Equivalent State Equations
description
Transcript of Equivalent State Equations
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President University Erwin Sitompul Modern Control 3/1
Dr.-Ing. Erwin SitompulPresident University
Lecture 3Modern Control
http://zitompul.wordpress.com
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President University Erwin Sitompul Modern Control 3/2
( )u t
1 F1
1 H
( )y t
Equivalent State EquationsChapter 4 State Space Solutions and Realizations
1x
2x 2x2x
State variables: • : inductor current iL• : capacitor voltage vC
1x2x
RRviR
LL
div Ldt
CC
dvi Cdt
1x
2x
2x2 1x u x
2 1 2x x x
2y x
1 1
2 2
0 1 1( )
1 1 0x x
u tx x
1
2
0 1x
yx
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President University Erwin Sitompul Modern Control 3/3
( )u t
1 F1
1 H
( )y t
Homework 2: Equivalent State EquationsChapter 4 State Space Solutions and Realizations
1x 2x
State variables: • : current of left loop• : current of right loop1x
2x
1. Prove that for the same system, with different definition of state variables, we can obtain a state space in the form of:
11
22
1 1 1( )
1 0 1xx
u txx
1
2
1 1x
yx
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( )u t
1 F1
1 H
( )y t
Homework 2: Equivalent State EquationsChapter 4 State Space Solutions and Realizations
1x 2x
State variables: • : loop current left• : loop current right1x
2x
1 2Cv x x y
1 1 2( ) 0u x x x
CC
dvi Cdt
2 1 2x x x
1 x 1 1 2x u x x
1 2 2( )u x x x
2 1x u x
11
22
1 1 1( )
1 0 1xx
u txx
1
2
1 1x
yx
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Homework 2: Equivalent State EquationsChapter 4 State Space Solutions and Realizations
2. Derive a state-space description for the following diagram
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Homework 2: Equivalent State EquationsChapter 4 State Space Solutions and Realizations
1x2x3x 1x2x3x
1 2x x
2 3x x
3 1 3x ax x by u
1 2y x cx 1 1
2 2
3 3
0 1 0 00 0 1 0
1 1
x xx x ux a b bc x
1
2
3
1 0 0x
y c x ux
1 3 1 2( )ax x b x cx u
1 2 3( )a b x bcx x u
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x = P x is called an equivalence transformation.
is said to be algebraically equivalent with the original state space equations.
Equivalent State EquationsChapter 4 State Space Solutions and Realizations
( ) ( ) ( )t t t x Ax Bu( ) ( ) ( )t t t y Cx Du
Consider an n-dimensional state space equations:
Let P be an nn real nonsingular matrix, and let x = P x. Then, the state space equations
( ) ( ) ( )t t t x Ax Bu
( ) ( ) ( )t t t y Cx Du
where1,A PAP ,B PB 1,C CP .D D
~
~
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Equivalent State EquationsChapter 4 State Space Solutions and Realizations
Proof:Substituting 1( ) ( )t tx P x
1 1( ) ( ) ( )t t t P x AP x Bu
1( ) ( ) ( )t t t y CP x Du
1( ) ( ) ( )t t t x PAP x PBu
A B
C D
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Equivalent State EquationsChapter 4 State Space Solutions and Realizations
From the last electrical circuit,State variables: • : inductor current iL• : capacitor voltage vC
1x2x
State variables: • : loop current left• : loop current right1x
2x
( )u t
1 F1
1 H
( )y t
The two sets of states can be related in the way:1 1
2 2
1 01 1
x xx x
or
11 1
2 2
1 01 1
x xx x
1 1
2 2
1 01 1
x xx x
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President University Erwin Sitompul Modern Control 3/10
Transfer Function and Transfer MatrixChapter 4 State Space Solutions and Realizations
( ) ( ) ( )t t u t x Ax B( ) ( ) ( )y t t Du t Cx
Consider a state space equations for SISO systems:
Using Laplace transform, we will obtain:( ) (0) ( ) ( )s s s U s X x AX B
( ) ( ) ( )Y s s DU s CX
For zero initial conditions, x(0) = 0,1( ) ( ) ( )s s U s X I A B
1( ) ( ) ( )Y s s D U s C I A B
1( )( ) ( )( )Y sG s s DU s
C I A B Transfer Function
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Realization of State Space EquationsChapter 4 State Space Solutions and Realizations
Every linear time-invariant system can be described by the input-output description in the form of:
( ) ( ) ( )Y s U s G s
If the system is lumped (i.e., having concentrated parameters), it can also be described by the state space equations
( ) ( ) ( )t t u t x Ax B
( ) ( ) ( )y t t Du t Cx
The problem concerning how to describe a system in state space equations, provided that the transfer function of a system, G(s), is available, is called Realization Problem.
G(s) A, B, C, D.
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Realization of State Space EquationsChapter 4 State Space Solutions and Realizations
Three realization methods will be discussed now: Frobenius Form Observer Form Canonical Form
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Chapter 4 Realization of State Space Equations
Frobenius Form( )( )( )Y sG sU s
1
1 1 01
1 1 0
m mm mn n
n
b s b s b s bs a s a s a
1
1 1 01
1
1 1 01
( ) ( ) ( ) ( )
( ) ( ) ( ) ( )
n n
nn n
m m
m mm m
d y t d y t dy ta a a y tdt dt dt
d u t d u t du tb b b b u tdt dt dt
Special Case: No derivation of input1
1 1 0 01
( ) ( ) ( ) ( ) ( )n n
nn n
d y t d y t dy ta a a y t b u tdt dt dt
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Chapter 4 Realization of State Space Equations
Frobenius Form We now define:
1( ) ( )x t y t2 ( ) ( )x t y t 1( )x t
3( ) ( )x t y t 2 ( )x t
( 1)( ) ( )nnx t y t 1( )nx t
1 1
2 2
0 1 1 0
( ) 0 1 0 ( ) 0( ) 0 0 1 0 ( ) 0
( )
( ) ( )n n n
x t x tx t x t
u t
x t a a a x t b
1
2
( )( )
( ) 1 0 0 0 ( )
( )n
x tx t
y t u t
x t
Frobenius Form,Special Case
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Chapter 4 Realization of State Space Equations
Frobenius Form General Case: With derivation of input
11 1 0
11 1 0
( )( )
m mm mn n
n
b s b s b s bY sU s s a s a s a
( )( )N sD s
m n
1 21 2 1 0
( )( )( )
n nn n
U sY s b s b s b s bD s
2 10 1 2 1
( ) ( ) ( ) ( )( )( ) ( ) ( ) ( )
n nn n
U s U s U s U sY s b b s b s b sD s D s D s D s
1( )X s 2 ( )X s 1( )nX s ( )nX s
If m = n–1 (largest possible value), then
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1( )( )( )
U sX sD s
Chapter 4 Realization of State Space Equations
Frobenius Form
11 1 0
( )n n
n
U ss a s a s a
1 21 1 1 2 1 1 1 0 1( ) ( ) ( ) ( ) ( ) ( )n n
ns X s a s X s a s X s a sX s a X s U s
1 21 1 1 2 1 1 1 0 1( ) ( ) ( ) ( ) ( ) ( )n n
ns X s U s a s X s a s X s a sX s a X s
( ) ( 1)1 1 1 2 1 1 1 0 1( ) ( ) ( ) ( ) ( ) ( )n n
nx t u t a x t a x t a x t a x t
1 2 3 1 2 0 1( ) ( ) ( ) ( ) ( ) ( )n n nx t u t a x t a x t a x t a x t
But
11 1( ) ( )x t X sL2 1( ) ( )x t x t
1 2( ) ( )n nx t x t
1( ) ( )n nx t x t
If m = n–1 (largest possible value), then
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Chapter 4 Realization of State Space Equations
Frobenius Form The state space equations can now be written as:
1 1
2 2
0 1 1
( ) 0 1 0 0 ( ) 0( ) 0 0 1 0 ( ) 0
( )
( ) ( ) 1n n n
x t x tx t x t
u t
x t a a a x t
1
20 1 2 1
( )( )
( ) 0 ( )
( )
n n
n
x tx t
y t b b b b u t
x t
Frobenius Form,General Case
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Chapter 4 Realization of State Space Equations
Observer Form1 2
1 2 1 01
1 1 0
( ) ,( )
n nn n
n nn
b s b s b s bY sU s s a s a s a
11 1 0
1 21 2 1 0
( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )
n nn
n nn n
s Y s a s Y s a sY s a Y sb s U s b s U s b sU s b U s
1n m
1 1 01
1 2 1 02 1
( ) ( ) ( )( )
( ) ( ) ( ) ( )
n n n
n n n n
Y s Y s Y sY s a a as s s
U s U s U s U sb b b bs s s s
1 1 2 2
0 0
1 1 1( ) ( ) ( ) ( ) ( )
1 ( ) ( )
n n n nY s b U s a Y s b U s a Y ss s s
b U s a Y ss
1( )X s
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1 0 01( ) ( ) ( )X s b U s a Y ss
Chapter 4 Realization of State Space Equations
Observer Form1 0 0( ) ( ) ( )x t b u t a y t
2 1 1 11( ) ( ) ( ) ( )X s bU s a Y s X ss
2 1 1 1( ) ( ) ( ) ( )x t b u t a y t x t
1 1 11( ) ( ) ( ) ( )n n n nX s b U s a Y s X ss 1 1 1( ) ( ) ( ) ( )n n n nx t b u t a y t x t
( ) ( )nY s X s ( ) ( )ny t x t
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The state space equations in observer form:
1 0 1 0
2 1 2 1
1 1
( ) 0 0 ( )( ) 1 0 ( )
( )
( ) 0 0 1 ( )n n n n
x t a x t bx t a x t b
u t
x t a x t b
1
2
( )( )
( ) 0 0 1 0 ( )
( )n
x tx t
y t u t
x t
Observer Form
Chapter 4 Realization of State Space Equations
Observer Form
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Chapter 4 Realization of State Space Equations
Canonical Form
( )( ) ( )( )N sY s U sD s
01
( )n
i
i i
rr U s
s
To construct state space equations in canonical form, we need to perform partial fraction decomposition to the respective transfer function.
In case all poles are distinct, we define:
1 1 1( ) ( ) ( )x t x t u t 11
1( ) ( )X s U ss
22
1( ) ( )X s U ss
1( ) ( )nn
X s U ss
2 2 2( ) ( ) ( )x t x t u t
( ) ( ) ( )n n nx t x t u t
1 1 2 2
0
( ) ( ) ( ) ( ) ( )n n
y t r x t r x tr x t r u t
1 1 2 2
0
( ) ( ) ( ) ( ) ( )n n
Y s r X s r X sr X s rU s
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Chapter 4 Realization of State Space Equations
Canonical Form The state space equations in case all poles are distinct:
1 1 1
2 2 2
( ) 0 0 ( ) 1( ) 0 0 ( ) 1
( )
( ) 0 0 ( ) 1n n n
x t x tx t x t
u t
x t x t
Canonical Form,Distinct Poles
1
21 2 0
( )( )
( ) ( )
( )
n
n
x tx t
y t r r r r u t
x t
• The resulting matrix A is a diagonal matrix.• The ODEs are decoupled, each of them can be
solved independently.
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Chapter 4 Realization of State Space Equations
Canonical Form In case of repeating poles, for example λ1 is repeated for p
times, the decomposed equation will be:1 111 12 2
0 21 1 1 2 1
( ) ( )( ) ( )
p n pp
n p
r rr r rY s r U ss s s s s
We define:
1 1 1( ) ( ) ( )x t x t u t 11
1( ) ( )X s U ss
2 21
1( ) ( )( )
X s U ss
2 1 2 1( ) ( ) ( )x t x t x t 11
1 ( )X ss
1
1( ) ( )( )p pX s U ss
1 1( ) ( ) ( )p p px t x t x t
11
1 ( )pX ss
•x1(t) coupled with x2(t)
•xp–1(t) coupled with xp(t)
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11 1 12 2
1 2 1
1 0
( ) ( ) ( ) ( ) ( ) ( ) ( )
p p p
n p n
y t r x t r x tr x t r x tr x t r u t
11 1 12 2
1 2 1
1 0
( ) ( ) ( ) ( ) ( ) ( ) ( )
p p p
n p n
Y s r X s r X sr X s r X sr X s rU s
1 2 1( ) ( ) ( )p px t x t u t 12
1( ) ( )pX s U ss
1( ) ( ) ( )n n p nx t x t u t 1
1( ) ( )nn p
X s U ss
Chapter 4 Realization of State Space Equations
Canonical Form
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Chapter 4 Realization of State Space Equations
Canonical Form The state space equations in case of repeating poles:
11 1
12 2
1
1
21 1
1
0 0( ) ( ) 11 0 0( ) ( ) 000 1( ) ( ) 0
0 0( ) ( ) 11
0 0( ) ( ) 1
p p
p p
n pn n
x t x tx t x t
x t x t ux t x t
x t x t
( )t
Canonical Form,Repeating Poles
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1
2
11 12 1 2 1 0
1
( )( )
( ) ( ) ( )( )
( )
p n p p
p
n
x tx t
y t r r r r r x t r u tx t
x t
The state space equations in case of repeating poles:
Canonical Form,Repeating Poles
Chapter 4 Realization of State Space Equations
Canonical Form
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Homework 3: Transfer Function to State SpaceChapter 4
Find the state-space realizations of the following transfer function in Frobenius Form, Observer Form, and Canonical Form.
Realization of State Space Equations
3 2
( ) 2( )( ) 8 19 12Y s sG sU s s s s
Hint: Learn the following functions in Matlab and use the to solve this problem: roots, residue, conv.