( ) ( )
A B K A BK t A BK t
= + −
= −
= −
x x x
x x
x x
ɺ ɺ ɺ
( )
( )t =eA BK t− (0)
Page : 9 EE406 Control Systems Lecture 18 : State Space Design
Controller Design via Pole Placement Method
• Thus, in state space, given the pair (A,B), we can
always determine the K (gain), to place all the
system closed-loop poles in the left-half of the
plane if and only if the system is controllable –
that is, if and only if the controllability matrix
C
Mis of full rank.
• And for a full rank C
Min a SISO system, it
implies that the C
Mmatrix is invertible.
Observer Design via Pole Placement Method
• In designing the observer using the
pole-placement method, all state variables are
assumed to be available for direct measurement
and feedback.
Page : 11 EE406 Control Systems Lecture 18 : State Space Design
Observer Design via Pole Placement Method
• Open-loop observer:
Observer Design via Pole Placement Method
Page : 13 EE406 Control Systems Lecture 18 : State Space Design
Observer Design via Pole Placement Method
• Exploded view of a closed-loop observer,
showing feedback arrangement to reduce
state-variable estimation error:
Observer Design via Pole Placement Method
Page : 15 EE406 Control Systems Lecture 18 : State Space Design
Observer Design via Pole Placement Method
• Consider a system defined by:
• And another estimator or observer is defined as:
• Subtracting gives:
A Bu y C = + = x x x ɺ A Bu C = + = x x y x ɺɶ ɶ ɶ ɶ
(
)
(
)
Ay y C
− = −
− = −
x x x x
x x
ɺ
ɺ ɶ ɶ
ɶ ɶ
error
…(i)
…(ii)
…(iii)
Observer Design via Pole Placement Method
• Note that the design of the observer is different from the design of the controller. Similar to the design of the controller vector K, the design of the observer consists of evaluating the constant vector L, so that the transient response of an observer is faster than the response of the controlled loop in order to yield a rapidly updated estimate of the state vector.
• The vector (matrix) L is the observer gain matrix and is to be determined as part of the observer design
Page : 17 EE406 Control Systems Lecture 18 : State Space Design
Observer Design via Pole Placement Method
• Let us now consider the following figure :
• The state equations are:
(
)
u y y y C= + + −
=
x Ax B L
x
ɺɶ ɶ ɶ
…(iv)
Observer Design via Pole Placement Method
• Now, subtracting equation (iv) from (i), results in:
• Re-arranging gives:
• Simplifying gives:
(
) (
)
(
)
A L y y
y y C
− = − − −
− = −
x x x x
x x
ɺ
ɺ ɶ ɶ ɶ
ɶ ɶ
(
)
(
)
(
)
A LC
y y C
− = − − −
− = −
x x x x x x
x x
ɺ
ɺ ɶ ɶ ɶ
ɶ ɶ
(
)(
)
(
)
A LC y y C− = − −
− = −
x x x x
x x
ɺ
ɺ ɶ ɶ
Page : 19 EE406 Control Systems Lecture 18 : State Space Design
Observer Design via Pole Placement Method
• Let the estimation error signal be denoted as e, hence:
• Take note that the dynamic behaviour of the error vector is determined by the eigenvalues of the matrix (A – LC).
• Also note that if the system is completely observable, then it is possible to choose matrix L such that (A – LC) has arbitrarily eigenvalues.
(
A LC
)
y
y
C
=
−
− =
x x
x
e
e
e
ɺ
ɶ
Approaches to State Space Design
• There are three methods to find the gain, K:
1. Direct approach.
2. Direct substitution approach. 3. Ackermann’s formula.
Page : 21 EE406 Control Systems Lecture 18 : State Space Design
Ackermann’s Formula
• Consider the following system:• We assume that the system is completely state
controllable. And furthermore, we also assume that the desired closed-loop poles are at s=µ1, µ2, …, µn. The control law says that:
• Which then gives, upon substitution:
A
Bu
y
C
=
+
=
x
x
x
ɺ
.
u
= −
Kx
(
A BK
)
=
−
x
ɺ
x
Ackermann’s Formula
• Next, let us define: , hence:• Thus, the eigenvalues are:
• Remember that the Cayley-Hamilton theorem says that every square matrix A satisfies its own characteristic equation. Then:
• We are going to use the above equation to derive Ackermann’s formula, and we choose n=3 for simplicity.
(
)
Aɶ= A BK−
x
ɺ
ɶ
=
A
ɶ
x
(
) (
1)(
2) (
n)
sI
− =
A
ɶ
sI
−
A
+
BK
= −
s
µ
s
−
µ
⋯
s
−
µ
( )
1 21 2 1
0
n n
n n n
φ
α
−α
−α
α
−
=
+
+
+ +
+
=
Page : 23 EE406 Control Systems Lecture 18 : State Space Design
Ackermann’s Formula
• Consider the following identities:• Let us multiply the above identity in order by
wherein , respectively and let us add all the results together to obtain:
(
)
(
)
2
2 2 2 2 2
factorize 3
3 3 2 2
=
= −
=
−
=
−
−
+
=
−
+
=
−
=
−
−
−
I
I
A
A BK
A
A BK
A
ABK
ABK
B K
A
ABK
BKA
A
A BK
A
A BK
ABKA BKA
ɶ
ɶ
ɶ
ɶ
ɶ
ɶ
3
,
2,
1,
0α α α α
01
α
=
( )
2 33 2 1 0
φ
A
ɶ
=
α
I
+
α
A
ɶ
+
α
A
ɶ
+
α
A
ɶ
Ackermann’s Formula
• On expansion, we obtain:
• On substitution, we obtain:
• Simplifying gives:
( )
(
)
(
) (
)
2 3
3 2 1 0 3 2
2 3 2 2
1
φ α α α α α α
α
= + + + = + −
+ − + + − − −
A I A A A I A BK
A ABK BKA A A BK ABKA BKA
ɶ ɶ ɶ ɶ
ɶ ɶ ɶ
( )
(
)
(
)
2(
)
33 2 1 0
1
φ
α
α
α
α
=
=
+
−
+
−
+
−
A
ɶ
I
A BK
A BK
A BK
( )
2 33 2 1 2 1 1
Page : 25 EE406 Control Systems Lecture 18 : State Space Design
Ackermann’s Formula
• We refer back to the equation (1) given in slide 19. This equation was obtain as a result of Cayley-Hamilton theorem and for n=3. We reproduce the equation here:
• And also:
• We next substitute equations (3) and (4) into equations (2), to get:
( )
3 21 2 3
0
φ
A
ɶ
=
A
ɶ
+
α
A
ɶ
+
α
A
ɶ
+
α
I
=
( )
3 21 2 3
0
φ
A
ɶ
=
A
+
α
A
+
α
A
+
α
I
≠
…(3)
…(4)
( )
2 33 2 1 2 1
( )
2 2
1
φ
φ α α α α α
α
= + + + − −
− − − −
A
A I A A A BK ABK
BKA A BK ABKA BKA
ɶ ɶ
ɶ ɶ ɶ
Ackermann’s Formula
• Simplifying leads us to:• Since , the above equation is reduced to:
• Next, we factorize the above expression to get:
( )
2 22 1 1
( )
φ
A
ɶ
=
φ
A
ɶ
−
α
BK
−
α
ABK
−
α
BKA
ɶ
−
A BK
−
ABKA BKA
ɶ
−
ɶ
( )
0
φ
A
ɶ
=
( )
2 22 1 1
φ
A
ɶ
=
α
BK
+
α
ABK
+
α
BKA
ɶ
+
A BK
+
ABKA BKA
ɶ
+
ɶ
( ) (
2)
(
)
22 1 1
Page : 27 EE406 Control Systems Lecture 18 : State Space Design
Ackermann’s Formula
• And because this is a matrix, then we can write equation (5) in the previous slide as:
• Note that since the system is completely controllable, then the inverse matrix exists. Pre-multiplying equation (5) with the inverse gives:
( )
2 2 1 2 1 controllability matrixα
α
φ
α
+
+
=
−
K
KA
KA
A
B
AB
A B
K
KA
K
ɶ
ɶ
ɶ
ɶ
( )
2 2 1 1 2 1α
α
φ
α
− + + = − K KA KA
B AB A B A K KA
K
ɶ ɶ
ɶ
…(6)
Ackermann’s Formula
• Now, note that the Ackermann’s formula is used to find the system gain, K. So, to find K, we pre-multiply
equation (6) by [0 0 1].
• Which actually, if re-written, gives:
[
]
( )
[
]
2 2 1 1 2 10 0 1 0 0 1
α α φ α − + + = − =
K KA KA
B AB A B A K KA K
K
ɶ ɶ ɶ
[
]
2 1( )
0 0 1 − φ
=
Page : 29 EE406 Control Systems Lecture 18 : State Space Design
Ackermann’s Formula
• Ladies and gentlemen, for an arbitrary integer
n, equation (7) can be generally written as:
• And this is what we call the Ackermann’s
formula for evaluating the system gain in
designing the controller in state space.
[
]
2 1 1( )
0 0 0 1 n− − φ
=
K ⋯ B AB A B ⋯ A B A
Page : 31 EE406 Control Systems Lecture 18 : State Space Design
Example 2 : Observer Design
Next Step
• Textbook reference : Chapter 12.
• Homework 17 has been posted on the course
website. Attempt them. You do not have to
submit Homework 17 as it will not be graded.
