1

Modeling

M. Sami FadaliProfessor of Electrical Engineering

University of Nevada

2

Transfer FunctionLTI System

a d cdt

a d cdt

a dcdt

a c

b d rdt

b d rdt

b drdt

b r

G sC sR s

b s b s b s ba s a s a s a

n

n

n n

n

n

m

m

m m

m

m

zero IC

mm

mm

nn

nn

1

1

1 1 0

1

1

1 1 0

11

1 0

11

1 0

Example

3

Response of LTI System• Convolution Theorem: = impulse response

L

• Impulse Input L

L

• The transfer function and the impulse response are Laplace transform pairs.

4

Example: Transfer Function of Point Massf t m x

F s m s X s

G s X sF s m s

( )

( ) ( )

( ) ( )( )

2

2

1

F s s

X sms s

x t tm

( )

( ) ( )

1

1 122

2

m

x

f

5

Analytical Modeling• Physical systems store and dissipate energy.• Lumped idealized elements: represent energy

dissipation and energy storage separately.• Physical elements may approximate the

behavior of the idealized elements. • Physical elements are not lumped. They

involve both energy dissipation and energy storage.

6

Relations

• Constitutive (elemental) Relations– Govern the behavior of the idealized elements.– Hold only approximately for physical elements.

• Connective Relations– Govern connections of elements.– Often derived from conservation laws.

7

Electrical Systems

• Ideal R: energy dissipation.• Ideal L: magnetic energy storage.• Ideal C: electrostatic energy storage.• Connective Relations: Kirchhoff’s Laws

– Current Law: conservation of charge– Voltage Law: conservation of energy

Constitutive Relations for R, L, C

8

Table 2.3Impedance Admittance

9

Mesh Analysis1. Replace passive elements with Z(s).2. Define clockwise current for each mesh.3. Write KVL in matrix form with source

voltages that drive clockwise current positive.

4.Use Cramer’s rule to solve for the transfer function.

Cramer’s Rule

• Provided that a solution exists

10

2221

1211

221

111

2

2221

1211

222

121

1 ,

aaaababa

x

aaaaabab

x

2

1

2

1

2221

1211

bb

xx

aaaa

211222112221

1211 aaaaaaaa

11

Example: Mesh Analysis

)()(

0)(

)()(

1111

)()()(

23

2

1

23

1

sIRsV

sVsIsI

sCsLRsCsCsCZ

sss

o

i

i

VIZ

211

112

2111

RRsLRsLR

RRsLZ

i C+

12

2231

3

23

111

)()(

sCsCsLRsCZsCR

VsIR

VVsG

ii

o

Cramer’s Rule

I s

Z sC VsC

Z sC sCsC R L s sC

i

2

1

1

3 2

11 0

1 11 1

( )

i

13

Node Analysis1.Replace passive elements with Y(s).2. For each node, define a node voltage relative

to a reference node.3. Write KCL in matrix form with source

currents that drive current into a node positive.

4.Solve for the transfer function using Cramer’s rule.

14

Example: Node Analysis

Change voltage source to

current source.

R1

L2 L1

Rs

Vs/Rs

R2 R3

15

Y V I( ) ( ) ( )s s s

00)(

)()()(

11

1111

11

231

21

2212

1

111

1

ss

c

b

a

sGsV

sVsVsV

sLGG

sLG

sLsLsLG

sL

GsLsL

GG

Node Equations

co VVRG ,/1 L2

Vo

R1

L1

Rs R2

Vs/Rs

R3 Vb

Va

16

G s VV

V

G GsL sL

VR

sLG

sL sL

GsL

G GsL sL

G

sLG

sL sL sL

GsL

G GsL

o

s

s

ss

s

s

( )

1

1 1

1 1 1 0

1 0

1 1

1 1 1 1

1 1

11 1

12

1 2

12

11 1

1

12

1 2 2

12

1 32

17

Translational Mechanical Systems

• Ideal Damper b: energy dissipation.• Ideal Spring k: potential energy storage.• Pure Mass m: kinetic energy storage.• Connective Relations: Newton’s 2nd Law

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a) Ideal Spring

• Elastic energy (neglect plastic deformation)• Linear element: force proportional to the

deformation .• No energy dissipation• No mass

k

x2x1

f

19

b) Ideal Viscous Damper

• Energy dissipation• No mass• No elastic deformation• Linear element: force proportional to rate of

deformation .

bx1 x2

f

20

c) Point Mass

•Perfectly rigid

•No dissipation

•Linear element

mf

x

Constitutive Relations for Spring, Mass, Damper

21 22

Ex. Mass-Spring-Damper

fkxxbxmxbxkf

fffxm ds

G sX sF s ms bs k

1

2

x

b k

f(t)

m

23

Example 2.11

m x b x b x x k x k x x f

m x b x b x x k x k x x

1 1 1 1 3 1 2 1 1 2 1 2

2 2 2 2 3 2 1 3 2 2 2 1 0

m s b b s k k b s kb s k m s b b s k k

XX

F12

1 3 1 2 3 2

3 2 22

2 3 2 3

1

2 0

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Transfer Function (Cramer’s Rule)

3232

2223

2321312

1

23

21312

1

2

01

kksbbsmksbksbkksbbsm

ksbFkksbbsm

F

FXsG

22332322

221312

1

23

ksbkksbbsmkksbbsmksb

3-D.O.F. Translational Mechanical System

25

Three equations of motion.

Rotational Mechanical Systems

26

27

Example 2.19

211111 KDJ

02

1

22

2

12

1

KsDsJKKKsDsJ

0122222 KDJ

28

Transfer Function (Cramer’s Rule)

KsDsJKKKsDsJ

KKsDsJ

sG

22

2

12

1

12

1

2 01

)(

22

221

21 KKsDsJKsDsJ

K

29

Example 2.20

00

3323233

1232222

211111

DDJKDJKDJ

J s Ds K KK J s D s K D s

D s J s D D s

12

1

22

2 2

2 32

2 3

1

2

3

0

000

30

Gears

Assume:1- No losses. 2- No inertia. 3- Perfectly rigid.Single velocity at point of contact Equal arc length

1 1

2 2

N1

N2

J

Energy Balance• Assume no losses

• Trade speed for torque

N2 > N1 : output side slower but delivers more torque

N2 < N1 : output side faster but delivers less torque

31 32

Energy Storage• Translation

• Mass

• Spring

• Rotation

• Inertia

• Spring

Energy Dissipation

• Power dissipated• Translation

• Rotation

33 34

Equivalent Inertia

Equivalent to inertia on

output side

2

2

12

1

2

2

1

222

211 2

121

NN

JJ

JJE

1 , 1

2 , 2

N1

N2

J

2

sourceofteethno.ndestinatioofteethno.

JJe

2

2

121

NNJJJe

35

Effect of Loading Output Side on Input Side

Inertia

2

2

2

1

1

1

12

11

1

2

22

sourceofteethno.ndestinatioofteethno.

J

NNJJ

NNJ

NN

J

e

1 , 1

2 , 2

N 1

N 2

J

1 , 1

2 , 2

1

236

Damper22

2

1

1

1

sourceofteethno.ndestinatioofteethno.

DNNDDe

22

2

1

1

1

sourceofteethno.ndestinatioofteethno.

KNNKKe

Damper and Spring

Spring

37

Example (Special Case)

)()()( 11

222

2 sTNNsTsKsDsJ ee

38

Example

,m Db

JL

2 L

K

N1

N2 DL

Jg Jb

• Two equations of motion.• Cannot simply add all rotational masses!

SolutionRedraw the schematic with (i) added “e” for elements (and variables) moved, and (ii) gears removed.

39

e,me Dbe

JL

2 L

K DL

Jbe Jge

40

Gear Train

12

12

12

1

22

32

4

3

2

1

12

3

1

2

1

1

KnKBnBJnJn

nNN

NN

NN

elelel

e

l

e

l

l

l

ll

2

l

N3

N2

N1

N4

1

3 l-1

41

TFs of Electromechanical Systems• Electrical Subsystem

– Varies with motor type– armature (rotor) conductors current – field (stator) conductors or permanent magnet

• Mechanical Subsystem– Varies with load.– Write equations of motion.

DC Motor

DC motor.

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National Instruments:http://zone.ni.com/devzone/cda/ph/p/id/52

DC motor armature (rotor)

Torque Equation• Magnetic flux Wb• Force

= conductor length= magnetic flux density = armature (rotor) current

Control• Vary torque by changing or

43

f

r

44

Field Control• Changing and fixing • Back EMF (Faraday’s Law)• Voltage induced in moving coil proportional to the

rate of cutting of lines of magnetic flux.

is only approximately constant through the use of high resistance (inefficient)

45

Armature Control• Changing and fixing

• Used in practice.

• KVL

ea vb

Za

46

Schematic

1

ea

J1, D1 J2

D2

2 L

vb

N1

K

Za

N2

47

Mechanical SubsystemEquations of motion for rotational system

01

''2

'2

'11111

LeLeLe

atLe

KDJiKKDJ

, 1

J2e

D2e

L

D1

Ke J1

48

00

0

0'1

22

2

12

1

a

a

L

eeee

tee

aab E

IKsDsJKKKKsDsJ

RsLsK

Matrix Form

01

''2

'2

'11111

LeLeLe

atLe

KDJiKKDJ

49

Transfer Function

0

000

01

)(

22

2

12

1

12

121

21'

eeee

tee

aab

e

te

aaab

a

a

L

a

L

KsDsJKKKKsDsJ

RsLsKK

KKsDsJRsLEsK

ENN

ENN

EsG

eeebteeeeeaa

te

KsDsJsKKKKsDsJKsDsJRsLKKNN

22

22

22

212

1

21

50

Evaluation of Motor ParametersDynamometer measure speed & torque for constant Dynamometer Test gives speed-torque curvesAssume supplied by manufacturer

0 10 20 30 40 500

100

200

300

400

500

Speed (rad/s)

Torq

ue (N

-m) ea= 100 V

=stall torque=no-load speed

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Solve for Parameters

• Stall torque:

• No-load speed:

ea vb

Ra

Transfer Function

52

T, m

JLe

DLe

Da Ja

batee

at

a

m

a

mbatatmee

KRKDsJRK

sEssG

RsKsEKsIKsDsJ

)()()(

)()()()(

Leae

Leae

DDDJJJ

53

Linearity(i) Homogeneity

(ii) Additivity

• Affine

Nonlinearities

54

55

Linearization1st order approximation (in the vicinity of )

for small 0 2 4 6 8

0

20

40

60

80

56

Equilibrium Point• System at an equilibrium stays there unless perturbed.

• Set all derivatives equal to zero for equilibrium.

= value of forcing function at equilibrium

• Cancel constants and

57

In the Vicinity of the Equilibrium

Special case: nonlinearity in output only

nidt

cddt

cdSimilarlydtdc

dtccd

dtcd

i

i

i

i

,,2,1,0

rrrcfcfdt

cdadt

cdadt

cdn

n

nn

n

0011

1

1 )()(

1st order approximation

rcdcdf

dtcda

dtcda

dtcd

cn

n

nn

n

0

11

1

1

Linearized Differential Equation

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d cdt

a d cdt

a d cdt

a c r a dfdc

n

n n

n

nc

1

1

1 1 0 00

,

G s CR s a s a s a s an

nn

nn( )

1

11

22

1 0

Linear: can Laplace transform to get the TF

59

Procedure

1. Determine the equilibrium point(s).2. Find the first order approximation of all

nonlinear functions.3. Rewrite the system differential equation in

terms of perturbations canceling the constants using Step 1.

60

Example: Pendulum

Moment of inertia

Equation of Motion

• Linearize about = 30o

• Equilibrium at = 30o sin( )

61

Linearization• Using Trigonometric Identity

sin sin sin sin cos

30 30 3030

o o o

o

dd

sin1cos2321

sin30coscos30sin30sin ooo

• Using 1st order approximation formula

23

2321 0

mglBJmglBJ

62

Potentiometer

• 10 turns• 1 turn = 2 rad• 20 V• Pot Gain = 20/(10 X 2 )

= (1/ ) V/rad

63

Fluid SystemsLinearized Modelh = R qConservation of Mass

qi

qo

h

H

inin

oin

Rqhdtdh

Rhq

dtdhC

AreaC

qQqQdt

dCh

1)()(

sR

sqsh

in