Lecture-Slides CHAPTER 02

ME464-Sys Dyn & Ctrl Spring-2013 Dr. Shaukat Ali


Mathematical Modelsof

Systems

Chapter 2


Introduction

Differential Equations of Physical Systems

Linear Approximation of Physical Systems

The Laplace Transform

The Transfer Function of Linear Systems

Block Diagram Models

Signal-Flow Graphs Models

Outline


To understand and control complex physical systems, we need their mathematical models.

To obtain mathematical models, we need the relationship between the system variables.

As the systems under consideration are dynamic in nature, then this relationship is in the form of

differential equations.

Introduction


In general, the linear differential equation of an nth-order system is written:

1st order linear ordinary differential equation:

2nd order linear ordinary differential equations:

In this course we treat only LINEAR ORDINARY DIFFERNTIAL EQUATIONS

Differential Equations

)()()()()(

11

1

1tftya

dt

tdya

dt

tyda

dt

tydon

n

nn

n

)()()(

tftyadt

tdyo

)()()()(

12

2

tftyadt

tdya

dt

tydo


Methods of Modeling Linear System:

Transfer Function Method (only linear systems)

State-Variable Method (both linear and nonlinear systems)

Most dynamic systems have nonlinear behavior:

Linearization by proper assumptions and approximations

Modeling of Physical Systems


The motion of a Mechanical system:

Translation

Rotation

Combination of above

Modeling of Mechanical Systems


Translational motion:

Newtons 2nd law of motion

Example: A mass M under the action of force f(t).


M

)(ty

)(tf

dt

tdvM

dt

tydMtMatf

)()()()(

2

2

MaFext


Linear Spring:

Hooks law:

Viscous Damper:


)()( tKytf

Bvdt

tdyBtf

)()(


Mass-spring-damper system


2

2)()(

)()(dt

tydM

dt

tdyBtKytf


Rotational motion

Eulers 2nd law of motion

Example: A body with inertia J under the action of a torque (t).


Jext

dt

tdJ

dt

tdJtJt

)()()()(

2

2


Torsional Spring:

Viscous damper:


)()( tKt

Bdt

tdBt

)()(


A disk in a viscous medium and supported by a shaft


)()()()( tJtttds

2

2)()(

)()(dt

tdJ

dt

tdBtKt

Jext


Resistor:

Inductor:

Capacitor:

Modeling of Electrical Systems

R

L

C

)(tv

)(tv

)(tv

)(tI

)(tI

)(tI

R

tvtI

)()(

dttvL

tI )(1

)(

dt

tdvCtI

)()(


Kirchhoff s laws:

Current law:

The algebraic sum of all currents entering a node is zero.

Voltage law:

The algebraic sum of all voltage drops around a complete closed loop is zero.

Example of RLC circuit:

Modeling of Electrical Systems

dttvLdt

tdvC

R

tvtr )(

1)()()(


Spring-Mass-Damper system:

RLC circuit:

Rotational motion:

This is called as velocity voltage analogy (force-current analogy)

Analogy

dt

tdJtBdttKt

)()()()(

dt

tdvMtBvdttvKtf

)()()()(

dt

tdvC

R

tvdttv

Ltr

)()()(

1)(


Linearization(Linear Approximation)


A linear system satisfies the following properties:

Superposition

Homogeneity

Example:

Test whether is linear.

Linearization

Linearsystem

)()(2211tuatua )()(

2211tyatya

5xf(x)


Examples of physical systems

Linearization

)()()()(

2

2

tutydt

tydLC

dt

tdyRC

)()()(

2

2

tudt

tdyB

dt

tydM

dt

tdyB

)(

M

)())(()()(

2

2

tutyfdt

tdyB

dt

tydM

B


Examples of physical systems

A nonlinear system can be described by a linear model for a small range of input values around an operating

point.

Linearization

)())(()()(

2

2

tutyfdt

tdyB

dt

tydM


To find the linear model of a nonlinear system f(y)

We expand f(y) into a Taylor series around the operating point or equilibrium point (yo, f(yo)):

If the variation around the operating point, is small, then we may neglect the higher-order terms:

This approximation results in a linear (straight line) relationship

Linearization

oo yy

o

yy

o

ody

fdyy

dy

dfyyyfyf

2

22

!2!1)()(

oyyy

ycyf )(


Linearization of differential equations

Example: Pendulum oscillator model

Linearization around the equilibrium point

This approximation is reasonably accurate for

Linearization

sin)( MgLT

)()()(oo

od

dTTT

o

o0

44

MgLT )(


Laplace Transform (LT)


LT is a mathematical tool that:

Transforms many time (t) domain functions f(t) into algebraic functions F(s) of a complex domain (s).

Provides an algebraic way to solve linear time invariant differential equations.

Can be used to predict the system performance without actually solving system differential equations.



Solution of differential equations

Obtain linearized differential equation.

Obtain the Laplace transform of the differential equation.

Solve the algebraic equation by the inverse Laplace transform.



Laplace Transform of a function f(t):

is the Laplace transform operator

(s) is a complex variable:

f(t) is a function of time (t) with f(t)=0 for t


Theorem 1: Multiplication by a constant

Theorem 2: Sum and differences

Theorem 3: Differentiation

Theorems of Laplace Transform

)()( skFtkf

)()()()(2121sFsFtftf

)0()()(

fssFdt

tdf

0

2

2

2)(

)0()()(

tdt

tdfsfsFs

dt

tfd


Step function:

Unit step function:

LTs of Simple Functions

0

00)(

tc

ttf

s

ce

s

cdtcedtetfsF

st

t

st

t

st

000

)(

s

csF )(

ssF

1)(


Ramp function:

Exponential function:

Sinusoidal function:

LTs of Simple Functions

0

00)(

tct

ttf 2)(

s

csF

0

00)(

te

ttf

at assF

1)(

22)(

s

sF

0sin

00)(

tt

ttf


Table of LTs


First order linear differential equation

Let

Now, as:

LT of equation (1) is:

LTs of Differential Equations

oo

ayytf ,)0(,0)(

)1()()()(

tftyadt

tdyo

)0()()(

)()(

yssYdt

tdy

sYty

s

ysY

o


Second order linear differential equation (Spring-Mass-Damper)

Let

LT of equation (1) is:

LTs of Differential Equations

)1()()()()(

2

2

tftKydt

tdyB

dt

tydM

0)(

,)0(,0)(

0

t

odt

tdyyytf

KBsMs

yBMssY

o

2


The inverse LT of F(s) is:

This is a complex integral and is rarely used.

For simple functions, we directly refer to the LTs table.

For complex functions, we first perform the partial-fraction expansion on F(s) and then use the LTs table.

Inverse Laplace Transform

j

j

stdsesF

jtfsF

2

11


Consider the following Laplace Transform function:

N(s) and D(s) are the polynomials of (s).

Characteristic equation:

Roots (s1, s2, sn) of this characteristic equation are called the poles of the system.

Distinct poles

Repeated poles

Partial-Fraction Expansion

sD

sNsG

o

n

n

n

n

nasasasassD

1

1

2

2

1

1

KBsMs

yBMssY

o

2

0sD


Case 1: Distinct poles

Consider the function

Write G(s) in terms of partial-fraction expansion:

Determine the coefficient k1 and k2


31

2

ss

s

sD

sNsG

2

11

1

1

ssD

sNsk

31

21

s

k

s

ksG

2

13

3

2

ssD

sNsk


The simplified function is:

Now taking the inverse LT:


32

1

12

1

sssG

tt eetg 35.05.0


Case 2: Repeated poles

Consider the function

Write G(s) in terms of partial-fraction expansion:

Determine the coefficient k1 and k2


21

2

s

s

sD

sNsG

11

1

2

2

ssD

sNsk

2

21

11

s

k

s

ksG

11

1

2

1

ssD

sNs

ds

dk


The simplified function is:

Now taking the inverse LT:


21

1

1

1

sssG

tt teetg


Solve the following 2nd order linear ODE

With us(t) as unit step function and following initial conditions:

Example

tuty

dt

tdy

dt

tyds

5232

2

2,10

0

tdt

tdyy


Transfer Function


The ratio of the Laplace Transform of the output variable to the Laplace Transform of the input variable, with all initial

conditions to be zero.

Consider the spring mass damper system: input is r(t), output is y(t).

Transfer Function

sRsKYsBsYsYMs 2

trtKy

dt

tdyB

dt

tydM

2

2

KBsMssR

sYsG

2

1

sInput

sOutputsG

tr


Write the transfer function of the following circuit, where:

Input: source voltage v1 Output: voltage drop across capacitor v2

Transfer Function


DC Motor

Converts DC electrical energy into rotational mechanical energy

Transfer Function DC Motor


DC Motor

Input: voltage (field, armature)

Output: speed of shaft, position of the shaft


current armature

current field

ntdisplacemerotor

voltagefield

voltagearmature

ti

ti

t

tv

tv

a

f

f

a

sV

ssG


Two types of control of dc motor

Field control

(variable field voltage and fixed armature voltage)

Armature control

(variable armature voltage and fixed field voltage)



Field control of dc motor

The angular displacement is proportional to the field voltage


voltagefield

ntdisplacemeangular

sV

ssG

f

sVsGsf

sVRsLBJss

Ks

f

ff

m


The air gap-flux is proportional to the field current

The motor torque Tm is assumed to be related linearly to and the armature current:

In case of field control, armature current is kept constant:

Km is the motor constant


tiKtff

titKtTam

1

titiKKtTaffm 1

tiKtTfmm

sIKsTfmm


Relating armature voltage to armature current:

In s-domain

Thus the motor torque is:


dt

tdiLtiRtv

f

ffff

sVsLR

sIf

ff

f

1

sVsLR

KsT

f

ff

m

m


The load torque in time domain:

In s-domain:

Now:


tBdt

tdJtT

L

sBJsssTL

ff

m

fRsLBJss

K

sV

ssG

sTsTsTdLm

dt

tdt


Transfer function

Block diagram


sLRsVsI

fff

f

1

m

f

mK

sI

sT

BJs

s

sTL

ss

s 1

sTsTsTdmL


Armature control of dc motor

The angular displacement is proportional to the armature voltage


voltageArmature

ntdisplacemeangular

sV

ssG

a

sVsGsa

sVKKBJssLRs

Ks

a

mbaa

m


The air gap-flux is proportional to the field current

The motor torque Tm is assumed to be related linearly to and the armature current:

In case of armature control, field current is kept constant:

Km is the motor constant


tiKtff

titKtTam

1

titiKKtTaffm 1

tiKtTamm

sIKsTamm


Relating armature voltage to armature current:

Where vb is the back emf voltage:

In s-domain

Thus the motor torque is:


tvdt

tdiLtiRtv

b

a

aaaa

sLR

sKsVsI

aa

ba

a

sKsVsLR

KsT

ba

aa

m

m

tKtvbb


Transfer function of armature controlled dc motor


mbaa

m

aKKBJssLRs

K

sV

ssG


Gear ratio:

Relate shaft torques:

Transfer Function Gear Trains

2

1

N

Nn

2

1

N

N

T

T

L

m


Tachometer

An electromechanical device that converts mechanical energy into electrical energy.

Input: shaft angular velocity

Output: voltage

Transfer Function Tachometer

t

s

sVsG K

2


Consider an incompressible fluid in a tank:

Determine the transfer function which relates head to inflow

Mass balance:

mass flow in mass flow out = accumulation rate of mass in tank

Transfer Function Fluid System

inletat rate flow volumetric:

in tank fluid of head:)(

outletat rate flow volumetric:)(q

fluid ofdensity :

area sectional-cross uniform:

o

iq

th

t

A

inflow

head

sQ

sHsG

i


Consider an incompressible fluid in a tank:

Determine the transfer function which relates head to inflow

Energy balance:energy in energy out = accumulation of energy in tank

Transfer Function Thermal System

peratueoutlet tem:

eratueinlet temp:

sourceheat fromheat :

inlet andoutlet at rate flow volumetric:

heat specific:

fluid ofdensity :

area sectional-cross uniform:

o

i

T

T

q

C

A

eTemperaturInlet

eTemperaturOutlet

sT

sTsG

i

o


Block Diagram Models (BDM)


So far:

Dynamic systems are represented by mathematical models:

Set of simultaneous differential equations in time domain.

Set of linear algebraic equations in the s-domain.

Transfer function:

Mathematically relating the output variable to the input variable in the s-domain.

Block Diagram Model (BDM)

Graphical technique for modeling control systems.

Graphical relationship between the variables of interest.

Introduction

s

ssG

Input

Output sG sInput sOutput


Block Diagram Model Usage

BDM provides a better understanding of the composition and interconnection of the components of a system.

BDM describes the input-output relationship throughout the system with the help of transfer functions.

Introduction

ControllerProcess

orPlant

Feedback

ActuatorRef.Input

ActualOutput

Measured output

Error

_

+ Actuatingsignal


Linear spring

Transfer function:

Block diagram model:

Introduction

K

sF

sXsG

K sF sX


Field control DC motor

Transfer function:

Block diagram model:

Introduction

ff

m

fRsLBJss

K

sV

ssG

sVf

s

ff

m

RsLBJss

K


Elements of BDM

Blocks

The rectangular box that contains a component of a system.

Signals

Arrowed lines from one block to another representing input/output variables.

Comparators (summing point)

Junction point for signals comparison.

Block Diagram Model

ControllerProcess

orPlant

Feedback

ActuatorRef.Input

ActualOutput

Measured output

Error

_

+ Actuatingsignal


Comparator (summing point)

To perform simple mathematical operations (addition or subtraction)

Block Diagram Algebra

+

+ sR

sY

sYsRsE

_

+ sR

sY

sYsRsE


Block

To represent the transfer function of a component of a system or the system as a whole.

Transfer function


sG sU sY

sU

sYsG

sUsGsY


Combining blocks in cascade


sXsGsGsX1213


Combining blocks in Parallel


sXsGsGsX1212

sX1

sX2

sGsG21

sX1

sX2


Feedback control system

Transfer function:

Forward-path TF:

Loop TF:


sHsG

sG

sR

sY

1

sHsGsL

sGsGsGsGpac

sGa

sGp sG c

sH


Eliminating a feedback loop

Unity feedback loop


G1

G

1X

2X

1X

2XG


Moving a summing point to the right of a block


213GXGXX 213 XXGX


Moving a summing point to the left of a block


213XGXX

G

XX

G

X2

1

3


Moving a takeoff point (pickoff point)to the left of a block


12GXX


Moving a takeoff point (pickoff point)to the right of a block


G

XX

2

1


Example: reduce the following block diagram and determine the transfer function

Block Diagram Reduction


Example: reduce the following block diagram and determine the transfer function



Multiple Inputs

1. Set all inputs except one equal to zero

2. Determine the output signal due to this one non-zero input

3. Repeat the above steps for each of the remaining inputs in turn

4. The total output of the system is the algebraic sum (superposition) of the outputs due to each of the inputs.



Signal-Flow Graphs (SFG)


Signal-flow graphs (SFG)

A graphical representation of control systems (a simplified version of Block diagram model)

The cause-and-effect relationship among the variables of a set of linear algebraic equations (like we have in case of linear

control systems)

A diagram consisting of nodes that are connected by several directed branches.

Introduction

sVsGsf


Node (junction point):

To represent the variables of the system

Branch (line segment):

To connect the nodes according to the cause-and-effect equations

Branch is a unidirectional line segment (from input toward the output)

Signal-Flow Graph Basic Elements

sVf

s sG

sVsGsf


Basic properties

SFG applies only to linear systems

The equations must be in algebraic form (in s-domain) in the form of cause-and-effect relationship.

Example:

For N equations;

Signal-Flow Graph Basic Properties

sYsGsYsGsY

sYsGsYsGsY

sYsGsYsGsY

3342244

4432233

3321122

NjsYsGsYkkj

N

k

j1,

1


Input node (Source)

A node that has only outgoing branches (example: node Y1)

Output node (Sink)

A node that has only incoming branches (example: Y4)

Path

A branch or a continuous sequence of branches that can be traversed from one node to another node

Forward path

A path that starts at an input node and ends at an output node with no node traversed more than once (example: Y1 to Y2 to Y3 )

Loop

A path that originates and terminates on the same node with no other node traversed more than once. (four loops in example)

Signal-Flow Graph Terms


Path gain The product of the branch gains encountered in traversing a

path.

For example, the path gain for the path Y1-Y2-Y3-Y4 is G12G23G34

Forward-path gain The path gain of a forward path

Loop gain The path gain of a loop

Non-Touching loops Two loops are non-touching if they do not have a common

node



2 Forward paths:

4 Loops:

Non-touching loops:

L1 and L3, L1 and L4, L2 and L3, L2 and L4


43211GGGGP

87652GGGGP

332HGL

221HGL

774HGL

663HGL


Series connection of branches

Parallel branches

Feedback control system

Signal Flow Graphs Algebra


Gain Formula

The linear dependence between input variable and output variable

Pk = Gain of the kth path from input variable to output variable

= Determinant of the SFG

k = Cofactor of the path Pk

N = the total number of forward paths between input and

output variable

Signal Flow Graphs Gain Formula

NN

N

k

kkPPP

P

T

22111

input

output


= 1 (sum of all loop gains)

+ (sum of the gain products of all combination of two non-touching loops)

(sum of the gain products of all combination of three non-touching loops)

+

k = with all the loops touching the kth forward path put to zero

Signal Flow Graphs Gain Formula

nontoching

pmn

pmn

nontoching

mn

mn

n

nLLLLLL

,,,

1


Find the system transfer function by using the gain formula of SFG

Signal Flow Graphs Example


Field-controlled motor

Draw the signal-flow graph for the above block diagram

Signal Flow Graph DC motor control


Armature-controlled motor

Signal Flow Graph DC motor control

sD

sD


Determine the transfer function of the system from the following signal flow graph, by using Masons Gain

formula.

Also draw the equivalent block diagram

Signal Flow Graphs Exercises (E2.22)

NN

N

k

kkPPP

P

sR

sYsT

22111


Determine the following transfer function:

Determine a relationship for the system that will make Y2(s)independent of R1(s)

Draw the equivalent block diagram

Signal Flow Graphs Problems (P2.33)

sR

sYsT

1

2

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Lecture-Slides CHAPTER 02