Eee3420 lecture01 rev2011

EEC3420 Industrial Control
Department of Electrical Engineering
│ Lecture 1 │
Introduction to Control Systems
© Vocational Training Council, Hong Kong. Week 1

EEE3420 Industrial Control
Learning objectives
 Understand the basic concept in control systems.
 Know what is a Transfer Function.
 Appreciate the PID control process.
 Know what criteria leading to a stable control system.

Open-loop control system
 An open-loop control system is one in which the
control signal of the process is independent of the
process output
 control accuracy is determined by the calibration of
the plant

Advantages & disadvantage of open-loop
control system
 Advantages
 simple and inexpensive
 no stability problem
 Disadvantages
 cannot compensate for any disturbances that
add to the controller’s driving signal

Closed-loop control system
 A closed-loop control system depends on the
output of the process to adjust the signal
controlling the closed loop
 process output is compared to the user command
and an output from the plant

Advantages & disadvantages of closed-loop
control system
 Advantages
 less sensitive to noise, disturbances and
changes in the environment
 transient response and steady-state error can
be controlled more conveniently and with
greater flexibility
 Disadvantages
 relatively expansive
 may be unstable if not properly designed

Architecture of a closed-loop control
system
 Controlled Variable
(CV).
 Set point.
 Error = set point –
current value of CV.
 Manipulated Variable.
 Feedback Loop.

Feedback control real-time scheduling
 Choices for control variables, manipulated variables, set
points
 Choice of appropriate control functions
 Stability problem of feedback control in the context of
real-time scheduling?
 How to tune control parameters?
 How significant is the overhead and how to minimize it?
 How to integrate a runtime analysis of time constraints
with scheduling algorithms?

Using negative feedback control system
 typically more stable
 less sensitive to variation in component values
 more immune to noise

Transfer function of a control system
 The transfer function of a control system is defined as
the ratio of the output to the input
 predict how the system will perform if the transfer
function is known
 output depends on both the present input and the past
history of the input, so the output c(t) is a convolution
product of the input r(t) and the system g(t)

 c(t) = r(t) * g(t)
= ò¥ - ×
0
r(t t ) g(t ) dt
 with the use of Laplace transform, we get
ò ò ò ¥ - ¥ ¥ × - úû ù
êë é
C(s) = L[c(t)] = c(t) × e st dt = r(t - t ) ×
g(t ) dt e st dt
0 0 0

 Let t’ = t – τ then t = t’ + τ
× úû ù
× = × úû ù
êë é
¥ ¥ - ¥ ¥ - +
ò ò ò ò
¥ - ¥ -
st s t
t
( ' )
êë é
C s r t g d e dt r t g d e dt
= - ×
ò ò
t t t t t
( ) ( ) ( ) ( ') ( ) '
0 0 0
r t e dt g e d R s G s
êë é
st '
s
0
× = úû ù
= × × ×
t
t t
( ') ' ( ) ( ) ( )
0 0
 the convolution product of r(t)*g(t) is now transformed
into an algebraic product of R(s).G(s)

 open-loop control system can be represented in
frequency domain as shown above
 the output C(s) is given by G(s) x R(s).
 time-domain output c(t) may be obtained by the inverse
Laplace operation
( ) 1[ ( )] 1
= - = × k j
ò + ¥
c t L C s ( )
k j
- ¥
C s est ds
j
p
2

PID Control System
 PID – Controller is the most widely used control strategy
in industry
 used for various control problems such as automated
systems or plants
 consists of three different elements
 P Proportional control
 I Integral control
 D Derivative control

PID Control System
for control loop to work properly, the PID loop must
be properly tuned

The PID transfer function
P K e(t) P = ´
I = K ´ òe t × dt I ( )
D K d e t D
= ´ ( ( ))
dt
P Proportional control,
I Integral control,
D Derivative control,

The total controller output,
Use the Laplace transform,
We get,
u t K e t K e t dt K d e t P I D
( ) ( ) ( ) ( ( ))
= + ´ ò × + ´
dt
x dt x
d s ; dx
; 1 ;
® ® ò ® ò ®
s
s
sx dt
dt
dt
K s 
G s K K
ù
C P × + + = = úû
( ) (1)
U s
( )
E s
( )
s
D
I
é
êë

2
G s K T T s T s
( ) = ( × × + × +1)
P I D I
C ×
Re-arrange to get, T s
I
 Where
 KP is the proportional gain
 TI is the integral time constant
 TD is the derivative time constant

 the three different adjustments (KP, TI, TD) interact with each
other
 it can be very difficult and time consuming to tune these
three values in order to get the best performance according
to the design specifications of the system.

A Thermal Control System
• electrical heater of heat capacity Ch & thermal
resistance Rho
• oven of heat capacity Co & thermal resistance Ro
• environment temperature Te, set-point temperature Ts
• temperature controller adjusts the power W by
comparing To with Ts

On-Off Control of the Thermal System
• the simplest form of control
• when the oven is cooler
than the set-point
temperature the heater is
turned on at maximum
power M
• once the oven is hotter
than the set-point
temperature the heater is
switched off completely
Red line: set-point temperature
Green line: actual temperature
Blue line: Delivered Power

Proportional Control of the Thermal
System
• proportional controller
attempts to perform
better than the On-Off
type by applying power
W to the heater in
proportion to the
difference in temperature
between the oven and
the set-point
• KP is known as the
proportional gain of the
controller
( ) P S O W = K ´ T -T

PD Control of the Thermal System
• add D-Control (proportional
to the time-derivative of the
error signal) to mitigate the
stability and overshoot
problems that arise from a
high gain proportional
controller
• adjust KD (the damping
constant) to achieve a
critically damped response
( ) ( )
W = K ´ T - T + K ´ d TS -
TO
dt
P S O D

PID Control of the Thermal System
• add I-Cotnrol (proportional to
the time-integral of the error
signal) to change the heater
power until the time-averaged
value of the temperature
error is zero
• KI is the integral gain
parameter
( ) ( ) ( )
W K T T K T T dt K d TS TO
= ´ - + ´ ò - + ´ -
dt
P S O I S D D

Analysis of a Control System using
Transfer Function
: ( ) 10 3 + 2 + +
=
s s s
The process G s P
given that, 6 11 6
and the feedback path: H(s) = 1

Transfer Function
for P-control only, if KP = 3, then GC(s) = KP =3, and
30
3 + 2 + +
G s G s
C P
G s G s s s s
6 11 36
= ×
( ) ( )
+ ×
1 ( ) ( )
=
C P
q
oq
i

Transfer Function
for PI-control, if KP = 2.7
and TI = 1.5, then
K T s
ö
æ
( ) 1 1 ( 1) = 2.7 (1.5 +1)
= × + ÷ ÷ø
ç çè
= +
C P T s
1.5
G s K
P I
I
×
and the transfer function is:
s
T s
I
×
= +
40.5 27
s
s
4 + 3 + 2 + +
s s s s
1.5 9 16.5 49.5 27
G s G s
= ×
( ) ( )
C P
G s G s
+ ×
1 ( ) ( )
C P
q
o
q
i

Transfer Function
for PID-control, if KP = 2,
TI = 0.9 and TD = 0.6, then
2 2 + + =
×
G s K T T s T s
= × × + × +
( ) ( 1) 2 (0.54 0.9 1)
P I D I
C T s
0.9
I
and the transfer function is:
2
s s
s s
s
= + +
10.8 18 20
4 3 2
s s s s
+ + + +
0.9 5.4 20.7 23.4 20
G s G s
= ×
( ) ( )
C P
G s G s
+ ×
1 ( ) ( )
C P
q
o
q
i

Transfer Function
The transfer functions give the
step responses as shown on
the right
• for P-control (the red curve) –
a steady state error occurs
• for PI-control (the blue curve)
– the response becomes more
oscillatory and needs longer to
settle, the error disappears
• for PID-control (the green
curve) – the overshoot and the
number of oscillatory cycles
are much reduced

Concluding remarks of the PID Effect
• In general, we may observe that
• P term is used to adjust the speed of response.
• I term provides zero error.
• D term introduces damping.

3. Stability of Control System
• a control system responds to an input by undergoing a
transient response before reaching a steady-state
• the total response of a system consists of two parts,
namely, the natural response and the forced response
• natural response describes the way the system
dissipates or acquires energy, the nature of this
response is dependent only on the system
• the nature of the forced response is dependent on the
input

• for a linear system, we can write
Total response = Natural response + Forced response
• for a control system to be useful, the natural response
must eventually approach zero, thus leaving only the
forced response
• a stable control system will always return to a stable
operating state
• in an unstable system, any disturbance will result in
oscillations building up until some parts fails

• Oscillatory System
• between the stable state and the unstable state
lies the conditionally stable system in which
oscillations neither increase nor decrease
• each cycle being identical to the previous one and
results in sustainable oscillation.

Stability of First Order System
Consider the response of a control
system with transfer function
(s+2)/(s+5) under a step input
with R(s) = 1/s
• a pole on the real axis
generates an exponential
response of the form e-αt, where
–α is the pole location on the
real axis.
• if α is positive, the transient
response will decay to zero
• if α is negative, then the
transient response will grow and
the system will be unstable.

Stability of Second Order System
As long as the poles of the
output function lies on the
left hand side of the
complex plane
• the system output will not
grow without bound
• it will be stable

Stability of Second Order System

Stability of Higher Order System

Stability of Higher Order System - Routh table
 The Routh table method will
yield the stability information of
a system without the need to
solve for the system poles.
 The method requires two steps:
(1) generate a Routh table
 For a fourth order system given
on the right,
 Construct the table for the
denominator by using the
formulae shown on the right
Similar formulae are used for system
with order higher than 4.
s4 a4 a2 a0
s3 a3 a1 0
s2 b1=(a2*a3-a1*a4)/a3 b2=(a0*a3- 0*a4)/a3 0
s1 c1=(a1*b1-a3*b2)/b1 0 0
s0 d1=(b2*c1-0*b1)/c1 0 0

(2) interpret the Routh table
 The number of poles that are
in the right half plane is
equal to the number of sign
change in the first column of
the Routh table.
Note: For the special cases such
as the element in the first
column is equal to zero or the
elements in the entire row
are equal to zero will not be
treated here, please refer to
descriptions in books dealing
with control theory.
s4 a4 a2 a0
s3 a3 a1 0
s2 b1=(a2*a3-a1*a4)/a3 b2=(a0*a3- 0*a4)/a3 0
s1 c1=(a1*b1-a3*b2)/b1 0 0
s0 d1=(b2*c1-0*b1)/c1 0 0

Example
 Determine the stability of the
system on the right by Routh
table method.
Solution
As there are two sign changes in
the first column of the Routh
table, so there are two poles
lying in the right half plane
and hence the system is not
stable.
s3 a3 = 1 a1 = 31 0
s2 a2 = 10 a0 = 1030 0
s1 b1=(31*10-
1*1030)/10
= -72
0 0
s0 c1 = (1030*(-72)-
0*10)/(-72)
= 1030
0 0

Summary of Introduction to Control
System
• Closed-loop control system is less sensitive to noise,
disturbances and changes in the environment.
• The transfer function of a control system is the ratio of the output
to the input.
• Proportional control is used to adjust the speed of response.
• Integral control provides zero error.
• Differential control introduces damping.
• If the poles of the output function lies on the left hand side of the
complex plane, the system will be stable.

Introduction to Control System
End of Lecture 1
 Revision
Norman S. Nise, Control Systems Engineering,
Fourth Edition, Johne Wiley & Sons, Inc., page
177 to page 183.

Eee3420 lecture01 rev2011

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