4.2 State Feedback for
Multivariable Systems
Basic idea：Suppose that a multivariable system (A, B) is
controllable, then the basic idea of its pole-placement is to
transform the problem into the pole-placement of single
variable systems. The steps are as follows.
1. Choose any
bIm(B) , b0
Then, find a state feedback gain matrix K1 such that the
single variable system
(A+BK1, b)
is controllable.
1
2. Assign the poles of the single variable system
(A+BK1, b) to the expected place.
A

BK

bk
1
14442 4443
3. Note that
with
A
b  Im B   L  R p1
b =BL
Therefore,
A  BK1  bk
 A  BK1  BLk  A  B(K

Lk
).
1
144442 44443
K
2
Theorem 4-4. Suppose that (A, B) is controllable. Then
for any nonzero vector b in the value field of B, there
exists a state feedback gain matrix K1 Rpn such that
(A+BK1 , b)
is controllable.
The proof includes the following steps:
3
1)Use the following conclusion.
Lemma. Suppose that (A, B) is controllable. Then we can
choose vectors u1, u2, …, un1 such that the n vectors x1,
x2, …, xn defined by the following equations are linearly
independent @p7:
x 1  b,
x  Ax  B u ,
1
1
 2
x 3  Ax 2  B u 2

L
x n 1  Ax n 2  B u n 2

x n  Ax n 1  B u n 1
where b is any nonzero vector in the value field of 4B ,
x Rn1 and u Rp1
2). Define K1Rpn:
K1  u1 u 2 L
un 1 0x1 x 2 L
x n n n
1
That is
K 1 x 1 x 2
L
x n   u1 u 2
L
u n 1
0n n
Note that
u1  K1x 1 ,
u  K x ,
2
1 2


L
u
n 1  K1x n 1


 0  K1x n
5
3). Prove that (A+BK1, b) is controllable @p5
It is easy to find that
x1  b
x 2  Ax1  Bu1  Ax1  BK1x1  ( A  BK1 )b
x 3  Ax 2  Bu 2  Ax 2  BK1x 2  (A  BK1 )x 2  (A  BK1 )2 b
M
x n 1  Ax n 2  Bu n 2  ( A  BK1 )x n 2  ( A  BK1 )n 2 b
x n  Ax n 1  Bun 1  ( A  BK1 )x n 1  ( A  BK1 )n 1b
6
, x nlinearly independent, we have
Since x 1 , x 2 ,L are
rank b ( A  BK1 )b ( A  BK1 )2 b L
( A  BK1)n 1b   n
 ( A  BK1, b) is controllable.
Q.E.D
7
Theorem 4-5. Suppose that the system (4-1) is
controllable. Then there exists a state feedback gain
matrix K such that the n eigenvalues of A+BK can be
arbitrarily assigned.
Proof . First of all, we choose a nonzero vector L such
that
b=BL
From Theorem 4-4, there exists K1 such that
(A+BK1, b)
is controllable. From the theorem about the pole
placement of single variable systems, we know that there
exists a n dimension vector k such that the poles of
8
A+BK1+bk
can be arbitrarily assigned. Note that
A+BK1+bk
=A+BK1+BLk
= A+B(K1+Lk)
Hence, one only needs to choose
K= K1+Lk
This completes the proof of Theorem 4-5.
Q.E.D
9
Proof of the Lemma. The proof is performed by
construction. Without loss of generality, suppose that the
columns of B=[b1, b2,…., bp] are linearly independent.
1) Construct n linearly independent columns in the
following way:
b1，Ab1，
L ，A n1 1b1
n1
b1，Ab1，
L，
A
b
(Until
can
1 be linearly expressed by
A n1 1b) 1
b 2，Ab 2，
L ，A n2 1b 2
n2
A
bcan
(Until
1 be linearly expressed by
n2 1
A )b 2
b1，Ab1，
L，
10
Continue the above procedure. Since (A, B ) is
controllable, there exists
n1 1
n2 1
nP 1
[b14444442
A b31 b14444442
A b 2 L b p Ab p L A b p ]
1 Ab1 L 444444
2 Ab 2 L 4444443
144444442 44444443
n1
n2
np
which forms a set of basis of the n dimensional space,
where ni = n.
11
2）Construct n linearly independent vectors xi as follows:
x 1  b1 ,


x  A x  b ,

1
1
 2
 b1
L ,



x n1  A x n1 1  b1 ,
x


A
x

b
 n1 1
n1
2



b2

L


x

n 1  n 2  A x n 1 n 2 1  b 2 

M



M
12
x1 ,Lare
, x nlinearly independent. In
It is easy to verify that
fact,
x1 x 2 L

x n1
, x n1 1 ,L , x n1 n 2 ,L , x n 
n 1 1
2
 [b
,
A
b
+
b
,
A
b
+
A
b
+
b
,
L
,
A
b1  L  Ab1 + b13,
1
1
1
1
1
14444444444444444444444444
144444444444444444444444442
n1个
A(An1 1b1  L  Ab1 + b1 )  b2 L

elementary transformation
b1, Ab1 , A 2b1 ,L , An1 1b1 , b 2 , Ab 2 , A 2b 2 , L , An 2 1b 2 , L , b p L , An p 1b p 
14444444444444444444444444444444442 4444444444444444444444444444444443

original basis
13
3). Determine ui. Let
u1  u 2  L  un1 1  e1  1 0 L
0  R p
T
un1  un1  2  L  un1 n 2 1  e2  0 1 L
0  R p
T
L
un1 n 2 L n p 1  un1 n 2 L n p 1 1
 L  un1 n 2 L n p 1 n p 1  e p   0 0 L
1  R p
T
Q.E.D
14
Example 1. Consider the system
1 1 0 
0 0 




&
x   0 1 0  x  1 0  u
0 0 1 
0 1 
L  1 1
T
Construct K1 such that (A+BK1 , b=BL) is
controllable.
Solution. Choose
x1 = BL =b=[0 1 1]T0, L=[1 1]T。
and consider
x1=b,
xk+1= A xk+ B uk (k=1, 2, … , n1)，
15
Noticing that
Ax1=[1 1 1]T
is linearly independent of x1, choose
x2= Ax1,
and
u1=[0 0]T
Because Ax2, x1 and x2 are linearly dependent, we can not
choose u2 as [0 0]T. Let
u2=[1 1]T，
Then, we have x3= Ax2+Bu2= [2 0 2] T.
16
From K1=[u1 u2 0][x1 x2 x3]1, we have
1
0 1 2 
0 1 0  
 1 1 1 

K1  
1 1 0





0
1
0

1
1

1
144442 44443 1 1 2




u1 u 2 0 14442 4443
x 1
x2
x3
1 1 0
A  BK1   1 1 1 


 1 1 1
0 1 1 
b ( A  BK1 )b ( A  BK1 ) 2 b   1 0 1

 

1 0 1 
17
It can be readily checked that (A+BK1 b) is
Example 2. Consider the following system:
0 1 0 0 
0 0 
0 0 1 0 
0 0 
x
u
x& 
0 0 1 0 
1 0 
0 0 0 1 
0 1 




Determine K such that the closed-loop system (A+BK)
has eigenvalues 2, 2,  1j.
Solution .
Choose L =[1 0]T,
u1=[-1 0]T,
u2=[0 0]T,
u3=[0 1]T,
x1=b1=[0 0 1 0]T;
x2=[0 1 0 0]T;
x3=[1 0 0 0]T;
x4=[0 0 0 1]T;
18
Then, we have
0
0

1
0
0
0


K1  
 1
0
0
1
0



0
0
1
0
0
1
0
0
0
0
0  0 0 1 0 

0  1 0 0 0 

1
It is clear that (A+BK1, b1) is controllable. Let k=[k1 k2
k3 k4], we have
0
0
A  BK1  b1k  
k1

1
1
0
k2
0
0
1
k3
0
0

0
k4 

1
19
whose characteristic polynomial is
s4 (1+k3)s3+(k3k2)s2+(k2k1)s+k1 k4,
The expected characteristic polynomial is
s4+6s3+14s2+16s+8,
Comparing the coefficients of the two polynomials, we
can obtain
k1 =37, k2=21, k3=7, k4=45
Hence, the feedback gain matrix can be selected as
 37 21 8 45
K  K1  Lk = 

1
0
0
0


20
In the above method, once L and ui are selected,
then k is determined. But L and ui are nonunique, which
implies that there exists a family of k to get the same
poles. This is one of the main differences between multiinput systems and single-input systems.
21
The other issue in pole-placement of multi-input
systems involves “nonlinear equations”. Denoting the
elements of K by kij , then the closed-loop polynomial
can be rewritten as
det[ sI  ( A  BK )]
 s n  f1 (K ) s n1  f 2 (K ) s n2 
 f n1 (K ) s  f n (K )
where fi (K)’s denote non-linear functions with kij as
variables. Let the expected polynomial be
s  a 1s
n
n 1
 a 2s
n 2
 L  a n 1s  a n
22
Comparing the coefficients, we have
f i (K )   i
(i  1, 2,L , n)
(S—3) @18
Example. Consider the multivariable system
1
0
A  BK=  0
0
 1 2
0  1 0 
k11



1   0 1  
k 21

3 0 0 
k12
k 22
k13 
k 23 
k11 k12  1 k13 
 k 21
k 22
k 23  1
 1
2
3 
It is clear that the coefficients of det(sIABK) are
nonlinear.
23
det(sIABK) is linear for single-input systems but
usually nonlinear for multi-input systems. Theorem 4-4
indicates that when the system is controllable, we can get
a set of linear equations of (sIABK) by abandoning
some free parameters of K.
24
Example 3. For the system in Example 2, compute K such
that fi(K)=i and the poles of the closed-loop system
(A+BK) are 2, 2, 1j.
Solution. Since
0
0
x& 
0
0

1
0
0
1
0
1
0
0
0
0
0
0
x
0
1
0
1 

0
0
u
0
1 
k1 k 2 k3 k 4 
K

k
k
k
k
 5 6 7 8
we have
0
0
A  BK  
k1

k5
1
0
0
1
k 2 1  k3
k6
k7
0 
0 
k4 

1  k8 
25
Scheme 1. Let k4=k5=k6=k7=0,
From
0
0
A  BK  
k1

k5
1
0
0
1
k 2 1  k3
k6 k7
1+k8= 2,
0 
0
0 
0 1
0 0

0 
1
0

 A  BK  
k4 
k1 k 2 1  k3
0 



1  k8 
0 1  k8 
0 0
(s  2)[(s  1)2  1]  s3  4s 2  6s  4
we have
That is,
k1  4,
k 2  6
1  k3  4
 4 6 5 0 
K

0
0
0

3


26
Scheme 2. Let
k1=k2= 0, k3= 1 , k4=1
0
0
A  BK  
k1

k5
1
0
0
1
k 2 1  k3
k6 k7
0 
0 
k4 

1  k8 
0
0
A  BK  
0

k5
0 
0
1
0 
0 1  (1) 1 

k6
k7
1  k8 
1
0
From
( s  2)2[( s  1)2  1]  s 4  6s3  14s 2  16s  8
we have
k5=8, k6=16 , k7=14, k8=7, i.e.
1 1 
0 0
K


8

16

14

7


27
Since the feedback gain matrix is nonunique, the
transfer function matrix of the closed-loop system is not
unique either. Hence, different response may be obtained.
We should choose K such that the closed-loop system
yields a better response.
The same as the single-input case, the condition that
the system is controllable in Theorem 4-4 is sufficient and
necessary for arbitrary assignment of the poles. But for a
given set of poles, it is only a sufficient condition.
28
5. Stabilization
1). Stabilization of state feedback systems. Consider
the following LTI system
x& Ax  Bu
If we can find a state feedback
u  v  Kx
such that all real parts of the poles of the closed-loop
system
x& Ax  Bu  ( A  BK )x  Bv
are negative, then the system is said to be stabilizable with
state feedback.
29
Theorem 4-6. The system (A, B) is stabilizable by state
feedback if and only if the real parts of all uncontrollable
modes are negative.
In fact, because the state feedback does not change the
uncontrollable modes：
 A1 A 2  B1 
A  BK  
    K1 K 2 

 0 A4   0 
 A1  B1K1 A 2  B1K 2 


0
A

4

 Re  ( A 4 )  0.
the system is stabilizable
30
2). System classification by stabilizability
The systems can be classified into:
The poles can
be assigned
arbitrarily
 Controllable 
 Stabilizable
 systems

Systems 
 Stabilizable: If the real parts of all

 uncontrollable modes are negative.
 Uncontrollabl 
 e systems
 Unstabilizable: If the real parts of
 some uncontrollable modes are
negative.
31
Conclusion
 The condition that pole assignment is possible: The
states can be measured.
 If the states of the systems can not be completely
measured, which is the common feature of most
control systems, then we can not use the state
feedback method to assign the poles directly. But the
result of pole placement is still important in theory
and engineering, and is a classical achievement of
linear system theory.
32