Download EE 529 Circuit and Systems Analysis Lecture 9

EE 529 Circuits and
Systems Analysis
Lecture 9
Mustafa Kemal Uyguroğlu
State vector
a listing of state variables in vector form
 x1 (t) 
 x (t) 
 2 
x(t)    


x
(t)
n

1


 xn (t) 
Eastern Mediterranean University
State equations
 x1 (t) 
 x (t) 
x (t)   2   A x(t)  Bu(t)
  
 
 xn (t) 
y (t)  C x(t)  Du(t)
System dynamics
Measurement
Read-out map
Eastern Mediterranean University
x:n-vector (state vector)
u:p-vector (input vector)
y:m-vector (output vector)
n
n A:nxn
p
n B:nxp
n
m C:mxn
p
m D:mxp
System matrix
Input (distribution) matrix
Output matrix
Direct-transmission matrix
Eastern Mediterranean University
Solution of state eq’ns
Consists of:
Free response & Forced sol’n
(Homogenous sol’n)
(particular sol’n)
Eastern Mediterranean University
Homogenous solution
Homogenous equation
x  Ax
has the solution
x(t)  Φ(t) x0
State transition matrix
Eastern Mediterranean University
X(0)
State transition matrix
An nxn matrix (t), satisfying
(t )  A(t ),
(0)  I
where I is n  n identity matrix.
x  0 =   0 x  0
Eastern Mediterranean University
Determination of (t): transform method
Laplace transform of the differential equation:
Eastern Mediterranean University
Determination of (t): transform method
At


  t   L  sI  A   e


1
Eastern Mediterranean University
1
Determination of (t): time-domain solution
Scalar case
x  ax
x t    t  x  0
 (t )  a. (t )

 (t )  e
at
where


e  1  at  ( at )  ......... 
at
1
2!
2

k 0
Eastern Mediterranean University
1
k!
k k
a t
Determination of (t): time-domain solution
For vector case, by analogy
 (t )  A. (t )

 (t )  e
At
where


e 1  At  ( At )  .........   A t
At
1
2!
2
k 0
Can be verified by substitution.
Eastern Mediterranean University
1
k!
k k
Properties of TM
(0)=I
-1(t)= (-t)
t0
Ф(t2-t1)Φ(t1-t0)= Φ(t2-t0)
[Φ(t)]k= Φ(kt)
t1
t2
Φ(t1-t0) Φ(t2-t0Φ(t
) 2-t1)
Φ(t) Φ(t) Φ(t)
Φ(kt)Φ(t) Φ(t) Φ(t)
Eastern Mediterranean University
General solution
Scalar case
Eastern Mediterranean University
General solution
Vector case
Eastern Mediterranean University
General solution: transform method
L{ x  Ax  Bu }

s xˆ(s)  x(0)  Axˆ(s)  Buˆ(s)

1
1
ˆ  (sI  A) x(0)  (sI  A) Bu(s)
ˆ
x(s)
Eastern Mediterranean University
Inverse Laplace transform yields:
x(t)  Φ(t) x(0)  Φ(t) * Bu(t)
t
x(t)  e x(0)   e
At
A(t - )
Bu( )d
0
Eastern Mediterranean University
For initial time at t=t0
x(t)  e
A(t -t0 )
t
x(t 0 )   e
A(t - )
t0
Eastern Mediterranean University
Bu( )d
The output
y(t)=Cx(t)+Du(t)
y (t)  Ce
t
A(t - t 0 )
x(t 0 )  C  e
A(t - )
Bu(t)d   Du(t)
t0
Zero-input
response
Zero-state response
Eastern Mediterranean University
Example
 Obtain the state transition matrix (t) of the
following system. Obtain also the inverse of the state
transition matrix -1(t) .
 x1   0 1   x1 
 x    2 3  x 
 2
 2 
For this system
0 1
A


2

3


the state transition matrix (t) is given by
(t )  eAt  L1[(sI  A)1 ]
since
 s 0  0 1   s 1 
sI  A = 
   2 3   2 s  3
0
s

 
 

Eastern Mediterranean University
Example
The inverse (sI-A) is given by
 s  3 1
1
( sI  A ) 
( s  1)( s  2)  2 s 
1
 s 21  s 1 2
( sI  A )   2
2

 s 1 s  2
1
 s 1 2 
1
2 

s 1
s2 
1
s 1
Hence
 2et  e2t
(t )  
t
2t
 2e  2e
et  e2t 

et  2e2t 
Noting that -1(t)= (-t), we obtain the inverse of transition matrix as:
1
 (t )  e
 At
 2et  e2t

t
2t
 2e  2e
et  e 2 t 

et  2e2t 
Eastern Mediterranean University
Exercise 1
Find x1(t) , x2(t)
The initial condition
Eastern Mediterranean University
Exercise 1 (Solution)
x = (t)x(0)
1
 (t )  L 1  sI  A  


 s 1 
sI  A  

5 s  4 
 s4
 s  4 1  s 2  4 s  5
1
1

 sI  A   2
5
s  4 s  5  5 s  
2
 s  4 s  5
Eastern Mediterranean University
1

s 2  4s  5 

s

2
s  4 s  5 
Example 2
Eastern Mediterranean University
Exercise 2
Find x1(t) , x2(t)
The initial condition
Input is Unit Step
Eastern Mediterranean University
Exercise 2 (Solution)
Eastern Mediterranean University
Matrix Exponential eAt
Eastern Mediterranean University
Matrix Exponential eAt
Eastern Mediterranean University
The transformation
1
 1

2
 1
2
22
P=  1


 
1n 1 2n 1
x = Pxˆ
1 
 n 

2
 n 

 
 nn 1 

where
1,2,…,n are distinct
eigenvalues of A. This
transformation will transform
P-1AP into the diagonal
matrix
0
1



2

P 1AP = 





n 
0
Eastern Mediterranean University
Example 3
Eastern Mediterranean University
 Method 2:
e At  L1[( sI  A) 1 ]
 s 0  0 1   s 1 
sI  A = 
  0 2   0 s  2 
0
s

 
 

1  1
1
 s  2 1  s s  s  2    s
1
1

( sI  A) 

s  
s  s  2   0
1  
0

  0
s2 

1

2 t 
1
1

e




2
At
1
1
e  L [( sI  A) ]  

1
0
e 2t 


2
Eastern Mediterranean University
11
1 



2  s s  2 

1


s2
Matrix Exponential eAt
Eastern Mediterranean University
Matrix Exponential eAt
Eastern Mediterranean University
Example 4
Eastern Mediterranean University
Laplace Transform
Eastern Mediterranean University
Eastern Mediterranean University
Eastern Mediterranean University