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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
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