Advanced Control Systems (ACS)
Lecture-6
State Space Modeling & Analysis
Dr. Imtiaz Hussain
email: imtiaz.hussain@faculty.muet.edu.pk
URL :http://imtiazhussainkalwar.weebly.com/
Lecture Outline
– Basic Definitions
– State Equations
– State Diagram
– State Controllability
– State Observability
– Output Controllability
– Transfer Matrix
– Solution of State Equation
Definitions
• State of a system: We define the state of a system at time t0 as
the amount of information that must be provided at time t0,
which, together with the input signal u(t) for t  t0, uniquely
determine the output of the system for all t  t0.
• State Variable: The state variables of a dynamic system are the
smallest set of variables that determine the state of the
dynamic system.
• State Vector: If n variables are needed to completely describe
the behaviour of the dynamic system then n variables can be
considered as n components of a vector x, such a vector is called
state vector.
• State Space: The state space is defined as the n-dimensional
space in which the components of the state vector represents
its coordinate axes.
Definitions
• Let x1 and x2 are two states variables that define the state
of the system completely .
dx
dt
x2
State (t=t1)
Velocity
State (t=t1)
State
Vector
x1
Two Dimensional State space
x
Position
State space of a Vehicle
4
State Space Equations
• In state-space analysis we are concerned with three types of
variables that are involved in the modeling of dynamic systems:
input variables, output variables, and state variables.
• The dynamic system must involve elements that memorize the
values of the input for t> t1 .
• Since integrators in a continuous-time control system serve as
memory devices, the outputs of such integrators can be
considered as the variables that define the internal state of the
dynamic system.
• Thus the outputs of integrators serve as state variables.
• The number of state variables to completely define the dynamics
of the system is equal to the number of integrators involved in the
system.
State Space Equations
• Assume that a multiple-input, multiple-output system involves 𝑛
integrators.
• Assume also that there are 𝑟 inputs 𝑢1 𝑡 , 𝑢2 𝑡 , ⋯ , 𝑢𝑟 𝑡 and 𝑚
outputs 𝑦1 𝑡 , 𝑦2 𝑡 , ⋯ , 𝑦𝑚 𝑡 .
• Define 𝑛 outputs of the integrators as state variables:
𝑥1 𝑡 , 𝑥2 𝑡 , ⋯ , 𝑥𝑛 𝑡 .
• Then the system may be described by
𝑥1 𝑡 = 𝑓1 (𝑥1 , 𝑥2 , ⋯ , 𝑥𝑛 ; 𝑢1 , 𝑢2 , ⋯ , 𝑢𝑟 ; 𝑡)
𝑥2 𝑡 = 𝑓2 (𝑥1 , 𝑥2 , ⋯ , 𝑥𝑛 ; 𝑢1 , 𝑢2 , ⋯ , 𝑢𝑟 ; 𝑡)
𝑥𝑛 𝑡 = 𝑓𝑛 (𝑥1 , 𝑥2 , ⋯ , 𝑥𝑛 ; 𝑢1 , 𝑢2 , ⋯ , 𝑢𝑟 ; 𝑡)
State Space Equations
• The outputs 𝑦1 𝑡 , 𝑦2 𝑡 , ⋯ , 𝑦𝑚 𝑡 of the system may be given as.
𝑦1 𝑡 = 𝑔1 (𝑥1 , 𝑥2 , ⋯ , 𝑥𝑛 ; 𝑢1 , 𝑢2 , ⋯ , 𝑢𝑟 ; 𝑡)
𝑦2 𝑡 = 𝑔2 (𝑥1 , 𝑥2 , ⋯ , 𝑥𝑛 ; 𝑢1 , 𝑢2 , ⋯ , 𝑢𝑟 ; 𝑡)
𝑦𝑚 𝑡 = 𝑔𝑚 (𝑥1 , 𝑥2 , ⋯ , 𝑥𝑛 ; 𝑢1 , 𝑢2 , ⋯ , 𝑢𝑟 ; 𝑡)
• If we define
𝑥1
𝑥2
𝒙 𝑡 = ⋮
𝑥𝑛
𝑓1 (𝑥1 , 𝑥2 , ⋯ , 𝑥𝑛 ; 𝑢1 , 𝑢2 , ⋯ , 𝑢𝑟 ; 𝑡)
𝑓 (𝑥 , 𝑥 , ⋯ , 𝑥𝑛 ; 𝑢1 , 𝑢2 , ⋯ , 𝑢𝑟 ; 𝑡)
𝒇 𝒙, 𝒖, 𝑡 = 2 1 2
⋮
𝑓𝑛 (𝑥1 , 𝑥2 , ⋯ , 𝑥𝑛 ; 𝑢1 , 𝑢2 , ⋯ , 𝑢𝑟 ; 𝑡)
𝑦1
𝑦2
𝒚 𝑡 = ⋮
𝑦𝑚
𝑔1 (𝑥1 , 𝑥2 , ⋯ , 𝑥𝑛 ; 𝑢1 , 𝑢2 , ⋯ , 𝑢𝑟 ; 𝑡)
𝑔2 (𝑥1 , 𝑥2 , ⋯ , 𝑥𝑛 ; 𝑢1 , 𝑢2 , ⋯ , 𝑢𝑟 ; 𝑡)
𝒈 𝒙, 𝒖, 𝑡 =
⋮
𝑔𝑚 (𝑥1 , 𝑥2 , ⋯ , 𝑥𝑛 ; 𝑢1 , 𝑢2 , ⋯ , 𝑢𝑟 ; 𝑡)
𝑢1
𝑢2
𝒖 𝑡 = ⋮
𝑢𝑟
State Space Modeling
• State space equations can then be written as
𝒙 𝑡 = 𝒇(𝒙, 𝒖, 𝑡)
𝒚 𝑡 = 𝒈(𝒙, 𝒖, 𝑡)
State Equation
Output Equation
• If vector functions f and/or g involve time t explicitly, then
the system is called a time varying system.
State Space Modeling
• If above equations are linearised about the operating
state, then we have the following linearised state
equation and output equation:
x (t )  A(t ) x(t )  B(t )u (t )
y (t )  C (t ) x(t )  D(t )u (t )
State Space Modeling
• If vector functions f and g do not involve time t explicitly then
the system is called a time-invariant system.
• In this case, state and output equations can be simplified to
x (t )  Ax(t )  Bu (t )
y (t )  Cx(t )  Du (t )
D
C
B
A
Example-1
• Consider the mechanical system shown in figure. We assume that
the system is linear. The external force u(t) is the input to the
system, and the displacement y(t) of the mass is the output. The
displacement y(t) is measured from the equilibrium position in the
absence of the external force. This system is a single-input, singleoutput system.
• From the diagram, the system equation is
𝑚𝑦(𝑡) + 𝑏𝑦(𝑡) + 𝑘𝑦(𝑡) = 𝑢(𝑡)
• This system is of second order. This means that
the system involves two integrators. Let us
define state variables 𝑥1 (𝑡) and 𝑥2 (𝑡) as
𝑥1 𝑡 = 𝑦(𝑡)
𝑥2 𝑡 = 𝑦(𝑡)
Example-1
𝑥1 𝑡 = 𝑦(𝑡)
𝑥2 𝑡 = 𝑦(𝑡)
𝑚𝑦(𝑡) + 𝑏𝑦(𝑡) + 𝑘𝑦(𝑡) = 𝑢(𝑡)
• Then we obtain
𝑥1 𝑡 = 𝑥2 (𝑡)
• Or
𝑏
𝑘
1
𝑥2 𝑡 = − 𝑦 𝑡 − 𝑦 𝑡 + 𝑢 (𝑡)
𝑚
𝑚
𝑚
𝑥1 𝑡 = 𝑥2 (𝑡)
𝑏
𝑘
1
𝑥2 𝑡 = − 𝑥2 𝑡 − 𝑥1 𝑡 + 𝑢 (𝑡)
𝑚
𝑚
𝑚
• The output equation is
𝑦 𝑡 = 𝑥1 𝑡
Example-1
𝑥1 𝑡 = 𝑥2 (𝑡)
𝑏
𝑘
1
𝑥2 𝑡 = − 𝑥2 𝑡 − 𝑥1 𝑡 + 𝑢 (𝑡)
𝑚
𝑚
𝑚
𝑦 𝑡 = 𝑥1 𝑡
• In a vector-matrix form,
 x1 (t )   0
 x (t )   k
 2   m
y (t )  1
1   x (t )   0 
b   1    1 u (t )
   x2 (t )  
m
m
 x1 (t ) 
0

x
(
t
)
 2

Example-1
• State diagram of the system is
𝑥1 𝑡 = 𝑥2 (𝑡)
𝑏
𝑘
1
𝑥2 𝑡 = − 𝑥2 𝑡 − 𝑥1 𝑡 + 𝑢 (𝑡)
𝑚
𝑚
𝑚
𝑦 𝑡 = 𝑥1 𝑡
-k/m
-b/m
𝑢(𝑡) 1/m
𝑥2
1/s
𝑥2 = 𝑥1
1/s
𝑥1
𝑦(𝑡)
Example-1
• State diagram in signal flow and block diagram format
-k/m
-b/m
𝑢(𝑡)
1/m
𝑥2
1/s
𝑥2 = 𝑥1
1/s
𝑥1
𝑦(𝑡)
Example-2
• State space representation of armature Controlled D.C Motor.
Ra
ea
La
ia
B 
eb
T
J

• ea is armature voltage (i.e. input) and  is output.
dia
ea  Ra ia  La
 eb
dt
T  J  B
Example-2
T  K t ia
eb  K b
J  B-K t ia  0
dia
La
 Ra ia  K b  ea
dt
• Choosing 𝜃, 𝜃 𝑎𝑛𝑑 𝑖𝑎 as state variables
0
1
𝐵
𝜃
0 −
𝑑
𝐽
𝜃 =
𝑑𝑡 𝑖
𝐾𝑏
𝑎
0 −
𝐿𝑎
0
0
𝐾𝑡
𝜃
0
𝐽
𝜃 + 1 𝑒𝑎
𝑅𝑎 𝑖𝑎
𝐿𝑎
−
𝐿𝑎
• Since 𝜃 is output of the system therefore output equation is given as
𝑦 𝑡 = 1 0
𝜃
0 𝜃
𝑖𝑎
State Controllability
• A system is completely controllable if there exists an
unconstrained control u(t) that can transfer any initial
state x(to) to any other desired location x(t) in a finite
time, to ≤ t ≤ T.
uncontrollable
controllable
State Controllability
• Controllability Matrix CM

CM  B
AB
A2 B  An1B
• System is said to be state controllable if
rank (CM )  n

State Controllability (Example)
• Consider the system given below
 1 0 
1
x  
x   u

 0  3
 0
y  1 2x
• State diagram of the system is
U (s ) 1
s -1
x1
1
1
2
 3 x2
s -1
Y (s)
State Controllability (Example)
• Controllability matrix CM is obtained as
CM  B AB
1
B 
0 
 1
AB   
0
• Thus
1  1
CM  

0 0 
• Since 𝑟𝑎𝑛𝑘(𝐶𝑀) ≠ 𝑛 therefore system is not completely
state controllable.
State Observability
• A system is completely observable if and only if there exists a finite
time T such that the initial state x(0) can be determined from the
observation history y(t) given the control u(t), 0≤ t ≤ T.
observable
unobservable
State Observability
• Observable Matrix (OM)
 C 
 CA 


2
Observabil ity Matrix OM   CA 


  
CAn 1 
• The system is said to be completely state observable if
rank (OM )  n
State Observability (Example)
• Consider the system given below
0 1  0 
x  
x   u

0  2 1
y  0 4x
• OM is obtained as
• Where
C 
OM   
CA
C  0 4
0 1 
CA  0 4
 0  12

0  2 
State Observability (Example)
• Therefore OM is given as
0 4 
OM  

0

12


• Since 𝑟𝑎𝑛𝑘(𝑂𝑀) ≠ 𝑛 therefore system is not completely state
observable.
4
U (s ) 1
s -1
x2 s -1
2
x1
Y (s)
Output Controllability
• Output controllability describes the ability of an external
input to move the output from any initial condition to any
final condition in a finite time interval.
• Output controllability matrix (OCM) is given as

OCM  CB CAB CA2 B  CAn 1 B

Home Work
• Check the state controllability, state observability
and output controllability of the following system
0 1
0
A
, B   , C  0 1

1 0
1
Transfer Matrix (State Space to T.F)
• Now Let us convert a space model to a transfer function model.
x (t )  Ax(t )  Bu (t )
(1)
y(t )  Cx(t )  Du (t )
(2)
• Taking Laplace transform of equation (1) and (2) considering initial
conditions to zero.
sX ( s )  AX ( s )  BU ( s )
Y ( s )  CX ( s )  DU ( s )
(3)
(4)
• From equation (3)
( sI  A) X ( s )  BU ( s )
X ( s )  ( sI  A) 1 BU ( s )
(5)
Transfer Matrix (State Space to T.F)
• Substituting equation (5) into equation (4) yields
1
Y ( s )  C ( sI  A) BU ( s )  DU ( s )
Y ( s )  C ( sI  A)
1

B  D U (s)
Example#3
• Convert the following State Space Model to
Transfer Function Model if K=3, B=1 and
M=10;
 x   0
     K
v   M
1   x  0 
B      1  f (t )
  v   
M
M 
y (t )  0
 x
1 
v 
Example#3
• Substitute the given values and obtain A, B, C
and D matrices.
0


x
 
3


 v  
   10
1   x  0 
1      1  f (t )
  v   
10 
10 
y (t )  0
 x
1 
v 
Example#3
 0
A 3

 10
1 
1
 
10 
C  0
0
B1
10 
1
Y ( s)
 C ( sI  A) 1 B  D
U ( s)
D0
Example#3
 0
A 3

 10
1 
1
 
10 
C  0
1
  s 0  0
Y (s)
 0 1 
 3

 0 s   
U (s)
 10

0
B1
10 
D0
1 
1 
 
10  
1
0
1
10 
Example#3
  s 0  0
Y (s)
 0 1 
 3

 0 s   
U (s)
 10

 s
Y (s)
 0 1  3

U (s)
 10
1 
1 
 
10  
1  
1 
s  
10  
1
1
0
1
10 
0
1
10 
1


 s  10 1  0 
Y (s)
1
1
 0 1


3
1
3
U (s)
s  10 
s(s  )   
10 10  10

Example#3
1


 s  10 1  0 
Y (s)
1
1
 0 1


3
1
3
U (s)
s  10 
s(s  )   
10 10  10

Y (s)
1

U (s) s(s  1 )  3
10 10
 3
 10
0

 1
s  
 10 
Y (s)
1
s

U ( s ) s ( s  1 )  3 10
10 10
Example#3
Y (s)
1
s

U ( s ) s ( s  1 )  3 10
10 10
Y (s)
s

U ( s ) s (10 s  1)  3
Home Work
• Obtain the transfer function T(s) from
following state space representation.
Answer
Forced and Unforced Response
• Forced Response, with u(t) as forcing function
 x1   a11 a12   x1   b1 
  u(t )
 x   a



 2   21 a22   x2  b2 
• Unforced Response (response due to initial conditions)
 x1   a11 a12   x1( 0)

 x   a

a
 2   21 22   x2 ( 0)
Solution of State Equations
• Consider the state equation given below
x (t )  Ax(t )
(1)
• Taking Laplace transform of the equation (1)
sX ( s )  x(0)  AX ( s )
sX ( s )  AX ( s )  x(0)
sI  AX ( s)  x(0)
X ( s )  sI  A x(0)
1
X (s) 
x(0)
sI  A
1
Solution of State Equations
1
X (s) 
x(0)
sI  A
• Taking inverse Laplace
x(t )  e At x(0)
 (t )  e
At
State Transition Matrix
Example-4
• Consider RLC Circuit obtain the state transition matrix ɸ(t).

vc   0
 i    1
 L 
L
1
  v   1 
c
C
  C u(t )



R  iL   
 
0
L
R  3, L  1 and C  0.5
vc  0
 i   
 L  1
 2 vc  2
i    u(t )

 3  L  0
iL
Vc
Vo
+
+
-
-
Example-4 (cont...)
vc  0
 i   
 L  1
 2 vc  2
i    u(t )

 3  L  0
• State transition matrix can be obtained as
  S
 (t )   1[( SI  A)1 ]   1  
  0
0  0  2  
 



S  1  3  

• Which is further simplified as
2
 S  3




1 ( S  1)( S  2 )
(
S

1
)(
S

2
)
 (t )   

1
S


 ( S  1)( S  2) ( S  1)( S  2) 
1
Example-4 (cont...)
2

 S  3



1 ( S  1)( S  2 )
)
2

S
)(
1

S
(
 (t )   

S
1


 ( S  1)( S  2) ( S  1)( S  2) 
• Taking the inverse Laplace transform of each element
( 2e  e ) ( 2e  2e )
 ( t )   t  2 t
t
 2t 
 ( e  e ) ( e  2e ) 
t
2t
t
2t
Home Work
• Compute the state transition matrix if
0
 1 0


A  0 2 0 
 0
0  3
Solution
 (t )   [( SI  A) ]
1
1
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)
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
Example-5
• For the RLC circuit of example-4 draw the state space trajectory
with following initial conditions.
vc (0) 1
  
 i L ( 0 )   2
• Solution
x(t )  (t )x(0)
vc  (2e t  e 2t ) (2e t  2e 2t ) 1
 i    t  2 t
t
 2t   
 L   (e  e ) (e  2e )  2
vc   3e  3e 
 i    t
 2t 
 L    e  3e 
t
2 t
vc  3e t  3e 2t
iL  e t  3e 2t
Example-5 (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-5 (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
Solution of State Equations
• Consider the state equation with u(t) as forcing function
x (t )  Ax(t )  Bu (t )
• Taking Laplace transform of the equation (1)
sX ( s )  x(0)  AX ( s )  BU ( s )
sX ( s )  AX ( s )  x(0)  BU ( s )
sI  AX ( s)  x(0)  BU ( s)
x(0)  BU ( s )
X (s) 
sI  A
(1)
Solution of State Equations
x(0)  BU ( s )
X (s) 
sI  A
x(0) BU ( s )
X (s) 

sI  A sI  A
• Taking the inverse Laplace transform of above equation.
t
x(t )   (t ) x(0)    (t   )u ( )dt
0
Natural Response
Forced Response
Example#6
• Obtain the time response of the following system:
 x1   0
 x    2
 2 
1   x1  0
  u (t )



 3  x2  1
• Where u(t) is unit step function occurring at t=0.
consider x(0)=0.
Solution
• Calculate the state transition matrix
 (t )   1[( SI  A) 1 ]
Example#6
• Obtain the state transition equation of the system
t
x(t )   (t ) x(0)    (t   )u ( )dt
0
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END OF LECTURE-6