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lecture 5
Dr. Ali Karimpour Sep 2015
Ali Karimpour
Associate Professor
Ferdowsi University of Mashhad
CONTROL IN
INSTRUMENTATION
References:
1- Modern Control Technologies: Components and Systems, 2nd Edition, by Kilian,
Delmar Publication Co, 2005
2- Principles and Practice of Automatic Process Control. 3rd Edition, by C. A. Smith
and A. B. Corripio
3- Power Points of Dr. Hamed Molla Ahmadian
lecture 5
Dr. Ali Karimpour Sep 2015
2
Lecture 5
Feedback Control DesignTopics to be covered include:
v Introduction.
v Tuning of PID controllers.
u Ziegler-Nichols Oscillation Method(Closed-loop).
u Ziegler-Nichols Reaction Curve Method(Open-Loop Case).
u Controller Synthesis Method (Dahlin Response).
v PID Controller Problems.
v Analysis of Some Common Loop.
lecture 5
Dr. Ali Karimpour Sep 2015
3
Feedback Control Design
v Introduction.
v Tuning of PID controllers.
u Ziegler-Nichols Oscillation Method(Closed-loop).
u Ziegler-Nichols Reaction Curve Method(Open-Loop Case).
u Controller Synthesis Method (Dahlin Response).
v PID Controller Problems.
v Analysis of Some Common Loop.
lecture 5
Dr. Ali Karimpour Sep 2015
4
Introduction
PID Control
lecture 5
Dr. Ali Karimpour Sep 2015
5
Tuning of PID Controllers
Because of their widespread use in practice, we present below several methods for tuning PID controllers.
In particular, we will study.
u Ziegler-Nichols Oscillation Method
u Ziegler-Nichols Reaction Curve Method
u Cohen-Coon Reaction Curve Method
u Controller Synthesis(Dahlin Response)
u Minimizing ISE or IAE
u Time Domain Design
u Frequency Domain Design
PIDتنظیم کنترلرهای
)1
1(_ sTsT
KControllerPIDd
i
p )1)(
11(_ '
'sT
sTKControllerPID
d
i
p
lecture 5
Dr. Ali Karimpour Sep 2015
6
Feedback Control Design
v Introduction.
v Tuning of PID controllers.
u Ziegler-Nichols Oscillation Method(Closed-loop).
u Ziegler-Nichols Reaction Curve Method(Open-Loop Case).
u Controller Synthesis Method (Dahlin Response).
v PID Controller Problems.
v Analysis of Some Common Loop.
lecture 5
Dr. Ali Karimpour Sep 2015
7
Ziegler-Nichols Oscillation Method(Closed-loop)
This procedure is only valid for open loop stable plants
and it is carried out through the following steps
u Set the true plant under proportional control, with a very
small gain.
u Increase the gain until the loop starts oscillating. Note that
linear oscillation is required and that it should be detected
at the controller output.
u Record the controller critical gain Kc and the oscillation period of the
controller output, T.
u Adjust the controller parameters according to Table
(حلقه بسته)نوسانیبروشنیکولززیگلرطراحی
lecture 5
Dr. Ali Karimpour Sep 2015
8
PI cK45.0
Kp Ti Td
T83.0
PID T5.0 T125.0cK6.0
P cK5.0
Ziegler-Nichols Oscillation Method(Closed-loop)
(حلقه بسته)نوسانیبروشنیکولززیگلرطراحی
)1
1(_ sTsT
KControllerPIDd
i
p
This method leads to
quarter decay ratio response
lecture 5
Dr. Ali Karimpour Sep 2015
Example1: Consider a plant with a model given by
Find the parameters of a PID controller using the Z-N
oscillation method. Obtain a graph of the response to
a unit step input reference.
9
Numerical Example مثال عددی
lecture 5
Dr. Ali Karimpour Sep 2015
10
Solution
Applying the procedure we find:
Kc = 8 and ωc = 3. T=3.62
Hence, from Table, we have
The closed loop response to a unit step in the reference at t
= 0 is shown in the next figure.
حل
4525.0125.081.15.08.46.0 TTTTKKdicp
)4525.081.1
11(8.4_ s
sControllerPID
lecture 5
Dr. Ali Karimpour Sep 2015
11
Response to step reference
0 5 10 150
0.5
1
1.5Step response for PID control
Time (sec)
Am
plit
ude
پاسخ سیستم به پله
ss
sCPID 17.265.2
8.4)(
117.201.0
17.265.28.4)(
s
s
ssCPID
lecture 5
Dr. Ali Karimpour Sep 2015
12
Feedback Control Design
v Introduction.
v Tuning of PID controllers.
u Ziegler-Nichols Oscillation Method(Closed-loop).
u Ziegler-Nichols Reaction Curve Method(Open-Loop Case).
u Controller Synthesis Method (Dahlin Response).
v PID Controller Problems.
v Analysis of Some Common Loop.
lecture 5
Dr. Ali Karimpour Sep 2015
13
Ziegler-Nichols Reaction Curve Method(Open-Loop Case)
For open-loop tuning, we first find the plant parameters by
applying a step input to the open-loop system.
The plant parameters K, TD and T1 are then found from the
result of the step test as shown in Figure.
حالت حلقه بازنیکولززیگلرطراحی
lecture 5
Dr. Ali Karimpour Sep 2015
14
PIDKT
T19.0
KP Ti Td
PIDDKT
T12.1
P
حالت حلقه بازنیکولززیگلرطراحی
DKT
T1
DT2
DT5.0
Ziegler-Nichols Reaction Curve Method(Open-Loop Case)
)1
1(_ sTsT
KControllerPIDd
i
p
DT33.3
This method leads to
quarter decay ratio response
?When
6.0/1.01 TT
D
lecture 5
Dr. Ali Karimpour Sep 2015
15
Numerical Example
Example2: Consider step response of an open-loop system as:
مثال عددی
s
esGTTCK
DsT
D201
40)(sec20sec,5,40 :So 1
lecture 5
Dr. Ali Karimpour Sep 2015
16
PID
KT
T1
9.0
Kp Ti Td
DT33.3
PIDD
KT
T1
2.1
PD
KT
T1
s
esGTTCK
DsT
D201
40)(sec20sec,5,40 :So 1
1.0)( sKP
ssKPI
0054.009.0)(
ss
sKPID 3.0012.0
12.0)(
Numerical Exampleمثال عددی
DT2
DT5.0
lecture 5
Dr. Ali Karimpour Sep 2015
17
Feedback Control Design
v Introduction.
v Tuning of PID controllers.
u Ziegler-Nichols Oscillation Method(Closed-loop).
u Ziegler-Nichols Reaction Curve Method(Open-Loop Case).
u Controller Synthesis Method (Dahlin Response).
v PID Controller Problems.
v Analysis of Some Common Loop.
lecture 5
Dr. Ali Karimpour Sep 2015
18
Controller Synthesis Method (Dahlin Response)
(داهلینبر اساس پاسخ )کنترلرتحلیلی طراحی
)()(1
)()(
)(
)()(
sGsG
sGsG
sR
sCsT
c
c
)(1
)(
)(
1)(
sT
sT
sGsG
c
Following response was suggested by Dahlin
1
1)(
sTsT
c
sTsGsG
c
c
1
)(
1)(
Tc is the time constant of the closed loop response and, being adjustable:
lecture 5
Dr. Ali Karimpour Sep 2015
19
Controller Synthesis Method (Dahlin Response)
(داهلینبر اساس پاسخ )کنترلرتحلیلی طراحی
sTsGsG
c
c
1
)(
1)(
Let G(s) be a constant process so:
controller integralpurea11
)(sTK
sGc
c
Let G(s) be an integral process so:
s
KsG )( controller alproportion purea
11)(
cc
c
KTsTK
ssG
Let G(s) be a FO process so:
1)(
Ts
KsG controller PIa)
11(
11)(
TsKT
T
sTK
TssG
cc
c
lecture 5
Dr. Ali Karimpour Sep 2015
20
Controller Synthesis Method (Dahlin Response)
(داهلینبر اساس پاسخ )کنترلرتحلیلی طراحی
sTsGsG
c
c
1
)(
1)(
Let G(s) be a second order process so:
Let G(s) be a FOTD process so:
)1)(1()(
21
sTsT
KsG )1)(
11(
1)1)(1()(
2
1
121
sTsTKT
T
sTK
sTsTsG
cc
c
LseTs
KsG
1)(
Ls
cc
Lsce
TsKT
T
sTKe
TssG )
11(
11)(
It is not realizable!
lecture 5
Dr. Ali Karimpour Sep 2015
21
Controller Synthesis Method (Dahlin Response)
(داهلینبر اساس پاسخ )کنترلرتحلیلی طراحی
)(1
)(
)(
1)(
sT
sT
sGsG
c
Let Ls
c
esT
sT
1
1)(
Let G(s) be a FOTD process so:
LseTs
KsG
1)(
Ls
cc
Lsce
TsKT
T
sTKe
TssG )
11(
11)(
It is not
realizable!
Ls
c
Ls
c
Ls
Lsc
esTK
Ts
esT
e
Ke
TssG
1
11
1
1)(
sL
sL
e Ls
21
21
)'1
21
)(1
1()(
)(sT
sL
TsLTK
TsG
c
c
)(2
'LT
LTT
c
c
lecture 5
Dr. Ali Karimpour Sep 2015
22
Feedback Control Design
v Introduction.
v Tuning of PID controllers.
u Ziegler-Nichols Oscillation Method(Closed-loop).
u Ziegler-Nichols Reaction Curve Method(Open-Loop Case).
u Controller Synthesis Method (Dahlin Response).
v PID Controller Problems.
v Analysis of Some Common Loop.
lecture 5
Dr. Ali Karimpour Sep 2015
23
Problem arise in D part of controller:
PID Controller Problems
lecture 5
Dr. Ali Karimpour Sep 2015
24
Solving the problem arise in D part of controller:
PID Controller Problems
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Dr. Ali Karimpour Sep 2015
25
Problem arise in Integral part of controller (Windup):
PID Controller Problems
Consider the
system:
But always in real systems we have limiter so:
lecture 5
Dr. Ali Karimpour Sep 2015
26
Problem arise in Integral part of controller (Windup):
PID Controller Problems
No limit case
With limiter
lecture 5
Dr. Ali Karimpour Sep 2015
27
Problem arise in Integral part of controller (Windup):
PID Controller Problems
With limiter
This is windup!
Removing windup
effect?
lecture 5
Dr. Ali Karimpour Sep 2015
28
Problem arise in Integral part of controller (Windup):
PID Controller Problems
This is controller with anti-windup.
How to choose Tt?
Choose TD < Tt < Ti A rule of thumb suggest Tt = √TDTi
lecture 5
Dr. Ali Karimpour Sep 2015
29
Problem arise in Integral part of controller (Windup):
PID Controller Problems
lecture 5
Dr. Ali Karimpour Sep 2015
30
PID Controller Problems
PID controller with Anti-windup.
lecture 5
Dr. Ali Karimpour Sep 2015
31
PID Controller Problems
Industrial PID controller with Anti-windup.
lecture 5
Dr. Ali Karimpour Sep 2015
32
Feedback Control Design
v Introduction.
v Tuning of PID controllers.
u Ziegler-Nichols Oscillation Method(Closed-loop).
u Ziegler-Nichols Reaction Curve Method(Open-Loop Case).
u Controller Synthesis Method (Dahlin Response).
v PID Controller Problems.
v Analysis of Some Common Loop.
lecture 5
Dr. Ali Karimpour Sep 2015
33
Flow Control
Applications: inlet flow control, outlet flow control of processes, reflux flow rate, pipeline flow rate,…,etc.
:Implementation
Remarks:
Due to the effect of turbulence and pressure fluctuation, the measurement is noisy.
No offset is allowed in flow rate control – integral action is necessary.
Flow rate process is fast – no need for derivative actions.
Conclusion: PI control is needed with low gain.
lecture 5
Dr. Ali Karimpour Sep 2015
34
Liquid Level
Applications: reactor volume control, buffer tank level control, reboiler level control accumulator level control, steam generator level control, …,etc.
:Implementation
Remarks:
The level process is basically noisy – fluctuation of the liquid level –low controller gain
In case of important levels such as reboiler level, accumulator, integral action is needed,
The level processes are basically a first order system.
lecture 5
Dr. Ali Karimpour Sep 2015
35
Liquid Level
Conclusion:
The tank level control is basically loose, for instance, to maintain the tank is not completely empty at low inlet flow rate and to maintain tank is not full at high inlet flow rate. Thus, a low gain P controller is frequently implemented.
However, in case of important level system such as reboiler level, accumulator, PI controllers should be used.
:Undesired behavior
lecture 5
Dr. Ali Karimpour Sep 2015
36
Air pressure control
Applications: Gas storage tank.
Remarks:
Vapor pressure control is not this case, it should be considered as temperature control in the next slide.
The measurement is not noisy.
Gas pressure system is fast – no derivative action is needed .
Conclusion:
Use high gain P-only controllers.
lecture 5
Dr. Ali Karimpour Sep 2015
37
Temperature control
Applications: Reactor temperature control, heat exchanger temperature control, temperature control of pre-heaters, vapor pressure control,…,etc.
Remarks:
The quality of sensor is crucial for this type of control, for instance, sensor noise and time lag will influence the control quality,.
Off set of the temperature is not allowed – integral action is needed,
The system is quite slow (heat transfer mechanism), derivative action is needed,
Manipulated variables :
Cooling water flow rate, steam flow rate.
،There exists an inherent upper bound of the controller gain, process stability is an issue.
Conclusion: Use a PID controllers.
lecture 5
Dr. Ali Karimpour Sep 2015
38
Composition control
Applications: pH control, reactor composition control, distillation
composition control, …,etc.
Remarks:In some cases, the measurements are noisy with time lag ,makes the control very difficult,
No offset is allowed – integral action.
The process is typically slow – derivative action.
Conclusion: Use a PID controllers, PI may be used in some cases
lecture 5
Dr. Ali Karimpour Sep 2015
39
Level control
Example 3:
a) Derive Tank Height Equations.
A=100m2, Fi=0.5 m3/sec, Fo=0.1h
dt
dhAFF oi tank,
dt
dhAFFR oivalve
dt
dhhFR ivalve 1001.0
1.0100)(
)()(
s
F
sR
shsg i
valve 1.0100
5.0
s
So open loop transfer function is:
)(1
)()(
skg
skgsT
b) Design a proportional controller that control the height.
ks
k
5.01.0100
5.0
?)10( kforess v Simulation file
lecture 5
Dr. Ali Karimpour Sep 2015
40
A simple example implemented by PLC
Example 4:
A batch process(filling a vat with a liquid, mixing
the liquid, and draining the vat) is automated
with a PLC.
2. The liquid in the vat is mixed for 3 min.
1. A fill valve opens and lets a liquid into a vat
until it’s full.
3. A drain valve opens and drains the tank.
lecture 5
Dr. Ali Karimpour Sep 2015
41
A simple example implemented by PLC
Example 4:
lecture 5
Dr. Ali Karimpour Sep 2015
Exercises
42
2- What problem is solved and what new problem is created with the addition of integral
feedback(windup)?
1- Derive a PID controller for following system according three mentioned method in the
lecture.
)13)(130)(110(
8.0)(
ssssG
. شده استنوسانیمودسیستم وارد 10و کنترل کننده تناسبی با بهره r=5در شکل زیر با فرض استفاده از -3.به روش دلخواهPIDطراحی کنترل کننده مطلوبست. به صورت شکل نشان داده شده استyدر این شرایط نمودار
G(s)PIDu y
r
0 0.5 1 1.5 2 2.5 30
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
lecture 5
Dr. Ali Karimpour Sep 2015
43
Exercises
در این فرآیند آبمیوه از سمت چپ شکل وارد و از سمت. یک سیستم کنترل دما در شکل نشان داده شده است-4هدف کنترل دمای آبمیوه. دیده است، از فرآیند حرارتی خارج می گرددحراراتراست آبمیوه ای که به وسیله بخار آب
.خروجی می باشد.حوزه زمان را با جزئیات تشریح نمائیدنیکولززیگلربه روش PIDآزمایش الزم جهت طراحی کنترل کننده ( الف.برای انتخاب شیر برقی مسیر بخار، توجه به دبی مورد نیاز جهت کنترل دمای آبمیوه خروجی ضروری است( ب
.یک پیشنهاد عملی جهت تعیین دبی شیر برقی انتخابی جهت کنترل مناسب دمای ماده خروجی ارائه نمائید
اگر تابع انتقال دمای ماده خروجی به فرمان اعمالی به شیر بخار به( پ
در طراحی کنترل کننده به روش. باشدصورت
مقداری از ضریب بهره تناسبی سیستم ازایبه نیکلززیگلرنوسانیدر این شرایط فرکانس نوسانات چقدر است؟. می شودنوسانی
3
1( )
(10 )G s
s