Post on 17-Jul-2015
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• Servo Design: Introduction of the
reference Input by Feed forward Control
• State Feedback with Integral Control
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.
bu(t) +Ax(t)=(t)x
kx(t)- =u(t)
r=cxs
Consider the completely controllable SISO LTI system with nth – order state variable model
We assume that all the n state variables can be accurately measured at all times. Implementation of appropriately designed Control Law of the form.
cx(t) =y(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 is a constant point in state space & is governed by the equations.
…………….…………………(i)
…………….…..………….…(ii)
sx
…………….…..………………………………(iii)
Continued…….
sx- x(t)=(t)x~
r -y(t) =(t)y~
u~b +x~A =x~
x~c=y~
ss bu +Ax=0
su-u(t)=(t)u~
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 .
sx
…………….………..………….…(iv)
suAssuming for the present that a exists that satisfies equns. (iii & iv)
.………..………….…..(v)
The shifted variables satisfy the equations
.………..……………...(vii)
.………..………….…..(vi)
su
0)(bk)x-(A s ss kxub
)()(x 1
sss kxubbkA
x~-k=u~
)()(cx 1
s ss kxubbkAcr
Nr)kx(u ss
Continued…….
This system possesses a time – invariant asymptotically stable control law
In terms of the original state variables , total control effort
or
The equation has a unique solutions for ; )( ss kxu
.………..………….…..(x)
.………..………….…..(viii)
skxsu -kx(t)u(t) .………..….……….…..(ix)
Nrkx(t)- u(t)
w
r u y
+ _
x
N
PLANT
K
Control Configuration of a Servo System
bbkAc 1-1 )(N
Continued…….
The control law (ix), therefore takes the form
Where N is a scalar feedforword gain, given by
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,
1
0b
00
10A
bu(t) +Ax(t)=(t)x
cx(t) =y(t)
Where
01c
Herex1(t)= Position x2(t)= Position )(t )(t
Continued…….
As the plant has integrating property, the steady- state value of the i/p must be zero(otherwise o/p cannot stay constant).
The reference i/p is a step function. The desired steady-state is,
For this case, the shifted regulator problem may be formulated as follows,
rrT
rsx 0
su
rxx 11~
22~ xx
Shifted regulator variables satisfy the equations.
u~b +x~A =x~
The state – feedback control
kx(t)- =u(t)
Continued…….
Therefore ,
x~bk)-(A =x~
As in previous example, we found eigenvalues of are placed at the desired location when,
bk)-(Aj44-
832kk 21 k
The control law expressed in terms of the original state variables is given as,
rkxkxkxkxku 122112211~~
rkkx 1
Continued…….
The control configuration for attitude control of the satellite is showing as following,
r
Attitude Control of a Satellite
State Feedback with Integral Control
For the system bu(t) +Ax(t)=(t)x
cx(t) =y(t)
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.
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
0
r)dt-(y(t)z(t)……….………….…(i)
Continued…….Since z(t) satisfies the differential equation,
rtcxrty )()((t)z ……….………….…(ii)
It is easily included by augmenting the original system as follows:
rub
z
x
c
A
z
x
1
0
00
0
Since “r” is constant, in the steady – state , provided that the system is stable.
0,0 zx
This means that the steady -state solutions & must satisfy the equation
ss zx , su
s
s
su
b
z
x
c
Ar
00
0
1
0
……….………….…(iv)
……….………….…(iii)
Continued…….Therefore,
From equation (iii) & (iv)
)(00
0s
s
suu
b
zz
xx
c
A
z
x
….………….…(v)
Now define new state variables as follows , representing the deviations from the steady - state:
s
s
zz
xxx~
)(~suuu
In terms of these variables,
….………….…(vi)
ubxAx ~~~….………….…(vii)
Continued…….Where,
0
0
c
AA
0
bb
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 as the desired state.0~x
We assume that an asymptotically stable solution to this problem exists & is given by-
x~-k=u
Partitioning “k” appropriately & using eqns. (vi)
][ ip kkk
Continued…….
s
s
ipszz
xxkkuu
)()( sisp zzkxxk
The steady-state terms must balance, therefore
t
ipip dtrtykxkzkxku0
))((
The control, thus consist of proportional state feedback & integral control of output error.
At steady-state, ; thereforeox~
0)(limt
tz rty )(limt
or
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.
Continued…….
ik
pk
r
w
u y
x
State feedback with integral control
Numerical Problem
Design the integral control system with a constant reference command signal.
Given
)3(
1
)(
)(
ssU
sY5n 5.0&with
Solution:-
Characteristic equation will be- 02552 ss
Plant equation will be- uxx 3xy
Augmenting the plant state “x” with the integral state “z” defined by the equation.
t
0
r)dt-(y(t)z(t)
Continued…….
ruz
x
z
x
1
0
0
1
01
03
We get
Since
s
s
zz
xxx~
)(~suuu
State equation becomes uxx ~
0
1~
01
03~
Continued…….
We can find “k” from
2550
1
01
03det 2 ssksI
255)3( 2
21
2 ssksksAfter solving
Therefore ][252 ip kkk
The control
t
ip dtrtyxzkxku0
))((252
Continued…….
r yw
25 s
1
2
State feedback with integral control
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”.