Modeling – Part of Control Design Process Fall ME451 - GGZ Week 3-4: Modeling Physical Systems...

2009 Fall ME451 - GGZ Week 3-4: Modeling Physical Systems

Derive mathematical models for

• Electrical systems

• Mechanical systems

• Electromechanical system

Modeling Modeling –– Part of Control Design ProcessPart of Control Design Process

1. Modeling

Mathematical modelMathematical model

2. Analysis

ControllerController

3. Design/Synthesis

4. Implementation

PlantPlantOutputOutput

ActuatorActuatorControllerController

SensorSensor

InputInput

Electrical Systems:

• Kirchhoff’s voltage & current laws

Mechanical systems:

• Newton’s laws


Modeling Physical Systems Modeling Physical Systems -- OverviewOverview

• Mathematical model is used to represent the input-output

(signal) relation of a physical system

• A model is used for the analysis, design and

simulation validation of control systems.

Physical Physical

systemsystem

ModelModel

ModelingModeling

InputInput OutputOutput

Transfer function obtained Transfer function obtained

via physical lawsvia physical laws


Modeling Modeling –– RemarksRemarks

• Modeling is the most important and difficult task in

control system design.

• No mathematical model exactly represents a physical

system.

• Do not confuse models with physical systems!

• In this course, we may use the term “system” or “plant”

to mean a mathematical model.

System Physical ModelMath

System Physical ModelMath

≈

≠


Modeling Modeling –– Transfer FunctionTransfer Function

• A transfer function is defined by

• A system is assumed to be at rest. (Zero initial condition)

Laplace transform Laplace transform

of system outputof system output

Laplace transform Laplace transform

of system inputof system input

)(

)(:)(

sU

sYsG =

)(sU )(sY)(sG


Modeling Modeling –– Impulse ResponseImpulse Response

• Suppose that u(t) is the unit impulse function and system is at rest.

• The output g(t) for the unit impulse input is called

impulse response.

• Since U(s)=1, the transfer function can also be defined

as the Laplace transform of impulse response:

SystemSystem

)]([:)()(

)()( tgLsY

sU

sYsG ===

)(tg

1)( =sU

)()( ttu δ=


Modeling Modeling –– Electrical ElementsElectrical Elements

ResistanceResistance CapacitanceCapacitanceInductanceInductance

Laplace Laplace

transformtransform

)(ti )(ti )(ti

)(tv )(tv )(tvR L C

)( )( tiRtv =

RsI

sV=

)(

)(

dt

tdiLtv

)()( =

)0)0(( =i

sLsI

sV=

)(

)(

∫+=

t

dttiC

vtv0

)(1

)0()(

sCsI

sV 1

)(

)(=

)0)0(( =v


Modeling Modeling –– ImpedanceImpedance

Capacitor

Inductor

Resistance

Impedances – domaint – domainElement

)resistance effective - )(( )( )()( domain,-s In sZsIsZsV =

)(sZ)(sI )(sV

)()( tRitv =

dt

tdiLtv

)()( =

∫=

t

diC

tv0

)(1

)( ττ

)()( sRIsV =

)( )( sIsLsV =

)(1

)( sIsC

sV =

R

sL

sC

1

L++= )()()( :eries in Impedance 21 sZsZsZs eff

L++=)(

1

)(

1)( :aralell in Impedance

21 sZsZsZp eff


Modeling Modeling –– KirchhoffKirchhoff’’s Voltage & Current Lawss Voltage & Current Laws

R

LC

Sv CvRv

Ri Ci

Li

• The algebraic sum of voltage drops around any loop is zero.

• The algebraic sum of currents into any junction is zero.

Zero sum voltage:

0=++− CRS vvv

0=++− LRS vvv

Zero junction current:

0=−−LCR

iii


Modeling Modeling –– Electrical Example 1Electrical Example 1

∫++=

t

i diC

tiRRv0

21 )(1

)()( ττ

Kirchhoff voltage law (zero

initial conditions):

1R

C

iv ovRv

i

2R

∫+=

t

o diC

tiRv0

2 )(1

)( ττ

Transfer Function

)(1

)()()( 21 sIsC

sIRRsVi ++= )(1

)()( 2 sIsC

sIRsVo +=

Laplace Transformation

CsRR

CsR

sIsC

sIRR

sIsC

sIR

sV

sVsG

i

o

)(1

1

)(1

)()(

)(1

)(

)(

)()(

21

2

21

2

++

+=

++

+

==


Modeling Modeling –– Electrical Example 1 (contElectrical Example 1 (cont’’d)d)

Kirchhoff voltage law in frequency domain (zero initial

conditions) or impedance approach:

1R

Cs

1iv ov

Rv

)(sI

2R

Transfer Function

)(1

)()()( 21 sIsC

sIRRsVi ++= )(1

)()( 2 sIsC

sIRsVo +=

CsRR

CsR

sIsC

sIRR

sIsC

sIR

sV

sVsG

i

o

)(1

1

)(1

)()(

)(1

)(

)(

)()(

21

2

21

2

++

+=

++

+

==


Modeling Modeling –– Electrical Example 2 (OP Amp)Electrical Example 2 (OP Amp)

Transfer Function

)()()( sIsZsV iii = )()()( sIsZsV ooo =

)(

)(

)()(

)()(

)(

)()(

sZ

sZ

sIsZ

sIsZ

sV

sVsG

i

o

ii

oo

i

o −===

--

++ )(sVo

)(sIo

)(sZo

)(sZ i)(sV

d

)(sId)(sIi

)(sVi

OP Amp Assumptions:

Infinity gain: 0)( =sVd

Infinity input impedance:

0)( =sId

)()( sIsI oi −=



Transfer Function

1)( RsZ i = )1(11

)( 22 +=+= CsRCsCs

RsZo

CsR

CsR

sZ

sZ

sV

sVsG

i

o

i

o

1

2 1

)(

)(

)(

)()(

+−=−==

--

++ )(sVo

)(sIo

)(sVd

)(sId)(sIi

)(sVi

Cs

1

2R

1R



Transfer Function

1

11

1

1

11

)(

1

R

sCRsC

RsZi

+=+=

2

221

)(

1

R

sCR

sZo

+=

1

1

)(

)(

)(

)()(

22

11

1

2

+

+⋅−=−==

sCR

sCR

R

R

sZ

sZ

sV

sVsG

i

o

i

o

--

++ )(sVo

)(sIo

)(sVd

)(sId)(sI i

)(sVi

sC2

1

2R

1R

sC1

1


Modeling Modeling –– Electrical ExamplesElectrical Examples

)(sVo)(sVi

2R1R

Cs

1

--

++)(sVo)(sVi

--

++ )(sVo)(sVi

--

++ )(sVo)(sVi

RfR

iRCs

1

)(sVo)(sVi

3R2R

1R Ls


Modeling Modeling –– Summary (Electrical System) Summary (Electrical System)

• Modeling

– Modeling is an important task!

– Mathematical model

– Transfer function

– Modeling of electrical systems

• Next, modeling of mechanical systems


Modeling Modeling –– Time invariant and varying systemsTime invariant and varying systems

• A system is called time-invariant (time-varying) if system

parameters do not (do) change in time.

• Example:

• For time-invariant systems:

• This course deals with time-invariant systems.

)()()( and )()( tftxtMtftxM == &&&&

SystemSystemTime shiftTime shift Time shiftTime shift


MassMass DamperDamperSpringSpring

M KK BB

Modeling Modeling –– Mechanical ElementsMechanical Elements

)(tf )(tx

)( )( txMtf &&=

0)0(

0)0(

=

=

x

x

&

)()( 2sXMssF =

)(tx)(tf )(tx)(tf

)( )( txKtf =

)()( sKXsF =

)( )( txBtf &=

0)0( =x

)()( sBsXsF =


M

KK BB

Modeling Modeling –– SpringSpring--MassMass--Damper SystemsDamper Systems

)(tf )(tx

)()()()( tftKxtxBtxM =++ &&&


M

KK BB

Modeling Modeling –– Free Body DiagramFree Body Diagram

)(tfK )(tfB )(tfK

)()( tKxtfK = )()( txBtfB&=

)(tx )(tf

)(tfB

)(tx

xMF

tx

&& :Law sNewton' Using

position resting springfor changent displaceme therepresents )( :Note

=

)()()()()()()( txBtKxtftftftftxMBK

&&& −−=−−=

)()()()( tftxBtKxtxM =++ &&&


• Equation of motion

• By Laplace transform (with zero initial conditions),

M

KK BB

(2(2ndnd order system)order system)

Modeling Modeling –– SpringSpring--MassMass--Damper System (contDamper System (cont’’d)d)

)(tf )(tx

)()()()( tftKxtxBtxM =++ &&&

)(1

)(2

sFKBsMs

sX++

=


M2

KK11 BB

KK22

M1Vehicle BodyVehicle Body

suspensionsuspension

wheelwheel

tiretire

Modeling Modeling –– Automotive Suspension SystemAutomotive Suspension System

)(tf

)(1 tx

)(2 tx

)())()(())()(()()(

))()(())()(()(

221211222

2112111

txKtxtxKtxtxBtftxM

txtxKtxtxBtxM

−−−−−=

−−−−=

&&&&

&&&&


Laplace transform with zero ICsLaplace transform with zero ICs

G2 G1

G3

Block diagramBlock diagram

Modeling Modeling –– Automotive Suspension System Automotive Suspension System (cont(cont’’d)d)

)())()(())()(()()(

))()(())()(()(

221211222

2112111

txKtxtxKtxtxBtftxM

txtxKtxtxBtxM

−−−−−=

−−−−=

&&&&

&&&&

)())()(())()(()()(

))()(())()(()(

22121122

2

2

211211

2

1

sXKsXsXKsXsXBssFsXsM

sXsXKsXsXBssXsM

−−−−−=

−−−−=

)()(1

)(

)()(

1

)(

21

2

2

1

)(

21

2

2

2

2

)(

1

2

1

1

1

32

1

sXKKBssM

KBssF

KKBssMsX

sXKBssM

KBssX

sGsG

sG

444 3444 21444 3444 21

44 844 76

+++

++

+++=

++

+=

)(sF)(2 sX )(1 sX


Moment of inertiaMoment of inertia FrictionFrictionRotational springRotational spring

J

KKBB

torquetorque

rotation anglerotation angle

Modeling Modeling –– Rotational MechanismRotational Mechanism

)(tθ

)(tτ )(tτ )(tτ

)()()(21

ttt θθθ −= )()()(21

ttt θθθ −=

)(tτ )(tτ

)()( tJt θτ &&=

)()( 2sJssT Θ=

)()( tKt θτ =

)()( sKsT Θ=

)()( tBt θτ &=

)()( sBssT Θ=

0)0(

0)0(

=

=

θ

θ

&0)0( =θ


J

KK

BB

friction between friction between

bob and airbob and air

Modeling Modeling –– TorsionalTorsional Pendulum System Pendulum System (Ex. 2.12)(Ex. 2.12)

)(tθ

)()()()( ttKtBtJ τθθθ =++ &&&

)(tτ


Newton’s law

JKK

Modeling Modeling –– Free Body DiagramFree Body Diagram

)(tk

τ

)(tθ

)()( tKtk

θτ =

)(tb

τ

)()()()()()()( tBtKtttttJbk

θθττττθ &&& −−=−−=

BB

)(tk

τ

)(tb

τ

)(tθ

)(tτ)()( tBtb

θτ &=

)()()()( ttKtBtJ τθθθ =++ &&&


• Equation of Motion

• By Laplace transform (with zero ICs),

)(tθ

J

KK

BB

friction between friction between

bob and airbob and air


Modeling Modeling –– TorsionalTorsional Pendulum System Pendulum System (cont(cont’’d)d)

)(tτ

)()()()( ttKtBtJ τθθθ =++ &&&

KBsJssT

ssG

++=

Θ=

2

1

)(

)()(


• By Newton’s law

• By Laplace transform (with zero ICs),

Modeling Modeling –– Example 4.7Example 4.7

)(tm

τ

Motor

mJ

)(tm

θm

B

K

)(tl

θ

Load

lJ

−−=

−−−=

))()(()(

))()(()()()(

ttKtJ

ttKtBttJ

mlll

lmmmm

θθθ

θθθτθ

&&

&&&

Θ−Θ−=Θ

Θ−Θ−Θ−=Θ

))()(()(

))()(()()()(2

2

ssKssJ

ssKssBsTssJ

mlll

lmmmmmm


• From second equation:

• From first equation:

G2

Block diagramBlock diagram

G1


(4(4thth order system)order system)

)()(

)(

2

1

sKsJ

Ks

m

sG

l

lΘ

+=Θ

43421

Modeling Modeling –– Example 4.7 (contExample 4.7 (cont’’d)d)

)(sTm

)(sm

Θ )(sl

Θ

)())((

)(

)(

23

2

2

sTKBsJJKsJBsJJs

KsJs

m

sG

mlmlmlm

l

m

4444444 34444444 21++++

+=Θ


ThrusterThruster

Double Double

integratorintegrator

•• BroadcastingBroadcasting

•• Weather forecastWeather forecast

•• CommunicationCommunication

•• GPS, etc.GPS, etc.

)(tθ

Modeling Modeling –– Rigid Body Satellite Ex 2.13 (contRigid Body Satellite Ex 2.13 (cont’’d)d)

)(tτ)(tτ

)()( tJt θτ &&=

2

1

)(

)()(

JssT

ssG =

Θ=


• Modeling of mechanical systems

– Translational

– Rotational

• Next, modeling of electromechanical

systems (DC-motor).

Modeling Modeling –– Summary (Mechanical System) Summary (Mechanical System)


• Quarter car model: Obtain a transfer function from R(s) to Y(s).

Road surfaceRoad surface

M1

xx (t)(t)

KKss BB

KKww

M2y(ty(t))

AnswerAnswer

Modeling Modeling –– Additional Exercises Additional Exercises


What is DC motor?

An actuator, converting electrical energy into rotational An actuator, converting electrical energy into rotational

mechanical energymechanical energy

(You will see DC motor during Labs 1 and 4.)(You will see DC motor during Labs 1 and 4.)

Modeling Modeling –– Electromechanical SystemsElectromechanical Systems


• Advantages:

– high torque

– speed controllability

– portability, etc.

• Widely used in control applications: robot, tape drives,

printers, machine tool industries, radar tracking system,

etc.

• Used for moving loads when

– Rapid (microseconds) response is not required

– Relatively low power is required

Modeling Modeling –– Why DC motor?Why DC motor?


ArmatureArmature

(from Dorf and Bishop book)

Modeling Modeling –– How Does DC Motor WorkHow Does DC Motor Work


J

BBmm

Armature circuitArmature circuit Mechanical loadMechanical load

InputInput

OutputOutput

Modeling Modeling –– Model of DC MotorModel of DC Motor

)(tea )(te

m

)(tia aR aL

)(tτ

)(tlτ

)(tθ

EMFback ""

current armature:

voltageapplied:

armature":"

m

i

e

a

a

a

fricton viscous:

inertiarotor :

locityangular ve:

positionangular :

B

J

ω

θ


• Armature circuit

• Connection between mechanical/electrical parts

– Motor torque

– Back EMF

• Mechanical load

• Angular position

Load torqueLoad torque

)()(

)()( tedt

tdiLtiRte

m

a

aaaa++=

Modeling Modeling –– Model of DC Motor in Time DomainModel of DC Motor in Time Domain

)()(

)()(

tKte

tiKt

mm

a

ω

ττ

=

=

)()()()( ttBttJl

τθτθ −−= &&&

)()( tt θω &=




– Motor torque

– Back EMF

• Mechanical load

• Angular position (zero initial condition)

Modeling Modeling –– Model of DC Motor in Model of DC Motor in ““ss”” DomainDomain

))()((1

)( sEsEsLR

sIma

aa

a−

+=

)()(

)()(

sKsE

sIKsT

mm

a

Ω=

=τ

)()( sss Θ=Ω

))()((1

)( sTsTBJs

sL

−+

=Ω


Feedback systemFeedback system

Tachometer Tachometer

(Page 46)(Page 46)

Encoder Encoder

(Page 44)(Page 44)

Modeling Modeling –– DC Motor Block DiagramDC Motor Block Diagram

)(sEa

sLRaa

+

1

)(sIa

τK

)(sT

)(sTl

-

-BJs +

1 )(sΩ )(sΘ

s

1

mK

)(sEm

0)( )),()(())((

)( =Ω−++

=Ω sTsKsEBJsRsL

Ks

lma

aa

τ

0)( )),()(()(

1)( =Ω

+−−

+=Ω sEs

RsL

KKsT

BJss

a

aa

m

l

τ


22ndnd order systemorder system

Modeling Modeling –– DC Motor Transfer FunctionDC Motor Transfer Function

)(:))((

))((1

))((

)(

)(1

sGKKBJsRsL

K

BJsRsL

KK

BJsRsL

K

sE

s

maa

aa

m

aa

a

=+++

=

+++

++=

Ω

τ

τ

τ

τ

)(:))((

)(

))((1

)(

1

)(

)(2

sGKKBJsRsL

RsL

BJsRsL

KK

BJs

sT

s

maa

aa

aa

ml

=+++

+−=

+++

+−

=Ω

ττ

)()()()()(21

sTsGsEsGsla

+=Ω

))()()()((1

)(1

)(21

sTsGsEsGs

ss

sla

+=Ω=Θ


Note:Note: In many cases In many cases LLaa<< << RRaa. Then, an approximated TF is . Then, an approximated TF is

obtained by setting obtained by setting LLa a = 0.= 0.

11stst order systemorder system22ndnd order systemorder system

Modeling Modeling –– DC Motor Transfer Function (contDC Motor Transfer Function (cont’’d)d)

+=

+=

+=

++≈

+++=

Ω

ττ

τ

τ

τ

τ

τ

KKBR

JRT

KKBR

KK

Ts

K

KKBJsR

K

KKBJsRsL

K

sE

s

ma

a

ma

mamaaa

,: 1

:

)())(()(

)(

)1()(

)(

+=

Θ

Tss

K

sE

s

a

ττKKBJsR

R

KKBJsRsL

RsL

sL

s

ma

a

maa

aa

l++

−≈

+++

+−=

Ω

)())((

)(

)(

)(


Negative feedback systemNegative feedback system

Memorize this!Memorize this!

Modeling Modeling –– Feedback System Transfer FunctionFeedback System Transfer Function

)(sR )(sY)(sG

)(sH

))()()()(()( sYsHsRsGsY −= )()()())()(1( sRsGsYsHsG =+

)()(1

)(

)(1

)(

)(

)(

sHsG

sG

sL

sF

sR

sY

g

g

+=

−=

−=

=

gain Loop)()()(

gain Forward)()(

sHsGsL

sGsF

g

g


• Modeling of DC motor

– What is DC motor and how does it work?

– Derivation of a transfer function

– Block diagram with feedback

• Next

– Block Diagram

Many systems can be represented as Many systems can be represented as

transfer functionstransfer functions!!

Modeling Modeling –– Summary (DC Motor)Summary (DC Motor)


• Suppose that u(t) is a unit impulse function and system is

at rest (zero initial conditions).

• The output g(t) for the unit impulse input is called

impulse response.

• Since U(s) = 1, the transfer function can also be defined as the Laplace transform of impulse response:

SystemSystem

Modeling Modeling –– Block Diagram (impulse response)Block Diagram (impulse response)

)()( ttu δ=

1)( =sU)(tg

)]([:)( tgLsG =


Series connection

Modeling Modeling –– Basic TF Block Diagram (1)Basic TF Block Diagram (1)

)(1 sG )(2 sG)(sR )(sZ )(sY

)()(

)(1 sG

sR

sZ= )(

)(

)(2 sG

sZ

sY=

)()()(

)(12 sGsG

sR

sY=

)(sR )(sY)()( 21 sGsG


Summation

+

-


)(1

sZ

)(2

sZ

)(sY

)()()(21

sZsZsY −=


Parallel connection

)(sY


)(1 sZ

)(2 sZ

)(sE)(

1sG

)(2

sG

)()(

)(1

1 sGsE

sZ=

)()(

)(2

2 sGsE

sZ=

)())()(()()()( 2121 sEsGsGsZsZsY +=+=

)()()(

)(21

sGsGsE

sY+=

)(sE )(sY)()(

21sGsG +


Negative feedback system

)(sE

Modeling Modeling –– TF Feedback ConnectionTF Feedback Connection

)(sR )(sY)(sG

)(sK

)()()()()( sEsGsKsRsE −=

)()()( sEsGsY =

)()(1

1

)(

)(

sGsKsR

sE

+=

)()(1

)(

)(

)(

sGsK

sG

sR

sY

+=


• Fg(s): Forward gain from R(s) to Y(s) G(s)

• Lg(s): Loop gain: G(s)K(s)(-1)

The loop gain is the product of

all transfer functions that form the loop

)(sG

Modeling Modeling –– TF Feedback Loop FormulaTF Feedback Loop Formula

)(sK

)(sE)(sR )(sY

)()(1

)(

)(

)(

sKsG

sG

sR

sY

+=


Compute transfer functions from Compute transfer functions from RR((ss)) to to YY((ss).).

)(sR )(sY

)(sH

)(sR )(sY)(sG)(sG)(sC

)(sR )(sY)(sG

)(sH)(sH

)(sY)(sR

)(1

sG

)(1

sG

Modeling Modeling –– Derivation of transfer functionDerivation of transfer function


BBmm

InputInput

OutputOutput

Modeling Modeling –– Model of DC Motor (Example 2)Model of DC Motor (Example 2)

)(tea )(te

m

)(tia aR aL

)(tτ

)(tlτ)(tθ

mgmg

R R

θ

)(tlτ)(tθ

θτ &BtB =)(

J

J

R R

θ

θsinmgmg

)(tlτ

θsinmgR

)(tτ




– Motor torque

– Back EMF

• Mechanical load

• Angular position

)()(

)()( tedt

tdiLtiRte

m

a

aaaa++=

Modeling Modeling –– Model of DC Motor in Time DomainModel of DC Motor in Time Domain (2)(2)

)()(

)()(

tKte

tiKt

mm

a

ω

ττ

=

=

)()()()( ttBttJl

τθτθ −−= &&&

)()( tt θω &=

)()()(sin)()( smallfor

2tmgRttmgRttmR

llθτθτθ

θ

−≈−=&&




– Motor torque

– Back EMF

• Mechanical load

• Angular position (zero initial condition)

Modeling Modeling –– Model of DC Motor in Model of DC Motor in ““ss”” DomainDomain (2)(2)

))()((1

)( sEsEsLR

sIma

aa

a−

+=

)()(

)()(

sKsE

sIKsT

mm

a

Ω=

=τ

)()( sss Θ=Ω

)),()((1

)(2

sTsTBsJs

sL

−+

=Θ )(1

)(22

sTmgRsmR

sl

+=Θ

)()(

1)(

22sT

mgRBssmRJs

+++=Θ


Modeling Modeling –– DC Motor Block Diagram DC Motor Block Diagram (2)(2)

)(sEa

sLRaa

+

1

)(sIa

τK

)(sT

)(sTl

-

-BJs +

1 )(sΩ

)(sΘs

1

sKm

)(sEm

mgRsmR +22

)(sEa

sLRaa

+

1

)(sIa

τK

)(sT

-mgRBssmRJ +++

22 )(

1

)(sΘ

sKm

)(sEm

BsJs

mgRsmR

BsJssG

+

+

+

+=

2

22

2

1)(

1

)),()((1

)(2

sTsTBsJs

sL

−+

=Θ

)(1

)(22

sTmgRsmR

s l+

=Θ

)()(

1)(

22sT

mgRBssmRJs

+++=Θ


Modeling Modeling –– DC Motor Transfer Function DC Motor Transfer Function (2)(2)

maa

aa

m

aa

a

KsKmgRBssmRJRsL

K

mgRBssmRJRsL

KsK

mgRBssmRJRsL

K

sE

s

τ

τ

τ

τ

+++++=

+++++

++++=

Θ

)))(((

)))(((1

)))(((

)(

)(

22

22

22

maaaKKBJsRsL

K

ssE

s

m

τ

τ

+++⋅=

Θ

=

))((

1

)(

)(

attached), pendulum (no 0 If


BB

Modeling Modeling –– TorsionalTorsional Motion Modeling Motion Modeling (Example 2)(Example 2)

1K

)(tf

)(tθR R

m)(tx

JR R

J

Free Body DiagramFree Body Diagram

1kτ )(tθBτ

2kf

22 )( KRxfk θ−=

11 Kk θτ =

θτ &BB =

θθθ

ττθ

&

&&

BKRKRx

RfJBkk

−−−=

−−=

12

12

)(

RxKRKKBJ2

2

21)( =+++ θθθ &&&

2K



)(tf

m)(tx

Free Body DiagramFree Body Diagram

2kf22 )()()()( KRxtftftfxm k θ−−=−=&&

)()(

)(2

21

2

2 sXRKKBsJs

RKs

+++=Θ

θRKtfxKxm 22 )( +=+&&

))()((1

)(2

2

2sRKsF

KmssX Θ+

+=


)(sF

2

2

1

Kms +

)(sΘ

RK2

+

222

221

22

2

2

221

22

2

222

221

22

2

2

))()((

))()((

))()((

1)(

RKRKKBsJsKms

RK

RKKBsJsKms

RK

RKKBsJsKms

RK

sG−++++

++++

++++=

−=


)( 2

21

2

2

RKKBsJs

RK

+++

)()(

)(2

21

2

2 sXRKKBsJs

RKs

+++=Θ))()((

1)(

2

2

2sRKsF

KmssX Θ+

+=

212

22

212

34

2

)]([ )(

KKBsKsRKKmJKmBsmJs

RKsG

++++++=


• Block diagrams

– Elementary diagrams

– Feedback connections

• Next

– Linearization

Modeling Modeling –– Summary (Block Diagram)Summary (Block Diagram)


• A system having Principle of Superposition

A nonlinear system does not satisfy the

principle of superposition.

SystemSystem

Modeling Modeling –– What Is a Linear System?What Is a Linear System?

)(tu )(ty

ℜ∈∀

+→+⇒

→

→

21

22112211

22

11

,

)()()()()()(

)()(

αα

αααα tytytututytu

tytu

linearnot is )0( )(function affineAn ≠+= bbaxtf


• Easier to understand and obtain solutions

• Linear ordinary differential equations (ODEs),

– Homogeneous solution and particular solution

– Transient solution and steady state solution

– Solution caused by initial values, and forced solution

• Add many simple solutions to get more complex ones (use

superposition!)

• Easy to check the Stability of stationary states (Laplace Transform)

Modeling Modeling –– Why Linear System?Why Linear System?


• Actual physical systems are inherently nonlinear. (Linear

systems do not exist!) Ex. f(t)=Kx(t), u(t)=Ri(t)

• TF models are only for Linear Time-Invariant (LTI) systems.

• Many control analysis/design techniques are available only

for linear systems.

• Nonlinear systems are difficult to deal with mathematically.

• Often we linearize nonlinear systems before analysis and design. How?

Modeling Modeling –– Why LinearizationWhy Linearization


• Nonlinearity can be approximated by a linear function for

small deviations δx around an operating point x0

• Use a Taylor series expansion at x0

Linear approximation

Nonlinear function

Operating point x0

Old coordinate

New coordinate

Modeling Modeling –– How to How to LinearizeLinearize It?It?

)(0

xxf δ+

)(xf

x

xδ


• Nonlinear system:

• Let u0 be a nominal input and let the resultant state be x0

• Perturbation:

• Resultant perturb:

• Taylor series expansion:

Modeling Modeling –– LinearizationLinearization

),( uxfx =&

)()()( 0 tututu δ+=

)()()( 0 txtxtx δ+=

...),(

),(),(),(

0

00

0

0

0

0

321≈=

=

=

=

+∂

∂+

∂

∂+=

TOHuu

uxf

xx

uxfuxfuxf

uuxx

uuxx

δ

δ


Note that ; hence

L. sys.L. sys.N. sysN. sys

Modeling Modeling –– Linearization (contLinearization (cont’’d)d)

uu

uxfx

x

uxfuxfxx

uuxx

uuxx

δδδ

0

0

0

0

),(),(),(

000

=

=

=

= ∂

∂+

∂

∂+=+ &&

),(000

uxfx =&

uu

uxfx

x

uxfx

uuxx

uuxx

δδδ

0

0

0

0

),(),(

=

=

=

= ∂

∂+

∂

∂=&

u x uδ

0u

),(000

uxfx =&

xδ

0x

xxx δ+=0uuu δ+=

0


• Motion of the pendulum

• Linearize it at

• Find u0

• New coordinates:

Modeling Modeling –– Pendulum LinearizationPendulum Linearization

)()(sin)(2tutmgLtmL =+ θθ&&)(tu

)(tθ

mg

L

0)()(sin

)(

),(

2=−+

44 344 21

&&

uf

mL

tu

L

tgt

θ

θθ

πθ =0

00sin

02

0=→=−+ u

mL

u

L

g ππ&&

uuuu δδ

δθπδθθθ

+=+=

+=+=

00

0


• Taylor series expansion of at

Modeling Modeling –– Pendulum Linearization (contPendulum Linearization (cont’’d)d)

),( uf θ .0 , == uπθ

L

g

L

guf

u

−==∂

∂

==

=πθ

πθ

θ

θ

θ cos),(

0

2

0

1),(

mLu

uf

u

−=∂

∂

=

=πθ

θ

0),(),(

00

=∂

∂+

∂

∂+

=

=

=

=

uu

ufuf

uu

δθ

δθθ

θθδ

πθπθ

&&

01

2=−− u

mLL

gδδθθδ &&


• TF derivation

• The more time delay is, the more difficult to control

(Imagine that you are controlling the temperature of your shower with a very long hose. You will either get burned

or frozen!)

(Memorize this!)

Modeling Modeling –– Time Delay Transfer FunctionTime Delay Transfer Function

delay time is where),()(dd

TTtxty −=

L

sTsT dd esX

sYsXesY

−−=⇒=

)(

)( )()(

1

11)0(

)(

0

0 +=⇒+=−

∂

∂+≈

−

=

= sTesTs

s

eee

d

sT

d

s

sT

s

sTsT d

d

dd


• Step One: identify the system model’s input r(t) and output y(t).

• Step Two: express model in form

• Step Three: define equilibrium operation point r0 and y0 ,

and setting all their corresponding derivatives to zero at

equilibrium point

• Perform Taylor expansion about (r0,c0) retaining only 1st

order derivatives

• Step Five: change variables to deviations:

• Step Six: rewrite the function defined in Step 5 in standard

ODE.

Modeling Modeling –– Summary: Steps of LinearizationSummary: Steps of Linearization

0,...),,...,,( =yyrrf &&

,...).0,0,...,0,0(0000

==== yyrr &&&&&&

,...).~,~,...,~,~(000000

yyyyyyrrrrrr &&&&&& −=−=−=−=


Modeling Modeling –– 6 Step Example 16 Step Example 1

)( :output system ),( :input System ttu θ

)()(sin)(2tutmgLtmL =+ θθ&&

0)()(sin

)(

),,(

2=−+

4444 34444 21

&&

&& uf

mL

tu

L

tgt

θθ

θθ

System ODE:System ODE:

Step One (input/output):Step One (input/output):

Step Two Step Two ((f f ((……)=0):)=0):

Step Three Step Three ((θθ00, u, u00))::

,0 and ,Let 000

=== θθπθ &&&

00sin

02

0=→=−+ u

mL

u

L

g πθ&&


Modeling Modeling –– 6 Step Example 1 6 Step Example 1 (cont(cont’’d)d)

)0(1

)0()(

)0(1

)0(1)(cos

)()()( ),,(),,(

2

2

000

0

000

0

0

0

0

0

0

0

0

0

−−−+−=

−−−⋅+−=

−∂

∂+−

∂

∂+−

∂

∂+≅

=

=

=

=

=

=

=

=

=

=

umLL

g

umLL

g

uuu

fffufuf

uuuuuu

θπθ

θπθθ

θθθ

θθθ

θθθθ

πθ

θθ

θθ

θθ

θθ

θθ

θθ

&&

&&

&&&&&&43421

&&&&

&&&&&&&&&&&&

Step Four (Taylor expansion):Step Four (Taylor expansion):

Step Five Step Five ((δδθθ00, , δδuu00))::

01

),,(2

=−+≅ umLL

guf δδθθδθθ &&&&

Step Six Step Six (rewrite(rewrite))::

mLgsmLsU

su

mLL

g

+=

∆

∆Θ⇒=+

222

1

)(

)(1δδθθδ &&



System ODE:System ODE:

2)( :force drag cAerodynami

)( :forcefriction Viscous

)( :input force Tractive

vKtff

vKtff

tff

add

vvv

mm

==

==

=

2vKvKffffvm

avmvdm−−=−−=&

mf

df

vf

mavfvKvKvm =++

2&

kgm

mNK

mNK

mv

a

v

1000

/sec0.1

sec/ 1.0

sec/ 30

22

0

=

−=

−=

=



)( :(velocity)output system ),( :force) (tractiveinput System tvtfm

Vehicle System:Vehicle System:



Step Three Step Three ((vv00,vdot,vdot00 uu00))::

000 find and 0, and 30Let m

fvv == &

NvKvKfavm

903301301.0 22

000 =×+×=+=

mavfvKvKvm =++

2&

0

),,(

2=−++

444 3444 21&

&vvff

mav

m

fvKvKvm


Modeling Modeling –– 6 Step Example 2 6 Step Example 2 (cont(cont’’dd--1)1)

0)()()2()(10

)()()( ),,(),,(

0000

000

0

000

0

0

0

0

0

0

0

0

0

=−+−⋅++−−=

−∂

∂+−

∂

∂+−

∂

∂+≅

=

=

=

=

=

=

=

=

=

vvmvvvKKff

vvv

fvv

v

fff

f

fvvffvvff

avmm

vvvv

ff

vvvv

ffmm

vvvv

ffm

mmmmmmmm

&&

&&&4434421

&&

&&&&&&


Step Five Step Five :)~

,~,~

(000

vvvvvvfffmmm

&&& −=−=−=


0~~)2(

~

1.6030121.0

0

1000

=−⋅++

=××+

mavfvvKKvm

43421&

1.601000

1

)(

)(~~1.60~

1000+

=⇒=+ssF

sVfvv

m

m&



1R

C

ivovRv

)(ti

)()()(1

)(0

0

0tvCtidi

Ctv

t

&=⇒= ∫ ττ

)()()( 3

22212tiRtiRtV

R+=

)()()()()( 0

3

22211 tvtiRtiRRtvi

+++=

)()()()()(0

3

0

3

220211tvtvCRtvCRRtv

i+++= &&

2

22211 /200 and ,1 ,1 ARRR Ω=Ω=Ω=

,1.0 FC = )()(2.0)(2.0)( 0

3

00 tvtvtvtvi

++= &&

VvVvi 5 and 10at it Linearize 0 ==



)( :tageoutput vol system ),( :ageinput volt System0

tvtvi

Circuit System:Circuit System:



Step Three Step Three ((vv00,vdot,vdot00 vvii))::

00 find and 5, and 10Let vvvi

&==

solution) real(only 81.2025 00

3

0 =⇒=−+ vvv &&&

)()(2.0)(2.0)(0

3

00tvtvtvtv

i++= &&

0)()()(2.0)(2.0

),,(

0

3

00

00

=−++44444 344444 21

&&

&vvvf

i

i

tvtvtvtv



0)81.2(938.4)5(1)10(10

)81.2()5()10( )81.2,5,10(),,(

0

6.02.0

0

0

81.2510

0

0

81.2510

0

81.2510

0

00

20

0

0

0

0

0

0

=−+−⋅+−⋅−=

−∂

∂+−

∂

∂+−

∂

∂+≅

×+

=

=

=

=

=

=

=

=

=

vvv

vv

fv

v

fv

v

ffvvvf

v

i

vvv

vvv

i

vvv

i

iiii

&321

&&4434421

&

&

&&&


Step Five Step Five :)81.2~

,5~,10~(000

−=−=−= vvvvvvii

&&


0~~~938.4

00=−+

ivvv&

1938.4

1

)(

)(~~~938.4 0

00+

=⇒=+ssV

sVvvv

i

i&



0)0(2.0)10(1)10(10

)0()10()10( )0,10,10(),,(

00

0

01010

0

0

01010

0

01010

0

00

0

0

0

0

0

0

=−⋅+−⋅+−⋅−=

−∂

∂+−

∂

∂+−

∂

∂+≅

=

=

=

=

=

=

=

=

=

vvv

vv

fv

v

fv

v

ffvvvf

i

vvv

vvv

i

vvv

i

iiii

&

&&43421

&

&&&


Step Five Step Five :)0~

,10~,10~(000

−=−=−= vvvvvvii

&&


0~~~2.0

00=−+

ivvv&

12.0

1

)(

)(~~~2.0 0

00+

=⇒=+ssV

sVvvv

i

i&

Step Three Step Three ((vvoo==1010 vvii==1010))::

solution) real(only 00 00

3

0 =⇒=+ vvv &&&


• Modeling of

– Nonlinear systems

– Systems with time delay

• Next

– Time response of physical systems

Modeling Modeling –– Linearization SummaryLinearization Summary

Modeling – Part of Control Design Process Fall ME451 - GGZ Week 3-4: Modeling Physical Systems...

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