Lecture-26-27-28-29
State Space Canonical forms
Dr. Imtiaz Hussain email: imtiaz.hussain@faculty.muet.edu.pk
URL : http://imtiazhussainkalwar.weebly.com/

– 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 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
Companion forms
Modal forms
• It is interesting to note that the dynamics properties of system remain unchanged whichever the type of representation is used.

• 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 n th order linear plant model described by the differential equation 𝑑 𝑛 𝑦 𝑑𝑡 𝑛
+ 𝑎
1 𝑑 𝑛−1 𝑦 𝑑𝑡 𝑛−1
+ ⋯ + 𝑎 𝑛−1 𝑑𝑦 𝑑𝑡
+ 𝑎 𝑛 𝑦 = 𝑢(𝑡)
• 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: 𝑥
1
= 𝑦, 𝑥
2
= 𝑦, 𝑥
3
= ⋯ , 𝑥 𝑛
= 𝑑 𝑛−1 𝑦 𝑑𝑡 𝑛−1

Phase Variable Canonical form 𝑥
1
= 𝑦, 𝑥
2
= 𝑦, 𝑥
3
= ⋯ , 𝑥 𝑛
= 𝑑 𝑛−1 𝑦 𝑑𝑡 𝑛−1
• Taking derivatives of the first n-1 state variables, we have
𝑥
1
= 𝑥
2
, 𝑥
2
= 𝑥
3
, 𝑥
3
= 𝑥
4
⋯ , 𝑥 𝑛−1
= 𝑥 𝑛
𝑥 𝑛
= −𝑎 𝑛 𝑥
1
− 𝑎 𝑛−1 𝑥
2
− ⋯ − 𝑎
1 𝑥 𝑛
+ 𝑢(𝑡)






 x x 

1
2 x  n

1 x 
 n







 








0
0

0 a n
1
0

0
 a n

1
0
1

0
 a n

3




 
0
0

1 a
1













 x x n

1 x x

1
2 n









0




0



0
1






 u

Phase Variable Canonical form 𝑥
1
= 𝑦, 𝑥
2
= 𝑦, 𝑥
3
= ⋯ , 𝑥 𝑛
= 𝑑 𝑛−1 𝑦 𝑑𝑡 𝑛−1
• Output equation is simply y


1 0  0 0







 x
1 x
2
 x n

1 x n








u ( t )
＋ y
( n

1 )  x n y
( n )
∫
＋
 a
1
∫
 a
2
∫ y ( n

2 )
 x n

1

 a n
… y
  x
2
∫ y
 x
1
8

u 1 x  n
1  n

1 s
1 s x  n

2
1 s
 
1
 
2
 
3

  n

1


1
 x
2
1 s
  n x
1
1 y
9

Phase Variable Canonical form (Example-1)
• Obtain the state equation in phase variable form for the following differential equation, where u(t) is input and y(t) is output.
2 𝑑 3 𝑦 𝑑𝑡 3
+ 4 𝑑 2 𝑦 𝑑𝑡 2
+ 6 𝑑𝑦 𝑑𝑡
+ 8𝑦 = 10𝑢(𝑡)
• The differential equation is third order, thus there are three state variables: 𝑥
1
= 𝑦 𝑥
2
= 𝑦 𝑥
3
= 𝑦
• And their derivatives are (i.e state equations)
𝑥
1
= 𝑥
2
𝑥
2
= 𝑥
3
𝑥
3
= −4𝑥
1
− 3𝑥
2
− 2𝑥
3
+ 5𝑢(𝑡)

Phase Variable Canonical form (Example-1)
𝑥
1
= 𝑥
2
𝑥
2
= 𝑥
3
𝑥
3
= −4𝑥
1
− 3𝑥
2
− 2𝑥
3
+ 5𝑢(𝑡)
• In vector matrix form 𝑥
1
= 𝑦 𝑥
2
= 𝑦 𝑥
3
= 𝑦



 x x
2 x
3
1










0
0
4 y ( t )


1 0
1
0

3
0




 x
1 x
2 x
3





0
1
2







 x x
2 x
1
3







0


0
5



 u ( t )
Home Work: Draw Sate diagram

Phase Variable Canonical form (Example-2)
• Consider the transfer function of a third-order system where the numerator degree is lower than that of the denominator.
𝑌(𝑠)
𝑈(𝑠)
= 𝑏 𝑜 𝑠 2 + 𝑏
1 𝑠 + 𝑏
2 𝑠 3 + 𝑎
1 𝑠 2 + 𝑎
2 𝑠 + 𝑎
3
• Transfer function can be decomposed into cascade form
𝑊(𝑠)
𝑈(𝑠) 1 𝑠 3 + 𝑎
1 𝑠 2 + 𝑎
2 𝑠 + 𝑎
3 𝑏 𝑜 𝑠 2 + 𝑏
1 𝑠 + 𝑏
2
𝑌(𝑠)
• Denoting the output of the first block as W(s) , we have the following input/output relationships:
𝑊(𝑠)
=
𝑈(𝑠)
1 𝑠 3 + 𝑎
1 𝑠 2 + 𝑎
2 𝑠 + 𝑎
3
𝑌(𝑠)
𝑊(𝑠)
= 𝑏 𝑜 𝑠 2 + 𝑏
1 𝑠 + 𝑏
2

Phase Variable Canonical form (Example-2)
𝑊(𝑠)
=
𝑈(𝑠)
1 𝑠 3 + 𝑎
1 𝑠 2 + 𝑎
2 𝑠 + 𝑎
3
𝑌(𝑠)
𝑊(𝑠)
= 𝑏 𝑜 𝑠 2 + 𝑏
1 𝑠 + 𝑏
2
• Re-arranging above equation yields 𝑠 3 𝑊 𝑠 = −𝑎
1 𝑠 2 𝑊 𝑠 − 𝑎
2 𝑠𝑊 𝑠 − 𝑎
3
𝑊(𝑠) + 𝑈(𝑠)
𝑌(𝑠) = 𝑏 𝑜 𝑠 2 𝑊(𝑠) + 𝑏
1 𝑠𝑊(𝑠) + 𝑏
2
𝑊(𝑠)
• Taking inverse Laplace transform of above equations.
𝑤 𝑡 = −𝑎
1 𝑤 𝑡 − 𝑎
2 𝑤 𝑡 − 𝑎
3 𝑤(𝑡) + u(𝑡) 𝑦(𝑡) = 𝑏 𝑜 𝑤 𝑡 + 𝑏
1 𝑤 𝑡 + 𝑏
2 𝑤(𝑡)
• Choosing the state variables in phase variable form 𝑥
1
= 𝑤 𝑥
2
= 𝑥
3

Phase Variable Canonical form (Example-1)
• State Equations are given as
𝑥
1
= 𝑥
2
𝑥
2
= 𝑥
3
𝑥
3
= −𝑎
3 𝑥
1
− 𝑎
2 𝑥
2
− 𝑎
1 𝑥
3
+ 𝑢(𝑡)
• And the output equation is 𝑦(𝑡) = 𝑏
2 𝑥
1
+ 𝑏
1 𝑥
2
+ 𝑏 𝑜 𝑥
3 𝑏 𝑜 𝑏
1 𝑏
2 𝑎
1 𝑎
2 𝑎
3

Phase Variable Canonical form (Example-1)
• State Equations are given as
𝑥
1
= 𝑥
2
𝑥
2
= 𝑥
3
𝑥
3
= −𝑎
3 𝑥
1
− 𝑎
2 𝑥
2
− 𝑎
1 𝑥
3
+ 𝑢(𝑡)
• And the output equation is 𝑦(𝑡) = 𝑏
2 𝑥
1
+ 𝑏
1 𝑥
2
+ 𝑏 𝑜 𝑥
3
−𝑎
1
−𝑎
2
−𝑎
3 𝑏 𝑜 𝑏
1 𝑏
2

Phase Variable Canonical form (Example-1)
• State Equations are given as
𝑥
1
= 𝑥
2
𝑥
2
= 𝑥
3
𝑥
3
= −𝑎
3 𝑥
1
− 𝑎
2 𝑥
2
− 𝑎
1 𝑥
3
+ 𝑢(𝑡)
• And the output equation is 𝑦(𝑡) = 𝑏
2 𝑥
1
+ 𝑏
1 𝑥
2
+ 𝑏 𝑜 𝑥
3
• In vector matrix form



 x x
2 x
1
3










0
0 a
3 y ( t )

 b
2 b
1
1
0
 a
2 b o




 x
1 x
2 x
3





0
1 a
1







 x x
2 x
1
3








 1
0
0



 u ( t )

Companion Forms
• Consider a system defined by n y
 a
1 n y

1
   a n

1 y   a n y
 b o n u
 b
1 n u

1
   b n

1 u   b n u
• where u is the input and y is the output.
• This equation can also be written as
𝑌(𝑠)
𝑈(𝑠)
= 𝑏 𝑜 𝑠 𝑛 + 𝑏
1 𝑠 𝑛−1 + ⋯ + 𝑏 𝑛−1 𝑠 + 𝑏 𝑛 𝑠 𝑛 + 𝑎
1 𝑠 𝑛−1 + ⋯ + 𝑎 𝑛−1 𝑠 + 𝑎 𝑛
• We will present state-space representations of the system defined by above equations in controllable canonical form and observable canonical form.

Controllable Canonical Form
𝑌(𝑠)
𝑈(𝑠)
= 𝑏 𝑜 𝑠 𝑛 + 𝑏
1 𝑠 𝑛−1 + ⋯ + 𝑏 𝑛−1 𝑠 + 𝑏 𝑛 𝑠 𝑛 + 𝑎
1 𝑠 𝑛−1 + ⋯ + 𝑎 𝑛−1 𝑠 + 𝑎 𝑛
• The following state-space representation is called a controllable canonical form:






 x 
2 x  n

1 x  x 

1 n
















0
0

0 a n
1
0

0
 a n

1
0
1

0
 a n

2




 
0
0

1 a
1













 x x
1
2 x n

1 x
 n







 


 1

0


 0



0



 u

Controllable Canonical Form
𝑌(𝑠)
𝑈(𝑠)
= 𝑏 𝑜 𝑠 𝑛 + 𝑏
1 𝑠 𝑛−1 + ⋯ + 𝑏 𝑛−1 𝑠 + 𝑏 𝑛 𝑠 𝑛 + 𝑎
1 𝑠 𝑛−1 + ⋯ + 𝑎 𝑛−1 𝑠 + 𝑎 𝑛 y

 b n
 a n b o b n

1
 a n

1 b o
 b
2
 a
2 b o b
1
 a
1 b o







 x
1 x
2
 x n

1 x n







 b o u

Controllable Canonical Form u ( t )
＋ v
( n )
∫
＋
 
1
∫ ∫
 
2

  n
… v
  x
2
∫ v
 x
1 d dt
 d dt d dt
…
 n
 n

1

1
＋

Controllable Canonical Form (Example)
𝑌(𝑠)
𝑈(𝑠)
= 𝑠 + 3 𝑠 2 + 3𝑠 + 2
• Let us Rewrite the given transfer function in following form
𝑌(𝑠)
𝑈(𝑠)
=
0𝑠 2 + 𝑠 + 3 𝑠 2 + 3𝑠 + 2 a
2

2 a
1

3 b
2

3 b
1

1 b o

0

 x  x 
2
1






0 a
2

1 a
1



 x x
2
1





0
1

 u

 x  x 
2
1






0
2

1
3



 x x
2
1





1
0

 u

Controllable Canonical Form (Example)
𝑌(𝑠)
𝑈(𝑠)
=
0𝑠 2 + 𝑠 + 3 𝑠 2 + 3𝑠 + 2 a
2

2 a
1

3 b
2

3 b
1

1 b o

0 y

 b
2
 a
2 b o b
1
 a
1 b o


 x x
1
2


 b o u y


 x x
2
1



Controllable Canonical Form (Example)
𝑌(𝑠)
𝑈(𝑠)
= 𝑠 + 3 𝑠 2 + 3𝑠 + 2
• By direct decomposition of transfer function
Y
U
( s
( s
)
)
 s
2 s


3 s
3

2
 s

2
P ( s ) s

2
P ( s )
Y ( s )
U ( s )

P ( s ) s

1

P ( s )

3
3 s

1
P ( s ) s

2

P ( s )
2 s

2
P ( s )
• Equating Y(s) with numerator on the right hand side and U(s) with denominator on right hand side.
Y ( s )
 s

1
P ( s )

3 s

2
P ( s ).........
.......( 1 )
U ( s )

P ( s )

3 s

1
P ( s )

2 s

2
P ( s ).........
.......( 2 )

Controllable Canonical Form (Example)
• Rearranging equation-2 yields
P ( s )

U ( s )

3 s

1
P ( s )

2 s

2
P ( s ).........
.......( 3 )
• Draw a simulation diagram using equations (1) and (3)
P ( s )

U ( s )

3 s

1
P ( s )

2 s

2
P ( s ) Y ( s )
 s

1
P ( s )

3 s

2
P ( s )
U(s)
-2
P(s)

2
-3
1/s x
2
 x 
1
1/s
1 x
1
3
Y(s)

Controllable Canonical Form (Example)
-2
U(s)
P(s)

2
-3
1/s x
2
  x
1
1/s x
1
3
Y(s)
1
• State equations and output equation are obtained from simulation diagram.
x 
1
 x
2

2

U ( s )

3 x
2

2 x
1
Y ( s )

3 x
1
 x
2

Controllable Canonical Form (Example) x 
1
 x
2

2

U ( s )

3 x
2

2 x
1
Y ( s )

3 x
1
 x
2
• In vector Matrix form

 x  x 
2
1






0
2

1
3



 x x
2
1





0
1

 f ( t ) y


3 1


 x
1 x
2



Observable Canonical Form
𝑌(𝑠)
𝑈(𝑠)
= 𝑏 𝑜 𝑠 𝑛 + 𝑏
1 𝑠 𝑛−1 + ⋯ + 𝑏 𝑛−1 𝑠 + 𝑏 𝑛 𝑠 𝑛 + 𝑎
1 𝑠 𝑛−1 + ⋯ + 𝑎 𝑛−1 𝑠 + 𝑎 𝑛
• The following state-space representation is called an observable canonical form:






 x x  n

1 x 
 x 

1
2 n









0




1



0
0
0
0

0
0





0
0

0
1



 n a n

1
 a a a
2
1













 x x
1
2 x n

1 x
 n














 b n b n

1 b
2 b
1




 a n b o a n

1 b o a a
2
1 b b o o






 u

Observable Canonical Form
𝑌(𝑠)
𝑈(𝑠)
= 𝑏 𝑜 𝑠 𝑛 + 𝑏
1 𝑠 𝑛−1 + ⋯ + 𝑏 𝑛−1 𝑠 + 𝑏 𝑛 𝑠 𝑛 + 𝑎
1 𝑠 𝑛−1 + ⋯ + 𝑎 𝑛−1 𝑠 + 𝑎 𝑛 y


0 0  0 1







 x
1 x
2
 x n

1 x n







 b o u

Observable Canonical Form (Example)
𝑌(𝑠)
𝑈(𝑠)
= 𝑠 + 3 𝑠 2 + 3𝑠 + 2
• Let us Rewrite the given transfer function in following form
𝑌(𝑠)
𝑈(𝑠)
=
0𝑠 2 + 𝑠 + 3 𝑠 2 + 3𝑠 + 2 a
2

2 a
1

3 b
2

3 b
1

1 b o

0

 x  x 
2
1





0
1

 x  x 
2
1





0
1

 a
2 a
1



 x x
2
1




 b b
2
1

 a
2 b o a
1 b o

 u


2
3



 x x
2
1





3

1
 u

Observable Canonical Form (Example)
𝑌(𝑠)
𝑈(𝑠)
=
0𝑠 2 + 𝑠 + 3 𝑠 2 + 3𝑠 + 2 a
2

2 a
1

3 b
2

3 b
1

1 b o

0 y


0 1


 x
1 x
2


 b o u y


0 1


 x
1 x
2



• 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 x  ( t )

Ax ( t )

Bu ( t ) y ( t )

Cx ( t )

Du ( t )
• Along with this, consider another state space model of the same plant x

( t )

A x ( t )

B u ( t ) y ( t )

C x ( t )

D u ( t )
• Here the state vector 𝑥 , say, represents the physical state relative to some other reference, or even a mathematical coordinate vector.

• 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, x ( t )

T x ( t )
• Where T is a nonsingular nxn transformation matrix.
• The transformed state 𝑥(𝑡) is written as x ( t )

T

1 x ( t )

• The transformed state 𝑥(𝑡) is written as x ( t )

T

1 x ( t )
• Taking time derivative of above equation x

( t )

T

1 x  (t) x

( t )

T

1

Ax ( t )

Bu ( t )

( t )

T

1

AT x ( t )

Bu ( t )
 x  ( t )

Ax ( t )

Bu ( t ) x ( t )

T x ( t )
( t )

T

1
AT x ( t )

T

1
Bu ( t )
( t )

A x ( t )

B u ( t )
A

T

1
AT B

T

1
B

• Consider transformed output equation y ( t )

C x ( t )

D u ( t )
• Substituting 𝑥 𝑡 = 𝑇 −1 𝑥(𝑡) in above equation y ( t )

C T

1 x ( t )

D u ( t )
• Since output of the system remain unchanged [i.e.
𝑦 𝑡 = 𝑦(𝑡) ] therefore above equation is compared with 𝑦 𝑡 =
𝐶𝑥 𝑡 + 𝐷𝑢(𝑡) that yields
C

CT D

D

• Following relations are used to preform transformation of coordinates algebraically
A

T

1
AT B

T

1
B
C

CT D

D

• Invariance of Eigen Values sI

A
 sI

T

1
AT
 sT

1
T

T

1
AT

T

1
T

I

T

1 sI

A T
 sI

A sI

A
 sI

A

Transformation to CCF
• Transformation to CCf is done by means of transformation matrix
P .
P

CM

W
• Where CM is controllability Matrix and is given as
𝐶𝑀 = 𝐵 𝐴𝐵 ⋯ 𝐴 𝑛−1 𝐵 and W is coefficient matrix
W







 a a n

1 n

2
 a
1
1 a n

2 a n

3

1
0




 a
1
1

0
0
1
0

0
0







Where the a i
’s are coefficients of the characteristic polynomial 𝑠𝐼 − 𝐴 = 𝑠 𝑛
+ 𝑎
1 𝑠 𝑛−1
+ 𝑎
2 𝑠 𝑛−2
+ ⋯ + 𝑎 𝑛−1 s+ 𝑎 𝑛

Transformation to CCF
• Once the transformation matrix P is computed following relations are used to calculate transformed matrices.
A

P

1
AP B

P

1
B C

CP D

D

Transformation to CCF ( Example )
• Consider the state space system given below.
𝑥
1 𝑥
2 𝑥
3
=
1 2 1
0 1 3
1 1 1 𝑥
1 𝑥
2 𝑥
3
+
1
0
1 𝑢(𝑡)
• Transform the given system in CCF.

Transformation to CCF ( Example ) 𝑥
1 𝑥
2 𝑥
3
=
1 2 1
0 1 3
1 1 1 𝑥
1 𝑥
2 𝑥
3
+
1
0
1
• The characteristic equation of the system is 𝑢(𝑡) 𝑠𝐼 − 𝐴 = 𝑠 − 1 −2 −1
0 𝑠 − 1 −3
−1 −1 𝑠 − 1
= 𝑠 3 − 3𝑠 2 − 𝑠 − 3 𝑎
1
= −3,
W




 a
2 a
1
1 a
1
1
0 𝑎
2
= −1,
1
0
0











1
1
3 𝑎
3
= −1

3
1
0
1
0
0





Transformation to CCF ( Example ) 𝑥
1 𝑥
2 𝑥
3
=
1 2 1
0 1 3
1 1 1 𝑥
1 𝑥
2 𝑥
3
+
1
0
1 𝑢(𝑡)
• Now the controllability matrix CM is calculated as
𝐶𝑀 = 𝐵 𝐴𝐵 𝐴 2 𝐵
𝐶𝑀 =
1 2 10
0 3 9
1 2 7
• Transformation matrix P is now obtained as
𝑃 = 𝐶𝑀 × 𝑊 =
1 2 10
0 3 9
1 2 7
−1 −3 1
−3 1 0
1 0 0
𝑃 =
3 −1 1
0 3 0
0 −1 1

Transformation to CCF ( Example )
• Using the following relationships given state space representation is transformed into CCf as
A

P

1
AP B

P

1
B
A

P

1
AP



0


0
3
1
0
1
B

P

1
B





0
0
1




0
1
3



 𝑠𝐼 − 𝐴 = 𝑠 3 − 3𝑠 2 − 𝑠 − 3

Transformation to OCF
• Transformation to CCf is done by means of transformation matrix
Q .
Q

( W

OM )

1
• Where OM is observability Matrix and is given as
𝑂𝑀 = 𝐶 𝐶𝐴 ⋯ 𝐶𝐴 𝑛−1 𝑇 and W is coefficient matrix
W







 a a n

1 n

2
 a
1
1 a n

2 a n

3

1
0




 a
1
1

0
0
1
0

0
0







Where the a i
’s are coefficients of the characteristic polynomial 𝑠𝐼 − 𝐴 = 𝑠 𝑛
+ 𝑎
1 𝑠 𝑛−1
+ 𝑎
2 𝑠 𝑛−2
+ ⋯ + 𝑎 𝑛−1 s+ 𝑎 𝑛

Transformation to OCF
• Once the transformation matrix Q is computed following relations are used to calculate transformed matrices.
A

Q

1
AQ B

Q

1
B C

CQ D

D

Transformation to OCF ( Example )
• Consider the state space system given below.
𝑥
1 𝑥
2 𝑥
3
=
1 2 1
0 1 3
1 1 1 𝑥
1 𝑥
2 𝑥
3
+ 𝑦(𝑡) = 1 1 0
• Transform the given system in OCF.
𝑥
1 𝑥
2 𝑥
3
1
0
1 𝑢(𝑡)

Transformation to OCF ( Example ) 𝑥
1 𝑥
2 𝑥
3
=
1 2 1
0 1 3
1 1 1 𝑥
1 𝑥
2 𝑥
3
+
1
0
1
• The characteristic equation of the system is 𝑢(𝑡) 𝑠𝐼 − 𝐴 = 𝑠 − 1 −2 −1
0 𝑠 − 1 −3
−1 −1 𝑠 − 1
= 𝑠 3 − 3𝑠 2 − 𝑠 − 3 𝑎
1
= −3,
W




 a
2 a
1
1 a
1
1
0 𝑎
2
= −1,
1
0
0











1
1
3 𝑎
3
= −1

3
1
0
1
0
0





Transformation to OCF ( Example ) 𝑥
1 𝑥
2 𝑥
3
=
1 2 1
0 1 3
1 1 1 𝑥
1 𝑥
2 𝑥
3
+
1
0
1 𝑢(𝑡) 𝑦(𝑡) = 1 1 0
• Now the observability matrix OM is calculated as 𝑥
1 𝑥
2 𝑥
3
𝑂𝑀 = 𝐶 𝐶𝐴 𝐶𝐴 2 𝑇
𝑂𝑀 =
1 1 0
1 3 4
5 6 10
• Transformation matrix Q is now obtained as
𝑄 = 𝑊 × 𝑂𝑀 −1 =
0.333
−0.166
0.333
−0.333
0.166
0.666
0.166
0.166
0.166

Transformation to CCF ( Example )
• Using the following relationships given state space representation is transformed into CCf as
A

Q

1
AQ B

Q

1
B C

CQ D

D
A

Q

1
AQ



0


1
0
0
0
1
3


1
3


B

Q

1
B





3
2
1




C

CQ


0 0 1


• 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)
= 𝑠 2 + 2𝑠 + 3 𝑠 3 + 5𝑠 2 + 3𝑠 + 2

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END OF LECTURES-26-27-28-29