Download ch1.3 relationship between IO and state space desicriptions

§1-3 Solution of a Dynamical Equation
1. Solution of a homogeneous differential equation
x  A (t )x
(1 - 50)
• A preliminary theorem for
x  A (t )x  f (t ),
x (t 0 )  x 0  R n
x , f  R n , A (t )  (a ij (t ))n n
(A1)
Preliminary Theorem: Let every element of A and f
be continuous over (, +). Then for any t0(, +)
and any constant vector x0, (A1) has a solution x(t)
defined over (, +), such that
x (t 0 )  x 0
(A 2)
Furthermore, the solution satisfying (A2) is unique.
Corollary 1: If (t) is a solution to the differential
equation
x  A (t )x
and for some t0, (t0)=0, then
Y (t )  0, t
Proof：It is clear that x(t)0, t is one of the solutions
of the above equation. On the other hand, since
(t0)=0
and (t) with the initial condition is unique, we have
Y (t )  x (t )  0, t
Q.E.D
Theorem 2: The set of all solutions of
x  A (t )x
(1-50)
forms an n-dimensional vector space over the field of
real numbers.
Outline of the proof.
a) All the solutions of (1-50) form a linear space;
b) The dimension of the linear space is n. More
specifically,
1) (1-50) has n linearly independent solutions;
2) Any solution can be expressed as the linear
combination of the n solutions.
Proof:
a) All the solutions form a linear space.
Let 1 and 2 be two arbitrary solutions of (1-50).
Then, for any real numbers 1 and 2, we have
d
(a 1Y 1  a 2 Y 2 )  a 1Y 1  a 2 Y 2  a 1A (t )Y 1  a 2 A (t )Y 2
dt
 A (t )(a 1Y 1  a 2 Y 2 )
b) The dimension of the space is n.
1). Let e1, e2,…, en be n linearly independent constant
vectors, and i(t), i=1, 2,…, n, are the solutions of (150) with
i(t0)=ei, i=1, 2,…, n
We then prove that i(t), i=1, 2,…, n, are linearly
independent over (, +).
The proof is by contradiction. If there exists a t0 (,
+), such that
i(t), i=1, 2,…, n
are linearly dependent. Then, there exists an n×1
nonzero vector , such that
[Y 1 (t ') Y 2 (t ')
a 1Y 1 (t ')  a 2 Y 2 (t ') 
Y n (t ')]a  0 
 a n Y n (t ')  0
Note that (t)0, t is a solution of (1-50) and
X (t ) : a 1Y 1 (t )  a 2 Y 2 (t ) 
 a n Y n (t )
is also a solution of (1-50) with
X (t ')  0
Hence, the uniqueness of the solution implies that
Y (t )  X (t )  a 1Y 1 (t )  a 2 Y 2 (t ) 
 a n Y n (t )  0 t
In particular, for t=t0,
[Y 1 (t 0 ) Y 2 (t 0 )
 [e1 e 2
Y n (t 0 )]a
e n ]a  0
which implies that the vectors e1, e2,…, en are linearly
dependent, a contradiction. Hence
i(t), i=1, 2,…, n,
are linearly independent over (, +).
2) To prove that any solution can be expressed as the
linear combination of the n solutions.
That is, all the solutions of (1-50) form an ndimensional vector space.
Let (t) be an arbitrary solution of (1-50) with
(t0)=e
Then, e can be uniquely expressed as
e  a1e 1  a 2e 2 
n
 a n e n   a ie i
i 1
n
i
a
Y
(is
t )also the solution of (1-50) with
Since  i
i 1
n
n
i 1
i 1
i
i
a
Y
(
t
)

a
e
 i
 i e
0
The uniqueness of the solution implies that
Y (t ) 
n
i
a
Y
 i (t )
i 1
Q.E.D
2. Fundamental matrix and state transition matrix
 Fundamental matrix
Definition: The matrix
[Y 1 (t ) Y 2 (t )
Y n (t )] : Y (t ), t  (, )
which is formed by n linearly independent solutions
of (1-50) is said to be a fundamental matrix of (1-50).
Some properties for a fundamental matrix:
Y  A (t )Y
Y (t 0 )  E
(1 - 51)
where E is a nonsingular constant matrix.
Theorem 3 The fundamental matrix of equation of (1-50)
is nonsingular for every t over (, +). .
Proof: By using Corollary 1 directly.
Corollary 1: If (t) is a solution to the differential
equation
x  A (t )x
and for some t0, (t0)=0, then
Y (t )  0, t
Theorem 4: Let 1 and 2 be two fundamental
matrixes of (1-50). Then, there exists a n×n
nonsingular constant matrix C, such that
1(t)=2(t)C
 State transition matrix
Definition 9: Let (t) be any fundamental matrix. Then
F t , t 0   Y (t )Y (t 0 )
1
is said to be the state transition matrix of (1-50), where
t, t0(, +).
Some important properties of state transition matrix:
1). F (t , t ) = I
2 ). F - 1 (t , t0 ) = Y (t0 )Y - 1 (t ) = F (t 0 , t )
3 ). F (t 2 , t0 ) = F (t 2 , t 1 )F (t 1 , t 0 )
4)
d F (t , t 0 )
 A (t )F (t , t 0 )
dt
F (t 0 , t 0 )  I
(1  52)
Under the condition x(t0)=x0, the solution of (1-50) is
x (t )   t ,t 0  x 0
（1－53）
Hence, (t, t0) can be considered as a linear
transformation which maps the state at t0 to the
state x(t) at time t. In fact, x(t) can always be expressed
as
x (t )  Y(t ) ,   0
In particular,
x (t 0 )  Y(t 0 )a  a  Y1 (t 0 )x (t 0 )
from which we can obtain the conclusion.
Example: Prove that the state transition matrix is
unique, i.e. the state transition matrix is regardless of
the choice of fundamental matrix.
3. Non-homogeneous equation
 Solutions of linear time-variant dynamical
equations
x  A (t )x  B (t )u
(1  49)
The solution to the state equation can be obtained by
using the method of variation of parameters. Let
x (t )  F t ,t 0  x(t )
Then, we have
(s  1)
Theorem 1-5 The solution of the state equation
is given by
x (t )  F t , t 0  x 0   F (t , t )B (t )u (t )d t
t
t0
where
F t , t 0  x 0
is called zero-input response, and

t
t0
F (t , t )B (t )u (t )d t
is called zero-state response.
(1  54)
 Relationship between I/O and state space
descriptions
Corollary 1-5 The output of the dynamical equation (1-34)
is
y (t )  C (t )F t , t 0  x 0
t
  C (t )F (t , t )B (t )u (t )d t  D (t )u (t )
t0
(1  57)
If x(t0)=0, the impulse response matrix is
t  t : G (t , t )  C (t ) (t , t )B (t )  D (t )d(t  t )
t  t : G (t , t )  0
 Solutions of a linear time-invariant dynamical
equation
Consider the following linear time-invariant
dynamical equation
x  A x  Bu
y  Cx  Du
(1  60)
where A, B, C and D are n×n, n×p, q×n and q×p real
constant matrices. From the corresponding
homogeneous equation, we have
Fundamental matrix: eAt
State transition matrix:
F t , t 0   e A t (e A t 0 )1  e A (t t 0 )  F t  t 0 
x (t )  e
y (t )  Ce
A (t t 0 )
t
x 0   e A (t  t )B u (t )d t
t0
A (t t 0 )
t
x 0   Ce A (t t )B u (t )d t  Du (t )
t0
Usually, we assume that t0=0,
t
x (t )  e x (0)   e A (t t )Bu (t )d t
At
0
t
(1  63)
y (t )  Ce x (0)   Ce A (t t )Bu (t )d t  Du (t )
At
0
(1  64)
The corresponding impulse response matrix is
t t
G (t  t )  Ce
t t
A (t t )
B  D d(t  t )
G (t , t )  0
Usually we write it as
G (t )  Ce B  D d(t ), t  0
At
(1  65)
The corresponding transfer function matrix of the above
equation is
G( s)  C ( sI  A)1 B  D
which is a rational function matrix.
(1  66)
 Equivalent dynamical equations
a) Time-invariant case
Definition 1-10: An LTI dynamical equation
x  A x  Bu
y  Cx  Du
(1  67)
is said to be equivalent to the dynamical equation (A,
B, C, D) if and only if there exists a nonsingular
matrix P, such that
A  PAP 1 , B  PB, C  CP 1 , D  D
where P is said to be an equivalence transfor-mation.
Equivalence transformation implies that the choice
of the state is not unique; different methods of
analysis often lead to different choices of the state.
Definition: Two time-invariant dynamical systems
are said to be zero-state equivalent if and only if they
have the same impulse response matrix or the same
transfer function matrix.
Theorem: Two equivalent LTI systems are zero-state
equivalent.
Example: Consider the two systems
x  3x  2u
and
 1 4 
1
x 
x    u , y  1 1 x

 4 1
1
which are not equivalent dynamical equations, but are
zero-state equivalent.