chapter 3 mathematical modeling(updated)

Chapter 3: Mathematical Modeling

Outline IntroductionTypes of Models

Theoretical Models Empirical Models Semi-empirical Models

LTI Systems State variables Models Transfer function Models

Block diagram algebraSignal flow graph and Masons gain formula

1

Introduction

A model is a mathematical representation of a physical , biological or information system.

2

Observations

Models ( analyses)

Predictions

The Conceptual WorldThe Real World

Phenomena

Figure 3.1 An elementary depiction of the scientific method that shows how our conceptual Models of the world are related to Observations made within that real world ( Dym & Ivey, 1980)

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Object/System

MODELVariables, Parameters

Model Predictions

Valid, Accepted Predictions

TEST

Use? How will we exercise the model?

Verified? Are the predictions good?

Valid? Are the predictions valid?

Improve? How can we improve the model?

How? How should we look at this model? Given? What do we know?Assume? What can we assume?

Predict? What will our model predict?

Why? What are we looking for?Find? What do we want to know?

Figure 3.2 A first-order view of mathematical modeling that shows how the questions asked in a principled approach to building a model relate to the development of that model( Carson & Cobelli, 2001)

Principles of Mathematical Modeling Introduction

Example showing importance of identifying WHY the MODEL is wanted

Suppose we want to estimate HOW MUCH POWER could be generated by a dam located on Gilgile-Gibe II River.

For a first estimate the avail power? height Estimate of river flow quantities Suppose we want to DESIGN THE ACUTAL DAM, what are the essential parameters?

All of the dams physical characteristics ( e.g., dimensions, materials, foundations etc.)

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1

2

Types of Models

Models can be classified based on how they are obtained.

[A] Theoretical (or White Box) Models Are developed using the physical and chemical laws of

conservation, such as mass balance , component balance,moment balance and energy balance.

Advantages: provide physical insight into process behavior. applicable over wide ranges of operating conditions

Disadvantage(s): Expensive & time-consuming to develop

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Types of Models

[B] Empirical (or Black Box) Models Are obtained by fitting experimental data.Advantages: easier to develop than theoretical models. applicable over wide ranges of operating conditions

Disadvantage(s):Typically dont extrapolate well!Caution!Empirical models should be used with caution foroperating conditions that were not included in theexperimental data used to fit the model

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Types of Models

[C] Semi-empirical (or Gray Box) Models Are a combination of the models in categories (a) &

(b). Used in situation where much physical insight is

available but certain information( parameter) orunderstanding is lacking.

Those unknown parameter(s) in a theoretical modelare calculated from experimental data.

Advantages:They incorporate theoretical knowledgeThey can be extrapolated over a wide range of

operating conditions.Require less development effort

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Theoretical ( White Box) Models

In this chapter we will be dealing models that are generated as a set of linear differential equations

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RResistance

(ohms)

LInductance(henrys)

CCapacitance

(farads)

Table 3.1a Linear Electrical elements

OR

OR

OR


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Table 3.1b Linear Mechanical elements ( Translational)

BViscous damper

(N.s/m)

KLinear Spring ( compliance)

(N/m)

MMass(Kg)

OR

Let

OR


In this chapter we will be dealing models that are generated as a set of linear differential equations

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BViscous Damper

(N.m.s/rad)

JMoment of

Inertia(Kg.m2)

KTortional stifness

(N.m/rad)

Table 3.1c Linear Mechanical elements(Rotational)


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Example 3.1(a) Armature Controlled DC Motor

12

w

va

TL


Example 3.1: Armature Controlled DC Motor the back emf (refer the previous schematic) is given by(3.1a)

Applying KVL to the armature circuit(3.1b)

Because of const. flux, the torque produced at theshaft by the armature current is(3.1c)

Assuming J and B for the motor, and TL load coupledto the shaft of the motor, the equation becomes

(3.1d)

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Example 3.1: Armature Controlled DC Motor By substitution of eqns. (3.1c) & (3.1d), we have

Substituting the above result into the armatureeqn.(and assuming constant TL).

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Example 3.1 :Armature Controlled DC Motor or equivalently,

Exercise 3.1: Field Controlled DC Motor 15

+--

+--

1/Lm kT

BRm

kbDC Motor Block


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Example 3.2: Gear train

gears are used to convert High Speed & SMALL Torqueinto Lower Speed & HighTorque or the converse.Let

- radius ,- Number of teeth ,- angular displacement , and- torque for gear


Example 3.2 : Gears trainKnown

the number of teeth on a gear is linearly proportionalto its radius, i.e.,

the linear distance traveled along the surfaces of bothgears are same, i.e.,

the linear forces developed at the contact point ofboth gears are equal ( Newtons 3rd law), i.e.,

combining (1), (2), & (3) gives

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1

2

3

4

Theoretical ( White Box) ModelsExample 3.3: Armature controlled DC motor driving a load through a gear train.

Let J1 be the total moment of inertia (including rotor,motor shaft, and gear 1) f1 - viscous friction coefficient on the motor shaft J2, & f2- are total moment of inertia & viscous friction

coefficient from load side respectively18


Example 3.3: Armature controlled DC motor driving a load through a gear train... the torque generated by the MOTOR must drive J1,overcome f1 and generate a torque T1 @ gear-1 to drive thesecond gear.Thus we have

(3.2) Torque T1, @ gear-1 generates a torque T2 @ gear-2,which in turn drives J2 and overcomes f2. Hence(3.3)

And using , we get

(3.4)19


Example 3.3: Armature controlled DC motor driving a load through a gear train... Substituting T2 (3.4) into T1 =(N1/N2) T2 and then into(3.2) gives

(3.5)

Where

(3.7) ,

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Example 3.4: Model of Automobile Suspension systemAssumptions: 1/4 model ( one of the four wheels) is used to simplifythe problem to 1D multiple spring-damper system.

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Body MassM1

SuspensionM2

K1

K2

y1

y2

FW


Objective: to study effect of road surface condition on the passenger.

22

M1 M2 FWK2

K1

Figure 3.3 Mechanical Network for Example 3.4

1

2

Developments of Block diagrams for control systems Example 3.5:Consider the control system shown in Fig 3.4. The load could be anantenna and is driven by an armature-controlled dc motor. Thesystem is designed so that the actual angular position of the loadwill follow the reference signal. The error e between the reference rand controlled signal y is detected by a pair of potentiometers withsensitivity k1. Develop the block diagram for this control system.

23Figure 3.4 Antenna position control system

Developments of Block diagrams for control systems

Example 3.5: contd

find the model for the Error detector shown in Fig 3.4

or

find the model of the armature controlled dc motor24


Example 3.5: contdModel of armature controlled DC motorLet J - be the total moment of inertia of the load, the shaftand the rotor of the motor. angular displacement of the load f viscous friction coefficient of the bearingThen, we have for armature controlled DC motor

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1

2

3


Example 3.5: contdModel of armature controlled DC motor

Equating (1), & (4) and taking Laplace Transform on theresulting equation and (3) gives

The elimination of Ia from equation (5), & (6) gives

26

5

6

4

7


Example 3.5: contd

Assumption used while reaching the above block diagram the armature inductance La is assumed to be zero. In this

case, eqn.(7), reduces to

27

r k1e

k2va

-+

LTI Systems

The set of ODEs drived so far are not suitable foranalysis and design, hence rearranged to a moresuitable form, i.e., State Space Model & TF Model

[1] SS-Model:Def: StateVariableA set of characterizing variables which give the totalinformation about the system under study at any timeprovided the initial state & the external input are known

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Set of ODEs

State Space Model Transfer Function Model

LTI Systems

[1] SS-Model

WhereA: system or dynamic matrix,B: input matrix ,C: output matrix,D: direct transfer matrix, 29

LTI Systems

Differential equation SS-ModelExample3.5:Derive the state-space model (i.e. find the A,B,C and Dmatrices) for each of the following differential equations.Take u(t) to be the input and y(t) to be the output.

(1)

(2)

Solution:(1) DefineSo we have the state equations:

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LTI Systems

Differential equation SS-ModelExample3.5

And the output equation:

The state-space model is then:

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LTI Systems

[2]TF-ModelIn general Transfer function is expressed as ( )

Differential equationTF modelExample 3.6:Find the transfer function for the system given inexample 3.5Solution:Taking LT on both sides of the equation gives ( assumingzero initial conditions)

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LTI Systems

Differential equationTF-ModelExample 3.6.

Exercise3.1: Find the TF-model for part (2) in example3.5Ans:

Exercise3.2: Find TF for an automobile suspensiondiscussed (refer pp 21-22) using Matlab SymbolicToolbox

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Block Diagram Algebra (Interconnection Rules)

[1] Series (Cascade) connection:

Note: This is only true if the connection of H2(s) toH1(s) doesnt alter the output of H1(s)-known as the no-loading condition

[2] Parallel Connection

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+

+


[3] Associative Rule:

[4] Commutative Rule:

35

+

+

+

+


[5]The closed-loopTF:[5.a] Unity Feedback:

From the block diagram:and

orRearranging:

36

++--Reference input

error

Controller Plant

Feedback path


[5.b] Feedback with sensor dynamic:

Similarly, in this case:

But now E(s) is the indicated error ( as opposed to theactual error):

So

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++--Reference input

error

Controller Plant

Indicated OutputActual output


[5.2] Feedback with sensor dynamic:Or

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Signal flow graph

is a diagram consisting of nodes that are connected by severaldirected branches and is a graphical representation of a setof linear relationships .

The signal can flow only in the direction of the arrow of thebranch and it is multiplied by a factor indicated along thebranch, which happens to be the coefficient of a modelequation(s).

Terminologies:Node: A node is a point representing a variable or signalBranch: A branch is a directed line segment between twonodes.The transmittance is the gain of a branch.Input node: An input node has only outgoing branches andthis represents an independent variable

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Signal flow graph

Terminologies

Output node: An output node has only incoming branchesrepresenting a dependent variable

Mixed node: A mixed node is a node that has both incomingand outgoing branches

Path: Any continuous unidirectional succession of branchestraversed in the indicated branch direction is called a path.Loop:A loop is a closed path

Loop gain: The loop gain is the product of the branchtransmittances of a loop

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Signal flow graph

TerminologiesNon-touching loops: Loops are non-touching if they do nothave any common node.Forward path: A forward path is a path from an input nodeto an output node along which no node is encounteredmore than once.Feedback path (loop): A path which originates andterminates on the same node along which no node isencountered more than once is called a feedback path.Path gain: The product of the branch gains encountered intraversing the path is called the path gain.

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Signal flow graph

Illustrative example:

Q. Write the equations for the system described by thesignal flow graph above. 42

Mixed nodes

input node

input node

output node

x3x1x2

x4

x3g12 g23

g43

g32

1

Fig. Signal flow graph

Signal flow graph

Properties of Signal flow graphs1. A branch indicates the functional dependence of onevariable on another.2. A node performs summing operation on all the incomingsignals and transmits this sum to all outgoing branches

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G(s)R(s) Y(s) Y(s)G(s)R(s)

R(s)

-H(s)

Y(s)G(s)E(s)1G(s)

H(s)

Y(s)E(s)R(s) ++

--

Fig: Block diagrams and corresponding signal flow graphs

Mansons gain formula

In a control system the transfer function between anyinput and any output may be found by Masons Gainformula. Masons gain formula is given by

Where

where

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45

Mansons gain formulaExample3.7: Find the closed loop transfer functionY(s)/R(s) using gain Mansons formula.

46

R(s) G1(s) G2(s) G3(s) x1(s)x3(s)x4(s)

-H2(s) -H1(s)

G4(s)

Y(s)

-1


Example 3.7Here we have two forward paths with gains

,And five individual loops with gains

Note for this example there are no non-touching loops,so for this graph is

47


Example 3.7

The value of 1is computed in the same way as byremoving the loops that touch 1st forward path M1 In this example, since path M1 touches all the five loops,1 is found as Proceeding the same way , we findTherefore, the closed loop transfer function betweenthe input R(s) and outputY(s) is given by,

48


Example 3.8Find Y(s)/R(s) for the system represented by the signalflow graph shown below.

49

G4X2X3

G2 G5 X1

G3

G1X4

-H1

-H2

G6

G7R(s) Y(s)


Example 3.8Observe from the signal flow graph , there are threeforward paths between R(s) andY(s)The respective forward path gains are:

There are four individual loops with gains:

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Example 3.8Since the loops L2 & L4 are the only non-touching loopsin the graph, the determinant will be given by:

Computing 1,which is computed by removing the loopsthat touch fist forward path M1

1=1Similarly , 2=1 andThus, the closed-loop TF is given byY(s)/R(s)

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chapter 3 mathematical modeling(updated)

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