Transcript Chapter 3 Feedback Control Systems

Review last lectures
centrifugal (fly-ball) governor
1788
Picture shows an operation principle of the fly-ball
(centrifugal) speed governor developed by James Watt.
simple feedback system
heat transfer
Qout
desired
temperature
_
Qin
thermostat
switch
air con
+
S
room
temperature
office room
closed-loop (feedback) system
error or
actuating signal
input or
reference
summing junction
or comparator
input filter
(transducer)
+
_
S
controller
disturbance
plant
control
signal
actuator
process
sensor or
output transducer
sensor noise
output or
controlled
variable
closed-loop system
advantages




high accuracy
not sensitive on
disturbance
controllable transient
response
controllable steady
state error
disadvantages





more complex
more expensive
possibility of instability
recalibration needed
need for output
measurement
open-loop system
input or
reference
disturbance
plant
output or
controlled
variable
control
signal
input filter
(transducer)
controller
actuator
process
open-loop system
advantages





simple construction
ease of maintenance
less expensive
no stability problem
no need for output
measurement
disadvantages
disturbances cause
errors
 changes in calibration
cause errors
 output may differ from
what is desired
 recalibration needed

Example 1: Liquid Level System
Goal: Design the input
valve control to maintain a
constant height regardless
of the setting of the
output valve
(input flow)
qi
Input valve
control
float
R
(height)
(resistance)
H
(output flow)
V
(volume)
Output
valve
qo
Example 2: Admission Control
Users
Goal: Design the controller
to maintain a constant
queue length regardless of
the workload
RPCs
Reference
value
Administrator
Tuning
control
Sensor
Controller
Server
Queue
Length
Server
Log
Why Control Theory

Systematic approach to analysis and
design
•
•
•
•

Transient response
Consider sampling times, control frequency
Taxonomy of basic controls
Select controller based on desired
characteristics
Predict system response to some input
• Speed of response (e.g., adjust to workload
changes)
• Oscillations (variability)

Approaches to assessing stability and limit
cycles
Controller Design Methodology
Start
System Modeling
Controller
Design
Block
diagram
construction
Controller
Evaluation
Transfer function
formulation and
validation
Objective
achieved
?
N
Model
Ok?
Y
N
Y
Stop
Control System Goals

Regulation
• thermostat, target service levels

Tracking
• robot movement, adjust TCP window to
network bandwidth

Optimization
• best mix of chemicals, minimize
response times
Approaches to
System Modelling

First Principles
• Based on known laws

Physics, Queuing theory
• Difficult to do for complex systems

Experimental (System ID)
• Statistical/data-driven models
• Requires data
• Is there a good “training set”?
Laplace transforms

The Laplace transform of a signal f(t) is defined
as

L f (t )  F (s)   f (t ) e
 st
dt
0


The Laplace transform is an integral transform
that changes a function of t to a function of a
complex variable s = s + jw
The inverse Laplace transform changes the
function of s back to a function of t
s  j
L F(s)  f (t )  2 j  F (s) e
1
1
s  j
st
ds
Laplace transforms of basic functions
f(t)
F(s)
Unit impulse
d (t)
1
Unit step
u (t)
Exponential
e−at
Sine wave
sin(t)
Cosine wave
cos(t)
Polynomial
tn
e−at x(t)
Note: f(t) = 0, t < 0
1
s
1
sa
w
s2  w2
s
s2  w2
n!
s n 1
X ( s  a)
Example
f (t)
Laplace transform:
signal
1
1
F (s) 
s 1
f (t )  e t
0
imagF (s  jw)
realF (s  jw)
w
s  s  jw
t
s
w
s
Properties of Laplace transforms

Linear operator:
if
then
L f1(t)  F1 (s) and L f 2(t)  F2 (s)
La1 f1(t)  a2 f 2(t)  a1F1 (s)  a2 F2 (s)
for any two signals f1(t) and f2(t) and any two constants a1 and a2

Time delay:
L f(t  t0 )  e
 st0
F ( s)
Properties of Laplace transforms cont.

Laplace transforms of derivatives:
if
L f (t )  F (s)
then
 df 
L    sF (s)  f (0)
 dt 
d 2 f  2
df
L  2   s F (s)  sf (0) 
dt
 dt 
t 0
d n f  n
n 1
n  2 df
L  n   s F (s)  s f (0)  s
dt
 dt 
d n1 f
 ...  n1
dt
t 0
t 0
Properties of Laplace transforms cont.

Laplace transform of integrals:
t
 1
L  f ( )d   F (s)
0
 s

The Laplace transform changes differential
equations in t into arithmetic equations in s
Laplace Transform Properties
Using Laplace transforms to solve ODEs


The Laplace transform can be used to solve
differential equations
Method:
1.
Transform the differential equation into the ‘Laplace
domain’ (equation in t → equation in s)
2.
3.
Rearrange to get the solution
Transform the solution back from the Laplace domain to
the time domain (signal in s → signal in t)

Usually the Laplace transform (step 1) and the
inverse transform (step 3) are done using a Table
of Laplace transforms
Example

Use Laplace transforms to find the unforced
response of a spring-mass-damper with initial
conditions x(0)  x0 , x (0)  0
x
k
m = 1 kg
k = 2 N/m
b = 3 Ns/m
f
m
b
x
Equation of motion
kx
bx
Free body diagram
m
f
mx  bx  kx  f (t )  0
Example – solution

Take Laplace transform of both sides of
equation of motion:
Lmx  bx  kx  Lf (t )
Lkx  kLx  kX (s)
Lf (t )  F (s)
Lbx  bLx  bsX (s)  x(0)
2
Lmx  mLx  ms X (s)  sx(0)  x(0)

Equation of motion in Laplace domain is
ms 2 X (s)  msx(0)  mx(0)  bsX (s)  bx(0)  kX (s)  F (s)
ms 2 X (s)  msx(0)  mx(0)  bsX (s)  bx(0)  kX (s)  F (s)

Rearrange:
ms
2

 bs  k X (s)  F (s)  msx(0)  mx(0)  bx(0)
External force

Initial conditions
The system can be in motion if
1. An external force is applied
2. The initial conditions are not an equilibrium state
(not zero)

Apply initial conditions:
X ( s) 
x(0)  x0 , x (0)  0
ms  b  x0
ms
2
 bs  k

s  3 x0
 s
2
 3s  2


Partial fraction expansion:
2 x0
x0
X ( s) 

s  1 s  2


Use tables to find inverse Laplace transform
System response (in time domain) is
x(t )  L X ( s)  2 x0e  x0e
t
-1
t
x(t )  (2e  e
 2t
) x0
2t
x(t)
x0
0
t
Partial fraction expansion

A partial fraction expansion can be used to find the inverse
transform of
cs  b1 s  b2 ...s  bm 
Y ( s) 
s  a1 s  a2 ...s  an 


This can be expanded as
Note that
kn
k1
k2
Y ( s) 

 ... 
s  a1  s  a2 
s  an 
s  a1 Y (s)s a
1

So
 s  a1 k1 s  a1 k 2 s  a1 k 3




 ...
 k1
s  a2  s  a3 
 s  a1 
 s a1
k i  s  ai Y (s)s ai
Example - Partial fraction expansion

Find the partial fraction expansion of
Y ( s) 

6
s  1s  3
This can be expanded as
k1
k2
Y ( s) 

s  1 s  3

So
k1  s  1Y ( s)s  1
 s  16 
6


3

 s  1s  3 s 1  1  3
k 2  s  3Y ( s)s  3
 s  36 
6


 3

 s  1s  3 s 3  3  1
3
3
Y ( s) 

s  1 s  3
Other examples
k0  1
1.
k
k2
s 2  2s  2 s 2  2s  2
Y ( s)  2

 k0  1 
s  1 s  2
s  3s  2 s  1s  2
2.
Y ( s) 
1
s  12 s  2


k3
k1
k2


s  1 s  12 s  2

 d  1 
 1 
d

2

k1   s  1 Y ( s) 
  

 1
2
 ds
 s 1  ds  s  2  s 1  s  2  s 1
3.
Y ( s) 
k
k2
1
 1
ss  2 s s  2
k1  sY ( s )s 0
 s 
1





s
s

2

 s 0 2
Insights from Laplace
Transforms

What the Laplace Transform says about
f(t)
• Value of f(0)

Initial value theorem
• Does f(t) converge to a finite value?

Poles of F(s)
• Does f(t) oscillate?

Poles of F(s)
• Value of f(t) at steady state (if it converges)

Limiting value of F(s) as s ---> 0
Transfer Function

Definition
• H(s) = Y(s) / X(s)



X(s)
H(s)
Y(s)
Relates the output of a linear system
(or component) to its input
Describes how a linear system
responds to an impulse
All linear operations allowed
• Scaling, addition, multiplication
Block Diagrams



Pictorially expresses flows and
relationships between elements in system
Blocks may recursively be systems
Rules
• Cascaded (non-loading) elements: convolution
• Summation and difference elements

Can simplify
Block Diagram of System
Disturbance
Reference Value
R(s)
+
S
E (s)
N (s )
G1 ( s )
Controller
U (s )
G2 (s)
S
–
B(s)
Transducer
H (s )
Plant
Y (s)
Combining Blocks
Reference Value
R(s)
+
S
E (s)
(G1 (s)  N (s))  G2 (s)
Combined Block
–
B(s)
Transducer
H (s )
Y (s)
Key Transfer Functions
Reference
R(s)
S
+
E (s)
G1 ( s )
Controller
G2 (s)
U (s )
Plant
–
Y (s)
H (s )
B(s)
Transducer
Feedforward:
Y ( s) Y ( s) U ( s)

 G1 ( s)G2 ( s)
E ( s) U ( s) E ( s)
B( s )
Open - Loop :
 G1 ( s)G2 ( s) H ( s)
E (s)
Feedback:
G1 (s)G2 ( s)
Y ( s)

R(s) 1  G1 (s)G2 (s) H ( s)
Class Home page:
http://saba.kntu.ac.ir/eecd/People/aliyari/