dynamic system analysis and simulation_19

8/12/2019 Dynamic System Analysis and Simulation_19

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National Chiao-Tung University Department of Electrical Engineering

Handout of

By Prof. Yon-Ping Chen

99

( ) ( ) ( ) ( ) ( )( )ttttt yyLBuxAx ++=& , ( ) 0x =0 (6)

( ) ( )tt xCy = (7)

where ( )tx is the estimated state, ( )ty is the estimated output, and the matrix Lis

purposely introduced as a key design element. Define the estimation error as

( ) ( ) ( )ttt xxx~ = (8)

then from (1) and (6) we have

( ) ( ) ( ) ( ) ( )tttt x~LCAx~CLx~Ax~ ==& , ( ) 00 xx~

= (9)

Clearly, if the eigenvalues of LCA are all located in the left-half complex plane,

then the estimation error will approach zero as t, i.e.,

( ) ( ) ( ) 0xxx~ = tt

ttt (10)

or

( ) ( )

tt

tt xx (11)

In other words, the estimated state is gradually approximated to the actual statex(t) as

time increases and thus, the state feedback in (4) can be replaced by the following

control

( ) ( )tt xKu = (12)

which will drive the system statex(t) fromx0to 0. The whole system block diagram is

shown in Figure-1.

Now, lets focus on the design of the matrixL. It is known that the eigenvalues

of LCA can be obtained by solving the following polynomial:

( ) 0= LCAI (13)

Since the determinant of a matrix His the same as the determinant of its transpose,

i.e., THH = , the polynomial (13) can be rewritten as


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Handout of


100

( ) ( ) 0== TT LCAILCAI T (14)

Based on the pole-placement concept given in (5), we can assign npoles {p1,p2,,pn}

first and solveLfrom the following equation:

( ) ( )( ) ( )nTTT ppp = L21LCAI (15)

by using the instruction L=(place(A,C,p)) in Matlab. One thing has to emphasize

that the existence ofKin (5) requires that the system must satisfies the condition (3).

Similarly, the existence of L in (15) requires that the system must satisfies the

following condition

[ ] nrank TnTTTTTT = CACACAC 12 L (16)

Due to the fact that ( ) ( )Trankrank HH = , (16) can be rearranged as

nrank

n

=

1

2

CA

CA

CA

C

M

(17)

which is the condition for a system to be observable.

+

A

B Cx

+

A

B Cx

L +

y

y

Ku

Figure-1


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Handout of


101

Lets consider the following example which has been introduced in the

state-feedback control:

( )

( )

( )

( )

( )

( )

( )

( )

( )

( )321

321321444 3444 21321&

&

&

&

&

+

=

t

tu

tu

t

tx

tx

tx

tx

t

tx

tx

tx

tx

u

BxAx

2

1

4

3

2

1

4

3

2

1

1

0

0

0

0

1

0

0

0110

1101

1010

0101

(18)

with initial condition [ ]T12450 =

x . It is easy to check that

412 = BABAABB nrank L (19)

which ensures the system can be controlled by state feedback u(t)=kx(t). Based on

the pole-placement method, we assign four eigenvalues for the matrixABK

[ ]42221 += jjp (20)

and solveKby the instruction K=place(A,B,p) in MATLAB which results in

= 3160

17011

K (21)

Hence, the control input is

( ) ( ) ( ) ( )

( ) ( ) ( ) ( )txtxtxtu

txtxtxtu

4322

4311

36

711

+=

= (22)

However, if the system state x(t) is not obtainable and only the following output is

measurable:

( ) [ ]

( )( )

( )

( )321

43421

=

t

tx

tx

tx

tx

ty

x

C4

3

2

1

1001 (23)

then we have to estimate the system state by the Luenberger observer as depicted in

Figure-1. To use the Luenberger observer, first we have to check the observability of

the system by the condition given in (17), which results in


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Handout of


102

4

3

2 =

CA

CA

CA

C

rank (24)

Clearly, the system is observable and thus we can estimate the system state by the use

of Luenberger observer in (6) and (7).

To determine the matrixL, lets choose four poles for the matrixALC, which are

[ ]855554 += jjq (25)

1

s

x4^

1

s

x4

1

s

x3^

1

s

x3

1

s

x2^

1

s

x2

1

s

x1^

1

s

x1

Scope1

Scope

-347.2

Gain8

3

Gain7

6

Gain6

7

Gain5

11

Gain4

82Gain3

20.8

Gain2370.2

Gain1

Figure-2


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Handout of


103

Further adopting the instruction L=(place(A,C,q)) in Matlab, we have

=

2347

82

820

2370

.

.

.

L (26)

By the use of SIMULINK, the block diagram of the whole system is shown in

Figure-2. The numerical result of the four state variables can be obtaine from the

Scope block and shown as below:

The numerical result of the four estimation error can be obtaine from the Scope1

block and shown as below:


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Handout of


104

Clearly, the system is stabilized since all the state variables are driven to approach the

origin under state-feedback control based on Luenberger observer.

P.1 Consider the following system:

( )

( )

( )

( )

( )

( )

( )

( ){

+

=

tu

t

tx

tx

tx

tx

t

tx

tx

tx

tx

bxAx

1

1

1

0

0110

1100

2001

0100

4

3

2

1

4

3

2

1

32144 344 21321&

&

&

&

&

( ) [ ]

( )

( )

( )( )321

43421

=

t

tx

tx

tx

tx

ty

x

C4

3

2

1

0101

with initial condition [ ]T12010 0 ==

xx . If the system statex(t) is not

measurable, please design thee state-feedback control based on the Luenberger

observer.

dynamic system analysis and simulation_19

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