State space analysis, eign values and eign vectors

By:
Shilpa Mishra
M.E., NITTTR Chandigarh
.
STATE SPACE ANALYSIS
Conversion of T.F. models to canonical state variable models and
concept of Eignvalues & Eignvectors.

Contents_______________
1. Introduction
2. Need of realization of transfer function into state variable models
3. Realization of transfer function into a state space model or
mathematical model in following possible representation:
a. First companion form (controllable form)
b. Second companion form (observable form)
c. Jordan canonical form
4. Eigenvalues and Eigen vectors
08/05/142 Shilpa Mishra ME IC 122509

Plant
Mathematical Model :
Differential equation
Linear, time invariant
Frequency Domain
Technique
Time Domain
Technique
Two approaches for analysis and design of control system:
1.Classical Technique or Frequency Domain Technique.(T.F. + graphical plots like
root locus, bode etc.)
2.Modern Technique or Time Domain Technique (State variable approach).
08/05/143
Shilpa Mishra ME IC 122509

Transfer Function form
Need of conversion of transfer function form into state space form:
1.A transfer function can be easily fitted to the determined experimental data in best
possible manner. In state variable we have so many design techniques available for system.
Hence in order to apply these techniques T.F. must be realized into state variable model.

08/05/145

BuAxx +=
DuCxy +=
x
x
y
u
A
B
C
D
= state vector
= derivative of the state vector with respect to time
= output vector
= input or control vector
= system matrix
= input matrix
= output matrix
= Feed forward matrix
State equation
output equation
General State Space form of Physical System

Deriving State Space Model from Transfer Function Model
 The process of converting transfer function to state space form is NOT
unique. There are various “realizations” possible.
 All realizations are “equivalent” (i.e. properties do not change). However,
one representation may have some advantages over others for a particular
task.
 Possible representations:
1. First companion form
2. Second companion form
3. Jordan canonical form

1. First Companion Form (SISO System)
If LTI SISO system is described by transfer function of the form;
Decomposition of transfer function:
.
012
23
3
01
2
2
)(
)(
)(
)(
asasasa
bsbsb
sR
sC
sU
sY
+++
++
==

( ) ( ) ( ) ( )sXbsbsbsCsY 101
2
2 ++==
( ) 10
1
12
1
2
2 xb
dt
dx
b
dt
xd
bty ++=
)2........(..........322110)( xbxbxbty ++=
( ) ( ) ( ) ( )sXasasasasRsU 101
2
2
3
3 +++==
( ) )(
)()()(
10
1
12
1
2
23
1
3
3 txa
dt
tdx
a
dt
txd
a
dt
txd
atu +++=
43322110)( xaxaxaxatu +++=
I.
II.
)1...().........(32211043 tuxaxaxaxa +−−−=
21 xx = 32 xx =1)( xtx = 43 xx =&
Select state
variables like :

21 xx = 32 xx =1)( xtx =
from equation (1) & (2) and state equation, block diagram realization in first
companion form of TF will be 
43 xx =

Again from equation (1) & (2) complete state model will be ;
)3).....((
3
/1
0
0
3
2
1
210
100
010
3
1
3
2
1
,
)(
3
1
3
3
2
2
3
1
1
3
0
43
)(32211043
tu
a
x
x
x
aaa
a
x
x
x
or
tu
a
x
a
a
x
a
a
x
a
a
xx
tuxaxaxaxa




































+
−−−
=
+−−−==
+−−−=




A
B

Equation (3)&(4) combining together gives the complete realization of the given
transfer function.
Matrix A has coefficients of the denominator of the TF preceded by minus sign in its
bottom row and rest of the matrix is zero except for the superdiagonol terms which are
all unity.
In matrix theory matrix with this structure is said to be in companion form therefore
this realization is called first companion form of realizing a TF.
[ ] )4.......(
3
2
1
210
)(
,
322110)(










=
++=
x
x
x
bbbty
or
xbxbxbty
C

Example :TF to State Space (constant term in numerator)
rcccc 2424269 =+++ 
cx =1 cx =2 cx =3
1. Inverse Laplace
2. Select state variables
21 xx =
32 xx =
rxxxx 2492624 3213 +−−−=
1xcy ==
( ) ( )
( )sD
sN
sG =
( )sN
( )sD
numerator
denominator

r
24
0
0
x
x
x
92624
100
010
x
x
x
3
2
1
3
2
1










+




















−−−
=













[ ]










=
3
2
1
001
x
x
x
y

Example:

2. Second Companion Form (SISO System)
In second companion form coefficient of the denominator of the transfer
function appear in one of the column of the A matrix.
This form can be obtained by the following steps:
Let the transfer function is,
n
nnn
n
nnn
asasas
bsbsbsb
sU
sY
sH
++++
++++
== −−
−−
......
........
)(
)(
)( 2
2
1
1
2
2
1
10
)5)].......(()([
1
....)]()([
1
)()(
0)]()([...)]()([)]()([,
)()....()().....(
110
11
1
0
1
10
1
1
sYasUb
s
sYasUb
s
sUbsY
sUbsYasUbsYassubsYsor
sUbsbsbsYasas
nnn
nn
nn
n
nn
n
nn
−++−+=
=−++−+−
+++=+++
−
−−
On dividing by and solving for Y(s);n
s

By equation (5) block diagram can be drawn as follows which is called second companion
form of realization:-

To get state variable model, output of each integrator is identified as
state variables starting at the left and preceding to the right.
The corresponding differential equations are,
ub
n
xy
u
n
bub
n
x
n
ax
u
n
bub
n
x
n
axx
ubub
n
xa
n
x
n
x
ubub
n
xa
n
x
n
x
0
isequationoutputtheand
)
0
(
1
1
)
0
(
112
2
)
0
(
221
1
)
0
(
11
+=
++−=
−
++
−
−=
++−
−
=
−
++−
−
=






Now from here state and output equations organized in matrix form are given
below:
BuAxx +=
DuCxy +=
[ ] 0
011
011
0
1
2
1
;1000
B;
100
010
001
000
bDC
bab
bab
bab
a
a
a
a
A nn
nn
n
n
n
==










−
−
−
=
















−
−
−
−
= −−−
−







•In this form of realizing a TF the poles of the transfer function form a string
along the main diagonal of the matrix A.
•In Jordan canonical form state space model will be like:-
3. Jordan canonical form (Non-repeated roots)

n
nnn
n
nnn
asasas
dsdsdsd
sU
sY
sH
++++
++++
== −−
−−
......
........
)(
)(
)( 2
2
1
1
2
2
1
10
Let general transfer function;
)(
)(
)(
)(
1
21
1
ssx
dt
tdx
L
xx
dt
tdx
xtx
if
=





==
=


urxλx
.
.
.
urxλx
urxλx
nnnn +=
+=
+=



2222
1111
nxxxudty ++++= ......)( 210
& output equation;
With the help of above equations jordan canonical state model can be obtained as;
State equation:
08/05/1422

Jordan Canonical Form
(Non-repeated roots): Block Diagram

Jordan canonical form: Example (Non repeated roots)

Jordan canonical form: Example (repeated roots):

08/05/1427

Eignvalues and Eignvector
Definition :
Given a linear transformation A, a non-zero vector x is defined to be an
eigenvector of the transformation if it satisfies the eigenvalue equation
for some scalar λ. In this situation, the scalar λ is called an eigenvalue of A
corresponding to the eigenvector x.
 It indicates that vector x has the property that its direction is not changed by the
transformation A, but that it is only scaled by a factor of λ.
 Only certain special vectors x are eigenvectors, and only certain special scalars λ
are eigenvalues.
 The eigenvector must be non-zero because the equation A0 = λ0 holds for every A
and every λ.
λxAx =

A acts to stretch the vector x, not change its direction, so x is an eigenvector of A.
The eigenvalue λ is simply the amount of "stretch" or "shrink" to which a vector
is subjected when transformed by A.
If λ = 1, the vector remains unchanged (unaffected by the transformation). A
transformation I under which a vector x remains unchanged, Ix = x, is defined as
identity transformation
If λ = −1, the vector flips to the opposite direction; this is defined as reflection.

Computation of eigenvalues & the characteristic
equation
 When a transformation is represented by a square matrix A, the eigenvalue equation
can be expressed as;
Ax = λx
(A − λI)x = 0
As x must not be zero, this can be rearranged to;
det(A − λI) = 0.
which is defined to be the characteristic equation of the n × n matrix A.
 Expansion of the determinant generates a polynomial of degree n in λ and may be
written as;
here λ1, λ2……λn are called eignvalues. For each eign value there exist eign vector x.

Example 1.
find the eigenvalues and eigenvectors of the matrix
Solution
So the eigenvalues are 3.421 and 0.3288.

Let
be the eigenvector corresponding to
Hence
.421x1 +1.5x2 = 0
.75x1 +2.671x2 = 0

If
then
The eigenvector corresponding to then is
The eigenvector corresponding to
is
Similarly, the eigenvector corresponding to
is
08/05/1433

Example 2.
Find the eigenvalues and eigenvectors of
Solution
The characteristic equation is given by
The roots of the above equation are
Note that there are eigenvalues that are repeated. Since there are only two distinct
eigenvalues, there are only two eigenspaces. But, corresponding to λ = 0.5 there should be
two eigenvectors that form a basis for the eigenspace
08/05/1434

To find the eigenspaces, let
Given
then
For ,
Solving this system gives

So
So the vectors and form a basis for the eigenspace for the eigenvalue .
For ,
Solving this system gives
The eigenvector corresponding to is ;
Hence the vector is a basis for the eigenspace for the eigenvalue of .
08/05/1436

Thanks for
Attention

State space analysis, eign values and eign vectors

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