§4-2 Decoupling by State Feedback
1. Problem formulation
1). Definition of decoupling systems
Consider a dynamic system given by
x& Ax  Bu ,
y  Cx
(4-32)
where A, B and C are n×n, n×p and p×n matrices,
respectively. Since p=q, this is a square matrix
decoupling problem.
The transfer function matrix of the system is
G(s )qp
q p
 C(sI  A)1 B
(4 - 33)
1
Definition. System (4-32) is said to be decoupled if the
transfer function matrix G(s)p=q is nonsingular and
diagonal.
 n 1 (s )
 d (s )
 1

G (s )  


 0

0
L
n 2 (s )
d 2 (s )
O
0
n i (s )
0
di (s )

0 




0 
n p (s ) 

d p (s ) 
2
Example. Consider the following system:
 1

0

  u1 
 y1 
 u1 
s 1
 
y  =G (s ) u   
1
 2
 2  0
 u 2 

s 2  2s  1 
From y=G(s)u, we have
1
y1 
u1
s 1
1
y2  2
u2
s  2s  1
It is clear that the system is decoupled. Now the
control problem becomes a single variable problem,
whose control law is easy to design.
3
The decoupling of multivariable systems is one of the most
important problems in multivariable systems control.
Consider the following system:
1
 1

 y1   s  1
s  4   u1 
 
y  =  1
1
 2 
 u 2 
 s  6 s 2  2s  1 
1
1


[u1 ] 
[u 2 ]


 y1 
s 1
s 4
 

1
y 2   1 [u ] 

[
u
]
 s  6 1 s 2  2s  1 2

Because of the coupling, it is difficult to design stable
control laws for multivariable systems with nice dynamic
4
performance.
2). Decoupling by state feedback
 p=q, i.e., the system is square;
 The state feedback control law is
u=Kx +Hv (H is nonsingular)
(4-34)
where
K  R pn is the state feedback matrix.
H  R p p is the nonsingular input
transformation matrix.
5
Hence, the closed-loop transfer function matrix with the
state feedback control law
u=Kx +Hv (H is nonsingular)
(4-34)
is
G f ( s, K , H)  C[ sI  ( A  BK )]1BH
6
Problem statement :
Find matrices K and H, such that
G f ( s, K , H)  C[ sI  ( A  BK )]1BH
is nonsingular and diagonal, i.e.,
0
g1 (s )

g2 (s )

G f (s ) 

 0

0
0 

 , g (s )  0
0  i

gp (s ) 
7
2. Preliminary lemmas
1). Relationship between the open-loop transfer
function matrix and the closed-loop transfer function
matrix.
Lemma 1: The relationship between the open-loop transfer
function matrix and the closed-loop transfer function
matrix is as follows:
G f ( s )  C[ sI  ( A  BK )]1 BH
=G ( s )[I  K ( sI  A  BK ) 1 B]H
 G( s)[I  K ( sI  A )1B]1 H
(4  37)
where Gf(s) and G(s) are the closed-loop and open-loop
8
transfer function matrices, respectively.
2). Nonnegative integers di and nonzero vectors Ei of the
open-loop transfer function matrix G(s)
Example. A realization of g(s)=1/(s3+3s2+2s+1) is
 0 1 0  0 
x&  0 0 1  x  0  u

  
 1 2 3 1 
y  1 0 0x
Express g(s) as
2
k 1
cb
cAb
c
A
b
cA
b
1
g (s )  c(s I  A) b   2  2 L  k  L
s
s
s
s
It can be verified that the first two elements of the above
progression are zeros and the third element is nonzero. 9
In the general case, let the ith row of C be denoted as ci and
the ith row of G(s) be denoted as Gi(s). Then,
 c1 (sI  A )1 B   G ( s ) 
1




1
 c2 (sI  A) B   G 2 ( s ) 
M
 M



G (s) 



1
 c (sI  A ) B   G i ( s ) 
 i
 

M
M

 


1 
G p ( s ) 

c
(
s
I

A
)
B

 p

Gi (s) can be expressed as
ci B ci AB
1
Gi (s )  ci (s I  A) B 
 2 
s
s
ci Adi 1B ci Adi B

 d 1 
di
s
si
144442 44443
di
10
If
ci B  ci AB 
 ci A di 1B  0
but
Ei： ci A di B  0，
then we get a nonzero vectors Ei and a nonnegative
di 0
di is the least integer such that
Ei : lims  s di 1Gi ( s)  lims  s di 1ci ( sI  A)1 B
11
From the definition of
di
and
Ei have
, we
Ei  lim s di 1G i ( s )  ci A di B  0
s 
di 1
di
c
B
c
A
B
c
A
B
di 1
i
i
i
 lim s s
(
L 
 d 1  L )  0
di
i
s
s
1444444442
44444444
3 s
0
(4-39)
The above analysis indicates that it is possible to
compute di and Ei from two different expressions G(s) and
(A,B,C).
12
Example 1. Compute di and Ei for the following G(s):

 s2
G(s)  

 s 2
s2
 2s  1
1
 2s  1
1

1 

2

G
(
s
)
c
(
s
I

A
)
B
s s2   1 
1





1
3
 G2 ( s)  c2 ( sI  A) B 
s 2  2s  4 
d1 =min{1,2}1=0 , d2 =min{2,2}1=1
Hence, from the definition
Ei  lim s di 1Gi (s )  ci Adi B  0
s 
we have
13
E1  lim sG1 ( s )  1
s 
0   c1A
d1 0
B
 E1c1B  0
E 2  lim s 2G 2 ( s )  1
s 
 c 2B = 0,
3  c 2 A d 2 1B
c 2 A1B  0
14
Example 4-5a. Consider the following system:@p33
 0 0 0  1 0 
x&  0 0 1  x  0 0  u

 

 1 2 3 0 1 
1 1 0 
y
x

0 0 1 
Compute di and Ei:
c1B=[1 0],
d1 =0;
E1=[1 0]
c2B=[0 1],
d2=0;
E2=[0 1]
15
3). Decoupling of the closed-loop tranfer function matrix
(1) Decoupling expression of the open-loop transfer
function matrix
d1

c1A B 
 E1 
 E   c A d2 B 
 2

2

E

 M
M 

E  
dp

 p  c p A B 
d1 1


c
A
 F1 
1
 F   c A d2 1 
 2

2

F

 M
M 

F  
d p 1


p
  c p A

The ith row of the open-loop transfer function
matrix can be expressed as
16
1
1
ci ( sI  A) B 
1
s
di 1
(ci A B  ci A
di
1
B L )
s
di 1
1
1
2 1
 d 1 [ci A B  ci A (I  A 2  A 3  L ) B]
s
s 444444444
s
s i 1442E 443 1442F 443 14444444442
3

di
di 1
i
i
( sI  A )1
1
s
1
[
E

F
(
s
I

A
)
B]
i
i
d 1
i
17
The open-loop transfer function matrix can be expressed
as @p21
 1

1
[
E

F
(
s
I

A
)
B
]
1
 s d1 1 1

 G1 (s )  

 G (s )   1 [E  F (s I  A ) 1 B] 
2
2

   s d 2 1 2

G (s ) 
 M  

M
G (s )  

 p   1

1
 d p 1 [Ep  Fp (s I  A ) B]
s

 1

 s d1 1



1

O
[
E

F
(
s
I

A
)
B]


1 
18


d p 1
(2) Decoupling of closed-loop transfer function matrix
From
G f ( s )  C[sI  ( A  BK )]1 BH
 G( s)[I  K ( sI  A  BK )1B]H
1
 G ( s )[I  K ( sI  A) B]1 H
(4  37)
19
The closed-loop transfer function matrix can be
expressed as
 1
 s d1 1

G f (s)  
O






1
1
[
E

F
(
s
I

A
)
B
][
I

K
(
s
I

A

BK
)
B]H

1 
(S-1)
d p 1 
s

or
 1
 s d1 1

G f (s)  



O



1
1
1
 [E  F ( sI  A ) B][I  K ( sI  A ) B] H
1 
d p 1 
s

(S-2)@p19
20
di nonzero vectors of Ei :
4). The nonnegative integers and
closed-loop systems
Ei
From the definition of di and Ei, we can get andd i of the
closed-loop systems. Note that
G f 1 (s ) 
 M 

G f (s )  
 M 
G (s ) 
 fp 
1
G fi (s )  ci [s I  ( A  BK )] BH=ci BH
s
1
1
k
ci ( A  BK )BH 2  L  ci ( A  BK ) BH k 1  L
s
s
21
1
di
Hence, there exist and
E
such
that:
i
Ei : ci ( A  BK )di BH  0,
ci ( A  BK )k BH  0, k  di
Lemma 2.
di  di , Ei  Ei H。
Proof. We only need to prove
ci ( A  BK ) di BH=ci A di BH  0,
and
ci ( A  BK )k BH  0 (k  di )
22
We first prove that
ci ( A  BK )k  ci Ak，k  0,1,2,L ,di
open - loop：
Ei : ci A di B  0
ci A k B  0, k  di
1) When
k  0,1,L ,,di  1
ci (A  BK )k BH  ci Ak BH  0
di
di
c
(
A

BK
)
BH

c
A
BH  Ei H:  Ei .
i
2) i
anddi  d.i
Q.E.D
23
Specially, if
 E1 
 
E2 

E：
=
,
 
 
E p 
 c1A d1 B 


d2
 c2 A B 
then, E  
H=EH.

 M 
c A d p B 
 p

where
Ei : ci ( A  BK )di BH  ci Adi BH=Ei H  0,
24
3. Sufficient and Necessary Condition for
Decoupling
Theorem 4-9. The system
x& Ax  Bu ,
y  Cx
(4-32)
can be decoupled by the state feedback u=Kx+Hv if and
only if the matrix
d1

c1A B 
 E1 
 E   c Ad2 B 
 2

2

E

 M
M 

E  
dp

 p  cp A B 
is nonsingular.
25
Proof. Necessity. We only need to prove that E is
nonsingular.
Suppose the system can be decoupled by the state
feedback u=Kx+Hv. Then Gf(s) is diagonal and
nonsingular:
n 1 (s )

g1 (s )  d (s )
1


G f (s )  



0

0
g2 (s ) 
L
n 2 (s )
d 2 (s )
) transfer functions.
where gi (sare
O
0






0

n p (s ) 

gp (s ) 
d p (s ) 
0
26
Ei that
Hence, there exist nonnegative integers anddi such
Ei：
 lim s di 1G fi ( s )
s 
 lim s
di 1
 K
ci
s 
 E  diag{c1 L

 0,K , 0,

L 0
ni ( s )
,0
di ( s )

L , 0

c p 1 c p }  E is nonsingular.
Since
E  EH
E is nonsingular.
27
Sufficiency. Substituting
K=E1F , H=E1
into (S-2) yields
 1
 s d1 1

G f (s)  
O




K=E1F
H=E1


1
-1
1
1 -1
[
E

F
(
s
I

A
)
B
][
I

E
F
(
s
I

A
)
B
]
E 

1 
d p 1 
s

1444
42 444
4
3
open -loop
28
 1
 s d1 1





O




1 
d p 1 
s

(4-49)
where
[I  E1F( sI  A)1B]1E1
 [E1E  E1F( sI  A)1B]1 E1
 [E  F( sI  A)1 B]1。
Q.E.D
29
Example 4-5. Transform the system of Example 4-5a into
an integrator decoupled system and verify that if the
decoupling contradicts the stability.
 0 0 0  1 0 
x&  0 0 1  x  0 0  u

 

 1 2 3 0 1 
1 1 0 
y 
x

0 0 1 
From the forgoing computation, di and Ei are as
follows@p16:
c1B=[1 0], d1 =0;
E1=[1 0]
1 0 
c2B=[0 1],
d2=0;
E

E2=[0 1]
0 1 
From Theorem 4-9,
K= E1F , H=E1, where,
 c1A d1 1   c1A 
F


c A 
d 2 1
 c2 A
  230 
根据例题4-5a的计算可知E是单位阵，故系统可解耦。
现采用定理4-9充分性证明中提供的(4-47)式将其化为
d1 1


c
A
 c1A 
1
1
1
积分器解耦系统。
K= E F , H=E ,
F


d 2 1
c
A
c2 A

  2–2 –3],

计算F阵,
F =c A=[0 0 1],
F =c A=[–1
1
1
2
2
故得
0 0 1
1 0 
u
 0Kx 0 Hv1 1 2 3  x  0 1  v
F





1 2 3
Hence,
1 0 
HE 
,

0 1 
1
0 0 1
K  E F  
,

1 2 3 
1
31
The feedback
control law is
0 0 1
1 0 
u  Kx  Hv  
x 
v


1 2 3 
0 1 
The dynamic equation of closed-loop system is
0 0 1
1 0 
x& ( A  BK )x  BHv  0 0 1  x  0 0  v
0 0 0 
0 1 
1 1 0 
y 
x

0 0 1 
The transfer function matrix of the closed-loop transfer
function matrix is
1
s
G f (s )  
0



0

1
s

The open-loop system is controllable and observable,
but the closed-loop system is unobservable, which
32
means that the state feedback used for decoupling in this
system changes the observability. Computing the
eigenvalues of the closed-loop system, we have
0 0 1
A  BK= 0 0 1   s 3  0
0 0 0 
which means that an instable mode has been cancelled and
the decoupling contradicts the stability.
33
4. A decoupling control law
Theorem 4-10. Suppose that the system can be decoupled
by state feedback, and
(d1  1)  (d 2  1)  L  (d p  1)
 d1  d 2  L  d p  p  n ,
Then, by using the state feedback:
H  E 1 ,
 c1 ( Ad1 1  k1d Ad1  L  k11A  k10I ) 
1


d 2 1
d2
c2 ( A
 k 2d 2 A  L  k 21A  k 20I ) 


K  E 1D  E 1 
M



di 1
di
c
(
A

k
A

L

k
A

k
I
)
i di
i1
i0
 i



M
34 
14444444444444442 4444444444444443
D
the closed-loop transfer function matrix can be
transformed into
 1
 f (s )
 1

G f (s )  



 0

fi (s )  s
di 1
0
1
f 2 (s )
0
 kidi s  kidi 1s
di

0 




0 
1 

f p (s ) 
di 1

 ki1s  ki 0
where kij are adjustable parameters which can be used to
assign the poles of the diagonal elements.
35
Proof. Consider
G f ( s)
1
 G ( s)[I  K ( sI  A) B]1 H
Substituting H=E1 and K= E1 D into the above
equation gives
1
G f ( s)  G ( s)[E  D( sI  A) B]1
Hence, we only need to prove that
(s di 1  kidi s di  kidi 1s di 1  L  ki 1s  ki 0 )Gi (s )
144444444444444442 44444444444444443
fi (s )
1
 [Ei  Di (s I  A) B]
36
where
di
di 1
di  2
c
A
B
c
A
B
c
A
B
1
i
i
i
Gi (s )  ci (s I  A) B  d 1  d 2  d 3  L
si
si
si
 c1 ( Ad1 1  k1d Ad1  L  k11A  k10I ) 
1


c 2 ( Ad 2 1  k 2d 2 Ad 2  L  k 21A  k 20I ) 


D
M



di 1
di
c
(
A

k
A

L

k
A

k
I
)
idi
i1
i0
 i



M
37
The proof is decomposed into the following steps:
 A)1 B
(1) Prove s di 1Gi (s )  Ei  ci Adi 1 (s I(A.1)
d
d
A)1 B
(2) Prove kidi s i Gi (s )  kidi ci A i (s I (A.2)
(3)
di
di 1
di  2
c
A
B
c
A
B
c
A
B
di 1
di 1 i
i
i
kidi 1s Gi (s )  kidi 1s ( d 1  d 2  d 3  L )
si
si
si
ci Adi 1B ci Adi B ci Adi 1B ci Adi 2B
 kidi 1 (



L )
2
3
4
s
s
s
14442s 4443
0
 kidi 1ci Adi 1 (s I  A)1 B,
(A.3)
38
(4) Generally, we have
1
kij s G i (s )  kij ci A (s I  A ) B,
j
j
(A.4)
j  di ,di  1,L ,0; i  1,2,L , p
(5) It follows from (A.1)(A.4) that
(s di 1  kidi s di  kidi 1s di 1  L  ki1s  ki 0 )Gi (s )
Ei  ci {Adi 1  kidi Adi  kidi 1Adi 1  L  ki 0 A 0 } 
1444444444444444442 444444444444444443
Di
(s I  A) 1 B  [Ei  Di (s I  A)1 B]
Q.E.D 39
Example. Consider the following dynamic equation:
 2 0 1
0
x&  0 3 1 x   1



 1 2 1
0



1 0 0 
y 
x

1 0 1 
1
1 u

1 
The problem is whether the closed-loop transform
function matrix can be transformed
into
d1
s  k10  s  1  d1  0
1


0
 s 1


G f (s)  
1
 0



2
s  3s  1 

k10  1
s d2 1  k 21d2 s d2  k 20  s 2  3s  1
 d 2  1, k 21  3, k 20  1
by using the state feedback u=Kx+Hv. If yes, find K and 40H.
c1B=[0 1], d1=0; c2B=[0 0], c2AB=[2 1],
 E1  0 1
E 

E
2
1

 2 
d2=1,
d1+d2 +p=0+1+2=3=n
Hence, the matrices in the state feedback can be
chosen as
1  1 1
HE  
2  2 0 
c1 ( A  I )   10.5 8 7.5
1 
K  E 


2
3
0
1
c
(
A

3
A

I
)

 2
 
1
41
5. Conclusion of decoupling problem ( p=q)
1). The system can be decoupled by u=Kx+Hv if
2). Other methods are required if
; E 0
E
. 0
3). If a system can be decoupled by u=Kx+Hv and
d1  d2  L  d p  p  n, then we can decouple it by using
Theorem 4-10, and the poles of its diagonal elements can
be arbitrarily assigned.
p  n there exist
If d1  d2  L  dp  then
n  (d1  d2  L  d modes
p  p) unobservable. If the decoupling
results in cancelled poles that do not lie in the left hand
side plane, we say the decoupling contradicts the stability.
42
Example. Consider the following dynamic equation:
1 1 0   1 1 
x& 0 2 0 x  1 1  u ,
0 1 3  0 0
1 0 0 
y 
x

0 1 1 
1) Whether the closed-loop system can be integrator
decoupled by state feedback? If yes, please give the
transfer function martrix after decoupling;
2) Whether the decoupling contradicts the stability?
43
c1B=[1 1], d1=0; c2B=[-1 1],
 E1   1
E 
E2  1
d2=0,
1 1 0 
F

0
3
3


1
1
Hence, the system can be decoupled by state feedback.
1 1 1 1 1 0 
K  E F   
2 1 1   0 3 3 
14442 4443144442 44443
1
E 1
1 1 2
 
2 1 4
F
3 
 1

1

3
2
2 



3  1
3

2 
14444442
2
2 3
444444
K
44
The decoupled transfer function matrix is
1 
 s 0

G f (s)  
0 1


 s
It is easy to verify that the original system is controllable
and observable. But after introducing the state feedback,
d1+d2 +p=0+0+2=2<3=n
0 0 0 
A  BK  0 1 3  det(sI  A  BK)  s 2 (s  2)
0 1 3 
which implies that a instable eigenvalue s=2 is
cancelled and therefore, the decoupling contradicts the 45