Lecture 05 Analysis (I)
Time Response and State Transition Matrix
5.1 State Transition Matrix
5.2 Modal decomposition --Diagonalization
5.3 Cayley-Hamilton Theorem
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d
x(t ) Ax(t ) Bu (t )
dt
y (t ) Cx (t ) Du (t )
The behavior of x(t) et y(t) :
1. Homogeneous solution of x(t)
2. Non-homogeneous solution of x(t)
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Homogeneous solution
x (t ) Ax(t ) 1 1
x(t ) L [( sI A) ]x(0)
sX ( s ) x(0) AX ( s )
e At x(0)
X ( s ) ( sI A) 1 x(0)
State transition matrix
(t ) e At L1[( sI A) 1 ]
x(t0 ) e At0 x(0)
At0
x(0) e x(t0 )
At At0 A( t t0 )
x(t ) e e x(t0 ) e x(t0 ) (t t0 ) x(t0 )
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Properties 1 1
(t ) e L [( sI A) ]
At
1. (0) I
1
2. (t ) (t )
3. x(0) (t ) x(t )
4. (t 2 t1 ) (t1 t0 ) (t 2 t0 )
k
5. (t ) (kt )
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Non-homogeneous solution
d
x(t ) Ax(t ) Bu (t )
dt
y (t ) Cx (t ) Du (t )
sX ( s ) x(0) AX ( s ) BU ( s )
( sI A) X ( s ) x(0) BU ( s )
1 1
X ( s ) ( sI A) x(0) ( sI A) BU ( s )
x(t ) L1[(sI A) 1 ]x(0) L1[( sI A) 1 BU ( s )]
t
x(t ) (t ) x(0) (t ) Bu ( )d Convolution
0
Homogeneous
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t
x(t ) (t ) x(0) (t ) Bu ( )d
0
t
x(t ) (t t0 ) x(t0 ) (t ) Bu ( )d
t0
t
y (t ) C (t t0 ) x(t0 ) C (t ) Bu ( )d Du (t )
t0
Zero-input response Zero-state response
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Example 1 x 1 0 1 x1 0
x 2 3 x 1u (t )
2 2
x(0) 0 0
T
let
2e t
e 2 t
e 1 e 2t
(t ) L1[(sI A) 1 ] e At t 2t
2 e 2 e e t 2e 2t
t
x(t ) (t ) x(0) (t ) Bu ( )d
0
x1 2e t e 2t
1 3
Ans: x 2 2 L1[( sI A) 1 BU ( s )]
2 2e 2e 2t
t
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x 1 0 1 x1 0
x 2 3 x 1u (t )
x 2 ( 0)
x1 (0)
s
2 2 s
x ( 0) 0 0
T
let
u 1 s 1 x2 s 1 x1 1 y
3
Using Maisons gain formula 2
1 3s 1 2 s 2
s 1 (1 3s 1 ) s 2 s 2
x1 ( s ) x1 (0) x2 (0) U (s)
2s 2 s 1 s 1
x2 ( s ) x1 (0) x2 (0) U ( s)
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How to find State transition matrix
1 1
(t ) e L [( sI A) ]
At
Methode 1: (t ) L1[( sI A) 1 ]
Methode 2: (t ) e At
Methode 3: Cayley-Hamilton Theorem
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1 1
Methode 1: (t ) L [( sI A) ]
x 1 0 1 0 x1 0 0
x 0 4 3 x 1 0 u1
2 2 u
x 3 1 1 2 x3 0 1 2
x1
y1 (t ) 1 0 0
y (t ) 0 0 1 x2
2 x
3
adj ( sI A)
( sI A) 1
sI A
s 2 6 s 11 s 2 3
1
3 s 2
2 3 s
s ( s 4)( s 2) 3 3s
s4 s 1 s 4 s
2
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Method 2: Diagonalization
Example 4.5 x 1 1 0 0 x1 1
x 0 2 0 x 1u diagonal matrix
2 2
x 3 0 0 3 x3 1
x1
y 6 6 1 x2
x3
e t 0 0
(t ) e At 0 e 2 t 0
0 0 e 3t
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Diagonalization via Coordinate Transformation
Plant: x Ax Bu A R n n
y Cx Du
Eigenvalue of A: i , satisfying Av i i vi , i 1, ,n
Assume that all the eigenvalues of A are distinct, i.e. 1 2 3 n
Then eigenvectors, v1, v2 , ,vn are independent.
Coordinate transformation matrix T [v1, v2 , ,vn ]
T 1 AT is a diagonal matrix.
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1 0
2
1
T AT
0 n
e At Te tT 1
e1t 0
e2t
where e t
nt
0 e
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New T -1 x
coordinate:
T 1 AT
A T 1 AT
1 1 0 0 0 1 B T 1 B
2 0 2 0 0 2 C CT
(4.1)
0
n 0 0 n n
i (t ) ii (t ), i 1, ,n
Solution of (4.1):
i (t ) ei ti (0), i 1, ,n
x (t ) T (t ) v1e1t 1 (0) v2e2t 2 (0) vn ent n (0), (0) T x (0)
1
The above expansion of x(t) is called modal decomposition.
Hence, system asy. stable all the eigenvales of A lie in LHP
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Example i distinct
2 1 1
x x , x ( 0 ) 2 1 1, 2 3
1 2
Find eigenvector
1 2 1 v11 v11 1
(1 I A)v1 0
1 1 2 v12 v12 1
3 2 1 v21 v1 1
(2 I A)v2 0
1 3 2 v22 v2 1
1 1 1 1 1 T 1 AT
1 0
T v1 v2
1
T 0 3
1 1 2 1 1
T 1 x
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1 0
(4.2)
0 3
1 11 1 (0)
(0) T x (0), (0)
1
3
2 2 2 ( 0 )
Solution of (4.1):
x (t ) T (t ) v1e1t 1 (0) v2e2t 2 (0)
3
1 2
(0) T x (0) (0)
1
2
3 t 1 3t
3 t 1 1 3t 1 2 e 2 e
x(t ) e e 3 1
2 1 2
1
e e
t 3 t
2 2
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In the case of A matrix is phase-variable form and
1 2 3 n
1 1 1
Vandermonde mat
P v1 v2 vn 1 2 n rix
for phase-variable
n 1 n 1 form
1 n 1
2 n
1
2
1
P AP
3
4
t 1
e Pe P
At
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Example: i distinct
Repeated eigenvalue case, i.e. i is not distinct
1 0 1 1 0 1
A 0 1 0 I A 0 1 0 ( 1)( 1)( 2)
0 0 2 0 0 2
1 2 0 0 1 v1
(1 I A)V1 0 0 0 v2 0
depend
0 0 1 v3
v1 1 v1 0
0v1 0v2 v3 0 v2 0 0v1 0v2 v3 0 v2 1
v3 0 v3 0
V1 V2
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3 2 1 0 1 v1
(3 I A)V3 0 1 0 v2 0
0 0 0 v3
v1 1
v1 0v2 v3 0 v2 0
v3 1
1 0 1 1 0 0
P V1 V2 V3 0 1 0 P 1 AP 0 1 0
0 0 1 0 0 2
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Case 3: i distinct Jordan form
1 2 3
P v1 v2 v3 P AP Jordan
1
form
Generalized eigenvectors
(1 I A)v1 0 1 1
(1 I A)v2 v1 1
P AP A 1 1
(1 I A)v3 v2 1
e 1t
te 1t t2
2e 1t
A t 1t
e e 1t te
e 1t
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Example:
3 1 3 1
A ( 2) 2
1 1 1 1
1 1 v11 v11 1
(1 I A)V1 0
1 1 v12 v12 1
1 1 v21 1 v21 1
(1 I A)V2
1 1 v22 1 v22 0
1 1 2 1
P V1 V2
1
P AP A
1 0 0 2
e 2 t
A t te 2t A t 1
e 2t
e Pe P
At
e
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Method 3: Cayley-Hamilton Theorem
Theorem: Every square matrix satisfies its char. equation .
Given a square matrix A, A R nn . Let f() be the char. polynomial of A.
Char. Equation:
f ( ) n an 1n 1 a1 a0 0
By Caley-Hamilton Theorem
f ( A) An an 1 An 1 a1 A a0 I 0
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An an 1 An 1 a1 A a0 I 0
An an 1 An 1 a1 A a0 I
An 1 an 1 An a1 A2 a0 A
an 1 ( an 1 An 1 a1 A a0 I ) a1 A2 a0 A
any f ( A) k0 I k1 A k 2 A2 k n An
f ( A) 0 I 1 A 2 A n 1 A
2 n 1
n 1
k 0
k A k
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Example: 1 2
A 100
? A
0 1
let f ( A) A100 0 I 1 A
1 2
( 1)( 2) 0 , 1 1, 2 2
0 2
f (1 ) 1
100
0 11 1
100
0 2 2100
f ( 2 ) 2
100
0 12 2 100
1 2100 1
1 0 1 2 1 2101
2
f ( A) A100 (2 2 )
100
(2 1)
100
0 1 0 1 0 1
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Example: 3 1
e ?
At
A
2 0
3 1
0 , 1 1, 2 2
2
f (1) e t 0 11 0 1 0 2e t e 2t
f (2) e 2t 0 12 0 1 2 1 e 2 t e t
t 2t 1 0 2t t 3 1
e 2e e
At
( e e )
0 1 2 0
2e 2 t e t e 2t e t
2t t 2t t
2e 2e e 2e
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Example:
2 0 1 1
x Ax bu 4 1 4 x 1u
2 0 1 0
y 1 0 1 x
1 1 0
0,1,1 v1 4, v 2 0, 1
2 1 0
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Example:
0 1 0 0
x Ax bu 0 0 1 x 0u
2 1 2 1
y 1 0 0 x
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