This document discusses state-space realizations of linear time-invariant (LTI) systems. It begins by introducing state-space representations using matrices A, B, C, and D. It then discusses the concept of equivalent state-space representations that have the same transfer function through transformations. The document also introduces the concepts of zero-state equivalence and companion forms. It concludes by discussing conditions for a transfer function to have a state-space realization and provides a method to obtain a realization using a block companion form.

1. INSPIRING CREATIVE AND INNOVATIVE MINDS
State-Space Realizations

2. Example 3.1
Consider:
x
(t)  Ax(t)  Bu(t)
y(t)  Cx(t)  Du(t)
x
(t)  Ax(t)  Bu(t)
y(t)  Cx(t)  Du(t)
x(t)  Px(t)
Equivalent
Transformation
Equivalent
Systems

3. Equivalent state equations
Given state-space description:
Let P be a nonsingular matrix s.t.:
x  Px  x  P-1
x
x
 Px  PAx  PBu  PAP1
x  PBu
y  Cx  Du  CP1
x  Du
A  PAP-1
, B  PB, C  CP-1
, D  D
x
(t)  Ax(t)  Bu(t); y(t)  Cx(t)  Du(t) (**)
 (*) and (**) are said to be equivalent to each other
and the procedure from (*) to (**) is called an
equivalent transformation.
x
(t)  Ax(t)  Bu(t) (*)
y(t)  Cx(t)  Du(t)
Let

4. ()  I  A
The feedforward matrix D between the input and output has nothing
to do with the state space and is not affected by the equivalent
transformation.
The characteristic equation for (*) is:
For (**), we have
()  I  A  I P1
AP  P1
P P1
AP
 P1
(λI)P P1
AP  P1
(λI  A)P
 P P1
λI  A  I A
Equivalent state equations have the same characteristic polynomial
and hence the same set of eigenvalues.
Equivalent state equations

5. Recall: A  PAP-1
and A are similar to each other
 They have same eigenvalues, same stability perf.
 Similar transfer functions!!
G(s)  C(sI  A)1
B  D G(s)  C(sI  A)1
B  D
and
G(s)  G(s)
To verify,
G(s)  C(sI  A)1
B  D
(XYZ)1
 Z1
Y1
X1
 CP-1
(sPP1
 PAP1
)1
PB  D
 CP-1
(P(sI - A)P-1
)-1
PB  D
 CP1
P(sI  A)1
P1
PB  D
 C(sI  A)1
B  D  G(s)
Equivalent state equations

6. Example 3.2
Consider again Ex. 3.1:
x1  x1
 2    2 
1x 
0 x1 
x  1
x1 

1
x(t)  Px(t)
From the circuit (via observation):
x2  (x1  x2 )1
Or
 
 
   
1
1
1
1 1
1 0
x2 
x 
x2 
x 
  
1
 
1
1
1x2 
0 x 

7. Two state equations are said to be zero-state equivalent
if they have the same transfer matrix or
D  C(sI  A)1
B  D  C(sI  A)1
B
Note that:
(sI  A)1
 s1
I  s2
A s3
A2

Then,
D  CBs1
 CABs2
 CA2
Bs3
 D  CBs1
 CABs2
 CA2
Bs3

Zero-State Equivalent

8. Zero-State Equivalent
matrix iff
Theorem 3.1
Two LTI state equations{A,B,C,D} and {A,B, C, D}
are zero-state equivalent or have the same transfer
D  D and
CAm
B  CAm
B ; m  0,1,2,...
- In order for two state equations to be equivalent, they
must have the same dimension.
- This is, however, not the case for zero-state equivalence.

9. Example 3.3:
Consider:
A  B  C  0
y(t)  0.5u(t) x(t)  x(t)
y(t)  0.5x(t)  0.5u(t)
A 1 ; B  0 ; C  0.5 ; D  0.5
D  D  0.5
CAm
B  CAm
B  0
Note that:
 The two systems are zero-state equivalent.~ Theorem 3.1

10. Companion Form
• Consider:
0
 3 2 1
 1 0 x(t)  0u(t)  Ax(t)  bu(t)
  
 4 3 1  1
x
(t)   2
• Then:
• It can be shown that:
• Can {b,Ab,A2
b} be used as the basis?
 YES!

11. Companion Form

5 


0
0 0 17 
0 15
1
A  1
• Then:
• Thus the representation ofA w.r.t. the basis {b,Ab,A2
b} is:

12. Companion Form


 0


0

0 
0 
 1
;

0
1
1  
 
3

 0
2   0 1
0
0 0 0 4  1 2 3 4 
 0 0 0
0 1 0 0
0 1 1


 0
0
;

1 

 0
0 
 0
 4
2
1
2
3
4

 0 1 0 0  1 1 0 0
0 1 0 1 
0 0 3 0 0 1
0 0
     
Transpose
All have similar
characteristic
polynomial:
2 3 4
1
4 3 2
()         
• Companion-form matrices:
• Any special?
 3
52
15 17  0
( 3)  2 1
2 ( 1) 0
 4 3 ( 1)
()  I  A 
A  1
(-1)
(-1)
0 0 17 
 0 15

0 1 5 

(-1)

13. Realizations
Q. What is "realization"?
space equation
– Implicitly implies LTI systems
– Shall start with multi-variable systems and will
sometimes specialize to single-variable systems
x
(t)  Ax(t)  Bu(t)
y(t)  Cx(t)  Du(t)
D(s)
– For a givenĜ(s) 
N(s)
, find a corresponding state-

14. Q. If Ĝ(s) is realizable, how many possible
realizations?
– Infinite ~ in view of equivalent transformations and the
possibility of adding un-controllable or un-observable
components
Q. Which one is the "good" realization?
- A “good” realization is the one with the minimal order
~ Irreducible realization
 Will be discussed in Topic 6
Realizations (cont.)

15. Realizations (cont.)
Q. Under what condition is Ĝ(s) realizable by an
LTI system?
-Recall that the transfer function of the dynamic equation is
Ĝ(s)  C(sI  A)1
B  D  C(sI  A)1
B  D
Ĝ(s) is realizable by a dynamic equation iff it is a
proper rational function (order of numerator  order of
denominator)
 In fact, the part contributed by C(sI - A)-1B is
strictly proper
Theorem 3.2

16. Realizations (cont.)
as:
Ĝ
sI  A
(s) : C(sI  A)1
B 
1
CAdj.(sI  A)B
sp
To determine the realization of Ĝ(s) , decompose it
Ĝ(s)  Ĝ()Ĝsp (s)
where
r r1
1
d(s)  s  s  r1s r
Let
i.e. least common denominator of all entries of Ĝsp (s).

17. Nr 1
y  N1 N2  Nr xĜ()u
 
 0 
 

0 

x   0 u













0

0
x
 
Ip 
p
Realizations (cont.)
~Block companion form
1Ip 2I p  r1Ip rIp 
 I 0  0 0 
Ip  0 0
  
0  Ip 0
(rprp) (rpp)
(q rp)
where Ip is the p p unit matrix and every 0 is a p p zero matrix.
Then the realization of Ĝ(s) is given by

18. Realizations (cont.)
Proof:
Define,
Z1 
Z 
Z :  2  : (sI  A)1
B

Z

 r 
   i
where Z is the (pp) and Z is a (rpp).
Simplifying,
Then,
(sI  A)Z  B  sZ  AZ B (*)
 
  







 


0 

Z   0 
0
   
 

 0
0

0



Ip 
Zr


p
sZr


sZ 
0 Z3    0 
2 
sZ3   
2 
sZ1  rIp Z1 
1Ip 2I p  r1Ip
 I 0  0
Ip  0
 
0 Ip

19. Realizations (cont.)
Using the shifting property of the companion form of A, we obtain
Substituting these into the 1st block of equation (*) yields
which implies
 Zr 1
sZ2  Z1 , sZ3  Z2 ,  , sZr
sZ1  1Z1 2Z2 rZr Ip
1
3
2 1 Z
1 r
1
sr 1
s2
s
Z 
1
Z , Z 
1
Z ,  , Z 
p
r
s
s





1
r1
2
1
 
     Z I
or
p
r
s
s





sr 1
1
r1
2
1
 Z 
d(s)
 I

s   

20. Realizations (cont.)
Thus we have,
Then,
ˆ
1
2
1
C(sI  A)1
B  Ĝ()  G()
r
r2
r1
 N s
Ns
d(s)
 N
1
Z1 
d(s)
Ip
sr 2
sr 1
, Z2 
d(s)
Ip ,  , Zr 
d(s)
Ip
 Ĝsp (s)Ĝ()
 Ĝ(s)

21. Special Case: p = 1
Realizations (cont.)
(For simplicity, let r = 4 and q =2.)
Then, the realization is
Consider a 21 proper rational matrix:



22 23
3
 21 24 
14

13
2
12
3
11
4
3
2
1 2
4 3
2
ˆ
 s   s2
  s  
s  
s  
 s  
s  s 
s  s 
d 
d1  1
G(s)    
d 
 2 
24 
23
22
 21
 
 

14 
x 
d1 
u
y 
11 12 13
  0
1
0  0
0   
 0
 0
0  0
x   u
1 2 3 4 
 1 0 0
1 0
0 1
x
 
Note: The
controllable-
canonical-form
realization can be
read out from the
coefficients of Ĝ(s).

22. Realizations (cont.)
• A multi-input LTI system is the sum of many single-input
LTI systems, so can realize each single-input subsystem
and form the sum:
1st and 2nd column of

23. 



 
3
12
0 0
0



2
(s  2)2


s 1 
(2s1)(s  2)
s  2 
2s 1
1
 Ĝ()Ĝsp (s)
Example 3.3:
3




(s  2)2



s 1 
(2s1)(s  2)
s  2 
,
4s 10
2s 1
1
Ĝ(s) 

Consider a proper rational matrix

 

  

(2s 1)(s  2) (s  2)2


s  2 
2s 1
1 s 1
4s 10 3
Ĝ(s) 

Determine a realization of this transfer matrix.
Solution:
Decompose Ĝ(s) into a strictly proper rational matrix
d(s)  (s 0.5)(s  2)2
 s3
 4.5s2
6s  2
The monic least common denominator of Ĝsp (s) is

24. 



1
ˆ
(s 1)(s0.5) 
0.5(s 2)
3(s 2)(s 0.5)
6(s 2)2
s3
4.5s2
6s 2
sp
G
Example (cont.)
Thus



 2
(s 1.5s 0.5)
3(s2
2.5s1) 
1 6(s2
4s 4)
d(s)  0.5(s 2)


 2
s 1.5s 0.5
3s2
7.5s3
1 6s2
24s 24
d(s)  0.5s 1

 
 


0.5
1.5  1
 0.5
24
 6 3
1 2
7.5
s
24 3 
s 
d(s)  0 1

25.  
 
 
  2 
1
0.5
 0 0 0u 
2 0 u
x
y 
6
Example (cont.)
Given,
 


  
  

 0 0
0 1
0 1 0 0
0 0 1 0 0 

0
0
0
1


 1
0
0 6 0 2 0  1 0
0 4.5 0 6 0 2
4.5
0u2 
0 0 0 0  0 0u1 
0 0 0 0  0
 
0  0 0
x
x
 



3




(s  2)2



s 1 
(2s1)(s  2)
s  2 
,
4s 10
2s 1
1
Ĝ(s) 

A B
3 24 7.5 24 3
1 0.5 C
1.5 1 D
Ĝ(s)
 A,B,C,D
B


C D
S 
A
Realization
the realization is,

26. Example 3.4:



 


(2s1)(s  2)

4s 10
2s 1
1

(s) 
c1
Ĝ
The 1st column is
Consider again the proper rational matrix in Ex. 3.3.





 3
(s  2)2


s 1
(2s 1)(s  2)
s  2 
4s 10
2s 1
1
Ĝ(s) 

1
 

 (2s 1)(s 2) 

(4s 10)(s  2)

 

 
 (2s 1)(s 2) 
   
1
2

2s2
5s 2 

2s 5s 2
4s2
2s 20
In Matlab: n1=[4 -2 -20;0 0 1]; d1=[2 5 2]; [a,b,c,d]=tf2ss(n1,d1)

27. Ex. 3.4 (cont.)
Using MATLAB: n1=[4 -2 -20;0 0 1];d1=[2 5 2];[a,b,c,d]=tf2ss(n1,d1)
yields the following realization for the 1st column of Ĝ(s):
1
1
1
c1 1 1 1 1 0
0.5 
 0
   
0  1 0 1
  
1 1 1 1 1 

12
x 
2
u
y  C x d u 
6
1
x 
1
u
x
  A x b u 
2.5
Similarly, the function tf2ss can generate the following realization for the
2nd column of Ĝ(s):
2
2
2
2
2 2 2 2 2
c2 2 2 2 2 0
1 1
   
  
 1 0  0

6
x 
0
u
y C x d u 
3
4
x 
1
u
x
  A x b u 
4

28. Example 3.4 (cont.)
Notice that:





 3
(s  2)2


s 1

(2s 1)(s  2)
s  2 
2s 1
1
4s 10
Ĝ(s) 


29. Ex. 3.4 (cont.)
 2   2  2   2  2 
x   0 A x   0 b u 
These two realizations can be combined as
x1

A1 0 x1 

b1 0 u1 
y  yc1 yc2  C1 C2 xd1 d2 u
or
u





 
x

 




x u
0 0
y   0
6 12 3 6 2 0
1
0 1
0 0 4 4
0 0 
0  0 0
0
1
1 0 0  1 0
0
2.5
x
  

 

 










2 

0
0 1
0  0 0
0  0
1
0
2
0  1 0
0 6 0 2
0 4.5 0 6 0
 4.5
0u 
0 0 0 0 0  0
x
1 0 0 0
x

 



  2 
0.5
 0
0 0 1 0 0
 
0 0 0 1 0 0 0
3 24 7.5 24
1 0.5 1.5 1 u


0u1 
3 
x 
2
y 
6
A
(4
0x4)
1
B
(

4
0x20)
0
C
.5 1
(2x4)
 
D
(2x2)
A
(6x6)
B0u1
(6x2)
C
(2x6)
0
D
0
(2x2)

30. Example 3.4 (cont.)
• Can also focus on the realizations of single-output systems. Then treat
LTI systems with multi-outputs as combinations of single-outputs
subsystems.
• i-th row of
• Overall realization:

0

0
  
cq

 dq


  
Aq

 Bq


x    u
x
 
Realize with {Ai, Bi, ci, di}
c1 0  d1 
0  B1 
A1
 x    u ;y   