Linear Systems Ch5 [Compatibility Mode]
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Transcript of Linear Systems Ch5 [Compatibility Mode]
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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)
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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−
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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
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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
− =
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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 − =
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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 −− ≠
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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= =
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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
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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
=
= −
= −
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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
=
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Time Invariant Case
12
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Time Invariant Case
13
1P AP− = Λ
Application
Standard Eigenvectors
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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
=
⋯
⋯
⋯ ⋯ ⋱ …
⋯
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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
−
= ≤ −
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Time Invariant Case
16
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Time Invariant Case
17
Proof
Condition for existence of scalar functions
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Cayley Hamilton Theorem
Proof
18
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Time Invariant Case
Taking Laplace Transform of both sides
where
19
{ }1 1( )Ate L sI A− −= −
Inverse Laplace Transform
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Time Invariant Case
Laplace Transform
20
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Time Invariant Case
Output equation
21
Change of variables
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Time Invariant Case
Impulse response
Laplace transform
22
Transfer function
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Periodic Case
23
Proof
Definition
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Periodic Case
Proof
24
Periodicity of P(t)
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Periodic Case
Proof
25
By uniqueness of solution
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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= ∫
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Periodic Case
Example
27
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Periodic Case
verification
28
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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
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Proof
Part 2
30
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Discrete-time: Two Important
Cases
Chapter-21
EE-826 Linear Control System
College of Electrical & Mechanical Engineering
National University of Sciences & Technology (NUST)
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Time Invariant Case
32
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Periodic Case
33