LECTURE (7)
STATE-SPACE REPRESENTATION
OF LTI SYSTEMS
Assist. Prof. Amr E. Mohamed
‫لما نصدق اننا ِن ْْقدَ ر‬
،،، ‫نِـ ْقـدَ ر‬
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
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
Definition of System State
u1 (t )
u2 (t )
ur (t )




y1 (t )
Inner state variables

x1 , x2 , xn

y2 (t )
y p (t )
State: The state of a dynamic system is the smallest set of variables
(𝒙𝟏 , 𝒙𝟐 , … … , 𝒙𝒏 ) (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
General State Representation


State equation:
Output equation:
x (t )  A x(t )  B u (t )
y (t )  C x(t )  D u (t )
 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
Dynamic equations
State-Space Equations (Model)


State equation:
Output equation:
State variable
 x1 (t ) 
 x (t ) 
x(t )   2 
  


x
(
t
)
 n  n1
State Vector

A  

nn




x (t )  A x(t )  B u (t )
Dynamic equations
y (t )  C x(t )  D u (t )
 y1 (t ) 
 y (t ) 
2

y (t )  
  


y
(
t
)
 p  p1
 u1 (t ) 
u (t )
u (t )   2 
  


u
(
t
)
 r  r1
Input Vector

B  

nr





C  

 x1 (0) 
 x (0) 
x(0)   2 
  


x
(
0
)
 n  n1
Output Vector
pn





D  

pr




6
Block Diagram Representation Of State Space Model
D
u (t )
B
x (t )
+
-
1
s
x(t )
+
C
+
y (t )
A
7
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
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.
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
State-Space Representation of nth-Order Systems of Linear
Differential Equations

Consider the following nth-order system:
(𝒏)



(𝒏−𝟏)
𝒚 + 𝒂𝟏 𝒚 + … + 𝒂𝒏−𝟏 𝒚 + 𝒂𝒏 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
State-Space Representation of nth-Order Systems of Linear
Differential Equations (Cont.)

Then the last Equation can be written as
12
State-Space Representation of nth-Order Systems of Linear
Differential Equations (Cont.)

Then, the stat-space state equation is

where
13
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
Example #1

From the diagram, the system Equation is
𝑀𝑦 + 𝐵𝑦 + 𝐾𝑦 = 𝑓(𝑡)
y
K
M
f(t)
B

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
1
1
−𝐵
𝐾
1
𝑥2 = 𝑦 =
−𝐵𝑦 + 𝐾𝑦 − 𝑓 𝑡 =
𝑥2 − 𝑥1 − 𝑓 𝑡
𝑀
𝑀
𝑀
𝑀
𝑀
15
Example #1 (Cont.)



Then, the State Space equation is
0
1 𝑥
0
𝑥1
1
= −𝐾 −𝐵 𝑥 + 1 𝑓(𝑡)
𝑥2
2
𝑀
𝑀
𝑀
The output Equation is
𝑥1
𝑦= 1 0 𝑥
2
The System Block diagram is
16
Example #2
+
Remark : the choice of states is not unique.
ei (t )
di(t ) 1 t
Ri (t )  L
  i(t )dt  ei (t )
dt
c 0
let
x1 (t )  i (t )
x2 (t )   i(t )dt
y(t )  i(t )
 R
 x1 



L
 x 

 2
 1

1 
LC 
0 
 x1 
y (t )  1 0  
 x2 
 x1   1 
 x    L  ei (t )
 2   0 
c
i (t )
-
let
+
L
R
ec (t )
-
xˆ1 (t )  i (t )
xˆ2 (t )  ec (t )
y(t )  i(t )
 R
R 
 xˆ1 
L
L
    1
0 
 L
 xˆ2 

 xˆ1 
y (t )  1 0  
 xˆ2 
 xˆ1   1 
 xˆ    L  ei (t )
 2   0 
17
y2
Example #3
K
M2
B3
y1
M1
f (t )
B1
B2
M 1 y1  B1 y1  B2 ( y1  y 2 )  K ( y1  y 2 )  f (t )
M 2 y2  B3 y 2  B2 ( y 2  y1 )  K ( y 2  y1 )  0
x1  y1
let
x2  y 2
x3  y1
x4  y 2
  
 x1  
  
 x2   
 x3  
  
 x4 






   
 x1   
   
 x2     f (t )
 x3   
   
 x4 
18
Example #4

Find the state space model for a system that described by the following
differential equation
c  9c  26c  24c  24r

Solution:

The system is 3rd order, then it has three states as follows
x1  c

differentiation
x1  x2
x2  c
x2  x3
x3  c
x3  24 x1  26 x2  9 x3  24r
The output equation is
y  c  x1
Example #4
1
0
 x1   0
 x    0

0
1
 2 

 x3   24  26  9
 x1 
y  1 0 0  x2 
 x3 
 x1   0 
x    0  r
 2  
 x3  24
State-Space Representations of
Transfer Function Systems
21
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
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
Observable Canonical Form

We consider the following state-space representation, being called an
observable canonical form, as
24
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
Diagonal Canonical Form (Cont.)

The diagonal canonical form of the state-space representation of this
system is given by
26
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
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
State Space model to Transfer
Function
29
State Space model to Transfer Function

The state space model
x  Ax  Bu
y  Cx  Du


by Laplace transform
sX s   AX s   BU s 
X s   sI  A BU s 
Y s   CX s   DU s 
Y s   C sI  A B  D U s 
Then, the transfer function is
Y s 
1
T s  
 C sI  A B  D
U s 
1

1

Example (2)


Find the transfer function from the following transfer function
Solution:
1
0
0
10
x   0
0
1  x   0  u
 1  2  3
 0 
y  1 0 0 x
s 1 0 
sI  A  0 s  1 
1 2 s  3
 ( s 2  3 s  2)
s3
1



1
s
(
s

3
)
s


2


s

(
2
s

1
)
s
adj ( sI  A)
1

sI  A 

det(sI  A)
s 3  3s 2  2 s  1
Example (2)
T s   C sI  A B  D
1
 ( s 2  3s  2)
s3
1

1 0 0   1
s ( s  3)
s
2


s

(
2
s

1
)
s


T (s) 
s 3  3s 2  2 s  1
10( s 2  3s  2)
T s   3
s  3s 2  2s  1
10
0
 
 0 
System Poles from State Space model

poles and check the stability of the following state space Example find the
System model
2
0
5 
x  
x   u

 1  3
0 

Solution:

Since
y  1 0 x
s  2 
sI  A  

1
s

3


sI  A 
s
2
 s ( s  3)  2  0

To find the poles 

Then the poles are {-1, -2 }, the system is stable
1 s3
State-Space Modeling with
MATLAB
34
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
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
State-Space Modeling with MATLAB

we can generate the observable and controllable models as follows:
37
State transition matrix
38
Introduction
d
x(t )  Ax (t )  Bu (t )
dt
y (t )  Cx(t )  Du(t )

The behavior of x(t) and y(t):
1) Homogeneous solution of x(t).
2) Non-homogeneous solution of x(t).
39
Homogeneous solution
x (t )  Ax (t )
sX ( s)  x(0)  AX ( s)
1
x(t )  L [( sI  A) ]x(0)
 e At x(0)
X ( s)  ( sI  A) 1 x(0)

1

1
1
 (t )  e  L [( sI  A) ]
State transition matrix
At
x(t0 )  e At 0 x(0)
x(0)  e
 At 0
x(t0 )
At  At 0
x(t )  e e
x(t0 )  e
A ( t t 0 )
x(t0 )  (t  t0 ) x(t0 )
40
State Transition Matrix Properties

1
1
(t )  e  L [(sI  A) ]
At
1.
 ( 0)  I
2.
 1 (t )   (t )
3.
x(0)   (t ) x(t )
4.
 (t 2  t1 )  (t1  t0 )   (t 2  t0 )
5.
 (t )   (kt )
k
41
Non-homogeneous solution
d
x(t )  Ax (t )  Bu (t )
dt
y (t )  Cx(t )  Du(t )
sX ( s)  x(0)  AX ( s)  BU ( s )
( sI  A) X ( s)  x(0)  BU ( s )
1
1
X ( s)  ( sI  A) x(0)  ( sI  A) BU ( s )
x(t )  L1[( sI  A) 1 ]x(0)  L1[( sI  A) 1 BU ( s)]
t
x(t )  (t ) x(0)   (t   ) Bu ( )d
Convolution
0
Homogeneous
42
Non-homogeneous solution (Cont.)
t
x(t )  (t ) x(0)   (t   ) Bu ( )d
0
t
x(t )  (t  t0 ) x(t0 )   (t   ) Bu ( )d
t0
t
y (t )  C(t  t0 ) x(t0 )   C(t   ) Bu ( )d  Du(t )
t0
Zero-input response
Zero-state response
43
Example 1
1   x1  0
 x1   0
 x    2  3  x   1u (t )
 2   
 2 
let
x(0)  0 0
u (t )  unit step
T
t
2t

2
e

e
(t )  L1[(sI  A) 1 ]  e At  
t
 2t

2
e

2
e

x(t )   (t ) x(0) 
Ans:
e 1  e 2t 
t
 2t 
 e  2e 
t
 (t   ) Bu( )d
0
1
1
 x1    e t  e  2t 
2
x    2
 2   e t  e  2t 
L1[(sI  A) 1 BU (s)]
44
How to find State transition matrix

(t )  e At  L1[(sI  A) 1 ]
Methode 1:
(t )  L1[(sI  A) 1 ]
Methode 2:
(t )  e At
Methode 3: Cayley-Hamilton Theorem
45
Method 1:
1
1
(t )  L [(sI  A) ]
1
0   x1  0 0
 x1   0
 x    0  4 3   x   1 0   u1 
 2 
 2  
 u 
 x3   1  1  2  x3   0 1  2 
 x1 
 y1 (t )  1 0 0  
 y (t )  0 0 1  x2 
x 
 2  
 3
 s 2  6 s  11 s  2
3 
adj ( sI  A)
1


1
2
( sI  A) 

3
s 2
3s 

sI  A
s ( s  4)( s  2)  3  3s
2
 s4


s

1
s

4
s


46
Method 2: (t )  e
At
0   x1  1
 x1   1 0
 x    0  2 0   x   1  u1 
 2 
  2    u 
 x3   0
0  3  x3  1  2 
 x1 
 y1 (t ) 
x 



6

6
1
 y (t )
 2
 2 
 x3 
e  t

At
(t )  e   0
0

0
e  2t
0
diagonal matrix
0 

0 
 3t 
e 
47
Diagonization
linear system by Meiling CHEN
48
Diagonization
linear system by Meiling CHEN
49
Case 1:
i  distinct
1
0
A


3

4



1
3 4
1  3
2  1
 (  3)(  1)
depend
 1   v1 
 3
(1 I  A)V1  
0



 3  3  4 v2 
 1  1  v1 
(2 I  A)V2  
0



 3 3  v2 
P  V1
 v1   1 
   
v2   3
 v1   1 
   
v2   1
1
1
 3 0 
1
V2   
 P AP  



3

1
0

1




50
In the case of A matrix is phase-variable form and
1  2  3  n
P  v1 v2
1
1 
 1





2
n 
 vn    1
 

 n 1
n 1
n 1 



2
n 
 1
1

2
1

P AP   



t
e  Pe P
At
3
Vandermonde matrix
for phase-variable form





4 
1
51
Case 2:
i  distinct
1 0  1
A  0 1 0 
0 0 2 
 1
I  A 
0
0
 1
1
0
0
0
 2
 (  1)(  1)(  2)
1  2
0 0 1   v1 
(1 I  A)V1  0 0 0  v2   0
0 0  1 v3 
 v1  1
0v1  0v2  v3  0  v2   0
v3  0
depend
 v1  0
0v1  0v2  v3  0  v2   1
v3  0
V1  V2
52
3  2
1 0 1  v1 
(3 I  A)V3  0 1 0 v2   0
0 0 0 v3 
 v1   1
v1  0v2  v3  0  v2    0 
v3   1 
P  V1 V2
1 0  1
1 0 0
V3   0 1 0   P 1 AP  0 1 0
0 0 1 
0 0 2
53
Case 3:
i  distinct
Jordan form
1  2  3
P  v1 v2
v3   P AP  Jordan
1
form
Generalized eigenvectors
(1 I  A)v1  0
(1 I  A)v2  v1
(1 I  A)v3  v2
e

Aˆ t
e 


1t
1 1



1
ˆ
P AP  A  
1 1 

1 
te
1t
e 1t
e 1t 
1t 
te 
e 1t 
t2
2
54
Example:
 3 1
A


 1 1
 3
1
1
 1
 (  2) 2
 1  1 v11 
(1 I  A)V1  
0



 1 1  v12 
v11   1 
   
v12   1
 1  1 v21 
1
(1 I  A)V2  
  



 1 1  v22 
 1
P  V1
v21  1
   
v22  0
 1 1
2 1 
1
ˆ
V2   
 P AP  A  



1
0
0
2




e 2t
e 

Aˆ t
te 2t 
At
Aˆ t 1
 e  Pe P
2t 
e 
55
Method 3:
56
An  an 1 An 1    a1 A  a0 I  0
An  an 1 An 1    a1 A  a0 I
An 1  an 1 An    a1 A2  a0 A
  an 1 (an 1 An 1    a1 A  a0 I )   a1 A2  a0 A
any
f ( A)  k0 I  k1 A  k2 A2    kn An  
f ( A)   0 I  1 A   2 A     n 1 A
2

n 1
n 1

k 0
k
A
k
57
Example:
100
A
?
1 2
A

0
1


f ( A)  A100   0 I  1 A
let
 1
2
0
 2
 (  1)(  2)  0 , 1  1, 2  2
f (1 )  
100
1
f (2 )  
100
2
  0  11  1
100
  0  12  2
100
 0  2  2100
1  2100  1
101
1
0
1
2

1
2
 2




100
100
100
f ( A)  A  (2  2 ) 
 (2  1) 




0
1
0
1
1 



 0
58
Example:
 3  1
A

2
0


e ?
At
 3 1
 0 , 1  1, 2  2
2 
f (1)  e t   0  11   0  1
 0  2e t  e 2t
f (2)  e 2t   0  12   0  1 2
1  e  2 t  e  t
t
e  2e  e
At
 2t
 1
1 0
 2t
t   3
0 1   (  e  e )  2

0




 2e  2t  e t

 2t
t

2
e

2
e

e  2 t  e t 
 2t
t 
 e  2e 
59
60
61
62
linear system by Meiling CHEN
63
linear system by Meiling CHEN
64
linear system by Meiling CHEN
65
Controllability and Observability
66
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
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
Motivation1: Controllability
 x1   2 1   x1  1
 x    0  1  x   0 u (t )
  2  
 2 
 x1 
y  1 0  
 x2 
uncontrollable
u
x2 (0)
s
s 1 x2
x 2
1
x1 (0)
s
1
x1
s 1
x1 1
y
2
1
controllable
69
Controllability and Observability

Plant:
x  Ax  Bu , x  R n
y  Cx  Du

Definition of Controllability
A system is said to be (state) controllable at time t 0 , if
there exists a finite t1  t0 such for any x(t0 ) and any x1 ,
there exist an input u[t0 ,t1 ] that will transfer the state x(t0 )
to the state x1 at time t1 , otherwise the system is said to
be uncontrollable at time t 0 .
70
Controllability Matrix

Consider a single-input system (u ∈ R):

The Controllability Matrix is defined as
 A, B  Controllable  rank (C )  n,

C ( A, B)  B



AB
A2 B  An 1 B
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

Example: Computing 𝐶(𝐴, 𝐵)

Let’s get back to our old friend:

Here,

Is this system controllable?
72
Controllability Matrix
 det(U )  0 if u  R

Controllability Matrix U  B
AB
A2 B  An1B

Example: An Uncontrollable System
 1 0 
1
x  
x   u

 0  3
 0
y  1 2x
※ State x2 is uncontrollable.
U (s) 1
s -1
x1
1
1
s
-1
Y (s)
x2 2
3
73
Proof of controllability matrix
xk 1  Axk  Buk
xk  2  Axk 1  Buk 1
xk  2  A( Axk  Buk )  Buk 1  A2 xk  ABuk  Buk 1
xk  n  An xk  An 1 Buk  An  2 Buk 1    ABuk  ( n  2 )  Buk  ( n 1)
xk  n  An xk  An 1 Buk  An  2 Buk 1    ABuk  ( n  2 )  Buk  ( n 1)
 uk 
  

xk  n  An xk  An 1 B  AB B 
uk  ( n  2 ) 


 uk  ( n 1) 


Initial condition
74
Motivation2: Observability
 x1   2 0   x1  3
 x    0  1  x   1 u (t )
  2  
 2 
 x1 
y  1 0  
 x2 
u
1
x2 (0)
s
s 1
x 2
x2
x1 (0)
s
x1
1
3
s 1
x1 1
y
2
observable
unobservable
75
Controllability and Observability

Plant:
x  Ax  Bu , x  R n
y  Cx  Du

Definition of Observability
A system is said to be (completely state) observable at
time t 0 , if there exists a finite t1  t0 such that for any x(t0 )
at time t 0 , the knowledge of the input u[t t ] and the
output y[t t ] over the time interval [t0 , t1 ] suffices to
determine the state x0 , otherwise the system is said to be
unobservable at t 0 .
0, 1
0, 1
76
Observability Matrix
 A, C  Observable  rank (V )  n
if y  R
 det(V )  0
 C 
 CA 


2
Observability M atrix V   CA 





CAn 1 
Example: An Unobservable System
0 1 
 0
x  
x   u

0  2 
1
y  0 4x
※ State x1 is unobservable.
U (s) 1
4
s -1
x2 s -1
x1
Y (s)
2
77
Proof of observability matrix
xk 1  Ax k  Bu k
yk  Cxk  Duk  (1)
yk 1  Cxk 1  Duk 1
yk 1  C ( Ax k  Bu k )  Duk 1  CAx k  CBu k  Duk 1  (2)
yk  n 1  CAn 1 xk  CAn  2 Bu k  CAn 3 Bu k 1    CBu k  ( n  2)  Duk  ( n 1)  (n)
 C 
 CA 
 xk
(1), (2), (n)  
  
 n 1 
CA 
 yk  Duk yk 1  CBu k  Duk 1   CABu k  ( n 3)  CBu k  ( n  2 )  Duk  ( n 1)

Inputs & outputs

78
Example

Plant:
x  Ax  Bu , x  R n
y  Cx  Du
0 1 
0 
A
, B   , C  0 1

1 0
1
0 1 
Controllability Matrix V  B AB   

1
0


 C  0 1 
Obervability Matrix
N  

CA
1
0
  


rank (V )  rank ( N )  2
Hence the system is both controllable and observable.
79
Controllability and Observability
Theorem I
xc (t )  Ac xc (t )  Bc u(t )
Controllable canonical form

Controllable
A system in Controller Canonical Form (CCF) is always controllable!!
Theorem II
xo (t )  Ao xo (t )  B o u (t )
y (t )  Co xo (t )
Observable canonical form

Observable
A system in Observable Canonical Form (OCF) is always controllable!!
80
Example
s2
T ( s) 
( s  1)(s  2)
Controllable canonical form
0 1

1

3


1
C   2
V  

CA

2

1
  

U  B


AB  
Observable canonical form
 2  2
U  B AB   

1

1


 C  0 1 
V  

CA 1  3
1
0
0 
xc  
xc   u

 2  3
1
y  2 1xc
rank[U ]  2  n
0  2 
 2
xo  
xo   u

1  3
1 
y  0 1xo
rank[U ]  1  n
rank[V ]  1  n
rank[V ]  2  n
81
Linear system (Analysis)
Theorem III
x (t )  Jx (t )  Bu (t )
Jordan form
y (t )  Cx(t )  Du(t )
Jordan block
 J1

J 
J2

C  C1 C2



J 3 
C3 
 B1 


B   B2 
 B3 
First column has no zero column
Least row
has no zero
row
82
Example
 2 1 0
b11 
x  0 2 0 x  b12 u
0 0 1
 1 
y  c11 c12 3x
If
b12  0
uncontrollable
If
c11  0
unobservable
83
1 1

1

1

x  
1

2




1
2
0

 1
 0
 
 0
 x  1
 
 0
1  0
 
2 


0 0
b11
0 0
b12
1 0
b13
0 1

1 2 u
1 0

0 1  b21



1 1 2 0 0 2 0
y  1 0 1 2 0 1 1  x
1 0 2 3 0 2 2
C11
C12 C13 C21
84
b11
C11
b12 b13L.I . 
b21L.I .
C13L.I .  C21L.I .
controllable
C12
observable
In the previous example
b11
C11
b12 b13L.I . 
b21L.I .
C13L.I .  C21L.D.
controllable
C12
unobservable
85
Example
2



x  




2
y  1
1
1
2
2
2
2 1 3
1 1 2
1 1 1
L.I.
  2
  2
 
  1
x   3

1 1   1

1  1
1  1
 1 1 1
0 0 0
0 0 0
1 1
1 1 
1 1
2 1u

0 0
0 1
0 0
L.I.
L.I.
L.D.
86
Kalman Canonical Decomposition
x  Ax  Bu
Diagonalization:

All the Eigenvalues of A are distinct, i.e. 1  2  3  n



&
y  Cx  Du

There exists a coordinate transform z  Tx such that
0
1
.
Am  T 1 AT where Am  



n 
 0
 bm1 
System in z-coordinate becomes
Bm  T 1B    
 
z  Am z  Bmu
bmn 
y  Cm z
Cm  CT  cm1  cmn 
Homogeneous solution of the above state equation is z(t )  v1e  t z1 (0)    vn e  t z n (0)
1
If bmi  0 and cmi  0, mode i is controllable and observable
n
87
How to construct coordinate transformation matrix for diagonalization

All the Eigenvalues of A are distinct, i.e. 1  2  3  n
Eigenvecto rs, v1, v2 , ,vn , are independent.

The coordinate matrix for diagonalization T  [v1, v2 , ,vn ]

Consider diagonalized system
z1  1 z1  bm1u
z2  2 z2  bm 2u

zn  λn zn  bmn u
y  cm1z1  cm2 z2    cmn zn
88
Transfer function is
n
H ( s)  
i 1
cmi bmi cm1bm1
c b

   mn mn
s  i s  1
s  n
If bmi  0 or cmi  0, mode i is uncontrollable or unobservable,

H(s) has pole-zero cancellation.
bm1



cm1
1
u (t )

bm 2



cm 2
2
∑
y (t )

bmn



n
cmn
89
Kalman Canonical Decomposition
SC O
u(t )
SCO
SC O
SCO


y (t )
SCO : Controllable, Observable Subsystem
SCO : Controllable, Unobservable Subsystem
SCO : Uncontrollable, Observable Subsystem
SCO : Uncontrollable, Unobservable Subsystem
90
Kalman Canonical Decomposition: State Space Equation
 xCO   ACO
 x   0
 CO   
 xC O   0
  
 xC O   0
y  CCO
0
0
ACO
0
0
AC O
0
0
0 CC O
0x
0   xCO   BCO 
0   xCO   BCO 


u
0   xC O   0 


AC O   xC O   0 
(5.X)
91
Example
Plant:
1
x  0

0
y  c11
0
2
0
c12
0
 b11 
0 x  b12 u

 
3
b13 
c13 x
If b13  0, mode 3 is uncontrollable.
If c13  0, mode 3 is unobservable.
The same reasoning may be applied to mode 1 and 2.
92
Pole-zero Cancellation in Transfer Function

From Sec. 5.2, state equation
x  Ax  Bu
y  Cx  Du

may be transformed to
n
H ( s)  
i 1
Cmi bmi Cm1bm1
C b

   mn mn
s  i
s  1
s  n
If bmi  0 , mode i is uncontrollable and vanishes in T.F..
If cmi  0, mode i is unobservable and vanishes in T.F..
Hence, the T.F. represents the controllable and observable
parts of the state variable equation.
93
Example
  4 0    2
x  
x   u

 1  2  1 

Plant:
1  2, 2  4
y  0 1x

Transfer Function
Y ( s)
1
 H ( s)  C sI  A B
U ( s)
0    2
s  2
1

 0 1  
s  4  1 
s  2s  4
 1
 2s  2
 0 1  

s  2s  4
 s  2  
1

s4

1
Mode"-2" vanishes in T.F..
94
Example 5.6
0 6    2 
x  
x   u

1  1  1 

Plant:
1  2, 2  -3
y  0 1x

Transfer Function
Y ( s)
1
1
 T ( s)  C sI  A B 
U ( s)
s3
Mode"2" vanishes in T.F..
95
Minimum Realization

Realization:
 Realize a transfer function via a state space equation.

Example
 Realization of the T.F. T ( s ) 
1
s3
 Method 1:
U (s) 1
Y ( s)
1
 T ( s) 
U ( s)
s3
 Method 2:
s -1
1
Y (s)
3
U (s) 1
Y ( s)
1 s2
 T ( s) 

U ( s)
s3 s2
s -1
3
s
 There is infinity number of realizations for a given T.F. .
1 Y (s)
1
-1
2
96
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. T ( s ) 
5
s3
Y ( s)
5
 T ( s) 
U ( s)
s3
U (s) 1
s -1
5
Y (s)
3
97
98