Linear system theory: linearization

7/29/2019 Linear system theory: linearization

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NATIONAL CHENG KUNG UNIVERSITY

Department of Mechanical Engineering

LINEAR SYSTEM

HOMEWORK 3

Instructor: Prof. Szu – Chi Tien

Student: Nguyen Van Thanh

Student ID: P96007019

Department: Inst. of Manufacturing & Information Systems

Class: 1001- N154000 – Linear System

October 26, 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 3 ..................................................................................................................... 19 Appendix A Matlab code for Problem 1 Part 4 ........................................................... 24 Appendix B Matlab code for Problem 1 Part 5 ........................................................... 25 Appendix C Matlab code for Problem 2 Part 2 ........................................................... 26 Appendix D Matlab code for Problem 2 Part 3 ........................................................... 27 Appendix E Matlab code for Problem 2 Part 4 ........................................................... 28




Problem 1

Consider the nonlinear state equations (which were studied extensively in class)

1. Find A and B of the linearized state equations evaluated at the above given nominal

condition, where

Is the system stable?

Solution

Rewrite the nonlinear state equations

Solve

To find equilibrium states. Take one equilibrium state with

, -




We can write as form

Where,

[

]

0 1

01 0 1 01

The polynomial characteristic equation of matrix A

The real part of the eigenvalues of matrix A are zero. Hence, the system is marginally

stable.

E.g. 0

1 , see Fig. 1.1.




Figure 1.1 Perturbed state responses.

Solution

0 1

Solution for that system differential equations is

* + 0 1} 0 1




01 The perturbed state responses are shown in Fig. 1.2.

Figure 1.2 Perturbed state responses with , -

Solution




0 1 01

By taking the Laplace transform, we obtain

0 1 01 0 1 0 1

The perturbed state responses are shown in Fig. 1.3.




Figure 1.3 Perturbed state responses with

, -

4. Under the conditions of Part 2, solve the full nonlinear state equations for x(t ) (using

numerical integration) for 0≤t≤ 20 sec. Compare graphically the results to those

obtained from the linearized state equations x(t )≈ xeq

+ δx(t ). Is the linear state model

representative of the nonlinear system? (Note: see supplementary reading for

writing your own function to solve the nonlinear equation).

Solution

The nonlinear and linearized states responses are shown in Fig.1.4.




Figure 1.4 The nonlinear and linearized state responses, under the conditions of Part 2

We can see that, the linearized state responses oscillate around the nonlinear state

responses with amplitude = 0.005. Hence, the linearized model can represent the

nonlinear system.

Note: please see the Appendix A for Matlab code

5. Under the conditions of Part 3, solve the full nonlinear state equations for x(t ) (using

numerical integration) for 0 ≤t ≤ 20 sec. Compare graphically the results to those

obtained from the linearized state equations x(t ) ≈ xeq

+ δx(t ). Is the linear state model

representative of the nonlinear system? (note: see supplementary reading for writing

your own function to solve the nonlinear equation).




Solution

Figure 1.5 The nonlinear and linearized state responses, under the conditions of Part 3

The nonlinear and linearized states responses are shown in Fig.1.5. We can see that,

when t becomes larger, the linearized state responses go far away from the nonlinear

state responses. Hence, the linear state model is not representative of the nonlinear

system.

Note: please see the Appendix B for Matlab code

6. Is the linearized system controllable?

Solution

The linearized system is




0 1 01

We will use the controllability matrix to test the system is controllable or not.

Controllability matrix

, - 0 1

Controllability matrix C has full rank. Hence, the system is controllable.

Problem 2

Given the following linear time-invariant system

I will use Matlab to solve this problem.

1. Is the system controllable? Explain.

Solution

The controllability matrix

, - 0 1

The controllability matrix C has full rank. Hence, the system is controllable.




2. Find a control input u(t) that bring the system states to () , -, where, . Confirm your results with simulation of the responses for 0 ≤ t

≤ 15 sec.

Solution

From Par1, the system is controllable, thus the controllability Grammian matrix is

invertible. The controllability Grammian matrix

Given x(0) and x(tf ), we choose a control input u(t)

, -

(

) . (

)/

( ) ( ) 0 1

( ) ∫




We can find the response through




( ) ∫




Figure 2.1 Input and state responses (tf = 10)

Fig. 2.1 Shows the control input and the state responses.

From the state responses, we have

0

1 So, we can find a input to make x(t=10) = xf, hence the system is controllable.

Note: the Matlab code is shown in the appendix C.




3. Repeat Part 2 where t f

=1 sec. Confirm your results with simulation of the responses

x(t) for 0≤ t ≤ 15 sec. How does this control input u(t ) compare with the previous

control obtained in Part 2? (i.e., higher or smaller control input).

Solution

Similar to Part 2, so I will directly show the figure.

Figure 2.2 Input and state responses (tf = 1)

From the state responses, we have

01




So, we can find a input to make x(t = 1) = xf , hence the system is controllable. And, we

can see that we need higher control input compare to Part 2 to control system. Hence, if

a system is controllable, we want to bring this to a given tf in a shorter time, we need

more input.

Note: the Matlab code is shown in the appendix D.

4. Find a totally different solution to the control input u(t ) that solves Part 2. That is,

the solution u(t )in Part 2 is NOT unique. For example, in digital control let

Where u1, u

2and t

1are constant parameters (to be determined) that provide another

control input u(t ) for the solution of Part 2.

Solution

If we can find a different input

to control the system as we want, and u1, u2, t1 are all constants, then the solution (or

responses) x(t) can be represented as

() . ( )/ ∫ . ( )/

∫ . ( )/




01 ∫ . ( )/

∫ . ( )/

() . ( )/

01

Using Matlab to solve this equation, choose t1 = 5 sec, we obtain

for

for




Figure 2.3 Input and state responses

A control input

for all , - and bring this system from initial sates to

final states, since x(t = tf = 10) = xf .

Note: the Matlab code is shown in the appendix E.




Problem 3

Consider the following linear time-invariant system,

1. Is the system controllable? Explain.

Solution

The controllability matrix

, -

Hence, the controllability Grammian is not invertible, so the system is uncontrollable.

2. Is the system stabilizable? Explain.

The characteristic polynomial equation of matrix A

There exists one that is positive, the system thus is unstable.

3. Identify the modes that are uncontrollable.

We will use the PHB rank test

, -




With

, -

This mode is uncontrollable.

With

, -

This mode is controllable.

With

, -

This mode is uncontrollable.

4. Find a set of orthonormal basis vectors for the controllable subspace.

Solution

The controllability matrix C has rank = 1, thus the controllable subspace is spanned by

a column of the matrix C. we select the first column of the matrix C

√ √

Hence, a set of orthonormal basis vectors for the controllable subspace is . Or by using

Matlab, with command: orth(C), we get




5.

Find a set of basis vectors for the uncontrollable subspace. Are there vectorsorthogonal to the controllable subspace?

Solution

From Part 4, the uncontrollable subspace must be of dimension . We can choose any two linearity independent vectors that are

not in the controllable subspace, for example

√ √√

So, a set of basis vectors for the uncontrollable subspace is √ √ √

We can check

These two vectors are orthogonal to the controllable subspace. But, it is not necessary.

6. Determine a state transformation P that separates the system into controllable and

uncontrollable subspace, where Solution

From state transformation formula




We choose

, - Where, are three eigenvectors of matrix A corresponding to three eigenvalues of

matrix A.

We obtain a new sate-space equation

7. Find the state model of the system transformed by the similarity transformation P

obtained in Part 6. Use this state model to determine which modes in the system are

controllable and uncontrollable.




Bring

From Part 6, we can find sate z1 is controllable mode, state z2 and z3 are

uncontrollable modes.




Appendix A Matlab code for Problem 1 Part 4

ODE function for the nonlinear system

function dx = xdiff(t,x) % nonlinear system u = 0; dx = zeros(2,1); dx(1) = x(2) - 2*x(1)*x(2); dx(2) = -x(1) + x(1)^2 + x(2)^2 + u; end

ODE function for perturbation

function ddx = dxdiff(dt,dx) % delta_x_diff = A*delta_x + B*delta_u; du = 0; ddx = zeros(2,1); ddx(1) = -dx(2); ddx(2) = dx(1) + du; end

m-file code

clc; clear all; close all; to = 0; tspan = 0.1; tfinal = 20; n = (tfinal - to)/tspan + 1;

% non-linear system xq = [1; 0]; [t,x] = ode45(@xdiff,to:tspan:tfinal,xq);

% perturbation dx0 = [0.01; 0.05]; [dt,dx] = ode45(@dxdiff,to:tspan:tfinal,dx0);

% linearized system x_linearized = zeros(n,2); x_linearized(:,1) = dx(:,1) + xq(1); x_linearized(:,2) = dx(:,2) + xq(2);

plot(t,x,dt,x_linearized); grid on; xlabel('Time, sec');

ylabel('States'); legend('x_1','x_2','Linearized x_1','Linearized x_2');




Appendix B Matlab code for Problem 1 Part 5

function dx = xdiff(t,x) % nonlinear system u = 0; dx = zeros(2,1); dx(1) = x(2) - 2*x(1)*x(2); dx(2) = -x(1) + x(1)^2 + x(2)^2 + u; end

ODE function for perturbation

function ddx = dxdiff(dt,dx) % delta_x_diff = A*delta_x + B*delta_u; du = 0.01*sin(dt); ddx = zeros(2,1); ddx(1) = -dx(2); ddx(2) = dx(1) + du; end

m-file code

clc; clear all; close all; to = 0; tspan = 0.1; tfinal = 20; n = (tfinal - to)/tspan + 1;

% non-linear system xq = [1; 0]; [t,x] = ode45(@xdiff,to:tspan:tfinal,xq);

% perturbation dx0 = [0; 0]; [dt,dx] = ode45(@dxdiff,to:tspan:tfinal,dx0);

% linearized system x_linearized = zeros(n,2); x_linearized(:,1) = dx(:,1) + xq(1); x_linearized(:,2) = dx(:,2) + xq(2);

plot(t,x,dt,x_linearized); grid on; xlabel('Time, sec'); ylabel('States');

legend('x_1','x_2','Linearized x_1','Linearized x_2');




Appendix C Matlab code for Problem 2 Part 2

close all; clear all; clc;

A = [-1 1; 0 -1]; B = [0;1]; x0 = [1; 1];

xf = [1; 1]; t0 = 0; tf = 10; syms t t1 s; Phitft = expm(A*(tf - t)); Phitft0 = expm(A*(tf - t0)); W = int(expm(A*s)*(B*B')*expm(A'*s),t0,tf - t0); u = B'*Phitft'*W^(-1)*(xf - Phitft0*x0); xt = expm(A*(t1 - t0))*x0 + int(expm(A*(t1 - t))*B*u,t0,t1);

t1 = 0:0.01:15; t = 0:0.01:15; ut = subs(u); x1 = subs(xt(1)); x2 = subs(xt(2));

subplot(211); plot(t,ut); title('Input ut'); xlabel('Time, (sec)'); ylabel('ut'); grid on; subplot(212); plot(t1,x1,t1,x2); title('State responses'); xlabel('Time, (sec)'); ylabel('States'); grid on; legend('x_1','x_2');




Appendix D Matlab code for Problem 2 Part 3


A = [-1 1; 0 -1]; B = [0;1]; x0 = [1; 1];

xf = [1; 1]; t0 = 0; tf = 1; syms t t1 s; Phitft = expm(A*(tf - t)); Phitft0 = expm(A*(tf - t0)); W = int(expm(A*s)*(B*B')*expm(A'*s),t0,tf - t0); u = B'*Phitft'*W^(-1)*(xf - Phitft0*x0); xt = expm(A*(t1 - t0))*x0 + int(expm(A*(t1 - t))*B*u,t0,t1);

t1 = 0:0.01:15; t = 0:0.01:15; ut = subs(u); x1 = subs(xt(1)); x2 = subs(xt(2));

subplot(211); plot(t,ut); title('Input ut'); xlabel('Time, (sec)'); ylabel('ut'); grid on; subplot(212); plot(t1,x1,t1,x2); title('State responses'); xlabel('Time, (sec)'); ylabel('States'); grid on; legend('x_1','x_2');



Appendix E Matlab code for Problem 2 Part 4


A = [-1 1; 0 -1];B = [0;1];

x0 = [1; 1]; xf = [1; 1]; t0 = 0; t1 = 5; tf = 10; syms t s; P1 = int(expm(A*(tf-s))*B,s,t0,t1); P2 = int(expm(A*(tf-s))*B,s,t1,tf); P = [P1 P2]; Q = xf - expm(A*(tf - t0))*x0; u = P^(-1)*Q; xt1 = expm(A*t)*x0 + int(expm(A*(t-s))*B*u(1),s,t0,t); % t = t0 ~ t1 xt2 = expm(A*t)*x0 + int(expm(A*(t-s))*B*u(1),s,t0,t1) + ...

int(expm(A*(t-s))*B*u(2),s,t1,t); %t = t1 ~ tf t = t0:0.01:t1; xt1 = subs(xt1); t = (t1 + 0.01):0.01:tf; xt2 = subs(xt2); xt11 = [xt1(1,:) xt2(1,:)]; xt22 = [xt1(2,:) xt2(2,:)]; t = t0:0.01:tf; u1 = subs(u(1)*ones(1,501)); u2 = subs(u(2)*ones(1,500)); subplot(211); plot(t(1:501),u1,t(502:1001),u2); title('Input ut'); xlabel('Time, (sec)'); ylabel('ut'); grid on; subplot(212); plot(t,xt11,t,xt22); title('State responses');

xlabel('Time, (sec)'); ylabel('States'); grid on; legend('x_1','x_2');

Linear system theory: linearization

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