Linear system theory: basics Algebra, Matrix

7/29/2019 Linear system theory: basics Algebra, Matrix

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NATIONAL CHENG KUNG UNIVERSITY

Department of Mechanical Engineering

LINEAR SYSTEM

HOMEWORK 2

Instructor: Prof. SzuChi Tien

Student: Nguyen Van Thanh

Student ID: P96007019

Department: Inst. of Manufacturing & Information Systems

Class: 1001- N154000Linear System

October 10, 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=


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Contents

Problem 1 ...............................................................................................................................

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

Problem 3 ............................................................................................................................. 1


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Problem 1

Consider the following linear system:

1.Find the eigenvalues and eigenvectors of the system matrix A (you may use Matlab). Is thsystem stable? Explain.

The characteristic polynomial equation of the matrixA is,

.0 1/

The eigenvectors ofA is defined by an expression:

0 1 01 - With :

- With : The system is stable since al the eigenvalues of matrixA those real part are negative.


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2.Find the modal matrix Am

= T1

AT , where columns of the similarity transformation

matrix T are the eigenvectors of the matrix A. Is the modal matrix unique? Explain.

, - 0 1

The modal matrix is not unique. It depends on how we specify the eigenvectors in the

similarity transformation matrix T. For example, if we change T:

, - 0 1

Note: one eigenvalue can have more than one eigenvector. Now, we change , and butstill keep their position in the similarity transformation matrix T. For example, , - SoAmdoesnt change. That means:

,

-

,

- ,

-

,

-


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3.Find the exponential matrix eAtusing four different methods:(a) Cayley Hamilton theorem (Finite series representation).We can express the exponential matrix as a form:

(3a.1Where the coefficients and are solutions of the differential equations 0 1

0

1 (3a.2

By using the Matlab commands:>> syms t;

>> S = dsolve('Dalpha0 = -4*alpha1',...

'Dalpha1 = alpha0 - 2*alpha1',...

'alpha0(0) = 1, alpha1(0) = 0');

>> alpha0 = S.alpha0;

>> alpha1 = S.alpha1;

>> pretty(alpha0)

>> pretty(alpha1)

We obtain the solution:

(3a.3

Another way for obtaining and is using the expression below. (3a.4

Where

are the eigenvalues of matrixA.


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(3a.5

From section 1, we already known the eigenvectors of matrixA, they are:

(3a.6By using the Eulers formula:

(3a.7 (3a.8

Now, we plug equations (1.2) and (1.3) into equation (1.1), we obtain.

(3a.9Plug equation (3a.3) (or (3a.9)),

0 1 0 1

(b)Resolvent matrix (Inverse Laplace transform of (sI A)1).


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Where,

Use the Leverrier-Fadeev Algorithm, we obtain

0 1 0 1 0 1 0 1

0 1 0 1 0 1

0 1

Taking the inverse Laplace transform of

(by looking up the table of Laplace transform

we obtain

./ ./ This can be verified by using the Matlab commands:

syms st;

Phi11 = (s+2)/(s^2+2*s+4);

Phi12 = 1/(s^2+2*s+4);

Phi21 = -4/(s^2+2*s+4);

Phi22 = s/(s^2+2*s+4);


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Phi = [Phi11 Phi12; Phi21 Phi22];

pretty(ilaplace(Phi))

(c)Modal transformation T where .

[ ./ ./ ]

(d)Use Maple or Mathematica (if you have accessed and familiar with one of these toolsnote: this part is optional).

In Maple, by using the commands:


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4.Find the limit of the exponential matrix as t .

(

)

Note that, all two eigenvalues of matrix A has negative real part . Sincethe system is modeled by equation (1) is stable.

5. Solve analytically for the time responses x(t) to initial conditions x(0) = [0 3]T and napplied input u(t)=0 for all t 0. Plot your time responses.


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Time responses x(t) to initial conditions x(0) = [0 3]T and no applied input u(t)=0 for all

0 are given by:

01

Time responses are shown in figure 5.1.

Figure 5.1 Time responses to initial conditions , -


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6. (a) Solve analytically for the time responses x(t) to zero initial conditions x(0) = [0 0]Tand with input u(t)=7 for all t 0.What are the steady-state values of x(t); i.e. ?

Time responsesx(t) to the initial conditionsx(0) = [0 0]T

and with the input u(t) = 7for all t >= 0 are given by

. / . / 01

Where, . By using the Hint or Maple, we obtain:

[ ]


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Time response are shown in figure 6a.1.

Figure 6a.1 Time response to input u(t) = 7

(b) Find the steady-values using the state-space model given in equation (1) withoutsolving for the time responses

In the steady-state, we have

0 1 01


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7. Solve analytically for the time responses x(t) to initial conditions in Part (5) and atthe same time with the applied input in Part (6). Plot your time responses.

The system is a LTI, so by applying the properties of the LTI, we use the superposition

property,

Time responses are shown in figure 7.1.

Figure 7.1 Time responses to x(0) = [0 3]T, and u(t) = 7


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Problem 2

Given the following matrix

, -1. (a) Find a set of orthonormal basis vectors for the range of A. Hint: Use Gram

Schmidt orthogonalization.

Recall that, if , the y is the range space of A, where x is any vector in R3.We have, From matrixA, the first column of matrix A = the third column the second column of

matrix A, that means

Hence, rank(A) = 2 (or we can using the Matlab command: >> rank(A)). So, the basis

vectors for the range ofA () are * + (because and are linearly independent).Now, we use Gram-Schmidt orthogonalization to orthogonalize the

* +.

||

| |

| |

A set of orthonormal basis vectors for the range space ofA is

* +(b) Can you find a solution x R

3

such that


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From , we can find a solution iff . However,the matrix A is not full rank, if , the solution x is not unique.We choose a set of basis vectors for the range space of matrixA is * +, where,

How we know is in the range space ofA or not? If we can find two coefficients such that,

are not simultaneously zero. Hence, is in the range space ofA. If we cannot find satisfying . Hence, is not in the range space ofA.(i) , -Find

satisfying,

Hence, we can find a solution

such that

. Because, the matrix A is not full

rank, so the solution x is not unique.

(ii) , -Find satisfying,


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Hence, we cannot find a solution

such that

2.Find a set of orthonormal basis vectors for the null space of A.Recall, then is in null space ofA.

Thus, the null space ofA is spanned by the vector

Hence, an orthonormal basis vector for the null space ofA is

| |

3. Use MATLAB to determine the singular value decomposition of A.Singular value decomposition:

Use the command [U, S, V] = svd(A) in Matlab, we obtain


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4. Using the results from the singular value decomposition in Part 3, answer thefollowing questions:

(a) What are the singular values (1, 2, 3) of A? Note that123.

(b) What is the rank of A? Verify your result using MATLAB rank(A).

Use the command rank(A) in Matlab (>> rank(A)), we obtain the ans = 2.

(c)What is the right singular vector associated with the singular value 3? Show theconnection of the right singular vector with the null space of A in Part 2.

, -, where, are the right singular vectors ofA. The right singularvector associated with the singular value 3 is

Recall from part 2, the null space ofA is spanned by the vector


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it is along the same direction as the null space ofA in Part 2.(d)What are the left singular vectors corresponding to the non-zero singular values of A?

Show the connection of the left singular vectors with the range space of A in Part 1.

, -, where, are the left singular vectors ofA. The left singularvectors corresponding to the non-zero singular values ofA is

We choose a set of basis vectors for the range space of matrixA is * +, where,

We have,

Hence, are in the range space ofA in Part 1. We can verify the result by usingthe command rank([u1 u2 U1 U2]) = 2.


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Problem 3

Given the following matrix

1.Find the eigenvalues and eigenvectors of the matrix ATA using Matlab. The

eigenvector matrix of ATA forms the unitary matrix V in the singular value

decomposition of A according to equation (5).

By using the following commands in Matlab,

>> A = [1 2; 2 7; 4 0];

>> [V1, D1] = eig(A'*A)

We obtain

0 1 0 1So, the eigenvalues of the matrixATA are

corresponding to two eigenvectors 0 1 012.Find the eigenvalues and eigenvectors of AATusing MATLAB. The eigenvector matrix

of AAT

forms the unitary matrix U in the singular value decomposition of A according

to equation (8).


>> A = [1 2; 2 7; 4 0];

>> [V2, D2] = eig(A*A')

We obtain


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So, the eigenvalues of the matrixAATare

(exactly the same to the eigenvalues of the matrix ATA) corresponding to two

eigenvectors

3.Finally, determine the singular values i(i =1, 2 rm) of the matrix A from the

square root of the eigenvalues of AT

A.


>> A = [1 2; 2 7; 4 0];

>> [V1, D1] = eig(A'*A);

>> SigmaMatrix = D1^(1/2)

We obtain

0 14. Compare the results obtained from the Matlab command svd(A) to the unitary

matrices U and V obtained in Parts (1) and (2).


>> A = [1 2; 2 7; 4 0];

>> [U, S, V] = svd(A)

We obtain


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0 1Recall, from Part 1:

0 1 0 1And from Part2:

Thus, the left singular vectors in Uare the same as the eigenvectors in Part2 and the

right singular vectors Vare the same as the eigenvectors in Part1. If we specify the

eigenvalues in given order (i.e. descending), hence,

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Linear system theory: basics Algebra, Matrix