Download Modern Control Design: Pole-Placement and Observer-Based State Feedback and more Summaries Design in PDF only on Docsity!
CHAPTER
MODERN
CONTROL
DESIGN
The classical design techniques of Chapters 6 and 7 are based on the root-locus and
frequency response that utilize only the plant output for feedback with a dynamic
controller. In this final chapter on design, we employ modern control designs that
require the use of all state variables to form a linear static controller. Modern control
design is especially useful in multivariable systems; however, in this chapter the ideas
of state-space design are illustrated using single-input, single-output systems.
One approach in modern control systems accomplished by the use of state feed-
back is known as pole-placement design. The pole-placement design allows all roots
of the system characteristic equation to be placed in desired locations. This results in
a regulator with constant gain vector K.
The state-variable feedback concept requires that all states be accessible in a
physical system, but for most systems this requirement is not met; i.e., some of the
states are inaccessible. For systems in which all states are not available for feedback,
a state estimator (observer) may be designed to implement the pole-placement design.
The other approach to the design of regulator systems is the optimal control
problem where a specified mathematical performance criterion is minimized.
8.1. Pole-Placement Design 171
8.1 Pole-Placement Design
The control is achieved by feeding back the state variables through a regulator with
constant gains. Consider the control system presented in the state-variable form
x _(t) = Ax(t) + Bu(t) (8.1)
y (t) = Cx(t)
Consider the block diagram of the system shown in Figure 8.1 with the following state
feedback control
u(t) = Kx(t) (8.2)
where K is a 1 n vector of constant feedback gains. The control system input r (t)
is assumed to be zero. The purpose of this system is to return all state variables to
values of zero when the states have been perturbed.
....................................................................................................................... ..... .
.........
......
.
................................................................................................................................................................................................................................................................................................................................................ ..... .
.........
......
.
Plant
................................................................................... ..... .
.........
......
.
. .. ... .. ... .. ... .. ... .. ... .. ... .. ... ..............................................................................................................................................................................................
......
..
........... .....
k
n
....................................................... ..
...
....
....
...
....
....
...
....
....
...
....
...
....
....
...
....
...
....
....
...
....
... ..... ...... ....
.. ... ... ... ... ..
. .. ... .. ... .. ... .. ... .. ... .. ... .. ... .. ... .. ... .. ... .. ... .. ... .. ... .. ... .. ... .. ... .. ... .. ... .. ... .. ... .. ... .. ... .. ... .. ... .. ... ..........................................................................................................................................................
......
..
........... .....
k
................................................................................................................................................ ..
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...... ... .... ......
... ... .. ... .. ...
. .. ... .. ... .. ... .. ... .. ... .. ... .. ... .. ... .. ... .. ... .. ... .. ... .. ... .. ... .. ... .. ... .. ... .. ... .. ... .. ... .. ... .. ... .. ... .. ... .. ... .. ... .. ... .. ... .. ... .. ... .. ... .. ... .. ... .. ... .. ... .. ... .. ... .. ... .. ... .. ... .. ... .. ... .. ... ......................................................................................................................
......
..
........... .....
k
........................................................................................................................................................................................................................................................................................................................................................................ ...
...
..
...
..
...
..
...
..
...
..
...
..
...
..
...
..
...
..
...
..
...
..
...
..
...
..
...
..
...
..
...
..
...
..
...
..
...
..
...
..
...
..
...
..
...
..
...
..
...
..
...
..
...
..
...
..
...
..
...
..
...
..
...
..
...
..
...
..
...
..
...
..
...
..
...
..
...
..
...
..
...
..
...
..
...
..
.. .. ... ... .. ... ..
. .. ... ... .. ... ..
r (t) = 0 u(t) y (t)
x
n
(t)
x
(t) x
(t)
FIGURE 8.
Control system design via pole placement.
Substituting (8.2) into (8.1), the closed-loop system state-variable representation
is
x _(t) = (A BK )x(t) = A
f
x(t) (8.3)
The closed-loop system characteristic equation is
jsI A + BKj = 0 (8.4)
Assume the system is represented in the phase variable canonical form as follows
2 6 6 6 6 6 6 4
x _
x _
x _
n 1
x _
n
3 7 7 7 7 7 7 5
2 6 6 6 6 6 6 4
a
a
a
: : : a
n 1
3 7 7 7 7 7 7 5 2 6 6 6 6 6 6 4
x
x
x
n 1
x
n
3 7 7 7 7 7 7 5
2 6 6 6 6 6 6 4
3 7 7 7 7 7 7 5
u(t)
8.1. Pole-Placement Design 173
y =
x
design a state feedback control to place the closed-loop pole at 3 j 4 and 8.
Obtain the initial condition response, given x
(0) = 1 , x
(0) = 1 and x
(0) = 1.
The following commands
A=[-1 0 0; -1 -2 0; 1 0 0];
B=[1; 0; 0];
C=[1 1 0 ]; D=0;
j=sqrt(-1);
P=[-3+j4 -3-j4 -8]; % desired closed-loop poles
[K,Af]=placepol(A,B,C,P); % returns gain K and closed-loop
% system matrix
% initial condition response
t=0:.02:2;
r=zeros(1,length(t)); % generates a row of zero input
x0=[1 1 -1]; % initial state
[y,x]=lsim(Af, B, C, D, r, t, x0); % initial state response
subplot(2,2,1), plot(t, x(:,1)),title('x_1(t)'), grid
subplot(2,2,2), plot(t, x(:,2)),title('x_2(t)'), grid
subplot(2,2,3), plot(t, x(:,3)),title('x_3(t)'), grid
subplot(2,2,4), plot(t, y),title('y(t)'), grid
subplot(111)
result in
Feedback gain vector K
Uncompensated Plant
Transfer function:
s^2 + s
s^3 + 3 s^2 + 2 s
Compensated system closed-loop
Transfer function:
s^2 + s
s^3 + 14 s^2 + 73 s + 200
Compensated system matrix A - B*K
The results are given in Figure 8.2.
174 8. Modern Control Design
x
(t)
x
(t)
x
(t)
y(t)
FIGURE 8.
Initial condition response for Example 8.1.
Example 8.
A single-input single-output plant has the transfer function
G(s) =
s(s + 1)(s + 4)
Obtain a state model for the plant and design a state feedback that will place the
closed-loop pole at 2 j 2 and 5. Also, obtain the closed-loop transfer function
for the controlled system.
The following commands
num=4; den=[1 5 4 0];
[A, B, C, D]=tf2ss(num, den) % converts transfer function
% to state model
j=sqrt(-1);
P=[-2+j2 -2-j2 -5]; % desired closed-loop poles
[K,Af]=placepol(A,B,C,P); % returns gain K & closed-loop
% system matrix
result in
A =
176 8. Modern Control Design
....................................................................................................... ........ ...
............
....
.......................................................................................................................................................................................................................................................................................... ........ ...
............
....
Plant
....................................................................................................... ........ ...
............
.... . .. ... .. ... .. ... .. ... .. ... .. ... .. ... .. ... .. ... .. ... .. ... .. ... .. ......
..
...
...
..
...
...
..
...
...
..
...
. .. ... .. ... .. ... .. ... .. .. ... ... .. ... .. .. ... ... .. ... .. .. ... ... .. ... .. .. ... .. ... ... .. .. ... .. ... ... ................................................................................................................................................................
..... ....... ...... ...
Observer
. ... .. .. ... ... .. ... .. ... .. ... .. ... .. ... .. ... .. ... .. ... .. ... .. ... ..............................................................
..... ........ ...... ..
Controller
....................................................................
...
..
...
..
...
..
...
..
...
..
...
..
...
..
...
..
...
..
...
..
...
..
...
..
...
..
...
..
...
..
...
..
...
..
...
..
...
..
...
..
...
..
...
..
...
..
...
..
...
..
...
..
...
..
...
..
...
..
...
..
...
..
...
..
.. .. ... .. ... ... ..
. .. ... .. ... ... ..
r (t) = 0 u(t) y (t)
Estimated
State x_
FIGURE 8.
State feedback design with an observer.
or estimator may be necessary. The observer is an estimator algorithm based on the
mathematical model of the system. The observer creates an estimate x^ (t) of the states
from the measurements of the output y (t). The estimated states, rather than the actual
states, are then fed to the controller. One scheme is shown in Figure 8.3.
Consider a system represented by the state and output equations
x _(t) = Ax(t) + Bu(t) (8.12)
y (t) = Cx(t) (8.13)
Assume that the state x(t) is to be approximated by the state ^x(t) of the dynamic
model
_
^x(t) = A ^x(t) + Bu(t) + G(y (t) y^ (t)) (8.14)
y ^ (t) = C ^x(t) (8.15)
Subtracting (8.14) from (8.12), and (8.15) from (8.13), we have
( x_(t)
_
^x(t)) = A(x(t) ^x (t)) G(y (t) y^ (t)) (8.16)
(y (t) y^ (t)) = C(x(t) ^x(t)) (8.17)
where x(t) ^x(t) is the error between the actual state vector and the estimated vec-
tor, and y (t) y^ (t) is the error between the actual output and the estimated output.
Substituting the output equation into the state equation, we obtain the equation for the
error between the estimated state vector and the actual state vector.
e _(t) = (A GC)e(t) = A
e
e(t) (8.18)
where
e(t) = x(t) ^x(t) (8.19)
If G is chosen such that eigenvalues of matrix A GC all have negative real parts,
then the steady-state value of the estimated state vector error e(t) for any initial con-
dition will tend to zero. That is, ^x(t) will converge to x(t). The design of the observer
is similar to the design of the controller. However, the eigenvalues of A GC must
8.3. Observer Design 177
be selected to the left of the eigenvalues of A. This ensures that the observer dynamic
is faster than the controller dynamic for providing a rapid updated estimate of the state
vector.
The estimator characteristic equation is given by
jsI A + GCj = 0 (8.20)
For a specified speed of response, the desired characteristic equation for the
estimator is
e
(s) = s
n
n 1
s
n 1
s +
Thus, the estimator gain G is obtained by equating coefficients of (8.20) and
(8.21). This is identical to the pole-placement technique, and G is found by the appli-
cation of Ackermann’s formula
G =
e
(A)
2
6
6
6
4
C
C A
C A
n 1
3
7
7
7
5
2
6
6
6
4
3
7
7
7
5
and the notation
e
(A) is given by
e
(A) = A
n
n 1
A
n 1
A +
I (8.23)
The function [G, A
e
] = observer(A, B, C, p
e
) is developed for the estimator. p
e
is
the desired estimator eigenvalues. This function returns the gain vector G and the
closed-loop system matrix A
f
The necessary condition for the design of an observer is that all the states can
be observed from the measurements of the output. This characteristic is known as
observability.
Example 8.
For the plant
x_
x _
x
x
u
y =
x
design a full-state observer such that the observer is critically damped with eigenval-
ues at 8 and 8.
The following commands
A=[ 0 1; 16 0];
B=[0; 1];
C=[1 0]; D=0;
Pe=[-8 -8]; % desired observer eigenvalues
[K,Ae]= observer(A,B,C,Pe); % returns gain K and closed-loop
result in
8.5. Combined Controller-Observer Design 179
Example 8.
For the system of Example 8.3, design a controller-observer system such that the
desired closed-loop poles for the system are at 1 j 2. Choose the eigenvalues of
the observer gain matrix to be p
e
= p
e
= 8.
The following commands
A=[ 0 1; 16 0];
B=[0; 1];
C=[1 0]; D=0;
j=sqrt(-1);
P=[-1+j2 -1-j2]; % desired regulator roots
Pe=[-8 -8]; % desired observer roots
[K,G,Af]= placeobs(A,B,C,P,Pe); % returns gain K,G & closed-loop
% system matrix
result in
Feedback gain vector K
Estimator gain vector G:
Open-loop Plant
Transfer function:
s^2 - 16
Controller-estimator
Transfer function:
496 s + 2192
s^2 + 18 s + 117
Controlled system closed-loop
Transfer function:
496 s + 2192
s^4 + 18 s^3 + 101 s^2 + 208 s + 320
Combined controller observer system matrix
From the above results the controller-observer transfer function is
G
ce
496 s + 2192
s
+ 18 s + 117
The closed-loop transfer function for the controlled plant is
T (s) =
C (s)
R (s)
496 s + 2192
s
+ 18 s
+ 101 s
+ 208 s + 320
180 8. Modern Control Design
8.6 Optimal Regulator Design
The object of the optimal regulator design is to determine the optimal control law
u
(x; t) which can transfer the system from its initial state to the final state (with zero
system input) such that a given performance index is minimized. The performance
index is selected to give the best trade-off between performance and cost of control.
The performance index that is widely used in optimal control design is known as the
quadratic performance index and is based on minimum-error and minimum-energy
criteria.
Consider the plant described by
x _(t) = Ax(t) + Bu(t) (8.27)
The problem is to find the vector K (t) of the control law
u(t) = K(t)x(t) (8.28)
which minimizes the value of a quadratic performance index J of the form
J =
Z
t
f
t
(x
Qx + u
Ru)dt (8.29)
subject to the dynamic plant equation in (8.27). In (8.29) Q is a positive semidefinite
matrix and R is a real symmetric matrix. Q is positive semidefinite if all its principal
minors are nonnegative. The choice of the elements of Q and R allows the relative
weighting of individual state variables and individual control inputs.
To obtain a formal solution, we can use the method of Lagrange multipliers.
The constraint problem is solved by augmenting (8.27) into (8.29) using an n-vector
of Lagrange multipliers , . The problem reduces to the minimization of the following
unconstrained function
L(x; ; u; t) = [x
Qx + u
Ru] +
[Ax + Bu x_] (8.30)
The optimal values (denoted by the subscript ) are found by equating the partial
derivatives to zero.
@ L
= AX
+ Bu
x_
= 0 ) x_
= AX
+ B u
@ L
@ u
= 2 Ru
B = 0 ) u
=
R
B (8.32)
@ L
@ x
= 2 x
Q+
_
A = 0 )
_
= 2 Qx
A
Assume that there exists a symmetric, time varying positive definite matrix p (t) satis-
fying
= 2 p(t)x
Substituting (8.34) into (8.32) gives the optimal closed-loop control law
u
(t) = R
B
p(t)x
182 8. Modern Control Design
Vector K(t) of the control law u(t)=−k(t)x(t)
Initial state response x
, x
, x
FIGURE 8.
Optimal feedback gain vector K (t) and initial condition response x (t).
function [A,B,Q,R,t0,tf,x0]=system(A,B,Q,R,t0,tf,x0)
A=[0 1 0; 0 0 1;0 -4 -5]; B=[0;0; 1];
Q=[4 0 0; 0 3 0;0 0 2]; R=.5;
t0=0; tf=15;
x0=[2 0 -2];
The above function is saved in an M-file named system.m. Then, the following com-
mands
[tt,p,k,t,x]=riccati
subplot(2,1,1), plot(tt,k),
title('Vector K(t) of the control law u(t)=-k(t)x(t)'),
grid,
subplot(2,1,2), plot(t,x),
title('Initial state response x_1, x_2, x_3'), grid
subplot(111)
result in the control law and the response is given in Figure 8.4.
Example 8.
For Example 8.5 use the MATLAB Control System Toolbox lqr2 to obtain the solution
to the algebraic Riccati equation.
The following commands
8.6. Optimal Regulator Design 183
A=[0 1 0; 0 0 1;0 -4 -5]; B=[0;0; 1];
Q=[4 0 0; 0 3 0;0 0 2];R=.5;
[K, p]=lqr2(A,B,Q,R)
result in
K =
p =
