#### Transcript State Space Trajectories

```Analysis of Control Systems in
State Space
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Introduction to State Space
• The state space is defined as the n-dimensional space in
which the components of the state vector represents its
coordinate axes.
• In case of 2nd order system state space is 2-dimensional space
with x1 and x2 as its coordinates (Fig-1).
 x1 
 a11
   
 x2 
 a 21
a12   x 1 
 b1 
      u(t)
a 22   x 2 
b2 
x2
x1
Fig-1: Two Dimensional State space
State Transition
• Any point P in state space represents the state of the system
at a specific time t.
x2
P(x1, x2)
x1
• State transitions provide complete picture of the system
x2
t0
t6
t1
t2
t3
t5
t4
x1
Forced and Unforced Response
• Forced Response, with u(t) as forcing function
 x1   a 11
   
 x 2   a 21
a 12   x 1   b1 
      u (t )
a 22   x 2   b 2 
• Unforced Response (response due to initial conditions)
 x1   a 11
   
 x 2   a 21
a 12   x 1 ( 0 ) 


a 22   x 2 ( 0 ) 
Solution of State Equations & State Transition Matrix
• Consider the state space model
x ( t )  Ax ( t )  Bu ( t )
• Solution of this state equation is given as
t
x ( t )   ( t ) ( 0 )    ( t   ) Bu ( ) d 
0
• Where  ( t ) is state transition matrix.
1
1
 ( t )   [( SI  A ) ]  e
At
Example-1
• Consider RLC Circuit
iL
Vc
+
+
Vo
-
C
dv c
dt
 u ( t )  iL
L
di L
-
  Ri L  v c
V o  Ri L
dt
• Choosing vc and iL as state variables
dv c
dt
 
1
C
iL 
1
C
u (t )
di L
dt

1
L
vc 
R
L
iL
Example-1 (cont...)

0
 v c 
   1
 iL 

L
1

1
v



c
C
  C u(t )



R  iL   


0
L
R  3, L  1 and
 v c 
0
   
1
 iL 
C  0 .5
 2  vc  2 
     u(t )
 3   iL   0 
Example-1 (cont...)
 v c 
0
   
1
 iL 
 2  vc  2 
     u(t )
 3   iL   0 
• State transition matrix can be obtained as
  S
1
1
1 
 ( t )   [( SI  A ) ]     
 0

 
0  0

S  1
 2 
 
 3 
• Which is further simplified as
S  3 


 1 ( S  1 )( S  2 )
 (t )   
1

 ( S  1 )( S  2 )


( S  1 )( S  2 ) 

S

( S  1 )( S  2 ) 
2
1





Example-1 (cont...)
S  3 


 1 ( S  1 )( S  2 )
 (t )   
1

 ( S  1 )( S  2 )


( S  1 )( S  2 ) 

S

( S  1 )( S  2 ) 
2
• Taking the inverse Laplace transform of each element
t
2 t
( 2 e  e )
 (t )   t
2t
 (e  e )
( 2e
(e
t
t
2 t
)

2t
 2e ) 
 2e
State Space Trajectories
• The unforced response of a system released from any initial
point x(to) traces a curve or trajectory in state space, with time
t as an implicit function along the trajectory.
• Unforced system’s response depend upon initial conditions.
x ( t )  Ax ( t )
• Response due to initial conditions can be obtained as
x(t )   (t ) x( 0 )
Example-2
• For the RLC circuit of example-1 draw the state space trajectory
with following initial conditions.
• Solution
 v c ( 0 )


 iL ( 0 ) 
1 
 
2
x(t )   (t ) x( 0 )
t
2 t
t
2 t
t
2 t
v



1
 c  ( 2e  e ) ( 2e  2e )  
 3e  3e 
  t
    t
2t
t
2t   
2t 
 iL   ( e  e ) (  e  2 e )   2    e  3e 
Example-2 (cont...)
• Following trajectory is obtained
State Space Trajectory of RLC Circuit
2
1.5
t-------->inf
iL
1
0.5
0
-0.5
-1
-1
-0.5
0
0.5
Vc
1
1.5
2
Example-2 (cont...)
State Space Trajectories of RLC Circuit
2
1.5
0 
 
1 
1
iL
0.5
1 
 
0 
  1
 
 0 
0
-0.5
 0 
 
  1
-1
-1.5
-2
-2
-1.5
-1
-0.5
0
Vc
0.5
1
1.5
2
Equilibrium Point
• The equilibrium or stationary state of the
system is when
x (t )  0
State Space Trajectories of RLC Circuit
2
1.5
1
iL
0.5
0
-0.5
-1
-1.5
-2
-2
-1.5
-1
-0.5
0
Vc
0.5
1
1.5
2
```