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Solutions of Autonomous Linear Systems
x t   A t x t ,
SOLUTION:
x (t 0 )  x 0 ,
x t   R ,
n
A t   R
n n
x t   Φ ( t , t 0 )  x t 0 
State transition matrix Φ ( t , t 0 ) describes how states evolve.
Properties of the State Transition Matrix
1-
Φ ( t 2 , t 0 )  Φ ( t 2 , t1 ) Φ ( t1 , t 0 ),
x (t 0 )  x 0
x ( t1 )  Φ ( t1 , t 0 ) x 0
x ( t1 )  x 1
x ( t 2 )  Φ ( t 2 , t1 ) x 1
t 0  t1  t 2
x ( t 2 )  Φ ( t 2 , t1 ) Φ ( t1 , t 0 ) x 0
x (t 2 )  Φ (t 2 , t 0 ) x 0
Φ ( t 2 , t 0 )  Φ ( t 2 , t1 ) Φ ( t1 , t 0 )
2-
Φ
1
(t , t 0 )  Φ (t 0 , t )
Φ ( t 2 , t 0 )  Φ ( t 2 , t1 ) Φ ( t1 , t 0 ),
Φ ( t 0 , t 0 )  Φ ( t 0 , t1 ) Φ ( t1 , t 0 ),
t 0  t1  t 2
t0  t2
I   ( t 0 , t1 ) Φ ( t1 , t 0 )
Φ
1
(t , t 0 )  Φ (t 0 , t )
3- State transition matrix satisfies the matrix
differential equation X = A × X
with initial condition Φ ( t 0 , t 0 )  I .
Solutions of Non-autonomous Linear Systems
x t   A t x t   B t u t ,
x t   R , A t   R
n
Assumption:
n n
x (t 0 )  x 0 ,
, B t   R
n p
x (t )  Φ (t , t 0 ) y (t )
d
 (t , t ) y (t )

x (t ) 
( Φ ( t , t 0 ) y ( t ))  Φ ( t , t 0 ) y ( t )  Φ
0
dt
 Φ ( t , t 0 ) y ( t )  A t Φ ( t , t 0 ) y ( t )
Substituting the solution in x t   A t x t   B t u t 
one gets A Φ ( t , t 0 ) y ( t )  Bu  Φ ( t , t 0 ) y ( t )  A Φ ( t , t 0 ) y ( t )
Bu  Φ ( t , t 0 ) y ( t )
1
y ( t )  Φ ( t , t 0 ) Bu
t
y (t ) 
Φ
t0
1
( , t 0 ) Bu ( ) d   y ( t 0 )
y (t 0 )  Φ
1
(t 0 , t 0 ) x 0  x 0
x (t )  Φ (t , t 0 ) y (t )
t
x (t )  Φ (t , t 0 ) x 0  Φ (t , t 0 )  Φ
1
( , t 0 ) Bu ( ) d 
t0
t
x (t )  Φ (t , t 0 ) x 0 
 Φ ( t ,  ) Bu ( ) d 
t0
For linear time-invariant systems
Φ (t , t 0 ) 
ˆ e
Φ (t ,0 )  e
x(t) = e A(t-t0 ) x 0 +
A ( t  t0 )
At
t
A(t-t )
e
Bu(t )d t
ò
t0
One should calculate e A ( t  t
0
)
in order to find the solution.
How to calculate e
1- Using its definition as a series
At
?
Taylor series expansion of f t  at t  0 :
f ( t )  f ( 0 )  f ( 0 ) t 
f  ( 0 )
t
2
Reminder
 ....... 
2!
f
(n)
t
n
 ....
n!
 x1 ( t )   x1 ( 0 )   x1 ( 0 ) 
 x1 ( 0 ) 

 
 



x2 ( 0 ) t 2
x 2 (t )
x2 (0)
x 2 ( 0 )
  

t  

x (t )  
 .....

 
 


 2!

 
 






x
(
t
)
x
(
0
)
x
(
0
)
x
(
0
)
 3
  3
  3

 3













x ( t ) t  0  Ax ( t ) t  0  x ( 0 )  Ax 0
x ( t )
 (t )
x ( 0 )  A x ( 0 )  A 2 x 0

A
x

t0
t0
x( t )
x ( t )
x( 0 )  A x ( 0 )  A 3 x 0

A

t0
t0



x
(k )
(t )
t0
 Ax
( k 1 )
(t )
t0
x ( t )  x ( 0 )  Ax ( 0 ) t 
1
2!
 x
(k )
( 0 )  Ax
A x ( 0 )t 
2
2
1
( k 1 )
(0)  A x 0
k
A x ( 0 ) t  ....
3
3
3!
1 2 2 1 3 3


x ( t )   I  A t  A t  A t  ....  x ( 0 )
2!
3!


x (t )  e x ( 0 )  e
At
At
1 2 2 1 3 3


  I  A t  A t  A t  .... 
2!
3!


How to calculate e
At
?
2- Jordan Canonical Form
By a similarity transformation, the matrix A can be upper-diagonalized.
A1



1
P AP  J  




A2
.
Ai
.








A k 
i

Ai  



1
i
For an eigenvalue  i with (algebraic) multiplicity k i
there are n i  k i Jordan blocks.
n i  dim(ker( A   i  I )) and it is called the geometric
multiplicity.
It is the number of independent eigenvectors that can be found
using ( A   i I ) p i  0 . Other (generalized) eigenvectors will be
found using ( A   i I ) p  p .
1
...



1

i 
P y ( t )  APy ( t ),
x ( t )  Ax ( t ),
x ( t ) ˆ Py ( t )
1
y ( t )  P APy ( t )
Columns of P are the .......................................
How can we find e
At
?
J
 0

2

 0
A  
 0
 0

  1
0
0
0
0
1
1
1
0
0
2
1
0
0
0
2
0
0
0
0
2
0
0
0
0
1 

1

0 

0 
0 

2 
Find e A t using Jordan canonical form!
>> A=[0 0 0 0 0 1;2 1 -1 -1 0 -1;0 0 2 1 0 0;0 0 0 2 0 0;0 0 0 0 2 0;-1 0 0 0 0 2]
A=
0 0 0 0 0 1
2 1 -1 -1 0 -1
0 0 2 1 0 0
0 0 0 2 0 0
0 0 0 0 2 0
-1 0 0 0 0 2
>> [P,J]=jordan(A)
P=
0 0 0 2
-1 0 2 -3
1 0 0 0
0 1 0 0
0 0 0 0
0 0 0 2
J=
2 1 0 0
0 2 0 0
0 0 1 1
0 0 0 1
0 0 0 0
0 0 0 0
-1
-2
0
0
0
1
0
0
0
0
1
0
0
0
0
1
1
0
0
0
0
0
0
2
>> expm(J*t)
ans =
[ exp(2*t), t*exp(2*t), 0,
0,
0,
0]
[
0, exp(2*t), 0,
0,
0,
0]
[
0,
0, exp(t), t*exp(t), (t^2*exp(t))/2,
0]
[
0,
0, 0, exp(t),
t*exp(t),
0]
[
0,
0, 0,
0,
exp(t),
0]
[
0,
0, 0,
0,
0, exp(2*t)]
>> pretty(ans)
3- Laplace Transform
Definition: Let f ( t ), t  0 be a continuous or

 t
piecewise continuous function. If
f
(
t
)
e
dt   ,   0


then the Laplace transform of f (t ) is given by
Pierre-Simon,
marquis de Laplace
1749-1827

F (s) 
ˆ

f (t ) e
 st
dt
0
F ( s )  L  f ( t )  is the Laplace transform of
f (t )  L
-1
f (t ) .
F ( s ) is the inverse Laplace transform.
Properties of the Laplace Transform
1- Uniqueness
Proof will be skipped..
2- Linearity
f1 ( t )  L  f1 ( t )  F1 ( s )
If c1 , c 2 are real constants, then
f 2 ( t )  L  f 2 ( t )   F2 ( s )
L c1 f1 ( t )  c 2 f 2 ( t )   c1 F1 ( s )  c 2 F2 ( s )
Proof: L c1 f 1 ( t )  c 2 f 2 ( t )  

 [ c1 f 1 ( t )  c 2 f 2 ( t )] e
 st
dt
0




c1 f 1 ( t ) e
 st
0
dt 

c 2 f 2 (t ) e
 st
dt
0

 c1  f 1 ( t ) e

 st
dt  c 2  f 2 ( t ) e
0
 c1 F1 ( s )  c 2 F 2 ( s )
0
 st
dt
3-
f ( t )  F ( s )  L  f ( t )
df ( t )
dt
Proof:
 df ( t ) 

 L
  sF ( s )  f ( 0 )
 dt 
 df ( t ) 
L
 
 dt 

e
0
 st
df ( t )
dt
dt
u
dv
 e
 st

f (t )



0
f ( t )(  se
0


  f ( 0 )  s  f (t ) e
0

  f ( 0 )  sF ( s )

 sF ( s )  f ( 0 )
 st
dt
 st
)dt
4-
f ( t )  F ( s )  L  f ( t )


 at
 at
fˆ ( t )  e
f ( t )  Fˆ ( s )  L e
f (t )  F ( s  a )
Proof:

L e

 at
 e
f (t ) 
 st
e
 at
f ( t )dt
0



e
( s  a )t
f ( t )dt
0

S 
ˆ s a


e
 St
f ( t )dt  F ( S )
0
 F (s  a)
5-
f ( t )  F ( s )  L  f ( t )
f ( t  T1 ) u ( t  T1 )  L  f ( t  T1 ) u ( t  T1 )   e
Proof:

0e
 st
dt 

f ( t  T1 ) e
T1
0

 ˆ t  T1
d 
ˆ dt
F (s)

T1
L  f ( t  T1 ) u ( t  T1 )  
 sT 1



f ( ) e
 s (   T1 )
d
T1

 e
 sT 1

f ( ) e
T1
 e
 sT 1
F (s)
 s
d
 st
dt
6-
f ( t )  F ( s )  L  f ( t )
t

1
f ( )d   L   f ( )d   
F (s)
s
0

t

0
Proof:
t

L   f ( )d   
0

 t

0
f ( )d  e
 st
dt
0
u
dv
t
 [  f ( )d  ]
0
u
e
 st
 s



0
1
s
F (s)
0
f (t )
0
 st
 s
dt
v
t
0

0

e
e

  [  f ( )d  ]
 [  f ( )d  ]
 s
 s

0
 0


e
0

 1

s




0
f (t ) e
 st
dt
7-
f ( t )  F ( s )  L  f ( t )
 s 
f ( at )  L  f ( at )  
F 
a
a
1
Proof:

L  f ( at )  

f ( at ) e
 st
dt
0
at 
ˆ p
adt 
ˆ dp
S 
ˆ
s




sp
a
f ( p )e

1
a


1
a
a
1
a
dp
a
0

1


f ( p )e
sp
a
dp
0


f ( p )e
 Sp
dp
0
F (S ) 
 s 
F 
a
a
1
8- f1 ( t )  L  f1 ( t )  F1 ( s )
f 2 ( t )  L  f 2 ( t )   F2 ( s )
t

0
t


f 1 ( ) f 2 ( t   ) d   L   f 1 ( ) f 2 ( t   ) d 

0





t


 L   f 1 ( t   ) f 2 ( ) d 

0





f1 ( t ) * f 2 ( t )
 F1 ( s ) F 2 ( s )
Some usefull formulas for Laplace Transform:
http://en.wikipedia.org/wiki/Laplace_transform
Finding the Zero-Input Solution using Laplace Transform
x  Ax ,
x (0 )  x 0
-
s X ( s )  x 0  AX ( s )
s X ( s )  AX ( s )  x 0
[ s I  A ]X ( s )  x 0
X ( s )  [ sI  A ]
1
x0
Φ (s)
x (t )  L
-1
Φ ( s )x 0
x (t )  e
At
x0
zero-input solution
x  Ax ,
x (0 )  x 0
2

A  1

 0
1
-
2
1
0
 1 



1 , x0  1



 0 
2 
Find the zero-input solution!
t
 h ( t   ) i ( ) d 
v o (t ) 
s
0
How can one find the output y (t ) for a given input u (t ) for linear timeinvariant systems.

input
u (t )
system
y (t )
h (t )

impulse response
y (t ) 
y ( t )  lim
output
 u ( k   ) h (t k   )   )
  0
N   k  

 u ( ) h ( t   ) d    u ( ) h (t   ) d  ˆ u (t ) * h (t )

0

Finding the Zero-State Solution using Laplace Transform

x  Ax  Bu , x ( 0 )  x 0
s X ( s )  x 0  AX ( s )  BU ( s )
0
s X ( s )  x 0  AX ( s )  BU ( s )
[ s I  A ] X ( s )  BU ( s )
X ( s )  [ sI  A ]
1
BU ( s )
Φ (s)
X ( s )  Φ ( s ) BU ( s )
x ( t )  L { Φ ( s ) BU ( s )}
-1
t
x (t ) 
 Φ ( t ,  ) Bu ( ) d 
0
zero-state solution
Finding the Total Solution using Laplace Transform

x  Ax  Bu , x ( 0 )  x 0
X ( s )  Φ ( s ) x 0  Φ ( s ) BU ( s )
t
x (t )  e
At
x0 
 Φ ( t ,  ) Bu ( ) d 
0
OUTPUT
y  Cx  Du
Y ( s )  CX ( s )  DU ( s )
Y ( s )  C [ Φ ( s ) x 0  Φ ( s ) BU ( s )]  DU ( s )
Y ( s )  C Φ ( s ) x 0  [ Φ ( s ) B  D ]U ( s )
t
y (t )  C e
At
x0  C
 Φ ( t ,  ) Bu ( ) d 
0
 Du ( t )
x  Ax  Bu ,
x (0 )  x 0
-
y ( t )  Cx ( t )
2

A  1

 0
1
2
1
0
0 

 
1 , x0  1

 
 0 
2 
0 
1
 
B  0 , C  
 
0
 1 
0
1
0  Find the output for the given LTI
 system.
1
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