Linear system theory: controllable, uncontrollable, observable and unobservable

7/28/2019 Linear system theory: controllable, uncontrollable, observable and unobservable.

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NATIONAL CHENG KUNG UNIVERSITY

Department of Mechanical Engineering

LINEAR SYSTEM

HOMEWORK 4

Instructor: Prof. Szu – Chi Tien

Student: Nguyen Van Thanh

Student ID: P96007019

Department: Inst. of Manufacturing & Information Systems

Class: 1001- N154000– Linear System

November 8, 2011

http://class-qry.acad.ncku.edu.tw/qry/classmate.php?syear=0100&sem=1&co_no=N164400&class_code=

http://class-qry.acad.ncku.edu.tw/qry/classmate.php?syear=0100&sem=1&co_no=N164400&class_code=



Linear System Theory Page 1

Contents

Problem 1 ....................................................................................................................... 2

Problem 2 ..................................................................................................................... 10




Problem 1

Given a linear time-invariant system for the three-mass-spring system described in

Example 3 of Section 5.5.3,

Note that in this problem we have equal forcesu(t) applied simultaneously to masses

m1 andm3 but in the opposite directions. The measured outputy(t)is the position of

massm2.

1. Is the system controllable? Explain.

Consider the controllability matrix

= [ 2 3 4 5]

CM =




Rank (CM) =2 <6. So the controllability matrix is not full rank, hence the system is

not controllable.

2. Identify the modes that are controllable and uncontrollable. Provide a physical

meaning to each of the controllable modes (if any).

In order to identify which modes are controllable or not. Firstly, we calculate the

eigenvalues of matrix A and then use PHB rank Test.

Eigenvalues of matrix A,

Called V1, V2, V3, V4, V5, V6 are corresponding to these six eigenvalues.

1 =

0.20410.3536 0.40820.70710.2041

0.3536

,2 =

0.20410.35360.40820.70710.2041

0.3536

,3 =

0.5774

0.00000.57740.00000.57740.0000

4 =

⎣0.50000.5000

0.0000.000 0.5000

0.5000 ⎦,5 =

⎣0.50000.5000

0.0000.000

0.50000.5000 ⎦

,6 =

0.57740.00000.57740.00000.57740.0000

PHB Test

([1 ]) = 5, this mode is uncontrollable.([2 ]) = 5, this mode is uncontrollable.

([3 ]) = 5, this mode is uncontrollable.

([4 ]) = 6, this mode is controllable.




([5 ]) = 6, this mode is controllable.

([6 ]) = 5, this mode is uncontrollable.

Two modes controllable, the eigenvectors are V4 and V5.

For V4, the displacement and the velocity of the mass 1 and the mass 3 are symmetric

about the mass 2 and the mass 1 and the mass 3 are going to close to the mass 1.

For V5, the displacement and the velocity of the mass 1 and the mass 3 are symmetric

about the mass 2 and the mass 1 and the mass 3 are going to be far away from the mass

1.

For the others, we cannot find any input that brings system to these states.

3. Can you find a control inputu(t) that bring the system initial states to the following

final states x(tf )?

The controllability matrix CM has rank 2, hencedim( ) = 2.

We pick up two independent column vectors of the controllability matrix, for example,

we pick up the first tow column vectors of the controllability matrix




1→2 =

01000

1

1000

1

0

(a) We can easily see that = 1, that means is in the range space of 1→2,(or in the controllable subspace) hence we can find a control input u(t) that bring

the system from initial state to this final states.

Assume that (0 = 0) = 0. A control input is given below,

ut = - ( ( i / ( 2*exp( conj ( t ) * i - ( 101*i ) / 10) ) - ( exp( conj ( t ) * i -

( 101*i ) / 10) *i ) / 2) *( 100*si n( 101/ 5) + 2020) ) / ( 50*cos( 101/ 5) -

25*cos( 101/ 5) 2 - 25*si n( 101/ 5) 2 + 10176) - ( ( 100*cos( 101/ 5) -

100) *( ( ( conj ( t ) * i - ( 101*i ) / 10) * i ) / ( 2*exp( conj ( t ) * i -

( 101*i ) / 10) *( conj ( t ) - 101/ 10) ) + ( exp( conj ( t ) *i -

( 101*i ) / 10) *( conj ( t ) * i - ( 101*i ) / 10) * i ) / ( 2*(conj ( t ) -

101/ 10) ) ) ) / ( 50*cos( 101/ 5) - 25*cos( 101/ 5) 2 - 25*s i n( 101/ 5) 2 +

10176) ;

Figure 1.1 A control input signal




Figure 1.2 Displacement of system from t0 to tf .

Figure 1.3 Sate of mass 1






Through fours figures, Fig.1.2 to Fig.1.5, we can see that, mass2 doesn’t move, mass 1

and mass 3 are symmetric about mass 1.




(b) Check ([1→2 () ] ) = 3 > 2, that means is not in the controllable

subspace, hence we cannot find a control input u(t) that bring the system from

initial state to this final states.

(c) Check ([1→2 () ] ) = 3 > 2, that means is not in the controllablesubspace, hence we cannot find a control input u(t) that bring the system from

initial state to this final states.

4. Can you find a similarity transformation that places the system into the controllable

form? If no, explain why not.

No, because the system is not controllable.

5. Can you find a similarity transformation that places the system into the controller


No, because the system is not controllable.

6. Is the system observable? Explain.

Rank (O) =4 <6, hence the system is unobservable.

7. Identify the modes that are observable and unobservable. Provide a physical

meaning to the unobservable modes (if any).




Use PHB rank test

Eigenvalues of matrix A,

PHB Test

1 = 6, this mode is observable.



4 = 5 < 6, this mode is unobservable.

5 = 5 < 6, this mode is unobservable.


Through the PHB rank test, we find that the modesλ 4=i and λ

5=-i are the

unobservable modes associated with eigenvectors

4 =

0.50000.50000.0000.000 0.5000

0.5000

,5 =

0.50000.50000.0000.000

0.50000.5000

In these modes, the masses m1and m3oscillate symmetrically about the mass m2

which stays stationary; hence the sensor ysen

(t) placed on mass m2perceives no

output, ysen

(t)=0.




8. Can you find a similarity transformation that places the system into the observable


No, because the system is unobservable.

9. Can you find a similarity transformation that places the system into the observer


No, because the system is unobservable.

Problem 2

Recall that all the units in Problem 1 are respectively m and m/sec for the mass

positions and velocities, and the applied force u(t) in N. For some odd reasons, your

high-level manager wants you to present the state-space model of the three-mass-spring

system given in Problem 1 to the company executives in the following units for the

states x(t), inputu(t) and outputy(t):

•x1(t)in units of inches, x2(t)in units of inch/sec.

•x3(t)in units of feet, x4(t)in units of ft/min.

•x5(t)in units of miles, x6(t)in units of mile/hr.

u(t)in units of lbs,•y(t)in units of feet.

1. What are your new state model matrices? , , ,?

Assume that, = , = , = . The transformation:

� = + = + → �−1 = −1 + −1−1 = −1 + −1 → � = −1 + −1 = −1 + −1




� = −1 + −1 = −1 + −1 → � = + = +

= −1

=

−1

= −1 = −1

Unit conversion:

1 m =1/0.0254 inches; 1 m/s =1/0.0254 inches/s; 1 m =3.2808 ft;

1 m/min =3.2808 * 60 ft/min; 1 m =1/1609 miles; 1 m/s =3600/1609 miles/hr

1 N =0.2248 lbs; 1 m =3.2808 ft;

Hence,

1 = 10.02541; 2 = 1

0.02542; 3 = 3.28083 4 = 3.2808 ∗ 604; 5 =1

1609.3442; 6 =

3600

1609.3446 = 0.2248; = 3.2808

→ =

1

0.02540.0000 0.0000 0.0000 0.0000 0.0000

0.0000

1

0.0254 0.0000 0.0000 0.0000 0.0000

0.00000.00000.00000.0000

0.00000.00000.0000

0.00000

3.28080.00000.0000

0.00000

0.00003.2808 ∗ 60

0.00000.00000

0.00000.0000

1

1609.3440.0000

0.00000.00000.00003600

1609.344

T is diagonal matrix.

= 0.2248;

= 3.2808

→ = −1 =

0.0000 1.0000 0.0000 0.0000 0.0000 0.00001.000 0.0000 12.000 0.0000 0.0000 0.00000.00004.99990.00000.0000

0.00000.00000.0000

0.00000

0.0000120.000.00000.6818

0.01670.00000.0000

0.00000

0.00003168000.00003600

0.00000.00000.00030.0000




= −1 =

0.00008.85040.00000.00000.0000

0.5029

= −1 = [0 0 1 0 0 0] = −1 = 0

2. What are the system eigenvalues of your new state model? Are they the same as the

original model?

The eigenvalues of matrix

:

They are the same to the eigenvalues of matrix A.

3. Is any of the system controllability properties changed? If yes, explain.

Yes. Since rank of the new controllability matrix has rank 3 more than the original

controllability matrix.

4. Is any of the system observability properties changed? If yes, explain.

No. Since rank of the new observability matrix has rank 4.

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Linear system theory: controllable, uncontrollable, observable and unobservable