control in...

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

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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.

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3

Feedback Control Design

v Introduction.







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4

Introduction

PID Control

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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

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v Introduction.







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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

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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

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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

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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

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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

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v Introduction.







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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

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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

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Numerical Example

Example2: Consider step response of an open-loop system as:

مثال عددی

s

esGTTCK

DsT

D201

40)(sec20sec,5,40 :So 1

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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

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v Introduction.







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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:

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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

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20



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!

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21



)(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

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v Introduction.







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Problem arise in D part of controller:

PID Controller Problems

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Solving the problem arise in D part of controller:


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Problem arise in Integral part of controller (Windup):


Consider the

system:

But always in real systems we have limiter so:

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No limit case

With limiter

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With limiter

This is windup!

Removing windup

effect?

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This is controller with anti-windup.

How to choose Tt?

Choose TD < Tt < Ti A rule of thumb suggest Tt = √TDTi

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PID controller with Anti-windup.

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Industrial PID controller with Anti-windup.

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v Introduction.







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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.

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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.

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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

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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.

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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.

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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

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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

control_in_inst_4_sim1.mdl

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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.

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A simple example implemented by PLC

Example 4:

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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

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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

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