1. Pole -Placement
Design
&
State Observer
Presented by:-
Er. Sanyam S.
Saini
ME(I&C) Regular
2. Outlines
• Servo Design: Introduction of the
reference Input by Feed forward Control
• State Feedback with Integral Control
3. Servo Design: Introduction of the
Reference Input by Feed Forward Control
The characteristics Equation of the Closed –Loop Control System is
chosen so as to give satisfactory Transient to disturbances.
However, no mention is made of a reference i/p or a design
consideration to yield good transient response with respect to
command changes.
In general, both of these considerations should be taken into account in
the designing of control system.
This can be done by the proper introduction of the references i/p into
the system equations.
4. Continued…….
Consider the completely controllable SISO LTI system with nth – order state
variable model
x(t) = Ax(t) + bu(t) …………….…………………(i)
y(t) = cx(t) …………….…..………….…(ii)
We assume that all the n state variables can be accurately measured at all
times. Implementation of appropriately designed Control Law of the form.
u(t) = - kx(t)
Let us now assume that for the system given by equs.(i & ii), the desired
steady- state value of the controllable variable y(t) is a constant reference
input r.
For this servo system, the desired equilibrium state x s is a constant point in
state space & is governed by the equations.
cx s = r …………….…..………………………………(iii)
5. Continued…….
We can formulate this command-following problem as a ”Shifted Regulator
Problem”, by shifting the origin of the state space to the equilibrium point x s .
0 = Axs + bu s …………….………..………….…(iv)
Assuming for the present that a u s exists that satisfies equns. (iii & iv)
us
~(t) = u(t) - u
u s
~(t) = x(t) - x
x s .………..………….…..(v)
~(t) = y(t) - r
y
The shifted variables satisfy the equations
~ = A~ + b~
x x u .………..………….…..(vi)
~ = c~
y x .………..……………...(vii)
6. Continued…….
This system possesses a time – invariant asymptotically stable control law
~ = -k~
u x .………..………….…..(viii)
In terms of the original state variables , total control effort
u(t) -kx(t) us kxs .………..….……….…..(ix)
(A - bk)x s b(us kxs ) 0
or
xs ( A bk ) 1 b(us kxs )
1
cx s r c( A bk ) b(us kxs )
The equation has a unique solutions for (us kxs ) ;
(u s kx s ) Nr .………..………….…..(x)
7. Continued…….
Where N is a scalar feedforword gain, given by
N -1 c( A bk ) 1 b
The control law (ix), therefore takes the form
u(t) - kx(t) Nr
w
r u y
N PLANT
+ _
x
K
Control Configuration of a Servo System
8. Numerical Problem
Consider a Attitude Control System for a rigid satellite . Design
the control Configuration for the give control system as given
follows. (Previous example taken in stability improvement by
state feedback)
Solution:-
The Plant equations are,
x(t) = Ax(t) + bu(t)
y(t) = cx(t)
Where
0 1 0
A b c 1 0
0 0 1
Here
x1(t)= Position (t ) x2(t)= Position (t )
9. Continued…….
The reference i/p r r
is a step function. The desired steady-state is,
T
xs r 0
As the plant has integrating property, the steady- state value of u s the i/p
must be zero(otherwise o/p cannot stay constant).
For this case, the shifted regulator problem may be formulated as follows,
~
x1 x1 r
~
x2 x2
Shifted regulator variables satisfy the equations.
~ = A~ + b~
x x u
The state – feedback control
u(t) = - kx(t)
10. Continued…….
Therefore ,
~ = (A - bk)~
x x
As in previous example, we found eigenvalues of (A - bk)are placed at
the desired location - 4 j4 when,
k k1 k2 32 8
The control law expressed in terms of the original state variables is given as,
u k1~1 k2 ~2
x x k1 x1 k2 x2 k1 r
kx k1 r
12. State Feedback with Integral Control
Need of state feedback with integral control:
The control configuration of previous problem produces a generalization of
Proportional & Derivative feedback but it dose not include integral control
unless special steps are taken in the design process.
For the system
x(t) = Ax(t) + bu(t)
y(t) = cx(t)
We can feedback the state “x” as well as the integral of the error in output
by augmenting the plant state “x” with extra “integral state” z,
defined by the equation t
z(t) (y(t)- r)dt
……….………….…(i)
0
13. Continued…….
Since z(t) satisfies the differential equation,
z(t) y(t ) r cx(t ) r ……….………….…(ii)
It is easily included by augmenting the original system as follows:
x A 0 x b 0
u r ……….………….…(iii)
z c 0 z 0 1
Since “r” is constant, in the steady – state
x
0, z 0, provided that the
system is stable.
This means that the steady -state solutions xs , zs & u s must satisfy the
equation
0 A 0 xs b
r us
1 c 0 zs 0 ……….………….…(iv)
14. Continued…….
Therefore,
From equation (iii) & (iv)
x A 0 x xs b
(u us )
z c 0 z zs 0 ….………….…(v)
Now define new state variables as follows , representing the deviations from
the steady - state:
~ x xs
x
z zs
~
u (u us ) ….………….…(vi)
In terms of these variables,
~
x A~
x ~
bu ….………….…(vii)
15. Continued…….
Where,
A 0 b
A b
c 0 0
The significance of this result is that by defining the deviation from steady-
state as state & control variables,
The design problem has been reformulated to be the Standard regulator
problem with ~ 0
x as the desired state.
We assume that an asymptotically stable solution to this problem exists & is
given by-
u = -k~
x
Partitioning “k” appropriately & using eqns. (vi)
k [k p ki ]
16. Continued…….
x xs
u us kp ki
z zs
kp (x xs ) ki ( z zs )
The steady-state terms must balance, therefore
t
u kpx ki z kpx ki ( y (t ) r )dt
0
The control, thus consist of proportional state feedback & integral control of
output error.
At steady-state,
~
x o ; therefore
lim z (t ) 0 or lim y(t ) r
t t
17. Continued…….
Thus, by integrating action , the output “y” is driven to the no- offset
condition. This will be true even in the presence of constant disturbances
acting on the plant.
w
r u y
ki
x
kp
State feedback with integral control
18. Numerical Problem
Design the integral control system with a constant reference
command signal.
Given Y ( s) 1
with 5 & 0.5
n
U ( s) (s 3)
Solution:-
Characteristic equation will be- s2 5s 25 0
Plant equation will be-
x 3x u
y x
Augmenting the plant state “x” with the integral state “z” defined by the
equation.
t
z(t) (y(t)- r)dt
0
19. Continued…….
x 3 0 x 1 0
We get u r
z 1 0 z 0 1
Since x xs
~
x
z zs
~
u (u us )
~ 3 0 ~ 1 ~
State equation becomes x x u
1 0 0
20. Continued…….
We can find “k” from
3 0 1
det sI k s2 5s 25
1 0 0
After solving s2 (3 k1 ) s k2 s2 5s 25
Therefore k 2 25 [k p ki ]
t
The control u k p x ki z 2 x 25 ( y (t ) r )dt
0
21. Continued…….
State feedback with integral control
w
r 1 y
25 s
2
This system will behave according to the desired closed-loop roots & will
exhibits the characteristics of integral control:
Zero steady-state error to a step “r” & zero steady –state error to a
constant disturbance “w”.
