State Space Canonical Forms - Dr. Imtiaz Hussain

Report
Feedback Control Systems (FCS)
Lecture-26-27-28-29
State Space Canonical forms
Dr. Imtiaz Hussain
email: [email protected]
URL :http://imtiazhussainkalwar.weebly.com/
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
Companion forms
Observable Canonical form
Diagonal Canonical form
Modal forms
Jordan Canonical Form
• It is interesting to note that the dynamics properties of system
remain unchanged whichever the type of representation is used.
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
 
−1 

+ 1 −1 + ⋯ + −1
+   = ()




• 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:
1 = ,
2 = ,
3 = ,
 −1 
⋯ ,  = −1

Phase Variable Canonical form
1 = ,
2 = ,
3 = ,
 −1 
⋯ ,  = −1

• Taking derivatives of the first n-1 state variables, we have
1 = 2 ,
2 = 3 ,
3 = 4 ⋯ , −1 = 
 = − 1 − −1 2 − ⋯ − 1  + ()
 x1   0

 
x 2
0

 
    

 
 x n 1   0
 x    a
n
 n  
1
0

0
1




0
0

 a n 1
 a n3

0   x1   0 

  
0
x2
0

  
        u

  
1   x n 1   0 
 a 1   x n   1 
Phase Variable Canonical form
1 = ,
2 = ,
3 = ,
 −1 
⋯ ,  = −1

• Output equation is simply
y  1
0

0
 x1 


x2


0   


 x n 1 
 x 
 n 
Phase Variable Canonical form
y
u (t )
y
( n 1 )
(n)
∫
+
 xn
∫
…
∫
y  x2
y  x1
∫
+
y
 a1
(n2)
 x n 1
 a2

 an
8
Phase Variable Canonical form
u
1
x n
1 x
1
n 1
x
s
s n2
1
x1  x 2
s

 1
1
1
s
y
x1
2
3

  n 1
n
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.
3
2

2 3 +4 2 +6
+ 8 = 10()



• The differential equation is third order, thus there are three
state variables:
1 = 
2 = 
3 = 
• And their derivatives are (i.e state equations)
1 = 2
2 = 3
3 = −41 − 32 − 23 + 5()
Phase Variable Canonical form (Example-1)
1 = 
1 = 2
2 = 
3 = 
2 = 3
3 = −41 − 32 − 23 + 5()
• In vector matrix form
 x1   0
  
x  0
 2 
 x 3    4
y ( t )  1
0
1
0
3
0   x1   0 
   
1
x  0 u (t )
 2   
 2   x 3   5 
 x1 
 
0 x2
 
 x 3 
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.
()
  2 + 1  + 2
= 3
()  + 1  2 + 2  + 3
• Transfer function can be decomposed into cascade form
()
1
 3 + 1  2 + 2  + 3
()
  2 + 1  + 2
()
• Denoting the output of the first block as W(s), we have the following
input/output relationships:
()
1
= 3
()  + 1  2 + 2  + 3
()
=   2 + 1  + 2
()
Phase Variable Canonical form (Example-2)
()
1
= 3
()  + 1  2 + 2  + 3
()
=   2 + 1  + 2
()
• Re-arranging above equation yields
 3   = −1  2   − 2   − 3 ()+ ()
() =   2 () + 1 () + 2 ()
• Taking inverse Laplace transform of above equations.
  = −1   − 2   − 3 ()+ u()
() =    + 1   + 2 ()
• Choosing the state variables in phase variable form
1 = 
2 = 
3 = 
Phase Variable Canonical form (Example-1)
• State Equations are given as
1 = 2
2 = 3
3 = −3 1 − 2 2 − 1 3 + ()
• And the output equation is
() = 2 1 + 1 2 +  3

1
2
1
2
3
Phase Variable Canonical form (Example-1)
• State Equations are given as
1 = 2
2 = 3
3 = −3 1 − 2 2 − 1 3 + ()
• And the output equation is
() = 2 1 + 1 2 +  3

1
2
−1
−2
−3
Phase Variable Canonical form (Example-1)
• State Equations are given as
1 = 2
2 = 3
3 = −3 1 − 2 2 − 1 3 + ()
• And the output equation is
() = 2 1 + 1 2 +  3
• In vector matrix form
 x1   0
  
x 
0
 2 
 x 3    a 3
y ( t )  b 2
b1
1
0
 a2
0   x1   0 
   
1
x  0 u (t )
 2   
 a 1   x 3   1 
 x1 
 
bo  x 2
 
 x 3 
Companion Forms
• Consider a system defined by
n
n 1
n
n 1
y  a 1 y    a n 1 y  a n y  b o u  b1 u    b n 1u  b n u
• where u is the input and y is the output.
• This equation can also be written as
()    + 1  −1 + ⋯ + −1  + 
= 
()
 + 1  −1 + ⋯ + −1  + 
• We will present state-space representations of the system
defined by above equations in controllable canonical form
and observable canonical form.
Controllable Canonical Form
()    + 1  −1 + ⋯ + −1  + 
= 
()
 + 1  −1 + ⋯ + −1  + 
• The following state-space
controllable canonical form:
 x1   0

 
x 2
0

 
    

 
 x n 1   0
 x    a
n
 n  
representation
1
0

0
1




0
0

 a n 1
 an2

is
called
0   x1   0 

  
0
x2
0

  
        u

  
1   x n 1   0 
 a 1   x n   1 
a
Controllable Canonical Form
()    + 1  −1 + ⋯ + −1  + 
= 
()
 + 1  −1 + ⋯ + −1  + 
y  b n  a n b o
b n 1  a n 1b o

b2  a 2 bo
 x1 


x2


b1  a 1b o     b o u


 x n 1 
 x 
 n 
Controllable Canonical Form
d
d
dt
dt
…

+
d
dt
u (t )
+
v
(n)
∫
∫
…
∫
+
1
2

n
v  x2
∫
v  x1
1
 n 1
n
Controllable Canonical Form (Example)
()
+3
= 2
()  + 3 + 2
• Let us Rewrite the given transfer function in following form
() 0 2 +  + 3
= 2
()  + 3 + 2
a2  2
a1  3
b2  3
b1  1
 x1   0
  
 x 2    a 2
1   x1   0 
     u
 a1   x 2   1 
 x1   0
  
 x 2    2
1   x1   0 
     u
 3  x2  1 
bo  0
Controllable Canonical Form (Example)
() 0 2 +  + 3
= 2
()  + 3 + 2
a2  2
a1  3
b2  3
y  b 2  a 2 b o
y  3
b1  1
bo  0
 x1 
b1  a 1b o    b o u
x2 
 x1 
1 
x2 
Controllable Canonical Form (Example)
()
+3
= 2
()  + 3 + 2
• By direct decomposition of transfer function
Y (s)
s3

s  3s  2
2
U (s)
Y (s)
U (s)

s
1

s
s
2
2
P (s)
P (s)
P (s)  3s
P (s)  3s
1
2
P (s)
P (s)  2s
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)  3s
U (s)  P (s)  3s
1
2
P ( s )......... .......( 1)
P (s)  2s
2
P ( s )......... .......( 2 )
Controllable Canonical Form (Example)
• Rearranging equation-2 yields
P (s)  U (s)  3s
•
1
P (s)  2s
2
P ( s )......... .......( 3 )
Draw a simulation diagram using equations (1) and (3)
P (s)  U (s)  3s
1
P (s)  2s
2
Y (s)  s
P (s)
1
P (s)  3s
-2
-3
x 2
U(s)
P(s)
1/s
x 2  x1
1/s
1
3
x1
Y(s)
2
P (s)
Controllable Canonical Form (Example)
-2
-3
x 2
U(s)
P(s)
1/s
x 2  x1
1/s
3
x1
Y(s)
1
•
State equations and output equation are obtained from simulation
diagram.
x1  x 2
x 2  U ( s )  3 x 2  2 x1
Y ( s )  3 x1  x 2
Controllable Canonical Form (Example)
x1  x 2
•
x 2  U ( s )  3 x 2  2 x1
Y ( s )  3 x1  x 2
In vector Matrix form
 x1   0
  
 x 2    2
y  3
1   x1   0 
      f (t )
 3  x2  1 
 x1 
1

x
 2
Observable Canonical Form
()    + 1  −1 + ⋯ + −1  + 
= 
()
 + 1  −1 + ⋯ + −1  + 
• The following state-space representation is called an
observable canonical form:
 x1   0

 
x 2
1

 
    

 

 x n 1   0
 x   0
 n  
0

0
0

0



0

0
0

1
 a n   x1   b n  a n b o 
 


x2
b n 1  a n 1b o
 a n 1
 


u

     
 


 a 2   x n 1   b 2  a 2 b o 
 a 1   x n   b1  a 1b o 
Observable Canonical Form
()    + 1  −1 + ⋯ + −1  + 
= 
()
 + 1  −1 + ⋯ + −1  + 
y  0
0

0
 x1 


x2


1    b o u


 x n 1 
 x 
 n 
Observable Canonical Form (Example)
()
+3
= 2
()  + 3 + 2
• Let us Rewrite the given transfer function in following form
() 0 2 +  + 3
= 2
()  + 3 + 2
a2  2
a1  3
 x1   0
  
 x 2   1
b2  3
b1  1
bo  0
 a 2   x1   b 2  a 2 b o 
u
   
 a 1   x 2   b1  a 1b o 
 x1   0
  
 x 2   1
 2   x1   3 
     u
 3   x 2  1 
Observable Canonical Form (Example)
() 0 2 +  + 3
= 2
()  + 3 + 2
a2  2
a1  3
b2  3
y  0
y  0
b1  1
 x1 
1   b o u
x2 
 x1 
1 
x2 
bo  0
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
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 , say, represents the physical state
relative to some other reference, or even a mathematical
coordinate vector.
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,
x (t )  T x (t )
• Where T is a nonsingular nxn transformation matrix.
• The transformed state () is written as
x (t )  T
1
x (t )
Similarity Transformations
• The transformed state () is written as
x (t )  T
1
x (t )
• Taking time derivative of above equation
x ( t )  T
x ( t )  T
x ( t )  T
x ( t )  T
A T
1
1
1
1
AT
1
x (t)
 Ax ( t )  Bu ( t ) 
x ( t )  Ax ( t )  Bu ( t )
 AT x ( t )  Bu ( t ) 
x (t )  T x (t )
AT x ( t )  T
1
Bu ( t )
B T
1
B
x ( t )  A x ( t )  B u ( t )
Similarity Transformations
• Consider transformed output equation
y (t )  C x (t )  D u (t )
• Substituting   =  −1 () in above equation
y (t )  C T
1
x (t )  D u (t )
• Since output of the system remain unchanged [i.e.   =
()] therefore above equation is compared with   =
  + () that yields
C  CT
D  D
Similarity Transformations
• Following relations are used to preform transformation of
coordinates algebraically
A T
1
C  CT
AT
B T
1
D  D
B
Similarity Transformations
• Invariance of Eigen Values
sI  A  sI  T
 sT
 T
1
1
1
AT
T T
1
AT
sI  A T
 sI  A
sI  A  sI  A
1
T T  I
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
 = 

−1 
⋯
and W is coefficient matrix
 a n 1

a
 n2
W   

 a1
 1
an2

a1
a n3

1



1

0
0

0
1

0



0
0 
Where the ai’s are coefficients of the characteristic polynomial
 −  =   + 1  −1 + 2  −2 + ⋯ + −1 s+
Transformation to CCF
• Once the transformation matrix P is computed following
relations are used to calculate transformed matrices.
A  P
1
AP
1
B  P B
C  CP
D  D
Transformation to CCF (Example)
• Consider the state space system given below.
1
1 2 1 1
1
2 = 0 1 3 2 + 0 ()
3
1 1 1 3
1
• Transform the given system in CCF.
Transformation to CCF (Example)
1
1 2 1 1
1
2 = 0 1 3 2 + 0 ()
3
1 1 1 3
1
• The characteristic equation of the system is
 − 1 −2
−1
 −  = 0
 − 1 −3 =  3 − 3 2 −  − 3
−1
−1  − 1
1 = −3,
a2

W  a1

 1
2 = −1,
a1
1
0
1   1
 
0  3
 
0   1
3 = −1
3
1
0
1

0

0 
Transformation to CCF (Example)
1
1 2 1 1
1
2 = 0 1 3 2 + 0 ()
3
1 1 1 3
1
• Now the controllability matrix CM is calculated as
 = 
1
 = 0
1
 2 
2 10
3 9
2 7
• Transformation matrix P is now obtained as
1
 =  ×  = 0
1
2 10 −1
3 9 −3
1
2 7
3 −1
= 0 3
0 −1
1
0
1
−3 1
1 0
0 0
Transformation to CCF (Example)
• Using the following relationships given
representation is transformed into CCf as
A  P
1
state
space
1
B  P B
AP
0

1
A  P AP  0

 3
1
0
1
0

1

3 
0 
 
1
B  P B  0
 
 1 
 −  =  3 − 3 2 −  − 3
Transformation to OCF
• Transformation to CCf is done by means of transformation matrix
Q.
1
Q  (W  OM )
• Where OM is observability Matrix and is given as
 = 

−1
⋯

and W is coefficient matrix
 a n 1

a
 n2
W   

 a1
 1
an2

a1
a n3

1



1

0
0

0
1

0



0
0 
Where the ai’s are coefficients of the characteristic polynomial
 −  =   + 1  −1 + 2  −2 + ⋯ + −1 s+
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.
1
1 2 1 1
1
2 = 0 1 3 2 + 0 ()
3
1 1 1 3
1
1
() = 1 1 0 2
3
• Transform the given system in OCF.
Transformation to OCF (Example)
1
1 2 1 1
1
2 = 0 1 3 2 + 0 ()
3
1 1 1 3
1
• The characteristic equation of the system is
 − 1 −2
−1
 −  = 0
 − 1 −3 =  3 − 3 2 −  − 3
−1
−1  − 1
1 = −3,
a2

W  a1

 1
2 = −1,
a1
1
0
1   1
 
0  3
 
0   1
3 = −1
3
1
0
1

0

0 
Transformation to OCF (Example)
1
() = 1 1 0 2
3
• Now the observability matrix OM is calculated as
1
1 2 1 1
1
2 = 0 1 3 2 + 0 ()
3
1 1 1 3
1
 = 

1 1
 = 1 3
5 6
2

0
4
10
• Transformation matrix Q is now obtained as
 =  × 
−1
0.333
= −0.333
0.166
−0.166
0.166
0.166
0.333
0.666
0.166
Transformation to CCF (Example)
• Using the following relationships given
representation is transformed into CCf as
A Q
1
AQ
0

1
A  Q AQ  1

 0
3
 
1
B Q B  2
 
 1 
B Q
0
0
1
1
B
state
C  CQ
D  D
3

1

3 
C  CQ  0
space
0
1
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
()
 2 + 2 + 3
= 3
()  + 5 2 + 3 + 2
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END OF LECTURES-26-27-28-29

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