Download LQR Design: A Comprehensive Guide with Examples and Applications and more Lecture notes Teaching method in PDF only on Docsity!
Updated 2 October 2016 Prof. Mohamed ZribiHandout 7: LQR DesignFall 2016Lumped Systems TheoryEE 510:
EE 510
Lumped Systems Theory
Prof. Mohamed Zribi
Design Using LQR
to Previously, compensators were designed
satisfy
requirements
for
steady
state
Now,not try to optimize any criteria.or closed loop locations, however we diderror, transient response, stability margins,
we
will
present
a
technique
that
is result in optimal solutions. This controller
called
Linear
Quadratic
Regulator
x Consider the following system,(LQR).
A
x
Bu
y
Cx
A
state
feedback
controller
will
be
and the control are zero. Therefore,designed such that the states, the output
0 as0 as0 as
x
t
y
t
u
t
The linear control law is such:
(^) u
Kx
where
K
(^) is the gain matrix.
EE 510
Lumped Systems Theory
Prof. Mohamed Zribi
2
variable w and a control input v such that: Consider the first-order plant with state Example 1:
w
w
v
the state We wish to design the control v to drive
(^) w (^) to 2.
In the steady state,
3
0
3
e
e
e
e
w
v
v
w
If
(^2)
6
d
d
w
v
.
Let
2
x
w
and
6
u
v
then the system
3
w
w
v
can be written as,
x
x
u
In
this
case,
the
desired
condition
is
0
x
u
.
EE 510
Lumped Systems Theory
Prof. Mohamed Zribi
Instead of trying to find
K
(^) of the controller
u
Kx
such
that
we
will
achieve
specified
closed
pole
locations
(pole
placement technique), we want to find a
K
criterion (cost function) that will minimize a specified performance
J
The performance criterion,
J,
(^) is such,
0
(^21)
)
(
dt
Ru
u
Qx
x
J
T
T
The
matrices
Q
and
R
are
symmetric
matrices.
and R
T
T
Q
Q
R
The matrix
Q
is usually called the state
The matrixweighting matrix.
Q
is positive semi-definite (a
matrix is positive semi-definite if
it has
The matrixnon-negative eignvalues).
R
is usually called the control
weighting matrix. The matrix
R
(^) is positive
definite (a matrix is positive definite if
it
has positive eignvalues).
EE 510
Lumped Systems Theory
Prof. Mohamed Zribi
6
x^ Consider the first order system,^ Example 3:
ax
bu
where
(^) a , b , u (^) and
(^) x (^) are scalars.
The cost function is such,
0
2
2
(^21)
)
(
dt
ru
qx
J
J
(^) represents the weighted sum of energy of
If the state and control.
r is very large compare to
q , then the
thatcontrol is penalized heavily, which means
the
control
will
diminish
at
the
If the control law.and amplifier gains needed to implementwill results in smaller motors, actuators,expense of larger values of the state. This
q is very large compare to
r , then the
Q of larger values of the control.that the state will diminish at the expensestate is penalized heavily, which means
and
R
represent respective weights on
different states and control channels.
EE 510
Lumped Systems Theory
Prof. Mohamed Zribi
x^ Consider the system,^ Example 4:
A
x
Bu
y
Cx
With
0
(^21)
)
(
dt
Ru
u
Qx
x
J
T
T
If
1
0
0
100
Q
and
R
r
Then,
x Qx
u Ru
x
x
u
T
T
(^12)
(^22)
2
controllingstate, we are putting more emphasis onBy putting a larger weight on the first
this
state
and
restricting
its
fluctuations.
EE 510
Lumped Systems Theory
Prof. Mohamed Zribi
8
x^ Consider the first order system,^ Example 5:
ax
bu
where
(^) a=-
(^) b=
u (^) and
(^) x (^) are scalars.
The cost function is such,
(^1)
(^2)
2
2 (^0) (
)
J
qx
u dt
(r=1)
Since
(^) u
Kx
, then
3
3
(
x
x
u
x
Kx
K
x
(^)
The solution of the above equation is:
(
( )
(0)
K
t
x t
e
x
Therefore,
(^1)
(^1)
1
(^2) (^2) (^2) (^2) (^2) (^2)
2
(^2)
(^2)
2
(^0)
(^0)
0
( ) ( ) ( )
J
qx
u dt
qx
K x
dt
q
K
x dt
(^1)
1
(^2)
2(
(^2)
(^2)
(^2)
2(
(^2)
2
(^0)
0
2
2
(
)
(0)
(
)
(0)
(0)
4(
K t
K t
J
q
K
e
x
dt
q
K
x
e
dt
q
K
x
K
where we assumed that
3
0
K
.
EE 510
Lumped Systems Theory
Prof. Mohamed Zribi
To minimize J for fixed q and
(0)
x
, we
compute
/
J
K
and set it to zero. This
gives,
(^2) 6
0
K
K
q
.
For
a
minimum,
we
require
that
(^2)
2
/
0
J
K
. Because q > 0, this implies
that
54
6
0
18
2 q
K
q
or
3
0
K (^)
Clearly, we got the condition
3
0
K
,
Note thatis stable.must be such that the closed-loop systemTherefore, the value of K that minimizes Jwhich is the condition for stability.
(^2)
6
0
K
K
q
2
(
9
0
3
9
K
q
K
q
condition Therefore, one of the roots will satify the
3
0
K
.
This
root
is
3
9
K
q
EE 510
Lumped Systems Theory
Prof. Mohamed Zribi
12
Assumption 1:
(^) The system is stabilizable
(weaker
condition
than
completely
x^ Consider the first order system,^ Example 7: controllable).
x
u
The performance index is such,
(^1)
(^2)
2
2 (^0) (
)
J
x
u dt
The controller
(^) u
Kx
This will lead to
(^) u (^)
(^) to drive x to zero.
This
is
a
problem
and
that’s
why
we
assume that R > 0.
EE 510
Lumped Systems Theory
Prof. Mohamed Zribi
x^ Consider the first order system,^ Example 8:
x
u
The performance index is such,
(^1)
(^2)
2
2 0 (
)
J
x
u dt
The controller
(^) u
Kx
This
will
lead
to
u (^)
to^
drive
x
to
Thisinfinity.
is
a
problem
and
that’s
why
we
assume that
0
Q (^)
EE 510
Lumped Systems Theory
Prof. Mohamed Zribi
14
Few
conditions
are
necessary
and
ofsufficient for the existence and uniqueness
the
optimal
controller
that
will
- The system is stabilizable.These conditions are:asymptotically stabilize the system.
and R
T
T
Q
Q
R
and
0 and
0
Q
R
- The pair
( , )
A F
is completely observable
root of(or detectable) with F being any square
Q
T
Q
F F
Example 9: a)
(^1)
0
(^1) (^0) then
(^0)
0
Q
F
q
q
is a choice for F.
b)
100
(^0) then
10
0
(^0) 0
Q
F
c)
(^0) (^0) then
(^0) 4
(^0) 16
Q
F
d)
10 (^3)
11/
13
3 / 13
then
(^3) (^1)
3 / 13
2 / 13
Q
F
EE 510
Lumped Systems Theory
Prof. Mohamed Zribi
selection of the design matricesThe closed loop poles will depend on theperformance index).the unstable states are “observed” by theappear in the performance index (i.e., all Condition 3 says that all the unstable states Remark:
Q
(^) and
R
the system is stabilizable,The poles will always be stable as long as
0 and Q
0
T
T
R
R
Q
with
( , )
A F
completely observable.
It
turns
out
that
the
controller
that
minimizes the cost function
J
(^) is such,
u
Kx
K
R B P
T
and
1
where
P
is the solution of the following
A algebraic riccati equation (ARE),
P
PA
PBR
B P
Q
T
T
1
symmetric matrix.Note that the matrix P is a positive definite
EE 510
Lumped Systems Theory
Prof. Mohamed Zribi
18
the matrix P should be constant and henceSince we are dealing with an LTI system,
0
P (^)
equation Therefore, the matrix Riccati differential
leads
to
the
Algebraic
Riccati
Equation
1
T
T
A
P
PA
Q
PBR
B
P
And
min
1
(0)
(0)
2
T
J
x
Px
The optimal gain Remark:
K
is found for a given
value of
Q
(^) and
R
, and the closed loop time
for responses are unsuitable, then new valuesresponses are found by simulation. If these
Q
and
R
are selected and the design is
repeated.
EE 510
Lumped Systems Theory
Prof. Mohamed Zribi
x^ Consider the first order system,^ Example 10:
u
zero We would like to keep the output x near
while
minimizing
the
performance
index
(^1)
(^2)
2
2 0 (
)
J
x
u
dt
The ARE implies that, Clearly, a=0, b=1, q=1 and r=1.
(^2)
(^2) 1
1
(
q
p
p
p
p
Then the controller is
1
T
u
Kx
R
B
Px
px
x
Note that the minimum value of J is such0; the closed loop pole is located at -1. Note that the open loop pole is located at
2
min
1
(0)
2
J
x
EE 510
Lumped Systems Theory
Prof. Mohamed Zribi
20
Consider the earlier first order system, Example 11:
3
x
x
u
zero We would like to keep the output x near
while
minimizing
the
performance
index
(^1)
(^2)
2
2 0 (
)
J
qx
u
dt
The ARE implies that, Clearly, a=-3, b=1, and r=1.
(^2)
2
3 3 6 0 3 9 (
p p q p p p q p q p
Then the controller is:minimize J.the one we obtained when we attempted to Note that the above equation is the same as
(^1)
( 3
9
)
T
u
Kx
R
B
Px
q
x
3; the closed loop pole is located at - Note that the open loop pole is located at -
(^9) q
.
Note that the minimum value of J is such
2
min
1 ( 3
9
)
(0)
2
J
q x
EE 510
Lumped Systems Theory
Prof. Mohamed Zribi
21
Consider the following system, Example 12:
,
x
Ax
Bu
y
Cx
where,
0
1
0
2
C
1
0
(0)
0
0
1
2
A
B
x
Design a controller,
u
Kx
, such that
the cost function,
0
2
(^12)
(^21)
)
(
dt
u
x
J
Q is minimized.
and
R
are easily obtained from the cost
function,
0
0
0
1
Q
and
R
r
The matrix
Q
(^) can be factored such that,
EE 510
Lumped Systems Theory
Prof. Mohamed Zribi
24
The controller u is such
(^1)
2
2
u
x
x
The minimum cost is: min
2
1
1
1
2
(0)
(0)
2
2
0
2
2
2
1
2
T
J
x
Px
The closed loop system matrix is such,
2
1
1
0
BK
A
A c
The charc. equation of
A
c is such,
det(
sI
A
c
s
s
(^2)
s^ Hence the closed loop poles are located at
j
1 2
,
[K,P,E]=lqr(A,B,Q,R)K=lqr(A,B,Q,R)The Matlab code is such:and it has a damping ratio of 0.707.Therefore the system has been stabilized
EE 510
Lumped Systems Theory
Prof. Mohamed Zribi
x^ Consider the following system,^ Example 13:
A
x
Bu
where,
1 0
0
0
1
0
B
Design a controller, A
u
Kx
, such that
the cost function,
(^1)
(^2)
(^2)
2
(^1)
2
2 0 (
)
J
x
qx
ru
dt
Q is minimized.
and
R
are easily obtained from the cost
function,
1
0
0
Q
q
and
R
r
The matrix
Q
(^) can be factored such that,
1
0
1
0
1
0
0
0
0
T
Q
F
F
q
q
q
EE 510
Lumped Systems Theory
Prof. Mohamed Zribi
26
theNow we need to check the conditions for
existence
and
uniqueness
of
the
1)The controllability matrix,optimal controller.
0
1
1
0
AB
B
C m
has
full
rank,
thus
the
system
is
- controllable.
The
matrix
R
R
is
symmetric
positive
definite.
The
matrix
Q
is
3)The observability matrix is such,symmetric positive definite.
1
0
0 0
1
0
0
m
F
q
O
FA
The
observability
matrix
has
full
rank,
thus
( , )
A F
is observable.
EE 510
Lumped Systems Theory
Prof. Mohamed Zribi
A unique optimal controller.Algebraic Riccati Equation will result in aAll the conditions are satisfied, thus the
P
PA
PBR
B P
Q
T
T
1
0
0
0
1
0
0
1
0
0
1
0
0
3
2
2
1
3
2
2
1
p
p
p
p
p
p
p
p
(^1)
(^2)
(^1)
(^2)
1
(^2)
(^3)
(^2)
3
p
p
p
p
r
p
p
p
p
Thus we get the following three equations,
(^2)
1
2
1
(^1)
(^2)
2 3
1
(^2)
3
r
r
p p r
p p
p
q
p
Therefore the matrix
P
(^) is such,
2
,
2
b
q
b
P
b
r
b
b
b
q
EE 510
Lumped Systems Theory
Prof. Mohamed Zribi
30
theNow we need to check the conditions for
existence
and
uniqueness
of
the
1)The controllability matrix,optimal controller.
0
1
1
1
m
C
B
AB
has
full
rank,
thus
the
system
is
- controllable.
The
matrix
R
R
is
symmetric
positive
definite.
The
matrix
Q
is
3)The observability matrix is such,symmetric positive definite.
1
0
0
1
0
1
0
1
m
F
O
FA
The
observability
matrix
has
full
rank,
thus
( , )
A F is observable.
EE 510
Lumped Systems Theory
Prof. Mohamed Zribi
31
A unique optimal controller.Algebraic Riccati Equation will result in aAll the conditions are satisfied, thus the
P
PA
PBR
B P
Q
T
T
1
(^1)
(^2)
(^1)
2
(^2)
(^3)
(^2)
3
0 0 0 1 1 0 1 1 0 1 0 0
p
p
p
p
p
p
p
p
(^1)
(^2)
(^1)
2
(^2)
(^3)
(^2)
3
p
p
p
p
p
p
p
p
Therefore the matrix
P
(^) is such,
2
1
1
1
P
The gain matrix
K
(^) is such,
(^1)
2
1
(^2)
3
(^2)
3
0
1
1
1
T
p
p
K R B P p p
p
p
The controller is such:
1
(^2) (^1)
(^3) (^2)
(^1)
2
(
)
u
Kx
r
p x
p x
x
x
It r=1, then
(^1)
2
2
u
x
q x
EE 510
Lumped Systems Theory
Prof. Mohamed Zribi
32
x^ Consider the following system,^ Example 15:
A
x
Bu
where,
0
1
0
1
1
1
A
B
Design a controller,
u
Kx
, such that
the cost function,
(^1)
(^2)
2
1
2 0 (
)
J
x
u
dt
We obtain: is minimized.
P
K
The closed loop poles are found to be: 1,
s
j
EE 510
Lumped Systems Theory
Prof. Mohamed Zribi
x^ Consider the following system,^ Example 16:
A
x
Bu
where,
0
1
0
0
0
0
1
0 ,
[1 0
0], D=
0
2
3
1
A
B
C
by Assume that the control signal u is given
(^1) (^1) (^2) (^2) (^3) (^3) (^1)
( )
u
k x
k x
k x
k r t
cost function, Design an optimal controller, such that the
(^1)
(^2)
(^2)
(^2)
2
(^1)
(^2)
3
2 0 (
)
J
x
x
x
u
dt
that the reference input r(t) is zero.In determining the control law, we assume is minimized.
EE 510
Lumped Systems Theory
Prof. Mohamed Zribi
36
EE 510
Lumped Systems Theory
Prof. Mohamed Zribi
LQR for Discrete Time Systems
Consider the following system, (
(
)
( )
x k
A x
k
Bu k
Design a controller,
( )
( )
u k
Kx
k
, such
that the cost function,
0
1 [ ( ) ( ) ( ) (
)]
2
T
T
k
x
k Qx
k
u
k Ru k
J
Few is minimized.
conditions
are
necessary
and
ofsufficient for the existence and uniqueness
the
optimal
controller
that
will
- The system is stabilizable.These conditions are:asymptotically stabilize the system.
and R
T
T
Q
Q
R
and
0 and
0
Q
R
- The pair
( , )
A F
is completely observable
root of(or detectable) with F being any square
Q
T
Q
F F
EE 510
Lumped Systems Theory
Prof. Mohamed Zribi
38
The optimal solution is: (
)
( )
u k
Kx k
with
1
[
]
T
T
K
B
PB
R
B
PA
where
P
is the solution of the discrete
algebraic Riccati equation (DARE):
1
[
]
T
T
T
T
A
PA
P
A
PB B
PB
R
B
PA
Q
EE 510
Lumped Systems Theory
Prof. Mohamed Zribi
Consider the following system, Example 17: (
( )
( )
x k
Ax
k
Bu k
( )
y
Cx
k
where,
0
1
0
10
0
2
3
1
A
B
C
Design a controller,
( )
( )
u k
Kx k
, such that the
cost function,
0
2
(^2)
2
(^1)
2
0
1 [ ( ) ( ) ( ) (
)]
2
1
[
( ) 10
( ) 100
( )]
2
T
T
k
k
x
k Qx
k
u
k Ru k
x k x k u k
J
[K,P,E]=dlqr (A,B,Q,R)Using the Matlab command is minimized.
