STate Space Analysis

It gives how states are representing in various canonical forms and how it it is different from transfer function approach. and finally test the system controllability and observability by kalman's test

1. Dr. K Hussain
Associate Professor and Head
Dept. of Electrical Engineering
SITCOE

2. Advantages of State Space Approach
Classical/Transfer function
Approach
State space approach/Modern
Control Approach
Transfer Function Approach
Linear Time Invariant System,
SISO
Laplace Transform, Frequency
domain
Only Input-output Description:
Less Detail Description on System
Dynamics
Doesn’t take all ICs
Taking specific inputs like step,
ramp and parabolic etc…
State Variable Approach
Linear Time Varying, Nonlinear,
Time Invariant, MIMO
Time domain
Detailed description of Internal
behaviour in addition to I-O
properties
Consider all ICs
Taking all inputs
2

3. CLASSICAL APPROACH (TRANSFER FUNCTION)
Theclassicalapproachorfrequencydomain
techniqueisbasedonconvertingasystem’s
differentialequationtoatransferfunction.
Itrelatesarepresentationoftheoutputtoa
representationoftheinput.
Itcanbeappliedonlytolinear,time-invariant
systems.
Itrapidlyprovidesstabilityandtransientresponse
information.
3

4. The state-space approach (also referred to as the modern, or
time-domain, approach) is a unified method for modeling,
analyzing, and designing a wide range of systems.
The state-space approach can be used to represent nonlinear
systems. Also, it can handle, conveniently, systems with
nonzero initial conditions and time varying.
The state-space approach is also attractive because of the
availability of numerous state-space software package for the
personal computer.
MODERN APPROACH (STATE-SPACE)
4

5. MODERN APPROACH (STATE-SPACE)
Itcanbeusedtomodelandanalyzenonlinear
(backlash,saturation),time-varying(missileswith
varyingfuellevels),multi-inputmulti-output
systems(i.e.,anairplane)withnonzeroinitial
conditions.
Butitisnotasintuitiveastheclassicalapproach.
Thedesignerhastoengageinseveralcalculations
beforethephysicalinterpretationofthemodelis
apparent.
5

6. STATE-SPACE REPRESENTATION
Selectaparticularsubsetofallpossiblesystem
variablesandcallthemstatevariables.
Forannth-ordersystem,writensimultaneous
first-orderdifferentialequationsintermsofstate
variables.
Ifweknowtheinitialconditionsofallstate
variablesatt
0andthesysteminputfortt
0,wecan
solvethesimultaneousdifferentialequationsfor
thestatevariablesfortt
0.
6

7. Concept of State Variables
State:The state of a dynamic system is the smallest set of
variables (called state variables) so that the knowledge of
these variables at t= t
0, together with the knowledge of the
input for tt
0, determines the behavior of the system for any
time tt
0.
State Variables:The state variables of a dynamic system are
the variables making up the smallest set of variables that
determine the state of the dynamic system.
State Vector:If nstate variables are needed to describe the
behavior of a given system, then the nstate variables can be
considered the ncomponents of a vector x. Such vector is
called a state vector.
State Space:The n-dimensional space whose coordinates
axes consist of the x
1axis, x
2axis, .., x
naxis, where x
1, x
2, .., x
n
are state variables, is called a state space.
7

8. Block Diagram Representation of Continuous
Time Control System in State space:
8

9. Time Varying/Invariant Systems
9
Time Varying System:
Time Invariant System:

10. State & Output Equation:
Matrices/Vectors
10
A(t)= State/System Matrix
B(t)= Input Matrix
C(t)=Output Matrix
D(t)=Direct Transmission Matrix

11. The general form of state and output equations for a linear,
time-invariant system can be written asDu+Cx=y Bu+Ax=x
Wherex is the n x 1 state vector,
uis r x 1 input vector,
yis the m x 1 output vector,
Ais nxn System Matrix,
Bis nxr Input Matrix,
C is mxn Output Matrix and
Dis mxr Transition Matrix.
11

12. RLC Circuit
12

13. 13

14. EXAMPLE:
+
i(t)
Lv(t)
RL
di
dt
Rivt
vtRit


()
()()( )
(state equation)
Output equation tLRtLR
L
R
L
R
eie
R
ti
s
i
ssR
sI
sVissIL
)/()/(
)0()1(
1
)(
)0(111
)(
)()]0()([















Assuming that v(t) is a unit step and knowing i(0), taking the LT of the state
equation
14

15. Writing the equations in matrix-vector form

i
v 0
i
v 0
v(t)
c
R
L
1
L
1
C c
1
L

























Assuming the voltage across the resistor as the output v(t)Ri(t)R0
i
v
R
c
 






15

16. Example: Given the electric network, find a state-space representation.
(Hint: state variables and , output)Cv Li )(
/1
0
0/1
/1)/(1
tv
Li
v
L
CRC
i
v
L
C
L
C





























  







L
C
R
i
v
Ri 0/1 Ri  0=D ; 0R=C
;
0
=B ;
0
--
=A ;
v
i
=x
L
R
1
L
1
L
R
c



















C
16

17. 17

18. Transfer function From State space :DuCxy
BuAxx


Given the state and output equations
Take the Laplace transform assuming zero initial conditions:
(1)
(2)
Solving for in Eq. (1),
or
(3)
Substitutin Eq. (3) into Eq. (2) yields
The transfer function is )()()(
)()()(
sss
ssss
DUCXY
BUAXX

 )(sX )()()( sss BUXAI  )()()(
1
sss BUAIX

 )()()()(
1
ssss DUBUAICY 
 )(])([
1
ss UDBAIC 
 DBAIC
U
Y

1
)(
)(
)(
s
s
s
18

19. Ex. Find the transfer from the state-space representationu























0
0
10
321
100
010
xx  x001y 




































321
10
01
321
100
010
00
00
00
)(
s
s
s
s
s
s
sAI
Solution:123
12
31
1323
)det(
)(
)(
23
2
2
2
1



















sss
sss
sss
sss
s
sadj
s
AI
AI
AI 123
)23(10
)(
)(
23
2



sss
ss
s
s
U
Y
19

20. Solution of State Equations:
20

21. 21

22. 22

23. State Transition Matrix
23

24. Laplace Transform Method (STM Computation)
24

25. 25

26. Total Response
26

27. Natural Modes
27

28. Diagonalization
28

29. 29

30. 30

31. STATE SPACE REPRESENTATION OF DYNAMIC
SYSTEMS
Consider the following n-th order differential equation in which the forcing
function does not involve derivative terms.dy
dt
a
dy
dt
a
dy
dt
aybu
Ys
Us
b
sas asa
n
n
n
n n n
n n
n n
  

  

 


1
1
1 1 0
0
1
1
1


()
()
Choosing the state variables asxy, xy, xy, , x
dy
dt
xx, xx, ,xx
x axax axbu
1 2 3 n
n1
n1
1 2 2 3 n-1 n
n n1 n12 1n 0
   
  
 



 
  




31

32. 





x
x
x
x
x
x
0 1 0 00 0
0 0 1 0 0 0
0 0 0100 0
0 0 0 001 0
0 0 0 0001
a a a a a
x
x
x
x
x
x
1
2
3
n2
n1
n n n1 n2 2 1
1
2
3
n2
n1
n








 

























    



































































0
0
0
0
0
b
u
y=[1 0 0 0 0]x
0

 32

33. State Space Representation of system with
forcing function involves Derivatives
Consider the following n-th order differential equation in
which the forcing function involves derivative terms.dy
dt
a
dy
dt
a
dy
dt
ay
b
du
dt
b
du
dt
b
du
dt
bu
Ys
Us
bsbs bsb
sas asa
n
n
n
n n n
n
n
n
n n n
n n
n n
n n
n n
  
   

  
  

 

 




1
1
1 1
0 1
1
1 1
0 1
1
1
1
1
1




()
()
33

34. Define the following n variables as a set of n state variablesuxu
dt
du
dt
ud
dt
ud
dt
yd
x
uxuuuyx
uxuuyx
uyx
nnnnn
n
n
n
n
n
n 11122
2
11
1
0
222103
11102
01

















34

35. 

.
.
.


. . . .
. . . .
. . . .
.
.
.
.
.
.
x
x
x
x a a a a
x
x
x
x
n
n n n n
n
n
n
1
2
1
1 2 1
1
2
1
1
2
1
0 1 0 0
0 0 1 0
0 0 0 1

 
 























   




















































 


n
n
u
y
x
x
x
u









































10 0
1
2
0

.
.
. 35

36. CONTROLLABLE CANONICAL FORMdy
dt
a
dy
dt
a
dy
dt
ay
b
du
dt
b
du
dt
b
du
dt
bu
Ys
Us
bsbs bsb
sas asa
n
n
n
n n n
n
n
n
n n n
n n
n n
n n
n n
  
   

  
  

 

 




1
1
1 1
0 1
1
1 1
0 1
1
1
1
1
1




()
()
36

37. Ys
Us
b
babs babbab
sas asa
YsbUsYs
Ys
babs babbab
sas asa
n
n n n n
n n
n n
n
n n n n
n n
n n
()
()
( ) ( )( )
() ()

()

()
( ) ( )( )

   
  
 

   
  

 



 


0
1 10
1
1 10 0
1
1
1
0
1 10
1
1 10 0
1
1
1




Us
Ys
babs babbab
Us
sas asa
Qs
n
n n n n
n n
n n
()

()
( ) ( )( )
()
()
1 10
1
1 10 0
1
1
1
   

  


 



 37

38. sQsasQs asQsaQsUs
YsbabsQs babsQs
babQs
n n
n n
n
n n
n n
() () () ()()

()( )()( )()
( )()
   
 




 
1
1
1
1 10
1
1 10
0


Defining state variables as follows:XsQs
XssQs
XssQs
XssQs
n
n
n
n
1
2
1
2
1
()()
()()
() ()
() ()







 sXsXs
sXsXs
sXsXs
n n
1 2
2 3
1
() ()
() ()
() ()




 


xx
xx
x x
n n
1 2
2 3
1





38

39. sXsaXs aXsaXsUs
x ax axaxu
YsbUsbabsQs babsQs
babQs
bUsbabXs
n n n n
n n n n
n
n n
n n
n
() () () ()()

() ()( )()( )()
( )()
()( )()
   
  
   

 



 
1 12 1
1 12 1
0 1 10
1
1 10
0
0 1 10



+
= 

 

    
 
 
( )()
( )()
( )( ) ( )
babXs
babXs
ybabxbabx babxbu
n n
n n
n n n n n
1 10 2
0 1
01 1 102 1 10 0
+ 39

40.   ub
x
x
x
babbabbaby
u
x
x
x
x
aaaax
x
x
x
n
nnnn
n
n
nnnn
n
0
2
1
0110110
1
2
1
121
1
2
1
.
.
.
1
0
.
.
.
0
0
.
.
.
1000
....
....
....
0100
0010
.
.
.





























































































































 40

41. BLOCK DIAGRAM
u +
b
0
+ + +
+ + + +
+
y
+
+
+
+
+
+
+
b
1-a
1b
0 b
2-a
2b
0 b
n-1-a
n-1 b
0 b
n-a
n b
0
 

a
1 a
2
a
n-1 a
n
x
n x
n-1
x
2 x
1
41

42. OBSERVABLE CANONICAL FORMsYsbUssaYsbUs
saYsbUsaYsbUs
YsbUs bUsaYs
bUsaYs bUsaYs
Xs
n n
n n n n
s
s n n s n n
n s
n n
[() ()][() ()]
[() ()]() ()
() ()[() ()]
[() ()][() ()]
()[
   
    
   
   


 
 

0
1
1 1
1 1
0
1
1 1
1
1 1
1
1
0
1



bUsaYsXs
Xs bUsaYsXs
Xs bUsaYsXs
Xs bUsaYs
YsbUsXs
n
n s n
sn n
sn n
n
1 1 1
1
1
2 2 2
2
1
1 1 1
1
1
0
() () ()]
()[() () ()]
()[() ()()]
()[() ()]
() ()()
 
  
  
 
 

 
 

42

43. sXsXsaXsbabUs
sXsXsaXsbabUs
sXsXsaXsbabUs
sXsaXsbabUs
xaxb
n n n
n n n
n n n n
nn n n
nn
() () ()( )()
() () ()( )()
()() ()( )()
() ()( )()
 (
  
  
   
 
 

 
  
1 1 1 10
1 2 2 2 20
2 1 1 1 10
1 0
1

n n
nn n n
n n n
n
abu
xxaxbabu
xxaxbabu
yxbu

  
 

  

0
2 1 1 1 10
1 1 1 10
0
)
 ( )
 ( )
 43

44. 

.
.
.


.. .
.. ..
.. ..
.
.
.
.
x
x
x
x
a
a
a
a
x
x
x
x
bab
ba
n
n
n
n
n
n
n n
n n
1
2
1
1
2
1
1
2
1
0
1
00 0
10 0
00
00 1



 













































































 
10
1 10
1
2
0
00 1
b
bab
u
y
x
x
x
bu
n
.
.
.
.
.
.
.










































 44

45. BLOCK DIAGRAM
b
0
a
1a
n-1
a
n
+
+
+
+
+
+
+
  
y
u
b
n-a
nb
0
b
n-1-a
n-1b
0
b
1-a
1b
0
x
1
x
2x
n-1 x
n
45

46. DIAGONAL CANONICAL FORM
Consider that the denominator polynomial involves only
distinct roots. Then,Ys
Us
bsbs bsb
spsp sp
b
c
sp
c
sp
c
sp
n n
n n
n
n
n
()
()( )( )( )

 
  







0 1
1
1
1 2
0
1
1
2
2



where, c
i, i=1,2, …, n are the residues corresponding p
i
46

47. YsbUs
c
sp
Us
c
sp
Us
c
sp
Xs
sp
UssXs pXsUs
Xs
sp
UssXs pXsUs
Xs
sp
UssXs pXsUs
n
n
n
n
n nn
() () () ()
() () () ()()
() () () ()()
() () () ()()
 







  


  


  
0
1
1
2
2
1
1
1 11
2
2
2 22
1
1
1

 47

48. 


() () () () ()
xpxu
x pxu
x pxu
YsbUscXscXs cXs
ycxcx cxbu
n nn
nn
nn
1 11
2 22
0 11 22
11 22 0

 
 
   



 48

49.  


.
.
.

.
.
.
.
.
.
.
.
.
.
.
x
x
x
p
p
p
x
x
x
u
ycc c
x
x
x
bu
n
n
n
n
n
1
2
1
2
1
2
1 2
1
2
0
0
0
1
1
1






























































































 49

50. BLOCK DIAGRAM
y
u
x
1
x
2
x
n
c
1
b
0
c
2
c
n
+
+
+
+
+1
1
sp 1
2
sp 1
sp
n

50

51. JORDAN CANONICAL FORM
Consider the case where the denominator polynomial involves multiple roots.
Suppose that the p
i’s are different from one another, except that the first three are
equal.Ys
Us
bsbs bsb
spspsp sp
Ys
Us
b
c
sp
c
sp
c
sp
c
sp
c
sp
n n
n n
n
n
n
()
()( )( )( )( )
()
() ( )( )

  
   











0 1
1
1
1
3
4 5
0
1
1
3
2
1
2
3
1
4
4



51

52. YsbUs
c
sp
Us
c
sp
Us
c
sp
Us
c
sp
Us
c
sp
Us
Xs
c
sp
Us
Xs
c
sp
Us
Xs
Xssp
Xs
c
sp
Us
Xs
X
n
n
() ()
( )
()
( )
()
() () ()
()
( )
()
()
( )
()
()
()
() ()
()
 












 




0
1
1
3
2
1
2
3
1
4
4
1
1
1
3
2
2
1
2
1
2 1
3
3
1
2
1
+ 
3 1
4
4
4
1
()
() ()
() ()
ssp
Xs
c
sp
Us
Xs
c
sp
Us
n
n
n






 52

53. sXs pXsXsx pxx
sXs pXsXsx pxx
sXs pXsUsx pxu
sXs pXsUsx pxu
sXs pXsUsx
n nn
1 11 2 1 11 2
2 12 3 2 12 3
3 13 3 13
4 44 4 44
() ()()
() ()()
() ()()
() ()()
() ()()
  
  
  
   
  

n nn
nn
nn
pxu
YsbUscXscXscXs cXs
ybucxcxcx cx
 
    
 
() () () () () ()
0 11 22 33
0 11 22 33

 53

54. BLOCK DIAGRAM
b
0
c
3
c
2
c
1
c
4
c
n
x
3
x
2 x
1
x
4
x
n
.
.
.
.
.
.1
1
sp 1
1
sp 1
4
sp 1
1
sp 1
sp
n

u
y
+
+
+
+
+
+
+
+
+
54

55. 




. .
.
x
x
x
x
x
p
p
p
p
p
x
x
x
x
x
u
ycc
n n n
1
2
3
4
1
1
1
4
1
2
3
4
1 2
1 0 0 0
0 1
0 0 0 0
0 0
0 0 0
0
0
1
1
1




 

















































































 

c
x
x
x
bu
n
n
1
2
0












 55

56. Example: Express in CCF/OCF/DCF
56
CCF
OCF
DCF

57. Converting a Transfer Function to State Space
A set of state variables is called phase variables, where each
subsequent state variable is defined to be the derivative of the
previous state variable.
Consider the differential equation
Choosing the state variablesubya
dt
dy
a
dt
yd
a
dt
yd
n
n
nn
n
0011
1
1 


  ubxaxaxa
dt
yd
x
dt
yd
x
x
dt
yd
x
dt
yd
x
x
dt
yd
x
dt
dy
x
x
dt
dy
xyx
nnn
n
nn
n
n 0121101
1
43
3
32
2
3
32
2
22
211











57

58. u
bx
x
x
x
x
aaaaaax
x
x
x
x
n
n
nn
n






















































































0
1
3
2
1
143210
1
3
2
1
0
0
0
0
100000
001000
000100
000010












  






















n
n
x
x
x
x
x
y
1
3
2
1
00001

 58

59. Converting a transfer function with constant term in numerator
Step 1. Find the associated differential equation.
Step 2. Select the state variables.
Choosing the state variables as successive derivatives.24269
24
)(
)(
23


ssssR
sC rcccc
sRsCsss
2424269
)(24)()24269(
23


 cx
cx
cx





3
2
1 1
3213
32
21
2492624
xy
rxxxx
xx
xx






 
59

60. Converting a transfer function with polynomial in numerator
Step1. Decomposing a transfer function.
Step 2. Converting the transfer function with constant term in numerator.
Step 3. Inverse Laplace transform.01
2
2
3
3
1
1
)(
)(
asasasasR
sX

 )()()()(
101
2
2 sXbsbsbsCsY  10
1
12
1
2
2)( xb
dt
dx
b
dt
xd
bty  01
2
2
3
3
1
1
)(
)(
asasasasR
sX

 )()()()(
101
2
2 sXbsbsbsCsY  322110)( xbxbxbty 
60

61. Ex. Find the transfer from the state-space representationu























0
0
10
321
100
010
xx  x001y 




































321
10
01
321
100
010
00
00
00
)(
s
s
s
s
s
s
sAI
Solution:123
12
31
1323
)det(
)(
)(
23
2
2
2
1



















sss
sss
sss
sss
s
sadj
s
AI
AI
AI 123
)23(10
)(
)(
23
2



sss
ss
s
s
U
Y
61

62. 62

63. 63

64. Output Eq. modifies to
64