Modern Control - Lec07 - State Space Modeling of LTI Systems

1. ‫ر‬َ‫ـد‬ْ‫ق‬‫ِـ‬‫ن‬،،،‫لما‬‫اننا‬ ‫نصدق‬ْْ‫ق‬ِ‫ن‬‫ر‬َ‫د‬
LECTURE (7)
STATE-SPACE REPRESENTATION
OF LTI SYSTEMS
Assist. Prof. Amr E. Mohamed

2. Agenda
 State Variables of a Dynamical System
 State Variable Equation
 Why State space approach
 Derive Transfer Function from State Space Equation
 Time Response and State Transition Matrix
2

3. Introduction
 The classical control theory and methods (such as root locus) that we
have been using in class to date are based on a simple input-output
description of the plant, usually expressed as a transfer function. These
methods do not use any knowledge of the interior structure of the
plant, and limit us to single-input single-output (SISO) systems, and as
we have seen allows only limited control of the closed-loop behavior
when feedback control is used.
 Modern control theory solves many of the limitations by using a much
“richer” description of the plant dynamics. The so-called state-space
description provide the dynamics as a set of coupled first-order
differential equations in a set of internal variables known as state
variables, together with a set of algebraic equations that combine the
state variables into physical output variables.
3

4. Definition of System State
 State: The state of a dynamic system is the smallest set of variables
(𝒙 𝟏, 𝒙 𝟐, … … , 𝒙 𝒏) (called State Variables or State Vector) such that knowledge of
these variables at 𝑡 = 𝑡0, together with knowledge of the input for 𝑡 ≥ 𝑡0 ,
completely determines the behavior of the system for any time t to t0 .
 The number of state variables to completely define the dynamics of the system is
equal to the number of integrators involved in the system (System Order).
 Assume that a multiple-input, multiple-output system involves n integrators (State
Variables).
 Assume also that there are r inputs u1(t), u2(t),……. ur(t) and p outputs y1(t),
y2(t), …….. yp(t).
4
Inner state variables
nxxx ,, 21 
)(1 tu
)(2 tu
)(tur
)(1 ty
)(2 ty
)(typ

5. General State Representation
 State equation:
 Output equation:
 𝑥 = 𝑆𝑡𝑎𝑡𝑒 𝑉𝑒𝑐𝑡𝑜𝑟
 𝑥 =
𝑑 𝑥(𝑡)
𝑡
= 𝐷𝑒𝑟𝑖𝑣𝑎𝑡𝑖𝑣𝑒 𝑜𝑓 𝑡ℎ𝑒 𝑆𝑡𝑎𝑡𝑒 𝑉𝑒𝑐𝑡𝑜𝑟
 𝑢 = 𝐼𝑛𝑝𝑢𝑡 𝑉𝑒𝑐𝑡𝑜𝑟
 𝑦 = 𝑂𝑢𝑡𝑝𝑢𝑡 𝑉𝑒𝑐𝑡𝑜𝑟
 𝐴 = 𝑆𝑡𝑎𝑡𝑒 𝑀𝑎𝑡𝑟𝑖𝑥 = 𝑆𝑦𝑠𝑡𝑒𝑚 𝑀𝑎𝑡𝑟𝑖𝑥
 𝐵 = 𝐼𝑛𝑝𝑢𝑡 𝑀𝑎𝑡𝑟𝑖𝑥
 𝐶 = 𝑂𝑢𝑡𝑝𝑢𝑡 𝑀𝑎𝑡𝑟𝑖𝑥
 𝐷 = 𝐹𝑒𝑒𝑑𝑏𝑎𝑐𝑘 𝑚𝑎𝑡𝑟𝑖𝑥
)()()( tuBtxAtx 
)()()( tuDtxCty 
Dynamic equations

6. State-Space Equations (Model)
 State equation:
 Output equation:
6
)()()( tuBtxAtx 
)()()( tuDtxCty 
Dynamic equations
1
2
1
)(
)(
)(
)(














nn tx
tx
tx
tx

1
2
1
)(
)(
)(
)(














rr tu
tu
tu
tu

1
2
1
)(
)(
)(
)(
















pp ty
ty
ty
ty

State Vector
State variable
Input Vector Output Vector










 nnA










 rnB










 npC
1
2
1
)0(
)0(
)0(
)0(














nnx
x
x
x











 rpD

7. Block Diagram Representation Of State Space Model
7
C
A
D
B
s
1
+
+
+
-
)(tu )(ty
)(tx)(tx

8. Input/Output Models vs State-Space Models
 State Space Models:
 consider the internal behavior of a system
 can easily incorporate complicated output variables
 have significant computation advantage for computer simulation
 can represent multi-input multi-output (MIMO) systems and nonlinear
systems
 Input/Output Models:
 are conceptually simple
 are easily converted to frequency domain transfer functions that are more
intuitive to practicing engineers
 are difficult to solve in the time domain (solution: Laplace transformation)
8

9. Some definitions
 System variable: any variable that responds to an input or initial
conditions in a system
 State variables: the smallest set of linearly independent system
variables such that the values of the members of the set at time t0
along with known forcing functions completely determine the value of
all system variables for all t ≥ t0
 State vector: a vector whose elements are the state variables
 State space: the n-dimensional space whose axes are the state variables
 State equations: a set of first-order differential equations with b
variables, where the n variables to be solved are the state variables
 Output equation: the algebraic equation that expresses the output
variables of a system as linear combination of the state variables and
the inputs.

10. General State Representation
1. Select a particular subset of all possible system variables, and call
state variables.
2. For nth-order, write n simultaneous, first-order differential equations
in terms of the state variables (state equations).
3. If we know the initial condition of all of the state variables at 𝑡0 as
well as the system input for 𝑡 ≥ 𝑡0, we can solve the equations

11. State-Space Representation of nth-Order Systems of Linear
Differential Equations
 Consider the following nth-order system:
𝒚
(𝒏)
+ 𝒂 𝟏 𝒚
(𝒏−𝟏)
+ … + 𝒂 𝒏−𝟏 𝒚 + 𝒂 𝒏 𝒚 = 𝒖
 where y is the system output and u is the input of the System.
 The system is nth-order, then it has n-integrators (State Variables)
 Let us define n-State variables
11

12. State-Space Representation of nth-Order Systems of Linear
Differential Equations (Cont.)
 Then the last Equation can be written as
12

13. State-Space Representation of nth-Order Systems of Linear
Differential Equations (Cont.)
 Then, the stat-space state equation is
 where
13

14. State-Space Representation of nth-Order Systems of Linear
Differential Equations (Cont.)
 Since, the output equation is
 Then, the stat-space output Equation is
 where
14

15. Example #1
 From the diagram, the system Equation is
𝑀 𝑦 + 𝐵 𝑦 + 𝐾𝑦 = 𝑓(𝑡)
 This system is of second order. This means that the system involves two
integrators (State Variables).
 Let us define the state variables
𝑥1 = 𝑦
𝑥2 = 𝑦
 Then, we obtain
𝑥1 = 𝑦 = 𝑥2
𝑥2 = 𝑦 =
1
𝑀
−𝐵 𝑦 + 𝐾𝑦 −
1
𝑀
𝑓 𝑡 =
−𝐵
𝑀
𝑥2 −
𝐾
𝑀
𝑥1 −
1
𝑀
𝑓 𝑡
15
y
K
M
B
f(t)

16. Example #1 (Cont.)
 Then, the State Space equation is
𝑥1
𝑥2
=
0 1
−𝐾
𝑀
−𝐵
𝑀
𝑥1
𝑥2
+
0
1
𝑀
𝑓(𝑡)
 The output Equation is
𝑦 = 1 0
𝑥1
𝑥2
 The System Block diagram is
16

17. Example #2
17
LR
c)(tei )(tec
+
- )(ti
+
-
 
t
i tedtti
cdt
tdi
LtRi
0
)()(
1)(
)(


dttitx
titx
let
)()(
)()(
2
1
)()( tity 
  



































2
1
2
1
2
1
01)(
)(
0
1
01
1
x
x
ty
teLx
x
LCL
R
x
x
i


)()(ˆ
)()(ˆ
2
1
tetx
titx
let
c

)()( tity 
  



























 






2
1
2
1
2
1
ˆ
ˆ
01)(
)(
0
1
ˆ
ˆ
01ˆ
ˆ
x
x
ty
teL
x
x
L
L
R
L
R
x
x
i

Remark : the choice of states is not unique.

18. Example #3
18
M2 M1
B3
B1B2
K
1y2y
)(tf
0)()(
)()()(
121222322
212121111


yyKyyByByM
tfyyKyyByByM


24
13
22
11
yx
yx
yx
yx
let






)(
4
3
2
1
4
3
2
1
tf
x
x
x
x
x
x
x
x



























































19. Example #4
 Find the state space model for a system that described by the following
differential equation
 Solution:
 The system is 3rd order, then it has three states as follows
 The output equation is
rcccc 2424269  
cx 1
cx 2
cx 3
21 xx 
32 xx 
rxxxx 2492624 3213 
1xcy 
differentiation

20. Example #4
 











3
2
1
001
x
x
x
y
r
x
x
x
x
x
x











































24
0
0
92624
100
010
3
2
1
3
2
1




21. State-Space Representations of
Transfer Function Systems
21

22. State-Space Representation in Canonical Forms
 We here consider a system defined by
 where u is the control input and y is the output. We can write this
equation as
 we shall present state-space representation of the system defined by
(1) and (2) in controllable canonical form, observable canonical form,
and diagonal canonical form.
22

23. Controllable Canonical Form
 We consider the following state-space representation, being called a
controllable canonical form, as
 Note that the controllable canonical form is important in discussing the
pole-placement approach to the control system design.
23

24. Observable Canonical Form
 We consider the following state-space representation, being called an
observable canonical form, as
24

25. Diagonal Canonical Form
 Diagonal Canonical Form greatly simplifies the task of computing the
analytical solution to the response to initial conditions.
 We here consider the transfer function system given by (2). We have the
case where the dominator polynomial involves only distinct roots. For
the distinct root case, we can write (2) in the form of
25

26. Diagonal Canonical Form (Cont.)
 The diagonal canonical form of the state-space representation of this
system is given by
26

27. Example #5
 Obtain the state-space representation of the transfer function system
(16) in the controllable canonical form.
 Solution: From the transfer function (16), we obtain the following
parameters: b0 = 1, b1 = 3, b2 = 3, a1 = 2, and a2 = 1. The resulting
state-space model in controllable canonical form is obtained as
27

28. Example #6
 Find the state-space representation of the following transfer function
system (13) in the diagonal canonical form.
 Solution: Partial fraction expansion of (13) is
 Hence, we get A = −1 and B = 3. We now have two distinct poles. For
this, we can write the transfer function (13) in the following form:
28

29. State Space model to Transfer
Function
29

30.  The state space model
 by Laplace transform
 Then, the transfer function is
     sBUsAXssX 
     sDUsCXsY 
     sBUAsIsX
1

BuAxx 
DuCxy 
      sUDBAsICsY 
1
   
 
  DBAsIC
sU
sY
sT 
1
State Space model to Transfer Function

31. Example (2)
 Find the transfer function from the following transfer function
 Solution:
uxx























0
0
10
321
100
010
  xy 001
 














321
10
01
s
s
s
AsI
 
)det(
)(1
AsI
AsIadj
AsI




123
)12(
)3(1
13)23(
23
2
2















sss
sss
sss
sss

32. Example (2)
    DBAsICsT 
1
 
123
)23(10
23
2



sss
ss
sT
 
123
0
0
10
)12(
)3(1
13)23(
001
)( 23
2
2

























sss
sss
sss
sss
sT

33. System Poles from State Space model
 poles and check the stability of the following state space Example find the
System model
 Solution:
 Since
 To find the poles 
 Then the poles are {-1, -2 }, the system is stable
uxx 













0
5
31
20
  xy 01
02)3(
31
2



 ss
s
s
AsI
  








31
2
s
s
AsI

34. State-Space Modeling with
MATLAB
34

35. State-Space Modeling with MATLAB
 MATLAB uses the controllable canonical form by default when converting from
a state space model to a transfer function. Referring to the first example
problem, we use MATLAB to create a transfer function model and then convert
it to find the state space model matrices:
35

36. State-Space Modeling with MATLAB
 Note that this does not match the result we obtained in the first example. See
below for further explanation. No we create an LTI state space model of the
system using the matrices found above:
36

37. State-Space Modeling with MATLAB
 we can generate the observable and controllable models as follows:
37

38. State transition matrix
38

39. Introduction
 The behavior of x(t) and y(t):
1) Homogeneous solution of x(t).
2) Non-homogeneous solution of x(t).
39
)()()(
)()()(
tDutCxty
tButAxtx
dt
d



40. Homogeneous solution
 State transition matrix
40
)0()()(
)()0()(
)()(
1
xAsIsX
sAXxssX
tAxtx




)0(
)0(])[()( 11
xe
xAsILtx
At

 
])[()( 11 

 AsILet At
)()()()()(
)()0(
)0()(
000
)(
0
0
0
00
0
0
txtttxetxeetx
txex
xetx
ttAAtAt
At
At






41. State Transition Matrix Properties
41
)()(.5
)()()(.4
)()()0(.3
)()(.2
)0(.1
020112
1
ktt
tttttt
txtx
tt
I
k






])[()( 11 

 AsILet At

42. Non-homogeneous solution
42
)()()(
)()()(
tDutCxty
tButAxtx
dt
d


 






t
dButxttx
sBUAsILxAsILtx
sBUAsIxAsIsX
sBUxsXAsI
sBUsAXxssX
0
1111
11
)()()0()()(
)]()[()0(])[()(
)()()0()()(
)()0()()(
)()()0()(
 Convolution
Homogeneous

43. Non-homogeneous solution (Cont.)
43
)()()()()()(
)()()()()(
)()()0()()(
0
0
00
00
0
tDudButCtxttCty
dButtxtttx
dButxttx
t
t
t
t
t









Zero-input response Zero-state response

44. Example 1
44
 T
xlet
tu
x
x
x
x
00)0(
)(
1
0
32
10
2
1
2
1




































 


tttt
ttt
At
eeee
eeee
eAsILt 22
212
11
222
2
])[()(
 
t
dButxttx
0
)()()0()()( 


















tt
tt
ee
ee
x
x
2
2
2
1
2
1
2
1
Ans: )]()[( 11
sBUAsIL 

stepunittu )(

45. How to find State transition matrix
45
Methode 1: ])[()( 11 
 AsILt
Methode 3: Cayley-Hamilton Theorem
Methode 2:
At
et  )(
])[()( 11 

 AsILet At

46. Method 1:
46
])[()( 11 
 AsILt







































































3
2
1
2
1
2
1
3
2
1
3
2
1
1
0
0
0
0
1
)(
)(
10
01
00
211
340
010
x
x
x
ty
ty
u
u
x
x
x
x
x
x




















 
ssss
ss
sss
ssssAsI
AsIadj
AsI
414
323
32116
33)2)(4(
1)(
)(
2
2
2
1

47. Method 2:
47
At
et  )(
 



































































3
2
1
2
1
2
1
3
2
1
3
2
1
166
)(
)(
1
1
1
300
020
001
x
x
x
ty
ty
u
u
x
x
x
x
x
x

















t
t
t
At
e
e
e
et
3
2
00
00
00
)(
diagonal matrix

48. linear system by Meiling CHEN 48
Diagonization

49. linear system by Meiling CHEN 49
Diagonization

50. 50
Case 1: distincti 
)1)(3(
43
1
43
10










 


A
1
3
2
1
































3
1
0
433
13
)(
2
1
2
1
11
v
v
v
v
VAI
























 

1
1
0
33
11
)(
2
1
2
1
22
v
v
v
v
VAI
depend
  














 
10
03
13
11 1
21 APPVVP

51. 51
n  321
In the case of A matrix is phase-variable form and
 













 11
2
1
1
21
21
111
n
n
nn
n
nvvvP





Vandermonde matrix
for phase-variable form













4
3
2
1
1




APP
1
 PPee tAt

52. 52
Case 2: distincti 
)2)(1)(1(
200
010
101
200
010
101














 
 



 AIA
0
100
000
100
)(
3
2
1
11 






















v
v
v
VAI
21  
depend












































0
1
0
000
0
0
1
000
3
2
1
321
3
2
1
321
v
v
v
vvv
v
v
v
vvv
21 VV 

53. 53
0
000
010
101
)(
3
2
1
33 





















v
v
v
VAI
23 






















1
0
1
00
3
2
1
321
v
v
v
vvv
 




















 
 
200
010
001
100
010
101
1
321 APPVVVP

54. 54
Case 3: distincti  Jordan form
321  
  formJordanAPPvvvP  1
321
Generalized eigenvectors
231
121
11
)(
)(
0)(
vvAI
vvAI
vAI

















1
1
1
1
1
1
ˆ



AAPP











t
tt
tttt
tA
e
tee
etee
e
1
11
1
2
11
2
ˆ




55. 55
Example:
2
)2(
11
13
11
13










 


A
























 

1
1
0
11
11
)(
12
11
12
11
11
v
v
v
v
VAI






























 

0
1
1
1
11
11
)(
22
21
22
21
21
v
v
v
v
VAI
  












 
20
12ˆ
01
11 1
21 AAPPVVP
1ˆ
2
22
ˆ 






 PPee
e
tee
e tAAt
t
tt
tA

56. 56
Method 3:

57. 57
AaAaIaAaAaa
AaAaAaA
IaAaAaA
IaAaAaA
n
nn
n
n
n
n
n
n
n
n
n
0
2
101
1
11
0
2
11
1
01
1
1
01
1
1
)(
0
















  n
n AkAkAkIkAf 2
210)(any







1
0
1
1
2
210)(
n
k
k
k
n
n
A
AAAIAf

 

58. 58







10
21
?100
AA
Example:
AIAAflet 10
100
)(  
2,1,0)2)(1(
20
21
21 





100
210
100
22
100
110
100
11
2)(
1)(




f
f
12
22
100
1
100
0









 













10
221
10
21
)12(
10
01
)22()(
101
100100100
AAf

59. 59





 

02
13
? AeAt
Example:
2,1,0
2
13
21 





2)2(
)1(
10210
2
10110






t
t
ef
ef
tt
tt
ee
ee




2
1
2
0 2
















 










tttt
tttt
ttttAt
eeee
eeee
eeeee
222
2
02
13
)(
10
01
2
22
22
22

60. 60

61. 61

62. 62

63. linear system by Meiling CHEN 63

64. linear system by Meiling CHEN 64

65. linear system by Meiling CHEN 65

66. Controllability and Observability
66

67. Introduction
 The main objective of using state-space equations to model systems is
the design of suitable compensation schemes to control these systems.
 Typically, the control signal u(t) is a function of several measurable
state variables. Thus, a state variable controller, that operates on the
measurable information is developed.
 State variable controller design is typically comprised of three steps:
 Assume that all the state variables are measurable and use them to design a
full-state feedback control law. In practice, only certain states or
combination of them can be measured and provided as system outputs.
 An observer is constructed to estimate the states that are not directly
sensed and available as outputs. Reduced-order observers take advantage of
the fact that certain states are already available as outputs and they don’t
need to be estimated.
 Appropriately connecting the observer to the full-state feedback control law
yields a state-variable controller, or compensator. 67

68. Introduction
 a given transfer function G(s) can be realized using infinitely many
state-space models
 certain properties make some realizations preferable to others
 one such property is controllability
68

69. Motivation1: Controllability
69
  
































2
1
2
1
2
1
01
)(
0
1
10
12
x
x
y
tu
x
x
x
x


1
s 1
s 1
1 2
u y1x2x
s
x )0(2
s
x )0(1
1 1x2x
1
controllable
uncontrollable

70. Controllability and Observability
 Plant:
 Definition of Controllability
70
DuCxy
RxBuAxx n

 ,
A system is said to be (state) controllable at time , if
there exists a finite such for any and any ,
there exist an input that will transfer the state
to the state at time , otherwise the system is said to
be uncontrollable at time .
0t
01 tt  )( 0tx 1x
][ 1,0 ttu )( 0tx
1x 1t
0t

71. Controllability Matrix
 Consider a single-input system (u ∈ R):
 The Controllability Matrix is defined as
 We say that the above system is controllable if its controllability matrix
𝐶(𝐴, 𝐵) is invertible.
 As we will see later, if the system is controllable, then we may assign
arbitrary closed-loop poles by state feedback of the form 𝑢 = −𝐾𝑥.
 Whether or not the system is controllable depends on its state-space
realization.
71
 
 BABAABBBAC
nCrankBA
n 12
),(
,)(leControllab,





72. Example: Computing 𝐶(𝐴, 𝐵)
 Let’s get back to our old friend:
 Here,
 Is this system controllable?
72

73. Controllability Matrix
73
1
1
)(sU
)(sY
1
-1
s
3
-1
s
2
Example: An Uncontrollable System
 xy
uxx
21
0
1
30
01
















1x
2x
※ State is uncontrollable.2x
0)det(  U
 BABAABBU n 12
MatrixilityControllab 
 
Ruif

74. Proof of controllability matrix
74
 

































)1(
)2(
1
)1()2(1
21
)1()2(1
21
1
2
12
112
1
)(
nk
nk
k
n
k
n
nk
nknkk
n
k
n
k
n
nk
nknkk
n
k
n
k
n
nk
kkkkkkk
kkk
kkk
u
u
u
BABBAxAx
BuABuBuABuAxAx
BuABuBuABuAxAx
BuABuxABuBuAxAx
BuAxx
BuAxx




Initial condition

75. Motivation2: Observability
75
  
































2
1
2
1
2
1
01
)(
1
3
10
02
x
x
y
tu
x
x
x
x


1
s 1
s 1
1 2
u y1x2x
s
x )0(2
s
x )0(1
1 1x2x
3
observable
unobservable

76. Controllability and Observability
 Plant:
 Definition of Observability
76
DuCxy
RxBuAxx n

 ,
A system is said to be (completely state) observable at
time , if there exists a finite such that for any
at time , the knowledge of the input and the
output over the time interval suffices to
determine the state , otherwise the system is said to be
unobservable at .
0t 01 tt  )( 0tx
][ 1,0 ttu
],[ 10 tt
0x
0t
0t
][ 1,0 tty

77. Observability Matrix
77
Example: An Unobservable System
 xy
uxx
40
1
0
20
10















※ State is unobservable.1x
1)(sU -1
s
-1
s 1x2x
2
4
)(sY
  nVrankCA  )(Observable, 0)det(  V

















1
2
MatrixityObservabil
n
CA
CA
CA
C
V

Ry if

78. Proof of observability matrix
78
 
 )1()2()3(11
1
)1()2(1
321
1
111
111
1
)(),2(),1(
)(
)2()(
)1(



























nknknkkkkkk
k
n
nknkk
n
k
n
k
n
nk
kkkkkkk
kkk
kkk
kkk
DuCBuCABuDuCBuyDuy
x
CA
CA
C
n
nDuCBuBuCABuCAxCAy
DuCBuCAxDuBuAxCy
DuCxy
DuCxy
BuAxx






Inputs & outputs

79. Example
 Plant:
 Hence the system is both controllable and observable.
79
 10,
1
0
,
01
10












 CBA
DuCxy
RxBuAxx n

 ,
 




















01
10
MatrixtyObervabili
01
10
MatrixilityControllab
CA
C
N
ABBV
2)()(  NrankVrank

80. Controllability and Observability
80
Theorem I
)()()( tuBtxAtx cccc 
Controllable canonical form Controllable
Theorem II
)()(
)()()(
txCty
tuBtxAtx
oo
oooo


Observable canonical form Observable
 A system in Controller Canonical Form (CCF) is always controllable!!
 A system in Observable Canonical Form (OCF) is always controllable!!

81. Example
81
  c
cc
xy
uxx
12
1
0
32
10














Controllable canonical form
 






















12
12
31
10
CA
C
V
ABBU
nVrank
nUrank


1][
2][
  o
oo
xy
uxx
10
1
2
31
20















Observable canonical form
 























31
10
11
22
CA
C
V
ABBU
nVrank
nUrank


2][
1][
)2)(1(
2
)(



ss
s
sT

82. Linear system (Analysis)
82
Theorem III
)()()(
)()()(
tDutCxty
tButJxtx


Jordan form
 321
3
2
1
3
2
1
CCCC
B
B
B
B
J
J
J
J























Jordan block
Least row
has no zero
row
First column has no zero column

83. Example
83
 xccy
ub
b
xx
3
1100
020
012
1211
12
11























If 012 b uncontrollable
If 011 c unobservable

84. 84
xy
uxx





























































2
1
0
203
102
200
201
101
211
100
010
211
100
010
001
000
1
1
1
2
2
2
1
1
1
1








11b
12b
13b
21b
11C 12C 13C 21C

85. 85
   
    ....
....
21131211
21131211
ILCILCCC
ILbILbbb

 controllable
observable
In the previous example
   
    ....
....
21131211
21131211
DLCILCCC
ILbILbbb

 controllable
unobservable

86. 86









 












































0
0
1
001
002
113
111
111
122
0
1
0
0
1
1
001
123
111
112
112
1
1
11
2
2
2
12
y
uxx
L.I.
L.I.
L.I. L.D.
Example

87. Kalman Canonical Decomposition
 Diagonalization: &
 All the Eigenvalues of A are distinct, i.e.
 There exists a coordinate transform such that
 System in z-coordinate becomes
 Homogeneous solution of the above state equation is
87
BuAxx  DuCxy 
n  321
Txz 
.
0
0
where
1
1










 
n
mm AATTA



zCy
uBzAz
m
mm


)0()0()( 11
1
n
t
n
t
zevzevtz n
 
 mnmm
mn
m
m
ccCTC
b
b
BTB


1
1
1











 
observableandlecontrollabismode0,and0If i mimi cb

88. How to construct coordinate transformation matrix for diagonalization
 All the Eigenvalues of A are distinct, i.e.
 The coordinate matrix for diagonalization
 Consider diagonalized system
88
][ 21 n,v,, vvT 
t.independenare,rs,Eigenvecto 21 n,v,, vv 
n  321
ubzλz
ubzz
ubzz
mnnnn
m
m







2222
1111


nmnmm zczczcy  2211

89. Transfer function is
 H(s) has pole-zero cancellation.
89
n
mnmnmm
n
i i
mimi
s
bc
s
bc
s
bc
sH
 




 

1
11
1
)(
le,unobservaborableuncontrollismode0,or0If i mimi cb
1
 1mc1mb
2
 2mc2mb
n
 mncmnb
∑

)(tu
)(ty








90. Kalman Canonical Decomposition
90
)(ty
OC
S

COS
OC
SOC
S

)(tu
SubsystembleUnobservale,Controllab:OC
S
SubsystemObservableable,Uncontroll:OC
S
SubsystemObservablele,Controllab:COS
SubsystembleUnobservaable,Uncontroll:OC
S

91. Kalman Canonical Decomposition: State Space Equation
91
 xCCy
u
B
B
x
x
x
x
A
A
A
A
x
x
x
x
OCCO
OC
CO
OC
OC
OC
CO
OC
OC
OC
CO
OC
OC
OC
CO
00
0
0
000
000
000
000























































(5.X)

92. Example
92
 xcccy
u
b
b
b
xx
311211
31
12
11
300
020
001























3mode0,If 13 b
3mode0,If 13 c
The same reasoning may be applied to mode 1 and 2.
Plant:
is uncontrollable.
is unobservable.

93. Pole-zero Cancellation in Transfer Function
 From Sec. 5.2, state equation
 may be transformed to
93
n
mnmnmm
n
i i
mimi
s
bC
s
bC
s
bC
sH
 




 

1
11
1
)(
Hence, the T.F. represents the controllable and observable
parts of the state variable equation.
BuAxx 
DuCxy 
T.F..invanishesandableuncontrollismode,0If imib
T.F..invanishesandleunobservabismode0,If imic

94. Example
 Plant:
 Transfer Function
94
  BAsICsH
sU
sY 1
)(
)(
)( 

  
 
  
 
 
 
4
1
2
22
10
42
1
1
2
41
02
10
42
1






























s
s
s
ss
s
s
ss
4,2 21  
 xy
uxx
10
1
2
21
04
















T.F..invanishes"-2"Mode

95. Example 5.6
 Plant:
 Transfer Function
95
uxx 













1
2
11
60

 xy 10
 
3
1
)(
)(
)( 1



s
BAsICsT
sU
sY
T.F..invanishes"2"Mode
-3,2 21  

96. Minimum Realization
 Realization:
 Realize a transfer function via a state space equation.
 Example
 Realization of the T.F.
 Method 1:
 Method 2:
 There is infinity number of realizations for a given T.F. .
96
3
1
)(


s
sT
1 1)(sU )(sY
3
1 1)(sU )(sY-1
s
3
2
-1
s
-1
s
1
3
1
)(
)(
)(


s
sT
sU
sY
2
2
3
1
)(
)(
)(





s
s
s
sT
sU
sY

97. Minimum Realization
 Minimum realization:
 Realize a transfer function via a state space equation with elimination of its
uncontrollable and unobservable parts.
 Example 5.8
 Realization of the T.F.
97
3
5
)(
)(
)(


s
sT
sU
sY
1 5)(sU )(sY
3
-1
s
3
5
)(


s
sT

98. 98