msw lecture 5

P(A SP) (A SP) T P (Q PSP) 0
PA c A c P Q c 0
T
PA A P PBR B P Q 0
Let A c A SP Q c Q PSP , then Riccati equation will be Note that
So the Riccati equation has a positive definite solution
p11 0 0 p11 p12 p11 p12 0 1 p11 p12 0 [1][0 1] 1 0 p p22 p12 p22 0 0 p12 p22 1 12 p12 2 2 1 p12 0 p11 p12 p22 0 2 p12 p22 0 p P 11 p12 p12 2 p22 1 2 1 p12 1 0 0 p22 0
The optimal control law: U * K * x ( 2 1) x
Example2
u
Controller
dt
x2
ቤተ መጻሕፍቲ ባይዱdt
x1
The model
x1 x2 x2 u


The matrix form
X AX BU

0 1 0 A B U u 0 0 1 The optimal control problem 1 min J ( X T QX U T U ) dt u 2 0
Choose the quadratic performance index 1 T J 0 (X Q X U T RU)dt 2 and the weighting matrices Q and R.
Check if the system controllable: Matrix D=[B AB A2B …An-1B] possesses full rank . If yes Solve the Riccati equation PA A T P PBR1 B T P Q 0 Calculate the feedback gain matrix K*=R-1BTP*
A c A SP A B(R 1 B T P) A BK
is the closed-loop system matrix.
Step 2. Select initial matrices P ( 0) then set
(1)
Ac Solve the equation
then obtain P (1) Step 3. Step k
Let
Explanation:
Suppose that there is a non-linear function
yi f ( x1 , x2 ,, xn )
f x1 x1
Xe
The Taylor Series expansion
f ( x1 , x2 , , xn ) f ( x1e , x2 e , , xne ) f x2 x 2

f (Xe ) 0
X ( x1 (t ), x 2 (t ) x n (t )) T X e ( x1e , x 2 e x ne ) T
Replacing xi(t) by
xi (t ) xi (t ) xie
df (X) f (X)
Replace derivations of functions by differentials of the functions, i.e.,
s.t
X AX BU
0 ( 0) R 1

Where
1 Q 0
Riccati Equation
PA AT P PBR 1 B T P Q 0
Judgment
rank D rankB AB 0 1 rank 2 1 0
Xe
f xn
x n
Xe
So
f ( x1 , x2 ,, xn ) f ( x1e , x2e ,, xne )
The above equation can also be written as:
where
K1 f x1 , K2
Xe
f ( xe ) f ( xe ) f ( xe ) x1 x2 xn x1 x2 xn
Modern Control Theory Lecture-5
Content:
I.5
Design of Linear Optimal Control
I.5 Design of Linear Optimal Control

I.5.1 The Flow-chart for Design I.5.2 Modeling of Linearized Control System I.5.3 Solution of Riccati Equation I.5.4 Optimal Fast Turbine Control for Large Turbo-generator
X f ( X ,U ) f (Xe ) 0 f ? Choose state variable X and control variable U
Step 3. Linearization (detailed in next page)

Step3 Linearization
X f (X )
P*
Design the optimal controller according to U*=-K*X(t)
Test the stability & dynamic performance of the controlled system
Example1
1 2 min J ( x u 2 )dt u 2 0 x xu
Step 1. Decomposition of Control Systems The controller and objective system.
r u Objective system x
controller
Step 2. Provide mathematical description to each component.
The flow chart
Input matrices A and B
Choose weighting matrices Q and R.
Given P(0) which is symmetric and positive definite with n×n dimension
Compute A(1)=A-BR-1BTP(1) Q(1)=Q+P(0)BR-1BTP(0) P(0)= P(1)
rankD=rank[B|AB|…|An-1B]=rank[1]=1 (A,B) is controllable, so Riccati Equation has a positive-definite solution In fact : P* 2 1
K * R 1 B T P * 2 1
X(t ) A(t ) X(t ) B(t )U(t ) X(t 0 ) X 0
2.The design steps of establishing the linearized state equations. (Detailed in next page)
2.The design steps of establishing the linearized state equations.
Solve the linear equation system P(1)Ac(1)+AcT(1)P(1) +Qc (1)=0 to obtain the solution P(1)
Tr P ( 0) Tr P (1) ? 10 6
Print P(1) Stop
2. Eigenvalue method.
X A S X T Λ Q A Λ
where S BR1BT P
Define Hamilton matrix
S A M Q AT

Proposition 1: The eigenvalues of Hamilton matrix M are symmetrical not only with respect to the real axis but also to the imaginary axis.

How to solve the optimal control law? Solution: Riccati equation: can be written as: A=(1) B=(1) Q=(1) R=(1)
PA AT P PBR 1 B T P Q 0
2P P 2 1 0
A SP( 0)
(1)
Qc
(1)
Q P ( 0) SP( 0)
(1)
P Ac
(1)
Ac
T (1)
P (1) Q c
0
P (1) Ac ,
( 2)
Qc( 2) P ( 2)
P ( k 2) P ( k 1)
Remark: The initial matrix A (c1) A SP(0) must be stable.
Xe
f xn
)T
Xe
I.5.3 Solution of the Riccati Matrix Equation T 1 T