Lec. 3Lec. 3Modeling of Dynamic System

2G 4G1G

4H

)(sY)(sR

3G

Quiz (10 mins)

Find the transfer function of the following block diagrams

2H

3H

1H

2

Solution:

2G 4G1G

4H)(sY

3G

)(sRA B

1

I1. Moving pickoff point A behind block

4G

H

1H

2H

3H4

1

G

4

1

G

4

3

G

H

4

2

G

H

3

2. Eliminate loop I and Simplify

II

443

432

1 HGG

GGG

+1G)(sY

B

H

)(sR

4

3

G

H

III

1H

4

2

G

H

II

332443

432

1 HGGHGG

GGG

++

I I I

4

142

G

HGH −

Not feedbackfeedback

4

)(sR )(sY

142

G

HGH −

332443

4321

1 HGGHGG

GGGG

++

3. Eliminate loop II & IIII

4G

143212321443332

4321

1 HGGGGHGGGHGGHGG

GGGG

sR

sY

−+++=)(

)(

5

• Static System: If a system does not change with time, it

is called a static system.

• The o/p depends on the i/p at the present time only

• Dynamic System: If a system changes with time, it is

called a dynamic system.

• The o/p depends on the i/p at the present time and the

past time. (integrator, differentiator).

6

• A system is said to be dynamic if its current outputmay depend on the past history as well as thepresent values of the input variables.

• Mathematically,

Time Input, ::

]),([)(

tu

tuty ≤≤= ττϕ 0

Time Input, :: tu

ExampleExampleExampleExample: A moving mass

MMMM

yyyyuuuu

ModelModelModelModel: Force=Mass XXXX Acceleration

uyM =ɺɺ

1. Mathematical.

2. State space.

9

A set of mathematical equations (e.g., differentialequs.) that describes the input-output behavior ofa system.

What is a model used for?What is a model used for?What is a model used for?What is a model used for?

• Simulation• Simulation

• Prediction/Forecasting

• Prognostics/Diagnostics

• Design/Performance Evaluation

• Control System Design

1. Electrical Systems1. Electrical Systems

Basic Elements of Electrical Systems

• The time domain expression relating voltage and current for the resistor is given by Ohm’s law i.e.

Rtitv RR )()( =

• The Laplace transform of the above equation is

RsIsV RR )()( =

Basic Elements of Electrical Systems

• The time domain expression relating voltage and current for the Capacitor is given as:

∫= 1dtti

Ctv cc ∫= )(

1)(

• The Laplace transform of the above equation (assuming there is no charge stored in the capacitor) is

)()( sICs

sV cc1=

Basic Elements of Electrical Systems

• The time domain expression relating voltage and current for the inductor is given as:

tdiLtv L )(

)( =dt

tdiLtv L

L)(

)( =

• The Laplace transform of the above equation (assuming there is no energy stored in inductor) is

)()( sLsIsV LL =

Component Symbol V-I Relation I-V Relation

Resistor Rtitv RR )()( =R

tvti R

R)(

)( =

15

Capacitor

Inductor

dt

tdiLtv L

L)(

)( =

dttiC

tv cc ∫= )()(1

dt

tdvCti c

c)(

)( =

dttvL

ti LL ∫= )()(1

The two-port network shown in the following figure has vi(t)as the input voltage and vo(t) as the output voltage. Find thetransfer function Vo(s)/Vi(s) of the network.

CCCCi(t)v ( t) v (t)

16

CCCCi(t)vi( t) vo(t)

∫+= dttiC

Rtitv i )()()(1

∫= dttiC

tv o )()(1

Taking Laplace transform of both equations, consideringinitial conditions to zero.

∫+= dttiC

Rtitv i )()()(1

∫= dttiC

tv o )()(1

)()()( sIRsIsV1+= )()( sIsV

1=

Re-arrange both equations as:

17

)()()( sICs

RsIsV i1+= )()( sI

CssV o

1=

)()( sIsCsV o =))(()(Cs

RsIsV i1+=

Substitute I(s) in equation on left

)()( sIsCsV o =))(()(Cs

RsIsV i1+=

))(()(Cs

RsCsVsV oi1+=

18

Cs

)()(

)(

CsRCssV

sV

i

o

11

+=

RCssV

sV

i

o

+=

1

1

)(

)(

The system has one pole at

RCssV

sV

i

o

+=

1

1

)(

)(

sRCs1

01 −=⇒=+

19

RCsRCs

101 −=⇒=+

Design an Electrical system that would place a pole at -4 ifadded to another system.

RCssV

sV

i

o

+=

1

1

)(

)(

CCCCi(t)vi( t) vo(t)

System has one pole at

Therefore,

20

RCs

1−=

14

RC− = − 250 1 mif R and C F= Ω =

iR(t)

+

IR(S)

+

TransformationTransformationTransformationTransformation

21

vR(t)

-

VR(S)

-

ZR = R

TransformationTransformationTransformationTransformation

iL(t)

vL(t)

+

IL(S)

VL(S)

+

ZL=LS

22

- -

ic(t)

vc(t)

+

Ic(S)

Vc(S)

+

ZC(S)=1/CS

23

- -

ZC(S)=1/CS

Consider following arrangement, find out equivalenttransform impedance.

L

CLRT ZZZZ ++=

24

C

R

CsLsRZT

1++=

CLRT ZZZZ

1111 ++=

CCCC

LLLL

CsLsRZ T 1

1111 ++=

25

CCCC

R

2. Mechanical Systems2. Mechanical Systems

• Part-I: Translational Mechanical System

• Part-II: Rotational Mechanical System

Mechanical Systems

• Part-III: Mechanical Linkages

27

Translational

Linear Motion

Rotational

Rotational Motion

28

Basic Elements of Translational Mechanical Systems

Translational SpringTranslational SpringTranslational SpringTranslational Spring

i)i)i)i)

Translational MassTranslational MassTranslational MassTranslational Mass

ii)ii)ii)ii)ii)ii)ii)ii)

Translational DamperTranslational DamperTranslational DamperTranslational Damper

iii)iii)iii)iii)

30

A translational spring is a mechanical element thatcan be deformed by an external force such that thedeformation is directly proportional to the forceapplied to it.

Translational SpringTranslational SpringTranslational SpringTranslational Spring

i)i)i)i)

Circuit SymbolsCircuit SymbolsCircuit SymbolsCircuit Symbols

31

Translational SpringTranslational SpringTranslational SpringTranslational Spring

i)i)i)i)

If F is the applied force

Then is the deformation if

Or is the deformation.

2x1x

02 =x1x

)( 21 xx −F

Or is the deformation.

The equation of motion is given as

Where is stiffness of spring expressed in N/m

32

)( 21 xx −

)( 21 xxkF −=

k

F

Translational Mass

Translational MassTranslational MassTranslational MassTranslational Mass

ii)ii)ii)ii)

• Translational Mass is aninertia element.

• A mechanical systemwithout mass does notexist.exist.

• If a force FFFF is applied to amass and it is displaced toxxxx meters then the relationb/w force anddisplacements is given byNewton’s law.

M)(tF

)(tx

xMF ɺɺ=33

Common Uses of DashpotsCommon Uses of DashpotsCommon Uses of DashpotsCommon Uses of DashpotsDoor Stoppers

Vehicle Suspension

Bridge SuspensionFlyover Suspension

34

Translational Damper

= −F = b x

• Where bbbb is damping coefficient (N/msN/msN/msN/ms----1111).

1 2( )F b x x= −

35

Example-1: Consider a simple horizontal spring-masssystem on a frictionless surface, as shown in figurebelow.

or

36

kxxm −=ɺɺ

0=+ kxxm ɺɺ

Consider the following system (friction is negligible)

k

F

xM

37

• Free Body Diagram

MF

kfMf

• Where and are force applied by the spring and inertial force respectively.

kf Mf

• Then the differential equation of the system is:

MF

kfMf

Mk ffF +=

38

• Then the differential equation of the system is:

kxxMF += ɺɺ

• Taking the Laplace Transform of both sides and ignoring initial conditions we get

)()()( skXsXMssF += 2

)()()( skXsXMssF += 2

• The transfer function of the system is

kMssF

sX

+=

2

1

)(

)(

Example-2

39

• if

12000

1000−=

=

Nmk

kgM

2

00102 +

=ssF

sX .

)(

)(

• The pole-zero map of the system is

2

00102 +

=ssF

sX .

)(

)(

Example-2

40Pole-Zero Map

40-1 -0.5 0 0.5 1-40

-30

-20

-10

0

10

20

30

Real Axis

Imag

inar

y A

xis

Consider the following system

k

F

2x

M1x B

41

• Mechanical Network

↑ M

k

BF

1x 2x

• Mechanical Network

↑ M

k

BF

1x 2x

42

)( 21 xxkF −=

At node 1x

At node 2x

22120 xBxMxxk ɺɺɺ ++−= )(

43

44

).()()( 10 eq =−+−+ ioioo xxkxxbxm ɺɺɺɺ

2 eq. iiooo kxxbkxxbxm +=++ ɺɺɺɺ

45

Taking Laplace Transform of the equation (2)

)()()()()( skXsbsXskXsbsXsXms iiooo +=++2

kbsms

kbs

sX

sX

i

o

+++=

2)(

)(

Car Body

46

Car Body

Bogie-2

Bogie

Frame

Bogie-1

WheelsetsPrimary

Suspension

Secondary

Suspension

47