control system - chapter 2

7/17/2019 Control System - Chapter 2

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CHAPTER

MATHEMATICAL MODELLING OF

DYNAMIC SYSTEM

1



INTRODUCTION

• A mathematical model of a dynamic system isdefined as a set of equation ,i.e differential equation,that represent the dynamic of the system accuratelyor at least fairly well.

• Note that the a mathematical model is not unique toa given system and may be represent in manydifferent ways, therefore, may have manymathematical model depending on one’s

perspective.

• In this chapter we obtain mathematical models thatrelate the outputs of the system to its input.

2



INTRODUCTION (cont.)

• To develop a mathematical model we

need to use the fundamental physical law

of science and engineering.

– To model electrical networks we need apply

Ohm’s law & kirchhoff’s law.

– To model mechanical system we need applyNewton’s Law.

3




• The relation between the input and output of the

system can be described by differential

equation.

Output, c(t)Input, r(t)

System

Figure 2.1

4




• In practice, the complexity of the system

requires some assumptions in the

determination of the mathematical model.

• The equations of the mathematical model

may be solved using mathematical tools

such as the Laplace Transform.

5



2.1 Review of Laplace Transform

6



• The Laplace transform method substitutes relatively

easily solved algebraic equations for the more difficult

differential equations.

• The Laplace transformation for a function of time, is;

LAPLACE TRANSFORM (cont.)

where s= σ+ j ω, a complex variable. While j = −1 , σ =Re {s} (real part)and ω = Im {s} (imaginary part).

• The Eq. 2.1 allows us to convert f(t) F(s).

0

)()( dt et f s F st =£{f(t)} Eq 2.1

7




EXAMPLE Eq 2.1:

Find the Laplace transform of

Sol:

)()( t u Aet f at

t a s st at st )(

a s

A

e

a s

A t t a s

0)(

0 0 0

8



• Table 2.1 shows the conversion of f(t) to F(s) for specific cases derived

using Eq.2.1

Table 2.1: Laplace Transform Table

9




• To transform F(s) to f(t), the inverse Laplace transform is

used.

• The inverse Laplace transform is written as:

1 j st

• However Eq. 2.2 requires complex integration to find f(t)

given F(s)

• The easy way to find f(t) given F(s) is by using Laplace

Transform Table, Table 2.1.

2 j j

10

q .




Example 2.2 (Find f(t) given F(s) using table ILT):

Find the inverse Laplace transform of 21

3

1)(

s s F

Solution:

From laplace table

)()()3(

1

)()(

1

3

2

2

t f t ute s

There fo re

t utea s

t

at

11

at m

m e

m

t

a s

!)(

1

I L TT a b l eF ro m: N o te

1



2.3 Laplace transform theorems

12




Laplace Transform theorems, Table 2.2 is addition to the Laplace

transform table, Table 2.1 to assist in transforming between f(t) and

F(s).

13




Table 2.1: Laplace Transform Theorems

14




2.3.1 Partial- Fraction Expansion method• Used to find the inverse Laplace transform of a complicated

system.

• Have 4 case:

Case1: Roots of Denominator of F(s) are Real and Distinct

)(..........

)()()(

)..().........()(

)(

)(

)(

)(

21

21

n

n

s s

N

s s

B

s s

A s F

s s s s s s

sC

s R

sC

s F

15




Case2: Roots of Denominator of F(s) are Real and Repeated

r

n s s s s s s

sC

s R

sC s F

)..().........()(

)(

)(

)()(

21

r

n s s

N

s s

B

s s

A s F

)(..........

)()()(

21

Case3: Roots of Denominator of F(s) are Complex or imaginary

16




Case 4:

Partial Fraction can be used when the order numerator

N(s) is less than the order denominator D(s). If the order

numerator is greater than or equal to the order

denominator, then the numerator must be divided by

denominator successively until the result has a

remainder whose numerator is of order less than its

denominator.

17




Note:

• Example all of the cases partial-fraction

expansion are shown on extra note .

• Example of the application of Laplace

transform, inverse Laplace transform

and partial-fraction expansion for solving differential equation are shown

on extra note given. ( extra note 2 )18



MODELLING OF PHYSICAL

SYSTEM

19



2.4 Modeling of Electrical System

Elements

20



Modeling of Electrical System Elements

• A mathematical model of an electrical circuit can be

obtained by Kirchhoff’s laws.

Figure Electronic Components

21



Modeling of Electrical System Elements. (cont)

Example 1 :

Obtain the transfer function of the RC network shown below.

22




Solution:

Network equation

V1(t) = i(t)R + V2(t)

)()( 2 t dV C t I

By using laplace transform, transfer function obtained is,

/1

/1

1

1

1

1

)(

)()(

1

2

s s RCs sV

sV sG

where = RC, the time constant of the network

23




Example 2 :

Construct the block diagram of the system and obtain the transfer

function of the RLC network shown below.

Fig : Series RLC network

24




Solution

• The network equations are

» eq1

)()( t it dvC C

»

» eq 2

)()(

)()( t vdt

t di L Rt it v C

)()()()( t v Rt it vdt t di L C

25




Laplace transformed the eq. 1 and eq. 2, we get

eq

eq 4

)()( s I sCsVc

LsI(s) = V(s) – I(s)R – V C (s)

Draw block diagram of the RLC network using eq 3 and eq 4

26




we get, Block diagram of the series RLC network

+

-

1

Ls

1

Cs

-

V(s)V C (s)+ LsI(s) I(s)

I(s)R V C (s)

The transfer function of the system is:

1

1

)(

)(2

RCs LCs sV

sV c

27




Exercise

Construct a block diagram of this system in the Figure below and

obtain the transfer function .

)()(

)(2

0

R Ls RLCs

R

sV

sV

i

solution

28




Example: Refer to extra note 3

.

29



Modeling of Electrical System Elements.

Potentiometer

potentiometer is used to measure a linear or rotational

displacement. . There are two types of potentiometers which is;

1. rotational potentiometers

2. linear potentiometers.

Fig: Rotational potentiometer

30




Potentiometer . (cont.)

For a rotational potentiometer

e(t) α θ(t)

Therefore, e(t) = Kpθ(t); where Kp = Potentiometer constant V/rads

Laplace transform of the equation is

E(s) = K pθ(s)

Block diagram of the system

K pE(s)θ(s)

31




Tachometer .

Tachometer produce a voltage proportional to the angular velocity of the

generator shaft.

et(t) m(t)

et(t) = K k m(t)

Inverse laplace transform, we get,

E t (s) = K k

m(s)

Where, Kk = tachometer constant

K k

E t (s)m(s)

Fig: Block diagram of a tachometer 32




Operational Amplifier (Op-Amps)

The output of the amplifier

Vo = KsVi(t)

Where, Ks = amplifier gain, and Vi = amplifier input

Laplace transform of the equation is

Vo(s) = KsVi(s)

Block diagram of the system

K sV 0 (s)V i (s)

Fig: Block diagram of an Op Amp

33



2.5 MODELLING OF MECHANICAL

SYSTEM

34



MODELLING OF MECHANICAL SYSTEM

The motion of mechanical elements can

be described in various dimensions, which

are:

. .

2. Rotational.

3. Gear System.

35



Translational Mechanical ystem

36



Modeling of Translational Mechanical System.

The motion of translation is defined as a

motion that takes place along or curved .

describe translational motion are

acceleration, velocity, and displacement.

37




(cont.)

Figure below shows the translational mechanical system

Figure: Translational Mechanical Components38




(cont.)

Example:

Find the transfer function of X(s)/F(s) , for system shown below

Figure : Spring-Mass-Damper System

39




(cont.)

Solution

Draw the free-body diagram of a system and assume the mass is

traveling toward the right.

Figure 2.4 a. Free-body diagram of mass, spring, and damper system;

b. transformed free-body diagram

40




(cont.)

From free-body diagram, write differential equation of motion using

Newton’s Law. Thus we get;

)()()()(

2

2

t f t Kxdt

t dx f

dt

t xd M

v

By taking the Laplace transform

)()()()(2 s F s KX s sX f s X Ms v

Therefore the transfer function of system is;

K s f Ms s F

s X sG

v

2

1

)(

)()(

41




(cont.)

Spring-mass system

Example : Obtain the transfer function between Xo(s)/Xi(s)

42

How to solve: Refer to extra note 4




(cont.)

Spring-mass with viscous frictional damping

Example : Obtain the transfer function between Xo(s)/Xi(s)

43




Modeling of Translational Mechanical System. (cont.)

Obtain the transfer function between X22(s)/F(s)

44




45




46




Rotational Mechanical ystem

47



Rotational Mechanical System. (Cont)

The rotational motion can be defined as motion about a fixed axis.

The extension of Newton’s Law of motion for rotational motion states

that the algebraic sum of moments or torque about a fixed axis is

equal to the product of the inertia and the angular acceleration about

the axis where,

T = Torque

θ = Angular Displacement

ω = Angular Velocity

where Newton’s second law for rotational system are,

o nacceleratiangular where J T Torque :,)(

48



Modeling of Rotational Mechanical System. (Cont)

49




Obtain the transfer function between θ2(s)/T(s)

50




51




52




Obtain the transfer function between θ2(s)/T(s)


53




Gear system

54



Gear system

Gear system is a mechanical device that

transmit energy from one part of the

system to another in such away that force,

, ,

altered. These devices can also be

regarded as matching devices used to

attain maximum power transfer.

55



Gear system. (Cont.)

Figure below show two gears are coupled together

Where,

T = torque

θ = angular displacement

N = teeth number

56




from front view

57




For ideal case, inertia and friction are neglected. The

relationship between T1 and T2, θ1 and θ2 and N1 and N2

are derived from the following facts;

1. The number of teeth on the surface of the gears is proportional

to the radius r 1 and r 2 of the ears that is

r 1N2 = r 2N1

2. The distance traveled along the surface of each gear is the same,

thus,

θ1r 1 = θ2r 2

3. The work done by one gears is same to that other since there areassumed to be no losses. thus,

T1θ1 = T2θ2

58




If the angular velocities of two gears ω1 and ω2 are brought into the

picture, Therefore

1

2

2

1

2

1

NT

59




60



Find transfer function between θ 1(s)/T 1(s)


61How to solve: Refer to extra note 6



Find transfer function between θ 2 (s)/T 1(s)


62





63



Find transfer function between θ 1(s)/T 1(s)


64



Find transfer function between θ 2 (s)/T 1(s)


65




2.6 Modelin of Electromechanical S stem Elements.

66



Modeling of Electromechanical System Elements.

(cont.)

Direct Current (DC) Motor

• Applications of DC motor e.g. tape drive, disk drive,

printer, CNC machines, and robot.

• The DC motor is represented by a circuits shown on

the next slide.

67




(cont.)

Magnetic flux

R Li a(t)

eb(t)v a(t)

Armature circuit

T L(t)

J m,

T m(t)

m

m(t)

rigid

Load

J L

68

BL

Bm




(cont.)

Let:

Ra = armature resistance

La = armature inductance

ia(t) = armature current m

Va(t) = armature input voltage

eb(t) = back emf

ωm(t) = motor angular velocity

θm(t) = motor angular displacementJm= moment of inertia of motor + loadBm=viscous frictional constant of motor + load

69




(cont.)

Two method to control DC motor, which are:

1. Armature control

2. Field control

70




(cont.) (Armature control)

Field current is fix and variable voltage

is applied to armature circuit.

For the given DC servo unit , find θθmm(s)/(s)/VVaa(s),(s), assume thatassume that KKtt is torqueis torque

constant andconstant and KKbb isis emf emf constantconstant..

71




(cont.) (Armature control)

There are 3 conditions of armature, which

are:

. n- oa con on

2. No-load condition

3. Combining with gear system

72




(cont.) (Armature control)(In load Condition)(In load Condition)

a) Find Armature circuit equation

s

b

73



b) Find force equation applied to the motor



L

2

L

74



)3()s(IK )s(T atm

3) Find relationship between torque(TTmm(s)(s)) and armature current(IIaa(s)(s))



4) Find relationship between motor emf(EEbb(s)(s)) and motor shaft velocity (ωωmm(s(s))))

)4()s(sK )s(E mbb

75



s

b

So now, we have four equations which are:

L

2

L



)3()s(IK )s(T atm

)4()s(sK )s(E mbb

From (3) and (2)

K

s

t

L

2

L

a

76



Assume :

Insert (4) and (5) to equation (1)

s

K

s

m

t

L

2

L

a



s

L

2

L

So that :

s

K

Z

b

t

a

m

sK K Z

K

)s(V

)s(

bta

t

a

m

77



t

a

m

K s

K

ANSWER :ANSWER :



t

a

m

K s

K

78



For No-load condition and Combining with


(cont.) (Armature control)(no load Condition)(no load Condition)

gear sys em p ease see e ex ra no e .

79



ExerciseExercise Armature control

Find θL(s)/Va(s)

L(t)

J m , Bm

T m(t)

m(t)

m(t)

Load

J L, BL

N1

N2

b

2

2

1

L

2

2

1

L

t

2

1

a

L

K

N

N

B

N

N

J s

K

N

N

ANSWER :ANSWER :

80



Exercise

Find transfer function between θ L(s)/E a(s)

81

2

2

1

2

2

1

2

1

)()(

)(

N

N B B Deq

N

N J J Jeq

KtKb Deq Jeqs Ra s

Kt N

N

s Ea

s

La

La

L

Ans:




(cont.) (Field control)

Armature current is fix and variable

voltage is applied to field circuit.

For the given DC servo unit , find

mm f f ,, f f

constantconstant..

82





83





1) Find Field circuit equation

f

84





2) Find force equation applied to the motor

L

2

L

m

2

m

or or

wherewhere

Lam JJJ Lam BBB andand85





)3()s(IK )s(T f f m

3) Find relationship between torque(TTmm(s)(s)) and field current(IIf f (s)(s))

86





m

2

m

So now, we have three equations which are:

)3()s(IK )s(T f f m

f

From (3) and (2)

K

s

f

m

2

m

f

From (4) and (1)

K

s

f

f

m

2

m

f

87



2

m

f

f

m

B

K

ANSWER :ANSWER :



2

m

f

f

m

B

K

Where :Where :

Lam

JJJ

Lam BBB

88



2.7 Modeling of speed and Position Control System

89



Modeling closed loop Position Control System

90




(cont.)

Figure below shows the closed loop position control system.

91




(cont.)

The objective of this system is to control the position of the

mechanical load in according with the reference position

The operation of this system is as follows:-

A pair of potentiometers acts as an error-measuring device. For input potentiometer, vi(t) = kpθi(t)

For the output potentiometer, vo(t) = kpθo(t)

The error signal,

Ve(t) = Vi(t) – Vo(t)

= kpθi(t) - kpθo(t) --------------------( 1

This error signal are amplified by the amplifier with gain constant, Ks.

Va(t) = K s Ve(t) -----------------------------( 2 )

92




(cont.)

Transforming equations ( 1 ) and ( 2 ):-

Ve(s) = Kpθi(s) - Kpθo(s) -----------------( 3 )

By using the mathematical models developed previously for motor and

gear the block diagram of the position control system is shown below:-

93




(cont.)

Transfer function

Let

bt eqeq

t

a

m

K K B s J R Ls

K

sV

s

))(()(

)(

11

J J ieq and B Bieq

Therefore

bt s

t

a

m

K K B Js R L

K

sV

s

))(()(

)(

For small motors, the armature inductance, L is small and can be neglected

Where, L = 0

94




(cont.)

Therefore,

)()(

)(

bt

t

a

m

K K BR sRJ

K

sV

s

K t

1

s K K BR

RJ

bt

bt

1

sm

m

where,

bt

t m

K K BR K K

bt

m K K BR

RJ

= motor again

= motor time constant

95




(cont.)

The simplified block diagram.

where,

sT

K

sV

s

m

m

a

m

1)(

)(

From the block diagram we get,

)1()(

sT s

n K K K sG

m

m s p

96




(cont.)

Let, n K K K K m s p

)1()(

sT s

K sG

m

so,

,

)1(1

)1(

)(

)(

sT s

K

sT s

K

s

s

m

m

i

o

mm

m

m

m

T

K s

T s

T

K

K s sT

K

K sT s

K

1

)1(

22

97




(cont.)

From the equation above, if θi(s) given, therefore,

θo(t)=

K s sT

s K

m

i

2

1 )(

£

From that, the value K are not fixed and can be change,

therefore the system response are depends on the value K.

98



Modeling closed loop Position Control sys em w ve oc y ee ac

99

Modeling closed loop Position Control system with



Modeling closed loop Position Control system with

velocity feedback

(cont.)

A tachometer is coupled to the motor shaft so that a voltage signal

which is proportional to motor speed can be feedback as shown

below

100

Modeling closed loop Position Control system




with velocity feedback

(cont.)

From block diagram above, the transfer function for

g m sm

m s

g m s

m

m s

e

m

K K K sT

K K

K K K

sT

K K

sV

s

11

)1(

)(

)(

m sT 1

Therefore the forward path transfer function of the system is,

)1()1(

)(

g m sm g m sm

m s p

K K K sT s

K

K K K sT s

n K K K sG

101





with velocity feedback

(cont.)

Therefore the transfer function of system is,

)1(1

)1(

)(

)(

g m sm

g m sm

i

o

K K K sT s

K

K

K K K sT s

K

s

s

g m sm s s

K s K K K s

K

g m sm

)1(2

102



Modeling closed loop Velocity Control System

103

M d li l d l V l it C t l S t




(cont.)

Figure below shows Schematic Diagram for speed control system.

K g

104

K s

R L

m(t)

Tachometer

+E

-E

K p

v e(t)v a(t)

v i (t)





(cont.)

• Input Voltage (Input Velocity)

• Precision control of the angular velocity of an inertia load driven directly by an

armature controlled dc motor can be achieved by comparing an input voltage,

Vi representing a demanded speed and derived from a source of constant

voltage Vs via a potentiometer with a feedback voltage Vg derived from a

iiiV

tachometer coupled to the motor shaft.

)()()( t V t V t V g ie

Therefore we get,

and,

)()( t V K t V e sa

105





(cont.)

The block diagram of the system are shown below.

106





(cont.)

Transfer function for

bt mm

t

a

m

K K B s J Ls R

K

sV

s

))(()(

)(

For small motors, the armature inductance L can be neglected where L=0

,

bt mm

t

a

m

K K B s J R

K

sV

s

)()(

)(

)( bt mm

t

K K RB s RJ

K

1

s K K RB

RJ

K K RB

K

bt m

m

bt m

t

107





(cont.)

Let

m

bt m

t K K K RB

K

= motor gain

m

bt m

m

K K RB

= motor time constant

Therefore,

m

m

a

m

s K

sV s

1)()(

108





(cont.)

The overall transfer function

g m s

m

m s

p

i

m

K K K

s

K K

K

s

s

1

)(

)(

m s 1

)1()(

)(

g m sm

m s p

i

m

K K K s

K K K

s

s

109



Did u know what is Speed Regulation?

(refer to extra note 8)

110



SEKIAN TERIMA KASIHEKIAN TERIMA KASIH

control system - chapter 2

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