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College of Engineering Department of Electrical Engineering
Lecture Notes 6
STATE FEEDBACK DESIGNT RACKING P ROBLEM
Prof. Rached Dhaouadi
E-mail: [email protected]
Tel: 06-515-2927
mailto:[email protected]:[email protected]:[email protected]:[email protected]
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STATE FEEDBACK DESIGNTracking Problem
Consider an LTI system represented by its state equations
u D X C y
u B X A X
y+u
+
++ B
X X C
A
D
State space representation of a system
y+
_
r +u
+
++
K
B
X X C
A
D
Closed loop with state feedback
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We will consider the case of tracking a reference input ( 0r
)and would like to study the steady state characteristic of
thesystem and see how well the output follows the reference.
For the tracking problem case, the control law is
X K r u .Substitution of this control law in the state equation
yields the
state-space representation of the closed-loop system
r B X K B A
X K r B X A X
where
K B A ACL , B BCL is the closed loop system matrix, and r is the
system input.
The output equation will be
r D X K DC X K r D X C Y where
K DC C CL , D DCL For good tracking performance we want r y )(
as t .We can check this performance in the frequency domain bythe
final value theorem:
)()(lim)(lim)(lim)( s R sG s s sY t y y CL0 s0 st
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for a unit-step reference s s R
1)(
)()(lim)( 0G sG y CLCL0 s
Therefore for good tracking performance the closed looptransfer
function from r to y should be approximately 1 at DC.
Closed loop transfer function:
CLCL1
CLCLCL D B A sI C sG
)(
Let 0 D , then
B AC B AC 0G 1CL1CLCL )( If 10GCL )( an extra gain N is used to
scale the referenceinput r so that
X K r N u
+
_
r u y+ +
++
K
N B X
C
A
D
X
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Now we have
r N B X K B A
X K r N B X A X
and the closed loop transfer function is
N B A sI C sG 1CLCL )( If we select
B BK AC B AC N
CL11
11
then, r y )( as t
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Example 1:
Consider the following system
2
1
2
1
2
1
x
x01 y
u1
0
x
x
23
11
x
x
1. Design a state feedback controller to place the closed loop
poles at -2, -3.
2. Design the required gain to have a good tracking
performance.
Solution:
From the previous section, first we found
25 K next
6
1111
B BK AC B AC N
CL
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Tracking Problem:
A formal way to compute N is to change the control law in
the new equivalent form
y _r
+ +++
++
u
+ X ss
N u
B X
C
A
D
K N x
U ss
X
x N is a vector gain used to modify the reference command rto
generate a feedforward state command ss X .
u N is a gain used to modify the reference command r togenerate
a feedforward control input to the system ssU .
u D X C y
u B X A X
In steady state we would like to have
r y
U u
X X
0 X
ss
ss
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ss ss
ss ss
U D X C r
U B X A0
Put these equations in matrix form:
r
0
U
X
DC
B A
ss
ss
The steady state values be computed as:
r 1
0
DC
B A
U
X 1
ss
ss
Let r N X x ss . and r N U u ss .
Then:
1
01
DC
B A
N
N
u
x
,
and the control law
KX r KN N
X r N K r N
X X K U u
xu
xu
ss ss
)(
)(
)(
xu N K N N
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Example 2:
Consider the system in example 1
2
1
2
1
2
1
x
x01 y
u1
0
x
x
23
11
x
x
1. Design a state feedback controller to place the closed loop
poles at -2, -3.
2. Design the required gain to have a good tracking
performance.
Solution:
From the previous section, first we found
25 K next
1
0
01
DC
B A
N
N
u
x
and the gains are
1,1
1 u x N N
xu N K N N 6 N
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Example 3:
Consider a system with the following transfer function
11
)( 2 s sG
1. Design a state feedback controller to stabilize the systemand
place the closed loop poles at -1, -2.
2. Design the required gain to have a good tracking
performance.
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Design of Servo Systems:
We consider the problem of designing a type-1 servo systemwhen
the plant involves an integrator.
- Assume that the control signal and the output signal
arescalars.
- Assume also that the output is equal to one of the
statevariables (for example y=x1).
Then, it can be shown that the tracking gain N is equal tok
1.
The state feedback control system can be arranged in
thefollowing configuration.
y=x 1
_r
+ +++
+
- u+
k 1 B X
C
A
D
K s
X
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X K r N u
n
2
1
n211
x
x
x
k k k r k u
n
2
1
n211
x
x
x
k k 0 xr k u
X K xr k u s11
Where,
n2 s k k 0 K The closed loop system is an asymptotically
stable
system. )( y will approach the constant value r (of thestep
input), and )( u will approach zero.
r k B )( X BK A0 )( X 1
r k B BK A )( X 11
