Chapter 3
Canonical Form and
Irreducible Realization of
Linear Time-invariant
Systems
1
§3-1Canonical form of systems
1. Canonical forms of single variable systems
x  Ax  bu , A  R nn , b  R n1
y  cx  du , c  R1n , d  R
The characteristic polynomial is
(s )  det(l I  A)  l
n
 a1l
n 1
 a 2l
n 2

 an
The controllable and observable matrices are
 c 
 cA 
  R nn
U  [b Ab A n1b]  R nn , V  


 n1 
cA
2
Steps: Find a equivalence transformation
x  Px
for a given ( A, b, c, d ), such that
x  Px
( A, b, c, d ) 
( A, b, c , d ) with canonical form.
where
A  PAP 1, b  Pb, c  cP 1, d  d
Two methods to compute the equivalence
transformations for canonical form:
1. Compute P firstly.
2. Compute P1 firstly.
3
1) Realization of controllable canonical forms
Theorem 3-1 Let the system (3-1) be controllable. Then,
we can transform it into the following controllability
canonical form by an equivalence transformation.
1
0
0
0 
 0
0 
 0

0 
1
0


 
x 
x   u
 0



0
0
0
1 
0

 
 an an 1 an 2
1 
a1 
y  [b n b n 1
b 2 b1 ]x  du
4
 The first method for computing controllability
canonical form: computing the transfer matrix P
firstly.
a) Compute the controllable matrix
n1
U=[b Ab A b];
1
U
b) Compute
and write its last row as h.
c) Construct the transform matrix
 h 
 hA 


P   hA 2  ；




hA n1 

 nn
5
d).
A  PAP 1, b  Pb, c  cP 1
A
 0
 h 

 hA 




2
 hA  A= 






hA n1 



  an
1
1
1
 an1  an2
P

 h 
  hA 


2 

hA




1 
 hA n1 
 a1 
where
hA n  anh  an1hA  an2hA 2 
Cayley-Hamilton theorem
 a1hA n1
6
Note that U1U  I
 
1
   [b Ab A n1b]  I= 0
 

h 
 0
0
0


1 
we have
hb  0, hAb  0, ，hA n2b  0, hA n1b  1
 h 
0
 hA 
0


 
2
 b = Pb =  hA  b = 0 


 


 
hA n1 


1




P
7
Question Is the matrix P nonsingular?
In order to prove that P is nonsingular, we should prove
that
 h 
 hA 


2 

 P   hA
0




hA n1 


 1h   2hA 
  nhA n1  0
  0
8
Consider that
1h   2hA 
  nhA n1  0
(*)
1)Mutipling the above equation by b, and noting that
hb  0, hAb  0, hAn2b  0, hA n1b  1，
we have
( 3- 4)
a n  0;
2) Multiplying Equation (*) by Ab, and noting
0 have
Equation (3-4) and  n ,we
a n 1  0
By the same token, we have
ai  0  a  0
9
 The second method for computing controllable
canonical form: compute the transfer matrix P1
firstly.
1). Let the base vectors be
an 1 an 2
a
n  2 a n 3

P - 1  [b Ab An 1b] 
 a
1
 1
 1
0
: [q1 q 2 q n ]
Note that
det( sI  A)  l n  a1l
n1
0
 a2l
a1 1 

1 0

0 0


0 
n2

 an
10
P - 1  [b Ab
: [q1 q 2
an 1 an 2
a
n  2 a n 3

An 1b ] 
 a
1
 1
 1
0
qn ]
0
a1 1 
1 0

0 0


0 
it is clear that
qn  b
and
q1  an 1b  an 2 Ab 
 An 1b
 an 1qn  A (an 2b 
 An 2b)  an 1qn  Aq 2
q2
 Aq 2  q1  an1q n
11
 A n2b
q 2  an2b  an3Ab 
 A n3b)  an2q n  Aq3
 an2q n  A (an3b 
q3
 Aq3  q 2  an2q n
With the same method, we have
qi =an -i qn +Aqi 1, i  1,2,
, n  1,
 Aqi 1  qi  an i qn , i  1,2,
At last, from
q1  an 1b  an 2 Ab 
,n  1
 An 1b
and Cayley-Hamilton theorem, we have
Aq1  an 1Ab  an 2 A 2b   a1An 1b  An b  an qn
12
Hence
A  PAP 1  AP 1 =P 1A 
AP 1  [q1 q 2
 [q1 q 2
 0
 0

qn ] 
 0

 an
q n ]A  [q1A q 2 A
1
0
0
1
0
0
an 1 an 2
0
q n A]
0 

0



1 
a1 
A
where we have used the following equations:
Aqi 1  qi  aniq n , i  1,2, , n  1;
Aq1  anq n
13
2)
b  Pb  b  P 1b
 [b Ab
an 1 an 2
a
a n 3
n

2

n 1
A b ]
 a
1
1

 1
0
1
c

cP
: [b n
3)
0
a1 1  0 
1 0  0 
 
0 0  0 
 
 
0  1 
b
b n1
b2
b1 ]
14
Argumentation
1) By using the uniqueness of the transformation, we have
 h  
 hA  

 
P   hA 2    [b Ab

 

 
hAn 1  

 
an 1 an 2
a
a n 3
n

2

An 1b] 
a
1
 1
 1
0
a1 1  


1 0 

0 0 


0

0  
P 1
15
1
2) Uniqueness of the transformation
Proposition Let (A, b) be controllable. If there exist two
nonsingular matrices P1 and P2, such that
A  P1AP11, b  P1b;
A  P2 AP21, b  P2b
Then, we have
P1=P2
Proof In fact, we have
[b Ab
A( n1)b]  P1[b Ab
 P1P21[b Ab
A ( n1)b]
A ( n1)b]
 (I  P1P21 )  0
 P1=P2 .
Q.E.D
16
Example Consider that following state-variable
equation
2
1
 2
0 
x   0
2
0  x  1  u


 
4
0 


 1

1 

Transform the equation into controllable canonical form.
First of all, we have to examine the controllability of
the system. If it is controllable, we can transform the
state-variable equation into controllable canonical
form.
17
3
14 
0
2

  1 2
U
b
Ab
A
b
4

 

11 

1 4

det U  0, hence, the system is controllable.
Now, we construct the matrix P.
18
U
1
2

 5


2
1
2
1
0

2

1

h  2
1
1
The matrix is
P  [hT (hA)T (hA2 )T ]T
i.e.
 2 1 1 
P   3 2 2 


 4 2 3 
2
P 1   1

 2
1
2
0
0
1

1 
19
 2 1 1   2 2 1  2 1 0 
A  PAP 1   3 2 2   0 2 0   1 2 1 




 4 2 3   1 4 0   2 0 1 
0 1 0
 0 0 1


 2 5 4 
 2 1 1  0  0 
b  Pb   3 2 2  1   0 

   
 4 2 3  1  1 
det( sI  A)  s 3  4s 2  5s  2
20
2) Observable canonical form
Theorem 3-2 Let the system (3-1) be observable. With
an equivalence transformation, we can transform it into
an observable canonical form as follows
0
1

x  0


0
y  [0
0
0
0
0
1
0
0
1
0
0
an   b n 
an 1  b n 1 
 

an 2   b n 1  u
 

 

a1   b1 
1]x  du
21
 The first method for computing observable canonical
form: computing the transfer matrix P1 firstly.
1) Compute the observability matrix
2) Compute
 c 
 cA 
;
V= 


 n1 
cA 
1
V
, and denote its last column as h;
3) Construct the transformation matrix
P 1  [h Ah
A n1h ]nn；
22
 The second method for computing observable
canonical form: computing the transfer matrix P firstly.
an 1 an 2
a
a n 3
n

2

P
 a
1
 1
 1
0
0
a1 1 
 c 

1 0 
 cA 

0 0 


 cA n 1 

0  
23
2. The canonical forms for multi-variable
systems
1) Luenberger controllablility canonical form
Consider the system
 x  Ax  Bu

 y  Cx  Du
(3  15)
Theorem 3-3 Let the system (3-15) be controllable. Then,
there exists an equivalence transformation which
transforms the system (3-15) into controllability canonical
form as follows
x  Ax  Bu
where
y  Cx  Du
(3-16)
24
 A11
A
 21
A


 A p1
0 1
0
1

Aii  
0

 
A1 p 





A pp 
A12
A 22




1
 
Bi
where A ii , A ij ,are
0 0
0 0

Aij  
0


 B1 
B 
 2

B




 B p 
0
0
0
0


 Bi  
0
0

0
  
0
0
0


0
1  
0
i  i , i  matrices,
j , i  p respectively.
25
0
0




0

0




A






0


p
mi  n 0

1


1
0
0
0
0
1







0
0
1
0
0
0
1
0
0







0
0
0
0
0
0
1
0
0
0
0








2
p
0





0


0








0


1

26 

0


0
1

0


B=  0

0


0


0
 0
0
0

0
0
1
0
0
0
0


0
 
0


0




0


0
1 


 m1




 m2


C=CP 1


 mp


27
2) Design steps
a)We assume that B=[b1 b2, … bp ] is full column rank;
b) Construct the controllability matrix:
A n1b1 A n1b 2 A n1b p ]
U  [b1 b 2 b p Ab1 Ab 2 Ab p
B
A n1B
AB
Select n linear independent vectors of U, and rearrange
them as follows
b1 Ab1
A
m1 1
It is clear that
b1 b 2 Ab 2
m1  m2 
A
m2 1
b2
 mp  n
b p Ab p
A
mP 1
bp
28
Remark
If Ab2 is linearly dependent with
Ab2  a 1b1  a 2b2 
b1，b2，
b,p，
i.e.A b1
 a pbp  a p1Ab1
k
A
bcannot
then, all
be elected for . k  1
2
A 2b2  a 1Ab1  a 2 Ab2   a p Ab p  a p1A 2b1
k
b2 be linearly expressed by the
which means that A can
forgoing vectors.
3) Let
1
P1

A m1 1b1 b 2 Ab 2
[b1 Ab1
m1
A m2 1b 2
m2
A mP 1b p ]
b p Ab p
mp
29
4) Compute P1. Let hi represent the
p
1, 1  2 , ,  i rows.


   m1
 
h  
 1 

P1   
 
 
  
 
  p
  
h p  
1
1 row
p
 i the nth row.
30
5) Construct the transform matrix
P2
h1




h1A





1 1 
h
A
 1



h2


 

 h A  2 1 
 2





h
p





 P 1 


 hp A




 1






p



Consider the nonsingular transformation
x  P2 x, ( x  P21 x )
we have
A
P2 AP21 , B
 P2B, C 
CP21
31
P2 is nonsingular:
a that
We have to prove that all the column vectors such
P2a  0  a  0
I.
From
P2
h1




h1A





1 1 
A
h

 1


h2


 
  0
 h A  2 1 

 2




h
p





 P 1 



 hp A
 hi A   0, i  1,2,
ji
, p; ji  0,1,
32
, i  1 (a  1)
In particular, we have
 h1 
h 
 2 0
 
h 
 p
pn
b) It is clear that the set of vectors of
b1 Ab1
A m1 2b1 b 2 Ab 2
A m2 2b 2
P1- 1
without
b p Ab p
A i 1bi
A mP 2b p
n( n  p )
T
T T
[
h
,
,
h
is a basis of the zero-space of
. p]
1
Hence, can be expressed as the linear combination of the
above vectors.
That is because (take p=2 for example)
33


 
 
1

h 

m2 1
 1  [b Ab A m1 1b b Ab
1 0
A b 2 ]  I  0
1
1
1 2
2



1
p
 
1
 0
0 0

 
  h1b1  0, h1Ab 2  0, , h1A 1 2b1  0, h1A 1 1b1  1
 
h 2  h b  0, h Ab  0, , h A 2 1b  0;
1 2
P1
1
2
1






1 
2
h 2b1  0, h 2 Ab1  0,
, h 2 A 1 1b1  0;
h 2b 2  0, h 2 Ab 2  0,
, h 2 A 2 2b 2  0, h 2 A 2 1b 2  1
which shows that the zero-space of
by the vectors above.
T
[h1 ,
T T
, h p ]be
can
formed
34
Generally, we have
hi A i 1bi  1, i  1,2,
and
j
hi A bk  0,
,p
if k  i , or k  i but j  i  1. (a  2)
c) Express aas follows
a 
m1  2
m2  2
0
0
i
a
A
 1i b1 
i
a
A
 2i b2 

mp  2
i
a
A
 pi b p
0
Multiplying the two sides of the equation by
a - 2)
(a -1) , we(have
noting the Equation
and
,hand
1A
35
h1Aa  0

m1  2
 a1i h1A
0
i 1
b1 
m2  2
 a 2i h1A
i 1
b2 

0
mp  2
 a pi h1A
i 1
bp
0
 a1m1 2 h1A m1 1b1  0  a1m1 2  0；
1
Then, multiplying the above equation by
,hwe
2 A have
a2 2 2h2 A 2 1b2  0  a2 2 2  0， ，
By the same token, we can prove that

. 0
Q.E.D
36
II. About B=P.2B
0


0
1

0


B=  0

0


0


0
 0
0
0

0
0
1
0
0
0
0


0
 
0


0




0


0
1 
37
   Take p=2 for example:

h1A 11[b1 b 2 ]  [1 ]
 
h1B  0 h1AB  0
 
1

h 

1


1


1
2
  [b Ab A 1 b b Ab
 0
1 0
A
b
]

I
1
1
1 2
2
2



1
p1
 0
 
0 0

 
 
 
Hence, we have
h
 2
1  2
1 1
h
B

0,
,
h
A
B

0
;
h
A
B  [1 ]
P1
1
1
1
h 2B  0,
, h 2 A 2 2B  0;
h 2 A 2 1B  h 2 A 2 1[b1 b 2 ]？
38






1 
If
h 2B  0,
, h2A
2  2
B  0; h 2 A
2 1
B  h2A
2 1
[b1 b 2 ]？
we first study how the set of base vectors is chosen.
b1 b 2 Ab1 Ab 2 Ab p
A 2 1b1 A 2 1b 2 A 2 1b p
2 1
A
b
1) If
is1 one of the vectors ,then
h 2 A 2 1b1; 0
 h 2 A m2 1B  h 2 A 2 [b1 b 2 ]=[0 1]
 1
b1not one of the vectors, it can be expressed as
2) If A 2 is
the linear combination of
b1 b 2 Ab1 Ab 2 Ab p A 2 2b1 A 2 2b 2
Hence, we still have
h 2 A 2 1b1  0
 h A m2 1B  h A 2 [b b ]=[0 1]
39
Generally, if the matrix (P1)1 is given as
b1 b 2
bp
Then, we
have
Ab1 Ab 2
0


0
1

0


B=  0

0


0


0

0
Ab p
0
0

0
0
1
0
0
0
A 2 1b1 A 2 1b 2
0


0


0


0




0


0
1

A 2 1b p
40
P.82 Example 3-2 Consider the dynamical system (A, B,
C), where
0
 0
 1
0
A
 22 11
 23 6

1
0 2 

4 0 
0 6 
0
0
0
B
0
1

0
0
0 0 0 1

 C
0 0 1 0 
1


3
Find the controllability canonical form.
Compute the controllability matrix
0
0
[B AB A 2B A 3B]  
0
1
m1  3, m2  1

Ab 2
A 2b1
1
3
6
0 2
6
13
1
4
6
0
0
3 6 18 25




41

It is clear that the first four linear independent
columns are the first, second, third and fifth columns,
respectively. Hence, 1=3,  2=1,
[b1
Ab1
A 2b1
 28 11 3
13 6 0
b 2 ]1  
2 1 0
0 0 1

1
0

0

0
h1=[2 1 0 0 ], h2=[0 0 1 0 ], and then we have
42
 h1   2
h A 
1
1

P2  
h1A 2   0

 
 h2  0
1
0
0
0
0
0
0
1
0

0

1

0
from which, we can figure out the controllability
canonical form.
A  P2 AP21, B  P2B, C  CP21
43
 0 1
 0 0
A
 6 11
 11 0

0

1 0

6 0 

0 4 
0
0
0
B
1
0

0

0
0 0 1 0

 C
0 0 0 1 
3



1
2) Observability canonical forms for multi-output
systems
Omitted.
44
Question
1. Can we write the Luenberger observability
canonical form?
Hint:
Vn 1
c1




c2






c
q


 c2 A 


 

 cq A 




 c A n 1 
 1





c A n 1
45
c1




c1A





 1 1 
 c1A



c2


P1  

 c A 2 1 
 2





c
q





 q 1 


 cq A

P11   q1
P 1  q1 Aq1 Aν1 1q1

qq
 qq 
 q2
Aqq
ν 1
A q qq 

46