linear control systems - personal...

Lecture 7

Dr. Ali Karimpour Dec 2017

LINEAR CONTROL

SYSTEMS

Ali Karimpour

Associate Professor

Ferdowsi University of Mashhad

Lecture 7


2

Lecture 7

Topics to be covered include:

Introduction. Various controller configurations.

Different kind of controllers.

Controller realization.

Controller design in time domain. PID controller design.

PD controller design.

PI controller design.

Lag controller design.

Lead controller design.

Time domain design of control systems




Lecture 7


3

Introduction

1 Control structure design.

2 Controller or compensator type.

3 Controller or compensator parameter tuning.

4 Checking the compensated system.

• Feedforward controller.

• Feedback or parallel compensation.

• Series or cascade compensation.

• Feedforward and series compensation.

• Series parallel compensation or cascade system.

• Series or cascade compensation.

• Lead controller or PD. • Lag controller or PI.

• Lead/lag controller or PID.

• Analytic controller design. • Graphic controller design.

• Controller design by table.

• Graphic controller design.• Analytic controller design.

Lecture 7


Series Compensation Structure

جبران سازی سریIn this Course we consider the

series or cascade compensation.

Controller

In This course

Different

kind of

controllers

PID controllers.

Lead lag controllers

H2 Controllers

H∞ Controllers

Intelligent Controllers

…………..Adaptive Controllers

NN Controllers

…………..

In Graduate courses

Lecture 7


PID and Operational Amplifiers

و تقویت کننده عملیاتی PIDکنترلر

55

)(

)(

)(

)()(

sZ

sZ

sV

sVsG

i

f

i

o

Lecture 7


6




Controller design in time domain. PID controller design.

PD controller design.

PI controller design.

Lag controller design.

Lead controller design.

Time domain design of control systems

Lecture 7


7

Time domain design طراحی حوزه زمانی

Cruise control of a vehicle.

Linear system is:

Transfer function ?

Closed loop and its specification?

We need zero steady state and settling time less than 1 sec?

Lecture 7


8


)25(

40

ss

)(sCk+

-

)(sR )(sE در سیستم مقابل آیا می توان کنترلر تناسبی یافت: 2مثال .کمتر گردد0/01که خطای شیب واحد در حالت دائم از

Example 2: Is it possible to design a P controller such that the steady

state error to unit ramp be less than 0.01?

)25(

40

ss

)(sCk+

-

)(sR )(sE

Example 3: Is it possible to design a P controller such that the steady

state error to unit ramp be less than 0.01 and damping ratio of complex

poles be 0.707 ?

What about other controller?

در سیستم مقابل آیا میتوان کنترلر تناسبی یافت که: 1مثال و کمتر0.707نسبت میرائی قطبهای مختلط سیستم بیشتر از

.از یک گردد

Example 1: Is it possible to design a P controller such that the damping

ratio of complex poles be larger than 0.707 and less than 1?

Lecture 7


9


Example 4: Design a controller such that the steady state error to unit

ramp be less than 0.01 and damping ratio of complex poles be 0.707?

در سیستم مقابل کنترلری طراحی کنید که ثابت خطای شیب: 4مثال گردد؟0.707و نسبت میرائی قطبهای مختلط سیستم 0/01واحد کمتر از

PD controller …..

PI controller …..

Lecture 7


10

Compare PI and PD controllers

PDو PIمقایسه کنترلرهای

)25(

40

ss

)(sC

s

kk I

P +

-

)(sR )(sE

9.1645.8 IP kk

)25(

40

ss

)(sC

skk 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)

Am

plit

ude

With PI controller

With PD controller

Lecture 7


11

Lag controller design procedure

رویه طراحی کنترلر پس فاز

)(sG)(sC

ps

zsk

+

-

)(sR )(sE

pz

1- Obtain the root-locus (without controller) and

determine the gain k0 to satisfy the desired

damping ratio or …...

2- Find the gain k to satisfy the desired steady

state error (without controller). If k is in

conflict with k0 continue.

3- Evaluate the needed controller gain

...or ratio damping desired hesatisfy t Gain to

error state-steady desired hesatisfy t Gain togain needed

0

k

k

4- Choose pole and zero of controller near origin such that:0

gain neededk

k

p

z

What is near?5 - Choose the controller as:ps

zsksGc

0)(

6 - Check the controller.

Why?

Lecture 7


12




و نسبت 0/01در حالت دائم کمتر از در سیستم مقابل کنترلری طراحی کنید که خطای شیب واحد: 5مثال .گردد0.707میرائی قطبهای مختلط سیستم

Lag controller …..

Lecture 7


13

Designing lag controller and its step response

)25(

40

ss

)(sC

1.0

8.08125.7

s

s+

-

)(sR )(sE

0 0.5 1 1.5 2 2.5 30

0.2

0.4

0.6

0.8

1

1.2

1.4Step Response

Time (sec)

Am

plit

ude

Lecture 7


14


Example 6: Design a controller such that the velocity constant be 50 and

percent overshoot be less than 16%.

% 16و درصد فراجهش کمتر از 50در سیستم مقابل کنترلری طراحی کنید که ثابت خطای شیب: 6مثال .باشد

Lag controller ….. )10)(5(

1

sss

)(sC

ps

zsk

+

-

)(sR )(sE

Lecture 7


15

Lag controller design

)10)(5(

1

sss

)(sC

01.0

2976.084

s

s+

-

)(sR )(sE

P.O. is not ok

Tune the controller to

derive the performance

0 2 4 6 8 10 120

0.2

0.4

0.6

0.8

1

1.2

1.4

Step Response

Time (sec)

Am

plit

ude

Lecture 7


16

Lag controller design

طراحی کنترلر پس فاز

When the design of lag controller is

not possible?

Lecture 7


17

Lead controller design procedure

رویه طراحی کنترلر پیش فاز

)(sG)(sC

ps

zsk

+

-

)(sR )(sE

pz

1- From the time-domain specifications

obtain the desired location of the closed-

loop dominant poles.

2- Select the zero of controller. Place the

zero on the real value of desired location of

the closed-loop dominant poles or on the

pole for pole-zero cancellation.

3- Locate the compensator pole so that the angle criterion is satisfied.

4- Determine the compensator gain k, such that the magnitude criterion is satisfied.

7 - If the overall response rise time, overshoot and settling time is not satisfactory, choose

another location of the closed-loop dominant poles.

5 - Choose the controller as:ps

zsksGc

)(

6 - Check the controller.

Lecture 7


18




و نسبت 0/01در حالت دائم کمتر از در سیستم مقابل کنترلری طراحی کنید که خطای شیب واحد: 7مثال .گردد0.707میرائی قطبهای مختلط سیستم

Lead controller …..

Lecture 7


19

Exercises

1 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(

40

ss

)(sC

)(sGPID

+

-

)(sR )(sE

)10)(5(

4

sss

)(sC

)(sGPID

+

-

)(sR )(sE2 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 7


20

Exercises

3 In the following system design

a PI controller such that that the



)5.48(

40

ss

)(sC

)(sGPID

+

-

)(sR )(sE

)10)(5(

4

sss

)(sC

)(sGPID+

-

)(sR )(sE

4 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 7


21

Exercises

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 in series with a

capacitor Cf .

a) Show that this choices lead to form a PID controller with high

frequency gain limit as;

5 Consider following structure:

b) Derive the parameters in the controller with respect to resistors and

capacitors.

Lecture 7


22

Exercises


a lag controller such that the



)5.48(

40

ss

)(sC)(sGlag

+

-

)(sR )(sE

)10)(5(

4

sss

)(sC)(sGlag

+

-

)(sR )(sE7 In the following system design a

lag controller such that the damping

ratio of complex poles be 0.6 and

ramp error constant be 80.

8 Design a lead controller for exercise 6.

9 Design a lead controller for exercise 7.

10- In the following system design a controller such that leads to zero

steady state error to step and 4.3% P.O.

(Final)

Lecture 7


23


a lag controller such that that the

settling time be less than 3 sec.

2

1

s

)(sC)(sGlag

+

-

)(sR )(sE


a lead controller such that that the

settling time be less than 3 sec.

2

1

s

)(sC)(sGlead

+

-

)(sR )(sE

Exercises


a Lead controller such that the


be 0.4.(Final 1390)

2

1

s

)(sC)(sGc

+

-

)(sR )(sE

Lecture 7


24

14- In the following system (Final 1395)

a) Draw root loci for K>0.)10(

81

ss

s )(sC

K+

-

)(sR )(sE

Exercises

b) Derive K for Ts=0.1s (Ts is Settling time)

c) Derive P.O. and velocity error constant for derived K in part “b”.

d) Derive a controller such that the velocity error constant multiplied

by 10 but P.O. and Ts remain the same as part “c”.

e) Derive a controller such that the velocity error be zero but P.O. and

Ts remain the same as part “c”.

f) Derive a controller such that the closed loop dominant poles are on

the -15±15j.

g) Derive step response of system for part “c” , “d” , “e” and “f”.

Lecture 7


25

Supplementary Exercises

15- In the following system design a

PID controller with

Ziegler-Nichols Oscillation Method

)5.48(

40

ss

)(sC

)(sGPID

+

-

)(sR )(sE

16- In the following system design a PID controller with Ziegler-Nichols

Oscillation Method)10)(5(

4

sss

)(sC

)(sGPID

+

-

)(sR )(sE

Lecture 7


26

Appendix: 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 7


27

Appendix: 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 7


28

PIDKT

T19.0

Kp Ki Kd

2

127.0

DKT

T

PIDDKT

T12.1

P

طراحی زیگلر نیکولز حالت حلقه باز

DKT

T1

2

16.0

DKT

T

K

T16.0

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

Lecture 7


29

Example 8: Following figure shows the step response of an

open-loop transfer function system. Design a PID controller.

s

esGTTCK

DsT

D201

40)(sec20sec,5,40 :So 1

Lecture 7


30

PI 09.09.0 1

DKT

T

Kp Ki Kd

0054.027.0

2

1 DKT

T

PID 12.02.1 1

DKT

T

P 1.01 DKT

T

012.06.0

2

1 DKT

T3.0

6.0 1 K

T

s

esGTTCK

DsT

D201

40)(sec20sec,5,40 :So 1

1.0)( sKP

ssKPI

0054.009.0)(

ss

sKPID 3.0012.0

12.0)(

Example 8: Following figure shows the step response of an

open-loop transfer function system. Design a PID controller.

Lecture 7


31

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

(حلقه بسته)طراحی زیگلر نیکولز بروش نوسانی

Lecture 7


32

PI cK45.0

Kp Ki Kd

T

Kc54.0

PIDT

Kc2.1 TKc075.0cK6.0

P cK5.0

Appendix: Ziegler-Nichols Oscillation Method(Closed-loop)

(حلقه بسته)طراحی زیگلر نیکولز بروش نوسانی

Lecture 7


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.

33

Example 9: Design a PID controller for following system.

Lecture 7


34

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 TKKT

KKKK cd

cicp

Lecture 7


35

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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