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MECH370: Modelling, Simulation and Analysis of
Physical SystemsChapter 6
Electrical SystemsVariables and relationship with translational mechanical elementsElement lawsInterconnection lawsObtaining the system model
Lecture Notes on Lecture Notes on MECH 370MECH 370 –– ModellingModelling, Simulation and Analysis of Physical Systems, Simulation and Analysis of Physical SystemsChapter 6Chapter 6 2
Course OutlineCourse OutlineModellingModelling (Ch. 2,3,4,5,6,9,10,11,12) Simulation (4) Analysis (7(Ch. 2,3,4,5,6,9,10,11,12) Simulation (4) Analysis (7,8),8)
1. Definition and classification of dynamic systems (chapter 1)
2. Translational mechanical systems (chapter 2) 3. Standard forms for system models (chapter 3)4. Block diagrams and computer simulation with
Matlab/Simulink (chapter 4)5. Rotational mechanical systems (chapter 5)6. Electrical systems (chapter 6)7. Analysis and solution techniques for linear systems
(chapters 7 and 8)8. Developing a linear model (chapter 9)9. Electromechanical systems (chapter 10)10.Thermal and fluid systems (chapters 11, 12)
Lecture Notes on Lecture Notes on MECH 370MECH 370 –– ModellingModelling, Simulation and Analysis of Physical Systems, Simulation and Analysis of Physical SystemsChapter 6Chapter 6 3
Electrical SystemsExample:
e - voltage (electric potential difference) in volts (V)
i - current in Amperes (A)
q - charge in Coulombs (C)
Variable:
Current is defined as the flow rate of charge
i
Lecture Notes on Lecture Notes on MECH 370MECH 370 –– ModellingModelling, Simulation and Analysis of Physical Systems, Simulation and Analysis of Physical SystemsChapter 6Chapter 6 4
Electrical Systems (cont’d)
dtdqi =
Current has a direction and magnitude-i i
=
iii == 21
Charge cannot accumulate in a component
A voltage at a point a is the electric potential difference from an arbitrary reference called a ground.
Grounde0=0
i1 Circuitelement
i2
Circuitelement
ea -+e1 e2
iCircuitelement
Lecture Notes on Lecture Notes on MECH 370MECH 370 –– ModellingModelling, Simulation and Analysis of Physical Systems, Simulation and Analysis of Physical SystemsChapter 6Chapter 6 5
Electrical Systems (cont’d):is element an across voltageThe ae
21 eeea −=
iep ⋅=
∫=1
0
t
tpdtE
The power supplied to an element is given by:
The energy supplied to the element is given by:
It has both a sign and magnitude-ea -+
=ea +-
Lecture Notes on Lecture Notes on MECH 370MECH 370 –– ModellingModelling, Simulation and Analysis of Physical Systems, Simulation and Analysis of Physical SystemsChapter 6Chapter 6 6
Relationship Between Electrical and Translational Elements
Notation Variable Notation Variable
Position Charge
Velocity Current
Acceleration Change in current
Force Voltage
Newton’s 2nd law Inductor law
Spring lawCapacitor law
Damper law Resistor law
)( xv &=
x
)( xva &&& ==
F
MaF =
KxF =
BvF =
q
qi &=
qdtdi
&&=
)(or ve
dtdiLe =
duuiC
ete t )(1)0()( 0∫=−
qC
e 1=
Rie =
Lecture Notes on Lecture Notes on MECH 370MECH 370 –– ModellingModelling, Simulation and Analysis of Physical Systems, Simulation and Analysis of Physical SystemsChapter 6Chapter 6 7
Relationship Between Electrical and Translational Elements(cont’d)
Notation Variable Notation Variable
Mass Inductance
Spring constant 1/Capacitance
Damping constant R Resistance
K
M
B
L
C1
Lecture Notes on Lecture Notes on MECH 370MECH 370 –– ModellingModelling, Simulation and Analysis of Physical Systems, Simulation and Analysis of Physical SystemsChapter 6Chapter 6 8
Element Laws
iRe ⋅=
QReRieip &====
22
R is the resistance in Ohms (Ω)
The energy is converted into heat.
Ohm’s Law:
i
ReResistor:
A resistor dissipates energy:
i
+ -e
R
http://en.wikipedia.org/wiki/
Lecture Notes on Lecture Notes on MECH 370MECH 370 –– ModellingModelling, Simulation and Analysis of Physical Systems, Simulation and Analysis of Physical SystemsChapter 6Chapter 6 9
Element Laws (cont’d)
Ceq =
∫+==t
tidτ
ctete
dtdeCi
0
1)()( , 0
Capacitor:
C is the capacitance in Farads (F)2
21 CeEp =
e
Cq
A capacitor stores energy in an electric field:
material dielectric theofty permittivi =⋅
= εε ;d
AC
i
+d
+
-A
-ee
http://en.wikipedia.org/wiki/
Lecture Notes on Lecture Notes on MECH 370MECH 370 –– ModellingModelling, Simulation and Analysis of Physical Systems, Simulation and Analysis of Physical SystemsChapter 6Chapter 6 10
Element Laws (cont’d)
∫ τ+==t
ted
Ltiti
dtdiLe
0
1)()( , 0
2
21 LiEp =
Inductor:
where L is the inductance in Henries (H).
e
i
+ -e
An inductor stores energy in a magnetic field:
dtdiLLi
dtddtde
==
=
)(
linkage)(flux i
+ -
L
http://en.wikipedia.org/wiki/
Lecture Notes on Lecture Notes on MECH 370MECH 370 –– ModellingModelling, Simulation and Analysis of Physical Systems, Simulation and Analysis of Physical SystemsChapter 6Chapter 6 11
Element Laws (cont’d)Notations:
Short circuit
Current source
Voltage source
Open circuit
circuit
e(t) circuit
i(t)
i(t)
i i=0
+ e=0e -
Lecture Notes on Lecture Notes on MECH 370MECH 370 –– ModellingModelling, Simulation and Analysis of Physical Systems, Simulation and Analysis of Physical SystemsChapter 6Chapter 6 12
Interconnection Laws
Kirchhoff’s Voltage Law (KVL):
e3
+
-
e4 +-
e1
+
-
e2 +-
loopany around ,0=∑ je
e.g.
00
2143
4321
=−−+=−−+
eeeeeeee
Lecture Notes on Lecture Notes on MECH 370MECH 370 –– ModellingModelling, Simulation and Analysis of Physical Systems, Simulation and Analysis of Physical SystemsChapter 6Chapter 6 13
Interconnection Laws (cont’d)
Kirchhoff’s Current Law (KCL):
Note: away from the node is +; into the node is –
nodeany around ,0=∑ jie.g.
0321 =+− iii
-
C
i1
A R
i2i3L
Lecture Notes on Lecture Notes on MECH 370MECH 370 –– ModellingModelling, Simulation and Analysis of Physical Systems, Simulation and Analysis of Physical SystemsChapter 6Chapter 6 14
Electrical Systems
RiedtdiLe
dtdeCi
RL ==
=
and
Example 6.1
From KVL:
0)(1)0(
0)(
0=−τ+++∴
=−++
∫ teidC
eRidtdiL
teeee
i
e
t
ce
e
iCRL
C
R
L
44 344 21
Since:
Lecture Notes on Lecture Notes on MECH 370MECH 370 –– ModellingModelling, Simulation and Analysis of Physical Systems, Simulation and Analysis of Physical SystemsChapter 6Chapter 6 15
Electrical Systems (cont’d)The above equation is not in state-variable or input-output form
)(1)0(0
teidC
eRidtdiL i
t
c =τ+++ ∫
: terms of ridget toateDifferenti ∫
dttdei
CdtdiR
dtidL i )(12
2
=++
Compare this equation with the one in Ex. 5.1 (textbook, pp.105) and in M-S-D system (next slide, or pp. 3-7)!
I/O form
It is interesting that the flow of current in this circuit is mathematically equivalent to motion of spring-mass system.
Lecture Notes on Lecture Notes on MECH 370MECH 370 –– ModellingModelling, Simulation and Analysis of Physical Systems, Simulation and Analysis of Physical SystemsChapter 6Chapter 6 16)(
)(
),(
)(
)(
2
2
tfKxxBxM
tfKxdtdxB
dtxdM
vdtdxtfKxBv
dtdvM
KxBvtf
fftfdtdvM
a
a
a
a
KBa
=++
=++
==++
−−=
−−=
&&&
Example:
Free body diagram (FBD):
Obtaining the System Model(Review)
Newton’s 2nd Law:
KM fa(t)
B x, v
M
fB
fK
fa(t)
fI
Force exerted by the dashpot
the applied force
Force exerted by the dashpot
the inertial force
x
y
D’Alembert’s Law(use the idea of an inertial force, fI)
)(0)(
)(0
tfaKxxBxMKxxBxMtfaffftfaf KBIi
=++⇒=−−−=
−−−==∑
&&&
&&&
dttdei
CdtdiR
dtidL i )(12
2
=++
Lecture Notes on Lecture Notes on MECH 370MECH 370 –– ModellingModelling, Simulation and Analysis of Physical Systems, Simulation and Analysis of Physical SystemsChapter 6Chapter 6 17
Obtaining the System Model (cont’d) (Review)
Example 5.1: Find the input-output and state-variable equations for the Example given in pp. 5.
( ):0 Law sAlembert'D'by or Law 2nd sNewton' =iiτΣ
(t)τKθBωωJBωKθ(t)τωJ
a
a
=++−−=
&
&
or
Very important step towards to obtaining correct system model!
Are the above equations input-output equation format?
(t)τKθBωωJ(t)τKθBωωJ
a
a
=++=+−−−
&
&
or 0
(t)τKθBωωJ(t)τKθBωωJ
a
a
=++=−++
&
&
or 0
+ +
Lecture Notes on Lecture Notes on MECH 370MECH 370 –– ModellingModelling, Simulation and Analysis of Physical Systems, Simulation and Analysis of Physical SystemsChapter 6Chapter 6 18
Obtaining the System Model (cont’d) (Review)
: iableoutput varfor equation output -Input θ
⎥⎦
⎤⎢⎣
⎡θω
=
τ⎥⎥
⎦
⎤
⎢⎢
⎣
⎡+⎥
⎦
⎤⎢⎣
⎡⎥⎥
⎦
⎤
⎢⎢
⎣
⎡ −−=⎥⎦
⎤⎢⎣
⎡
=
τ+−−=
1] [0
0
1
01
1
y
)t(Jθω
JK
JB
θω
ωθ
))t(KθBω(J
ω
a
a
&&
&
&
⎟⎠⎞
⎜⎝⎛ ====++ θω,θ
dtdθω(t)τKθθBθJ a
&&&&&&&
[ ] ⎥⎦
⎤⎢⎣
⎡=
τ⎥⎥⎦
⎤
⎢⎢⎣
⎡+⎥
⎦
⎤⎢⎣
⎡
⎥⎥⎦
⎤
⎢⎢⎣
⎡−−=⎥
⎦
⎤⎢⎣
⎡
τ+−−=
=
ωθ
y
)t(Jω
θ
JB
JK
ωθ
)t(J
ωJBθ
JKω
ωθ
a
a
01
1010
1
&
&
&
&
State-variable equations:
or
or
(t)τKθBωωJBωKθ(t)τωJ
a
a
=++−−=
&
&
or
Lecture Notes on Lecture Notes on MECH 370MECH 370 –– ModellingModelling, Simulation and Analysis of Physical Systems, Simulation and Analysis of Physical SystemsChapter 6Chapter 6 19
Electrical Systems (cont’d)(1))( or LLteeRi
dtdiL iC +−−=
From the element law for the capacitor
(2)LLidt
deC C =
(1) and (2) are in the state-variable form, or
1
)(11
iCdt
de
teL
eL
iLR
dtdi
C
iC
=
+−−=)(
0
1
01
1
teLei
C
LLR
dtdedtdi
iCC ⎥
⎥
⎦
⎤
⎢⎢
⎣
⎡+⎥
⎦
⎤⎢⎣
⎡
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡ −−=
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
If we are interested in ec as output variable, then
[ ] ⎥⎦
⎤⎢⎣
⎡==
CC e
iey 10
[ ] ⎥⎦
⎤⎢⎣
⎡===
CRR e
iRRieye 0 then iable,output var as or
or
Lecture Notes on Lecture Notes on MECH 370MECH 370 –– ModellingModelling, Simulation and Analysis of Physical Systems, Simulation and Analysis of Physical SystemsChapter 6Chapter 6 20
Obtain the Input-Output Model
e+
-i1 i2
There are two main approaches:1) Loop equation method (mesh analysis)
We assume a number of fictitious loop currents (i1, i2, …)
Then apply Kirchhoff’s voltage law and write the voltage in terms of the currents using element laws.2) Node equation method (node analysis)
i3
i2i1
e1 e2
Lecture Notes on Lecture Notes on MECH 370MECH 370 –– ModellingModelling, Simulation and Analysis of Physical Systems, Simulation and Analysis of Physical SystemsChapter 6Chapter 6 21
Obtain the Input-Output Model (cont’d)• Select an arbitrary ground node
• Label the remaining node voltages (e1,e2, …)
• Label unknown currents (i1, i2, …)
• Apply Kirchhoff’s current law for nodes
• Express currents in terms of node voltages using element laws.
*We will focus on this method.
Lecture Notes on Lecture Notes on MECH 370MECH 370 –– ModellingModelling, Simulation and Analysis of Physical Systems, Simulation and Analysis of Physical SystemsChapter 6Chapter 6 22
Obtain the Input-Output Model (cont’d)Example: Find I/O equation for input ei(t) and output eo.
KCL:
( )
( ) O) (Node 01)0(
A) (Node 0)(0 :O Node0 :A Node
0 0022
0
12
0
1
22
121
=τ+++−
−
=+−
+−
−∴
=++−=++−
∫t
LA
AAAi
LCR
CRR
deL
ieCR
ee
eCR
eeR
eteiiiiii
&
&
ei(t)+
-
R1 R2C1 L C2
ei(t) eA eO
A O
Lecture Notes on Lecture Notes on MECH 370MECH 370 –– ModellingModelling, Simulation and Analysis of Physical Systems, Simulation and Analysis of Physical SystemsChapter 6Chapter 6 23
Obtain the Input-Output Model (cont’d)
O) (Node 0)0( (2)
A) (Node 0)(2 (1)get to1H ,F1 ,Ω1Let
0
2121
=−τ+++
=−−+=====
∫ A
t
OLOO
iOAA
edeiee
teeeeLCCRR
&
&
To get the input-output equation for ei, eO differentiate (2) to get
0 (3) =−++ AOOO eeee &&&&
2)(
(3)'0)1()3(
0)()2()1(2
+−
=∴
=−++⇒
=−−+⇒
petee
peepp
teeep
oiA
AO
iOA
LL
Use p operator to get
Lecture Notes on Lecture Notes on MECH 370MECH 370 –– ModellingModelling, Simulation and Analysis of Physical Systems, Simulation and Analysis of Physical SystemsChapter 6Chapter 6 24
Obtain the Input-Output Model (cont’d)
To get state-variable equations leave the inductor current
LOAO
iOAA
ieeeteeee
−−=⇒++−=⇒
&
&
(2))(2(1)
(2)0)0(
(1)0)(2
0LL
4434421&
LL&
=−τ+++
=−−+
∫ A
i
t
OLOO
iOAA
edeiee
teeee
Lfrom inductor element law
Substitute into (3)’ to get
)(223 )()223(
))(()1)(2(
))((2
)1
23
2
2
teeeeetpeeppp
etepeppp
etep
pep(p
iOOOO
iO
OiO
OiO
&&&&&&& =+++∴=+++
+=+++
++
=++
I/O form
OL e
dtdiL =
Lecture Notes on Lecture Notes on MECH 370MECH 370 –– ModellingModelling, Simulation and Analysis of Physical Systems, Simulation and Analysis of Physical SystemsChapter 6Chapter 6 25
Obtain the Input-Output Model (cont’d)Example 6.4: Find the input-output equation for eO and ei(t).
KCL:
(2)0(1)0
22
231
LL
LL
=++=−++
CR
RRL
iiiiiii
(1) + (2): 0321 =++++ CRRRL iiiii
Lecture Notes on Lecture Notes on MECH 370MECH 370 –– ModellingModelling, Simulation and Analysis of Physical Systems, Simulation and Analysis of Physical SystemsChapter 6Chapter 6 26
Obtain the Input-Output Model (cont’d)
Differentiate (3) to get: )()( 12 teteee OL −+=∴
[ ] [ ] [ ] 01)(1)()(1)()(1
22
312\
112 =++++−++−+ OOOOO eCe
Rtee
Rtetee
Rtetee
L&&&&&&&&
Collecting terms gives:
[ ])()(1)(11)(11111212
311
1321
teteL
teRR
teR
eL
eRRR
eC OOO −+⎟⎟⎠
⎞⎜⎜⎝
⎛+−=+⎟⎟
⎠
⎞⎜⎜⎝
⎛+++ &&&&&
[ ] [ ]
[ ]
(3)01)(1
)()(1)()(1)0(
23
1
22
3
121
0 12
LL&
3214434421
444 3444 2144444 344444 21
=++++
+−++τ−++ ∫
C
RR
RL
iO
i
O
i
O
i
O
i
t
OL
eCeR
teeR
teteeR
dteteeL
i
0)()( 112 =−−+ teetee RO0)()( 12 =−−+ teetee LO
Lecture Notes on Lecture Notes on MECH 370MECH 370 –– ModellingModelling, Simulation and Analysis of Physical Systems, Simulation and Analysis of Physical SystemsChapter 6Chapter 6 27
Obtain the State-Variable Model
:form followinghave willequations variable-state The . e.g. voltagescapacitor
of in terms voltagesnode expressingby avoided becan assuch termsCapacitor terms.integral avoid to variablestate as choosecan
one example, previous theoftion representa variable-state obtain the To
C
C
L
eCeC
i
&
&
LL
cc
eLdt
di
iC
e
1
1
=
=&
. )( outputsfor equations algebraic Write(5)
.1 :currentsinductor for equations state Write(4)
.1 :voltagescapacitor for equations state Write(3)
).,( inputs and , variables-state of in terms currentscapacitor for Solve (2)
. voltagescapacitor of in terms voltagesnode Write(1):Procedures
K
&
K
,,ie
eLdt
di
iC
e
(t) (t), ieiei
e
OO
LL
CC
ii
LCc
C
=
=
Lecture Notes on Lecture Notes on MECH 370MECH 370 –– ModellingModelling, Simulation and Analysis of Physical Systems, Simulation and Analysis of Physical SystemsChapter 6Chapter 6 28
Obtain the State-Variable Model (cont’d)Example 6.7: Find the state-variable and output (eO) equations.
CLL
iCLC
iCLC
LCCi
eL
eLdt
di
tieR
iC
e
tieR
ii
ieR
iti
11
))(1(1
)(1
01)(
==
+−−=
+−−=
=+++−
&
or
[ ] ⎥⎦
⎤⎢⎣
⎡=
⎥⎥
⎦
⎤
⎢⎢
⎣
⎡+⎥
⎦
⎤⎢⎣
⎡
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡ −−=
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
L
C
iL
C
L
C
ie
e
tiCie
L
CRC
dtdidt
de
01
)(0
1
01
11
0
KCL:
Lecture Notes on Lecture Notes on MECH 370MECH 370 –– ModellingModelling, Simulation and Analysis of Physical Systems, Simulation and Analysis of Physical SystemsChapter 6Chapter 6 29
Obtain the State-Variable Model (cont’d)Example 6.8: Find the state-variable equations
KCL:
B) (Node 01)(1
A) (Node 0)(1))((1
2221
2111
32
21
=+++−−
=−++−−
CCLCC
CCCCi
eR
iieeR
eeR
ieteR
2
2122
2111
11
11111
)(111111
32211
122111
CLL
CCLCC
iCCCC
eL
eLdt
di
eRR
eR
iC
iC
e
teR
eR
eRRC
iC
e
==
⎥⎦
⎤⎢⎣
⎡⎟⎟⎠
⎞⎜⎜⎝
⎛+−+−==
⎥⎦
⎤⎢⎣
⎡++⎟⎟
⎠
⎞⎜⎜⎝
⎛+−==
&
&
Lecture Notes on Lecture Notes on MECH 370MECH 370 –– ModellingModelling, Simulation and Analysis of Physical Systems, Simulation and Analysis of Physical SystemsChapter 6Chapter 6 30
Obtain the State-Variable Model (cont’d)or
)(00
1
010
11111
01111
11
2
1
232222
12211
2
1
teCR
iee
CRRCCR
CRRRC
dtdidt
dedt
de
i
B
L
C
C
A
L
C
C
32144444444 344444444 21
⎥⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢⎢
⎣
⎡
+⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
⎥⎥⎥⎥⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢⎢⎢⎢⎢
⎣
⎡
−⎟⎟⎠
⎞⎜⎜⎝
⎛+−
⎟⎟⎠
⎞⎜⎜⎝
⎛+−
=
⎥⎥⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢⎢⎢
⎣
⎡
( )
( )
)(100
001
1111010
)(1
11y
1
2
1
1
212
1
23
212
2
1
2 te
Ri
ee
R
RRR
eteR
eR
ieeR
e
iie
i
L
C
C
Ai
CLCC
C
R
C
B
⎥⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢⎢
⎣
⎡
+⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
⎥⎥⎥⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢⎢⎢⎢
⎣
⎡
−
−⎟⎟⎠
⎞⎜⎜⎝
⎛+−=
⎥⎥⎥⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢⎢⎢⎢
⎣
⎡
−
−−−=⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡=
Lecture Notes on Lecture Notes on MECH 370MECH 370 –– ModellingModelling, Simulation and Analysis of Physical Systems, Simulation and Analysis of Physical SystemsChapter 6Chapter 6 31
Obtain the State-Variable Model (cont’d)Example 6.9: Find the state-variable equations with output eO.
))(263(51
2 (2)
))(3(51
0222)((2)(1)(2)B node 02 (1)A node 0)(2))(( :KCL
0
teeii
eii
teeie
eieeteeeeii
ieeteee
iCLC
OLC
iCLO
OCCOiCO
LC
CCOiCO
+−=
+=⇒
+−−=
=++++−+⇒+∴=++−
=+++−+LL
LL
Lecture Notes on Lecture Notes on MECH 370MECH 370 –– ModellingModelling, Simulation and Analysis of Physical Systems, Simulation and Analysis of Physical SystemsChapter 6Chapter 6 32
Obtain the State-Variable Model (cont’d)
)(11
))(263(101 C
DOLL
iCL
eeL
eLdt
di
teeie
−==
+−=∴ &
LDDL ieei5105:D nodefor KCL =⇒=+−
))(3(51
))(32(52
))(32(51
teei e
teeidtdi
teeie
iCLO
iCLL
iCLL
+−−=
+−−=
+−−=∴
Lecture Notes on Lecture Notes on MECH 370MECH 370 –– ModellingModelling, Simulation and Analysis of Physical Systems, Simulation and Analysis of Physical SystemsChapter 6Chapter 6 33
Obtain the State-Variable Model (cont’d)Dependent state variables
0)(:KVL =−+ teee iBA 0)(:KCL =++− BAi iiti
State variable are not independent for these problems:In this case, we will get one state equation for each independent state variable.
Algebraic relationship
B
Lecture Notes on Lecture Notes on MECH 370MECH 370 –– ModellingModelling, Simulation and Analysis of Physical Systems, Simulation and Analysis of Physical SystemsChapter 6Chapter 6 34
Obtain the State-Variable Model (cont’d)
.1 ,
variable,-state theas choose figure hand-left above for theequation variable-state theFind
11
A CAiB
A
iC
eeee
e
=−= &
KCL: (Node A)
Example 6.10:
0))((11))((21
2 1=−+++−−
444 3444 21
&&
ABi
iAACAi teeR
eR
ieteC
⎥⎦
⎤⎢⎣
⎡++⎟⎟
⎠
⎞⎜⎜⎝
⎛+−−=
++⎟⎟⎠
⎞⎜⎜⎝
⎛+−−=∴
)(1)(111or
)(1)(11
22
212
1
22
2121
teR
teCeRR
eCC
e
teR
teCeRR
eCi
iiAAA
iiAAC
&&&
&&
Ae•
B
Lecture Notes on Lecture Notes on MECH 370MECH 370 –– ModellingModelling, Simulation and Analysis of Physical Systems, Simulation and Analysis of Physical SystemsChapter 6Chapter 6 35
Obtain the State-Variable Model (cont’d)
⎪⎭
⎪⎬⎫
⎪⎩
⎪⎨⎧
⎥⎦
⎤⎢⎣
⎡⎟⎟⎠
⎞⎜⎜⎝
⎛+⎟⎟
⎠
⎞⎜⎜⎝
⎛+−+⎟⎟
⎠
⎞⎜⎜⎝
⎛+−
+=∴
+−=
⎥⎦
⎤⎢⎣
⎡+⎟⎟
⎠
⎞⎜⎜⎝
⎛+−
+=
+−
⎥⎦
⎤⎢⎣
⎡++⎟⎟
⎠
⎞⎜⎜⎝
⎛+−
+=
)(111111
)(let and
)(1111)(
)(1)(111
21
2
2122121
21
2
2212121
2
22
2121
teCC
CRRR
xRRCC
x
teCC
Cex
teR
eRRCC
teCC
Ce
teR
teCeRRCC
e
i
iA
iAiA
iiAA
&
&&
&&
Solve for
Note that one state variable is sufficient to represent the system dynamics.
Lecture Notes on Lecture Notes on MECH 370MECH 370 –– ModellingModelling, Simulation and Analysis of Physical Systems, Simulation and Analysis of Physical SystemsChapter 6Chapter 6 36
Reading and ExerciseReading and Exercise• Reading
Chapter 6: 6.1-6.6
• Assignment #3Ex. 6.2, 6.19, 6.28, 7.6, 7.11, 7.15 (to be
handed in)
Due: Friday, 27/7/07 at lecture.
Ex. 6.3, 6.6, 6.29, 7.3, 7.20 (for your practice, no need to hand in)
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