[hw3 ls ee571] akhtar rasool

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EE 571 Linear Systems Home Work 3 / 30

EE 571 Linear Systems Spring 2014

Home Work # 3

Due Date: 02 April; 2014

Submitted By: Akhtar Rasool

Student ID: 17197

Faculty of Engineering and Natural Sciences,

Sabanci University, Istanbul

Turkey




Contents Page No.

Question # 01 3

Question # 02 5

Question # 03 6

Question # 04 7

Question # 05 9

Question # 06 18

Question # 07 19

Question # 08 24

Question # 09 26

Question # 10 29




Question # 01

Let

∗ be a Hermitian-symmetric matrix (i.e.

∗

= where

∗ represents complex conjugate

operation). Prove that the matrix is always diagonalizable even if it has repeated eigenvalues.

Solution-Proof:

In this we will show that every Hermitian matrix (even if it has repeated eigenvalues) can be diagonalized

by a unitary matrix. Equivalently, every Hermitian matrix has an orthonormal basis. If is real symmetric,

then the basis of eigenvectors can be chosen to be real. Therefore this will show that every real symmetric

matrix can be diagonalized by an orthogonal matrix.

The argument we present works whether or not has repeated eigenvalues and also gives a new proof

of the fact that the eigenvalues are real.

To begin we show that if

is any

× matrix (not necessarily Hermitian), then there exists a unitary

matrix such that × is upper triangular. To see this start with any eigenvector of with eigenvalue. (Every matrix has at least one eigenvalue.) Normalize so that ‖‖ = 1. Now choose an orthonormal

basis , … … , for the orthogonal complement of the subsace spanned by , so that , , … , form

an orthonormal basis for all of ℂ. Form the matrix with these vectors as columns. Then using ∗ to

denotes a number that is not necessarily 0, we have

∗ =

⎣⎢⎢⎢⎡ ⋮⋮

⎦⎥⎥⎥⎤ [ ⋯ ⋯ ]

= ⎣⎢⎢⎢⎡ ⋮⋮ ⎦⎥⎥⎥⎤ [ ⋯ ⋯ ] = ‖‖ ∗⟨, ⟩ ∗ ⋯ ∗⋯ ∗⋮ ⋮⟨, ⟩ ∗ ⋮ ⋮⋯ ∗

= ∗0 ∗ ⋯ ∗⋯ ∗⋮ ⋮0 ∗ ⋮ ⋮⋯ ∗

= ∗0 ⋯ ∗⋮0

Here is an ( − 1) × ( − 1) matrix.

Repeating the same procedure with we can construct an ( − 1) × ( − 1) unitary matrix with




∗ = ∗0 ⋯ ∗

⋮0

Now we use the ( − 1) × ( − 1) unitary matrix to construct an × unitary matrix

= 1 00 ⋯ 0⋮0

Then it is not hard to see that

∗∗ = ⎣⎢⎢⎢⎡ ∗ ∗ ∗ ∗0 ∗ ⋯ ∗0⋮0 0⋮0 ⎦⎥⎥⎥⎤

Continuing in this way, we find unitary matrices , , … , so that

∗ ⋯ ∗∗ ⋯ =

⎣⎢⎢⎢⎡ ∗ ∗ ∗ ∗0 ∗ ⋯ ∗0⋮

0

0⋮

0 ⋱

⎦⎥⎥⎥⎤

Define = … .

Then is unitary, since the product of unitary matrices is unitary, and ∗ = ∗ … ∗∗. Thus the

equation above says that ∗ is upper triangular, i.e;

∗ =⎣⎢⎢⎢⎡ ∗ ∗ ∗ ∗0 ∗ ⋯ ∗0⋮0

0⋮0 ⋱ ⎦⎥⎥⎥⎤

Notice that if we take the adjoint of this equation, we get

∗ ∗ =⎣⎢⎢⎢⎡ ∗ ∗ ∗ ∗0 ∗ ⋯ ∗0⋮0

0⋮0 ⋱ ⎦⎥⎥⎥⎤

Now lets return to the case where is Hermitian. Then ∗ = so that the matrices appearing in the

previous two equations are equal. Thus




⎣⎢⎢⎢⎡ ∗ ∗ ∗ ∗0 ∗ ⋯ ∗0⋮0 0⋮0 ⋱ ⎦⎥⎥⎥

⎤ = ⎣⎢⎢⎢

⎡ ∗ ∗ ∗ ∗0 ∗ ⋯ ∗0⋮0 0⋮0 ⋱ ⎦⎥⎥⎥

⎤

This implies that all the entries denoted ∗ must actually be zero. This also shows that = for every .

In other words, Hermitian matrices can be diagonalized by a unitary matrix, and all the eigenvalues are

real. Hence the matrix is always diagonalizable even if it has repeated eigenvalues.

Question # 02

Using the definition of for vectors, show that the induced norm of a matrix × is given by‖ ‖ = ∑ || (i.e. the largest column absolute sum).

Solution-Proof: The norm is;

‖ ‖ = ||

Let = (, ∈ ×) and so

=

And

‖ ‖ = ‖‖ = || =

Now

≤ ≤

Thus

‖ ‖ ≤

Changing the order of summation then gives

‖ ‖ ≤

≤

where ∑ is the sum of absolute values in column .




Each column will have a sum, = ∑ , ≤ ≤ ≤ , where is the column with the

largest sum. So

‖ ‖ ≤ ≤ = ‖ ‖

And hence ‖ ‖‖ ‖ ≤ , Where is the maximum column sum of absolute values. This is true for all non-zero . Hence, ‖ ‖ ≤

We need to determine now if

is the maximum value of

‖ ‖‖ ‖ , or whether it is merely an upper bound.

In other words, is it possible to pick an ∈ for which we reach ?

Try by choosing the following;

= [0,…,0,1,0,…,0 ] ℎ = 0 ≠ 1 =

Now, substituting this particular into = implies = ∑ = and thus

‖ ‖ = ‖‖ =

= ||

=

So, ‖ ‖‖ ‖ = =

This means that the bound can be reached with a suitable and so

‖ ‖ = ‖ ‖ ‖ ‖‖ ‖ =

Where is the maximum column sum of absolute values. Hence, is not just an upper bound, it is

actually the maximum!

Question # 03

Using the definition of for vectors, show that the induced norm of a matrix × is given by‖ ‖ = ∑ || (i.e. the largest row absolute sum).

Solution: Let ‖‖ = then by definition || ≤ || = for some . Then

‖ ‖ =

≤ ≤ ∙

≤ =




Hence the maximum row sum is always greater than or equal to the infinity vector norm of .

Conversely, suppose the maximum row sum is obained from row

of the matrix

. Then choose the vector

defined by;

= ≥ − <

Then ‖‖ = and

‖ ‖ ≥ ‖ ‖ ≥

= =

Hence here we have a specific vector of length for which the vector norm dominates the maximum

row sum. Therefore

‖ ‖ =

Question # 04 (Problem 3.18 on page 82 from Chen’s Book)

Find the characteristic polynomials and the minimal polynomials of the following matrices:

=

=

=

=

Solution:

The characteristic polynomial of is ;

∆() = ( − ) = − 0001 − 00

01 − 0000 −

= ( − )( − )

Where determinant is calculated by expanding column 1.

The minimal polynomial of is ;

=⎣⎢⎢⎢⎡00⋯0

10⋯001⋯0

⋮⋮⋮⋮⋮000⋯⎦⎥⎥

⎥⎤

() = ( − ) ℎ = ≤ = =




where is the largest order of all Jordan blocks associated with .

() = ( − )

( − ) ; =


∆() = ( − ) = − 000−1 − 00

0−1 − 0000 −

= ( − )


The minimal polynomial of

is ;

=⎣⎢⎢⎢⎡00⋯0

10⋯001⋯0

⋮⋮⋮⋮⋮000⋯⎦⎥⎥

⎥⎤ ( )

() = ( − ) ℎ = ≤ = =


() = ( −

)

;

=


∆() = ( − ) = − 000−1 − 00

00 − 0000 −

= ( − )



= ⎣⎢⎢⎢⎡0⋯00 1⋯00 ⋮⋮⋮⋮⋮ 00⋯0 00⋯0⎦⎥⎥⎥⎤ ( )

() = ( − ) ℎ = ≤ = =





() = ( − ) ; =


∆() = ( − ) = − 0000 − 00

00 − 0000 −

= ( − )



= ⎣⎢⎢⎢⎡

⋯000 ⋮⋮⋮⋮⋮ 0⋯00 0⋯000⋯00⎦⎥⎥⎥⎤ ( )

() = ( − ) ℎ = ≤ = =

where is the largest order of all Jordan blocks associated with .() = ( − ) ; =


Use two different methods to compute for and in Problem 3.13

= , = − −

Solutions for : − :

= 1 4 1 00 2 00 0 3


∆() = ( − ) = − 1 − 4 − 1 00 − 2 00 0 − 3 = ( − 1)( − 2)( − 3)

Since ∆() is a diagonal matrix so the eigenvalues of are 1,2,3. They all are distinct so the − of will be diagonal.

The associated wtih = 1 is any non-zero solution of




( − 1) = 0 4 1 00 1 00 0 2

= 0

=≫ 0 4 1 00 1 00 0 2 = 000

Here we set = , now the above system can be written as;4 + 10 = 0, = 0,2 = 0 =≫ = 0, = 0Hence = ( ) = ( 0 0) = (1 0 0)

Where

is any arbitrary number. We may also write as follows;

= (1 0 0)


( − 2) = − 1 4 1 00 0 00 0 1 = 0

=≫ − 1 4 1 00 0 00 0 1 = 000

Here we set = , now the above system can be written as;

− + 4 + 10 = 0,2 = 0 =≫ = 0, = , = 4Hence = ( ) = (4 0) = (4 1 0)

Where is any arbitrary number. We may also write as follows; = (4 1 0)


( − 3) = − 2 4 1 00 − 1 00 0 0 = 0

=≫ − 2 4 1 00 − 1 00 0 0 = 000

Here we set = , now the above system can be written as;−2 + 4 + 10 = 0, − = 0




=≫ = , = 0, = 5Hence

= ( ) = (5 0 ) = (5 0 1)

Where is any arbitrary number. We may also write as follows; = (5 0 1)

Now we have;

= [ ] =

|| = 1 4 50 1 00 0 1 = 1 1 00 1 = 1

= ()|| = 11 ⎣⎢⎢⎢⎢⎡ 1 00 1 − 0 00 1 0 10 0− 4 50 1 1 50 1 − 1 40 0

4 51 0 − 1 50 0 1 40 1 ⎦⎥⎥⎥⎥⎤

= 1 0 0− 4 1 0− 5 0 1 = 1 − 4 − 50 1 00 0 1

Thus the − of with respect to { } is;

=

= 1 − 4 − 50 1 00 0 1

1 4 1 00 2 00 0 3

1 4 50 1 00 0 1 =

1 − 4 − 50 1 00 0 1

1 8 1 50 2 00 0 3

=≫ = =

So

= = − −

=≫

=

=

− −

= ( − ) ( − )

− : The characteristic polynomial of is;

∆() = ( − ) = − 1 − 4 − 1 00 − 2 00 0 − 3 = ( − 1) − 2 00 − 3 (∴ )

=≫ ∆() = ( − ) = ( − 1)( − 2)( − 3)




. ; = , = , =

Let

ℎ() = + +

on the spectrum of , we have () = ℎ() (1) = ℎ(1) , = + + …… () (2) = ℎ(2) , = + 2 + 4 …… () (3) = ℎ(3) , = + 3 + 9 …… ()

Solving

(), () () simultaneously, we proceed as follows;

From (), = − − substituting in () & (), we get a new set of equations with twounknowns as; − = + 3 …… () − = 2 + 8 …… ()

Solving () & (), we get;

= − 52 + 4 − 32 ; = 12 ( − 2 + )

Substituting these values in equation

(), we obtained the last unknown

;

= 3 − 3 +

∴ = 3 − 3 + ; = − 52 + 4 − 32 ; = 12 ( − 2 + )

Now, = + +

=≫ = (3 − 3 + ) 1 0 00 1 00 0 1

+ − 52 + 4 − 32 1 4 1 00 2 00 0 3

+ 12 ( − 2 + ) 1 4 1 00 2 00 0 3

ℎ =≫ = 1 0 00 1 00 0 1 + 1 4 1 00 2 00 0 3 + 1 4 1 00 2 00 0 3

=≫ = 1 0 00 1 00 0 1 + 1 4 1 00 2 00 0 3 + 1 1 2 4 00 4 00 0 9 = + + 1 2 4 00 00 0 9




=≫ = + + 4 + 12 10 + 400 + 2 + 4 00 0 + 3 + 9

= 4( − ) 5 ( − )0 00 0

=≫ = ( − ) ( − )

− :

It can also be calculated using Inverse Laplacean Transformation as the formula is as under; = ( − )

( − ) = 1 0 00 1 00 0 1 −

1 4 1 00 2 00 0 3

= − 1 − 4 − 1 0

0 − 2 00 0 − 3

| − | = − 1 − 4 − 1 00 − 2 00 0 − 3 = ( − 1) − 2 00 − 3 = ( − 1)( − 2)( − 3)

( − ) =⎣⎢⎢⎢⎢⎡ − 2 00 − 3 − 0 00 − 3 0 − 20 0 − − 4 − 1 00 − 3 − 1 − 1 00 − 3 − − 1 −40 0 − 4 − 1 0 − 2 0 − − 1 −100 0 − 1 −40 − 2⎦⎥⎥

⎥⎥⎤

( − ) = ( − 2)( − 3) 0 04( − 3) ( − 1)( − 3) 010( − 2) 0 ( − 1)( − 2)

( − ) = ( − 2)( − 3) 4( − 3) 10( − 2)0 ( − 1)( − 3) 00 0 ( − 1)( − 2)

( − ) = ( − )| − |

( − ) = 1( − 1)( − 2)( − 3) ( − 2)( − 3) 4( − 3) 10( − 2)

0 ( − 1)( − 3) 00 0 ( − 1)( − 2)

( − ) =⎣⎢⎢⎢⎢⎡ 1 − 1 4( − 1)( − 2) 10( − 1)( − 3)0 1 − 2 0

0 0 1 − 3 ⎦⎥⎥⎥⎥⎤




= {( − )} = ⎩⎪⎨⎪⎧

⎣⎢⎢⎢⎢⎡ 1 − 1 4( − 1)( − 2) 10( − 1)( − 3)

0 1 − 2 00 0 1 − 3 ⎦⎥⎥⎥⎥⎤

⎭⎪⎬⎪⎫

=> ⎩⎨⎧ 4( − 1)( − 2) = 4( − 2) − 4 − 1 10( − 1)( − 3) = 5( − 3) − 5 − 1⎭⎬

⎫

=

( −

)

=

( − ) ( − )

∴

− =

− = .

Solutions for : − :

= 0 4 30 2 0 1 60 − 2 5 − 2 0The characteristic polynomial of is ;

∆() = ( − ) = −4 −30 − 2 0 − 1 60 2 5 + 2 0 =

Clearly has only one disctinct eigenvalue 0 with multiplicity 3, ( − 0) = 3− 2 = 1

Thus has only one independent eigenvector associated with 0 ; =

=≫ 0 4 30 2 0 1 60 − 2 5 − 2 0

= 00

0

Here we set = 1, = 0 now the above system can be written as;4 + 3 = 0 =≫ = 0Hence = ( ) = ( )

We can compute generalized eigenvectors of from equations below; =




=≫ 0 4 30 2 0 1 60 − 2 5 − 2 0

= 100

4 + 3 = 1,20 + 16 = 0

Solving both of the above equations gives; = ( − )

Now, =

=≫ 0 4 30 2 0 1 60 − 2 5 − 2 0

= 04−5

4 + 3 = 0,20 + 16 = 4

Solving both of the above equations gives; = ( − )

Now we have;

= [ ] = − −

|| = 1 0 00 4 − 30 − 5 4 = 1 4 −3−5 4 = 1

= ()|| = 11 ⎣⎢⎢⎢⎢⎡ 4 −3−5 4 − 0 −30 4 0 40 −5− 0 0−5 4 1 00 4 − 1 00 −5

0 04 −3 − 1 00 −3 1 00 4 ⎦⎥⎥⎥⎥⎤

= 1 0 00 4 50 3 4 = 1 0 00 4 30 5 4

Thus the repre of with respect to the basis { } is ;

=

= 1 0 00 4 30 5 4

0 4 30 2 0 1 60 − 2 5 − 2 0

1 0 00 4 − 30 − 5 4 =

1 0 00 4 30 5 4

0 1 00 0 40 0 − 5

=≫ = =

Since the Jordan form have repeated eigenvalues with multiplicity 3 so we can determine as

follows;




=

=

So

= = 1 0 00 4 − 30 − 5 4 1 20 1 0 0 1 1 0 00 4 30 5 4

=≫ = = 1 0 00 4 − 30 − 5 4

1 4 + 52 3 +2

0 4 + 5 3 + 40 5 4 = 1 52 + 4 2 + 3

0 1 + 20 160 −25 1 − 20

=≫ = = 1 52 + 4 2 + 30 1 + 20 160 −25 1 − 20

− :

The characteristic polynomial of is;

∆() = ( − ) = −4 −30 − 2 0 − 1 60 25 + 20

= − 2 0 − 1 625 + 20

= ( − 400 + 400) =

Let ℎ() = + +

on the spectrum of , we have () = ℎ() (0) = ℎ(0) , = + + =≫ = =≫ = 1 (0) = ℎ(0), = + 2 =≫ () = + 2(0) =≫ =

(0) = ℎ(0), = 2 =≫ () = 2 =≫ =

2

Solving above equations we have;

= 1; = ; = 2

= + +




=≫ = 1 0 00 1 00 0 1

+ 0 4 30 2 0 1 60 − 2 5 − 2 0

+

2 0 4 30 2 0 1 6

0 − 2 5 − 2 0

=≫ = 1 0 00 1 00 0 1 + 0 4 30 20 160 −25 −20 + 2 0 4 30 2 0 1 60 − 2 5 − 2 0

=≫ = 1 4 30 1 + 20 160 − 2 5 1 − 2 0 + 2 0 5 40 0 00 0 0 = 1 4 + 5 2 3 +4 20 1 + 20 160 −25 1 − 20

=≫ = + +

+ − −

− :

It can also be calculated using Inverse Laplacean Transformation as the formula is as under; = ( − )

( − ) = 1 0 00 1 00 0 1 − 0 4 30 2 0 1 60 − 2 5 − 2 0 = −4 −30 − 2 0 − 1 60 2 5 + 2 0

| − | = −4 −30 − 2 0 − 1 60 2 5 + 2 0 = − 2 0 − 1 625 + 20 = {( − 20)( + 20) +400} =

( − ) =⎣⎢⎢⎢⎢⎡ − 2 0 − 1 625 + 20 − 0 − 1 60 + 20 0 − 200 25

− − 4 − 32 5 + 2 0 −30 + 20 − −40 25 −4 −3 −20 −16 − −30 − 1 6 −40 − 20⎦⎥⎥

⎥⎥⎤

( − ) = 0 04 + 5 ( + 20) −253 + 4 16 ( − 20)

( − ) = + + ( + ) − ( − )

( − ) = ( − )| − |




( − ) = 1

4 + 5 3 + 40 ( + 20) 160 −25 ( − 20) = ⎣⎢⎢⎢⎢⎡1 4 + 5 3 + 4

0 1 + 20 160 − 25 1 − 20 ⎦⎥⎥⎥⎥⎤

= ( − ) = ⎩⎪⎨⎪⎧

⎣⎢⎢⎢⎢⎡ + + + − − ⎦⎥

⎥⎥⎥⎤⎭⎪⎬⎪⎫

= ( − ) = +

+ + − − ∴

= = .


An oscillation can be generated by;

= − …… ()

Show that its solution is:

() = − ()

Solution-Proof:

Let

= 0 1−1 0

So (1) can be written as follows;

() = () = ()

The characteristic polynomial of

is;

∆() = ( − ) = −11 = ( + 1) = ( − ) = ( + )( − )

. ; = , = −

Let ℎ() = +

on the spectrum of , we have




Computation of :

() = ℎ()

( ) = ℎ( ) , = + … … () (−) = ℎ(−) , = − … … ()

Equations () & () can also be written as follows; = cos() + ∙ () = + … … () = cos() − ∙ () = − … … ()

Solving equations () & (), we get the values;

= cos() & = sin()

Now,

= + = cos() 1 00 1 +sin() 0 1−1 0 = () ()−() ()

Computation of (Alternative Method):

= ( − ) = 1 + 1 1−1 = + 1 1 + 1−1 + 1 + 1

= {( − )} = + 1 1 + 1 −1 + 1 + 1 = () ()−() ()

() = () = () = () ()−() ()()

Hence;

() = () ()−() ()()


Find the companion-form and modal-form equivalent equations of;

= − − − + , = [ − ]

Solution: Let




= − 2 0 01 0 10 − 2 − 2

, = = 101

, = [1 − 1 0], = [0]

Determining of Companion Form Equivalent Equations:

= − 2 0 01 0 10 − 2 − 2 101 = −22−2 , = ( ) = − 2 0 01 0 10 − 2 − 2 −22−2 = 4−40

= [ ] = − − −

= ( )

|| = 1 − 2 40 2 − 41 − 2 0 = 1(0 − 8) − 0 + 1(8 − 8) = −8 (∴ )

= || = 1−8 ⎣⎢⎢⎢⎢⎡ 2 −4−2 0 − 0 −41 0 0 21 −2− −2 4−2 0 1 41 0 − 1 −21 −2−2 42 −4 − 1 40 −4 1 −20 2 ⎦⎥⎥

⎥⎥⎤

= − 18 − 8 − 4 − 2− 8 − 4 00 4 2 =⎣⎢⎢⎢⎢⎡1 12 141 12 00 − 12 − 14⎦⎥⎥

⎥⎥⎤

= || = . . − . . − . = ( )

Now

= = 1 1 00 . 5 0 . 5 − 0 . 50 . 2 5 0 − 0 . 2 5 − 2 0 01 0 10 − 2 − 2 1 − 2 40 2 − 41 − 2 0

= = −1 0 1− 0 . 5 1 1 . 5−0.5 0.5 0.5 1 − 2 40 2 − 41 − 2 0 = 0 0 − 41 0 − 60 1 − 4

Similarly we can find , ;

= = 1 1 00 . 5 0 . 5 − 0 . 50 . 2 5 0 − 0 . 2 5 101 = 100 = = [1 − 1 0] 1 − 2 40 2 − 41 − 2 0 = [1 − 4 8] & = = 0




= = − −

− ; = =

;

= = [ − ]; = =

Finally we can write the Companion form equivalent equations as under;

= − − − + , = [ − ]

Determining of Model Form Equivalent Equations:

= − − −

+

, = [ − ] ( )

Characteristic Polynomial = | − | = + 2 0 0− 1 − 10 2 + 2 = ( + 2){( + 2) + 2}

| − | = ( + 2)( + 2 + 2) = ( + 2)( + 2 + 1 − 1 + 2) = ( + 2){( + 1) − } | − | = ( + )( + − )( + + ) =≫ = −, = − + , = − −

The associated wtih = −2 is any non-zero solution of;

( + 2) = 0 0 01 2 10 − 2 0

= 0 =≫ 0 0 01 2 10 − 2 0

= 00

0

Here we set = , now the above system can be written as; + 2 + = 0, = 0 =≫ = −Hence = ( ) = ( 0 − ) = (1 0 −1)

Where is any arbitrary number. We may suppose = 1/√ 2 ; = (1/√ 2 0 − 1 /√ 2)

The

associated wtih

= −1 + is any non-zero solution of;

( − (−1 + )) = −1 − 0 01 − 1 + 10 −2 −1 − = 0 =≫ −1 − 0 01 − 1 + 10 −2 −1 − = 000

Here we set = , now the above system can be written as; + (− 1 + ) + = 0, = 0 =≫ = (1 − )

Hence = ( ) = (0 ( 1 − ) ) = (0 1 1 − )




Where is any arbitrary number. We may suppose = 1/√ 3 ;

= (0 1/√ 3 1 − √ 3 )

The associated wtih = −1 − is any non-zero solution of

( − (−1 − )) = −1 + 0 01 1 + 10 −2 −1 + = 0 =≫ −1 + 0 01 1 + 10 −2 −1 + = 000

Here we set = , now the above system can be written as; + (1 + ) + = 0, = 0 =≫ = (−1−)

Hence

= ( ) = (0 ( − 1 − ) ) = (0 1 − 1 − )

Where is any arbitrary number. We may suppose the same since complex conjugate eigenvalue,

= √ ;

= (0 1√ 3 − 1 − √ 3 )

Now for the modal form can be written as;

= [ () ()] = [() () ] =⎣⎢⎢⎢⎢⎢⎡

√ √ − √ − √ − √ ⎦⎥⎥

⎥⎥⎥⎤

|| = 0 0 1√ 20 1√ 3 0

− 1√ 3 −

1√ 3 −

1√ 2

= 1√ 2 0 1√ 3− 1√ 3 − 1√ 3 = 1√ 2 13 = 13√ 2




= = || = 113√ 2

⎣⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎡

1

√ 3 0

− 1√ 3 − 1√ 2 − 0 0

− 1√ 3 − 1√ 2 0 1

√ 3

− 1√ 3 − 1√ 3− 0 1√ 2− 1√ 3 − 1√ 2 0 1√ 2− 1√ 3 − 1√ 2 − 0 0− 1√ 3 − 1√ 3

0 1√ 21√ 3 0 − 0 1√ 20 0 0 00 1√ 3 ⎦⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎤

= 3√ 2⎣⎢⎢⎢⎢⎢⎡− 1√ 3 1√ 2 0 13− 1√ 3 1√ 2 1√ 3 1√ 2 0− 1√ 3 1√ 2 0 0⎦⎥⎥

⎥⎥⎥⎤

= −√ 3 0 √ 2−√ 3 √ 3 0−√ 3 0 0 = −√ −√ −√ √ √

= = −√ 3 −√ 3 −√ 30 √ 3 0

√ 2 0 0

− 2 0 01 0 10 − 2 − 2

⎣⎢⎢⎢⎢⎢⎡ 0 0 1√ 20 1√ 3 0

− 1√ 3 − 1√ 3 − 1√ 2⎦⎥⎥⎥⎥⎥⎤

= = −√ 3 −√ 3 −√ 30 √ 3 0√ 2 0 0 ⎣⎢⎢⎢⎢⎡

0 0 −√ 2− 1√ 3 − 1√ 3 02√ 3 0 √ 2 ⎦⎥⎥⎥⎥⎤ = − 1 1 0− 1 − 1 00 0 − 2

Similarly we can find , ;

= = −√ 3 −√ 3 −√ 3

0 √ 3 0√ 2 0 0 101 =

−2√ 30√ 2

= = [1 − 1 0]⎣⎢⎢⎢⎢⎢⎡ 0 0 1√ 20 1√ 3 0− 1√ 3 − 1√ 3 − 1√ 2⎦⎥⎥

⎥⎥⎥⎤

= 0 − 1√ 3 1√ 2 & = = 0




⎩⎪⎨⎪⎧ = = − − −

− ; = = −√

√ ;

= = − √ √ ; = = ⎭⎪⎬⎪⎫

So now we can write the Modal form equivalent equations as under;

= − − − − + −√ √ , = − √ √


Are the two sets of state equations;

= + , = [ − ] & = − + , = [ − ]equivalent? Are they zero-state equiavlent?

Solution: Note that;

(A) the equivalent state equations have the same characteristic polynomial and consequently

the same set of eigenvalues and same transfer matrix.

(B) Two state equations are said to be Zero-state equivalent if they have the same transfer

matrix.

So here we will determine and compare the characteristic polynomial and tranfer mater in both of

the set of equations to know about equivalency;

(i) For the 1st set of Equations:

() = ( − ) = [1 − 1 0] − 2 −1 −20 − 2 −20 0 − 1 110

. = | − | = − 2 −1 −20 − 2 −20 0 − 1

= ( − 2) − 2 −20 − 1

= ( − 1)( − 2)

( − ) = ( − )| − | = 1( − 1)( − 2)⎣⎢⎢⎢⎢⎡ − 2 −20 − 1 − 0 −20 − 1 0 − 20 0 − − 1 − 20 − 1 − 2 −20 − 1 − − 2 −10 0

− 1 − 2 − 2 −2 − − 2 −20 −2 − 2 −10 − 2⎦⎥⎥⎥⎥⎤

( − ) = 1( − 1)( − 2) ( − 2)( − 1) 0 0( − 1) ( − 2)( − 1) 02 + 2 − 4 2( − 2) ( − 2)




( − ) = ⎣⎢⎢⎢⎢⎢⎡ 1 − 2 0 0

1( − 2) 1 − 2 02( −1)( − 1)( − 2) 2( − 1)( − 2) 1 − 1⎦⎥⎥⎥⎥⎥⎤

= ⎣⎢⎢⎢⎢⎢⎡ 1 − 2 1( − 2) 2( − 2)

0 1 − 2 2( − 1)( − 2)0 0 1 − 1 ⎦⎥⎥⎥⎥⎥⎤

Substituting the corresponding values, we get as under,

() = ( − ) = [1 − 1 0]

⎣⎢⎢⎢⎢⎢⎡

1 − 2 1( − 2) 2( − 2)0 1 − 2 2( − 1)( − 2)

0 0 1

− 1 ⎦⎥⎥⎥⎥⎥⎤ 110

() = ( − ) = [1 − 1 0]⎣⎢⎢⎢⎡ 1 − 2 + 1( − 2)1 − 20 ⎦⎥⎥

⎥⎤

() = ( − ) = 1 − 2 + 1( − 2) − 1 − 2 = − 2 + 1 − + 2( − 2) = 1( − 2)

() = ( − ) =

( − )

(ii) For the 2st set of Equations: Similar to above, we can proceed as follows;

() = ( − ) = [ − ] − − − − − +

. = | − | = − 2 −1 −10 − 2 −10 0 + 1 = ( − 2) − 2 −10 + 1 = ( + 1)( − 2)

( − ) = ( − )| − | = 1( + 1)( − 2) ⎣⎢⎢⎢⎢⎡ − 2 −10 + 1 − 0 −10 + 1 0 − 20 0 − − 1 − 10 + 1 − 2 −10 + 1 − − 2 −10 0 − 1 − 1 − 2 −1 − − 2 −10 −1 − 2 −10 − 2⎦⎥⎥⎥⎥

⎤

( − ) = 1( + 1)( − 2) ( − 2)( + 1) 0 0( + 1) ( − 2)( + 1) 01 + − 2 ( − 2) ( − 2)




( − ) = ⎣⎢⎢⎢⎢⎢⎡ 1 − 2 0 0

1( − 2) 1 − 2 0 − 1( + 1)( − 2) 1( + 1)( − 2) 1 + 1⎦⎥⎥⎥⎥⎥⎤

= ⎣⎢⎢⎢⎢⎢⎡ 1 − 2 1( − 2) − 1( + 1)( − 2)

0 1 − 2 1( + 1)( − 2)0 0 1 + 1 ⎦⎥⎥⎥⎥⎥⎤

Substituting the corresponding values, we get as under,

() = ( − ) = [1 − 1 0]

⎣⎢⎢⎢⎢⎢⎡

1 − 2 1( − 2) − 1( + 1)( − 2)0 1 − 2 1( + 1)( − 2)

0 0 1

+ 1 ⎦⎥⎥⎥⎥⎥⎤ 110

() = ( − ) = [1 − 1 0]⎣⎢⎢⎢⎡ 1 − 2 + 1( − 2)1 − 20 ⎦⎥⎥

⎥⎤

() = ( − ) = 1 − 2 + 1( − 2) − 1 − 2 = − 2 + 1 − + 2( − 2) = 1( − 2)

() = ( − ) =

( − )

onclusion:

. = | − | = ( − 1)( − 2). = | − | = ( + 1)( − 2) Since at least the characteristic polynomials of the 1st and 2nd set of equations are not equal so both the

systems are not equivalent. However since;

() = () = ( − ) = & = =

so both of the givens sets of state equations are zero-state equivalent.


Find fundamental matrices and state transition matrices for

() = & () = − −

Solution ():




() () = ()()

Writing each state as a separate equation, re-arranging and integrating both sides, we see

() = ()() = () =≫⎩⎪⎨⎪⎧ () = () ()()

= ⎭⎪⎬

⎪⎫ =≫⎩⎪⎨⎪⎧() = ()

+ (0)ln () = 2 + ⎭⎪⎬

⎪⎫

=≫ () = = = (0). ℎ (0) =

=≫ () = ()

+ ()() = ().

() (0) = 10 =≫ () = & () (0) = 01 =≫ () = . .

Both of the initial states are taken to be linearly independent so;

() = . . = ()

Now = (, ) = () ()

So,

() = ()| ()| = 1. . − . 0 1 = 1 −. .

0 .

(, ) = ()

() = 1 .

0 . 1 −. .

0 .

=≫ (, ) = 1 −. . + . .

0 . .

=≫ (, ) = 1 . − . + .

0 .




=≫ (, ) = 1 . . + .

0 .

∴ ′

=≫ (, ) = () () = . . . = ()

Solution ():

() () = − − ()()


() = −() + ()() = −() =≫ ⎩⎪⎨⎪⎧ () = [−() + ()] ()() = − 1

⎭⎪⎬⎪⎫

=≫ () = [−() + ()]() = =≫ () = [−() + ()]

() = (0) ; (0) =

=≫ () = [−() + ()] = [−() + (0)]

= [−() + (0)]

=≫ () = −() + ()

Since the above equation has the General function () and its derivative () so its derivative can

be coputed by multiplying both sides by and then solving by integration by parts from → .

=≫ () = − ()

+ () + ()

=≫ () = − ()

− ()

+ (0) + (0)

=≫ () = − ()( − 1) − ()( − 1) + 12 (0)( − 1) + (0)

=≫ () = −()( − 1) + ()

− () + 12 (0)( − 1) + (0)

=≫ 0 = −()( − 1) − () + 12 (0)( − 1) + (0)




=≫ 0 = −() + 12 (0)( − 1) + (0) =≫ () = 12 (0)( − 1) + (0)

=≫ () = 12 (0)( − ) + (0)

=≫ () = .()( − ) + () () = ()

() (0) = 10 =≫ () = & () (0) = 01 =≫ () = . ( − ) Both of the initial states are taken to be linearly independent so;

() = . ( − )

= ()

Now

= (, ) = () ()

So,

() = ()| ()| = 1 −0.5( − )0 = −0.5( − )0

(, ) = () () = 0.5( − )0 −0.5( − )0

=≫ (, ) = −0.5( − ) +0.5( − )0 =≫ (, ) = () () = () .() − () = ()Question # 10 (Problem 4.20 on page 120 from Chen’s Book)

Find the state transition matrix of = − −

Solution: Let () () = − − ()()




() = −()

() = −() =≫ ⎩⎪⎨⎪⎧()() = −()

()() = −()⎭⎪⎬⎪⎫

=≫ ⎩⎪⎨⎪⎧ ()()

= −()

()() = −() ⎭⎪⎬⎪⎫

=≫ ln () = () −(0)+ln () = −() −(0)+ =≫ () = ()() = () =≫ () = (). () = ().

=≫ () = (0). ()() = (0). () ℎ = (0) = = (0) =

() (0) = 10 =≫ () = () & () (0) = 01 =≫ () = () Both of the initial states are considered to be linearly independent so;

() = () () =

Now we need

= (, ) = () ()

So,

(

) = ()

| ()| = 1

() () () 0

0 () =

1() 00 1()

(, ) = () 00 () 1() 00 1()

= ()() 00 ()()

=≫ (, ) = () () = ()() ()() =

[hw3 ls ee571] akhtar rasool

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