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Transfer Matrix (State Space to T.F) Substituting equation (5) into equation (4) yields
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Feedback Control Systems (FCS)
Dr. Imtiaz Hussainemail: [email protected]
URL :http://imtiazhussainkalwar.weebly.com/
Lecture-36-37Transfer Matrix and solution of state equations
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)
)()()( sBUsAXssX
)()()( sBUsXAsI
)()()( 1 sBUAsIsX (5)
Transfer Matrix (State Space to T.F)• Substituting equation (5) into equation (4) yields
)()()()( 1 sDUsBUAsICsY
)()()( 1 sUDBAsICsY
DBAsICsUsY
1)()()(
Example#1
• Convert the following State Space Model to Transfer Function Model if K=3, B=1 and M=10;
)(tfMv
x
MB
MK
vx
1010
vx
ty 10)(
Example#1
• Substitute the given values and obtain A, B, C and D matrices.
)(1010
101
103
10tf
vx
vx
vx
ty 10)(
Example#1
101
103
10A
10C
1010
B
0D
DBAsICsUsY
1)()()(
Example#1
101
103
10A
10C
1010
B
0D
1010
101
103
10
00
10)()(
1
ss
sUsY
Example#1
1010
101
103
10
00
10)()(
1
ss
sUsY
1010
101
103
110
)()(
1
ss
sUsY
1010
103
1101
103)
101(
110)()(
s
s
sssUsY
Example#1
1010
103
1101
103)
101(
110)()(
s
s
sssUsY
1010
103
103)
101(
1)()( s
sssUsY
10103)
101(
1)()( s
sssUsY
Example#1
10103)
101(
1)()( s
sssUsY
3)110()()(
sss
sUsY
Example#2
• 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)
)(tubb
xx
aaaa
xx
2
1
2
1
2221
1211
2
1
)()(00
2
1
2221
1211
2
1
xx
aaaa
xx
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-3• Consider RLC Circuit obtain the state transition matrix ɸ(t).
Vc
+
-
+
-
VoiL
)(0
1
1
10tuCi
v
LR
L
Civ
L
c
L
c
5013 ., CandLR
)(tuiv
iv
L
c
L
c
02
3120
Example-3 (cont...)
)(tuiv
iv
L
c
L
c
02
3120
1
111
3120
00
])[()(S
SASIt
))(())((
))(())(()(2121
121
221
31
SSS
SS
SSSSS
t
• State transition matrix can be obtained as
• Which is further simplified as
Example-3 (cont...)
))(())((
))(())(()(2121
121
221
31
SSS
SS
SSSSS
t
• Taking the inverse Laplace transform of each element
)()()()()(
tttt
tttt
eeeeeeee
t22
22
2222
Example#4• Compute the state transition matrix if
300020001
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
2xP(x1, x2)
1x
2xt0
t1
t2
t3
t4t5
t6
Example-5• For the RLC circuit of example-3 draw the state space trajectory
with following initial conditions.
• Solution
21
00)()(
L
c
iv
21
)2()()22()2(
22
22
tttt
tttt
L
c
eeeeeeee
iv
)()()( 0xttx
tt
tt
L
c
eeee
iv
2
2
333
ttL
ttc
eei
eev2
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
01
01
10
10
Equilibrium Point• The equilibrium or stationary state of the system
is when0)(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
AsIsBUxsX
)()0()(
Solution of State Equations
• Taking the inverse Laplace transform of above equation.
AsIsBUxsX
)()0()(
AsIsBU
AsIxsX
)()0()(
dtutxttxt
)()()0()()(0
Natural ResponseForced 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.
)(10
3210
2
1
2
1 tuxx
xx
Solution
• Calculate the state transition matrix])[()( 11 ASIt
Example#6• Obtain the state transition equation of the system
dtutxttxt
)()()0()()(0
END OF LECTURES-36-37
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