
Homework#3 : 3-5, 3-7, 3-15, 3-17

Due Date: May. 19

Midterm Exam: Apr. 28th 10:00~11:30

Content: Chapter 1, 2
Modern Control Theory
Lecture 8
1
Chapter 3: Controllability & Observability
3.1 Controllability of linear continuous time system
3.1.1 time-varying system
x  A(t ) x(t )  B(t )u (t )
Definition: If there exist u (t ), t0  t f   such that for any
x( t0 ), x(t f )  0 . The system is said to be controllable at t0
Theorem 3-1 The time-varying system is controllable in the time
slot [ t0 , t f ] iff the Gram matrix Wc( t 0 ,t f ) is non-singular.
tf
Wc( t 0 ,t f )    (t0 , ) B( )BT ( ) T (t0 , )d
t0
Modern Control Theory
Lecture 8
2
u(t )  BT (t )T (t0 , t )Wc1 (t0 , t f ) x(t0 ) ,
Proof: if part let
we can get
tf
x(t f )   (t f , t0 ) x(t0 )    (t f , ) B ( )u ( )d
t0
tf
  (t f , t0 ) x(t0 )    (t f , ) B ( ) B T ( ) T (t0 , )Wc1 (t0 , t f ) x(t0 )d

t0
tf

  (t f , t0 ) x(t0 )    (t0 , ) B ( ) B T ( ) T (t0 , )d  Wc1 (t0 , t f ) x(t0 )  0
t0
only if part assume controllable & Wc( t 0 ,t f ) is singular, there
exists non-zero vector  such that Wc( t0 ,t f )  0. Since
Wc( t 0 ,t f )   (t0 , )B( )(t0 , ) B( )T d  0
tf
T
t0
We get (t0 , t ) B(t )  0 . Let x(t0 )   T , we have
tf
x(t f )   (t f , t0 ) x(t0 )    (t f , ) B( )u ( )d
t0
tf
T

  (t f , t0 )     (t0 , ) B( )u ( )d   0


t0
tf
and 0      (t0 , ) B( )u ( )d
T
t0
tf
or 0      (t0 , ) B( )u ( )d
T
t0
  T  0
Modern Control Theory
Lecture 8
3
0 t 
0 
x (t )  
x
(
t
)


1u (t )
0
0


 
Example 3-1
0 t 2 
A(t1 ) A(t 2 )  A(t 2 ) A(t1 )  

0 0 
2
0  
1  t 0   
 (0, t )  I   
d    

d   
0 0
0 0
0
0
2
!


 
 
1 2

1

t

2 
0
1 

1 

0
 1
t 1
 t 2  0 

1
Wc (0, t f )  
2
10 1  t 2 1 dt
0
0

 2

1  

t
1 2
 1 4
t

t 
t
4
2
 
 dt
0
1 2
 t
1 
 2

1 3
 1 5
t

tf 
f
 20
6


1 3
 t f
tf 
 6

1 6 1 6
det Wc (0, t f ) 
t f  t f  0 for t f  0.
20
36
Modern Control Theory
Lecture 8
4
Sufficient condition test For x  A(t ) x(t )  B(t )u (t )
Let B1 (t )  B(t )
Bi (t )   A(t ) Bi 1 (t )  B i 1 (t ), i  2,3,  , n
Qc (t )  B1 (t ) B2 (t )  Bn (t )
If rank Qc (t f )  n for some t f  0, the system is controllable.
For example 3-1 B (t )  B(t )  0
1
 
B2 (t )   A(t ) B1 (t )  B1 (t )
1
0 t  0    t 
 
 



0 0 1  0 
0  t 
Qc (t )  B1 (t ) B2 (t )  

1 0 
det Qc (t )  t  0, rank Qc (t )  2.
Modern Control Theory
Lecture 8
5
Controllability of output x  A(t ) x(t )  B(t )u (t )
y (t )  C (t ) x(t )
Definition: If there exist u (t ), t0  t f   such that for any
given y f & y ( t0 )  0 , y (t f )  y f . The system is said to be
output controllable at t0
Theorem 3-2 The time-varying system is output controllable at t0
iff there exists t f  t0 and for any   t0 , t f  , the rows of the
impulse response matrix G (t , ) are linearly independent.
Proof: if part Assume the rows of G (t , ) are linearly independent
t
W (t0 , t f )   G (t , )G T (t , )d is non-singular, let
t
f
0
u( )  G (t , )W 1(t0 ,t f )y f
T
tf
y (t f )   G(t , )u ( )d
t0
tf
  G(t , )G T (t , )W 1(t 0 ,t f )d  y f  y f
t0
Modern Control Theory
Lecture 8
6
Only if part: Assume G (t , ) are linearly dependent. there
exists non-zero vector  such that G (t , )  0 . Then
tf
y (t f )   G (t , )u ( )d  0
t0
y (t f )  0
3.1.2 Controllability of LTI system: x  Ax(t )  Bu (t )
Theorem 3-3 The following statements are equivalent:
(1) LTI system is controllable.
 At
(2) Matrix e B is row linearly independent.
(3) Matrix ( sI  A) 1 B is row linearly independent.
t
A
T A
(4) Gram matrix Wc   e B  B e d is non-singular.
0
T
(5)

rank B

AB  An1B  n.
Modern Control Theory
Lecture 8
7
Proof: (1)~(4) can be easily get from Theorem 3-1. From
x(t f )  e
A ( t f t 0 )
tf
x(t0 )   e
A ( t f  )
t0
Bu ( )d
let t0  0 & x(t f )  0 we can get
tf
x(0)    e
 A
0
n 1
Bu ( )d   
t f n 1
i

(

)
A
Bu ( )d
 i
0
i 0

  A B   i ( )u ( )d   B
i
i 0
tf
0
 f 0 (t f ) 
 f (t ) 
1 f

AB  An 1 B 
  


f
(
t
)
 n 1 f 

n 1
For given x(0) , iff when rank B AB  A B  n.
We can find solution f i (t f ) ( input signal u (t ) )
Modern Control Theory
Lecture 8
8
Example 3-2
1 3 2
2 1
x (t )  0 2 0 x(t )   1 1 u (t )
0 1 3
 1  1
3
2
5
4
2 1
U c  B AB A2 B   1 1
2
2
4
4 
 1  1  2  2  4  4
rankU c  2  n


NOT controllab le.
Output Controllability
x  Ax(t )  Bu (t )
y (t )  Cx(t )  Du (t )
Theorem 3-4 LTI system is output controllable iff


rank D CB CAB CA2 B  CAn1B  m.
Modern Control Theory
Lecture 8
9
Proof: From
tf
 f 0 (t f ) 
 f (t ) 
1 f
n 1

AB  A B 
  


 f n 1 (t f )


x(0)    e  A Bu ( )d   B
0
y (0)  Cx(0)  Du (0)
We can get
 f 0 (t f ) 
 f (t ) 
1 f
n 1
  Du (0)
  CB CAB  CA B 
  


 f n 1 (t f )
  u (0) 
 f (t ) 
 0 f 
  D CB CAB  CAn 1 B  f1 (t f ) 





 f n 1 (t f )






For given x(0) , iff rank D CB CAB CA2 B  CAn1B  m.
We can find solution f i (t f ) ( input signal u (t ) )
Modern Control Theory
Lecture 8
10
Example 3-3
  4 5
 5
x (t )  
x
(
t
)

u (t )



 1 0
1
y (t )  1  1x(t )
rank CB CAB  rank  6 30  1  m, output controllab le
rank B
 5 25 
AB  rank 
 1  n, NOT state controllab le.

 1  5
3.1.3 Controllability test of canonical system
Theorem 3-5 If A is diagonal, system is controllable iff the
elements of the same row in matrix B are not all zero.
Theorem 3-6 If A is Jordan, system is controllable iff
(1) the elements of the same row in matrix B corresponding
to mutual different eigenvalues are not all zero.
(2) The elements of the row in matrix B corresponding to
the last row in each Jordan block are not all zero.
Modern Control Theory
Lecture 8
11
Example 3-4
0
 7 0
0 1 
(1) x (t )   0  5 0  x(t )  4 0u (t )
 0
7 5
0  1
0
 7 0
0 1 
(1) x (t )   0  5 0  x(t )  4 0u (t )
 0
7 5
0  1
  3 1 0
0 0 
(2) x (t )   0  3 0 x(t )  2  1u (t )
 0
0 3 
0 1
0
3 0
 2
(3) x (t )  0  1 0  x(t )  1 u (t )
0 0  2
0
0
 4 1
 4 2
(4) x (t )   0  4 0  x(t )  0 0u (t )
 0
3 0
0  2
System (1), (2) are controllable and (3), (4) are not
controllable.
Modern Control Theory
Lecture 8
12
3.2 Observability of linear continuous time system
3.2.1 time-varying system
x  A(t ) x(t )  B (t )u (t )
y (t )  C (t ) x(t )
x(t0 )
Definition: If any initial state
can be uniquely
y (t ), t  [t0 , t f ]
determined by the output0
. The system is
t
said to be observable at
Theorem 3-7 The time-varying system is observable in
the time slot [ t0 , t f ] iff the Gram matrix Wo( t0 ,t f ) is
non-singular.
tf
Wo( t 0 ,t f )   T (t0 , )C T ( )C ( )(t0 , )d
t0
Modern Control Theory
Lecture 8
13
Proof: Observability is independent of input and we can ignore
input signal.
if part x(t )  (t , t0 ) x(t0 )
y (t )  C (t ) x(t )  C (t ) (t , t0 ) x(t0 )

tf
t0
tf
 T ( , t0 )C T ( ) y ( )d    T ( , t0 )C T ( )C (t ) (t , t0 ) x(t0 )d
t0
 W0 (t0 , t f ) x(t0 )
tf
x(t0 )  W (t0 , t f )   T ( , t0 )C T ( ) y ( )d
1
0
t0
only if part assume observability & Wo( t 0 ,t f ) is singular.
y (t )  C (t ) (t , t0 ) x(t0 )

tf
t0
tf
y T ( ) y ( )d   xT (t0 ) T ( , t0 )C T ( )C (t ) (t , t0 ) x(t0 )d
t0
 xT (t0 )W0 (t0 , t f ) x(t0 )
there exists non-zero initial state
x(t0 )
x (t0 )W0 (t0 , t f ) x(t0 )  0
such that
T
y T (t ) y (t )  0 or y (t )  C (t )(t0 , t f ) x(t0 )  0
Modern Control Theory
Lecture 8
14
Sufficient condition test let
C1 (t )  C (t )
Ci (t )  Ci 1 (t ) A(t )  C i 1 (t ), i  2,3,  , n
 C1 (t ) 
R (t )    
Cn (t )
If rank R(t f )  n for some t f  0, the system is observable.
t 1 0 
Example 3-6 Let


A(t )  0 t 0 , C (t )  1 0 1
0 0 t 2 
C1 (t )  C (t )  1 0 1
C2 (t )  C1 (t ) A(t )  C1 (t )  t 1 t 2
C3 (t )  C2 (t ) A(t )  C 2 (t )  t 2  1 2t t 4  2t




0
1 
 C1 (t )   1
R (t )  C2 (t )   t
1
t 2 
C3 (t )  t 2  1 2t t 4  2t 
rank R (t )  3  n, system is observable for t  0.
Modern Control Theory
Lecture 8
15
3.2.2 Observability of LTI system:
Theorem 3-8 The following statements are equivalent:
(1) LTI system is observable.
(2) Matrix Ce  At is column linearly independent.
1
C
(
sI

A
)
(3) Matrix
is column linearly independent.
t
A  T
A
(4) Gram matrix Wo  t e C Ce d is non-singular
f
(5)

0
rank C
T
T
AC
T

T
A 
T n 1
C
T
  n.
T
Proof: Ignore (1)~(4) and input signal , let t0  0
y (t )  Cx(t )  Ce At x(0)
n 1
 C   i (t ) Ai x(0)
i 0
 C 
 CA 
 x(0)
  0 (t ) 1 (t )   n 1 (t ) 
  
 n 1 
CA 
Modern Control Theory
Lecture 8
16
3.2.3 Observability test of canonical system
Theorem 3-9 If A is diagonal, system is observable iff
the elements of the same column in matrix C are
not all zero.
Theorem 3-10 If A is Jordan, system is observable iff
(1) the elements of the same column in matrix C
corresponding to mutual different eigenvalues are
not all zero.
(2) The elements of the column in matrix C
corresponding to the first column in each Jordan
block are not all zero.
Modern Control Theory
Lecture 8
17
Example 3-8
  2 0

(1) x(t )  
 x(t )
0
5


y (t )  1 3x(t )
2 1 0
0 2 0
(2) x (t )  
0 0 3

0 0 0
0 1 1
y (t )  
0 1 1
 3 1
(3) x (t )   0  3
 0
0
0
0
x(t )
1

3
0
x(t )
1
0
0 x(t )
1
1 0 0 
y (t )  
 x(t )
0
0

1


System (1), (3) are observable , system (2) is not observable .
Modern Control Theory
Lecture 8
18