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1Chapter 1Mathematical Modeling
of Dynamic Systems in
State Space
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2Introduction to State Space
analysis Two approaches are available for the
analysis and design of feedback
control systems
Classical or Frequency domain technique
Modern or Time domain technique
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3Introduction to State Space
analysis Classical technique is based on converting a
systems differential equation to a transfer function
Disadvantage
Can be applied only to Linear Time Invariant system
Restricted to Single Input and Single output system
Advantage
Rapidly provide stability and transient response information
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4Introduction to State Space analysis Modern technique or state space approach is
a unified method for modeling, analyzing and
designing a wide range of systems
Advantages :
Can be used to nonlinear system
Applicable to time varying system
Applicable to Multi Input and Multi Output system
Easily tackled by the availability of advanced digital computer
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5Time varying A time-varying control system is a
system in which one or more of the
parameters of the system may vary as a
function of time
Dynamic system: input, state, output and initial condition
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6The state variables of a dynamic
system
The state of a system is a set of variables whose values, together with the input signals and the equations describing the dynamics , will provide the future state and output of the system
The state variables describe the present configuration of a system and can be used to determine the future response, given the excitation inputs and the equations describing the dynamics.
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7The State Space Equations
)()()(
)()()(
tDutCxty
tButAxtx
matrixfeedfowardD
matrixoutputC
matrixinputB
matrixsystemA
vectorcontrolofinputtu
vectoroutputty
vectorstatetx
vectorstatetheofderivativetx
_
_
_
_
___)(
_)(
_)(
____)(
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8Two types of equation
State equation
Output equation
)()()( tButAxtx
)()()( tDutCxty Saturday, September
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Terms State equations: a set of n simultaneous,
first order differential equations with n
variables, where the n variables to be
solved are the state variables
State space: The n-dimensional space whose axes are the state variables
State space representation: A mathematical model for a system that
consists of simultaneous, first order
differential equations and output equation
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Terms
State variables: the smallest set of linearly independent system variables
such that the value of the members of the
set
State vector: a vector whose elements are the state variables
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11
Modeling of Electrical NetworksVoltage-current, voltage-charge, and
impedance relationships for capacitors,
resistors, and inductors
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12
An RLC circuit
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13
State variable characterization The state of the RLC system described
a set of state variables x1 and x2
X1 = capacitor voltage = vc(t)
X2 = inductor current = iL(t)
This choice of state variables is intuitively satisfactory because the
stored energy of the network can be
described in terms of these variables
22
2
1
2
1cL CvLiE
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Utilizing Kirchhoffs current law
At the junction
First order differential equation
Describing the rate of change of capacitor voltage
Lc
c itudt
dvCi )(
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15
Utilizing Kirchhoffs voltage law Right hand loop
Provide the equation describing the rate of change of inductor current
Output of the system, linear algebraic equation
cLL vRi
dt
diL
)(tRiv Lo Saturday, September
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State space representation A set of two first order differential equation and
output signal in terms of the state variables x1 and x2
2
1
2
1
2
1
2
212
21
.0
.
0
1
.1
10
)()(
1
)(11
x
xRy
uCx
x
L
R
L
C
x
x
Rxtvty
xL
Rx
Ldt
dx
tuC
xCdt
dx
o
x
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17
Example 1 : RL serial network
Figure below shows an RL serialnetwork with an input voltage vi(t) and
voltage drop at inductance, L as an
output voltage vo(t). Form a state space
model for this system using the current
i(t) in the loop as the state variable.
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Modeling of Electrical Networks
RL serial network first order system
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RL serial network Write the loop equation for the system
using Kirchhoffs voltage law,
dt
tdiLRtitV
RtitV
tVdt
tdiLtV
tVtVtVtVtV
i
R
oL
oRLRi
)()()(
)()(
)()(
)(
)()()()()(
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RL serial network State variable is given only one, therefore
the system is a first order system
A state equation involving i is required
)(1
)()(
)(1
)()(
)()()(
)()()(
tVL
tiL
Rt
tVL
tiL
R
dt
tdi
tVRtidt
tdiL
dt
tdiLRtitV
i
i
i
i
i
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RL serial network
The output equation,
)(1)()()()()(
)()()(
)()()()()(
tVtiRty
tVRtitV
tVtVtV
tVtVtVtVtV
i
io
iRo
oRLRi
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22
Example 2 : RC serial network Figure below shows an RC circuit with
input voltage vi(t) and output voltage at
resistor ie vo(t). Form a state space model
for this system using the voltage vc(t)
across the capacitor as the state variable
Vi
R
CVR
VCi
V0
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RC serial network Write the equations for the system using
Kirchhoffs voltage law,
)3()()(
__
)2()(
)(
__
)1()()()()()(
Rtitv
resistorthefor
dt
tdvCti
capacitorthefor
tvtvtvtvtv
o
c
occRi
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RC serial network State variable is given only one
Therefore the system is a first order system
Therefore a state equation involving vc is required
Combine equation (2) and (3) yields
)4()(
)(
)()(
)(
dt
tdvRCtv
dt
tdvCti
R
tv
co
co
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RC serial network
Eliminate vo(t) from equation (4) and combine with equation (1) and rearrange
gives
)5()(1
)(1
)()(
)()()(
)()()(
)()()(
tvRC
tvRC
tvdt
tdv
tvtvdt
tdvRC
dt
tdvRCtvtv
tvtvtv
iccc
icc
cci
oci
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RC serial network
Output of the system
Rearrange equation (5) and (6) in matrix form yields
)6()()()( tvtvtv ico
)(1)(1)(
)(1
)(1
)(
tvtvty
tvRC
tvRC
tv
ic
icc
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RC serial network
Where,
1__
1_
1_
1_
)()(_)(
)(_)(
)(__)(
)(_)(
matrixontransmissidirectD
matrixouputC
RCmatrixinputB
RCmatrixstateA
tvtvvectoroutputty
tvvectorinputtu
tvvectorstatederivativetx
tvvectorstatetx
ro
i
c
c
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Modeling of Electrical Networks
Consider RLC serial network
RLC serial network second order system
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State Variables and output
Select two state variables,
)()(
)()(
)()(
)()(
2
1
tVtuinput
tVtyoutput
titx
tqtx
i
L
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Loop equation
Using Kirchoffs Voltage Law,
)()(1
)()(
)()()()(
tvdttiC
tRidt
tdiL
tvtvtvtv
i
cLRi
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Converting to charge
Using equation,
)()(1)()(
)()(
2
2
tvtqCdt
tdqR
dt
tqdL
dt
tdqti
i
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Derivatives of state vector
dt
tditx
titx
txtidt
tdqtx
tqtx
)()(
)()(
)()()(
)(
)()(
2
2
21
1
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State equation First state equation
Second state equation, using
)()()(
)( 21 txtidt
tdqtx
)(1
)()(1
)(
)()()()(
)()(1
)()(
)()(
212 tuL
txL
Rtx
LCtx
L
tv
L
tRi
LC
tq
dt
tdi
tvdttiC
tRidt
tdiL
dttitq
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State equation in matrix form
)(10
)(
)(1
10
)(
)(
)(
)(10
)(
)(1
10
)(
)()(
)()()(
2
1
2
1
tv
Lti
tq
L
R
LCdt
tdidt
tdq
tx
tu
Ltx
tx
L
R
LCtx
txtx
tButAxtx
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Output equation Output system is VL
)()()(1
)(
)()()(1
)(
)()()(1
)(
)()()()(
)()()()(
21 tutRxtxC
tV
tvtRitqC
tV
tvRtidttiC
tV
tvtVtVtV
tvtVtVtV
L
iL
iL
iRCL
iCRL
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Output equation in matrix form
)(1)(
)(1)(
)(1)(
)(1)(
)()()(
2
1
tvti
tqR
CtV
tutx
txR
Cty
tDutCxty
L
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Change State Variables but
output still same
)()(
)()(
)()(
)()(
2
1
tVtu
tVty
tVtx
tVtx
i
L
C
R
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Voltage formula for R, L and C
dt
tdiLtV
dttiC
tV
RtitV
L
C
R
)()(
)(1
)(
)()(
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Derivative of first state equation
)()()()(
)()()()(
)()()()()(
)(
)()(
21`1
1
`1
1
tuL
Rtx
L
Rtx
L
Rtx
tvL
RtV
L
RtV
L
Rtx
tVtVtvL
R
dt
tdiR
dt
tdVtx
tVtx
CR
CRR
R
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Derivative of second state
equation
)(1
)(
)(1
)(1)(
)(
)()(
12
2
2
txRC
tx
tVRC
tiCdt
tdVtx
tVtx
RC
C
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State equation in matrix form
)(
0)(
)(
01)(
)(
)(
)(
0)(
)(
01
)(
)()(
)()()(
2
1
2
1
tvL
R
tV
tV
RC
L
R
L
R
dt
tdVdt
tdV
tx
tuL
R
tx
tx
RC
L
R
L
R
tx
txtx
tButAxtx
C
R
C
R
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Output equation
)()()()(
)()()()(
)()()()(
21 tutxtxty
tvtVtVtV
tvtVtVtV
CRL
CRL
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Output equation in matrix form
)(1)(
)(11)(
)(1)(
)(11)(
)()()(
2
1
tvtV
tVtV
tutx
txty
tDutCxty
C
R
L
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Example 3 : 2 loop
Find a state space representation if the output is the current through the resistor.
State variables VC(t) and iL(t)
Output is iR(t)
Input is Vi(t)
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Electrical network LRCL
R C
Vi
iL
iR
iC
node 1
VL
VR
VC
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Solution : Step 1
Label all of the branch currents in the network.
iL(t), iR(t) and iC(t)
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Solution : Step 2 Select the state variables by writing
the derivative equation for all
energy-storage elements i.e.
inductor and capacitor
)2()(
)(
)1()(
)(
)(1
)(
dt
tdiLtV
dt
tdVCti
dttiC
tV
LL
CC
CC
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Solution : Step 3 Apply network theory, such as Kirchoffs
voltage and current laws to obtain iC(t)
and VL(t) in terms of the state variable
VC(t) and iL(t)
At node 1,
Around the outer loop,
)3()()(1
)(
)()()(
)()()(
titVR
ti
tititi
tititi
LCC
RLC
CRL
)4()()()(
)()()(
tVtVtV
tVtVtV
iCL
CLi
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Solution : Step 4 Substitute the result of equation (3) and
equation (4) into equation (1) and (2)
Rearrange
)8()()()(
)7()()(1)(
tVtVdt
tdiL
titVRdt
tdVC
iCL
LCC
)10()(1
)(1)(
)9()(1
)(1)(
tVL
tVLdt
tdi
tiC
tVRCdt
tdV
iCL
LCC
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Solution : Step 5
Find the output equation
)11()(1
)( tVR
ti CR
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Solution : Step 6
State space representation in vector matrix form are
)13()(
)(.0
1)(
)12()(10
)(
)(.
01
11
)(
)(
ti
tV
Rti
tv
Lti
tV
L
CRC
dt
tdidt
tdV
L
C
R
L
C
L
C
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Example 4 : 2 loop
Find the state space representation of the electrical network shown in figure below
Input vi(t)
Output vo(t)
State variables x1(t) = vC1(t), x2(t) = iL(t) and x3(t) = vC2(t)
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RLC two loop network
Identifying appropriate variables on the circuit yields
DC
C1
node R
C2
L
VoVi
iC1
iL
iR
iC2
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RLC two loop network
Represent the electrical network shown in figure in state space where
Output is v0(t)
Input is vi(t)
State variables :-X1(t) = vC1(t)
X2(t) = iL(t)
X3(t) = vC2(t)
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Solution Writing the derivative relations for
energy storage elements i.e. C1, C2 and
L
)()(
)()(
)()(
22
2
11
1
tidt
tdVC
tvdt
tdiL
tidt
tdvC
CC
LL
CC
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Solution Using Kirchhoffs current and voltage
laws
))()((1
)()(
)()()(
))()((1
)()(
)()()(
)()()(
22
1
21
21
1
tvtvR
titi
tvtvtv
tvtvR
titi
tititi
tititi
CLRC
iCL
CLLC
cLC
RLC
D
C
C1
node R
C2
L
VoVi
iC1
iL
iR
iC2
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Solution Substituting these relations and
simplifying yields the state equations as
2
2
2
2
1
2
2
1
1
2
11
1
1
1
111
11
1111
Co
iCCC
iCL
iCLCC
vv
vRC
vRC
vRCdt
dv
vL
vLdt
di
vRC
vRC
iC
vRCdt
dv
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Solution Putting the equations in vector matrix
form
xy
v
RC
L
RC
x
RCRC
L
RCCRC
x i
100
1
1
1
10
1
001
111
2
1
22
111
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Tutorial 1 : Number 1
Represent the electrical network shown in figure in state space where
Output is v0(t) and Input is vi(t)
State variables :-
x1 = v1
x2 = i4
x3 = v0
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Electrical network 1 Add the branch current and node
voltages to the network
Vi
R1 = 1 Ohm R
2 = 1 Ohm R
3 = 1 Ohm
C1 = 1 F C
2 = 1 F
L = 1 H Vo
V1
V2
i1
i3
i2
i5
i4
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Solution
Write the differential equation for each energy storage element
FCbecauseidt
dv
HLbecausevdt
di
FCbecauseidt
dv
1_;
1_;
1_;
250
24
121
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Solution Therefore the state vector is ,
Derivative state vector is ,
ov
i
v
x
x
x
x 4
1
3
2
1
ov
i
v
x
x
x
x 4
1
3
2
1
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Solution
Now obtain i2, v2 and i5 in terms of the state variables,
o
o
ii
vivv
Therefore
vivvviiviv
vvvvvvviii
2
1
2
1
2
1
,
2)(
412
042104352
21211312
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Solution
Substituting v2 in i2,
o
i
vivi
vngsubstituti
ivviii
also
vivvi
2
1
2
1
2
1
,_
,
2
1
2
1
2
3
415
2
421435
0412
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Solution
Therefore rearrange i2, v2 and i5 in matrix form yields
o
i
oo
v
i
v
y
v
v
i
v
i
v
i
v
i
v
x
x
x
x
4
1
4
1
5
2
2
4
1
3
2
1
.100
0
0
1
.
2
1
2
1
2
12
1
2
1
2
12
1
2
1
2
3
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Tutorial 1 : Number 2
Represent the electrical network shown in figure in state space where
Output is iR(t)
Input is vi(t)
State variables :-
x1 = i2
x2 = vC
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Electrical network 2 Add the branch currents and node
voltages to the schematic and obtain
DC
Vi
R1 = 1 Ohm
R2=1 Ohm
C = 1F
L = 1H4V
1
i4
i1
i2
i3
iR
node V2
node V1
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Solution
Write the differential equation for each energy storage element
FCbecauseidt
dv
HLbecausevdt
di
c 1_:
1_;
3
12
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Solution
Therefore the state vector is,
cv
i
x
xx
2
2
1
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Solution
Now obtain v1 in terms of the state variables
ic
ic
c
c
Rc
c
vviv
vivvvv
viivv
vivv
ivv
vvv
2
1
2
1
2
1
4
4)(
4
21
1211
1211
131
1
21
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Solution
Now obtain i3 in terms of the state variables
ic
ici
i
vvii
ivvivi
ivvi
iii
2
3
2
1
2
3
2
1
2
1
2
1
23
223
213
213
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Solution
Now obtain the output iR in terms of the state variables
icR
R
vvii
vii
2
1
2
3
2
1
4
2
13
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Solution
Hence the state space representation
i
c
i
cc
vv
iy
vv
i
i
v
v
ix
2
1.
2
3
2
1
2
32
1
.
2
1
2
32
1
2
1
2
2
3
12
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Tutorial 1 : Number 3 Find the state space representation of the
network shown in figure if
Output is v0(t)
Input is vi(t)
State variables :-
x1 = iL1
x2 = iL2
x3 = vC
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Electrical network 3 Add the branch currents and node
voltages to the schematic and obtain
DC
Vi
Vo
L1 = 1H L2 = 1H
R2=1 Ohm
C = 1F
i1
i2
i3
R3 = 1 Ohm
node
Vi
node
Vo
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Solution
Write the differential equation for each energy storage element
21
22
11
iidt
dv
ivdt
di
vvdt
di
c
cL
cL
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Solution
where,
L1 is the inductor in the loop with i1 L2 is the inductor in the loop with i2 iL1 = i1 i3 iL2 = i2 i3 Now,
i1 i2 = ic = iL1 iL2 -----------------(1)
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78
Solution
Also writing the node equation at vo,
i2 = i3 + iL2 ----------------------(2)
Writing KVL around the outer loop yields
i2 + i3 = vi -----------------------(3)
Solving (2) and (3) for i2 and i3 yields
)5(2
1
2
1
)4(2
1
2
1
23
22
iL
iL
vii
vii
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-
79
Solution
Substituting (1) and (4) into the state equations.
To find the output equation,
vo = -i3 + vi Using equation (5),
iLo viv2
1
2
12
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-
80
Solution
Summarizing the results in vector matrix form
i
C
L
L
o
i
C
L
L
C
L
L
v
v
i
i
vy
v
v
i
i
dt
dvdt
didt
di
x
x
x
x
2
1.0
2
10
02
11
.
011
12
10
100
2
1
2
1
2
1
3
2
1
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81
Tutorial 1 : Number 4
An RLC network is shown in figure. Define the state variable as :-
X1 = i1 X2 = i2 X3 = Vc Let voltage across capacitor, Vc is the
output from the network. Input of the
system is Va and Vb
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-
82
Tutorial 1 : Number 4
Determine the state space representation of the RLC network in matrix form
Determine the range of resistor R in order to maintain the systems stability, if C = 0.1 F and L1=L2=0.1 H. The characteristic
equation of the system is,
0100020010 23 RsRssSaturday, September
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-
83
RLC network with 2 input
DC
DC
Vb
Va
RL
1L
2
i2
iC
i1
VC
-
+
C
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-
84
Solution State variables and their derivatives
c
b
a
cc
vy
vu
vu
dt
dvxvx
dt
dixix
dt
dixix
2
1
33
2222
1111
Saturday, September
29, 2012
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-
85
Solution
The derivatives equations for energy storage elements
)3(
)2(
)1(
22
2
11
1
CC
L
L
idt
dvC
vdt
diL
vdt
diL
Saturday, September
29, 2012
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-
86
Solution
For loop (1) ;
For loop (2) ;
)4(11
11
CaL
CLa
vRivv
vvRiv
)5(2
2
CbL
CLb
vvv
vvv
Saturday, September
29, 2012
PMDRMFRCIED
DC
DC
Vb
Va
RL
1L
2
i2
iC
i1
VC
-
+
C
-
87
Solution For current iC ;
Substituting equation (4), (5) and (6) into equation (1), (2) and (3) yields
)6(21 iiiC
)7(11
11
1
3
1
1
1
11
11
1
1
1
11
1
a
aC
Ca
vL
xL
xL
R
dt
dix
vL
vL
iL
R
dt
di
vRivdt
diL
Saturday, September
29, 2012
PMDRMFRCIED
DC
DC
Vb
Va
RL
1L
2
i2
iC
i1
VC
-
+
C
-
88
Solution
Substituting equation (4), (5) and (6) into equation (1), (2) and (3) yields
)8(11
11
2
3
2
22
22
2
22
b
bC
Cb
vL
xLdt
dix
vL
vLdt
di
vvdt
diL
Saturday, September
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-
89
Solution
Substituting equation (4), (5) and (6) into equation (1), (2) and (3) yields
)9(11
11
213
21
21
xC
xCdt
dvx
iC
iCdt
dv
iidt
dvC
C
C
C
Saturday, September
29, 2012
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-
90
Solution Rewrite equation (7), (8) and (9) in state
space representation matrix form
3
2
1
2
1
3
2
1
2
11
3
2
1
.100
.
00
10
01
.
011
100
10
x
x
x
y
v
v
L
L
x
x
x
CC
L
LL
R
x
x
x
xb
a
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-
91
Solution Characteristic equation
Routh Hurwitz table
s3 1 200 0
s2 10R 1000R 0
s1 0
s0 200
0100020010 23 RsRss
R
RR
10
)10002000(
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-
92
Solution
For stability, all coefficients in first column of Routh Hurwitz table must be positive ;
0
01000
010
)10002000(
R
R
R
RR
Saturday, September
29, 2012
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-
93
Modeling of Mechanical
Networks Mass
massM
forcetf
ntdisplacemety
velocitytv
onacceleratita
dt
tdvMtf
dt
tydMtf
taMtf
)(
)(
)(
)(
)(.)(
)(.)(
)(.)(
2
2
M f(t)
y(t)
Saturday, September
29, 2012
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-
94
Modeling of Mechanical
Networks Linear Spring
tconsspringK
ntdisplacemety
forcetf
tyKtf
tan_
)(
)(
)(.)(
K y(t)
f(t)
Saturday, September
29, 2012
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-
95
Modeling of Mechanical
Networks Damper
frictionalviscousB
ntdisplacemety
forcetf
dt
tdyBtf
_
)(
)(
)(.)(
B y(t)
f(t)
Saturday, September
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-
96
Modeling of Mechanical
Networks
Inertia
InertiaJ
ntdisplacemeangulart
velocityangulart
TorquetT
dt
tdJtT
dt
tdJtT
_)(
_)(
)(
)(.)(
)(.)(
2
2
J
T(t))(t
Saturday, September
29, 2012
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-
97
Force-velocity, force-displacement, and
impedance translational relationships
for springs, viscous dampers, and mass
Saturday, September
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-
98
Torque-angular velocity, torque-angular
displacement, and impedance rotational
relationships for springs, viscous dampers,
and inertia
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-
99
Example 5 Determine the state space
representation of the mechanical
system below if the state variables are
y(t) and dy(t)/dt. Input system is force
f(t) and output system is y(t)
B
y(t)
f(t)
K
M
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-
100
Example 5
a. Mass, spring, and damper system; b. block diagram
Saturday, September
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-
101
State variables, input and output
)(
)(
)()()(
)()(
12
1
tyyoutput
tfuinput
dt
tdx
dt
tdytx
tytx
Saturday, September
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-
102
Mass, spring and damper
system Draw the free body diagram
f(t)
y(t)
dt
tdyB
tKy
dt
tydM
)(
)(
)(2
2
M
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29, 2012
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-
103
Mass, spring and damper system
a. Free-body diagram of mass, spring, and damper system;
b. transformed free-body diagram
Saturday, September
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-
104
Mass, spring and damper
system The force equation of the system is
Rearranged the equation yields
)(.)(
.)(
.)(2
2
tyKdt
tdyB
dt
tydMtf
)(.1
)(.)(
.)(
2
2
tfM
tyM
K
dt
tdy
M
B
dt
tyd
Saturday, September
29, 2012
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-
105
Mass, spring and damper
system State equations and output equation
)()(
)(.1
)(.)(.)(
)()(
1
212
21
txty
tfM
txM
Btx
M
Ktx
txtx
Saturday, September
29, 2012
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-
106
Mass, spring and damper
system State space representation in vector
matrix form are
)(
)(.01)(
)(.10
)(
)(.
10
)(
)(
2
1
2
1
2
1
tx
txty
tf
Mtx
tx
M
B
M
K
tx
tx
Saturday, September
29, 2012
PMDRMFRCIED
B
y(t)
f(t)
K
M
-
Example: The mechanical
system Consider the mechanical system shown in
Figure below by assuming 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 and single output system.
Saturday, September
29, 2012
PMDRMFRCIED 107
-
Mechanical system diagram
Saturday, September
29, 2012
PMDRMFRCIED 108
-
Mechanical system diagram
From the diagram, the system equation is
The system is of second order. This means that the system involves two
integrators. Define the state variables x1(t)
and x2(t) as
Saturday, September
29, 2012
PMDRMFRCIED 109
ukyybym
)()(
)()(
2
1
tytx
tytx
-
Then we obtain,
Saturday, September
29, 2012
PMDRMFRCIED 110
1
212
2
21
1
11
xy
um
xm
bx
m
kx
um
ybkym
x
xx
-
111
Mass, spring and damper system a. Two-degrees-of-freedom translational
mechanical system
b. block diagram
Saturday, September
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-
112
Mass, spring and damper system a. Forces on M1 due only to motion of M1
b. forces on M1 due only to motion of M2c. all forces on M1
Saturday, September
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113
Mass, spring and damper system a. Forces on M2 due only to motion of M2;
b. forces on M2 due only to motion of M1;
c. all forces on M2
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114
Exercise 1
Figure below shows a diagram for a quarter car model (one of the four wheels) of an automatic suspension system for a long distance express bus. A good bus suspension system should have satisfactory road handling capability, while still providing comfort when riding over bumps and holes in the road. When the coach is experiencing any road disturbance, such as potholes, cracks, and uneven pavement, the bus body should not have large oscillations, and the oscillations should be dissipate quickly.
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115
Exercise 1
(i). Draw the free-body diagrams of the system
(ii). Determine the state space representation
of the quarter car system by considering the
state vector
And the displacement of bus body mass M1 as
the output of the system.
T
txtxtxtx
)()()()( 2121z(t)
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-
116Saturday, September
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-
117
Constant value
Bus body mass, M1 = 2500 kg
Suspension mass, M2 = 320 kg
Spring constant of suspension system, K1 = 80,000 N/m
Spring constant of wheel and tire, K2 = 500,000 N/m
Damping constant of suspension system, B1= 350 Ns/m
Damping constant of wheel and tire, B2 = 15,020 Ns/m
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-
118
Solution Free body diagram for M1
Forces on M1 due to motion of M1
Forces on M1 due to motion of M2
All forces on M1
M1 u
K1X1M1s
2X1B1sX1
M1
K1X2
B1sX2
M1
K1X1M1s
2X1B1sX1
K1X2u
B1sX2
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-
119
Solution Free body diagram for M2
Forces on M2 due to motion of M2
Forces on M2 due to motion of M1
All forces on M2
M2K1X2B1sX2
K2X2M2s
2X2B2sX2
M2
M2
(K1+K2)X2M2s
2X2(B1+B2)sX2
K1X1
B1sX1
K1X1
B1sX1
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-
120
Solution
State variables
Derivative state variables
24132211 ;;;
xzxzxzxz
2413422311 ;;;
xzxzzxzzxz
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-
121
Solution
Total force for M1
uzzzzz
udt
dx
dt
dxxx
dt
xd
dt
dxB
dt
xdMxK
dt
dxBxKu
0004.014.014.03232
0004.014.014.03232
43213
21212
1
2
112
1
2
1112
121
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-
122
Solution
Total force for M2
43214
21212
2
2
221
2
2
22211
111
031.48094.15.1812250
031.48094.15.1812250
)()(
zzzzz
dt
dx
dt
dxxx
dt
xd
dt
dxBB
dt
xdMxKK
dt
dxBxK
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-
123
Solution
State space representation
4
3
2
1
4
3
2
1
0001
0
0004.0
0
0
031.48094.15.1812250
14.014.03232
1000
0100
z
z
z
z
y
u
z
z
z
z
z
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-
124
Tutorial 1 : Number 5
Figure shows a mechanical system consisting of mass M1 and M2, damper constant B, spring stiffness K1 and K2. When force f(t) acts on mass M1, it moves to position x1(t) while mass M2moves to position x2(t). Find the state space representation of the system using x1(t), x2(t) and their first derivatives as state variables. Let x2(t) be the output.
Saturday, September
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-
125
Mechanical system consist of 2
mass, 2 spring and 1 damper
B
f(t)
K1
M1
M2
K2
X1
X2
Saturday, September
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-
126
Mechanical system consist of 2
mass, 2 spring and 1 damper
State variables and their derivatives :-
)()(
)()(
)()()()(
)()()()()(
)()()()(
)()()()()(
2
2424
42323
1212
21111
txtyoutput
tftuinput
txtztxtz
tztxtztxtz
txtztxtz
tztxtztxtz
Saturday, September
29, 2012
PMDRMFRCIED
-
127
Mechanical system consist of 2
mass, 2 spring and 1 damper
Draw the free body diagram
1
11
11
xB
xM
xK
2
22
22
xB
xM
xK
2
)(
xB
tf
1xB
M1
M2
Saturday, September
29, 2012
PMDRMFRCIED
-
128
Mechanical system consist of 2
mass, 2 spring and 1 damper
Differential equation in mass M1
Differential equation in mass M2
)1()()(
)(
112111
211111
xKxxBxMtf
xBxKxBxMtf
)2()(0
0
221222
122222
xKxxBxM
xBxKxBxM
Saturday, September
29, 2012
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-
129
Mechanical system consist of 2
mass, 2 spring and 1 damper Substitute all state variables and their first
derivatives in equation (1) and (2) yields
)6(
)5(
)4(
)3(1
)(
43
21
3
2
22
2
4
2
4
2
2
21
2
2
2
2
1
1
1
14
1
2
1
2
1
1
1
12
1
1
1
1
zz
zz
zM
Kz
M
Bz
M
Bz
xM
Kx
M
Bx
M
Bx
uM
zM
Kz
M
Bz
M
Bz
M
tfx
M
Kx
M
Bx
M
Bx
Saturday, September
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-
130
Mechanical system consist of 2
mass, 2 spring and 1 damper
Rearrange equation 3, 4, 5 and 6 in matrix form
4
3
2
1
1
4
3
2
1
22
2
2
111
1
4
3
2
1
.0100
0
0
10
.
0
1000
0
0010
z
z
z
z
y
uM
z
z
z
z
M
B
M
K
M
B
M
B
M
B
M
K
z
z
z
z
z
Saturday, September
29, 2012
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131
Tutorial 1 : Number 6 Represent the translational mechanical
system shown in figure in state space
where x3(t) is the output and f(t) is the
input.
B1
f(t)
K1
M1
M2
K2
X1
X2
M3
B2
X3
Saturday, September
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132
Example : 3M, 2K and 2B
Represent the translational mechanical system shown in figure in state space
where x3(t) is the output and f(t) is the
input.
Saturday, September
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133
Example : 3M, 2K and 2B
K1 = K2 = 1 N/m
M1 = M2 = M3 = 1 kg
B1 = B2 = 1 N-s/m
Find the state space representation of the system using x1, x2, x3 and their first
derivatives as state variables.
363524
231211
;;
;;;
xzxzxz
xzxzxz
Saturday, September
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PMDRMFRCIED
-
134
Example : 3M, 2K and 2B Draw the free body diagram
)(
21
tf
xB
22
21
22
xK
xB
xM
M1
M2
M3
11
11
11
xK
xB
xM
32
32
33
xK
xB
xM
32
11
xK
xB
22xK
Saturday, September
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-
135
Example : 3M, 2K and 2B
Writing the equations of motion
)3(
)2(
)1()(
22323233
3211222122
21111111
xKxKxBxM
xKxBxKxBxM
tfxBxKxBxM
Saturday, September
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-
136
Example : 3M, 2K and 2B
Substitute the value of K, M and B.
Rearrange equation (1), (2) and (3)
2333
32212
2111
xxxx
xxxxx
fxxxx
Saturday, September
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-
137
Example : 3M, 2K and 2B From the state variables
53
3563636
63535
53422424
42323
4121212
21111
zxy
zzzxzxz
zxzxz
zzzzxzxz
zxzxz
fzzzxzxz
zxzxz
Saturday, September
29, 2012
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-
138
Example : 3M, 2K and 2B
In vector matrix form
zy
tfzz
010000
)(
0
0
0
0
1
0
110100
100000
011110
001000
001011
000010
Saturday, September
29, 2012
PMDRMFRCIED
-
139
Modeling of Electro-Mechanical System
NASA flight simulator robot arm with electromechanical control system components
Saturday, September
29, 2012
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-
140
Modeling of Electro-Mechanical System
Armature Controlled DC Motor
Saturday, September
29, 2012
PMDRMFRCIED
-
Armature Controlled DC Motor
141Saturday, September
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-
142
DC motor armature control
The back electromotive force(back emf), VB
tconsemfBackK
dt
tdKtV
dt
tdtV
B
mBB
mB
tan__
)1()(
.)(
)()(
Saturday, September
29, 2012
PMDRMFRCIED
-
143
DC motor armature control Kirchoffs voltage equation around the
armature circuit
a
a
m
a
mbaaa
baaa
Lignore
ceresisarmatureR
armaturetheofntdisplacemeangular
currentarmaturei
dt
tdKRtite
tVRtite
_
tan_
____
_
)2()(
)()(
)()()(
Saturday, September
29, 2012
PMDRMFRCIED
-
144
DC motor armature control
The torque, Tm(t) produced by the motor
motorthebydensityviscousequivalentD
motorthebyinertiaequivalentJ
tconsTorqueK
dt
dD
dt
dJtT
tiKtT
titT
m
m
t
mm
mmm
atm
am
_____
____
tan_
)3()(
)()(
)()(
2
2
Saturday, September
29, 2012
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-
145
DC motor armature control
Solving equation (3) for ia(t)
)4()(2
2
dt
d
K
D
dt
d
K
Jti m
t
mm
t
ma
Saturday, September
29, 2012
PMDRMFRCIED
-
146
DC motor armature control
Substituting equation (4) into equation (2) yields
)5(..)(
)(
2
2
2
2
dt
dK
K
DR
dt
d
K
JRte
dt
dK
dt
d
K
D
dt
d
K
JRte
mb
t
mam
t
maa
mb
m
t
mm
t
maa
Saturday, September
29, 2012
PMDRMFRCIED
-
147
DC motor armature control
Define the state variables, input and ouput
Substituting equation (6) into equation (5) yields
m
a
m
m
y
teu
bdt
dx
ax
1.0
)(
)6(
)6(
2
1
)7(..)( 22
xK
K
DR
dt
dx
K
JRte b
t
ma
t
maa
Saturday, September
29, 2012
PMDRMFRCIED
-
148
DC motor armature control Solving for x2 dot yields,
)8()(..1
.)(.
.)(
22
22
2
2
teJR
Kx
R
KKD
Jdt
dx
xJR
KK
J
Dte
JR
K
dt
dx
K
JR
xKK
DRte
dt
dx
a
ma
t
a
tbm
m
ma
tb
m
ma
ma
t
t
ma
b
t
maa
Saturday, September
29, 2012
PMDRMFRCIED
-
149
DC motor armature control
Using equation (6) and (8), the state equations are written as
)(..1
22
21
teJR
Kx
R
KKD
Jdt
dx
xdt
d
dt
dx
a
ma
t
a
btm
m
m
Saturday, September
29, 2012
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150
DC motor armature control Assuming that the output o(t) is 0.1 the
displacement of the armature m(t) as x1.
Hence the output equation is
State space representation in vector matrix form are
11.0 xy
2
1
2
1
2
1
.01.0
)(.
0
.10
10
x
xy
te
JR
Kx
x
R
KKD
Jx
xa
ma
t
a
btm
m
Saturday, September
29, 2012
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-
151
Tutorial 1 : Number 7
The representation of the positioning system using an armature-controlled dc motor is shown in figure.
The input is the applied reference voltage, r(t) and the output is the shafts angular position, o(t).
The dynamic of the system can be described through the Kirchoff equation for the armature circuit, the Newtonian equation for the mechanical load and the torque field current relationship.
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152
Figure : DC motor armature control
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153
Example : ex-exam question
The Newtonian equation for the mechanical load is
The back e.m.f voltage induced in the armature circuit, eb(t) is proportional to
the motor shaft speed,
)()()( tttJ oo
obb Ke Saturday, September
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154
Example : ex-exam question
A potentiometer was installed to measure the motor output position. Its output
voltage, v(t) is then compared with the
system reference input voltage, r(t)
through an op-amp.
Determine the complete state-space representation of the system by
considering the following state variables.
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Example : ex-exam question
State variables :-
State variables derivative
x1(t) ia (t)
x2(t) o(t)
x3(t)
o(t)
x
1(t)
dia (t)
dt i
a
x
2 (t) do(t)
dt
o(t)
x
3(t)
d2o(t)
dt 2
o(t)
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156
Example : ex-exam question
Mechanical load
Jo
(t) o
(t) (t) K tia
J x
3 x3 K tx1
x
3 K tJx1
Jx3 (1)
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Example : ex-exam question Electrical (armature) circuit
Using Kirchoff Voltage Law
)2(1
)(
311
311
uL
xL
Kx
L
Rx
xKRxxLu
KRidt
diLu
givenKe
but
eRidt
diLu
b
b
obaa
obb
baa
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158
Example : ex-exam question
From the state variable defination
)4(
___
)3(
2
32
2
xKru
Kru
vru
partinputtheFor
xx
x
s
os
o
o
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159
Example : ex-exam question
Substituting (4) into (2)
)5(1
)(1
3211
2311
rL
xL
Kx
L
Kx
L
Rx
xKrL
xL
Kx
L
Rx
bs
sb
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160
Example : ex-exam question
Writing equations (1), (3) and (5) in the vector matrix form gives :-
rL
x
x
x
JJ
K
L
K
L
K
L
R
x
x
x
T
bs
0
0
1
.
0
100
3
2
1
3
2
1
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161
Example : ex-exam question
The output
3
2
1
010
x
x
x
y o
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162
Modelling of Electro-Mechanical
System
Field Controlled DC Motor
Gelung Angker
Tetap
+
-
Lf
ef (t)
Gelung Medan
Ba
Ja
ia
ea
Ra L
a
if (t)
Tm(t)
)(tm
TL(t)
+
-
Rf
RAJAH 7.11 : MOTOR SERVO A.T. TERUJA BERASINGAN
DALAM KAWALAN MEDAN
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DC motor field control
For field circuit
For mechanical load, torque
)1()( dt
diLRite
f
fff
)2()(2
dt
dB
dt
dJtT oo
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164
DC motor field control
For torque and field current relationship
Define the state variables, input and output
)3()()(
)()(
tiKtT
titT
ft
f
)(
)(
)6()(
)5()(
)4()(
3
2
1
ty
teu
tix
dt
tdx
tx
o
f
o
o
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DC motor field control
From equation (4) and (5), we can determine the first state equation as :
Another two state equations are :
)7()()( 21
dt
dtxtx o
)9(
)8(
3
2
2
2
dt
dix
dt
dx
f
o
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DC motor field control
Substituting x3 and x3 dot into equation (1) yields
Substituting equation (3) into equation (2) yields
33)( xLRxte ff
ftoo iK
dt
dB
dt
dJ
2
2
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DC motor field control
Substituting x2 dot, x2 and x3, hence
Rewrite equations
322 xKBxxJ t
322
33
1)(
xJ
Kx
J
Bx
Lte
L
Rxx
t
ff
f
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DC motor field control
Matrix form
xy
u
L
x
L
RJ
K
J
Bx
f
f
f
t
001
10
0
00
0
010
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Block diagrams
The block diagram is a useful tool for simplifying the representation of a
system.
Simple block diagrams only have one feedback loop.
Complex block diagram consist of more than one feedback loop, more than 1 input
and more than 1 output i.e. inter-coupling
exists between feedback loopsSaturday, September
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-
Block diagrams
Integrator
Amplifier or gain
Summer
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x1 dtxx 12
Kx1 x2 = Kx1
x1
x2
x3
+
+
-x4 = x1-x2+x3
-
Signal flow graphs Having the block diagram simplifies the
analysis of a complex system.
Such an analysis can be further simplified by using a signal flow graphs (SFG) which
looks like a simplified block diagram
An SFG is a diagram which represents a set of simultaneous equation.
It consist of a graph in which nodes are connected by directed branches.
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-
Signal flow graphs
The nodes represent each of the system variables.
A branch connected between two nodes acts as a one way signal multiplier: the
direction of signal flow is indicated by an
arrow placed on the branch, and the
multiplication factor(transmittance or
transfer function) is indicated by a letter
placed near the arrow.Saturday, September
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-
Signal flow graphs
A node performs two functions:
1. Addition of the signals on all incoming
branches
2. Transmission of the total node signal(the
sum of all incoming signals) to all outgoing
branches
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-
Signal flow graphs There are three types of nodes:
1. Source nodes (independent nodes) these represent independent variables and have
only outgoing branches. u and v are source
nodes
2. Sink nodes (dependent nodes) - these
represent dependent variables and have
only incoming branches. x and y are source
nodes
3. Mixed nodes (general nodes) these have both incoming and outgoing branch. W is a
mixed node.Saturday, September
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-
Signal flow graphs
x2 = ax1
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x1 a x2 = ax1
-
Signal flow graphs
w = au + bv
x = cw
y = dw
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u
v
x
y
wa c
b d
-
Signal flow graphs
x = au + bv +cw
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u
w
v b
ca
1x x
Mixed
node
Sink
node
-
Signal flow graphs A path is any connected sequence of
branches whose arrows are in the same
direction
A forward path between two nodes is one which follows the arrows of successive
branches and in which a node appears
only once.
The path uwx is a forward path between the nodes u and x
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-
Signal flow graphs Series path (cascade nodes) series path
can be combined into a single path by
multiplying the transmittances
Path gain the product of the transmittance in a series path
Parallel paths parallel paths can be combined by adding the transmittances
Node absorption a node representing a variable other than a source or sink can be
eliminated
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-
Signal flow graphs
Feedback loop a closed path which starts at a node and ends at the same
node.
Loop gain the product of the transmittances of a feedback loop
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-
Signal flow graphs
simplification
Original graph Equivalent graph
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x
a
y z
b
x z
ab
-
Signal flow graphs
simplification
Original graph Equivalent graph
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x
a
b
y
x y
(a+b)
-
Signal flow graphs
simplification
Original graph Equivalent graph
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w
x y
z
b
ac
w
x
ac
bc
z
-
Block diagram of feedback
system
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CER
B
G
H
-
Block diagram of feedback system
R=reference input
E=actuating signal
G=control elements and controlled system
C=controlled variable
B=primary feedback
H=feedback elements
C = GE
B = HC
E = R-B
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-
Successive reduction of SFG
first
4 nodes
second
Node B eliminated
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-1
1 E G C
H
B
R R 1 E G C
-H
-
Successive reduction of SFG
third
Node E eliminated, self loop of value -GH
fourth
Self loop eliminated
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R CG R C
G/(1+GH)
-GH
-
188
SIGNAL FLOW GRAPHS OF
STATE EQUATIONS demonstrate how to draw signal flow
graphs from state equations.
Consider the following state and output equations:
)1(2352 3211 arxxxx
)1(5226 3212 brxxxx
)1(743 3213 crxxxx
y 4x1 6x2 9x3 (1d)Saturday, September
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189
SIGNAL FLOW GRAPHS OF
STATE EQUATIONS Step 1 : Identify three nodes to be the
three state variables, , and three nodes,
placed to the left of each respective
state variables. Also identify a node as
the input, r, and another node as the
output, y.
R(s)
sX (s)3
X (s)3
X (s)2 X (s)1sX (s)2
sX (s)1
Y(s)
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SIGNAL FLOW GRAPHS OF
STATE EQUATIONS
Step 2 : Interconnect the state variables and their derivatives with the defining
integration, 1/s.
R(s)
sX (s)3
X (s)3
X (s)2 X (s)1sX (s)2
sX (s)1
Y(s)
1s
1s
1s
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191
SIGNAL FLOW GRAPHS OF
STATE EQUATIONS
Step 3 :Using Eqn (1a), feed to each node the indicated signals.
R(s)
sX (s)3
X (s)3 X (s)2
X (s)1sX (s)2
sX (s)1
Y(s)
1s
1s
1s
2
3
2
-5
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192
SIGNAL FLOW GRAPHS OF
STATE EQUATIONS
Step 4 :Using Eqn (1b), feed to each node the indicated signals.
R(s)
sX (s)3
X (s)3 X (s)2 X (s)1
sX (s)2 sX (s)1
Y(s)
1s
1s
1s
2
3
2
-52
-6
5
-2
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193
SIGNAL FLOW GRAPHS OF
STATE EQUATIONS
Step 5 :Using Eqn (1c), feed to each node the indicated signals.
R(s)sX (s)
3 X (s)3 X (s)2X (s)
1sX (s)
2 sX (s)1
Y(s)
1s
1s
1s
2
3
2
-52
-6
5
-2
7
-4
-3
1Saturday, September
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194
SIGNAL FLOW GRAPHS OF
STATE EQUATIONS
Step 6 :Finally, use Eqn (1d) to complete the signal flow graph.
R(s)sX (s)
3 X (s)3 X (s)2X (s)
1sX (s)
2sX (s)
1
Y(s)
1s
1s
1s
2
3
2
-52
-6
5
-2
7
-4
-3
1
-4
9
6
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Example 7
Draw a signal-flow graph for each of the following state equations :
3
2
1
3
2
1
.011)(
)(
1
0
0
.
642
100
010
)(
x
x
x
ty
tr
x
x
x
tx
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196
Solution
State and output equations
)()()(
)()(6)(4)(2)(
)()(
)()(
21
3213
32
21
txtxty
trtxtxtxtx
txtx
txtx
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197
Solution Signal flow graph
r yx
3x
2x
1
1/s 1/s1/s 111
-2
-4
-6
1
1
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198
Example 8
Draw a signal-flow graph for each of the following state equations :
)(021)(
)(
1
1
0
)(
543
130
010
)(
txty
trtxtx
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199
Solution
State and output equations
)(2)()(
)()(5)(4)(3)(
)()()(3)(
)()(
21
3213
322
21
txtxty
trtxtxtxtx
trtxtxtx
txtx
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200
Solution Signal flow graph
r y1/s 1/s1/s
x3
x2
x1
111 1
-3
-4
-5
21
-3
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201
Example 9
Draw a signal-flow graph for each of the following state equations :
)(231)(
)(
1
2
1
)(
201
123
017
)(
txty
trtxtx
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202
Solution
State and output equations
)(2)(3)()(
)()(2)()(
)(2)()(2)(3)(
)()()(7)(
321
313
1212
211
txtxtxty
trtxtxtx
trtxtxtxtx
trtxtxtx
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203
Solution Signal flow graph
r yx
1x
3x
2
1/s 1/s1/s1 11-1
-1
2
-3
2 7
1
2 3
2
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204
Q1 For the circuit shown in figure, identify
a set of state variables
Answer : one possible set of state variables is the current iL2 via L2, the
voltage VC2 across C2 and the current
iL1 via L1
VC1 the voltage across C1 can replace iL1 via L1 as the third state variable
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205Saturday, September
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206
Q1 For the circuit shown in figure,
determine the state space
representation if :
(a). Input are V1 and V2, output is VC2and state variables are define as x1=iL2,
x2=VC2 and x3=iL1
(b). Input are V1 and V2, output is VC2and state variables are define as x1=iL2,
x2=VC2 and x3=VC1
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207
Q2 Use state variable model to describe
the circuit of the figure.
Choose x1=VC and x2=i as state variables.
Determine the state equation only.
i
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208
Tips
in
in
in
C
inC
Vxx
V
L
x
L
R
L
Cx
VL
xL
xL
Rx
xC
x
idtC
V
VVRidt
diL
10
0
4010
10000
10
1
10
11
1
1
122
21
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209
Q3 Determine a state variable differential matrix
equation for the circuit shown in the figure.
Choose x1=v1 and x2=v2 as state variables.
Two inputs are u1=va and u2=vb. The output
is y=v0=v2V1 V2
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210
Tips
b
a
b
a
V
V
CR
CR
V
V
CRCRCR
CRCRCRx
R
VV
R
VVVC
node
R
VV
R
VVVC
node
.1
0
01
.111
111
2_
1_
23
11
2
1
222322
121211
2
21
3
222
2
12
1
111
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211
Q4An RLC circuit is shown in figure,
(a). identify a suitable set of state variables
(b). obtain the set of first order differential equations in terms of the state variables x1=i and x2=VC(c). write the state differential equation.
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212
Tips
VLV
i
C
LL
R
x
idtC
V
VL
iL
RV
Ldt
di
C
C
C
0
1
.
01
1
1
11
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213
Q5 Determine the state equation of the
figure. State variables are define as
x1=iL and x2=Vc. Input V1 and V2. Draw
the corresponding block diagram and
signal flow graph of the system
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214
Q6 Determine the state space differential
equation of the figure. Define the state
variables as x1=iL and x2=Vc. System
input v1 and v2. The output system is
iR. Use KVL around the outer loop and
KCL at the node.
iR
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215
Tips
2
1
2
1
2
12
.1
0
11
.11
10
0
V
V
RC
LLV
i
RCC
L
x
x
R
V
R
Vi
iidt
dVC
VVVdt
diL
C
L
CR
RLC
CL
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216
Q7 Determine the state variable matrix
equation for the circuit shown in the
figure. Defined state variables as x1=v1,
x2=v2 and x3=iL=i
System input are Vi and iS
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217
Tips
s
i
L
L
L
iL
i
v
i
v
v
x
vvdt
di
iv
idt
dv
vvi
dt
dv
equationNode
00
20000
01
.
0500500
200020
400001
0002.0
01000
00005
04000
00025.0
_
2
1
12
322
11
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218
Q8 Determine the state variable matrix
differential equation for the circuit
shown in the figure. The state
variables are x1=i, x2=v1 and x3=v2. The
output variable is vo(t) and input is V.
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219
Tips
VCR
V
V
i
CRCRCRC
RCRR
L
x
R
ViVV
Rdt
dVC
VVR
VVRdt
dVC
Vdt
diL
0
10
.
1111
1110
100
01
01
)(1
11
2
1
2322222
2121
3
212
2
22
21
2
1
1
11
2
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220
Q9 Determine the state equation for the
two input and one output circuit shown
in the figure where state variables are
define as x1=iL and x2=Vc the output is
y=i2 i1 i3
iC
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221
Tips
3
23
21121
12
12211
111
32
R
VVi
iRVViRR
iii
VViRiR
RiVdt
diL
iidt
dVC
C
LC
L
C
L
C
Saturday, September
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