Lec. 3
Mathematical Modeling of Dynamic System
Dr. Tamer Samy Gaafar

www.tsgaafar.faculty.zu.edu.eg

Email:
tsgaafar@yahoo.com
Quiz
(10 mins)
Find the transfer function of the following block diagrams
H4
R(s)
Y (s )
G1
G3
G2
G4
H3
H2
H1
3
Solution:
1. Moving pickoff point A behind block
G4
I
H4
R(s)
Y (s )
G1
G3
G2
H3
H2
H3
G4
H2
G4
A
G4
B
1
G4
1
G4
H1
4
2. Eliminate loop I and Simplify
R(s)
G2G3G4
1  G3G4 H 4
G1
II
Y (s )
B
H3
G4
H2
G4
III
H1
II
feedback
G2G3G4
1  G3G4 H 4  G2G3 H 3
III
Not feedback
H 2  G4 H 1
G4
5
3. Eliminate loop II & IIII
R(s)
G1G2G3G4
1  G3G4 H 4  G2G3 H 3
Y (s )
H 2  G4 H 1
G4
G1G2G3G4
Y (s)

R( s ) 1  G2G3 H 3  G3G4 H 4  G1G2G3 H 2  G1G2G3G4 H1
6
•
•
•
•
Static System: If a system does not change with time,
it is called a static system.
The o/p depends on the i/p at the present time only
Dynamic System: If a system changes with time, it is
called a dynamic system.
The o/p depends on the i/p at the present time and
the past time. (integrator, differentiator).
7
• 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
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
9
•
•
•
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.
10
A set of mathematical equations (e.g., differential equs.)
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
1. Electrical Systems
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 )
Componen
t
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
16

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)
i(t)
C
vo(t)
1
v i ( t )  i( t ) R 
 i( t )dt
C
1
vo ( t ) 
 i( t )dt
C
17

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 )
18

1
Vi ( s )  I ( s )( R 
)
CsVo ( s )  I ( s )
Cs
Substitute I(s) in equation on left
1
Vi ( s )  CsVo ( s )( R 
)
Cs
Vo ( s )

Vi ( s )
1
1
Cs( R 
)
Cs
Vo ( s )
1

Vi ( s ) 1  RCs
19
Vo ( s )
1

Vi ( s ) 1  RCs

The system has one pole at
1  RCs  0
s
1
RC
20

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)
C
i(t)
System has one pole at
v2(t)
1
s
RC
Therefore,
1

 3
RC
if
R  1 M
and
C  333 pF
21
iR(t)
IR(S)
+
+
Transformation
vR(t)
-
ZR = R
VR(S)
-
22
iL(t)
IL(S)
+
vL(t)
-
ZL=LS
+
VL(S)
LiL(0)
-
23
ic(t)
Ic(S)
+
vc(t)
-
ZC(S)=1/CS
+
Vc(S)
-
24

Consider following arrangement,
equivalent transform impedance.
ZT  Z R  Z L  Z C
1
Z T  R  Ls 
Cs
find
out
L
C
R
25
1
1
1
1



ZT
Z R Z L ZC
1
1
1
1
 

1
ZT
R Ls
Cs
L
C
R
26
2. Mechanical Systems
Mechanical Systems
•
Part-I: Translational Mechanical System
•
Part-II: Rotational Mechanical System
•
Part-III: Mechanical Linkages
28

Translational
◦ Linear Motion

Rotational
◦ Rotational Motion
29
Basic Elements of Translational
Mechanical Systems
i)
ii)
iii)
Translational
Spring
Translational
Mass
Translational
Damper
31

A translational spring is a mechanical
element that can be deformed by an
external force such that the deformation is
directly proportional to the force applied to
it.
i)
Translational
Spring
Circuit Symbols
Translational Spring
32

If F is the applied force
x1
x2

Then
x1 is the deformation if
x2  0
F

Or ( x1  x 2 )

The equation of motion is given as
is the deformation.
F
F  k ( x1  x2 )

Where k is stiffness of spring expressed in
N/m
33
Translational Mass
• Translational Mass
inertia element.
is
an
ii)
Translational
Mass
• A
mechanical
system
without mass does not
exist.
• If a force F is applied to a
mass and it is displaced to
x meters then the relation
b/w
force
and
displacements is given by
Newton’s law. F  M
x
x(t )
F (t )
M
34
Translational Damper
• When the viscosity or drag is
not negligible in a system, we
often model them with the
damping force.
• All the materials exhibit the
property of damping to some
extent.
iii)
Translational
Damper
• If damping in the system is not
enough then extra elements
(e.g. Dashpot) are added to
increase damping.
35
Common Uses of Dashpots
Door Stoppers
Bridge Suspension
Vehicle Suspension
Flyover Suspension
36
Translational Damper
b
F  bx
b
F  b( x1  x2 )
• Where b is damping coefficient (N/ms-1).
37

Example-1: Consider a simple horizontal spring-mass
system on a frictionless surface, as shown in figure
below.
mx   kx
or
mx  kx  0
38

Consider the following system (friction is
negligible)
k
x
M
F
• Free Body Diagram
fk
M
fM
F
• Where f k and f M are force applied by the
spring and inertial force respectively.
39
fk
M
fM
F
F  fk  fM
• Then the differential equation of the system is:
F  Mx  kx
• Taking the Laplace Transform of both sides
and ignoring initial conditions we get
F ( s )  Ms 2 X ( s )  kX( s )
40
Example-2
F ( s )  Ms 2 X ( s )  kX( s )
• The transfer function of the system is
X (s)
1

F(s)
Ms 2  k
• if
M  1000 kg
k  2000 Nm 1
X (s)
0.001
 2
F(s)
s 2
41
Example-2
X (s)
0.001
 2
F(s)
s 2
• The pole-zero map of the system is
Pole-Zero Map
40
30
Imaginary Axis
20
10
0
-10
-20
-30
-40
-1
-0.5
0
Real Axis
0.5
1
42

Consider the following system
x2
x1 k
B
F
M
• Mechanical Network
x1
F
↑
k
x2
M
B
43
• Mechanical Network
x1
F
At node
↑
k
x2
M
B
x1
F  k ( x1  x 2 )
At node
x2
2
0  k ( x2  x1 )  Mx2  Bx
44
45
46
mxo  b( x o  xi )  k ( xo  xi )  0
(eq .1)
mxo  bxo  kxo  bxi  kxi
eq. 2
Taking Laplace Transform of the equation (2)
2
ms X o ( s )  bsX o ( s )  kXo ( s )  bsX i ( s )  kXi ( s )
X o (s)
bs  k

X i ( s ) ms 2  bs  k
47
Car Body
Bogie-2
Bogie-1
Secondary
Suspension
Wheelsets
Primary
Bogie
Frame
Suspension
48
49