lecture- 2 introduction mathematical modeling mathematical...

Automatic Control Systems

Lecture- 2Introduction Mathematical Modeling

Mathematical Modeling of Mechanical Systems

1

Lecture Outline

• Introduction to Modeling

– Ways to Study System

– Modeling Classification

• Mathematical Modeling of Mechanical Systems

– Translational Mechanical Systems

– Rotational Mechanical Systems

– Mechanical Linkages

2

Model

• A model is a simplified representation or

abstraction of reality.

• Reality is generally too complex to copy exactly.

• Much of the complexity is actually irrelevant in

problem solving.

3

What is Mathematical Model?

A set of mathematical equations (e.g., differential eqs.) thatdescribes the input-output behavior of a system.

What is a model used for?

• Simulation

• Prediction/Forecasting

• Prognostics/Diagnostics

• Design/Performance Evaluation

• Control System Design

Ways to Study a System

5

System

Experiment with actual System

Experiment with a model of the System

Physical Model Mathematical Model

Analytical Solution

Simulation

Frequency Domain Time Domain Hybrid Domain

Black Box Model

• When only input and output are known.

• Internal dynamics are either too complex orunknown.

• Easy to Model

6

Input Output

Basic Types of Mechanical Systems

• Translational

– Linear Motion

• Rotational

– Rotational Motion

7

Basic Elements of Translational Mechanical Systems

Translational Spring

i)

Translational Mass

ii)

Translational Damper

iii)

Translational Spring

i)

Circuit Symbols

Translational Spring• A translational spring is a mechanical element that

can be deformed by an external force such that thedeformation is directly proportional to the forceapplied to it.

Translational Spring

Translational Spring• If F is the applied force

• Then is the deformation if

• Or is the deformation.

• The equation of motion is given as

• Where is stiffness of spring expressed in N/m

2x1x

02 x1x

)( 21 xx

)( 21 xxkF

k

F

F

Translational Mass

Translational Mass

ii)

• Translational Mass is an inertiaelement.

• A mechanical system withoutmass does not exist.

• If a force F is applied to a massand it is displaced to x metersthen the relation b/w force anddisplacements is given byNewton’s law.

M)(tF

)(tx

xMF

Translational Damper

Translational Damper

iii)

• Damper opposes the rate ofchange of motion.

• All the materials exhibit theproperty of damping to someextent.

• If damping in the system is notenough then extra elements (e.g.Dashpot) are added to increasedamping.

Common Uses of Dashpots

Door StoppersVehicle Suspension

Bridge SuspensionFlyover Suspension

Translational Damper

xCF

• Where C is damping coefficient (N/ms-1).

)( 21 xxCF

Example-1

• Consider the following system (friction is negligible)

15

• Free Body Diagram

MF

kf

Mf

k

F

xM

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

kf Mf

Example-1

16

• Then the differential equation of the system is:

xMkxF

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

MF

kf

Mf

Mk ffF

)()()( skXsXMssF 2

17

)()()( skXsXMssF 2

• The transfer function of the system is

kMssF

sX

2

1

)(

)(

• if

12000

1000

Nmk

kgM

2

00102

ssF

sX .

)(

)(

Example-1

18

• The pole-zero map of the system is

2

00102

ssF

sX .

)(

)(

Example-2

-1 -0.5 0 0.5 1

0

𝑗 2

Pole-Zero Map

Real Axis

Ima

gin

ary

Axis

−𝑗 2

Example-2

• Consider the following system

19

• Free Body Diagram

k

F

xM

C

MF

kf

Mf

Cf

CMk fffF

Example-3

20

Differential equation of the system is:

kxxCxMF

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

)()()()( skXsCsXsXMssF 2

kCsMssF

sX

2

1

)(

)(

Example-3

21

kCsMssF

sX

2

1

)(

)(

• if

1

1

1000

2000

1000

msNC

Nmk

kgM

/

1000

00102

sssF

sX .

)(

)(-1 -0.5 0 0.5 1

-2

-1.5

-1

-0.5

0

0.5

1

1.5

2

Pole-Zero Map

Real Axis

Imagin

ary

Axis

Example-4

• Consider the following system

22

• Mechanical Network

k

F

2x

M

1x B

↑ M

k

BF

1x 2x

Example-4

23

• Mechanical Network

↑ M

k

BF

1x 2x

)( 21 xxkF

At node 1x

At node 2x

22120 xBxMxxk )(

Example-5

• Find the transfer function X2(s)/F(s) of the following system.

1M 2M

k

B

Example-6

25

k

)(tf

2x

1M4B3B

2M

1x

1B2B

↑ M1k 1B)(tf

1x 2x3B

2B M24B

Example-7

• Find the transfer function of the mechanical translationalsystem given in Figure-1.

26

Free Body Diagram

Figure-1

M

)(tf

kf

Mf

Bf

BMk ffftf )(kBsMssF

sX

2

1

)(

)(

Example-8

27

• Restaurant plate dispenser

Example-9

28

• Find the transfer function X2(s)/F(s) of the following system.

Free Body Diagram

M1

1kf

1Mf

Bf

M2

)(tF

1kf

2Mf

Bf2kf

2k

BMkk fffftF 221

)(

BMk fff 11

0

Example-10

29

1k

)(tu

3x

1M

4B3B

2M

2x

2B 5B

2k 3k

1x

1B

Basic Elements of Rotational Mechanical Systems

Rotational Spring

)( 21 kT

21

Basic Elements of Rotational Mechanical Systems

Rotational Damper

21

)( 21 CT

T

C

Basic Elements of Rotational Mechanical Systems

Moment of Inertia

JT

TJ

Example-11

1

T 1J

1k1B

2k

2J

2 3

↑ J1

1k

T

1 31B

J2

2

2k

Example-12

↑ J1

1k

1BT

1 32B

3B J24B

2

1

T 1J

1k

3B

2B4B

1B

2J

2 3

Example-13

1T

1J

1k

2B 2J

22k

Example-14

Gear

• Gear is a toothed machine part, suchas a wheel or cylinder, that mesheswith another toothed part totransmit motion or to change speedor direction.

37

Fundamental Properties• The two gears turn in opposite directions: one clockwise and

the other counterclockwise.

• Two gears revolve at different speeds when number of teethon each gear are different.

Gearing Up and Down

• Gearing up is able to convert torque tovelocity.

• The more velocity gained, the more torquesacrifice.

• The ratio is exactly the same: if you get threetimes your original angular velocity, youreduce the resulting torque to one third.

• This conversion is symmetric: we can alsoconvert velocity to torque at the same ratio.

• The price of the conversion is power loss dueto friction.

Why Gearing is necessary?

40

• A typical DC motor operates at speeds that are far too

high to be useful, and at torques that are far too low.

• Gear reduction is the standard method by which a

motor is made useful.

Gear Trains

41

Gear Ratio• You can calculate the gear ratio by using

the number of teeth of the driverdivided by the number of teeth of thefollower.

• We gear up when we increase velocityand decrease torque.Ratio: 3:1

• We gear down when we increase torqueand reduce velocity.Ratio: 1:3

Follower

Driver

𝐺𝑒𝑎𝑟 𝑟𝑎𝑡𝑖𝑜 =𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑡𝑒𝑒𝑡ℎ 𝑜𝑓 𝑖𝑛𝑝𝑢𝑡 𝑔𝑒𝑎𝑟

𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑡𝑒𝑒𝑡ℎ 𝑜𝑓 𝑜𝑢𝑝𝑢𝑡 𝑔𝑒𝑎𝑟=

𝐼𝑛𝑝𝑢𝑡 𝑇𝑜𝑟𝑞𝑢𝑒

𝑂𝑢𝑝𝑢𝑡 𝑇𝑜𝑟𝑞𝑢𝑒=𝑂𝑢𝑡𝑝𝑢𝑡 𝑆𝑝𝑒𝑒𝑑

𝐼𝑛𝑝𝑢𝑡 𝑆𝑝𝑒𝑒𝑑

Example of Gear Trains• A most commonly used example of gear trains is the gears of

an automobile.

43

Mathematical Modeling of Gear Trains

• Gears increase or descrease angular velocity (whilesimultaneously decreasing or increasing torque, suchthat energy is conserved).

44

2211 NN

1N Number of Teeth of Driving Gear

1 Angular Movement of Driving Gear

2N Number of Teeth of Following Gear

2 Angular Movement of Following Gear

Energy of Driving Gear = Energy of Following Gear

Mathematical Modelling of Gear Trains

• In the system below, a torque, τa, is applied to gear 1 (withnumber of teeth N1, moment of inertia J1 and a rotational frictionB1).

• It, in turn, is connected to gear 2 (with number of teeth N2,moment of inertia J2 and a rotational friction B2).

• The angle θ1 is defined positive clockwise, θ2 is defined positiveclockwise. The torque acts in the direction of θ1.

• Assume that TL is the load torque applied by the load connectedto Gear-2.

45

B1

B2

N1

N2

Mathematical Modelling of Gear Trains

• For Gear-1

• For Gear-2

• Since

• therefore

46

B1

B2

N1

N2

2211 NN

11111 TBJa Eq (1)

LTBJT 22222 Eq (2)

12

12

N

N Eq (3)

Mathematical Modelling of Gear Trains

• Gear Ratio is calculated as

• Put this value in eq (1)

• Put T2 from eq (2)

• Substitute θ2 from eq (3)

47

B1

B2

N1

N2

22

11

1

2

1

2 TN

NT

N

N

T

T

22

11111 T

N

NBJa

)( La TBJN

NBJ 2222

2

11111

)( La TN

N

N

NB

N

NJ

N

NBJ

2

12

2

121

2

12

2

11111

Mathematical Modelling of Gear Trains

• After simplification

48

)( La TN

N

N

NB

N

NJ

N

NBJ

2

12

2

121

2

12

2

11111

La TN

NB

N

NBJ

N

NJ

2

112

2

2

11112

2

2

111

La TN

NB

N

NBJ

N

NJ

2

112

2

2

1112

2

2

11

2

2

2

11 J

N

NJJeq

2

2

2

11 B

N

NBBeq

Leqeqa TN

NBJ

2

111

Mathematical Modelling of Gear Trains

• For three gears connected together

49

3

2

4

3

2

2

12

2

2

11 J

N

N

N

NJ

N

NJJeq

3

2

4

3

2

2

12

2

2

11 B

N

N

N

NB

N

NBBeq

Example-15

• Drive Jeq and Beq and relation between appliedtorque τa and load torque TL for three gearsconnected together.

50

J1 J2 J3

1

3

2

τa

1N

2N

3N

1B2B

3B

LT