Control System Modelling Case Studies_2014

8/11/2019 Control System Modelling Case Studies_2014

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2

Chapter Introduction

Other Considerations

The three main objectives of control system analysis and design have already been

enum erated. However, other important considerations must be taken into account. For

example, factors affecting hardware selection, such as motor sizing to fulfill power

requiremen ts and choice of sensors foraccuracy, must be considered early in the design.

Finances are another consideration. Control system designers cannot create

designs without considering their economic impact. Such considerations as budget

allocations and competitive pricing must guide the engineer. For example, if your

product is one of a kind, you may be able to create a design that uses more expensive

components w ithout appreciably increasing totalcost.However, if your design will be

used for many copies, slight increases in cost per copy can translate into m any more

dollars for your company to propose during contract bidding and to outlay before sales.

Another consideration is robust design. System parameters considered con

stant during the design for transient response, steady-state errors, and stability

change over time when the actual system is built. Thus, the performance of the

system also changes over time and will not be consistent with your design. Un

fortunately, the relationship between parameter changes and their effect on per

formanceisnot linear. In some cases, even in the same system, changes in p arameter

values can lead to small or large changes in performance, depending on the system s

nominal operating point and the type of design used. Thus, the engineer wants to

create a robust design so that the system will not be sensitive to parameter changes.

We discuss the concept of system sensitivity to pa ram eter changes in Chapters7and

8. This concept, then, can be used to test a design for robustness.

Introduction to a Case Study

Now that our objectives are stated, how do we meet them? In this section we will

look at an example ofafeedback control system. The system introduce d he re will

be used in subsequent chapters as a running case study to demonstrate the

objectives of those chapters. A colored background like this will identify the

case study section a t the end of each chapter. Section 1.5, which follows this first

case study, explores the design process that will help us build our system.

Antenna Azimuth: An Introduction to Position Control Systems

A position control system converts a position input command to a position output

response. Position control systems find widespread applications in antennas, robot

arms, and computer disk drives. The radio telescope antenna in Figure 1.8 is one

example of a system thatusesposition control systems. In this

section,

we willlook in

detail at an antenna azimuth position control system that couldbeused to position a

radio telescope an tenna. We will see how the system works and how we can effect

changes in its performance. The discussion he re willbe on aqualitative level, with the

objective of getting an intuitive feeling for the systems with whichwe willbe dealing.

An antenna azimuth position control system is shown in Figure

1.9 a),

with a

more detailed layout and schematic in Figures 1.9 6) and

1.9 c),

respectively.

Figure

1.9 d)

shows

afunctional blockdiagram

of the system. The functions are

shown above the blocks, and the requ ired hardw are is indicated inside the blocks.

Parts of Figure 1.9 are repeated on the front endpapers for future reference.

FIGURE1.8 Thesearch for

extraterrestrial lifeis being

carried outwithradio antennas

like the

one

pictured

here.

A

radio antennais anexampleof

a system with position

controls.


3/9

Case Study

The purpose of this system is to have the azimuth angle output of the antenna,

9

0

t), follow th e input an gle of the poten tiom eter, 0,-(f). Let us look at Figure1.9 d)

and describe how this system works. The input command is an angular displace

ment. The potentiometer converts the angular displacement into a voltage.

Antenna

Desired

azimuth ang

input

a)

Potentiometer

Antenna

Differential amplifier

and power amplifier

Potentiometer

Potentiometer

Amplifiers

Motor

Differential

and

power

amplifier

K

Armature

resistance

Armature

Fixed field

Gear

Gear

0,, 0

Potentiometer-.

Inertia Viscous

damping

Gear

0

FIGURE1.9 Antenn a azimuth

position control system:

a. system concept;

b.

detailed

layout; c. schematic;

{figure continues


4/9

14

Chapter1 Introduction

FIGURE1.9

Continued)

d. functional block diagram

Input

transducer

Angular

input

Potentiometer

Voltage Error

proportional Summing or

t 0

junction Actuating

input + ,


5/9

1.5 Th e Design Proc ess

We have discussed the transient response of the position control system. Let us

now direct our attention to the steady-state position to see how closely the output

matches the input after the transients disappear.

We define steady -state error as the difference b etwe en the input and the ou tput

after the transients have effectively disappeared. The definition holds equally well

for step, ram p, and oth er types of inputs. Typically, the steady-state error decreases

with an increase in gain and increases with a decrease in gain. Figure 1.10 shows

zero error in the steady-state response; that is, after the transients have disap

peared, the outp ut posit ion equals the comm anded in put posit ion. In some systems,

the steady -state error will not b e zero; for these systems, a simple gain adjustm ent

to regulate the transient response is either not effective or leads to a trade-off

between the desired transient response and the desired steady-state accuracy.

To solve this prob lem , a controller with a dynam ic response, such as an electrical

filter, is used along with an amplifier. With this type of controller, it is possible to

design both the required transient response and the required steady-state accuracy

without the trade-off required by a simple setting of gain. However, the controller

is now m ore com plex. The filter in this case is called a comp ensator. Man y system s

also use dynamic elements in the feedback path along with the output transducer to

improve system performance.

In summary, then, our design objectives and the system s performa nce revolve

around the transient response, the steady-state error, and stability. Gain adjust

ments can affect performance and sometimes lead to trade-offs between the

performance criteria. Compensators can often be designed to achieve performance

specifications without the need for trade-offs. Now that we have stated our

objectives and some of the methods available to meet those objectives, we describe

the orderly progression that leads us to the final system design.

l 5 The Design Process

In this section, we establish an orderly sequence for the design of feedback control

systems that will be followed as we progress thro ugh th e rest of the boo k. Figure 1.11

shows the desc ribed process as well as the ch apters in which the steps are discussed.

The antenna azimuth position control system discussed in the last section is

representative of control systems that must be analyzed and designed. Inherent in

Step Step 2

Step 3 Step 4

Step 5

Determine

a physical

system and

specifications

from the

requirements.

Draw a

functional

block

diagram.

Transform

the physical

system into

a schematic.

Analog: Chapter 1

Digital:

FIGURE 1 .11 Th e con t ro l sys tem des ign p roce ss

Use the

schematic

to obtain a

block diagram,

signal-flow

diagram,

or state-space

representation.

Chapters 2,3

Chapter 13

If multiple

blocks, reduce

the block

diagram to a

single block or

closed-loop

system.

Chapter 5

Chapter 13

Step 6

Analyze,

design, and test

to see that

requirements

and

specifications

are met.

Chapters 4, 6-12

Chapter 13


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9 4

Chapter 2 Modeling in the Frequency Dom ain

Case Studies

Antenna Control: Transfer Functions

This chapter showed that physical systems can be modeled mathematically with

transfer functions. Typically, systems are composed of subsystems of different

types, such as electrical, mechanical, and electromechanical.

The first case study uses our ongoing example of the antenna azimuth position

contro l system to show how to represent each subsystem as a transfer function.

PROBLEM: Find the transfer function for each subsystem of the antenna

azimuth position control system schematic shown on the front endpapers. Use

Configuration 1.

SOLUTION: First, we identify the individual subsystems for which we must find

transfer functions; they are summarized in Table 2.6. We proceed to find the

transfer function for each subsystem.

TABLE 2.6

Subsystems of the anten na azimuth position control system

Subsystem

Input potent iometer

Pre a mp

Power amp

Motor

Output potent iometer

Input

Angular rotation from user, #,(*)

Voltage from potentiom eters ,

v

e

(t) = v,{t) - v

0

{t)

Voltage from preamp,

v

p

{t)

Voltage from power amp, e

a

(t)

Angular rotation from load,0Q(()

Output

Voltage to pream p,

Vj(t)

Voltage to power amp,

v

p

(t)

Voltage to motor,

e

l

t)

Angular rotation to load,

0o(t)

Voltage to preamp,

VQ(0

Input Potentiometer; Output Potentiometer

Since the input and output potentiome ters are configured in the same way, their

transfer functions will be the same.

We

neglectthe dynamics for the potentiometers

and simply find the relationship between the output voltage and the input angular

displacement. In the center position the output voltage is zero. Five turns toward

either the positive 10 volts or the negative 10 volts yields a voltage change of 10

volts. Thus, the transfer function, V,-

(s)/0,;(s),

for the potentiome ters is found by

dividing the voltage change by the angular displacement:

Vt s)

m

10

lOJr

2.202)

Preamplifier; Power Amplifier

The transfer functions of the amplifiers are given in the problem statement. Two

phenomena are neglected. First, we assume that saturation is never reached.

Second, the dynamics of the preamplifier areneglected,since its speed of response

is typically much greater than that of the power amplifier. The transfer functions of

both amplifiers are given in the problem statement and are the ratio of the Laplace

transforms of the output voltage divided by the input voltage. Hence, for the


7/9

Case Studies

95

preamplifier,

and for the power amplifier,

E

a

(s)

V

P

(s)

I)

100

5

+ 100

2.203)

2.204)

Motor and Load

The motor anditsload are next. The transfer function relating the armature displace

ment to the armature voltage is given in Eq. (2.153). The equivalent inertia, /,,is

J - ^ + ' r f J g )

=

0.02

1 ^

=

0.03

2.205)

where

JL

= lis the load inertia at9 .The equivalent viscous damping, D

m

, at the

armature is

D

B f l

+ 1 > i

( p ) = 0.01 + 1 ^ = 0.02 (2.206)

where

D

L

is the load viscous damping at9Q.From the problem statement,

K

t

= 0.5

N-m/A,Kb

=

0.5 V-s/rad, and the arm ature resistance

R

a

=8

ohms. These quantit

ies along with

J

m

and

D

m

are substituted into Eq. (2.153), yielding the transfer

function of the mo tor from the armature voltage to the arma ture displacement, or

9

m

{s) _ K

t

/{R

a

J

m

)

2.083

E

a

(s)

_L

n

K

Kb

s+ [D

m

+

5(5 + 1.71)

T l J?

To complete the transfer function of the motor, we multiply by the gear ratio to

arrive at the transfer function relating load displacement to armature voltage:

Oo(s) 6

m

(s)

0 2083

EW =

0A

EM-4^vn)

(2

-

208)

The results are summarized in the block diagram and table of block diagram

parameters (Configuration 1) shown on the front endpapers.

CHALLENGE: We now give you a problem to test your knowledge of this chapter's

objectives; Referring to the antenna azimuth position control system schematic

shown on the front endpapers, evaluate the transfer function of each subsystem.

Use C onfiguration

2.

Record y our results in the table of block diagram param eters

shown on the front en dpapers forusein subseq uent chap ters' case study challenges.

Transfer Function of a Human Leg

In this case study we find the transfer function of a biological system. The system is

a hum an leg, which pivots from the hip joint. In this problem, the component of

weight is nonlinear, so the system requires linearization before the evaluation of

the transfer function.


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96

Chapter 2 Model ing in the Frequency Dom ain

Hip

joint

FIGURE 2. 51 Cy lin de r mo de l of a

human leg .

M

D

Tt

Tntt)

PROBLEM : Th e t r a ns fe r func t ion o f a hu m a n l e g re l a t e s t he ou tpu t a n gu la r

ro t a t ion a bou t t he h ip jo in t t o t he inpu t t o rque s upp l i e d by the l e g mus c le . A

s impl i f i ed m od e l fo r t he l e g is s ho wn in F igu re 2 .51 . Th e mo de l

assumes

a n

a p p l i e d m u s c u l a r t o r q u e ,

T

m

(t),

v i s c o u s d a m p i n g ,

D,

a t t h e h ip jo in t , a n d

ine r t i a ,

J,

a r o u n d t h e h i p j o i n t .

1 5

Als o , a c ompone n t o f t he we igh t o f t he l e g ,

Mg,

w h e r e

M

is the mass of the leg and

g

i s t he a c c e le ra t ion due to g ra v i ty ,

c r e a t e s a n o n l i n e a r t o r q u e . I f w e

assume

tha t the leg is of un ifo rm dens i ty ,

the we igh t c a n be a pp l i e d a t L /2 , wh e re L i s t he l e n g th o f t he l e g

(Milsum,

1966).

D o the fo l lowing :

a . E v a l u a t e th e n o n l i n e a r t o r q u e .

b .

F ind the t r a ns fe r func t ion ,

9(s)/T

m

(s),

fo r s ma l l a ng le s o f ro t a t ion ,

w h e r e

9{s)

i s t he a ngu la r ro t a t i on o f t he l e g a bo u t t he h ip jo in t .

SOLUTION: Firs t , ca lc ula t e the to rq ue du e to the weight . Th e to t a l weig ht of

the l e g is

M g

a c t ing ve rt i c a l ly . Th e c o m po ne n t o f t h e we igh t i n the d i re c t io n

o f ro t a t ion i s

Mg

s in

9.

Th i s fo rc e is a pp l i e d a t a d i s t a nc e L / 2 f rom the h i p

j o i n t . H e n c e t h e t o r q u e i n t h e d i r e c t io n o f r o t a t i o n ,

Tw(t), is Mg(L/2)

si n

9.

Ne xt , d ra w a f re e -bod y d ia g r a m o f the l e g , s how ing the a pp l i e d to rq ue ,

T

m

{t), t h e t o r q u e d u e t o t h e w e i g h t , T

w

(t), a n d t h e o p p o s i n g t o r q u e s d u e t o

ine r t i a a nd v i s c ous da mping ( s e e F igu re 2 .52 ) .

S u m m i n g t o r q u e s , w e g e t

T

w

(t)

J

&9_

dt

2

d9 I

D-^ +

Mg-sm9 = T

m

(t)

(2 .209)

FIGURE 2.52 Free-body diagram of W e l ine arize th e sys tem a bo ut th e eq ui l ibr ium p oin t ,

9 = 0,

the ve r t i c a l

l eg mode l pos i t i o n o f t h e l e g . Us i ng Eq . (2 .182 ) , we ge t

s i n # - s i n 0 = ( co s0 )< 5# ( 2 .2 1 0 )

f rom whic h , s in

9 = 89.

A l s o ,

J d

2

9/dt

2

= J d

2

89/dt

2

a n d

D d9/dt = D d89/dt.

H e n c e E q . (2 . 2 09 ) b e c o m e s

r

d

2

89

n

d8 9

mr

L _ , , ,

(2 .211)

N o t i c e t h a t t h e t o r q u e d u e t o t h e w e i g h t a p p r o x i m a t e s a s p r i n g t o r q u e o n t h e l e g .

Ta k ing the La p la c e t r a ns fo rm wi th z e ro in i t i a l c ond i t ions y i e ld s

Js

2

+ Ds + Mg^\89{s) = T

m

(s)

(2 .212)

f rom whic h the t r a ns fe r func t ion i s

89(s) 1/7

T

m

(s)

9

2

+%S +

MgL

2 /

(2 .213)

15

For em phasis,J is not around the center of mass, as we previously assumed for inertia in mechanical

rotation.


9/9

Review Questions

9 7

for small excursions about the equilibrium point,9= 0.

CHALLENGE:

We

now introduce a case study challenge to test your

knowledge of this chapter's objectives. Although the physical

system is different from a human leg, the problem demonstrates

the same principles: linearization followed by transfer function

evaluation.

Given the no nlinear electrical network shown in Figure

2.53,

find

the transfer function relating the output nonlinear resistor voltage,

V

r

s), to the input source voltage, V(s).

v

r

O =

2/: /)

FIGURE2.53 No nlin ear electric circuit

^ Summary ^

In this chapter, we discussed how to find a mathematical model, called a transfer

function,

for linear, time-invariant electrical, mechanical, and electromechanical

systems. The transfer function is defined as G(s) = C (s)/R(s), or the ratio of the

Laplace transform of the ou tput to the Laplace transform of the input. This relation

ship is algebraic and also adapts itself to modeling interconnected subsystems.

We realize that th e physical world consists of more systems than we illustrated

in this chap ter. For example, we could apply transfer function modeling to hydrau lic,

pneumatic, heat, and even economic systems. Of course, we must assume these

systems to be linear, or make linear approximations, in order to use this modeling

technique.

Now that we have our transfer function, we can evaluate its response to a

specified inpu t. System respon se

will

be covered in Chapter

4.

For those pursuing the

state-space approach, we continue our discussion of modeling in C hapter 3, where

we use the time domain rather than the frequency domain.

Review Questions ^

1. What mathematical model permits easy interconnection of physical systems?

2.

To what classification of systems can the transfer function be best applied?

3. What transformation turns the solution of differential equations into algebraic

manipulations?

4.

Define the transfer function.

5.

What assumption is made concerning initial conditions when dealing with

transfer functions?

6. What do we call the mechanical equations written in order to evaluate the

transfer function?

7. If we understand the form the mechanical equations take, what step do

we

avoid

in evaluating the transfer function?

8. Why do transfer functions for mechanical networks look identical to transfer

functions for electrical networks?

Control System Modelling Case Studies_2014

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