LINEAR CONTROL SYSTEMS

LINEAR CONTROLLINEAR CONTROL SYSTEMS SYSTEMS

Ali KarimpourAssociate Professor

Ferdowsi University of Mashhad

Lecture 16

Dr. Ali Karimpour May 2013

2

Lecture 16

Topics to be covered include: Design of controller in time domain.

Various controller configurations.Different kind of controllers. Controller realization.

Time domain design of the PID controllers. Design of PID controllers. Design of PD controllers. Design of PI controllers.

Time domain design of control systems

Lecture 16


3

Various controller configurations. ساختارهای کنترلیمتفاوت

)(sGp)(sGc+-

r(t) e(t) u(t) c(t)1 Series or cascade compensation. ساختار کنترلی سری1

2 State-feedback control. کنترل فیدبک حالت2

K

+-

r(t) u(t) x(t))(sGp

c(t)D

3 Forward compensation with series compensation. (Two degree of freedom) جبران سازی پیش رو با جبران سازی سری )دو درجه آزادی(3

+

-

r(t) e(t) u(t) c(t))(sGp)(1 sGc)(2 sGc

Lecture 16


44

Various controller configurations. ساختارهای کنترلیمتفاوت

4 Feed forward compensation.(Two degree of freedom)

کنترلر پیش خور 4)دو درجه آزادی(

)(2 sGc

+r(t) e(t) u(t) c(t))(sGp+

+)(1 sGc-

Controller

Controller

Controlledprocess

Lecture 16


Series Compensation Structure

In this Course we consider the series or cascade compensation.

جبران سازی سری

Controller

In This course

Different kind of

controllers

PID controllers.

Lead lag controllers

H2 Controllers

H∞ Controllers

Intelligent Controllers

………….. Adaptive ControllersNN Controllers…………..

In Graduate courses

Lecture 16


66

PID Controllers

PID has become almost universally used in industrial control. These controllers have proven to be robust and extremely beneficial in the control of many important applications.PID stands for: P (Proportional)

I (Integral)D (Derivative)

The standard form PID are:

An Alternative form for PID

کنترلر PID

Proportional only:

Proportional plus Integral:

Proportional plus derivative:Proportional, integral and derivative: sK

sKKsG d

ipPID )(

sKKsG i

pPI )(

sKKG dpPD

pP KsG )(

1sKsK

d

d

11)(

sKsK

sKKsG

d

dipPIDseries

Lecture 16


77

Lead-lag Compensators

Closely related to PID control is the idea of lead-lag compensation. The transfer function of these compensators is of the form:

If a<1 , then this is a lag network. Or (z>p) in other form.

If a>1 this is a lead network. Or (z<p) in other form.

کنترلر پیش فاز پس فاز

11)(

ssaksG

pszsksG

)( or

Lecture 16


PID and Operational Amplifiers کنترلرPID و تقویت کننده عملیاتی

88)()(

)()()(

sZsZ

sVsVsG

i

f

i

o

Lecture 16


99


pin

f

in

out KRR

sVsVsG )()()(

sKRCsCs

RsVsVsG D

in

out /1)(

)()(

sKRCsRCs

sVsVsG i

in

out //1/1)()()(

)()(

)()()(

sZsZ

sVsVsG

i

f

i

o

Lecture 16


1010


)()(

)()()(

sZsZ

sVsVsG

i

f

i

o

sCR

sCR

sVsVsG

i

o

11

22

11

1

)()()(

sCRsCRCR

CRCRsG 122121

1122 1)()(

Lecture 16


11

0 5 10 15 20 25 300

1

2Step Response

Time (sec)

Amplitu

de

0 5 10 15 20 25 30-1

0

1Step Response

Time (sec)Am

plitu

de

0 5 10 15 20 25 30-1

-0.5

0

0.5Impulse Response

Time (sec)

Amplitu

de

)(tc

)(te

)(te

Effects of the PD control on the time response.

بر پاسخ زمانیPDتاثیر کنترلر

)2(

2

n

n

ss

)(sCPK

sKD

+

+

+-

)(sR )(sE

PD controller

Derivative part can improve the oscillation.

جمله مشتق می تواند رفتار گذرا را بهبود بخشد.

Lecture 16


12

Effects of the PI control on the time response. بر پاسخ زمانیPIتاثیر کنترلر

)2(

2

n

n

ss

)(sCPK

sK I

+

++ -

)(sR )(sE

PI controller

Loop transfer function without controller

)2()(

2

n

n

sssG

)2(

)()(2

n

nIPc sss

KsKsGsG

Loop transfer function with controller

PI controller can improve error by increases the type of system by one

Lecture 16


13

Tuning of PID Controllers

Because of their widespread use in practice, we present below several methods for tuning PID controllers. Actually these methods are quite old and date back to the 1950’s. Nonetheless, they remain in widespread use today.In particular, we will study.

Ziegler-Nichols Oscillation Method Ziegler-Nichols Reaction Curve Method Cohen-Coon Reaction Curve Method Time domain design Frequency domain design

PIDتنظیم کنترلرهای

Lecture 16


14

Ziegler-Nichols Design

طراحی زیگلر نیکولز This procedure is only valid for open loop stable plants.

Open-Loop Tuning Closed-Loop Tuning

According to Ziegler and Nichols, the open-loop transfer function of a system can be approximated with time delay and single-order system, i.e.

where TD is the system time delay and T1 is the time constant.

Lecture 16


15

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.

طراحی زیگلر نیکولز حالت حلقه باز

Lecture 16


16

PIDKTT19.0

Kp Ki Kd

2127.0

DKTT

PIDDKTT12.1

P

طراحی زیگلر نیکولز حالت حلقه باز

DKTT1

216.0

DKTT

KT16.0

Ziegler-Nichols Reaction Curve Method(Open-Loop Case)

Lecture 16


17

Numerical Example

Consider step response of an open-loop system as:

مثال عددی

sesGTTCK

DsT

D 20140)(sec20sec,5,40 :So 1

Lecture 16


18

PI 09.09.0 1 DKTT

Kp Ki Kd

0054.027.021

DKTT

PID 12.02.1 1 DKTT

P 1.01 DKT

T

012.06.021

DKTT

3.06.0 1 K

T

sesGTTCK

DsT

D 20140)(sec20sec,5,40 :So 1

1.0)( sKP

ssKPI

0054.009.0)(

ss

sKPID 3.0012.012.0)(

Numerical Exampleمثال عددی

Lecture 16


19

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

Set the true plant under proportional control, with a very small gain.

Increase the gain until the loop starts oscillating. Note that linear oscillation is required and that it should be detected at the controller output.

Record the controller critical gain Kc and the oscillation period of the controller output, T.

Adjust the controller parameters according to Table

طراحی زیگلر نیکولز بروش نوسانی)حلقه بسته(

Lecture 16


20

PI cK45.0

Kp Ki Kd

TKc54.0

PID TKc2.1 TKc075.0cK6.0

P cK5.0

Ziegler-Nichols Oscillation Method(Closed-loop) طراحی زیگلر نیکولز بروش نوسانی)حلقهبسته(

Lecture 16


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.

21

Numerical Exampleمثال عددی

Lecture 16


22

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

17.2075.065.22.18.46.0 TKKTKKKK cd

cicp

Lecture 16


23

Response to step reference

0 5 10 150

0.5

1

1.5Step response for PID control

Time (sec)

Ampl

itude

پاسخ سیستم به پله

ss

sCPID 17.265.28.4)(

117.201.017.265.28.4)(

ss

ssCPID

Lecture 16


24

Time domain design طراحی حوزه زمانی

)25(40ss

)(sCk

+-

)(sR )(sE

را بگونه ای تنظیم کرد که kدر سیستم زیر آیا می توان گردد؟0.707نسبت میرائی قطبهای مختلط سیستم

Is it possible to set the value of k such that the damping ratio of complex poles be 0.707?

-25 0

k

k

0k0k

45

45

?40 k25.12

8125.7 k

Yes

بله

25.12

1

Lecture 16


25


)25(40ss

)(sCk

+-

)(sR )(sE

را بگونه ای تنظیم کرد که ثابت kدر سیستم زیر آیا می توان گردد؟100خطای شیب معادل

Is it possible to set the value of k such that ramp error constant be 100?

100vk

100)25(

40lim0

ss

kss

250040 k

5.62 k

Yes

بله

Lecture 16


26


)25(40ss

)(sCk

+-

)(sR )(sE

را بگونه ای تنظیم کرد که kدر سیستم زیر آیا می توان ثابت خطای و 0.707نسبت میرائی قطبهای مختلط سیستم

؟ گردد100شیب معادل

Is it possible to set the value of k such that the damping ratio of complex poles be 0.707 and ramp error constant be 100 ?

8125.7k

5.62k

Clearly the design is not possible???!!!???

Other controllers

Lecture 16


27

نسبت در سیستم ضرایب کنترلر را بگونه ای تنظیم کنید که ثابت خطای شیب و 0.707میرائی قطبهای مختلط سیستم

؟ گردد100معادل

Determine the controller coefficient such that the damping ratio of complex poles be 0.707 and ramp error constant be 100 ?

)25(40ss

)(sCskk DP

+-

)(sR )(sE

10025

40)25(

40)(lim0

P

sDPv

kss

skksk 5.62 Pk

0250025

401 2

ss

skD

Tuning PD controller طراحی کنترلر PD

Lecture 16


28

)25(40ss

)(sCskk DP

+-

)(sR )(sE

0250025

401 2

ss

skD

Tuning kD by graphical method.


Dk

-12.5

48.4

0Dk

0Dk

45Dk40-50 -35

22 4.135.22 22 4.835.22 22 3535

1426.1Dk

Lecture 16


29

)25(40ss

)(sCskk DP

+-

)(sR )(sE

0250025

401 2

ss

skD

Tuning kD by mathematical method

0250025

401 2

ss

skD 02500)4025(2 sks D

02: withCompare 22 nnss 5025002 nn

1425.140

7.457.7040252 DDn kk


Lecture 16


30

)25(40ss

)(sCskk DP

+-

)(sR )(sE 1425.15.62 DP kk

0 0.02 0.04 0.06 0.08 0.1 0.12 0.14 0.160

0.2

0.4

0.6

0.8

1

1.2

1.4Step Response

Time (sec)

Ampl

itude

100Clearly vk


Why P.O. > 4.3%

25007.70)5.621425.1(40

)()(

2

sss

sRsC

Lecture 16


31

Tuning PI controller طراحی کنترلر PI

)25(40ss

)(sCPk

skI

+

+

+-

)(sR )(sE

)25(40ss

)(sC

sksk IP +

-

)(sR )(sE

Clearly type of system is 2 so: vk

نسبت در سیستم ضرایب کنترلر را بگونه ای تنظیم کنید که ثابت خطای شیب و 0.707میرائی قطبهای مختلط سیستم

؟ گردد100معادل

Determine the controller coefficient such that the damping ratio of complex poles be 0.707 and ramp error constant be 100 ?

Lecture 16


32

Tuning PI controller (Continue) )ادامه(PIطراحی کنترلر

)25(40ss

)(sC

sksk IP +

-

)(sR )(sE

نسبت میرائی قطبهای مختلط سیستم حال نیاز داریم که گردد.0.707

We now need damping ratio of complex poles be 0.707.

Root loci with proportional controller

k

k

0k0k 45

45

0)25(

401

ss

kP

0)25(

40)/(1

ssskksk PI

P

5.12/Let PI kk

lociroot in changeimportant No -25 0

2/ PI kk

Pk4025.12 25.12

45.8Pk

22 5.125.10

25.12

Lecture 16


33

Tuning PI controller (Continue) )ادامه(PIطراحی کنترلر

را تعیین کنیم. kPحال نیاز داریم که We now need to set kP.

Root loci with PI controller

k

k

0k0k 45

45

0)25(

40)2(1

sssskP

-25 0

-10.9 -4.6-11.5

Lecture 16


34

)25(40ss

)(sC

skk I

P +

-

)(sR )(sE9.16245.8 PIP kkk

vkClearly

Tuning PI controller طراحی کنترلر PI

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.60

0.2

0.4

0.6

0.8

1

1.2

1.4Step Response

Time (sec)

Am

plitu

de

Why P.O. > 4.3%

67633825)2(338

)()(

23

sss

ssRsC

j5.123.11,37.2 :are Poles

Lecture 16


35

Compare PI and PD controllers مقایسه کنترلرهایPI PDو

)25(40ss

)(sC

skk I

P +

-

)(sR )(sE

9.1645.8 IP kk

)25(40ss

)(sCskk DP

+-

)(sR )(sE

1425.15.62 DP kk

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.60

0.2

0.4

0.6

0.8

1

1.2

1.4Step Response

Time (sec)

Ampl

itude

With PI controller

With PD controller

Lecture 16


36

Exercises تمرینها1- In the following system design a PID controller with Ziegler-Nichols Oscillation Method

)5.48(40ss

)(sC)(sGPID

+-

)(sR )(sE

2- In the following system design a PID controller with Ziegler-Nichols Oscillation Method

)10)(5(4

sss

)(sC)(sGPID

+-

)(sR )(sE

Lecture 16


37

Exercises تمرینها3 In the following system design a PD controller such that that the damping ratio of complex poles be 0.6 and ramp error constant be 80.

)5.48(40ss

)(sC)(sGPID

+-

)(sR )(sE

)10)(5(4

sss

)(sC)(sGPID

+-

)(sR )(sE

4 In the following system design a PD controller such that that the damping ratio of complex poles be 0.6 and ramp error constant be 80.

Lecture 16


38

Exercises تمرینها5 In the following system design a PI controller such that that the damping ratio of complex poles be 0.6 and ramp error constant be 80.

)5.48(40ss

)(sC)(sGPID

+-

)(sR )(sE

)10)(5(4

sss

)(sC)(sGPID

+-

)(sR )(sE

6 In the following system design a PI controller such that that the damping ratio of complex poles be 0.6 and ramp error constant be 80.

Lecture 16


39

Exercises تمرینها

Let the input impedance be generated by a resistor R2 be in series with a resistor R1 and a capacitor C1) that are in parallel, and let the feedback impedance be generated by a resistor Rf and a capacitor C f .a) Show that this choices lead to form a PID controller with high frequency gain limit as;

7 Consider following structure:

b) Derive the parameters in the controller.

LINEAR CONTROL SYSTEMS

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