Control Systems (CS)
Lecture-3
Introduction to Mathematical Modeling
&
Mathematical Modeling of Electrical Systems
Dr. Imtiaz Hussain
Associate Professor
Mehran University of Engineering & Technology Jamshoro, Pakistan
email: imtiaz.hussain@faculty.muet.edu.pk
URL :http://imtiazhussainkalwar.weebly.com/
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Types of Systems
•
•
Static System: If a system does not change
with time, it is called a static system.
Dynamic System: If a system changes with
time, it is called a dynamic system.
2
Dynamic Systems
• A system is said to be dynamic if its current output may depend on
the past history as well as the present values of the input variables.
• Mathematically,
y(t )  [u( ),0    t ]
u : Input, t : Time
Example: A moving mass
y
u
Model: Force=Mass x Acceleration
My  u
M
Ways to Study a System
System
Experiment with a
model of the System
Experiment with actual
System
Mathematical Model
Physical Model
Analytical Solution
Simulation
Frequency Domain
Time Domain
Hybrid Domain
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Model
•
•
•
A model is a simplified representation or
abstraction of reality.
Reality is generally too complex to copy
exactly.
Much of the complexity is actually irrelevant
in problem solving.
5
What is Mathematical Model?
A set of mathematical equations (e.g., differential eqs.) that
describes the input-output behavior of a system.
What is a model used for?
• Simulation
• Prediction/Forecasting
• Prognostics/Diagnostics
• Design/Performance Evaluation
• Control System Design
Black Box Model
• When only input and output are known.
• Internal dynamics are either too complex or
unknown.
Input
Output
• Easy to Model
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Grey Box Model
• When input and output and some information
about the internal dynamics of the system is
known.
u(t)
y(t)
y[u(t), t]
• Easier than white box Modelling.
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White Box Model
• When input and output and internal dynamics
of the system is known.
u(t)
dy(t )
du(t ) d 2 y(t )
3

dt
dt
dt 2
y(t)
• One should know have complete knowledge
of the system to derive a white box model.
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Basic Elements of Electrical Systems
• The time domain expression relating voltage and current for the
resistor is given by Ohm’s law i-e
v R (t )  iR (t )R
• The Laplace transform of the above equation is
VR ( s )  I R ( s ) R
Basic Elements of Electrical Systems
• The time domain expression relating voltage and current for the
Capacitor is given as:
1
vc (t )   ic (t )dt
C
• The Laplace transform of the above equation (assuming there is no
charge stored in the capacitor) is
1
Vc ( s ) 
Ic (s)
Cs
Basic Elements of Electrical Systems
• The time domain expression relating voltage and current for the
inductor is given as:
diL (t )
v L (t )  L
dt
• The Laplace transform of the above equation (assuming there is no
energy stored in inductor) is
VL ( s )  LsI L ( s )
V-I and I-V relations
Component
Resistor
Symbol
V-I Relation
I-V Relation
v R (t )  iR (t )R
v R (t )
iR (t ) 
R
Capacitor
dvc (t )
1
vc (t )   ic (t )dt ic (t )  C
C
dt
Inductor
diL (t )
v L (t )  L
dt
1
iL (t )   v L (t )dt
L
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Example#1
• The two-port network shown in the following figure has vi(t) as
the input voltage and vo(t) as the output voltage. Find the
transfer function Vo(s)/Vi(s) of the network.
vi( t)
v i ( t )  i( t ) R 
i(t)
C
vo(t)
1
 i( t )dt
C
1
vo ( t ) 
 i( t )dt
C
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Example#1
1
1
vo ( t ) 
v i ( t )  i( t ) R 
 i( t )dt
 i( t )dt
C
C
• Taking Laplace transform of both equations, considering initial
conditions to zero.
1
Vi ( s )  I ( s ) R 
I (s)
Cs
1
Vo ( s ) 
I (s)
Cs
• Re-arrange both equations as:
Vi ( s )  I ( s )( R 
1
)
Cs
CsVo ( s )  I ( s )
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Example#1
1
Vi ( s )  I ( s )( R 
)
Cs
• Substitute I(s) in equation on left
CsVo ( s )  I ( s )
1
Vi ( s )  CsVo ( s )( R 
)
Cs
Vo ( s )

Vi ( s )
1
1
Cs( R 
)
Cs
Vo ( s )
1

Vi ( s ) 1  RCs
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Example#1
Vo ( s )
1

Vi ( s ) 1  RCs
• The system has one pole at
1  RCs  0
1
s
RC
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Example#2
• Design an Electrical system that would place a pole at -3 if
added to another system.
Vo ( s )
1

Vi ( s ) 1  RCs
vi( t)
i(t)
C
v2(t)
• System has one pole at
1
s
RC
• Therefore,
1

 3
RC
if
R  1 M
and
C  333 pF
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Example#3
• Find the transfer function G(S) of the following
two port network.
L
vi(t)
i(t)
C
vo(t)
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Example#3
• Simplify network by replacing multiple components with
their equivalent transform impedance.
L
Z
Vi(s)
I(s)
C
Vo(s)
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Transform Impedance (Resistor)
iR(t)
IR(S)
+
+
Transformation
vR(t)
-
ZR = R
VR(S)
-
21
Transform Impedance (Inductor)
iL(t)
IL(S)
+
vL(t)
-
ZL=LS
+
VL(S)
LiL(0)
-
22
Transform Impedance (Capacitor)
ic(t)
Ic(S)
+
vc(t)
-
ZC(S)=1/CS
+
Vc(S)
-
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Equivalent Transform Impedance (Series)
• Consider following arrangement, find out equivalent
transform impedance.
ZT  Z R  Z L  Z C
L
C
1
Z T  R  Ls 
Cs
R
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Equivalent Transform Impedance (Parallel)
1
1
1
1



ZT
Z R Z L ZC
1
1
1
1
 

1
ZT
R Ls
Cs
L
C
R
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Equivalent Transform Impedance
• Find out equivalent transform impedance of
following arrangement.
L2
L2
R1
R2
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Back to Example#3
L
Z
Vi(s)
I(s)
C
Vo(s)
1
1
1


Z ZR ZL
1 1
1
 
Z R Ls
RLs
Z
1  RLs
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Example#3
Z
RLs
1  RLs
L
Z
Vi(s)
I(s)
1
Vi ( s )  I ( s )Z 
I (s)
Cs
C
Vo(s)
1
Vo ( s ) 
I (s)
Cs
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Operational Amplifiers
Vout
Z2

Vin
Z1
Vout
Z2
 1
Vin
Z1
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Example#4
• Find out the transfer function of the following
circuit.
Vout
Z2

Vin
Z1
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Example#5
• Find out the transfer function of the following
circuit.
v1
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Example#6
• Find out the transfer function of the following
circuit.
v1
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Example#7
• Find out the transfer function of the following
circuit and draw the pole zero map.
100kΩ
10kΩ
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END OF LECTURE-3
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