lecture-2-_mathematical_modeling_of_dynamic_systems.pdf

CS Slide on Mathametical modeling of Dynamic System

1. Dr. Imtiaz Hussain
Associate Professor
Mehran University of Engineering & Technology Jamshoro, Pakistan
email: [email protected]
URL :http://imtiazhussainkalwar.weebly.com/
Note:Idonotclaimanyoriginalityintheselectures.Thecontentsofthispresentationare
mostlytakenfromthebookofOgatta,NormanNise,BishopandBC.Kuoandvariousother
internetsources. 1
Control Systems (CS)
6
th
Semester 14ES (SEC-I)

2. Outline
•Introduction
•Types of Models
•Mathematical Modeling of
–Electrical Systems
–Electronic Systems
–Mechanical Systems
–Electromechanical Systems
2

3. Types of Systems
•StaticSystem:Ifasystemdoesnotchange
withtime,itiscalledastaticsystem.
•DynamicSystem:Ifasystemchangeswith
time,itiscalledadynamicsystem.
3

4. Dynamic Systems
•Asystemissaidtobedynamicifitscurrentoutputmaydependon
thepasthistoryaswellasthepresentvaluesoftheinputvariables.
•Mathematically, Time Input, ::
]),([)(
tu
tuty   0
Example: A moving mass
M
y
u
Model: Force=Mass x AccelerationuyM

5. Ways to Study a System
5
System
Experiment with a
model of the System
Experiment with actual
System
Physical Model
Mathematical Model
Analytical Solution
Simulation
Frequency Domain Time Domain Hybrid Domain

6. Model
•Amodelisasimplifiedrepresentationor
abstractionofreality.
•Realityisgenerallytoocomplextocopy
exactly.
•Muchofthecomplexityisactuallyirrelevant
inproblemsolving.
6

7. What is Mathematical Model?
Asetofmathematicalequations(e.g.,differentialeqs.)that
describestheinput-outputbehaviorofasystem.
What is a model used for?
•Simulation
•Prediction/Forecasting
•Prognostics/Diagnostics
•Design/Performance Evaluation
•Control System Design

8. Black Box Model
•Whenonlyinputandoutputareknown.
•Internaldynamicsareeithertoocomplexor
unknown.
•Easy to Model
8
Input Output

9. Grey Box Model
•Wheninputandoutputandsomeinformation
abouttheinternaldynamicsofthesystemis
known.
•Easier than white box Modelling.
9
u(t) y(t)
y[u(t), t]

10. White Box Model
•Wheninputandoutputandinternaldynamics
ofthesystemisknown.
•Oneshouldknowhavecompleteknowledge
ofthesystemtoderiveawhiteboxmodel.
10
u(t) y(t)2
2
3
dt
tyd
dt
tdu
dt
tdy )()()(


11. Modeling of Electrical Systems
11

12. Basic Elements of Electrical Systems
12
Component Symbol
V-I Relation
(Time Domain)
V-I Relation
(Frequency
Domain)
Resistor
Capacitor
Inductordt
tdi
Ltv
L
L
)(
)( dtti
C
tv
cc  )()(
1 Rtitv
RR )()( RsIsV
RR )()( )()( sI
Cs
sV
cc
1
 )()( sLsIsV
LL

13. Example#1
•Thetwo-portnetworkshowninthefollowingfigurehasv
i(t)as
theinputvoltageandv
o(t)astheoutputvoltage.Findthe
transferfunctionV
o(s)/V
i(s)ofthenetwork.
13
Ci(t)v
i( t) v
o(t) dtti
C
Rtitv
i )()()(
1  dtti
C
tv
o )()(
1

14. Example#1
•TakingLaplacetransformofbothequations,consideringinitial
conditionstozero.
•Re-arrangebothequationsas:
14 dtti
C
Rtitv
i )()()(
1  dtti
C
tv
o )()(
1 )()()( sI
Cs
RsIsV
i
1
 )()( sI
Cs
sV
o
1
 )()( sIsCsV
o ))(()(
Cs
RsIsV
i
1


15. Example#1
•SubstituteI(s)inequationonleft
15)()( sIsCsV
o ))(()(
Cs
RsIsV
i
1
 ))(()(
Cs
RsCsVsV
oi
1
 )(
)(
)(
Cs
RCs
sV
sV
i
o
1
1

 RCssV
sV
i
o


1
1
)(
)(

16. Example#1
•Thesystemhasonepoleat
16RCssV
sV
i
o


1
1
)(
)( RC
sRCs
1
01 

17. Example#2
•Find the transfer function G(S) of the following
two port network.
17
i(t)v
i(t) v
o(t)
L
C

18. Example#2
•Simplifynetworkbyreplacingmultiplecomponentswith
theirequivalenttransformimpedance.
18
I(s)V
i(s) V
o(s)
L
C
Z

19. Transform Impedance (Resistor)
19
i
R(t)
v
R(t)
+
-
I
R(S)
V
R(S)
+
-
Z
R= R
Transformation

20. Transform Impedance (Inductor)
20
i
L(t)
v
L(t)
+
-
I
L(S)
V
L(S)
+
-
Li
L(0)
Z
L=LS

21. Transform Impedance (Capacitor)
21
i
c(t)
v
c(t)
+
-
I
c(S)
V
c(S)
+
-
Z
C(S)=1/CS

22. Equivalent Transform Impedance (Series)
•Considerfollowingarrangement,findoutequivalent
transformimpedance.
22
L
C
RCLRT ZZZZ  Cs
LsRZ
T
1


23. Equivalent Transform Impedance (Parallel)
23CLRT ZZZZ
1111

C
L
RCs
LsRZ
T
1
1111


24. Equivalent Transform Impedance
•Findoutequivalenttransformimpedanceof
followingarrangement.
24
L
2
L
2
R
2
R
1

25. Back to Example#2
25
I(s)V
i(s) V
o(s)
L
C
ZLRZZZ
111
 LsRZ
111
 RLs
RLs
Z


1

26. Example#2
26
I(s)V
i(s) V
o(s)
L
C
ZRLs
RLs
Z


1 )()()( sI
Cs
ZsIsV
i
1
 )()( sI
Cs
sV
o
1


27. Modeling of Electronic Systems
27

28. Operational Amplifiers
281
2
Z
Z
V
V
in
out
 1
2
1
Z
Z
V
V
in
out


29. Example#3
•Find out the transfer function of the following
circuit.
291
2
Z
Z
V
V
in
out


30. Example#4
•Find out the transfer function of the following
circuit.
30
v
1

31. Example#5
•Findoutthetransferfunctionofthefollowing
circuit.
31
v
1

32. Example#6
•Findoutthetransferfunctionofthefollowing
circuitanddrawthepolezeromap.
32
10kΩ
100kΩ

33. Modeling of Mechanical Systems
33
•Part-I: Translational Mechanical System
•Part-II: Rotational Mechanical System
•Part-III: Mechanical Linkages

34. Basic Types of Mechanical Systems
•Translational
–Linear Motion
•Rotational
–Rotational Motion
34

35. TRANSLATIONAL MECHANICAL SYSTEMS
Part-I
35

36. Basic Elements of Translational Mechanical Systems
Translational Spring
i)
Translational Mass
ii)
Translational Damper
iii)
36

37. Translational Spring
i)
Circuit Symbols
Translational Spring
•Atranslationalspringisamechanicalelementthat
canbedeformedbyanexternalforcesuchthatthe
deformationisdirectlyproportionaltotheforce
appliedtoit.
Translational Spring
37

38. Translational Spring
•IfFistheappliedforce
•Thenisthedeformationif
•Or isthedeformation.
•Theequationofmotionisgivenas
•WhereisstiffnessofspringexpressedinN/m2x 1x 0
2x 1x )(
21xx )(
21xxkF  k F F
38

39. Translational Mass
Translational Mass
ii)
•TranslationalMassisaninertia
element.
•Amechanicalsystemwithout
massdoesnotexist.
•IfaforceFisappliedtoamass
anditisdisplacedtoxmeters
thentherelationb/wforceand
displacementsisgivenby
Newton’slaw.
M)(tF )(tx xMF 
39

40. Translational Damper
Translational Damper
iii)
•Whentheviscosityordragisnot
negligibleinasystem,weoften
modelthemwiththedamping
force.
•Allthematerialsexhibitthe
propertyofdampingtosome
extent.
•Ifdampinginthesystemisnot
enoughthenextraelements(e.g.
Dashpot)areaddedtoincrease
damping.
40

41. Common Uses of Dashpots
Door Stoppers
Vehicle Suspension
Bridge Suspension
Flyover Suspension
41

42. Translational DamperxCF
•Where Cis damping coefficient (N/ms
-1
).)(
21xxCF 

43. Example-7
•Consider the following system (friction is negligible)
43
•Free Body Diagram
MF kf Mf k F x M
•Where and are forces applied by the spring and
inertial force respectively. kf Mf

44. Example-7
44
•Then the differential equation of the system is:kxxMF 
•Taking the Laplace Transform of both sides and ignoring
initial conditions we get
MF kf Mf MkffF  )()()( skXsXMssF 
2

45. 45)()()( skXsXMssF 
2
•The transfer function of the system iskMssF
sX


2
1
)(
)(
•if1
2000
1000



Nmk
kgM 2
0010
2


ssF
sX .
)(
)(
Example-7

46. 46
•The pole-zero map of the system is2
0010
2


ssF
sX .
)(
)(
Example-7

47. Example-8
•Consider the following system
47
•Free Body Diagramk F x M C
MF kf Mf Cf CMk fffF 

48. Example-8
48
Differential equation of the system is:kxxCxMF  
Taking the Laplace Transform of both sides and ignoring
Initial conditions we get)()()()( skXsCsXsXMssF 
2 kCsMssF
sX


2
1
)(
)(

49. Example-8
49kCsMssF
sX


2
1
)(
)(
•if1
1
1000
2000
1000





msNC
Nmk
kgM
/ 2
001.0
)(
)(
2


sssF
sX -1 -0.5 0 0.5 1
-2
-1.5
-1
-0.5
0
0.5
1
1.5
2
Pole-Zero Map
Real Axis
Imaginary Axis

50. Example-9
•Consider the following system
50
•Mechanical Networkk F 2x M 1x B
↑ Mk B F 1x 2x

51. Example-9
51
•Mechanical Network
↑ Mk B F 1x 2x )(
21xxkF 
At node 1x
At node 2x 22120 xBxMxxk  )(

52. Example-10
52k )(tf 2x 1M 4B 3B 2M 1x 1B 2B
↑ M
1k 1B )(tf 1x 2x 3B 2B
M
24B

53. ROTATIONAL MECHANICAL SYSTEMS
Part-I
53

54. Basic Elements of Rotational Mechanical Systems
Rotational Spring)(
21kT 2 1
54

55. Basic Elements of Rotational Mechanical Systems
Rotational Damper2 1 )(
21

CT T C
55

56. Basic Elements of Rotational Mechanical Systems
Moment of Inertia

JT  T J
56

57. Example-111 T 1J 1k 1B 2k 2J 2 3
↑ J
11k T 1 3 1B
J
22 2k
57

58. Example-12
↑ J
11k 1B T 1 3 2B 3B
J
24B 2 1 T 1J 1k 3B 2B 4B 1B 2J 2 3
58

59. Example-131 T 1J 1k 2B 2J 2 2k
591B

60. MECHANICAL LINKAGES
Part-III
60

61. Gear
•Gearisatoothedmachinepart,such
asawheelorcylinder,thatmeshes
withanothertoothedpartto
transmitmotionortochangespeed
ordirection.
61

62. Fundamental Properties
•Thetwogearsturninoppositedirections:oneclockwiseand
theothercounterclockwise.
•Twogearsrevolveatdifferentspeedswhennumberofteeth
oneachgeararedifferent.
62

63. Gear Ratio
•Youcancalculatethegearratiobyusing
thenumberofteethofthedriver
dividedbythenumberofteethofthe
follower.
•Wegearupwhenweincreasevelocity
anddecreasetorque.
Ratio:3:1
•Wegeardownwhenweincreasetorque
andreducevelocity.
Ratio:1:3
Gear Ratio = # teeth input gear / # teeth output gear
= torque in / torque out = speed out / speed in
Follower
Driver
63

64. Mathematical Modeling of Gear Trains
•Gearsincreaseorreduceangularvelocity(while
simultaneouslydecreasingorincreasingtorque,such
thatenergyisconserved).
642211  NN  1N
Number of Teeth of Driving Gear1
Angular Movement of Driving Gear2N
Number of Teeth of Following Gear2
Angular Movement of Following Gear
Energy of Driving Gear = Energy of Following Gear

65. Mathematical Modeling of Gear Trains
•Inthesystembelow,atorque,τ
a,isappliedtogear1(with
numberofteethN
1,momentofinertiaJ
1andarotationalfriction
B
1).
•It,inturn,isconnectedtogear2(withnumberofteethN
2,
momentofinertiaJ
2andarotationalfrictionB
2).
•Theangleθ
1isdefinedpositiveclockwise,θ
2isdefinedpositive
clockwise.Thetorqueactsinthedirectionofθ
1.
•AssumethatT
Listheloadtorqueappliedbytheloadconnected
toGear-2.
65
B
1
B
2
N
1
N
2

66. Mathematical Modeling of Gear Trains
•ForGear-1
•ForGear-2
•Since
•therefore
66
B
1
B
2
N
1
N
22211 NN 11111 TBJ
a  

Eq(1)LTBJT 
22222 

Eq(2)1
2
1
2 
N
N

Eq(3)

67. Mathematical Modeling of Gear Trains
•GearRatioiscalculatedas
•Putthisvalueineq(1)
•PutT
2fromeq(2)
•Substituteθ
2fromeq(3)
67
B
1
B
2
N
1
N
22
2
1
1
1
2
1
2
T
N
N
T
N
N
T
T
 2
2
1
1111 T
N
N
BJ
a  
 )(
La TBJ
N
N
BJ 
2222
2
1
1111 
 )(
2
2
1
21
2
1
2
2
1
1111 La T
N
N
B
N
N
J
N
N
BJ  


68. Mathematical Modeling of Gear Trains
•Aftersimplification
68)(
2
2
1
21
2
1
2
2
1
1111 La T
N
N
B
N
N
J
N
N
BJ  
 La T
N
N
B
N
N
BJ
N
N
J
2
1
12
2
2
1
1112
2
2
1
11 

















 
 La T
N
N
B
N
N
BJ
N
N
J
2
1
12
2
2
1
112
2
2
1
1 

































 
 2
2
2
1
1 J
N
N
JJ
eq 







 2
2
2
1
1 B
N
N
BB
eq 







 Leqeqa T
N
N
BJ
2
1
11  


69. Mathematical Modeling of Gear Trains
•Forthreegearsconnectedtogether
693
2
4
3
2
2
1
2
2
2
1
1 J
N
N
N
N
J
N
N
JJ
eq 
























 3
2
4
3
2
2
1
2
2
2
1
1 B
N
N
N
N
B
N
N
BB
eq 


























70. Modeling of Electromechanical
Systems
70

71. 71
D.C Drives
•VariableVoltagecanbeappliedtothearmatureterminalsoftheDC
motor.
•Anothermethodistovarythefluxperpoleofthemotor.
•Thefirstmethodinvolveadjustingthemotor’sarmaturewhilethe
lattermethodinvolvesadjustingthemotorfield.Thesemethodsare
referredtoas“armaturecontrol”and“fieldcontrol.”

72. oltageback-emf vwhere e,e
dt
di
LiRu
bb
a
aaa  Mechanical SubsystemBωωJT
motor 
Input: voltage u
Output: Angular velocity 
Electrical Subsystem(loop method):
Example-14: Armature Controlled D.C Motor
u
i
a
T
R
a L
a
J

B
e
b
72

73. Torque-Current:
Voltage-Speed:atmotor iKT 
•Combingpreviousequationsresultsinthefollowingmathematical
model:
Power Transformation:ωKe
bb 






0
at
baa
a
a
i-KBωJ
uωKiR
dt
di
L

Example-14: Armature Controlled D.C Motor
u
i
a
T
R
a L
a
J

B
e
b
73

74. TakingLaplacetransformofthesystem’sdifferentialequationswith
zeroinitialconditionsgives:
Eliminating I
ayields the input-output transfer function 
btaaaa
t
KKBRsBLJRJsL
K
U(s)
Ω(s)


2  
 





0(s)IΩ(s)-KBJs
U(s)Ω(s)K(s)IRsL
at
baaa
Example-14: Armature Controlled D.C Motor
74

75. Reduced Order Model
Assuming small inductance, L
a0
Example-14: Armature Controlled D.C Motor
75)(
btaa
t
KKBRsJR
K
U(s)
Ω(s)



76. If output of the D.Cmotor is angular position θthen we know
Which yields following transfer function
Example-14: Armature Controlled D.C Motor)()( sssor
dt
d


 
u
i
a
T
R
a L
a
J
θ
B
e
b
76)]([
btaa
t
KKBRsJRs
K
U(s)
(s)




77. Applying KVLat field circuit
Example-15: Field Controlled D.C Motor
i
f
T
m
R
f
L
f J
ωB
R
a L
a
e
a
e
fdt
di
LRie
f
ffff

Mechanical SubsystemBωωJT
m 
77

78. Torque-Current:ffm
iKT
Combing previous equations and taking Laplace transform (considering
initial conditions to zero) results in the following mathematical model:
Power Transformation:






)()()(
)()()(
sIKsBsJs
sIsLsIRsE
ff
fffff
where K
f: torque constant
Example-15: Field Controlled D.C Motor
78

79. If angular position θis output of the motor
Eliminating I
f(S)yields
Example-15: Field Controlled D.C Motor  )(
ff
f
f RsLBJs
K
(s)E
Ω(s)


i
f
T
m
R
f
L
f J
θB
R
a L
a
e
a
e
f  )(
ff
f
f RsLBJss
K
(s)E
(s)



79

80. +
k
p
-
J
L
_
i
a
e
b
R
a
L
a
+
T
r c
e
a
_
+
e
_
+
N
1
N
2
B
L
θ
i
f = Constant
J
M
B
M
80
Example-16: Angular Position ControlSystem

81. Numerical Values for System constants
r =angular displacement of the reference input shaft
c =angular displacement of the output shaft
θ=angular displacement of the motor shaft
K
1=gain of the potentiometer shaft = 24/π
K
p=amplifier gain = 10
e
a=armature voltage
e
b=back emf
R
a=armature winding resistance = 0.2 Ω
L
a =armature winding inductance = negligible
i
a=armature winding current
K
b=back emfconstant = 5.5x10
-2
volt-sec/rad
K
t=motor torque constant = 6x10
-5
N-m/ampere
J
m= moment of inertia of the motor = 1x10
-5
kg-m
2
B
m=viscous-friction coefficients of the motor = negligible
J
L=moment of inertia of the load = 4.4x10
-3
kgm
2
B
L=viscous friction coefficient of the load = 4x10
-2
N-m/rad/sec
n=gear ratio = N
1/N
2= 1/10
81

82. Example-16: Angular Position Control System
82
•Transfer function of the armature controlled D.C motor with load connected to it
is given by
•Where
�
??????�=�
??????+
??????
1
??????
2
2
�
??????=1×10
−5
+
1
10
2
×4.4×10
−3
=5.4×10
−5
�
??????�=�
??????+
??????
1
??????
2
2
�
??????=
1
10
2
×4×10
−2
=4×10
−4
??????(�)
??????
�(�)
=
6
�(1.08�+8.33)
??????(�)
??????
�(�)
=
�
??????
��
??????��
�+�
??????��
��+�
??????��
�+�
??????�
�

83. Example-16: Angular Position Control System
83
??????�=�
1��−??????(�)
•Error is difference between reference input ????????????and out calculated ????????????and can be
calculated as
•Output of amplifier is
•Merging eq(a) and eq(b) yields
•Relation between angular position of motor ??????and angular position of load ??????is given as
•or
E�=
24
??????
��−�(�)=7.64��−�(�)
??????
��=�
�??????�=10??????(�)
??????
��=76.4��−�(�)
(a)
(b)
��=
1
10
??????(�)
10��=??????(�)

84. Example-16: Angular Position Control System
84
•Final Closed loop transfer function of the system can now be written as
10�(�)
76.4��−�(�)
=
6
�(1.08�+8.33)
�(�)
��
=
42.3
�
2
+7.69�+42.3
??????(�)
??????
�(�)
=
6
�(1.08�+8.33)

85. END OF LECTURE-2
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85