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State space analysis concept, state space model to transfer function model in first and second companion forms jordan canonical forms, Concept of eign values eign vector and its physical meaning,characteristic equation derivation is presented from the control system subject area.
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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
04/09/232 Shilpa Mishra ME IC 122509
Plant
Mathematical Model : Differential equation
Linear, time invariant
Frequency DomainTechnique
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).
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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.
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BuAxx DuCxy
xx
y
uA
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
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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
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1. First Companion Form (SISO System)
If LTI SISO system is described by transfer function of the form;
Decomposition of transfer function:
.
01223
3
012
2
)(
)(
)(
)(
asasasa
bsbsb
sR
sC
sU
sY
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sXbsbsbsCsY 1012
2
101
121
2
2 xbdt
dxb
dt
xdbty
)2........(..........322110)( xbxbxbty
sXasasasasRsU 1012
23
3
)()()()(
101
121
2
231
3
3 txadt
tdxa
dt
txda
dt
txdatu
43322110)( xaxaxaxatu
I.
II.
)1...().........(32211043 tuxaxaxaxa
21 xx 32 xx 1)( xtx 43 xx &
Select state variables like :
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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
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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
13
3
22
3
11
3
043
)(32211043
tu
ax
x
x
aaaa
x
x
x
or
tua
xa
ax
a
ax
a
axx
tuxaxaxaxa
A B
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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
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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
sNsG
sN
sD
numerator
denominator
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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
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Example:
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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,
nnnn
nnnn
asasas
bsbsbsb
sU
sYsH
......
........
)(
)()(
22
11
22
110
)5)].......(()([1
....)]()([1
)()(
0)]()([...)]()([)]()([,
)()....()().....(
110
111
0
110
11
sYasUbs
sYasUbs
sUbsY
sUbsYasUbsYassubsYsor
sUbsbsbsYasas
nnn
nnnn
nnn
nnn
On dividing by and solving for Y(s);ns
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By equation (5) block diagram can be drawn as follows which is called second companion form of realization:-
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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,
ubn
xy
un
bubn
xn
ax
un
bubn
xn
axx
ububn
xan
xn
x
ububn
xan
xn
x
0
isequation output theand
)0
(1
1)
0(
112
2)
0(
221
1)
0(
11
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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
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•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)
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nnnn
nnnn
asasas
dsdsdsd
sU
sYsH
......
........
)(
)()(
22
11
22
110
Let general transfer function;
)()(
)(
)(
1
21
1
ssxdt
tdxL
xxdt
tdx
xtx
if
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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:
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Jordan Canonical Form(Non-repeated roots): Block Diagram
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Jordan canonical form: Example (Non repeated roots)
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Jordan canonical form: Example (repeated roots):
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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
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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 transformationIf λ = −1, the vector flips to the opposite direction; this is defined as reflection.
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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.
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Example 1. find the eigenvalues and eigenvectors of the matrix
75.075.0
5.13][A
Solution
75.075.0
5.13][][ IA
0)5.1)(75.0()75.0)(3()det( IA
0125.1375.025.2 2 0125.175.32
)1(2
)125.1)(1(4)75.3()75.3( 2
2
092.375.3
3288.0,421.3 So the eigenvalues are 3.421 and 0.3288.
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3288.0,421.3 21 Let
2
1][x
xX
be the eigenvector corresponding to 421.31
Hence 0]])[[]([ 1 XIA
010
01421.3
75.075.0
5.13
2
1
x
x
0
0
671.275.0
5.1421.0
2
1
x
x
.421x1 +1.5x2 = 0 .75x1 +2.671x2 = 0
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If sx 1
then
sx
xs
2808.0
05.1421.0
2
2
The eigenvector corresponding to 421.31 then is
s
sX
2808.0][
2808.0
1s
The eigenvector corresponding to 421.31
is
2808.0
1
Similarly, the eigenvector corresponding to 3288.02
is
781.1
1
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Example 2. Find the eigenvalues and eigenvectors of
005.0
5.05.05.0
105.1
][A
Solution
The characteristic equation is given by 0])[]det([ IA
0
05.0
5.05.05.0
105.1
det
0)]5.0)(5.0()0)(5.0)[(1()]0)(5.0())(5.0)[(5.1(
025.12 23 The roots of the above equation are
0.1,5.0,5.0 Note that there are eigenvalues that are repeated. Since there are only two distinct eigenvalues, there are only two eigenspaces. But, corresponding to = 0.5 there should be two eigenvectors that form a basis for the eigenspace
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To find the eigenspaces, let
3
2
1
][
x
x
x
X
Given 0][)][( XIA
then
0
0
0
05.0
5.05.05.0
105.1
3
2
1
x
x
x
For 5.0 ,
0
0
0
5.005.0
5.005.0
101
3
2
1
x
x
x
Solving this system gives axbxax 321 ,
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So
a
b
a
x
x
x
3
2
1
0
0
0 b
a
a
0
1
0
1
0
1
ba
So the vectors
1
0
1
and
0
1
0
form a basis for the eigenspace for the eigenvalue 5.0 .
For 1 ,
0
0
0
105.0
5.05.05.0
105.0
3
2
1
x
x
x
Solving this system gives axaxax 5.0,5.0, 321
The eigenvector corresponding to 1 is
5.0
5.0
1
5.0
5.0 a
a
a
a
;
Hence the vector
5.0
5.0
1
is a basis for the eigenspace for the eigenvalue of 1 .
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Thanks
for
Attentio
n04/09/2337 Shilpa Mishra ME IC 122509
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