Feedback Control Systems (FCS)
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Transcript of Feedback Control Systems (FCS)
Feedback Control Systems (FCS)
Dr. Imtiaz Hussainemail: [email protected]
URL :http://imtiazhussainkalwar.weebly.com/
Lecture-26-27-28-29State Space Canonical forms
Lecture Outlineβ Canonical forms of State Space Models
β’ Phase Variable Canonical Form
β’ Controllable Canonical form
β’ Observable Canonical form
β Similarity Transformations
β’ Transformation of coordinates
β Transformation to CCF
β Transformation OCF
Canonical Formsβ’ Canonical forms are the standard forms of state space models.
β’ Each of these canonical form has specific advantages which makes it convenient for use in particular design technique.
β’ There are four canonical forms of state space modelsβ Phase variable canonical formβ Controllable Canonical formβ Observable Canonical formβ Diagonal Canonical formβ Jordan Canonical Form
β’ It is interesting to note that the dynamics properties of system remain unchanged whichever the type of representation is used.
Companion forms
Modal forms
Phase Variable Canonical form
β’ The method of phase variables possess mathematical advantage over other representations.
β’ This type of representation can be obtained directly from differential equations.
β’ Decomposition of transfer function also yields Phase variable form.
Phase Variable Canonical formβ’ Consider an nth order linear plant model described by the
differential equation
β’ Where y(t) is the plant output and u(t) is the plant input.
β’ A state model for this system is not unique but depends on the choice of a set of state variables.
β’ A useful set of state variables, referred to as phase variables, is defined as:
ππ π¦ππ‘π
+π1ππβ 1π¦ππ‘πβ 1 +β―+ππβ1
ππ¦ππ‘ +ππ π¦=π’(π‘)
π₯1=π¦ , π₯2=οΏ½ΜοΏ½ , π₯3=π¦ ,β― , π₯π=ππβ1 π¦ππ‘πβ1
Phase Variable Canonical form
β’ Taking derivatives of the first n-1 state variables, we have
π₯1=π¦ , π₯2= οΏ½ΜοΏ½ , π₯3=π¦ ,β― , π₯π=ππβ1 π¦ππ‘πβ1
οΏ½ΜοΏ½1=π₯2 , οΏ½ΜοΏ½2=π₯3 , οΏ½ΜοΏ½3=π₯4β― , οΏ½ΜοΏ½πβ1=π₯π
οΏ½ΜοΏ½π=βππ π₯1βππβ1π₯2ββ―βπ1π₯π+π’(π‘)
u
xx
xx
aaaaxx
xx
n
n
nnnn
n
10
00
1000
01000010
1
2
1
131
1
2
1
Phase Variable Canonical form
β’ Output equation is simply
π₯1=π¦ , π₯2= οΏ½ΜοΏ½ , π₯3=π¦ ,β― , π₯π=ππβ1 π¦ππ‘πβ1
n
n
xx
xx
y
1
2
1
0001
8
β« β« β« β«
1a
2a
na
1xy 2xy
nn xy )1(
)(ny
1)2(
nn xy
β¦
)(tu
οΌ οΌ
Phase Variable Canonical form
9
Phase Variable Canonical form
yu s1
s1
s1
s1
1 1
1
2
3
1 n
n
1x
21 xx nx 1nx2nx
β’ Obtain the state equation in phase variable form for the following differential equation, where u(t) is input and y(t) is output.
β’ The differential equation is third order, thus there are three state variables:
β’ And their derivatives are (i.e state equations)
2 π3 π¦ππ‘3 +4 π2 π¦
ππ‘2 +6 ππ¦ππ‘ +8 π¦=10π’ (π‘)
π₯1=π¦ π₯2=οΏ½ΜοΏ½ π₯3= οΏ½ΜοΏ½
οΏ½ΜοΏ½1=π₯2
οΏ½ΜοΏ½2=π₯3
οΏ½ΜοΏ½3=β4 π₯1β3π₯2β2π₯3+5π’ (π‘)
Phase Variable Canonical form (Example-1)
Phase Variable Canonical form (Example-1)
β’ In vector matrix form
π₯1=π¦ π₯2=οΏ½ΜοΏ½ π₯3= οΏ½ΜοΏ½οΏ½ΜοΏ½1=π₯2
οΏ½ΜοΏ½2=π₯3
οΏ½ΜοΏ½3=β4 π₯1β3π₯2β2π₯3+5π’ (π‘)
3
2
1
3
2
1
3
2
1
001)(
)(500
234100010
xxx
ty
tuxxx
xxx
Home Work: Draw Sate diagram
β’ Consider the transfer function of a third-order system where the numerator degree is lower than that of the denominator.
β’ Transfer function can be decomposed into cascade form
β’ Denoting the output of the first block as W(s), we have the following input/output relationships:
Phase Variable Canonical form (Example-2)
π (π )π (π )
=πππ 2+π1π +π2
π 3+π1π 2+π2π +π3
1π 3+π1π 2+π2π +π3
πππ 2+π1π +π2π (π ) π (π )π (π )
π (π )π (π )
= 1π 3+π1π
2+π2π +π3
π (π )π (π )
=πππ 2+π1π +π2
β’ Re-arranging above equation yields
β’ Taking inverse Laplace transform of above equations.
β’ Choosing the state variables in phase variable form
Phase Variable Canonical form (Example-2)
π (π )π (π )
= 1π 3+π1π
2+π2π +π3
π (π )π (π )
=πππ 2+π1π +π2
+
π (π )=πππ 2π (π )+π1π π (π )+π2π (π )
+
π¦ (π‘)=πποΏ½ΜοΏ½ (π‘ )+π1 οΏ½ΜοΏ½ (π‘ )+π2π€(π‘ )
π₯1=π€π₯2=οΏ½ΜοΏ½π₯3=οΏ½ΜοΏ½
β’ State Equations are given as
β’ And the output equation is
οΏ½ΜοΏ½1=π₯2 οΏ½ΜοΏ½2=π₯3 οΏ½ΜοΏ½3=βπ3π₯1βπ2π₯2βπ1π₯3+π’(π‘ )
Phase Variable Canonical form (Example-1)
π¦ (π‘ )=π2π₯1+π1π₯2+ππ π₯3
ππ
π2
π1
π1π2
π3
β’ State Equations are given as
β’ And the output equation is
οΏ½ΜοΏ½1=π₯2 οΏ½ΜοΏ½2=π₯3 οΏ½ΜοΏ½3=βπ3π₯1βπ2π₯2βπ1π₯3+π’(π‘ )
Phase Variable Canonical form (Example-1)
π¦ (π‘ )=π2π₯1+π1π₯2+ππ π₯3
ππ
π2
π1
βπ1
βπ2
βπ3
β’ State Equations are given as
β’ And the output equation is
β’ In vector matrix form
οΏ½ΜοΏ½1=π₯2 οΏ½ΜοΏ½2=π₯3 οΏ½ΜοΏ½3=βπ3π₯1βπ2π₯2βπ1π₯3+π’(π‘ )
3
2
1
12
3
2
1
1233
2
1
)(
)(100
100010
xxx
bbbty
tuxxx
aaaxxx
o
Phase Variable Canonical form (Example-1)
π¦ (π‘ )=π2π₯1+π1π₯2+ππ π₯3
Companion Forms
β’ Consider a system defined by
β’ where u is the input and y is the output. β’ This equation can also be written as
β’ We will present state-space representations of the system defined by above equations in controllable canonical form and observable canonical form.
ububububyayayay nn
nn
onn
nn
1
1
11
1
1
π (π )π (π )
=πππ π+π1π πβ1+β―+ππβ1π +ππ
π π+π1π πβ1+β―+ππβ1π +ππ
Controllable Canonical Form
β’ The following state-space representation is called a controllable canonical form:
π (π )π (π )
=πππ π+π1π πβ1+β―+ππβ1π +ππ
π π+π1π πβ1+β―+ππβ1π +ππ
u
xx
xx
aaaaxx
xx
n
n
nnnn
n
10
00
1000
01000010
1
2
1
121
1
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1
Controllable Canonical Form
π (π )π (π )
=πππ π+π1π πβ1+β―+ππβ1π +ππ
π π+π1π πβ1+β―+ππβ1π +ππ
ub
xx
xx
babbabbabbaby o
n
n
ooonnonn
1
2
1
112211
Controllable Canonical Form
β« β« β« β«
1
2
n
1xv 2xv )(nvβ¦
)(tu
οΌ οΌ n
1n
1dtd
dtd
dtd
β¦
οΌ
Controllable Canonical Form (Example)π (π )π (π )
= π +3π 2+3π +2
0 1 3 3 2 1212 obbbaa
π (π )π (π )
=0π 2+π +3π 2+3π +2
β’ Let us Rewrite the given transfer function in following form
uxx
aaxx
1010
2
1
122
1
uxx
xx
10
3210
2
1
2
1
Controllable Canonical Form (Example)
0 1 3 3 2 1212 obbbaa
π (π )π (π )
=0π 2+π +3π 2+3π +2
ubxx
babbaby ooo
2
11122
2
113xx
y
Controllable Canonical Form (Example)π (π )π (π )
= π +3π 2+3π +2
β’ By direct decomposition of transfer function
)()(
233
)()(
2
2
2 sPssPs
sss
sUsY
)(2)(3)()(3)(
)()(
21
21
sPssPssPsPssPs
sUsY
β’ Equating Y(s) with numerator on the right hand side and U(s) with denominator on right hand side.
)1.......().........(3)()( 21 sPssPssY
)2.......().........(2)(3)()( 21 sPssPssPsU
Controllable Canonical Form (Example)β’ Rearranging equation-2 yields
)3.......().........(2)(3)()( 21 sPssPssUsP
β’ Draw a simulation diagram using equations (1) and (3)
)(3)()( 21 sPssPssY )(2)(3)()( 21 sPssPssUsP
1/s 1/sU(s) Y(s)
-2
-3
P(s)
2x
12 xx 1x3
1
Controllable Canonical Form (Example)
β’ State equations and output equation are obtained from simulation diagram.
213)( xxsY
122 23)( xxsUx
1/s 1/sU(s) Y(s)
-2
-3
P(s)
2x
12 xx 1x3
1
21 xx
Controllable Canonical Form (Example)
β’ In vector Matrix form
213)( xxsY 122 23)( xxsUx 21 xx
)(10
3210
2
1
2
1 tfxx
xx
2
113xx
y
Observable Canonical Form
β’ The following state-space representation is called an observable canonical form:
π (π )π (π )
=πππ π+π1π πβ1+β―+ππβ1π +ππ
π π+π1π πβ1+β―+ππβ1π +ππ
u
babbab
babbab
xx
xx
aa
aa
xx
xx
o
o
onn
onn
n
n
n
n
n
n
11
22
11
1
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1
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1
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1
100000
001000
Observable Canonical Form
π (π )π (π )
=πππ π+π1π πβ1+β―+ππβ1π +ππ
π π+π1π πβ1+β―+ππβ1π +ππ
ub
xx
xx
y o
n
n
1
2
1
1000
Observable Canonical Form (Example)π (π )π (π )
= π +3π 2+3π +2
0 1 3 3 2 1212 obbbaa
π (π )π (π )
=0π 2+π +3π 2+3π +2
β’ Let us Rewrite the given transfer function in following form
ubabbab
xx
aa
xx
o
o
11
22
2
1
1
2
2
1
10
uxx
xx
13
3120
2
1
2
1
Observable Canonical Form (Example)
0 1 3 3 2 1212 obbbaa
π (π )π (π )
=0π 2+π +3π 2+3π +2
ubxx
y o
2
110
2
110xx
y
Similarity Transformationsβ’ It is desirable to have a means of transforming one state-space
representation into another.
β’ This is achieved using so-called similarity transformations.β’ Consider state space model
β’ Along with this, consider another state space model of the same plant
β’ Here the state vector , say, represents the physical state relative to some other reference, or even a mathematical coordinate vector.
)()()( tButAxtx
)()()( tDutCxty
)()()( tuBtxAtx
)()()( tuDtxCty
Similarity Transformationsβ’ When one set of coordinates are transformed into another
set of coordinates of the same dimension using an algebraic coordinate transformation, such transformation is known as similarity transformation.
β’ In mathematical form the change of variables is written as,
β’ Where T is a nonsingular nxn transformation matrix.
β’ The transformed state is written as
)( )( txTtx
)( )( 1 txTtx
Similarity Transformationsβ’ The transformed state is written as
β’ Taking time derivative of above equation )( )( 1 txTtx
(t) )( 1 xTtx
)()( )( 1 tButAxTtx
)( )( txTtx
)()()( tButAxtx
)()( )( 1 tButxATTtx
)()()( 11 tBuTtxATTtx )()()( tuBtxAtx
ATTA 1 BTB 1
Similarity Transformationsβ’ Consider transformed output equation
β’ Substituting in above equation
β’ Since output of the system remain unchanged [i.e. ] therefore above equation is compared with that yields
)()()( tuDtxCty
)()()( 1 tuDtxTCty
CTC DD
Similarity Transformations
β’ Following relations are used to preform transformation of coordinates algebraically
CTC DD
ATTA 1 BTB 1
Similarity Transformationsβ’ Invariance of Eigen Values
ATTsIAsI 1
ITTATTTsT 111
TAsIT 1
AsI
AsIAsI
Transformation to CCFβ’ Transformation to CCf is done by means of transformation matrix
P.
β’ Where CM is controllability Matrix and is given as
and W is coefficient matrix
Where the aiβs are coefficients of the characteristic polynomial
WCMP
πΆπ=[π΅ π΄π΅ β― π΄πβ1 π΅ ]
0001001
011
1
32
121
a
aaaaa
Wnn
nn
s+
Transformation to CCFβ’ Once the transformation matrix P is computed following
relations are used to calculate transformed matrices.
CPC DD APPA 1 BPB 1
Transformation to CCF (Example)β’ Consider the state space system given below.
β’ Transform the given system in CCF.
[π₯1
π₯2
π₯3]=[1 2 1
0 1 31 1 1] [π₯1
π₯2
π₯3]+[101 ]π’(π‘ )
Transformation to CCF (Example)
β’ The characteristic equation of the system is
[π₯1
π₯2
π₯3]=[1 2 1
0 1 31 1 1] [π₯1
π₯2
π₯3]+[101 ]π’(π‘ )
|π πΌβ π΄|=|π β1 β2 β10 π β1 β3β1 β1 π β1|=π 3β3π 2βπ β3
π1=β3 ,π2=β1 ,π3=β1
001013131
001011
1
12
aaa
W
Transformation to CCF (Example)
β’ Now the controllability matrix CM is calculated as
β’ Transformation matrix P is now obtained as
[π₯1
π₯2
π₯3]=[1 2 1
0 1 31 1 1] [π₯1
π₯2
π₯3]+[101 ]π’(π‘ )
πΆπ=[π΅ π΄π΅ π΄2 π΅ ]
πΆπ=[ 1 2 100 3 91 2 7 ]
π=πΆπΓπ=[1 2 100 3 91 2 7 ] [β1 β3 1
β3 1 01 0 0 ]
π=[3 β1 10 3 00 β1 1]
Transformation to CCF (Example)β’ Using the following relationships given state space
representation is transformed into CCf as
APPA 1 BPB 1
313100010
1APPA
100
1BPB
|π πΌβ π΄|=π 3β3π 2βπ β3
Transformation to OCFβ’ Transformation to CCf is done by means of transformation matrix
Q.
β’ Where OM is observability Matrix and is given as
and W is coefficient matrix
Where the aiβs are coefficients of the characteristic polynomial
1)( OMWQ
ππ=[πΆ πΆπ΄ β― πΆπ΄πβ1 ]π
0001001
011
1
32
121
a
aaaaa
Wnn
nn
s+
Transformation to OCFβ’ Once the transformation matrix Q is computed following
relations are used to calculate transformed matrices.
CQC DD AQQA 1 BQB 1
Transformation to OCF (Example)β’ Consider the state space system given below.
β’ Transform the given system in OCF.
[π₯1
π₯2
π₯3]=[1 2 1
0 1 31 1 1] [π₯1
π₯2
π₯3]+[101 ]π’(π‘ )
π¦ (π‘)= [1 1 0 ] [π₯1
π₯2
π₯3]
Transformation to OCF (Example)
β’ The characteristic equation of the system is
[π₯1
π₯2
π₯3]=[1 2 1
0 1 31 1 1] [π₯1
π₯2
π₯3]+[101 ]π’(π‘ )
|π πΌβ π΄|=|π β1 β2 β10 π β1 β3β1 β1 π β1|=π 3β3π 2βπ β3
π1=β3 ,π2=β1 ,π3=β1
001013131
001011
1
12
aaa
W
Transformation to OCF (Example)
β’ Now the observability matrix OM is calculated as
β’ Transformation matrix Q is now obtained as
[π₯1
π₯2
π₯3]=[1 2 1
0 1 31 1 1] [π₯1
π₯2
π₯3]+[101 ]π’(π‘ )
ππ=[πΆ πΆπ΄ πΆπ΄2 ]π
ππ=[1 1 01 3 45 6 10]
π=(π Γππ )β 1=[ 0 .333 β0.166 0.333β0.333 0.166 0.6660.166 0.166 0.16 6]
π¦ (π‘)= [1 1 0 ] [π₯1
π₯2
π₯3]
Transformation to CCF (Example)β’ Using the following relationships given state space
representation is transformed into CCf as
310101300
1AQQA
123
1BQB
CQC DD AQQA 1 BQB 1
100CQC
Home Work
β’ Obtain state space representation of following transfer function in Phase variable canonical form, OCF and CCF by β Direct Decomposition of Transfer Functionβ Similarity Transformationβ Direct Approach
π (π )π (π )
=π 2+2 π +3
π 3+5π 2+3 π +2
END OF LECTURES-26-27-28-29
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