Transcript Automatic control theory

Automatic control theory
A Course ——used for analyzing and
designing a automatic control system
Chapter 1 Introduction
21 century — information age, cybernetics(control theory), system
approach and information theory , three science theory mainstay(supports)
in 21 century.
1.1 Automatic control
A machine(or system) work by machine-self, not by manual operation.
1.2 Automatic control systems
1.2.1 examples
1) A water-level control system
* Operating principle……
* Feedback control……
Figure 1.1
Chapter 1 Introduction
Another example of the water-level
control is shown in figure 1.2.
* Operating
principle……
* Feedback
control……
f l oat
l ever
wat er
ent r ance
2) A temperature Control system
(shown in Fig.1.3)
W
at er exi t
Fi gur e 1. 2
cont ai ner
+
t her mo
met er
uf
ur
ampl i f i er
e
ua=k( ur - uf )
M
* Operating principle…
* Feedback
control(error)…
Gear
assembl y
Fi gur e 1. 3
Chapter 1 Introduction
3) A DC-Motor control system
+
Uk=k( ur - uf )
ur

ua
e
DC
mot or
M
r egul at or
l oad
t r i gger
-
r ect i f i er
Uf ( Feedback)
M
* Principle…
Fi g. 1. 4
* Feedback control(error)…
t echomet er
+
Chapter 1 Introduction
4) A servo (following) control system
ser vopot ent i omet er
out put
T c
I nput
T r
l oad
ser vomechani sm
M
+
Fig. 1.5
* principle……
* feedback(error)……
ser vomodul at or
ser vo mot or
Chapter 1 Introduction
5) A feedback control system model of the family planning
(similar to the social, economic, and political realm(sphere or field))
excess
procreat e
Desi re
popul at i on
-
+
government
( Fami l y pl anni ng commi t t ee)
popul at i on
soci et y
Pol i cy or
st at ut es
census
* principle……
* feedback(error)……
Fig. 1.6
Chapter 1 Introduction
1.2.2 block diagram of control systems
The block diagram description for a control system : Convenience
x
x1
x3
+ + e
x2
Example:
Si gnal
( var i abl e)
xxx
Adder s ( compar i son)
e=x1+x3- x2
Fig. 1.7
Component s
( devi ces)
Chapter 1 Introduction
For the Fig.1.1, The
water level control
system:
Figure 1.1
resistance comparator
Desired
water level
Input
amplifier
Error
Actuator
Motor
Gearing
Water
container
Process
controller
Float
Feedback
signal
Valve
Fig. 1.8
measurement
(Sensor)
Actual
water level
Output
Chapter 1 Introduction
For the Fig. 1.4, The DC-Motor control system
comparat or
Desi red
rot at e speed n
Act uat or
uk
e
Regul at or
Ref erence
i nput ur
Tri gger
a
ua
Rect i f i er
Error
Process
cont rol l er
Techomet er
Feedback si gnal
DC
mot or
uf
measurement ( Sensor)
Fi g. 1. 9
Act ual
rot at e speed n
Out put n
Chapter 1 Introduction
1.2.3 Fundamental structure of control systems
1) Open loop control systems
Di st ur bance
( Noi se)
I nput r ( t )
Ref er ence
desi r ed out put
uk
Cont r ol l er
uact
Act uat or
Cont r ol
si gnal
Pr ocess
Out put c( t )
( act ual out put )
Act uat i ng
si gnal
Fi g. 1. 10
Features: Only there is a forward action from the input to the
output.
Chapter 1 Introduction
2) Closed loop (feedback) control systems
Di st ur bance
( Noi se)
I nput r ( t )
Ref er ence
desi r ed out put
+
e( t ) =
r ( t ) - b( t )
Cont r ol l er
( +)
uk
uact
Act uat or
Cont r ol
si gnal
Pr ocess
Out put c( t )
( act ual out put )
Act uat i ng
si gnal
Feedback si gnal b( t )
measur ement
Fi g. 1. 11
Features:
not only there is a forward action , also a backward action
between the output and the input (measuring the output and
comparing it with the input).
1) measuring the output (controlled variable) . 2) Feedback.
Chapter 1 Introduction
Notes: 1) Positive feedback; 2) Negative feedback—Feedback.
1.3 types of control systems
1) linear systems versus Nonlinear systems.
2) Time-invariant systems vs. Time-varying systems.
3) Continuous systems vs. Discrete (data) systems.
4) Constant input modulation vs. Servo control systems.
1.4 Basic performance requirements of control systems
1) Stability.
2) Accuracy (steady state performance).
3) Rapidness (instantaneous characteristic).
Chapter 1 Introduction
1.5 An outline of this text
1) Three parts: mathematical modeling; performance analysis ;
compensation (design).
2) Three types of systems:
linear continuous; nonlinear continuous; linear discrete.
3) three performances: stability, accuracy, rapidness.
in all: to discuss the theoretical approaches of the control
system analysis and design.
1.6 Control system design process
shown in Fig.1.12
Chapter 1 Introduction
1. Establish control goals
6. Describe a controller and select
key parameters to be adjusted
2. Identify the variables to control
3. Write the specifications
for the variables
7. Optimize the parameters and
analyze the performance
Performance
meet the
specifications
Performance does not
4. Establish the system configuration Meet the specifications
Identify the actuator
Finalize the design
5. Obtain a model of the process,
the actuator and the sensor
Fig.1.12
Chapter 1 Introduction
1.7 Sequential design example: disk drive read system
A disk drive read system Shown in Fig.1.13
Rotation
of arm
Spindle
Disk
Track a
Track b
Actuator
motor
◆
Configuration
◆ Principle
Arm
Head slider
Fig.1.13 A disk drive read system
Chapter 1 Introduction
Sequential design:
here we are concerned with the design steps 1,2,3, and 4 of Fig.1.12.
(1) Identify the control goal:
Position the reader head to read the date stored on a track on the disk.
(2) Identify the variables to control: the position of the read head.
(3) Write the initial specification for the variables:
The disk rotates at a speed of between 1800 and 7200 rpm and the read head
“flies” above the disk at a distance of less than 100 nm.
The initial specification for the position accuracy to be controlled:
≤ 1 μm (leas than 1 μm ) and to be able to move the head from track a to track b
within 50 ms, if possible.
Chapter 1 Introduction
(4) Establish an initial system configuration:
It is obvious : we should propose a closed loop system , not
a open loop system.
An initial system configuration can be shown as in Fig.1.13.
Desired
head
position
error
Control
device
Actuator
motor
Read
arm
Actual
head
position
sensor
Fig.1.13 system configuration for disk drive
We will consider the design of the disk drive further in the aftermentioned chapters.
Chapter 1 Introduction
Exercise:
E1.6, P1.3, P1.13
Chapter 2 mathematical models of systems
2.1 Introduction
2.1.1 Why？
1) Easy to discuss the full possible types of the control systems—in terms of the
system’s “mathematical characteristics”.
2) The basis — analyzing or designing the control systems.
For example, we design a temperature Control system :
Disturbance
Input r(t)
+
e(t)=
r(t)-b(t)
Controller
desired output (－)
temperature
Feedback signalb(t)
uk
uac
Actuator
Control
signal
Output T(t)
Process
Actuating
signal
temperature
measurement
Fig. 2.1
The key — designing the controller → how produce uk.
actual
output
temperature
Chapter 2 mathematical models of systems
Different characteristic of the process — different uk:
T(t)
T2
T1
Ⅰ
Ⅱ
uk
uk11 uk12
uk21
Ⅰ uk 11
For T1 
Ⅱ uk 12
 Ⅰ uk 21
For T1 
Ⅱ uk 22  
2.1.2 What is ？
Mathematical models of the control systems—— the mathematical
relationships between the system’s variables.
2.1.3 How get？
1) theoretical approaches 2) experimental approaches
3) discrimination learning
Chapter 2 mathematical models of systems
2.1.4 types
1) Differential equations
2) Transfer function
3) Block diagram、signal flow graph
4) State variables(modern control theory)
2.2 Input-output description of the physical systems — differential
equations
The input-output description—description of the mathematical
relationship between the output variable and the input variable of the
physical systems.
2.2.1 Examples
Chapter 2 mathematical models of systems
Example 2.1 : A passive circuit
R
ur
define: input → ur
we have：
L
i
C
uc
output → uc。
di
du
 uc  ur i  C c
dt
dt

d 2uc
duc
LC 2  RC
 uc  ur
dt
dt
Ri  L
L
d 2uc
duc
make: RC  T1
 T 2  T1T2 2  T1
 uc  ur
R
dt
dt
Chapter 2 mathematical models of systems
Example 2.2 : A mechanism
Define: input → F ，output → y. We have:
F
dy
d2y
F  ky  f
m 2
dt
d t

d2y
dy
m 2 f
 ky  F
dt
dt
k
m
y
f
If we m ake:
we have:
f
 T1,
k
m
 T2
f
d2y
dy
1
T1T2
 T1
y F
2
dt
k
dt
Compare with example 2.1: uc→y; ur→F ─ analogous systems
Chapter 2 mathematical models of systems
Example 2.3 : An operational amplifier (Op-amp) circuit
R2
R3
C
i2
ur
R1 i 1
R1
R4
+
i3
Input →ur
output →uc
1
(i3  i2 )dt  R4 (i3  i2 )......(1)

C
u
i2  i1   r ...........................................(2)
R1
uc
1
i3 
(uc  R2i2 ).....................................(3)
R3
uc  R3i3 
R2 R3


du
dur
R

R
c
2
3
(2)→(3); (2)→(1); (3)→(1)： R C
 uc   R (
 R4 )C
 ur 
4
1
dt
dt
 R2  R3

m ake: R4C  T ;
we have: T
R2  R3
R2  R3
 k; (
 R4 )C  
R1
R2  R3
duc
du
 uc  k ( r  ur )
dt
dt
Chapter 2 mathematical models of systems
Example 2.4 : A DC motor
La
Ra
ua
( J1, f 1)
w1
ia
M
( J2, f 2)
w2
( J3, f 3)
w3
Input → ua， output → ω1
dia
La
 Raia  Ea  ua ....(1)
dt
M  Cmia .........................(2)
Ea  Ce1.........................(3)
d1
M M J
 f 1.....(4)
dt
Mf
i1
i2
(4)→(2)→(1) and (3)→(1):
La J 
La f
Ra J 
R f
1  (

) 1  ( a  1)1
CeCm
CeCm CeCm
CeCm

1
L
R
 ua  a M  a M
Ce
CeCm
CeCm
Chapter 2 mathematical models of systems
J
J
J  J1  22  2 32 ......equivalent m om ent of inertia
i1 i1 i2
f
f
here : f  f1  22  2 32 ......equivalent friction coefficient
i1 i1 i2
Mf
M
..........................equivalenttorque
i1i2
(can be derived from : 1  i12  i1i23 )
Make: Te  La ............electric - m agnetic tim e- constant
Ra
Ra J
Tm 
.......m echanical- electric tim e - constant
CeCm
Tf 
Ra f
.......friction - electric tim e - constant
CeCm
Chapter 2 mathematical models of systems
The differential equation description of the DC motor is:


TeTm 1  (TeT f  Tm ) 1  (T f  1)1

1
1

ua  (TeTm M  Tm M )
Ce
J
Assume the motor idle: Mf = 0, and neglect the friction: f = 0,
we have:
d 2
d
1
TeTm

T



u
m
a
2
dt
Ce
dt
Chapter 2 mathematical models of systems
Example 2.5 :
A DC-Motor control system
R2
+
ur R1
-
R3
DC
mot or
R3
ua
-
M
w l oad
uk
R1
t r i gger
Uf
-
r ect i f i er
M
t echomet er
+
Input → ur，Output → ω; neglect the friction:
R2
uk 
(ur  u f )  k1(ur  u f )........................................(1)
R1
u f  .....................(2)
d 2
ua  k2uk ......................(3)

d
1
1
TeTm
 Tm
 
ua  (TeTm M  Tm M )......(4)
2
dt
Ce
J
dt
Chapter 2 mathematical models of systems
（2）→（1）→（3）→（4），we have：
d 2

d
1
Tm
1
TeTm 2  Tm
 (1  k1k2 C )  k1k2
ur 
(Te M  M )
e
dt
Ce
J
dt
2.2.2 steps to obtain the input-output description (differential
equation) of control systems
1) Determine the output and input variables of the control systems.
2) Write the differential equations of each system’s components in
terms of the physical laws of the components.
* necessary assumption and neglect.
* proper approximation.
Chapter 2 mathematical models of systems
3) dispel the intermediate(across) variables to get the input-output
description which only contains the output and input variables.
4) Formalize the input-output equation to be the “standard” form:
Input variable —— on the right of the input-output equation .
Output variable —— on the left of the input-output equation.
Writing polynomial—— according to the falling-power order.
2.2.3 General form of the input-output equation of the linear
control systems—A nth-order differential equation:
Suppose:
input → r ，output → y
y (n )  a1 y ( n 1)  a2 y ( n 2)      an 1 y (1)  an y
 b0r (m)  b1r (m1)  b2r (m 2)      bm1r (1)  bmr.........n  m
Chapter 2 mathematical models of systems
2.3 Linearization of the nonlinear components
2.3.1 what is nonlinearity？
The output is not linearly vary with the linear variation of the
system’s (or component’s) input → nonlinear systems (or
components).
2.3.2 How do the linearization？
Suppose: y = f(r)
The Taylor series expansion about the operating point r0 is:
f ( r )  f ( r0 )  f
(1)
f ( 2) ( r0 )
f (3) ( r0 )
2
( r0 )(r  r0 ) 
( r  r0 ) 
( r  r0 )3    
2!
3!
 f ( r0 )  f (1) ( r0 )(r  r0 )
make: y  f (r)  f (r0 ) and : r  r  r0
wehave: y  f ' (r0 )r ............linearization equation
Chapter 2 mathematical models of systems
Examples:
Example 2.6 :
Elasticity equation
suppose: k  12.65;   1.1;
F ( x)  kx
operatingpointx0  0.25
F ' ( x)  kx 1  F ' ( x0 )  12.65 1.1  0.250.1  12.11
we have:
thatis :
F ( x )  F ( x0 )  12.11( x  x0 )
ΔF  12.11x..............linearization equation
Example 2.7 : Fluxograph equation
Q( p)  k p
Q —— Flux;
p —— pressure difference
Chapter 2 mathematical models of systems
because: Q ' ( p ) 
k
2 p
k
thus: Q 
p...........linearization equation
2 p0
2.4 Transfer function
Another form of the input-output(external) description of control
systems, different from the differential equations.
2.4.1 definition
Transfer function: The ratio of the Laplace transform of the
output variable to the Laplace transform of the input variable,with
all initial condition assumed to be zero and for the linear systems,
that is:
Chapter 2 mathematical models of systems
C ( s)
G( s) 
R( s)
C(s) —— Laplace transform of the output variable
R(s) —— Laplace transform of the input variable
G(s) —— transfer function
Notes:
* Only for the linear and stationary(constant parameter) systems.
* Zero initial conditions.
* Dependent on the configuration and the coefficients of the
systems, independent on the input and output variables.
2.4.2 How to obtain the transfer function of a system
1) If the impulse response g(t) is known
Chapter 2 mathematical models of systems
G( s)  Lg (t )
We have:
Because:
G( s) 
Then:
C ( s)
, if r (t )   (t )  R( s)  1
R( s )
G( s)  C( s)  Lg (t )
Example 2.8 :
g (t )  5  3e
 2t
5
3
2( s  5)
 G( s)  

s s  2 s ( s  2)
2) If the output response c(t) and the input r(t) are known
We have:
Lc(t )
G( s) 
Lr (t )
Chapter 2 mathematical models of systems
Example 2.9:
Then:
1
r (t )  1(t )  R(s)  ........Unit step function
s
1
1
3
 3t
c(t )  1  e
 C ( s)  

s s  3 s( s  3)
.........Unit step response
C ( s) 3 s( s  3)
3
G( s) 


R( s )
1s
s3
3) If the input-output differential equation is known
•Assume: zero initial conditions;
•Make: Laplace transform of the differential equation;
•Deduce: G(s)=C(s)/R(s).
Chapter 2 mathematical models of systems
Example 2.10:



2 c ( t )  3 c ( t )  4c ( t )  5 r ( t )  6r ( t )

2 s 2C ( s )  3sC ( s )  4C ( s )  5sR ( s )  6 R( s )

C(s)
5s  6
G(s) 
 2
R(s) 2 s  3s  4
4) For a circuit
* Transform a circuit into a operator circuit.
* Deduce the C(s)/R(s) in terms of the circuits theory.
Chapter 2 mathematical models of systems
Example 2.11:
R1
ur
For a electric circuit:
R2
C1
R1
ur ( s)
uc
C2
R2
1/ C1s
1/ C2s
1
1
1
// ( R2 
)
sC1
sC2
sC2
U c ( s) 
U r ( s) 
1
1
1
R1 
// ( R2 
)
R2 
sC1
sC2
sC2

1
2
U r ( s)
T1T2 s  (T1  T2  T12 ) s  1
U ( s)
1
G( s)  c

U r ( s ) T1T2 s 2  (T1  T2  T12 ) s  1
here : T1  R1C1;
T2  R2C2;
T12  R1C2
uc( s)
Chapter 2 mathematical models of systems
Example 2.12: For a op-amp circuit
R2
ur R1
R1
+
R2
C
uc
ur R1
R1
+
1/ Cs
uc
1
R2 
U c ( s)
sC   R2Cs  1
G( s) 

U r ( s)
R1
R1Cs
1
 k (1  )..................PI-Controller
s
R
here : k  2 ;   R2C......Integral im
t e constant
R1
Chapter 2 mathematical models of systems
5) For a control system
• Write the differential equations of the control system, and Assume
zero initial conditions;
• Make Laplace transformation, transform the differential equations
into the relevant algebraic equations;
• Deduce: G(s)=C(s)/R(s).
Example 2.13 the DC-Motor control system in Example 2.5
R2
+
ur R1
-
R3
DC
mot or
R3
ua
-
w l oad
uk
R1
t r i gger
Uf
M
-
r ect i f i er
M
t echomet er
+
Chapter 2 mathematical models of systems
In Example 2.5, we have written down the differential equations
as:
R2
uk 
(ur  u f )  k1(ur  u f )..................................(1)
R1
u f  ....................(2)
ua  k2uk ...................(3)

d 2
d
1
T
TeTm 2  Tm
 
ua  m (Te M  M )......(4)
dt
Ce
J
dt
Make Laplace transformation, we have:
U k ( s )  k1[U r ( s )  U f ( s )]...................................................(1)
U f ( s )  ( s )...............(2)
(TeTm s 2  Tm s  1)( s ) 
U a ( s )  k2U k ( s )..............(3)
1
T T s  Tm
Ua ( s)  e m
M ( s )......(4)
Ce
J
Chapter 2 mathematical models of systems
(2)→(1)→(3)→(4), we have:
1
1
TeTm s  Tm
[TeTm s  Tm s  (1  k1k2 )]( s)  k1k2 U r ( s) 
M ( s)
Ce
Ce
J
k1k2 1
( s )
Ce
G( s) 

U r ( s ) T T s 2  T s  (1  k k  1 )
e m
m
1 2
Ce
2
La
here : Te 
...........electric  m agnetic tim e - constant
Ra
Ra J
Tm 
......m echanical electric tim e - constant
CeCm
Chapter 2 mathematical models of systems
2.5 Transfer function of the typical elements of linear systems
A linear system can be regarded as the composing of several
typical elements, which are:
2.5.1 Proportioning element
Relationship between the input and output variables:
c(t )  kr(t )
C ( s)
k
Transfer function:
R( s )
Block diagram representation and unit step response:
G( s) 
R( s)
C( s)
k
r( t )
Examples:
C( t )
amplifier, gear train,
tachometer…
k
1
t
t
Chapter 2 mathematical models of systems
2.5.2 Integrating element
Relationship between the input and output variables:
t
1
c(t ) 
r(t )dt..........TI : integral tim e constant

TI
0
C ( s)
1
G( s) 

R( s) TI s
Transfer function:
Block diagram representation and unit step response:
R( s)
1
TI s
r( t )
1
t
C( s)
1
Examples:
C( t )
TI
Integrating circuit, integrating
motor, integrating wheel…
t
Chapter 2 mathematical models of systems
2.5.3
Differentiating element
Relationship between the input and output variables:
dr (t )
c(t )  TD
dt
C ( s)
Transfer function:
G( s) 
 TD s
R( s )
Block diagram representation and unit step response:
R( s)
TDs
r( t )
1
C( s)
Examples:
C( t )
differentiating amplifier, differential
valve, differential condenser…
TD
t
t
Chapter 2 mathematical models of systems
2.5.4 Inertial element
Relationship between the input and output variables:
dc (t )
T
 c(t )  kr (t )
dt
C ( s)
k
Transfer function:
G( s) 

R( s ) Ts  1
Block diagram representation and unit step response:
R( s)
k
Ts  1
r( t )
1
t
C( s)
Examples:
C( t )
inertia wheel, inertial load (such as
temperature system)…
k
T
t
Chapter 2 mathematical models of systems
2.5.5 Oscillating element
Relationship between the input and output variables:
T
2
d
c( t )
2
2
dt
dc(t )
 2T
 c(t )  kr(t )
dt
0  1
C ( s)
k
 2 2
Transfer function: G( s) 
R( s) T s  2Ts  1
Block diagram representation and unit step response:
R( s)
r( t )
1
C( s)
T 2 s 2  2Ts  1
C( t )
Examples:
oscillator, oscillating table,
oscillating circuit…
k
1
t
0  1
t
Chapter 2 mathematical models of systems
2.5.6
Delay element
Relationship between the input and output variables:
c(t )  kr(t   )
C ( s)
Transfer function:
G( s) 
 kes
R( s )
Block diagram representation and unit step response:
R( s)
C( s)
ke s
Examples:
gap effect of gear mechanism,
threshold voltage of transistors…
C( t )
r( t )
k
1
t

t
Chapter 2 mathematical models of systems
2.6
block diagram models (dynamic)
Portray the control systems by the block diagram models more
intuitively than the transfer function or differential equation models.
2.6.1 Block diagram representation of the control systems
Si gnal
( var i abl e)
X( s)
Component
( devi ce)
G( s)
X3( s)
Adder ( compar i son)
E( s) =x1( s) +x3( s) - x2( s)
X1( s)
+
+
E( s)
-
Examples:
X2( s)
Chapter 2 mathematical models of systems
Example 2.14
For the DC motor in Example 2.4
In Example 2.4, we have written down the differential equations as:
dia
La
 Raia  Ea  ua ....(1) M  Cmia .........................(2)
dt
d
Ea  Ce .........................(3) M  M  J
 f  .....(4)
dt
Make Laplace transformation, we have:
U a ( s )  Ea ( s )
La sI a ( s )  Ra I a ( s )  Ea ( s )  U a ( s )  I a ( s ) 
.............(5)
La s  Ra
M ( s )  Cm I a ( s )......................................................................................(6)
Ea ( s )  Ce( s ).......................................................................................(7)
M ( s )  M ( s )  J s( s )  f ( s )  ( s ) 
1
[ M ( s )  M ( s )]......(8)
Js  f
Chapter 2 mathematical models of systems
Draw block diagram in terms of the equations (5)～(8):
M (s )
Ua( s)
-
1
La s  Ra
I a( s)
Cm
Ea( s)
M( s)
-
(s )
1
Js  f
Ce
M (s )
Consider the Motor as a whole:
Ua( s)
1
Ce
TeTm s 2  (Tm  TeT f ) s  T f  1
1
(TeTm s  Tm )
J
TeTm s 2  (Tm  TeT f ) s  T f  1
-
(s )
Chapter 2 mathematical models of systems
Example 2.15
The water level control system in Fig 1.8:
1
s
Ce
k1 TeTm s2  Tm s  1
Desi r ed
wat er l evel
I nput hi
e
ampl i f i er
ua
Mot or

k2e
s
Gear i ng
Feedback si gnal hf
Tm
(Te s  1)
J

M ( s)
2
TeTm s  Tm s  1
k3
T1s  1

k4
T2 s  1
Act ual
wat er l evel
Q Wat er Out put h
Val ve
cont ai ner
Fl oat

Chapter 2 mathematical models of systems
The block diagram model is: