Lecture 13 - State Feedback

Report
Disturbance
Controller
Plant/Process
S
kr
S
u
d
x  Ax  Bu
dt
y  Cx  Du
Output
y
Prefilter
-K
x
State Feedback
State Controller
Professor Walter W. Olson
Department of Mechanical, Industrial and Manufacturing Engineering
University of Toledo
State Feedback
Outline of Today’s Lecture
 Review
 Reachability
 Testing for Reachability
 Control System Objective
 Design Structure for State Feedback
 State Feedback
 2nd Order Response
 State Feedback using the Reachable Canonical Form
Reachability
 We define reachability (often times called controllability) by the
following:
 A state in a system is reachable if for any valid states of the system,
say, initial state at time t=0, x0 , and a state xf , there exists a solution
for t>0 such that x(0) = x0 and x(t)=xf.
 There are systems which we can not control
 the states are not reachable with our input.
 There in designing control systems, it is important to know if the
system is controllable.
 This is closely linked with the concept of ergodicity of the system
in which we ask the question whether or not it is possible to with
some measure of our system to measure every possible state of the
system.
Reachability
 For the system,
,
d
x  Ax  Bu
dt
and y  Cx  Du
all of the states of the system are reachable if and only if Wr is
invertible where Wr is given by
Wr   B
AB
An 1 B 
Canonical Forms
 The word “canonical” means prescribed
 In Control Theory there a number transformations that can
be made to put a system into a certain canonical form where
the structure of the system is readily recognized
 One such form is the Controllable or Reachable Canonical
form.
Reachable Canonical Form
 A system is in the reachable canonical form if it has the structure
 z1   a1
z   1
2
d   
 z3    0
dt   
 ...   ...
 zn   0
y  c1
 a2
0
1
...
0
 a3
0
0
... an   z1  1
... 0   z2  0
   
0   z3   0 u
   
...   ...  0
1
0   zn  0
0
c3 ... cn  z  Du
c2
Such a structure can be represented by blocks as
D
u
S
-1

S
S
c1
c2
z1

…
…
z2
a1
a2
S
S

S
S
cn-1
cn
zn-1
an-1
…
S

zn
an
y
Control System Objective
Given a system with the dynamics and the output
d
x  Ax  Bu
dt
y  Cx  Du
Design a linear controller with a single input which is
stable at an equilibrium point that we define as xe  0
Our Design Structure
Disturbance
Controller
Plant/Process
Input
r
kr
S
S
u
d
x  Ax  Bu
dt
y  Cx  Du
Prefilter
-K
x
State Feedback
State Controller
Output
y
Our Design Structure
Disturbance
Controller
Plant/Process
Input
r
S
kr
S
u
d
x  Ax  Bu
dt
y  Cx  Du
Output
y
Prefilter
-K
x
State Feedback
State Controller
Ignoring the disturbance
u   Kx  k r r
d
x  Ax  BKx  Bk r r  ( A  BK ) x  Bk r r
dt
Therefore, our new system equations are
d
 x  ( A  BK ) x  Bk r r
 dt
 y   C  DK  x  Dk r r

d
 x  ( A  BK ) x  Bk r r
If D  0 as it mostly is, then  dt

y  Cx
Restated Control System Objective
(Eigenvalue Assignment Problem)
Given a system with the dynamics and the output
d
x  Ax  Bu
dt
y  Cx  Du
Design a linear controller with a single input which is
stable at an equilibrium point that we define as xe  0
with a state feedback controller such that
d
 xe  ( A  BK ) xe  Bkr r
 dt

y  Cxe

where the eigenvalues of ( A  BK ) all have negative real components
Note that kr does not affect stability, is a scalar, and can be chosen as
kr  
1
C ( A  BK )1 B for ye=r
Example
 Design a controller that will
control the angular position to a
given angle, 0
 3998.35 0
 30
A 
B

0
1
0

 
 119950
 30 119950
AB  
Wr   B AB   

30 
 30 
0
0.0333 133.3 
Wr 1  
 states are reachable
0.0333
 0.
 3998.35 0  30
 3998.35  30K1
A  BK  

K
K



1
2

1
0  0 
1


The characteristic polynomial for the system is
 I  A  30K 2   3998.35  30K1     2  n 2  2n    2
n  30 K 2  
From Lecture 5 and with values:
 bRa  KK b
d    
JRa

dt   
1


 K 
0   
 JRa Va


 


0    0 
 
y  0 1  
 
Ra  0.2
K b  5.5  10
b  4  10
2
2
V-s/rad
N-m/rad/s
2 30 K 2
If n  1 and   .6 then
2 30 K 2   3998.35
1
 0.0333 and K1 
 133.185
30
30
K   133.185 0.0333
K2 
kr  
J  1  105 kg-m 2
K  6  105 N-m/amp
3998.35  30 K1
kr  
1

C ( A  BK ) 1 B
1
1
-1.3500 -1.0000 30
0 1 
0   0 
 1.0000
1
1.0000  30
 0
0 1 
 
 -1.0000 -1.3500  0 
 0.0333
30K 2 
0 
Example
 Design a controller that will
control the angular position to a
given angle, 0
K   133.185 0.0333
k r  0.0333
This results in the system
 d     -1.3500 -1.0000    1 

r
  
0     0.
 dt  e   1.0000

 

y  0 1  

 e 

2nd Order Response
 As the example showed, the characteristic equation for which the roots are the
eigenvalues allow us to design the reachable system dynamics
 When we determined the natural frequency and the damping ration by the
equation
 I  A  30K2  3998.35  30K1     2  n2  2n   2
we actually changed the system modes by changing the eigenvalues of the
system through state feedback
n=1
0.6
Im()
0.1
1
0.4 x xx
0.6 x
1
1

x
0
Re()
x
-1
0.6
x
0.4 x xx
0.1
0
-1
Im()
x
n=4
1
n=2 x
n=1 x
n
-1
n=2 x
n=4 x
x n=1
-1
Re()
State Feedback Design with the
Reachable Canonical Equation
 Since the reachable canonical form has the coefficients of the
characteristic polynomial explicitly stated, it may be used for
design purposes:
 z1   a1
z   1
2
d   
 z3    0
dt   
 ...   ...
 zn   0
 a2
0
1
...
0
 a3
0
0
0
y  Cz
 a1  k 1
 z1  
z   1
2
d    0
 z3   
dt    ...
 ...   0
 zn  
K   k 1
k2
... an   z1  1 

... 0   z2  0 
   
0   z3   0 u 
    
...   ...  0 
1
0   zn  0 


 a2  k 2
0
1
...
0
 a3  k 3
0
0
...
...
0
1
an  k n   z1   kr 
   
0   z2   0 
0   z3    0  r
   
...   ...   0 
0   zn   0 
y  Cz
k 3 ... k n 
Since K is applied to z  Tx the control action is KTx or
K  KT
Example: Inverted Pendulum
 Design a controller that will
stabilize the Segway forward
velocity at a given position, r0
Using the model developed in Lecture 5 for the inverted pendulum
0


0



 x 
J  ml 2 
m2l 2 g



0 
0  
  v   J ( M  m )  Mml 2 
J ( M  m )  Mml 2

F

0
0
1   
0





 
mlg ( M  m )
ml
0
0    

2
J ( M  m )  Mml
 J ( M  m )  Mml 2 

M  10kg
 x
v 
m  80kg
y  0 1 0 0   with
l  1m
 
 
J  100kg  m 2 / s 2
 
0
x
  
  0
d v 

dt   0
  
  0


1
0
From Lecture 12:
The characteristic polynomial of A is
 I  A   4  7.205 2
0 
1 0 7.2051
0 1.
0
7.2051


Wr 
0
0.
1.
0




0.
1. 
0 0.
 z1 
 z1  0 7.205 0 0   1
z2
  
0
0 0   0
d  z2   1
  z3     u

1
0 0   0
dt  z3  0
  
 ...  
0
1 0   0
 z4   0
 zn 
y  Cz
The eigenvalues are 2.684, -2.684, 0, 0
Note the system is unstable and that the only pair of
eigenvalues is at 0 (undamped!)
Our objective then would be to stabilize the system
and add some damping, say  =0.6
Example
Starting with the characteristic equation
we can write it as
 I  A   2   2  7.205
If we regard this a combination of two
2nd order systems in which one acts extremely
slow ( 2 ) and the other faster but unstable

2
 7.205 then we want to dampen the slow
system with eigenvalues in the unit circle to
maintain speed but with a damping ratio of about  =0.6:
 
Re( )

0.6
 0.6 
 .62  x 2  1  x  .4  0.632
1
We have one eigenvalue (-2.684) which is acceptible but
we need to move the other to the approximately -3 so that
 I  A     1  i 0.632    1  i 0.632   +2.684    3
 I  A   4  7.683 3  20.81 2  24.05  11.26
K  [7.683  0 20.81  7.205 24.05  0 11.26  0]
K  7.683 28.015 24.05 11.26
0
0
122.5
 0
 0
0
122.5
0 

K  7.683 28.015 24.05 11.26 
12.491
0
28.10 
 0


0.
28.10
0. 
 12.491
K   140.6, 300.4, 3748, 1617 Using T developed from Lecture 12
1
 7.0864
C ( A  BK ) 1 B
The new controlled system is
kr  
d
x  ( A  BK ) x  Bk r r
dt
1.
0.
0. 
0.
 0.


 2.5834 5.5177 62.442 29.702 
 0.1302 
d
x
r
x
0.
0.
1. 
0.
 0.


dt




 1.1482 2.452 23.393 13.201
 0.05785
Summary
 Control System Objective
 Design Structure for State Feedback
Disturbance
Controller
 State Feedback
Plant/Process
Input
r
d
 x  ( A  BK ) x  Bkr r
 dt
 y   C  DK  x  Dkr r

S
kr
S
u
d
x  Ax  Bu
dt
y  Cx  Du
Output
y
Prefilter
x
-K
State Feedback
State Controller
 2nd Order Response
 State Feedback using the Reachable Canonical Form
 a1  k 1
 z1  
z   1
2
d    0
 z3   
dt    ...
 ...   0
 zn  
 a2  k 2
a3  k 3
0
1
...
0
0
0
0
... an  k n   z1   kr 
   
...
0   z2   0 
0   z3    0  r
   
...   ...   0 
1
0   zn   0 
y  Cz
Next: State Observers

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