3.Mathematical Modelling

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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

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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