Chapter 2-3.doc

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Mathematical Modeling of Systems

Concept

Given a physical system, find a model (mathematical

representation) that accurately predicts the behavior (the output

for a given input) of the system.

Key Points

To analyze and design control systems, we need quantitative

mathematical models of systems. The dynamic behavior of

systems can be described using differential equations.

If the equations can be linearized, then the aplace transform

can be used to simplify the solution.The input!output relationship for linear components and sub"

systems can be described in the form of transfer functions.

System Representation and Modeling

In order to analyze (and subsequently control) most dynamic

systems, it is essential to attain a reasonable understanding of

how a system functions. To achieve this ob#ective, we formulate

mathematical models that help us describe the behavior ofsystems.

$athematical models generally serve two purposes%

a. They are used in con#unction with analytical techniques to

develop control schemes for the systems that they represent.

b. They are used as a design tool in computer simulation studies.

In this conte&t, the model is used as the control ob#ect to test and

on which to evaluate possible control schemes. This procedure

is generally more efficient, cheaper and less time consumingthan to test the control schemes on the actual system.

'ere we will lay the groundwor for the formulation of

mathematical models for energetic (dynamic) systems. f

paramount importance in our methodology are the concept of

state of the system and the selection of a set of state variables

that describe the state of the system and adequately serve our

modeling ob#ectives.

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State Variables and State Equations

-onsider a single input, single output (I) dynamic system,

as shown schematically in figure *. ince the system is dynamic,

the output y(t) at time t/ is a function of the past as well as the

present inputs% u0/,t1. 2otice that u0/,t1 represents the entire

history of the input variable u( ) from 34 / to 34 t. It is

obviously a problem to eep trac of the entire system history in

order to be able to predict the future system output. Thus, it is

important to determine what the minimum amount of

information needed, at time t, to predict the output of the system

immediately after time t.

Figure 1 S!S" dynamic system

The state variables are the minimum set of variables such thatnowledge of these variables at time , together with the current

(present) state of the system, is sufficient to determine the future

state and output of a system.

The state variables represent the minimum amount of

information that needs to be retained at any time t in order to

determine the future behavior of a system. 5lthough the number

of state variables is unique (that is, it has to be the minimum and

necessary number of variables), for a given system, the choiceof state variables is not. -hoosing state variables and deriving

models is in general a non"trivial problem. In many cases, there

is no universal or general procedure. 6or engineering energetic

systems, the choice of state variables still depends on the

problem. 'owever, there e&ists a fairly systematic procedure for

selecting state variables and deriving models.

The order of the system designates the number n of state

variables needed (minimum and necessary) to describe thesystem.

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The set of n differential or difference first"order equations that

govern the relationship between the input to a system and the n

state variables are the state equations. Together with the output

equation, they constitute the model of the system.

8enoting the n state variables of a system as &*, &9, :, &nand

defining dtdxx !=

the state equations can be written as%

))(),(),...,(),(()(

))(),(),...,(),(()(

))(),(),...,(),(()(

9*

9*99

9***

tutxtxtxftx

tutxtxtxftx

tutxtxtxftx

nnn

n

n

=

=

=

2ote that the right" hand side of the state equations is only a

function of the state variables and the input at time t. 8efine the

nthorder vectors%

=

nx

x

x

...

9

*

# and

=

nf

f

f

...

9

*

f

;e can write the system of nstate equations in vector form as%

))(),(()( tutt ## f=

;hen the dynamic system is linear, we can write the above

equation as follows%

)()()()()( tutttt b#$# +=

where $(t) is a nby nmatri& and b(t) is an n by one vector.

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=

)(...)(

......

)(...)(

)(

*

***

tata

tata

t

nnn

n

$ and

=

)(

...

)(

)(

)(9

*

tb

tb

tb

t

n

b

If $and bare constants, then the system is an nthorder inearTime Invariant (TI) system.

)()()( tutt b$## +=

The output equation is given by the algebraic equation%

))(),(()( tuthty #=

;hen the system is TI, the output equation can be written as%

)()()( tdutty += c#

where [ ]nccc ...9*=c is a one by nvector and d is a scalar.

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Mass Spring Damper Example

-onsider the mass"spring"damper e&ample shown in figure 9,

wherexis the position of the mass,

=xv is the velocity of the

mass, u is the input force applied to the mass, is the coefficient

of elasticity of the spring, and b is the viscous damping

coefficient of the damper.

Figure % Mass&Spring&'amper System

5ssuming that the system is linear and time invariant, a model

of the system can be

derived as follows%

a. 8efine the state of the system to be the position and the

velocity of the mass,

[ ]Tvx=# or

=

v

x# .

b. >se 2ewton?s laws to derive the equations of motion for thesystem%

)(tuffvm vk ++=

, wherefkandfvare the forces e&erted by the

spring and damper respectively.

c. >se the constitutive relations for the spring and damper,bvfkxf vk == and

d. -ombine these equations and write them in matri& form%

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)(!*

/

)(

)(

!!

*/

)(

)(tu

mtv

tx

mbmktv

tx

dt

d

+

=

[ ]

=

)(

)(/*)(

tv

txty

2ote that this is not the only state equation model that could

have been used to describe the system.

Modeling of dynamic systems( )eneral approach

5 dynamic (or energetic) system is a collection of energy

storage elements, power dissipative elements, power sources,

transformers and transducers. 6or successful modeling ofenergetic systems, it is important to now the characteristics of

each of these elements.

In this class, we will model energetic systems as lumped

parameter systems. That is, we will model the system such that

each point in the system embodies the properties of the region

immediately surrounding it. umped parameter systems are

described by a finite number of state variables.

Example of a Lumped System: Mass-Spring System

Figure * Mass&spring system

/=+

kxxm

The properties of the surrounding region are lumped(concentrated) at each point. The state equations describing

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lumped parameter systems can be written as finite order

ordinary differential equations.

$any energetic systems cannot be modeled as lumped

parameter systems. These systems must be modeled as

distributed parameter systems.

Example of a Distributed System: The Cantilever Beam

Figure + Cantile,er beam

/),(*),(9

9

99

9=

ttxu

axtxu

@ach little AchunB of mass acts as an elastic segment. The

properties of mass and elasticity cannot be separated from each

other. The state equations that describe distributed parameter

systems are partial differential equations.

In the ne&t pages, we will present a unified approach for

modeling mechanical, electrical, fluid and!or thermal lumpedparameter systems. The first step in modeling energetic systems

is to brea the system up into elements, then find the basic

relations that describe the individual elements that form the

system. Chysical laws are generally used to obtain these

relations.

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Mehanial systems: !e"ton#s La"s

!e"ton$s %irst La" of Motion &La" of 'nertia(:Every body

continues in its state of rest or of uniform speed in a straight

line unless it is compelled to change that state by forces acting

on it.

!e"ton$s Seond La" of Motion: The acceleration of an object

is directly proportional to the net force acting on it and is

inversely proportional to its mass.

The direction of the acceleration is in the direction of the

applied net force.

!e"ton$s Third La" of Motion &La" of )tion-*eation(:

Whenever one object exerts a force on a second object, the

second exerts an equal and opposite force on the first.

Mehanial Systems: Momentum

The La" of the Conservation of Momentum: The total

momentum of an isolated system of bodies remains constant.

Mehanial Systems: Momentum

The La" of the Conservation of Momentum: The total

momentum of an isolated system of bodies remains constant.

Eletri Systems: +irhhoff#s *ules

33,oint *ule: The algebraic sum of the currents toward any

branch point is zero./=I

= /IE

33Loop *ule: The algebraic sum of the potential differences in

any loop, including those associated with @$6s and those of

resistive elements, must equal zero.

Energeti )pproah to System Modeling

To describe the elements that comprise energetic systems, we

use a pair of variables. In our unified nomenclature, we willrefer to one of the variables as the effort (or level) variable, e,

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and to the other as the flow (or rate) variable, f. The product of

these two variables represents the instantaneous power being

transmitted to the element.fe! &=

The fundamental linear ideal elements are given in the followingtable. D"elements dissipate power, while I"elements and -"

elements store energy.

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Energy Storage

$echanical systems store energy in two forms, inetic and

potential.

pring%9

9kxE=

$ass%9

9mvE=

@lectrical systems store energy as either capacitive or inductive.

Inductor%9

9"iE=

-apacitor%9

9#vE=

Energy Dissipation

8amper% =t

dttbvE/

9 )(

Desistor% dt

tvE

t

=/

9 )(

2ote that energy storage elements have integral constitutive

relation while dissipative elements have a static constitutive

relation.

Causality &'nputs and utputs(

;e did not specify in the fundamental equations which of a pair

of variables (e and f) is the element input or output. et usconsider the causality of the D"elements, I"elements and -"

elements.

6or D"elements, since the fundamental relation between e and f

is static, the causality is reversibleF meaning either e or f can be

the input or output.

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6or I"elements and -"elements, to avoid unrealistic situations,

the integral (or natural) causality must be used, as e&plained

below.

6or I"elements%

iindicates the static dependence offuponp.

6or -"elements%

2ote% the differential causality for I"elements and -"elements isunrealizable, as the output in the differential causality would

depend on future inputs.

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Gear pump% flow source

'deal transformers and transduers

6or transformers and transducers, we need two pairs of effort (e)

and flow (f) variables.

6or each pair, the power is defined as the product of e and f.

Cower is conserved for ideal transformers and transducers.

-ausality determination depends on other elements in the

system.

Ideal Transformer (two ApowersB are in the same media, e.g.

mechanical)

That is, 99** $'$' = 3(power is conserved).

Ideal Transducer (two ApowersB are in different media, e.g.

mechanical and fluid)

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Example of t&o dependent energy storage elements%

tate variables are dependent if there is an algebraic relationrelating them, and must be combined. The above system is

equivalent to%

Derivation of the state e/uations

The number of state variables sufficient to describe the basic

motion of an energetic system is equal to the number of

independent energy storage elements. ome e&tra state variables

might be needed if we are interested in a particular aspect of the

system behavior. 6or e&ample, if we have a system formed of amass, and we are interested on the position of the mass and its

velocity, it is convenient to select two state variables for the

mass% position and velocity.

The procedure to derive state equations for an energetic system

is as follows%

a. Identify independent energy storage elements and select state

variables.

b. ;rite the fundamental relation (dynamic or static) for eachelement in the system.

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c. 5ssemble the relations obtained in step (b) in terms of state

variables and input variables. In this step you will need to select

sign conventions (positive or negative spring force, for

e&ample), tae into account geometric considerations, and use

physical laws such as continuity, force balance etc: 5lso,

mae sure that there is no conflict in causal alignment between

connected elements. 6or e&ample, a parallel combination of a

voltage and capacitor is a bad model

Example 0: Mehanial system

a. Identify energy storage elements and select state variables% the

mass and the spring, with associated state variables Jmand 6b. ;rite the fundamental relation for each element%

i. $ass% mm $m

'dt

d *= (*)

ii. pring% kk k'$dt

d= (9)

iii. 8amper% bb b'$ =

(+)c. Tae$and$b to be positive when the spring and damper are

in compression.

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>se 2ewton?s third law%

bk $$$m = (=)

6rom geometric considerations%

mink ''' = (E)

and

bm '' = ()

>sing the fundamental relations (*"+) and substituting in (="),

we obtain%

[ ]mkm b'$m

'dt

d=

*()

[ ]mink ''k$dt

d= (7)

@quations () and (7) are the state equations. ;ritten in matri&

form, we have%

in

m

k

m

k'

k

'

$

mbm

k

'

$

dt

d

/

!!*

/

+

=

*E=


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If we are interested in 'mas the output%

[ ]

=

m

k

out'

$y */ is the output equation.

Example 1: Eletrial system


inductance and the capacitor, with associated state variables Ii

and #v

b. ;rite the fundamental relation for each element%

Inductor% II v

Idt

di *= (*)

-apacitor% ## i

#dt

dv *= (9)

Desistor% iv = (+)

c. >se Kirchhoff?s current law at node * (continuity equation)%

#I iii += (=)

>se Kirchhoff?s voltage law%

I#in vvv += (E)# vv = ()

*EE


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6rom (*) and (E),

[ ]#in

I vvIdt

di=

*

r

in#I v

Iv

Idt

di **+= ()

6rom (9), (+), (=) and (),

[ ] #I#

II# v

#i

#

vi

#ii

#dt

dv ****=

== (7)

() and (7) are the state equations. In matri& form,

in

i

#

I

#

Iv

I

v

i

##

Ii

v

i

dt

d

/

!*

!*!*

!*/

+

=

If we are interested in v-as the output%

[ ]

= #

I

outv

i

y */ is the output equation.

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Example 2: %luid System

The pipe is very long so that fluid inertance has a significant

effect.


tan and the fluid inertance, with associated state variables C-

and LI.

b. ;rite the fundamental relation for each element%

Tan% ## (

)

g

dt

d! = (*)

Cipe% II !

"

a

dt

d(

= (9)

rifice% (! = (+)

c. >se the continuity law%

Iin# ((( = (=)

I (( =

(E)

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6rom equations (*) and (=),

[ ]inIIin

# ()

g(

)

g((

)

g

dt

d! +== ()

6rom equations (9), (+), (E) and (),

[ ] I#I#I (

"

a!

"

a(!

"

a

dt

d(

== (7)

() and (7) are the state equations. In matri& form,

in

I

## ()g

(

!

"a"a

)g

(I

!

dtd

/

!

!!

!/

+

=

If (outis the system output, the output equation is%

[ ]

=

I

#

out(

!t( */)(

Example 3: DC Motor 4 Transduer

The figure below represents an armature controlled 8- motor

with a fi&ed field being used to drive a rotational inertial load.

The motor armature circuit is being driven by a voltage source,

v. The torque generated by the motor is proportional to the

armature current%

*iT=

(*)

The bac electromotive force voltage, va, is proportional to the

motor speed%

ma *v = (9)

@quations (*) and (9) imply that ma Tiv = 33that is

electromechanical power conservation. They are the ideal

transducer equation (the motor is a real transducer).

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;e derive the state equations assuming that all elements in the

system are linear and ideal.


inductor and the inertia, with associated state variables Ii and3

+

.b. ;rite the fundamental relation for each element%

$otor%aa *iT = (*)aa *v = (9)

Inductor% II v

Idt

di *= (+)

Desistor% iv = (=)

Inertia% ++ T

+dt

d *=

(E)

Hearing% ,, ,T = ()

c. 5pply Kirchhoff?s voltage law%

ainI vvvv = ()

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5pply Kirchhoff?s current law%

aI iii == (7)

6rom the torque balance%

,a+ TTT = (

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