Z Transform (1) Hany Ferdinando Dept. of Electrical Eng. Petra Christian University.
State Space Analysis Hany Ferdinando Dept. of Electrical Engineering Petra Christian University.
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State Space Analysis Hany Ferdinando Dept. of Electrical Engineering Petra Christian University
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Transcript of State Space Analysis Hany Ferdinando Dept. of Electrical Engineering Petra Christian University.
State Space Analysis
Hany FerdinandoDept. of Electrical Engineering
Petra Christian University
State Space 2 - Hany Ferdinando 2
OverviewState Transition MatrixTime ResponseDiscrete-time evaluation
State Space 2 - Hany Ferdinando 3
State Transition Matrix
The solution of
t
o
tt deet )()0()( )( Buxx AA
t
o
dttt )()()0()()( Buxxis
If the initial condition x(0), input u() and the state transition matrix (t) are known the time response of x(t) can be evaluated
State Space 2 - Hany Ferdinando 4
State Transition Matrix
The (t) is inverse Laplace Transform of (s) and 1)( AIss
When the input u(t) is zero, then
)0()()( xX ss
State Space 2 - Hany Ferdinando 5
State Transition Matrix
)0()()( xX ss
From the equation above we can expand the matrix into (for example, two elements)
)0(
)0(
)()(
)()(
)(
)(
2
1
2221
1211
2
1
x
x
ss
ss
sX
sX
State Space 2 - Hany Ferdinando 6
State Transition Matrix
The 11(s) can be evaluated from the relation between X1(s) and x1(0), the 12(s), 21(s) and
22(s) can be evaluated with the same procedure
)0(
)0(
)()(
)()(
)(
)(
2
1
2221
1211
2
1
x
x
ss
ss
sX
sX
State Space 2 - Hany Ferdinando 7
Time Response
It is the time response of X(t).
First, find (t) from (s). It is simply the inverse Laplace Transform of (s). Do the inverse Laplace Transform for each element of (s).
)0()()( xX tt
State Space 2 - Hany Ferdinando 8
Example
i(t)
LC
C itidt
dvCi )( LC
LL Riv
dt
diLu
State Space 2 - Hany Ferdinando 9
Example
LC
LC
iC
tiC
v
iC
tiCdt
dv
1)(
1
1)(
1
LCL
LCL
iL
RvL
i
iL
RvLdt
di
1
1
If x1 = vC and x2 = iL then
21
1)(
1x
Cti
Cx 212
1x
L
Rx
Lx
State Space 2 - Hany Ferdinando 10
Example
)(0
1
1
10
2
1
2
1 tiCx
x
L
R
L
Cx
x
)(0
2
31
20
2
1
2
1 tix
x
x
x
For R = 3, L = 1 and C = 0.5,
State Space 2 - Hany Ferdinando 11
Example
I(s) V(s)s-1 s-11/C
-1/C
1/L
-R/LX1(s) X2(s)
R
x1(0)/s x2(0)/s
State Space 2 - Hany Ferdinando 12
Example
s-1 s-1
-1/C
1/L
-R/LX1(s) X2(s)
x1(0)/s
x2(0)/s
When U(s) = 0
State Space 2 - Hany Ferdinando 13
Example11(s) is transfer function of X1(s)/x1(0). Here, use the Mason Gain Formula to get 11(s)
s
xs
sX
)0()(.1
)(
11
1
1
111
1
1 )(.1)(
)0(
)( sss
x
sX
State Space 2 - Hany Ferdinando 14
Example
1(s) is path cofactor of , is 1 + 3s-1 + 2s-2
1(s) = 1 + 3s-1
1
111
)(.1)(
sss
23
3
231
)31()(
221
11
11
ss
s
ss
sss
State Space 2 - Hany Ferdinando 15
Example
With the same procedures, find the 12(s), 21(s) and 22(s)!
2323
123
2
23
3
)(
22
22
ss
s
ss
ssss
s
s
State Space 2 - Hany Ferdinando 16
Example
31
2
s
ss AI
s
s
ssss
1
23
2)3(
1)( 1AI
2323
123
2
23
3
)(
22
22
ss
s
ss
ssss
s
s
State Space 2 - Hany Ferdinando 17
Example
tttt
tttt
eeee
eeeet
22
22
2
222)(
Then the X(t) can be calculated with
)0()()( xX tt
State Space 2 - Hany Ferdinando 18
Discrete-time Evaluation
For discrete-time, use the approximation
T
tTt )()( xxx
)()()()(
)()()()(
)()()()(
tTtTTt
tTttTTt
ttT
tTt
BuxIAx
BuxAxx
BuAxxx
State Space 2 - Hany Ferdinando 19
Discrete-time Evaluation
)()()()1(
)()()()1(
kTktk
kTTkTTTk
Buxx
BuxIAx
)()( IA Tt
State Space 2 - Hany Ferdinando 20
ExampleWith the same example above and T = 0.2s,
)(0
2
31
20
2
1
2
1 tix
x
x
x
)(2.0)()2.0()1( kkk BuxIAx
)(0
4.0
4.02.0
4.01
2
1
2
1 tix
x
x
x
State Space 2 - Hany Ferdinando 21
Matlab
Use function expm to calculate the (t)
A = [0 -2; 1 -3]; T = 0.2
psy = expm(A*T)
