EEE 317: Control System I

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Nov 15, 2013 (3 years and 7 months ago)

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EEE 317: Control System I


SYLLABUS

Introduction to control systems. Linear system models: transfer function, block diagram and
signal flow graph (SFG), State variables: SFG to state variables, transfer function to state
variables and state variable to tra
nsfer function. Feedback control System: closed
-
loop systems,
parameter sensitivity, transient characteristics of control systems, effect of third pole and zero on
the system response and system types and steady
-
state error. Routh stability criterion, Anal
ysis
of feedback control system: root
-
locus method and frequency response method. Design of
feedback control system: controllability and observability, root locus, frequency response and
state
-
variable methods. Digital control systems: introduction, sample
d data systems, stability
analysis in z
-
domain.


REFERENCE BOOKS

1.

Control systems engineering


I. J. Nagrath

2.

Modern control engineering


Katsuhiko Ogata

3.

Control systems engineering


Norman S. Nise

4.

Analog and digital control system


Chen

5.

Advanced control

engineering


Roland S. Burns


TOPICS COVERED



Introduction to control systems



Mathematical

Models of Physical systems



Feedback characteristics of control systems



Time
-
domain analysis of control systems



Concept of stability



The root
-
locus technique



Frequen
cy response analysis



Control system design by root
-
locus method



Control system design by frequency response



Sampled
-
data control systems


OBJECTIVE OF THE COURSE

The objective of this course
is to highlight the following:

1.

Representing a system using ordina
ry differential equations and Laplace transforms.

2.

Analysis of single input/ single output systems and their components considering their
input/ output relationship is linear.

3.

Design controllers for single input/ single output systems that meet design requi
rements.






1

Introduction to Control Systems


The goal of control engineering is to design and build real physical systems to perform given
tasks.

A
n engineer is asked to design an install a heat exchanger to control the temperature and
humidity of a lar
ge building.
He determines the required capacity of the exchanger and then
proceed to install the system. After installation the exchanger is found to be
insufficiently
powerful to control the buildings environment. He has to replace the unit and place a p
owerful
one. This is an example of
empirical method
.

Consider again the task of sending astronauts in moon and bringing them back safely. It cannot
be carried out by empirical method. In this case
analytical method

is indispensable. This method
consists o
f the following steps:

-

modeling

-

setting up mathematical equations

-

analysis and design

Empirical methods may be expensive and dangerous. Analytical methods are simulated in
computers to see the result. If the design is satisfactory, the system is implemente
d using
physical devices.


A control system is an interconnection of components or devices so that the output of the overall
system will follow as closely as possible a desired signal. The reasons of designing control
systems include:

-

Automatic control

(e.
g.
, control of room temperature
)

-

Remote control (e.g., antenna position control)

-

Power amplification(e.g., control system will generate sufficient power to turn
the heavy antennas)

Control system components can be mechanical, electrical, hydraulic, pneumat
ic, thermal, or it
may be a computer program. It plays an important role in the development of modern
civilization. We use heating and air
-
conditioning in domestic domain for comfortable living. It
has found application in quality control of manufacturing
products, machine
-
tool control,
weapon systems, power systems, robotics and many other places.


Classification of Control Systems


Control systems are basically classified as




Open
-
loop control system



Closed
-
loop control system

In open
-
loop system the con
trol action is independent of output. In closed
-
loop system control
action is somehow dependent on output. Each system has at least two things in common, a
controller and an actuator

(final control element)
.
The input to th
e

controller is called reference
input. This signal represents the desired system output.

Open
-
loop control system is
used
for very simple applications

where inputs are known ahed of
time and there is no disturbance
. Here the output is sensitive to the changes in
disturbance

inputs.
Distu
rbance inputs

are undesirable inputs that tend to deflect
the plant outputs from their
desired values. They must be calibrated and adjusted at regular intervals to ensure proper
operation.



Closed
-
loop systems are also called feedback control systems. Fee
dback is the property of the
closed
-
loop systems which permits the output to be compared with the input of the system so

2

that appropriate control action may be formed as a function of inputs and outputs. Feedback
systems has the following features:

-

reduced

effect of nonlinearities and distortion

-

Increased accuracy

-

Increased bandwidth

-

Less sensitivity to variation of system parameters

-

Tendency towards oscillations

-

Reduced effects of external disturbances

The general block diagram of a control system is shown

below.


Figure: Closed
-
loop control system


Some Definitions

Reference

input



It is
the actual signal input to the control system
.

Output (
Controlled variable
)



It is the actual response obtained from a control system.

Actuating

error

signal



It is
the difference between the reference input and feedback signal.

Controller



It is a component required to generate control signal to drive the actuator.

Control signal



The signal obtained at the output of a controller is called control signal.

Actuator



It is a power device that produces input to the plant according to the control signal,
so that output signal approaches the reference input signal.

Plant


The combination of object to be controlled and the actuator is called the plant.

Feedback Eleme
nt



It is the element that provides a mean for feeding back the output quantity
in order to compare it with the reference input.

Servomechanism



It is a feedback control system in which the output is mechanical position,
velocity, or acceleration.


Exam
ple of Control Systems

Toilet tank filling system:



Figure: Toilet tank filling system




3

Position control system:

[
antenna
]


Figure: Position control system


Velocity

control system:

[audio/ video recorder]


Figure: Velocity control system


Clothes Dr
yer:


Figure: Automatic dryer


4

Temperature control system:

[oven, refrigerator, house]


Figure: Temperature control system


Computer numerically controlled (CNC) machine tool:


(a)


(b)

Figure: CNC machine tool control system



5

Control System Design

In
order to design and implement a control system, we need
knowledge about the following
things:



Knowledge of desired value, (performance specification)



Knowledge of the output value, (feedback sensor, its resolution and dynamic
response)



Knowledge of contro
lling device,



Knowledge of actuating device,



Knowledge of the plant.

With all of this knowledge and information available for the control system designer, he can
start the design steps shown below in the flow diagram.



6

Mathematical Model
s

of Physical Sys
tem
s

We use mathematical models of physical systems to design and analyze control systems.
Mathematical models are described by ordinary differential equations. If the coefficients of the
describing differential equations are function of time, then the mat
hematical model is linear
time
-
varying. On the other hand, if the coefficients describing differential equations are
constants, the model is linear time
-
invariant.

The differential equations describing a LTI system can be reshaped into different forms
for

the convenience of analysis. For transient response or frequency response analysis of single
-
input
-
single
-
output linear systems, the transfer function representation is convenient. On the
other hand, when the system has multiple inputs and outputs, the ve
ctor
-
matrix notation may be
more convenient.


Powerful mathematical tools like Fourier and Laplace transforms are available for linear
systems. Unfortunately no physical system in nature is perfectly linear. Certain assumptions
must always be made to get a

linear model. In the presence of strong nonlinearity or in presence
of distributive effects it is not possible to obtain linear models.


A commonly adopted approach is to build a simplified linear model by ignoring certain
nonlinearities and other physica
l properties that may be present in a system and thereby get an
approximate idea of the dynamic response of the system. A more complete model is then built
for more complete analysis.


Transfer function

The transfer function of an LTI system is the ratio o
f Laplace transform of the output variable to
the Laplace transform of the input variable assuming zero initial conditions.

Following are some
examples of how transfer functions can be determined for some dynamic system elements.


[See chapter 3 of Ogata for details]


The insertion of an isolation amplifier between the two RC
-
circuits will produce no loading
effect.





2
( )
1
( ) 1
o
i
E s
E s LCs RCs

 

2
1 1 2 2 1 1 2 2 1 2
( )
1
( ) ( ) 1
o
i
E s
E s RC R C s RC R C RC s

   


7

A.
Transfer Function of Armature Controlled DC Motor








In Laplace domain,


2
0
( ) ( )
( ) ( ) ( ) ( )
( ) ( ) ( )
b b
a a a b
T a
E s K s s
L s R I s E s E s
Js f s s K I s





  


 

;


0
( )
( )
( ) ( )( )
T
a a T b
K
s
G s
E s s R sL Js f K K

  
  

Neglecting
a
L
,
2
0
//
( )
(/) ( ) ( 1)
T a T a m
T b a m
K R K R K
G s
Js s f K K R s Js f s s

  
   
;

where,
0
/
T b a
f f K K R
 

and
/; /.
m T a m
K K R f J f

 

m
K
and
m

are called the motor gain and time constant respectively. These two parameters are
usually supplied by the manufacturer.


The block diagram model is,




B. Transfer
F
unction of a
F
ield
-
controlled
DC

M
otor


'
1 1
2
2
T
M a f f a f
f
f f f
M T f
T K i K K i i K i
di
L R i e
dt
d d
J f T K i
dt dt

 

  



 




  


;
2
( ) ( ) ( )

( ) ( ) ( ) ( )
f f f
M T f
L s R I s E s
Js fs s T s K I s

 





  




We obtain,

1
;
f f M f f a T a
K i T K K i i K i

  

;
a
b b a a a b
di
d
e K L R i e e
dt dt

   

2
0
2
M T a
d d
J f T K i
dt dt
 
  

( )
( )
( ) ( )( )
T
f f
K
s
G s
E s s L s R Js f


 
 


8

Mechanical System

Translational system and rotational system







Mechanical Acclerometer


For a constant input acceleration
y

becomes constant.

Th
en,
x k y


. Taking Laplace transform

of the previous equation
,
2
( ) ( ) ( )
X s s c s k Y s
 
  
;


( ) ( )
2
1
X s Y s
s c s k
 
 
 


Rotary Potentiometer


2
2
1 2 1 2 2 1
( );( )
m
d s
du d x
u ma m m
dt dt
u k u u u k x x
  
   

2
2
1 1 2 2 1 2
;
( );( )
J
d s
d d
T J J
dt dt
T k T k
 
   
 
   

2
1 2
2
( ) ( )
( )
d y t dy t
m k k y t u
dt dt
  

max
max
max max
( )
( ) ( )
or, ( ) ( )
o
o p
V
t
V t V t
V t k t


 

   
 
   
   



9

Example
:
Draw the block diagram of the following system.


Example:

Control of flap
s in airplane





10

Example:

Figure below shows a reduction gearbox being driven by a motor that develops a torque
( )
m
T t
. It
has a gear reduction ratio of ‘n

= b / a
’. Find a differential equation relating the motor
torque
( )
m
T t
and the output angular position
( )
o
t

.


2
2
( ) ( )
m m
a m m m
d d
T t T t I C
dt dt
 
  
;
2
0
2
( )
o o
b o
d d
T t I C
dt dt
 
 
;
( )
1
( )
a
b
T t
a
T t b n
 
;


m
o
b
n
a


 

From above,
2 2
0
2 2
( )
o o o o
m m m o
d d d d
n T t nI nC I C
dt dt dt dt
   
 
   
 
 
;

N.B.
2 2
2 2
;
m o m o
d d d d
n n
dt dt dt dt
   
 
 
 
 


Gearbox parameters:

,
,

From above,
2 2
( ) ( ) ( )
o m o o m o m
I n I C n C nT t
 
    

Inputting parameters,
0.0225 0.3 50 ( )
o o m
T t
 
 




[Ans]

Example:


Gear train and its equivalence



11



max
( ) ( ) ( )
/
p r o
p batt
V s k s s
k V
 

 

1 1 1 1
motor eq eq
T J f
 
 

Tachometers


Error detector




Example


Example

Sketch the analogus electrical circuit of the following mechanical system.



( ) ( ) ( )
t t
V s k s k s s
 
 



max
( ) ( ) ( )
/
p r o
p batt
V s k s s
k V
 

 


Error detector using op amp


e = r
-

v
w


Obtain the transfer function X(s)/E(s) for the
electromechanical system shown left assuming
that the coil has a back emf
1
b
dx
e K
dt

and the
coil current
i
2

produces a force
2 2
c
F k i

on the
mass M.

2
4 3
2
1 2 1 2
( )
( ) ( ) (2
) ( 2 ) 2
K
X s
E s RLCMs L M RCB s RC LK
k k s RB LK k k s RK

  
    


12

Example

Draw the analogous electric circuit of the system below using
f
-
i
analogy.







[Ans]

Synchros



Figure: Synchro error detector

( ) sin
r r c
v t V t


;
1
2
3
( ) sin cos( 120 )
( ) sin cos
( ) sin cos( 240 )
s r c
s r c
s r c
v t KV t
v t KV t
v t KV t
 
 
 
 

 

;
1 2
2 3
3 1
( ) 3 sin sin( 240 )
( ) 3 sin sin( 120 )
( ) 3 sin sin
s s r c
s s r c
s s r c
v t KV t
v t KV t
v t KV t
 
 
 
 
 


When
0


,
3 1
( ) 0
s s
v t

and maximum voltage is
induced on S
2

coil. This position of the rotor is
defined as the “
electrical zero
” of the transmitter and used as reference position of the rotor.


The output of the synchro transmitter is applied to the stator winding of a “
synchro control
transformer

.
Circulating current of the same phase but of different magnitude flows through
the two sets of stator coil. The pair acts as an
error detector.

The voltage induced in the control transformer rotor is proportional to the cosine of the angle
between the two
rotors and is given by,



( ) sin cos; where, 90
r c
e t K V t
    

   
.


( ) sin cos(90 ) sin sin( ) ( )sin
r c r c r c
e t KV t KV t KV t
        
  
       

…(01)

The
equation

above

hold
s

for small angular displacement.

Thus the synchro transmitter
-
control transformer pair acts as an error detector which gives a
voltage signal

at the rotor terminal of the control transformer proportional to the angular
difference between the shaft positions.


A synchro is an electromagnetic
transducer that is used to convert
angular shaft position into an
electric signal. The basic element
of a synchro is a synchro
transmitter whose construction is
ver
y similar to that of the 3
-


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-
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-

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a扯癥.


13

Equation (01) is represented graphically in figure below for an arbitrary time variation of
(
 

).







It is s
een that the output of the
synchro error detector is a
modulated signal, where the
ac
signal applied to the rotor of
synchro transmitter acts as carrier
and the modulating signal is,

( ) ( );
m s s r
e t K K KV
 

   







A.C. Servomotor

For low power applicat
ion a.c. motors are preferred, because of their light weight, ruggedness
and no brush contact.

Two phase induction motors are mostly used

in control system
.



The torque speed characteristic of the motor is different from conventional motor. X / R ratio i
s
low and the curve has negative slope for stabilization.


The torque
-
speed curve is not linear. But we assume it as linear for the derivation of transfer
function.
The troque is afunction of both speed and the r.m.s. control voltage, ie.,
(,)
M
T f E


.

The motor has two stator windings
displaced 90 elec
trical degree apart.
The voltages applied to the
windings are not balanced. Under
normal operating conditions a fixed
voltage from a constant voltage
source is applied to one phase. The
other phase, called control phase, is
energized by a voltage of variab
le
magnitude which is 90


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灨p獥 眮w⹴⸠ 瑨攠 癯汴a来 潦 晩fe搠
灨p獥⸠


14

Using Tailor series expansion about the normal operating point
0 0 0
(,,)
M
T E

we get,


0 0
0 0
0 0 0
( ) ( )
M M
M M
E E E E
T T
T T E E
E
 
 
 
 
 
 
    

For position control system,
0
0
E

,
0
0


,
0
M
T

Thus, the above equation may be simplified as,


0
;
M
T kE m J f
  
   


where,
0
0
M
E E
T
k
E






and
0
0
M
E E
T
m






.

Performing Laplace transform,
2
0
( ) ( ) ( ) ( )
kE s ms s Js s f s
  
  

Or,

2
0
( )
( )
( ) ( ) ( 1)
m
m
K
s k
G s
E s Js f m s s s


  
  
; where,
0
m
k
K
f m


and
0
m
J
f m



.

Since ‘m’ is negative the transient part is decaying as
m

is positive. If ‘m’ would positive and
0
m f

the transient part will increase with time and the s
ystem would be unstable.

k
and
m

are
the slope of the torque
-
voltage and torque
-
speed curve.


A.C. Position Control System



The reference motor phase and the synchro transmitter rotor coil are excited from the same
carrier supply. The carrier voltage dr
iving the control phase is amplified and a phase shift of
90

is produced by use of two RC networks. The signal flow graph of the control system is
shown above. The overall transfer function of the system is,

2
( )
( ) (1 )
c m a s
m m a t m a s
s K K K n
R s s K K K s K K K n



  

m
K
=motor gain constant,
m

=motor time constant,
s
K
=synchro sensitivity in volts/rad.