Lecture – 21
Pole Placement Control Design
Dr. Radhakant Padhi
Asst. Professor
Dept. of Aerospace Engineering
Indian Institute of Science - Bangalore
Pole Placement Control Design
Assumptions:
ƒ
The system is completely state controllable.
ƒ
The state variables are measurable and are
available for feedback.
ƒ
Control input is unconstrained.
ADVANCED CONTROL SYSTEM DESIGN
Dr. Radhakant Padhi, AE Dept., IISc-Bangalore
2
Pole Placement Control Design
Objective:
The closed loop poles should lie at
‘desired locations’.
μ1 ,… , μn , which are their
Difference from classical approach:
Not only the “dominant poles”, but “all poles” are forced to lie at
specific desired locations.
Necessary and sufficient condition:
The system is completely state controllable.
ADVANCED CONTROL SYSTEM DESIGN
Dr. Radhakant Padhi, AE Dept., IISc-Bangalore
3
Closed Loop System Dynamics
X = AX + BU
The control vector U is designed in the following
state feedback form
U = − KX
This leads to the following closed loop system
X = ( A − BK ) X = ACL X
where ACL
( A − BK )
ADVANCED CONTROL SYSTEM DESIGN
Dr. Radhakant Padhi, AE Dept., IISc-Bangalore
4
Philosophy of
Pole Placement Control Design
The gain matrix K is designed in such a
way that
sI − ( A − BK ) = ( s − μ1 )( s − μ2 )
where μ1 ,
( s − μn )
, μn are the desired pole locations.
ADVANCED CONTROL SYSTEM DESIGN
Dr. Radhakant Padhi, AE Dept., IISc-Bangalore
5
Pole Placement Design Steps:
Method 1 (low order systems, n ≤ 3)
z
Check controllability
z
Define K = [ k1 k2 k3 ]
z
Substitute this gain in the desired
characteristic polynomial equation
sI − A + B K = ( s − μ 1 )
z
(s − μ n )
Solve for k1 , k2 , k3 by equating the like
powers on both sides
ADVANCED CONTROL SYSTEM DESIGN
Dr. Radhakant Padhi, AE Dept., IISc-Bangalore
6
Pole Placement Control Design:
Method – 2
X = AX + Bu
u = −K X ,
K = [ k1 k2
kn ]
Let the system be in first companion (controllable canonical) form
⎡ 0
⎢ 0
⎢
A=⎢
⎢
⎢ 0
⎢⎣ −an
1
0
0
0
1
0
0
0
0
−an−1 −an−2
−an−3
0 ⎤
⎡0⎤
⎢0⎥
0 ⎥⎥
⎢ ⎥
⎥,B = ⎢ ⎥
⎥
⎢ ⎥
1 ⎥
⎢0⎥
⎢⎣1⎥⎦
−a1 ⎥⎦
ADVANCED CONTROL SYSTEM DESIGN
Dr. Radhakant Padhi, AE Dept., IISc-Bangalore
7
After applying the control, the closed loop system
dynamics is given by
X = ( A − BK ) X = ACL X
1
0
⎡ 0
⎢ 0
0
1
⎢
⎢
ACL = ⎢
⎢
⎢ 0
0
0
⎢
⎢⎣ − a n − a n −1 − a n − 2
0
1
⎡
⎢
0
0
⎢
⎢
0
0
ACL = ⎢
⎢
⎢
0
0
⎢
⎢⎣( −an − k1 ) ( −an −1 − k2 )
0 ⎤ ⎡0
0 ⎥⎥ ⎢⎢ 0
⎥ ⎢
⎥−⎢
1 ⎥ ⎢
0 ⎥ ⎢0
⎥ ⎢
− a1 ⎥⎦ ⎢⎣ k1
0 0
1
0
0
1
0
0
0
0
0
0
0
k2
0
k3
0
1
ADVANCED CONTROL SYSTEM DESIGN
Dr. Radhakant Padhi, AE Dept., IISc-Bangalore
⎤
⎥
0
⎥
⎥
0
⎥
0
⎥
⎥
1
⎥
a
k
−
−
( 1 n )⎥⎦
0⎤
0 ⎥⎥
⎥
⎥
⎥
0⎥
⎥
k n ⎥⎦
(1)
8
Pole Placement Control Design:
Method – 2
If μ1 , , μ n are the desired poles. Then the desired
characteristic polynomial is given by,
( s − μ 1 ) ( s − μ n ) = s n + α 1 s n −1 + α 2 s n − 2 + + α
⎡
This characteristic
⎢
polynomial, will lead to
⎢
⎢
the closed loop system
⎢
matrix as
ACL = ⎢
State space form
0
0
1
0
0
1
⎢
⎢
0
⎢ 0
⎢ −α −α
−α n − 2
n −1
⎣ n
ADVANCED CONTROL SYSTEM DESIGN
Dr. Radhakant Padhi, AE Dept., IISc-Bangalore
0⎤
0 ⎥⎥
⎥
⎥
⎥
1⎥
⎥
⎥
−α1 ⎥⎦
n
( 2)
9
Pole Placement Control Design:
Method – 2
Comparing Equation (1) and (2), we arrive at:
⎡ an + k1 = αn ⎤ ⎡ k1 = (αn − an ) ⎤
⎥
⎢
⎥ ⎢
a
k
α
k
α
a
+
=
=
−
(
⎢
−
−
−
n
1
2
n
1
2
n
1
n −1 ) ⎥
⎢
⎥⇒
⎥
⎢
⎥ ⎢
⎥
⎢
⎥ ⎢
⎣ a1 + kn = α1 ⎦ ⎢⎣ kn = (α1 − a1 ) ⎥⎦
K = (α − a )
( Row vector form)
ADVANCED CONTROL SYSTEM DESIGN
Dr. Radhakant Padhi, AE Dept., IISc-Bangalore
10
What if the system is not given
in the first companion form?
Define a transformation
X = TXˆ
Xˆ = T −1 X
Xˆ = T −1 ( AX + Bu )
(
)
(
)
Xˆ = T −1 AT Xˆ + T −1B u
Design a T such that T-1AT will be in first companion form.
Select
where
T = MW
M
⎡⎣ B AB
An−1B ⎤⎦ is the controllability matrix
ADVANCED CONTROL SYSTEM DESIGN
Dr. Radhakant Padhi, AE Dept., IISc-Bangalore
11
Pole Placement Control Design:
Method – 2
⎡ an −1
⎢a
⎢ n−2
W =⎢
⎢
⎢ a1
⎢⎣ 1
an − 2
a1
1
0
1⎤
0 ⎥⎥
⎥
⎥
⎥
0 ⎥⎦
Next, design a controller for the transformed system (using the
technique for systems in first companion form).
u = − Kˆ Xˆ = − ( Kˆ T
−1
)X = −KX
Note: Because of its role in control design as well as the use of M
(Controllability Matrix) in the process, the ‘first companion
form’ is also known as ‘Controllable Canonical form’.
ADVANCED CONTROL SYSTEM DESIGN
Dr. Radhakant Padhi, AE Dept., IISc-Bangalore
12
Pole Placement Design Steps:
Method 2: Bass-Gura Approach
z
z
z
z
z
Check the controllability condition
Form the characteristic polynomial for A
| sI − A |= s n + a1s n −1 + a2 s n − 2 + + an −1s + an
find ai’s
Find the Transformation matrix T
Write the desired characteristic polynomial
( s − μ1 ) ( s − μn ) = s n + α1s n−1 + α 2 s n−2 + + α n
and determine the αi’s
The required state feedback gain matrix is
K = [(α n − an ) (α n −1 − an −1 )
(α1 − a1 )] T −1
ADVANCED CONTROL SYSTEM DESIGN
Dr. Radhakant Padhi, AE Dept., IISc-Bangalore
13
Pole Placement Design Steps:
Method 3 (Ackermann’s formula)
Define A = A − BK
desired characteristic equation is
sI − ( A − BK ) = ( s − μ1 )
( s − μn )
| sI − A |= s n + α1s n −1 + α 2 s n −2 +
+ α n −1s + α n = 0
Caley-Hamilton theorem states that every matrix A
satisfies its own characteristic equation
φ ( A) = An + α1 An−1 + α 2 An−2 +
+ α n −1 A + α n = 0
For the case n = 3 consider the following identities.
ADVANCED CONTROL SYSTEM DESIGN
Dr. Radhakant Padhi, AE Dept., IISc-Bangalore
14
Pole Placement Design Steps:
Method 3: (Ackermann’s formula)
Multiplying the identities in order by α3, α2, α1 respectively and
adding we get
……….(1)
From Caley-Hamilton Theorem for A
And also we have for A
ADVANCED CONTROL SYSTEM DESIGN
Dr. Radhakant Padhi, AE Dept., IISc-Bangalore
15
Pole Placement Design Steps:
Method 3 (Ackermann’s formula)
Substituting ϕ ( A) and ϕ ( A) in equation (1) we get
0
Since system is completely controllable inverse of the controllability
matrix exists we obtain
………(2)
ADVANCED CONTROL SYSTEM DESIGN
Dr. Radhakant Padhi, AE Dept., IISc-Bangalore
16
Pole Placement Design Steps:
Method 3 (Ackermann’s formula)
Pre multiplying both sides of the equation (2) with [0 0 1]
z
For an arbitrary positive integer n ( number of states)
Ackermann’s formula for the state feedback gain
matrix K is given by
K = [0 0 0
1 ] ⎡⎣ B
AB
w h e re φ ( A ) = A n + α 1 A n − 1 +
2
A B
A B ⎤⎦
+ α n −1 A + α n I
n −1
−1
φ ( A)
α i ' s a re th e c o e ffic ie n ts o f th e
d e s ire d c h a ra c te ris tic p o ly n o m ia l
ADVANCED CONTROL SYSTEM DESIGN
Dr. Radhakant Padhi, AE Dept., IISc-Bangalore
17
Choice of closed loop poles :
Guidelines
z
Do not choose the closed loop poles far away
from the open loop poles, otherwise it will
demand high control effort
z
Do not choose the closed loop poles very
negative, otherwise the system will be fast
reacting (i.e. it will have a small time constant)
•
In frequency domain it leads to large bandwidth, and
hence noise gets amplified
ADVANCED CONTROL SYSTEM DESIGN
Dr. Radhakant Padhi, AE Dept., IISc-Bangalore
18
Choice of closed loop poles :
Guidelines
z
Use “Butterworth polynomials”
n +1
2n
n +1
⎛ ( j ( 2 k + 1) π ) ⎞
s
2n
= ⎜e
(
) = ( − 1)
k = 0,1, 2 - - ⎟
wo
−1
⎝
⎠
w o = a constant (like " natural frequency")
n = system order (num ber of closed loop poles)
choose only stable poles.
Im
Example: 1
let n = 1 only one pole
use k = 1
-ω0
s=ω0 (cos π + j sin π ) = −ω0
ADVANCED CONTROL SYSTEM DESIGN
Dr. Radhakant Padhi, AE Dept., IISc-Bangalore
Re
19
Choice of closed loop poles :
Guidelines
Example 2:
Let n = 2 we know ( cos θ + j sin θ ) m =( cos mθ + j sin mθ )
n+1 3
=
2n
4
Im
S2
s = ω 0 [cos ( (2 k + 1)3π / 4 ) + j sin ( (2 k + 1)(3π / 4) )]
Im
S1
Re
k = 0 ⇒ s1 = ω 0 [cos ( 3π / 4 ) + j sin ( (3π / 4) )]
stabilizing: acc ep t .
Re
k = 1 ⇒ s 2 = ω 0 [cos ( 9π / 4 ) + j sin ( (9π / 4) )]
Re
Destabiliz i ng: reject.
k = 2 ⇒ s 3 = ω 0 [cos (15π / 4 ) + j sin ( (15π / 4) )]Im
Im
S3
Re
Destabilizing: reject.
k = 3 ⇒ s 4 = ω 0 [cos ( 21π / 4 ) + j sin ( (21π / 4 ) )]
S4
stabilizi ng: accept .
ADVANCED CONTROL SYSTEM DESIGN
Dr. Radhakant Padhi, AE Dept., IISc-Bangalore
20
ADVANCED CONTROL SYSTEM DESIGN
Dr. Radhakant Padhi, AE Dept., IISc-Bangalore
21
Example: Inverted Pendulum
⎡ 0
⎢ 20.601
A=⎢
⎢ 0
⎢
⎣ −0.4905
1
0
0
0
0
0
0
0
0⎤
⎡0 ⎤
⎢ −1 ⎥
0 ⎥⎥
⎡1 0 0 0 ⎤
; B = ⎢ ⎥ ;C = ⎢
⎥
⎢0 ⎥
1⎥
0
1
0
0
⎣
⎦
⎥
⎢ ⎥
0⎦
⎣0.5⎦
ADVANCED CONTROL SYSTEM DESIGN
Dr. Radhakant Padhi, AE Dept., IISc-Bangalore
22
Step 1: Check controllability
Rank = 4 ; Controllable
M ≠0
Hence, the system is controllable.
ADVANCED CONTROL SYSTEM DESIGN
Dr. Radhakant Padhi, AE Dept., IISc-Bangalore
23
Step 2: Form the characteristic equation and get ai’s
ADVANCED CONTROL SYSTEM DESIGN
Dr. Radhakant Padhi, AE Dept., IISc-Bangalore
24
Step 3: Find Transformation T = MW and its inverse
ADVANCED CONTROL SYSTEM DESIGN
Dr. Radhakant Padhi, AE Dept., IISc-Bangalore
25
Step 4: Find αi’s from desired poles μ1 , μ2 , μ3 , μ4
ADVANCED CONTROL SYSTEM DESIGN
Dr. Radhakant Padhi, AE Dept., IISc-Bangalore
26
Step 5: Find State Feed back matrix K and input u
ADVANCED CONTROL SYSTEM DESIGN
Dr. Radhakant Padhi, AE Dept., IISc-Bangalore
27
Time Simulation: Inverted Pendulum
ADVANCED CONTROL SYSTEM DESIGN
Dr. Radhakant Padhi, AE Dept., IISc-Bangalore
28
Choice of closed loop poles :
Guidelines
z
Do not choose the closed loop poles far away
from the open loop poles. Otherwise, it will
demand high control effort.
z
Do not choose the closed loop poles very
negative. Otherwise, the system will be fast
reacting (i.e. it will have a small time constant)
In frequency domain it leads to a large bandwidth,
which in turn leads to amplification of noise!
ADVANCED CONTROL SYSTEM DESIGN
Dr. Radhakant Padhi, AE Dept., IISc-Bangalore
29
Multiple input systems
z
z
The gain matrix is not unique even for
fixed closed loop poles.
Involved (but tractable) mathematics
X = AX + BX
U = − KX
⎡ k11 k12
⎢
K =⎢
⎢
⎢
⎣ k m1 k m 2
k1 n ⎤
⎥
⎥
⎥
⎥
k mn ⎦
ADVANCED CONTROL SYSTEM DESIGN
Dr. Radhakant Padhi, AE Dept., IISc-Bangalore
30
Multiple input systems:
Some tricks and ideas
z
Eliminate the need for measuring some xj by appropriately
choosing the closed loop poles.
E xam ple :
u = μ 1 x1 + ( μ 2 − β ) x 2
select μ 2 = β provided β < 0
z
z
z
Relate the gains to proper physical quantities
⎡ x⎤
⎡ux ⎤ ⎡ g11 0 g13 0 ⎤ ⎢⎢ y ⎥⎥
⎢u ⎥ = ⎢
⎥ ⎢ x⎥
0
0
g
g
y
22
24
⎣
⎦
⎣ ⎦
⎢ ⎥
⎣ y⎦
Shape eigenvectors: “Eigen structure assignment control”
Introduce the idea of optimality: “optimal control”
ADVANCED CONTROL SYSTEM DESIGN
Dr. Radhakant Padhi, AE Dept., IISc-Bangalore
31
Control Allocation:
ADVANCED CONTROL SYSTEM DESIGN
Dr. Radhakant Padhi, AE Dept., IISc-Bangalore
32
References
z
K. Ogata: Modern Control Engineering,
3rd Ed., Prentice Hall, 1999.
z
B. Friedland: Control System Design,
McGraw Hill, 1986.
ADVANCED CONTROL SYSTEM DESIGN
Dr. Radhakant Padhi, AE Dept., IISc-Bangalore
33
ADVANCED CONTROL SYSTEM DESIGN
Dr. Radhakant Padhi, AE Dept., IISc-Bangalore
34