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3.0MATHEMATICAL MODELLING
3.1 INTRODUCTION
In studying control systems the reader must understand how to model dynamic systems and analyze
dynamic characteristics. A mathematical model of a dynamic system is defined as a set of equations
that represents the dynamics of the system accurately or, at least, fairly well. Note that a mathematical
model is not unique to a given system. A system may be represented in many different ways and,
therefore, may have many mathematical models, depending on one's perspective.
The dynamics of many systems, whether they are mechanical, electrical, thermal, economic,
biological, and so on, may be described in terms ofdifferential equations. Such differential equations
may be obtained by using physical laws governing a particular system, for example, Newton's lawsfor mechanical systems and Kirchhoff's laws for electrical systems. We must always keep in mind
that deriving reasonable mathematical models is the most important part of the entire analysis of
control systems.
A mathematical model, is a description of a system in terms of equations.Mathematical Modelling is a
procedure of constructing a model of a dynamic system based on the physical laws ( Newtons lawand so forth) that the system elements and their interconnections are known to obey.
Dynamic systems are systems for which the variables are time-dependent. Not only that the responses
are not instantaneously proportional to the excitations but at any instant the derivatives of one and
more variables depend on the values of the system variables at that instant. Dynamic systems can
respond to input signals, disturbance signals, or initial conditions.
In many cases, signals produced by the system can be measured and be used to contruct a model. The
procedure of constructing a model based on measured data is known as system identification.
The process of using the mathematical model to determine certain features of the systems cause-and-
effect relationships is reffered as solving the model or simulation.
3.1.1 STEPS IN MODELLING AND SIMULATION OF DYNAMIC SYSTEMS
modification
Actual Physicalsystem
EngineersPerception
MathematicalRepresentation
Performance
Analysis
Calculated
Response
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3.1.2 THE ADVANTAGES OF MODELLING AND SIMULATION
Allows predicting the behavior of the system before it is built. This is sometimes called virtual
prototyping
We can analyze the performance of an existing system with the intent of improving its dynamic
behavior.
We can determine what might happen to the system with an unusual input or condition without
exposing the actual system to risky conditions.
Prediction or forecasting are areas closely related to modeling and simulation, as attempts to
predict future events is limited by the accuracy of the model.
The engineer must realize that the model being analyzed is only an approximate mathematical
description or the system and not the physical system itself.
3.2 MODELS FOR DYNAMICS SYSTEMS
Mathematical models are derived from the conservation laws of physics and the engineering properties
of each system component.
The prefered forms are:
i. Configuration form
ii. State-space representation
iii. Input-output equation (single nth order differential equation)
iv. Transfer Function
3.2.1 Configuration form
In compact form, the configuration model can be written as
=(, , ) with initial conditions (0) and (0)A convenient form is the standard second-order matrix form
+ + =Where , , and is the mass, damping and stiffnes matrix respectively, is the vector of thegeneralized coordinates, and is vector of the generalized external forces.
3.2.2 State-space representation
= + State equation = + Output equationWhere
is the state variables vector,
is the output vector, and
is the input vector
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3.2.3 Transfer function
Transfer function is defined as the ratio of the Laplace transform of the output and the input by
assuming that the initial conditions are all zero.
= ()() = 0 + 11+ . . . + + 01+ . . . +1 + 3.3 MODELLING OF MECHANICAL SYSTEMS
Mechanical systems are either in translational or rotational motion or both. Mechanical elements
include inertia, spring, and damperelements for both translational and rotational motion.
3.3.1 MASS & INERTIA ELEMENT
Newtons 2nd Law for planar motion of a rigid body can be written as;
In Linear motion;
= = Alternatively, in Rotational motion
=
...But = + 2 = + 2
= o
=
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3.3.2 SPRING ELEMENT
For a linear spring, its load is directly proportional to its deflection, with the stiffnessconstant, k. For a linear torsional spring, the torque and the deflection angles of twist are
linearly related.
= = (2 1) =
= (2 1)3.3.3 DAMPER ELEMENT
A damperor dashpot, is a device that generates a force in proportion to the difference in the
two end points of the device. The elemental equation for translational and rotational dampers
can be written as;
= = (2 1)
=
= (2 1)3.4 GEARS
Most of machines and physical systems are operate rotational mechanism and thus the development of
so called gear train to transmit energy and movement from one point to another rapidly advances to
cater and provide efficient and effective function ability of a particular system or equipment. The
concept of this gear train is quite simple and highlighted as follows;
2 1
12
12
2 1
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The two gears rotate opposing each other and give relationships of speed ratio, tangential velocity,
tangential force and etc.
3.4.1 EQUIVALENT SYSTEMS FOR A SYSTEM WITH GEARS
3.5 DEGREES OF FREEDOM
The number of degrees of freedom of a dynamic system is defined as the number of independentgeneralized coordinates that specify the configuration of the system.
Generalized coordinates is a set of independent coordinates that completely describes the motion of a
system. For a given system, this set of coordinates is not unique.
1 ,1 2 ,2
2 = 1 2 =2 =2
1 ,1 2 = 1 1 =2 1 1 =11 = 1
2++
1
1 221 = 11 = 22Speed ratio : = 21Tangential Velocity :
Tangential Force : = 11 = 22 12 21
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3.6 DERIVATION OF MATHEMATICAL MODELS FOR MECHANICAL SYSTEMS
EXAMPLE 3.1
Consider the 2 DOF system shown below. Express the governing equations in configuration form.
The Free body Diagram of the system
Applying Newtons 2nd Law
= 11 = 22 1 + 22 1 11 1122 = 22 1 22 1The configuration form is
1 = 11 22 1 + 22 1 11 112 = 12 22 1 + 22 1
2()1()
2
2
1
1
22 1
22 1
11
11
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EXAMPLE 3.2
Suppose the system in Example 3.1 is subjected to an applied force() as shown below. Express thesystems equation of motion in second-order matrix form.
The Freebody Diagram of the system
The expression in second-order matrix form;
1 00 2 12 + 1 + 2 22 2 12 + 1 + 2 22 2 12 = 0()
EXAMPLE 3.3
Obtain the state-space from of the system shown in Example 3.2.
From the configuration form of;
1 = 11 22 1 + 22 1 11 112 = 12 22 1 + 22 1
2211
22 1
22 1
11
11
2()1()
()
()
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Let
State variables Derivative of State variables1 = 1 1 = 1 = 2
2 =
1
2 =
1 =
1
1 2
2
1
+
2
4
3
1
3
1
1
3 = 2 3 = 2 = 44 = 2 4 = 2 12 23 1 + 24 2State equation = +
12
3
4
=
0 (1 + 2)10
22
2
1210
22
2
0 (1 + 2)10
22
2
02211
22
2123
4
+
0001
2()
where
= 1234, = 0 (1+2)10222
1210222
0 (1+2)10 222
02211 222
, = 1234, = 00012 and =()
Output equation
=
+
= 1 00 0
0 01 0
1234, where = 1 00 0 0 01 0, = 1234 = 0
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3.7 MODELING OF ELECTROMECHANICAL SYSTEMS
Electromechanical systems are constructed by combining electrical and mechanical elements. Such
electromechanical systems include potentiometers, galvanometer, microphone and loudspeakers, and
motors and generators. They can be categorized according to the type of electrical element involved in
the coupling: mechanically varying a resistance, moving a current-carrying conductor in a magneticfield, or varying capacitance between plates.
3.7.1 COMMON ELECTRICAL ELEMENTS
=
=
=
= + + 1 = 1
=
+
+
1
= = 1 + + 1 = 1 + 2 + 1 =12 + + 1
Inductance, L (Henry)
Resistance, R (Ohm)
Capacitance, C (Farad)
LRC circuitApplying Kirchoffs Law
The rate of change at terminal is
Taking Laplace Transform
iV
L
iV
R
i
V
C
R L
C i
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The Relationship of Voltage-Current with the major elements of electrical system, i.e. R, L and C can
be summarized as table below
Table 3.1: Voltage-current, voltage-charge, and impedance relationships for capacitors, resistors, and inductors (Nise 4th)
3.7.2 RESISTIVE COUPLING
A variable resistance can be controlled by mechanical motion by continuously moving an
electrical contact. One terminal is fixed and the other terminal, known as wiper, is free to slide
along the bar while maintaining a good electrical contact. The resistance between the two
terminal is
)(txA
R
)(trA
R
= resistivity = cross section areaSince resistors cannot store energy, this method of coupling does not involve mechanical forces
that depend on electrical variables. A very useful device known as a potentiometer is obtained
by adding a third terminal to the other end of the variable resistor. The two end terminals are
normally connected across a voltage source, and the voltage of the wiper is considered the
output. The voltage divider rule gives
- e +
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)()(1
max
tetxx
e io
)(
21
2 teRR
Re io
3.7.3 COUPLING BY A MAGNETIC FIELD
The physical laws governing the movement of current-carrying wires within a magnetic field
state that: (1) a wire in a magnetic field that carries a current will have a force exerted on it,and (2 ) a voltage will induced in a wire that moves relative to the magnetic field.
The differential force on a conductor of differential length L carrying a current, in amagnetic flux density B is
= (x)For straight conductors that are perpendicular to a unidirectional magnetic field, the scalar
relationship is
= where the direction of the force follows the right-hand rule.
= thumbThe voltage induced a conductor of differential length L moving with velocity v in a fieldflux densityB is
= x. = induced voltage
If the three vectors are mutually perpendicular, the scalar relationship is
i
fe
B
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= v
= thumb indicate +ve direction of induced voltageThe force and the induced voltage associated with a wire moving perpendicularly to a
magnetic field can be incorporated in a schematic representation of translational
electromechanical system shown below. The induced voltage is represented by a source in the
electrical circuit, while the magnetically induced force is shown acting on the mass M towhich the conductor is attached.
3.7.4 DC MOTORS
There two common types used in almost all industries, namely direct current(dc) motors and
alternating current (ac) motors. Within the dc motor category there are the armature-
controlled motor and the field-controlled motor.
+
-
em
i
M
v
fe+
-
em
i
M
v
fe
v
B
+
-
em
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EXAMPLE 3.4
The schematic diagram shown in Figure E3.4 represents a DC servo system whereae is the input and
L is the output. The motor drives an inertia load ofJL kg-m
2through a gear train as shown below.
Figure E3.4
The torque generated by the motor is given by
atiKT 1
and the back electro-magnetic force (e.m.f) is related to the rotational velocity by the following
expression
1bb
KV .
where
bK is the electromagnetic force constant of the motor.
tK is the torque constant of the motor
1T is the torque induced by motor
1 is the angular speed of the motor
R is the electric resistance of the motor
a
e is the voltage applied to the motor
ai is the current flowing through the motor
bV is the back e.m.f generated by the motor
aJ is the moment of inertia of the motor
aD is the damping constant of the motor
nN is the number of teeth for gear train (n = 1, 2,. . .)
LJ is the moment of inertia of the rotational system
LD is the damping constant of the rotational system
N1 = 20
N2 = 100
L(t)
+ea
-
Motor
Ja = Ja Kg-m2
Da = Da N-m-s/rad
JL
JL = JL Kg-m2
DL = DL N-m-s/rad
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Ifae and
L are the input and output for the system respectively,
i) Derive the mathematical representation for the system
ii) Obtain the state-space model for the system
Answer
From Differential equations of
baa VRite )( (1.1)
111 )( ee DJtT (1.2)
atiKT 1 (1.3)
1bb KV (1.4)
Substituting equations (1.2, 1.3, and 1.4) into (1.1) yields;
t
bteea
K
KKDJRte 111)(
(1.5)
The relationship between motor and load is given by;
)(1
)/(
)(
)(1
R
KKDJss
RJK
sE
s
bte
e
et
a
(1.6)
The equivalent inertia,m
J at input shaft is
2
2
1
N
NJJJ Lam (1.7)
Similarly the equivalent viscous damping, mD at input shaft
2
2
1
N
NDDD Lam (1.8)
These quantities are substituted into the motor equation, yielding the transfer function of the motor
from the armature voltage to the armature displacement. The gear ratio to arrive at the transfer
function relating load displacement to armature voltage is;
)(
)(
)(
)(
2
1
sE
s
N
N
sE
s
a
m
a
L
(1.9)
)(1
)/(2.0
)(
)(
R
KKDJss
RJK
sE
s
bt
m
m
mt
a
L
(1.10)
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State-space representation
Taking inverse Laplace Transform equation (1.10) and rearrange to
)/(2.0
)(1
)(mt
L
a
bt
m
m
L
aRJK
R
KKDJ
te
(1.11)
L
a
bt
m
m
La
m
t
R
KKDJ
teRJ
K )(
1)(
2.0 (1.11)
Defining state variable as;
Lx 1
Differentiating the statevariables
21 xx L
Lx 2
22 )(12.0
xR
KKDJ
eRJ
Kx
a
bt
m
m
a
m
t
L
Hence the state equation;
a
m
t
a
bt
m
m
e
RJ
Kx
x
R
KKDJx
x
2.00
)(1
0
10
2
1
2
1
Output equation;
2
101x
xy
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EXAMPLE 3.5
For the given electromechanical system shown below, obtain the transfer function.
Figure 3.2: Electromechanical System taken from Control Systems Engineering, Fourth
Edition by Norman S. Nise Copyright 2004 by John Wiley & Sons. All rights reserved.
Solution:
Referring to electromechanical system above, we know that the system is a combination of electrical
and mechanical system. So, derivation will start with DC motor and the output of this will drive the
mechanical system.
The armature voltage,ae is;
)()( tvIRte baaa (1)
Back eletromagnetic field, or is proportional to angular speed,() produced by rotor,Hence,
dt
tdKtv mbb
)()(
(2)
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Meanwhile, the relationship between induced and mechanical system (torque, ) is;)(tiKt atm (3)
Where
is torque motor constant
Rearrange Eqn (3);
t
maK
tti )( (4)
Subtituting Eqn (1) and (4) into Eqn (2);
dt
tdK
K
tRte mb
t
maa
)()(
(5)
The mechanical loading on motor is;
dt
tdD
dt
tdJt mm
mmm
)()(2
2 (6)
For rotational mechanical load, the induced angular speed is transmitted though a gear transmission
ratio where 1 is number of teeth for driver gear which torque generator (rotor) and 2 is number ofteeth for driven gear, generated torque (load).
mJ is the mass moment of inertia for rotational shaft of motor and load, where the relationship is;
2
2
1
N
NJJJ lam (7)
2
2
22.12
1000
100.700.5 mkgmkgmkgJm
Similarly, , is dashpot constant for electrical and mechanical system, where the relationship is;2
2
1
N
NDDD lam (8)
rad
Nms
rad
Nms
rad
NmsDm 10
1000
1008002
2
Hence, eqn (6) become;
dt
td
dt
tdt mmm
)(10
)(12
2
2 (9)
Take Laplace Transform of Eqn (5) and (9);
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)()( ssKK
RTsE mbt
ama (10)
)(1012)( 2 ssssT mm (11)
Substituting Eqn (11) into Eqn (10) yields;
)()(1012)( 2 ssKK
RssssE mbt
ama
)(1012)( ssKsK
RsE mb
t
aa
(12)
Thus, The transfer function of the system;
sKsK
RsE
s
b
t
aa
m
1012
1
)(
)((13)
The Torque-speed curves gives relationship between mechanical torque,m
T and speed,dt
td m )(
From Eqn (5), the following relationship exists when the motor is operating at steady state with a dc
voltage input;
mb
t
ama KKRTe (14)
Solving for yieldsa
a
tm
a
tbm e
R
K
R
KKT (15)
Refering to Torque-speed curve, when is equal to zero, Torque
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a
a
tm eR
KT (16)
And at this condition,
is known as
And when torque is zero, 0mT the speed, m yields
a
a
tm
a
tb eR
K
R
KK 0
a
a
tm
a
tb eR
K
R
KK
b
a
m K
e
(17)
And at this condition, m is known as loadno
Hence, by referring to torque-speed curve,
= =
=
(18)
= = Substituting all parameters in (18) into (13);
sssEs
a
m
210125
1
1
)(
)(
sssEs
a
m
667.1
125
)(
)(
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EXAMPLE 3.6
Find the relationship between output voltage and the angular orientation of the mechanical rotorattached to the wiper.
The schematic diagram:
Ro is load / external and usually very high resistance
Solution:
Let 21 RRRT
)(max
2 tR
R T
(1)
Using voltage divider rule,
)()(1
tERR
Rte
eq
eq
o
(2)
Where,
+
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2
RR
RRR
o
oeq
(3)
Substituting eqn (3) into eqn (2) give;
)()(
1
2
2
2
2
tER
RR
RR
RR
RR
te
o
o
o
o
o
)()(2112
2 tERRRRRR
RRte
oo
oo
)()(21
2 tERRRR
RRte
To
oo
Assuming that To RR
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