state space
DESCRIPTION
best stateTRANSCRIPT
State Space Modeling & Analysis
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 .
4
1x
2x
Two Dimensional State space
State (t=t1)
State Vector x
dt
dx
State space of a Vehicle
Velocity
Position
State (t=t1)
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
𝑥 1 𝑡 = 𝑓1(𝑥1, 𝑥2, ⋯ , 𝑥𝑛; 𝑢1, 𝑢2, ⋯ , 𝑢𝑟; 𝑡)
𝑥 2 𝑡 = 𝑓2(𝑥1, 𝑥2, ⋯ , 𝑥𝑛; 𝑢1, 𝑢2, ⋯ , 𝑢𝑟; 𝑡)
𝑥 𝑛 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.
• If we define
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)
𝒙 𝑡 =
𝑥1𝑥2⋮𝑥𝑛
𝒇 𝒙, 𝒖, 𝑡 =
𝑓1(𝑥1, 𝑥2, ⋯ , 𝑥𝑛; 𝑢1, 𝑢2, ⋯ , 𝑢𝑟; 𝑡) 𝑓2(𝑥1, 𝑥2, ⋯ , 𝑥𝑛; 𝑢1, 𝑢2, ⋯ , 𝑢𝑟; 𝑡)
⋮𝑓𝑛(𝑥1, 𝑥2, ⋯ , 𝑥𝑛; 𝑢1, 𝑢2, ⋯ , 𝑢𝑟; 𝑡)
𝒚 𝑡 =
𝑦1𝑦2⋮𝑦𝑚
𝒈 𝒙, 𝒖, 𝑡 =
𝑔1(𝑥1, 𝑥2, ⋯ , 𝑥𝑛; 𝑢1, 𝑢2, ⋯ , 𝑢𝑟; 𝑡) 𝑔2(𝑥1, 𝑥2, ⋯ , 𝑥𝑛; 𝑢1, 𝑢2, ⋯ , 𝑢𝑟; 𝑡)
⋮𝑔𝑚(𝑥1, 𝑥2, ⋯ , 𝑥𝑛; 𝑢1, 𝑢2, ⋯ , 𝑢𝑟; 𝑡)
𝒖 𝑡 =
𝑢1𝑢2⋮𝑢𝑟
State Space Modeling
• State space equations can then be written as
• If vector functions f and/or g involve time t explicitly, then the system is called a time varying system.
𝒙 𝑡 = 𝒇(𝒙, 𝒖, 𝑡)
𝒚 𝑡 = 𝒈(𝒙, 𝒖, 𝑡)
State Equation
Output Equation
State Space Modeling
• If above equations are linearised about the operating state, then we have the following linearised state equation and output equation:
)()()()()( tutBtxtAtx )()()()()( tutDtxtCty
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
)()()( tButAxtx )()()( tDutCxty
B
D
A
C
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, single-
output 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
𝑚𝑦 (𝑡) + 𝑏𝑦 (𝑡) + 𝑘𝑦(𝑡) = 𝑢(𝑡)
• Then we obtain
• Or
• The output equation is
𝑥1 𝑡 = 𝑦(𝑡) 𝑥2 𝑡 = 𝑦 (𝑡)
𝑥 1 𝑡 = 𝑥2(𝑡)
𝑥 2 𝑡 = −𝑏
𝑚𝑦 𝑡 −
𝑘
𝑚𝑦 𝑡 +
1
𝑚𝑢 (𝑡)
𝑥 1 𝑡 = 𝑥2(𝑡)
𝑥 2 𝑡 = −𝑏
𝑚𝑥2 𝑡 −
𝑘
𝑚𝑥1 𝑡 +
1
𝑚𝑢 (𝑡)
𝑦 𝑡 = 𝑥1 𝑡
Example-1
)(10
)(
)(10
)(
)(
2
1
2
1tu
mtx
tx
m
b
m
ktx
tx
)(
)(01)(
2
1
tx
txty
• In a vector-matrix form,
𝑥 1 𝑡 = 𝑥2(𝑡) 𝑥 2 𝑡 = −𝑏
𝑚𝑥2 𝑡 −
𝑘
𝑚𝑥1 𝑡 +
1
𝑚𝑢 (𝑡) 𝑦 𝑡 = 𝑥1 𝑡
Example-1
• State diagram of the system is
𝑥 1 𝑡 = 𝑥2(𝑡)
𝑥 2 𝑡 = −𝑏
𝑚𝑥2 𝑡 −
𝑘
𝑚𝑥1 𝑡 +
1
𝑚𝑢 (𝑡)
𝑦 𝑡 = 𝑥1 𝑡
1/s 1/s 𝑢(𝑡) 𝑦(𝑡)
-k/m
-b/m 𝑥 2 1/m
𝑥2 = 𝑥 1
𝑥1
Example-1
• State diagram in signal flow and block diagram format
1/s 1/s 𝑢(𝑡) 𝑦(𝑡)
-k/m
-b/m 𝑥 2 1/m
𝑥2 = 𝑥 1
𝑥1
Example-2 • State space representation of armature Controlled D.C Motor.
• ea is armature voltage (i.e. input) and is output.
ea
ia T
Ra La
J
B
eb
ba
aaaa edt
diLiRe
BJT
atiKT bb Ke
abaaa
a
at
eKiRdt
diL
i-KBJ
0
Example-2
• Choosing 𝜃, 𝜃 𝑎𝑛𝑑 𝑖𝑎 as state variables
• Since 𝜃 is output of the system therefore output equation is given as
𝑑
𝑑𝑡
𝜃𝜃
𝑖𝑎
=
0 1 0
0 −𝐵
𝐽
𝐾𝑡𝐽
0 −𝐾𝑏𝐿𝑎
−𝑅𝑎𝐿𝑎
𝜃𝜃
𝑖𝑎
+
001
𝐿𝑎
𝑒𝑎
𝑦 𝑡 = 1 0 0
𝜃𝜃
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.
controllable
uncontrollable
State Controllability • Controllability Matrix CM
• System is said to be state controllable if
BABAABBCM n 12
)( nCMrank
State Controllability (Example) • Consider the system given below
• State diagram of the system is
xy
uxx
21
0
1
30
01
1
1
)(sU
)(sY1
-1s
3-1s
2
1x
2x
State Controllability (Example) • Controllability matrix CM is obtained as
• Thus
• Since 𝑟𝑎𝑛𝑘(𝐶𝑀) ≠ 𝑛 therefore system is not completely
state controllable.
ABBCM
00
11 CM
0
1 B
0
1A B
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)
• The system is said to be completely state observable if
1
2MMatrix ity Observabil
nCA
CA
CA
C
O
nOMrank )(
State Observability (Example) • Consider the system given below
• OM is obtained as
• Where
xy
uxx
40
1
0
20
10
CA
COM
40C
12020
1040
CA
State Observability (Example) • Therefore OM is given as
• Since 𝑟𝑎𝑛𝑘(𝑂𝑀) ≠ 𝑛 therefore system is not completely state
observable.
120
40MO
1)(sU -1s-1s 1x2x
2
4
)(sY
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
BCABCACABCBCM n 12O
Home Work
• Check the state controllability, state observability and output controllability of the following system
10,1
0,
01
10
CBA
Transfer Matrix (State Space to T.F)
• Now Let us convert a space model to a transfer function model.
• Taking Laplace transform of equation (1) and (2) considering initial conditions to zero.
• From equation (3)
)()()( tButAxtx (1)
)()()( tDutCxty (2)
)()()( sBUsAXssX (3)
)()()( sDUsCXsY (4)
)()()( sBUsXAsI
)()()( 1 sBUAsIsX (5)
Transfer Matrix (State Space to T.F) • Substituting equation (5) into equation (4) yields
)()()()( 1 sDUsBUAsICsY
)()()( 1 sUDBAsICsY
Example#3
• Convert the following State Space Model to Transfer Function Model if K=3, B=1 and M=10;
)(tf
Mv
x
M
B
M
Kv
x
1010
v
xty 10)(
Example#3
• Substitute the given values and obtain A, B, C and D matrices.
)(
10
10
10
1
10
310
tfv
x
v
x
v
xty 10)(
Example#3
10
1
10
310
A
10C
10
10
B
0D
DBAsICsU
sY 1)(
)(
)(
Example#3
10
1
10
310
A
10C
10
10
B
0D
10
10
10
1
10
310
0
010
)(
)(1
s
s
sU
sY
Example#3
10
10
10
1
10
310
0
010
)(
)(1
s
s
sU
sY
10
10
10
1
10
31
10)(
)(1
s
s
sU
sY
10
10
10
3
110
1
10
3)
10
1(
110
)(
)(
s
s
sssU
sY
Example#3
10
10
10
3
110
1
10
3)
10
1(
110
)(
)(
s
s
sssU
sY
10
10
10
3
10
3)
10
1(
1
)(
)(s
sssU
sY
10
10
3)
10
1(
1
)(
)( s
sssU
sY
Example#3
10
10
3)
10
1(
1
)(
)( s
sssU
sY
3)110()(
)(
ss
s
sU
sY
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
• Unforced Response (response due to initial conditions)
)(tub
b
x
x
aa
aa
x
x
2
1
2
1
2221
1211
2
1
)(
)(
0
0
2
1
2221
1211
2
1
x
x
aa
aa
x
x
Solution of State Equations • Consider the state equation given below
• Taking Laplace transform of the equation (1) )()( tAxtx (1)
)()0()( sAXxssX
)0()()( xsAXssX
)0()( xsXAsI
)0()(1xAsIsX
)0(1
)( xAsI
sX
Solution of State Equations
• Taking inverse Laplace
)0(1
)( xAsI
sX
)0()( xetx At
Atet )( State Transition Matrix
Example-4 • Consider RLC Circuit obtain the state transition matrix ɸ(t).
Vc
+
-
+
-
Vo iL
)(tuCi
v
L
R
L
Ci
v
L
c
L
c
0
1
1
10
5013 ., CandLR
)(tui
v
i
v
L
c
L
c
0
2
31
20
Example-4 (cont...)
)(tui
v
i
v
L
c
L
c
0
2
31
20
1
111
31
20
0
0
S
SASIt ])[()(
))(())((
))(())(()(
2121
121
2
21
3
1
SS
S
SS
SSSS
S
t
• State transition matrix can be obtained as
• Which is further simplified as
Example-4 (cont...)
))(())((
))(())(()(
2121
121
2
21
3
1
SS
S
SS
SSSS
S
t
• Taking the inverse Laplace transform of each element
)()(
)()()(
tttt
tttt
eeee
eeeet
22
22
2
222
Home Work
• Compute the state transition matrix if
300
020
001
A
])[()( 11 ASIt
Solution
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.
• Response due to initial conditions can be obtained as
)()( tAxtx
)()()( 0xttx
State Transition • Any point P in state space represents the state of the system
at a specific time t.
• State transitions provide complete picture of the system
1x
2x
P(x1, x2)
1x
2xt0
t1
t2
t3
t4 t5
t6
Example-5 • For the RLC circuit of example-4 draw the state space trajectory
with following initial conditions.
• Solution
2
1
0
0
)(
)(
L
c
i
v
2
1
)2()(
)22()2(22
22
tttt
tttt
L
c
eeee
eeee
i
v
)()()( 0xttx
tt
tt
L
c
ee
ee
i
v2
2
3
33tt
L
tt
c
eei
eev
2
2
3
33
Example-5 (cont...) • Following trajectory is obtained
-1 -0.5 0 0.5 1 1.5 2-1
-0.5
0
0.5
1
1.5
2
Vc
iLState Space Trajectory of RLC Circuit
t-------->inf
Example-5 (cont...)
-2 -1.5 -1 -0.5 0 0.5 1 1.5 2-2
-1.5
-1
-0.5
0
0.5
1
1.5
2
Vc
iLState Space Trajectories of RLC Circuit
0
1
0
1
1
0
1
0
Equilibrium Point
• The equilibrium or stationary state of the system is when
0)(tx
-2 -1.5 -1 -0.5 0 0.5 1 1.5 2-2
-1.5
-1
-0.5
0
0.5
1
1.5
2
Vc
iL
State Space Trajectories of RLC Circuit
Solution of State Equations • Consider the state equation with u(t) as forcing function
• Taking Laplace transform of the equation (1) )()()( tButAxtx (1)
)()()0()( sBUsAXxssX
)()0()()( sBUxsAXssX
)()0()( sBUxsXAsI
AsI
sBUxsX
)()0()(
Solution of State Equations
• Taking the inverse Laplace transform of above equation.
AsI
sBUxsX
)()0()(
AsI
sBU
AsI
xsX
)()0()(
dtutxttx
t
)()()0()()(0
Natural Response Forced Response
Example#6
• Obtain the time response of the following system:
• Where u(t) is unit step function occurring at t=0. consider x(0)=0.
)(1
0
32
10
2
1
2
1tu
x
x
x
x
Solution
• Calculate the state transition matrix
])[()( 11 ASIt
Example#6
• Obtain the state transition equation of the system
dtutxttx
t
)()()0()()(0
END OF LECTURE