
Solutions of x (t )  A(t ) x (t )
Definition:
An n × n matrix Ψ(t) is said to be a fundamental matrix of

x (t )  A(t ) x (t ) if and only if the n columns of Ψ(t) consists

of n linearly independent solutions of
x (t )  A(t ) x (t )
Example:
Find the fundamental matrix for the system
1 0 
x
 x (t )
0
2
t



Solution: Let us define Ψ(t) = [Ψ1(t)
Ψ2(t)].
Possible solutions for Ψ1(t) and Ψ2(t) can be obtained by
directly solving the equations
1 0 
i  
i i  1 and 2

 0 2t 



since x1  x1 , and x 2  2tx2 have solutions
101
x1(t)=x1(0)et
and
x2(t)=x2(0) et
2
Now two linearly independent solutions are
Ψ1(t) = [et
2
et ]T
0]T, and Ψ2(t) = [0
Thus
e t
 (t )  
0
0
2
et 
Note that Ψ(t) is not unique since by changing the initial
conditions and by interchanging Ψ1 and Ψ2 we will get
different Ψ

Theorem: The set of all solutions of x(t )  A(t ) x(t )
for an n
× n A(t) forms an n-dimensional vector space over the field of
real numbers R.
Theorem: Every fundamental matrix Ψ(t) is nonsingular for all
t in (-∞, ∞).
102
Definition: Let Ψ( . ) be any fundamental matrix of

x(t )  A(t ) x(t )
. Then
Ф(t, to) = Ψ(t)Ψ-1(to) for all t, to in (-∞, ∞) is said to be state

transition matrix of x(t )  A(t ) x(t )
.
Properties of the state transition matrix:
1.
Ф(t, t) = I
2.
Ф-1(t, to) = Ψ(to)Ψ-1(t) = Ф(to, t)
3.
Ф(t2, to) = Ф(t2, t1) Ф(t1, to)
Since
Ф(t2, to) = Ψ(t2)Ψ-1(to) = Ψ(t2) Ψ-1(t1) Ψ(t1)Ψ-1(to)
=
Ф(t2, t1) Ф(t1, to)
Question
Ф( , ) is defined in term of Ψ( . ) which is not unique. Is Ф( ,
) unique?
The answer is yes, Ф( , ) is uniquely determined by A.
103
To show the claim let Ψ1 and Ψ2 be two fundamental matrices

of
x (t )  A(t ) x (t ) . Since the columns of Ψ1 form a basis for
the solution space over R.
Then
Ψ2(t) = Ψ1(t)[α1 … αn] = Ψ1(t) P
P is invertible.
Now Ф(t, to) = Ψ2(t)Ψ2-1(to) = Ψ1(t)P (Ψ1(to)P)-1
= Ψ1(t)P P-1 Ψ1-1(to)= Ψ1(t)Ψ1-1(to)
Ф( , ) is unique
Remark Ф(t, to) is the unique solution of the matrix equation

(t, to )  A(t )(t, to )
*
with the initial condition Ф(to, to) = I
To show *
104

(t )  A(t ) (t )

(t )  1 (t o )  A(t ) (t ) 1 (t o )
d
[  (t )  1 (t o )]  A(t )  (t )  1 (t o )
dt
d
 (t , to )  A(t ) (t , to )
dt
Evaluation of the State Transition Matrix
1.
If A(t) has the following commutative property
t
t
to
to
A(t )(  A( )d )  (  A( )d )A(t )
for all t and to, then
t
 (t , to )  exp(  A( )d )
to
2.
From
Ф(t, to) = Ψ(t)Ψ-1(to)
1 0 
x

Example: Find Ф(t, to) for the system
 0 2t  x ( t )



105
Solution:
e t
 (t )  
0
0
2
et 
Then Ф(t, to) = Ψ(t)Ψ-1(to)
 e t 0   e  to
=
t 2 
0
e

 0
0   e t  to
2 
e  to   0
0 
2 2 
e t  to 

The solution of
Is given by
x (t )  A(t ) x (t ) ,
x(to)=xo
x(t) = Ф(t, to) x(to), where
Ф(t, to) is the state transition matrix.
Solution of

x(t )  A(t ) x(t )  B(t )u(t ),
x(to )  xo *
Theorem: The solution of

x(t )  A(t ) x(t )  B(t )u(t ),
x(to )  xo
Is given by
t
x (t )   (t , to ) x (to )    (t , )B( )u( )d
to
106
Proof: Let z(t) = Ф (to, t)x(t)
x(t) = Ф(t, to)z(t)
Or



x(t )  (t, to ) z(t )  (t, to ) z(t )

Using * and the fact that (t , to )  A(t )(t , to ) we will have

A(t ) x(t )  B(t )u(t )  A(t )(t, to )z(t)  (t, to ) z (t)

A(t )(t, to )z(t)  B(t )u(t )  A(t )(t, to )z(t)  (t, to ) z (t)
or

(t, to ) z (t)  B(t )u(t )

z (t)  (to , t ) B(t )u(t )
By integrating both sides
t 
t
to
to
 z (t )dt    (to , )B( )u( )d
or
t
z (t )  z (to )    (to , )B( )u( )d
to
t
(to , t ) x(t )  (to , to ) x(to )   (to , )B( )u ( )d
to
t
x (t )   (t , to ) x (to )   (t , to )   (to , )B( )u( )d
to
107
t
x (t )   (t , to ) x (to )    (t , )B( )u( )d
to
Zero-input
zero-state
The output
y(t)= C(t)x(t)+D(t)u(t)
t
y(t )  C (t )(t , t o ) x(t o )  C (t )  (t , )B( )u ( )d  D(t )u (t )
to
Recall: For a system which is linear and is relaxed at t = to
t
y (t )   G (t , )u ( )d
to
What is G(t,τ) for

x(t )  A(t ) x(t )  B(t )u (t ),
y(t)  C(t)x(t)  D(t)u(t)
y (t )   C (t ) (t , )B ( )u ( )d  D (t )u (t )
t
to
  [C (t ) (t , )B ( )  D ( ) (t   )]u ( )d
t
to
G (t , )  C (t ) (t , ) B( )  D( ) (t   )
108
Time – Invariant Case

x(t )  Ax(t )  Bu (t ),
y(t)  Cx(t)  Du(t)
Claim:
x(t )  e A(t  t o ) x(to )

Is a solution to x(t )  Ax (t )
given x(to)
Proof:
d
d
x(t )  (e A( t  t o ) x(to ))
dt
dt
 e A( t  t o ) Ax(to )
But A and the function of A commute. Thus

x(t )  e A(t t o ) Ax(to )

x(t )  Ax(t )
Q.E.D
109
From the definition of the state transition matrix then
(t , to )  e A(t to )  (t  to )
Thus the solution for the time-invariant system is
t
x(t )  e A(t to ) x(t o )   e A(t  ) B( )u ( )d
to
If we take to = 0, then
t
x(t )  e xo   e A( t  ) Bu ( )d
At
0
and
y (t )  Ce
At
t
xo   Ce A(t  ) Bu( )d  Du(t )
0
The impulse response matrix is
G (t , to )  G (t  to )  Ce A(t  ) B  D (t   )
Which is commonly written as
G(t )  Ce At B  D (t )
Transfer function
G(s)=C(sI-A)-1B+D
110