Two Important Cases

Chapter-5

EE-826 Linear Control System

College of Electrical & Mechanical Engineering

National University of Sciences & Technology (NUST)

Preliminaries

Eigenvalues and Eigenvectors

Ax xλ=

( )A I xλ−

� ‘x’ lies in the null space of ( )A Iλ−

3

� ‘x’ lies in the null space of

� Null space is orthogonal to the space spanned by the rows

of a matrix

� Null space exists when a matrix is rank deficient

( )A Iλ−

Preliminaries

Example

1 2 3

1 0 1

0 1 0 , 1, 1, 2

0 0 2

A λ λ λ−

= = = =

Repeated eigenvalues

4

Repeated eigenvalues

Eigenvectors

1,2

0 0 1

( ) 0 0 0

0 0 1

A Iλ−

− =

Rank = 1

Dimension of null space = 2

Preliminaries

Example

1 2

1 0

0 , 1

0 0

x x

= =

1 0 1− −

5

Rank = 2

Dimension of null space = 1

3

1 0 1

( ) 0 1 0

0 0 0

A Iλ− − − = −

3

1

0

1

x

− =

Preliminaries

Diagonalization

1 2 3

1

[ ]P x x x

P AP−

=

= Λ

1 1 2

λ λ λ = = = =

Case 2

6

Rank = 2

Dimension of null space = 1

1 2 30 1 3 , 1, 1, 2

0 0 2

A λ λ λ = = = =

1,2

0 1 2

( ) 0 0 3

0 0 1

A Iλ − =

Preliminaries

Generalized Eigenvectors

2

0 0 5

( ) 0 0 3

0 0 1

A Iλ − =

Rank = 1 Dimension of null space = 2

7

Concept

There exists a vector ‘u’ such that but

2( ) 0A I uλ− = ( ) 0A I uλ− ≠

2

1 ( )

u u

u A I uλ

=

= −

Example

There exists a vector ‘u’ such that but

( ) 0k

A I uλ− = ( ) 10

kA I uλ −− ≠

Preliminaries

2

0

1

0

u u

= =

3 1 1 1 0 0−

1

0 1 2 0 1

( ) 0 0 3 1 0

0 0 1 0 0

u A I uλ = − = =

Example

8

3 1 1 1 0 0

1 1 1 1 0 0

0 0 2 0 1 1

0 0 0 2 1 1

0 0 0 0 1 1

0 0 0 0 1 1

A

− − −

= − −

1,2,3,4,5 62, 0λ λ= =

Preliminaries

1 1 1 1 0 0

1 1 1 1 0 0

0 0 0 0 1 12

0 0 0 0 1 1

0 0 0 0 1 1

0 0 0 0 1 1

A I

− − − −

− = − −

−

−

Rank = 4 Dimension of null space = 2

9

0 0 0 0 1 1−

( )2

0 0 2 2 0 0

0 0 2 2 0 0

0 0 0 0 0 02

0 0 0 0 0 0

0 0 0 0 2 2

0 0 0 0 2 2

A I

− = −

− −

Rank = 2 Dimension of null space = 4

Preliminaries

( )3

0 0 0 0 0 0

0 0 0 0 0 0

0 0 0 0 0 02

0 0 0 0 0 0

0 0 0 0 4 4

0 0 0 0 4 4

A I

− = −

−

Rank = 1 Dimension of null space = 5

10

0 0 0 0 4 4

−

Grade 3 eigenvector ‘u’ such that

( )

( )( )

3

2

2 0

2 0

2 0

A I u

A I u

A I u

− =

− ≠

− ≠

( )

( )

3

2

2

1

2

2

u u

u A I u

u A I u

=

= −

= −

Preliminaries

Grade 2 eigenvector ‘v’ such that( )( )

2

22 0, . 0

2 0

A I v u v

A I v

− = =

− ≠

( )2

1 2

v v

v A I v

=

= −

1 2 3 2 1[ ]Q u u u v v w=

11

1Q AQ J− =

2 1 0 0 0 0

0 2 1 0 0 0

0 0 2 0 0 0

0 0 0 2 1 0

0 0 0 0 2 0

0 0 0 0 0 0

J

=

Time Invariant Case

12

Time Invariant Case

13

1P AP− = Λ

Application

Standard Eigenvectors

Time Invariant Case

1A P P−= Λ

1At te Pe PΛ −=

Generalized Eigenvectors1Q AQ J− =

14

Q AQ J=1At Jte Qe Q−=

1

2

0 0

0 0

0 0 l

J

JJ

J

=

⋯

⋯

⋯ ⋯ ⋱ …

⋯

Time Invariant Case1

2

0 0

0 0

0 0 l

J

J

Jt

J

e

ee

e

=

⋯

⋯

⋯ ⋯ ⋱ …

⋯

k k kJ I Nλ= +

15

k k kJ t It N te e e

λ=k k kJ t t N t

e e eλ=

,( )!

k

j iN t

ij

te i j

j i

−

= ≤ −

Time Invariant Case

16

Time Invariant Case

17

Proof

Condition for existence of scalar functions

Cayley Hamilton Theorem

Proof

18

Time Invariant Case

Taking Laplace Transform of both sides

where

19

{ }1 1( )Ate L sI A− −= −

Inverse Laplace Transform

Time Invariant Case

Laplace Transform

20

Time Invariant Case

Output equation

21

Change of variables

Time Invariant Case

Impulse response

Laplace transform

22

Transfer function

Periodic Case

23

Proof

Definition

Periodic Case

Proof

24

Periodicity of P(t)

Periodic Case

Proof

25

By uniqueness of solution

Periodic Case

Comment about ‘R’

( )( )

,

t

A d

t eτσ σ

τ∫

Φ =

( ,0)RTe T= Φ

26

0

( )

T

A dRTe e

σ σ∫=

1( )

t

R A dT τ

σ σ= ∫

Periodic Case

Example

27

Periodic Case

verification

28

Proof

Part 1

29

0

0

( )

0

( )

0

( )R t T t

R t t RT

z t T e z

e e z

+ −

−

+ =

=

Definition

Proof

Part 2

30

Discrete-time: Two Important

Cases

Chapter-21

EE-826 Linear Control System

College of Electrical & Mechanical Engineering

National University of Sciences & Technology (NUST)

Time Invariant Case

32

Periodic Case

33