Feedback Control Systems (FCS)

Dr. Imtiaz Hussainemail: imtiaz.hussain@faculty.muet.edu.pk

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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