Feedback Control Systems (FCS)
Dr. Imtiaz Hussainemail: [email protected]
URL :http://imtiazhussainkalwar.weebly.com/
Lecture-26-27-28-29State Space Canonical forms
Lecture Outline– Canonical forms of State Space Models
• Phase Variable Canonical Form
• Controllable Canonical form
• Observable Canonical form
– Similarity Transformations
• Transformation of coordinates
– Transformation to CCF
– Transformation OCF
Canonical Forms• Canonical forms are the standard forms of state space models.
• Each of these canonical form has specific advantages which makes it convenient for use in particular design technique.
• There are four canonical forms of state space models– Phase variable canonical form– Controllable Canonical form– Observable Canonical form– Diagonal Canonical form– Jordan Canonical Form
• 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
• The method of phase variables possess mathematical advantage over other representations.
• This type of representation can be obtained directly from differential equations.
• Decomposition of transfer function also yields Phase variable form.
Phase Variable Canonical form• Consider an nth order linear plant model described by the
differential equation
• Where y(t) is the plant output and u(t) is the plant input.
• A state model for this system is not unique but depends on the choice of a set of state variables.
• A useful set of state variables, referred to as phase variables, is defined as:
𝑑𝑛 𝑦𝑑𝑡𝑛
+𝑎1𝑑𝑛− 1𝑦𝑑𝑡𝑛− 1 +⋯+𝑎𝑛−1
𝑑𝑦𝑑𝑡 +𝑎𝑛 𝑦=𝑢(𝑡)
𝑥1=𝑦 , 𝑥2=�̇� , 𝑥3=𝑦 ,⋯ , 𝑥𝑛=𝑑𝑛−1 𝑦𝑑𝑡𝑛−1
Phase Variable Canonical form
• Taking derivatives of the first n-1 state variables, we have
𝑥1=𝑦 , 𝑥2= �̇� , 𝑥3=𝑦 ,⋯ , 𝑥𝑛=𝑑𝑛−1 𝑦𝑑𝑡𝑛−1
�̇�1=𝑥2 , �̇�2=𝑥3 , �̇�3=𝑥4⋯ , �̇�𝑛−1=𝑥𝑛
�̇�𝑛=−𝑎𝑛 𝑥1−𝑎𝑛−1𝑥2−⋯−𝑎1𝑥𝑛+𝑢(𝑡)
u
xx
xx
aaaaxx
xx
n
n
nnnn
n
10
00
1000
01000010
1
2
1
131
1
2
1
Phase Variable Canonical form
• Output equation is simply
𝑥1=𝑦 , 𝑥2= �̇� , 𝑥3=𝑦 ,⋯ , 𝑥𝑛=𝑑𝑛−1 𝑦𝑑𝑡𝑛−1
n
n
xx
xx
y
1
2
1
0001
8
∫ ∫ ∫ ∫
1a
2a
na
1xy 2xy
nn xy )1(
)(ny
1)2(
nn xy
…
)(tu
+ +
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
• Obtain the state equation in phase variable form for the following differential equation, where u(t) is input and y(t) is output.
• The differential equation is third order, thus there are three state variables:
• And their derivatives are (i.e state equations)
2 𝑑3 𝑦𝑑𝑡3 +4 𝑑2 𝑦
𝑑𝑡2 +6 𝑑𝑦𝑑𝑡 +8 𝑦=10𝑢 (𝑡)
𝑥1=𝑦 𝑥2=�̇� 𝑥3= �̈�
�̇�1=𝑥2
�̇�2=𝑥3
�̇�3=−4 𝑥1−3𝑥2−2𝑥3+5𝑢 (𝑡)
Phase Variable Canonical form (Example-1)
Phase Variable Canonical form (Example-1)
• In vector matrix form
𝑥1=𝑦 𝑥2=�̇� 𝑥3= �̈��̇�1=𝑥2
�̇�2=𝑥3
�̇�3=−4 𝑥1−3𝑥2−2𝑥3+5𝑢 (𝑡)
3
2
1
3
2
1
3
2
1
001)(
)(500
234100010
xxx
ty
tuxxx
xxx
Home Work: Draw Sate diagram
• Consider the transfer function of a third-order system where the numerator degree is lower than that of the denominator.
• Transfer function can be decomposed into cascade form
• Denoting the output of the first block as W(s), we have the following input/output relationships:
Phase Variable Canonical form (Example-2)
𝑌 (𝑠)𝑈 (𝑠 )
=𝑏𝑜𝑠2+𝑏1𝑠+𝑏2
𝑠3+𝑎1𝑠2+𝑎2𝑠+𝑎3
1𝑠3+𝑎1𝑠2+𝑎2𝑠+𝑎3
𝑏𝑜𝑠2+𝑏1𝑠+𝑏2𝑈 (𝑠) 𝑌 (𝑠)𝑊 (𝑠)
𝑊 (𝑠)𝑈 (𝑠)
= 1𝑠3+𝑎1𝑠
2+𝑎2𝑠+𝑎3
𝑌 (𝑠)𝑊 (𝑠)
=𝑏𝑜𝑠2+𝑏1𝑠+𝑏2
• Re-arranging above equation yields
• Taking inverse Laplace transform of above equations.
• Choosing the state variables in phase variable form
Phase Variable Canonical form (Example-2)
𝑊 (𝑠)𝑈 (𝑠)
= 1𝑠3+𝑎1𝑠
2+𝑎2𝑠+𝑎3
𝑌 (𝑠)𝑊 (𝑠)
=𝑏𝑜𝑠2+𝑏1𝑠+𝑏2
+
𝑌 (𝑠)=𝑏𝑜𝑠2𝑊 (𝑠 )+𝑏1𝑠𝑊 (𝑠)+𝑏2𝑊 (𝑠)
+
𝑦 (𝑡)=𝑏𝑜�̈� (𝑡 )+𝑏1 �̇� (𝑡 )+𝑏2𝑤(𝑡 )
𝑥1=𝑤𝑥2=�̇�𝑥3=�̈�
• State Equations are given as
• And the output equation is
�̇�1=𝑥2 �̇�2=𝑥3 �̇�3=−𝑎3𝑥1−𝑎2𝑥2−𝑎1𝑥3+𝑢(𝑡 )
Phase Variable Canonical form (Example-1)
𝑦 (𝑡 )=𝑏2𝑥1+𝑏1𝑥2+𝑏𝑜 𝑥3
𝑏𝑜
𝑏2
𝑏1
𝑎1𝑎2
𝑎3
• State Equations are given as
• And the output equation is
�̇�1=𝑥2 �̇�2=𝑥3 �̇�3=−𝑎3𝑥1−𝑎2𝑥2−𝑎1𝑥3+𝑢(𝑡 )
Phase Variable Canonical form (Example-1)
𝑦 (𝑡 )=𝑏2𝑥1+𝑏1𝑥2+𝑏𝑜 𝑥3
𝑏𝑜
𝑏2
𝑏1
−𝑎1
−𝑎2
−𝑎3
• State Equations are given as
• And the output equation is
• In vector matrix form
�̇�1=𝑥2 �̇�2=𝑥3 �̇�3=−𝑎3𝑥1−𝑎2𝑥2−𝑎1𝑥3+𝑢(𝑡 )
3
2
1
12
3
2
1
1233
2
1
)(
)(100
100010
xxx
bbbty
tuxxx
aaaxxx
o
Phase Variable Canonical form (Example-1)
𝑦 (𝑡 )=𝑏2𝑥1+𝑏1𝑥2+𝑏𝑜 𝑥3
Companion Forms
• Consider a system defined by
• where u is the input and y is the output. • This equation can also be written as
• 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
𝑌 (𝑠)𝑈 (𝑠 )
=𝑏𝑜𝑠𝑛+𝑏1𝑠𝑛−1+⋯+𝑏𝑛−1𝑠+𝑏𝑛
𝑠𝑛+𝑎1𝑠𝑛−1+⋯+𝑎𝑛−1𝑠+𝑎𝑛
Controllable Canonical Form
• The following state-space representation is called a controllable canonical form:
𝑌 (𝑠)𝑈 (𝑠 )
=𝑏𝑜𝑠𝑛+𝑏1𝑠𝑛−1+⋯+𝑏𝑛−1𝑠+𝑏𝑛
𝑠𝑛+𝑎1𝑠𝑛−1+⋯+𝑎𝑛−1𝑠+𝑎𝑛
u
xx
xx
aaaaxx
xx
n
n
nnnn
n
10
00
1000
01000010
1
2
1
121
1
2
1
Controllable Canonical Form
𝑌 (𝑠)𝑈 (𝑠 )
=𝑏𝑜𝑠𝑛+𝑏1𝑠𝑛−1+⋯+𝑏𝑛−1𝑠+𝑏𝑛
𝑠𝑛+𝑎1𝑠𝑛−1+⋯+𝑎𝑛−1𝑠+𝑎𝑛
ub
xx
xx
babbabbabbaby o
n
n
ooonnonn
1
2
1
112211
Controllable Canonical Form
∫ ∫ ∫ ∫
1
2
n
1xv 2xv )(nv…
)(tu
+ + n
1n
1dtd
dtd
dtd
…
+
Controllable Canonical Form (Example)𝑌 (𝑠)𝑈 (𝑠 )
= 𝑠+3𝑠2+3𝑠+2
0 1 3 3 2 1212 obbbaa
𝑌 (𝑠)𝑈 (𝑠 )
=0𝑠2+𝑠+3𝑠2+3𝑠+2
• 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
𝑌 (𝑠)𝑈 (𝑠 )
=0𝑠2+𝑠+3𝑠2+3𝑠+2
ubxx
babbaby ooo
2
11122
2
113xx
y
Controllable Canonical Form (Example)𝑌 (𝑠)𝑈 (𝑠 )
= 𝑠+3𝑠2+3𝑠+2
• By direct decomposition of transfer function
)()(
233
)()(
2
2
2 sPssPs
sss
sUsY
)(2)(3)()(3)(
)()(
21
21
sPssPssPsPssPs
sUsY
• 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)• Rearranging equation-2 yields
)3.......().........(2)(3)()( 21 sPssPssUsP
• 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)
• 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)
• 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
• The following state-space representation is called an observable canonical form:
𝑌 (𝑠)𝑈 (𝑠 )
=𝑏𝑜𝑠𝑛+𝑏1𝑠𝑛−1+⋯+𝑏𝑛−1𝑠+𝑏𝑛
𝑠𝑛+𝑎1𝑠𝑛−1+⋯+𝑎𝑛−1𝑠+𝑎𝑛
u
babbab
babbab
xx
xx
aa
aa
xx
xx
o
o
onn
onn
n
n
n
n
n
n
11
22
11
1
2
1
1
2
1
1
2
1
100000
001000
Observable Canonical Form
𝑌 (𝑠)𝑈 (𝑠 )
=𝑏𝑜𝑠𝑛+𝑏1𝑠𝑛−1+⋯+𝑏𝑛−1𝑠+𝑏𝑛
𝑠𝑛+𝑎1𝑠𝑛−1+⋯+𝑎𝑛−1𝑠+𝑎𝑛
ub
xx
xx
y o
n
n
1
2
1
1000
Observable Canonical Form (Example)𝑌 (𝑠)𝑈 (𝑠 )
= 𝑠+3𝑠2+3𝑠+2
0 1 3 3 2 1212 obbbaa
𝑌 (𝑠)𝑈 (𝑠 )
=0𝑠2+𝑠+3𝑠2+3𝑠+2
• 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
𝑌 (𝑠)𝑈 (𝑠 )
=0𝑠2+𝑠+3𝑠2+3𝑠+2
ubxx
y o
2
110
2
110xx
y
Similarity Transformations• It is desirable to have a means of transforming one state-space
representation into another.
• This is achieved using so-called similarity transformations.• Consider state space model
• Along with this, consider another state space model of the same plant
• 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 Transformations• 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.
• In mathematical form the change of variables is written as,
• Where T is a nonsingular nxn transformation matrix.
• The transformed state is written as
)( )( txTtx
)( )( 1 txTtx
Similarity Transformations• The transformed state is written as
• 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 Transformations• Consider transformed output equation
• Substituting in above equation
• Since output of the system remain unchanged [i.e. ] therefore above equation is compared with that yields
)()()( tuDtxCty
)()()( 1 tuDtxTCty
CTC DD
Similarity Transformations
• Following relations are used to preform transformation of coordinates algebraically
CTC DD
ATTA 1 BTB 1
Similarity Transformations• Invariance of Eigen Values
ATTsIAsI 1
ITTATTTsT 111
TAsIT 1
AsI
AsIAsI
Transformation to CCF• Transformation to CCf is done by means of transformation matrix
P.
• Where CM is controllability Matrix and is given as
and W is coefficient matrix
Where the ai’s are coefficients of the characteristic polynomial
WCMP
𝐶𝑀=[𝐵 𝐴𝐵 ⋯ 𝐴𝑛−1 𝐵 ]
0001001
011
1
32
121
a
aaaaa
Wnn
nn
s+
Transformation to CCF• 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)• Consider the state space system given below.
• Transform the given system in CCF.
[𝑥1
𝑥2
𝑥3]=[1 2 1
0 1 31 1 1] [𝑥1
𝑥2
𝑥3]+[101 ]𝑢(𝑡 )
Transformation to CCF (Example)
• The characteristic equation of the system is
[𝑥1
𝑥2
𝑥3]=[1 2 1
0 1 31 1 1] [𝑥1
𝑥2
𝑥3]+[101 ]𝑢(𝑡 )
|𝑠𝐼− 𝐴|=|𝑠−1 −2 −10 𝑠−1 −3−1 −1 𝑠−1|=𝑠3−3𝑠2−𝑠−3
𝑎1=−3 ,𝑎2=−1 ,𝑎3=−1
001013131
001011
1
12
aaa
W
Transformation to CCF (Example)
• Now the controllability matrix CM is calculated as
• Transformation matrix P is now obtained as
[𝑥1
𝑥2
𝑥3]=[1 2 1
0 1 31 1 1] [𝑥1
𝑥2
𝑥3]+[101 ]𝑢(𝑡 )
𝐶𝑀=[𝐵 𝐴𝐵 𝐴2 𝐵 ]
𝐶𝑀=[ 1 2 100 3 91 2 7 ]
𝑃=𝐶𝑀×𝑊=[1 2 100 3 91 2 7 ] [−1 −3 1
−3 1 01 0 0 ]
𝑃=[3 −1 10 3 00 −1 1]
Transformation to CCF (Example)• Using the following relationships given state space
representation is transformed into CCf as
APPA 1 BPB 1
313100010
1APPA
100
1BPB
|𝑠𝐼− 𝐴|=𝑠3−3𝑠2−𝑠−3
Transformation to OCF• Transformation to CCf is done by means of transformation matrix
Q.
• Where OM is observability Matrix and is given as
and W is coefficient matrix
Where the ai’s are coefficients of the characteristic polynomial
1)( OMWQ
𝑂𝑀=[𝐶 𝐶𝐴 ⋯ 𝐶𝐴𝑛−1 ]𝑇
0001001
011
1
32
121
a
aaaaa
Wnn
nn
s+
Transformation to OCF• 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)• Consider the state space system given below.
• Transform the given system in OCF.
[𝑥1
𝑥2
𝑥3]=[1 2 1
0 1 31 1 1] [𝑥1
𝑥2
𝑥3]+[101 ]𝑢(𝑡 )
𝑦 (𝑡)= [1 1 0 ] [𝑥1
𝑥2
𝑥3]
Transformation to OCF (Example)
• The characteristic equation of the system is
[𝑥1
𝑥2
𝑥3]=[1 2 1
0 1 31 1 1] [𝑥1
𝑥2
𝑥3]+[101 ]𝑢(𝑡 )
|𝑠𝐼− 𝐴|=|𝑠−1 −2 −10 𝑠−1 −3−1 −1 𝑠−1|=𝑠3−3𝑠2−𝑠−3
𝑎1=−3 ,𝑎2=−1 ,𝑎3=−1
001013131
001011
1
12
aaa
W
Transformation to OCF (Example)
• Now the observability matrix OM is calculated as
• Transformation matrix Q is now obtained as
[𝑥1
𝑥2
𝑥3]=[1 2 1
0 1 31 1 1] [𝑥1
𝑥2
𝑥3]+[101 ]𝑢(𝑡 )
𝑂𝑀=[𝐶 𝐶𝐴 𝐶𝐴2 ]𝑇
𝑂𝑀=[1 1 01 3 45 6 10]
𝑄=(𝑊 ×𝑂𝑀 )− 1=[ 0 .333 −0.166 0.333−0.333 0.166 0.6660.166 0.166 0.16 6]
𝑦 (𝑡)= [1 1 0 ] [𝑥1
𝑥2
𝑥3]
Transformation to CCF (Example)• 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
• Obtain state space representation of following transfer function in Phase variable canonical form, OCF and CCF by – Direct Decomposition of Transfer Function– Similarity Transformation– Direct Approach
𝑌 (𝑠)𝑈 (𝑠 )
=𝑠2+2 𝑠+3
𝑠3+5𝑠2+3 𝑠+2
END OF LECTURES-26-27-28-29
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