1 chap 6 the compensation of the linear control systems p553

1

Chap 6 The Compensation of the linear control systems

P553

2

Chap 6 The Compensation of the linear control systems

§6-1 Introduction 6.1.1 definition of compensation 6.1.2 types of compensation§6-2 The basic controller operation analysis 6.2.1 PI D controller ---active compensation 6.2.2 phase-lead controller 6.2.3 phase-lag controller 6.2.4 phase lag-lead controller§6-3 Cascade compensation method of Root loci §6-4 Cascade compensation method of frequency- Domain§6-5 Feedback compensation

passive compensation controller

3

6.1 Introduction

6.1.1 What is compensation or correction of a control system ?

))(()()( :example For

11

sTss

KsHsG

stable be can system loop-closed this make

:get can wecriterion, Hurwitz-Routh to According

0)T 0(K

TT

TK

11

1

)()()( :if But

12

Tss

KsHsG

.or varying only stable benot can system

loop-closed thisCriterion, Hurwitz-Routh to ding Accor

TK

solution

4

TτTss

sKsHsG

)1(

)1()()( :make we If

2

This closed-loop system can be stable. We make the system stable by increasing a component.

6.1 Introduction

This procedureis called the compensation or correction.

Definition of the compensation: increasing a component ,which makes the system’s performance to be improved, other than only varying the

system’s parameters, this procedure is called the compensation or correction of the system.

5

Compensator:

The compensator is an additional component or circuit that is inserted into a control system to compensate for a deficient performance.

r.compensato a is 1s stable, be can system the

component, 1s increase to ,)(

)()( :Example

12 Tss

KsHsG

6.1 Introduction

6.1.2 Types of the compensation

:types severalget can we system, the of

structure the in )( of location the to according and ),(

as designated isr compensato the of functiontransfer The

sGsG cc

6

(1) Cascade(or series) compensation

(2) Feedback compensation

(3) Both series and feedback compensation

(4) Feed-forward compensation

(1) Cascade(or series) compensation

Features : simple but the effects to be restricted.

6.1 Introduction

7

(2) Feedback compensationR(s) C(s)

)(0 sG

CG

- -R(s) C(s)

10G 20G

CG

- -

Features: complicated but noise limiting, the effects are more

than the cascade compensation.

(3) Both cascade and feedback compensation -

R(s) C(s)

-1CG

2CG

0G

Features: have advantages both

of cascade and feedback compensation.

6.1 Introduction

8

(4) Feed-forward compensation

Features: theoretically we can make the error of a system to be zero and no effects to the transient performance of the system.

6.1 Introduction

R(s) C(s)+

-

CG

10G 20GC(s)

F(s)

R(s)

-+

CG

10G 20G

For input For disturbance(voice)

Demonstration:

9

R(s) C(s)+

-

CG

10G 20G

For input

6.1 Introduction

0)( 1

:

)(1

1

)(1

1

)()()(

20

2010

20

2010

202010

sEG

GMake

sRGG

GG

sRGG

GGGG

sCsRsE

C

C

C

But no effect to the characteristic equation: 1+ G10G20 = 0

Question: actually the could not be easy implemented

especially maybe the G20 is variable.20

1

GGC

10

6.1 Introduction

C(s)

F(s)

R(s)

-+

CG

10G 20G

For disturbance(voice)01

1

1

1

10

2010

1020

2010

202010

)( :

)()(

)(

)()(

sEG

GMake

sRGG

GGG

sRGG

GGGG

sCsE

NC

C

C

FF

Also no effect to the characteristic equation: 1+ G10G20 = 0

Question: actually the could not be easy implemented

especially maybe the G20 is variable. And the F(s) could not be easy measured.

10

1

GGC

11

example

6.1 Introduction

C(s)

N(s)

R(s)

-

-

CNG

10G 20GE(s)

GCR

+

+

Fig.6.1.7

For the system shown in Fig.6.1.7:

Solution :

Determine GCR and GCN , makeE(s) to be zero.

Where:

150

2

5

20

10

sG

sG

.

).(. ; . 150501

201

2010 s

GGs

GG CRCN

Thinking: if r(t) = n(t) = t , Determine GCR and GCN , make ess to be zero — as a exercise.

12

6.2 Operation analysis of the basic compensators

6.2.1 Active Compensation

PID controller － active “compensator”.

Transfer function:

DpDI

pIDIp

DI

pc

KKK

KsKs

KK

sτsτ

K(s)G

;

)(

1

11

stability. improvingcontroller aldifferenti --D

. clearingcontroller gintegratin -- I

y.sensitivit promoting controller alproportion --P

sse

13

PD controller )( :functiontransfer sKKsG Dpc

C(s)G(s)

R(s)－

＋pK

sK D

＋＋

)(sGC

)2()(

:Assuming

2

n

nss

sG

)2(

)()()( :is system

dcompensate the of functiontransfer loopopen The 2

n

DPnc ss

sKKsGsG

D

P

K

Ks :at zero loopopen

a adding to equivalent is controller PD thethat showsIt

14

Effects of PD controller:

2) PD controller improve the system’s stability (to increase damping and reduce maximum overshoot);

3) PD controller reduce the rise time and settling time;

4) PD controller increase BW(Band Width) and

improve GM(Kg),PM(γc), and Mr .

1) PD controller does not alter the system type;


－ bring in the noise !

15

PI controller s

KKsG Ipc1

)( :functionTransfer

)2()( :Assuming

2

n

n

sssG

C(s)G(s)

R(s)－

＋pK

sK I

1

＋＋

)(sGC


)2(

)(

)2(

)1

()()(

:is system dcompensate the of functiontransfer loopopen The

2

22

n

IPn

n

IPnc

ss

KsK

sss

KKsGsG

0s :at pole a and :at zero

loopopen a adding to equivalent is controller PI thethat showsIt

P

IK

Ks

16

Effects of PI controller:

1) Increase the system’s type － clear the steady-state error ;

2) reduce BW(Band Width) and GM(Kg), PM(γc) and Mr ;


beneficial to the noise limiting ,not beneficial to the system’s stability.

17

G(s)R(s) C(s)

- +pK

sKI

1

sKD)(sGC

PID controller


Transfer function: sKs

KK(s)G DIpc 1

PID controller have advantages both of PI and PD.

18

Circuits of PID

_

+

C

R1ur

u0

PI controller

R2

_

+C

R1ur

u0

PD controller

R2

_

+C1

R1ur

u0

PID controller

C2R2

)()()(

CsRR

RUU

sR

s

21

20 11

)()()(

sCRR

RUU

sR

s11

1

20 1

?)()(

sR

sUU0

19

For example:

Disk driver control system

:onspecifcati following the

satisfy to system the make values, : DP , KKDetermine

)(for 250mst 5%% s ttr

20

:figure following in shown is 0:

for system the of loci-root the , choose design, of type one

D

DP

K

KK

))((

)()()()(

: functiontransfer loopopen

100020

500021

sss

zsKsGsGsG p

c . :p

D

K

Kzhere

solution

How to get？Shown in 6.3 detail.

21

6.2.2 Passive compensation controllers

ab

c

c

c

,

s1

s1

s1

s1(s)G controller lead-lag phase 3)

s1

s1(s)G controller lag-phase 2)

1 s1

s1(s)G controller lead-phase )

11

1

1

a

a

b

b

Types of passive compensation controller

22

1 )( :functionTransfer

controller lead-Phase )

ps

zs

τs

τssGc

1

1

1

1

τ p

τz

11

Zero and pole


23

τj

jjωGc

1

1)(

response Frequency

1

occurs whichat frequency the

and, phase, the of value Maxmum

mm

m

zpm

m

m

m

sin

sin

sin

:get we which from

tantan

)()(

1

1

1

11

11

jjGc

Effects are similar to PD.Compensation ideal:

make ωm to be ωc !

Bode plotz pzp

24

Circuit of the phase-lead controller

11

121

2

11

21

2

1

τsτs

CsRRR

R

CsR

RR

R(s)c G

21

21 RR

RCR

25

1 1

1s

1s)( :functionTransfer

controller lag-Phase )

ps

zssGc

2

βτ p

τz

11

Zero and pole


26

jω

jωjωGc

1

1)(

response Frequency

Effects are similar to PI.

Compensation ideal:

Make 1/τto be in the lower frequency-band and far from ωc !

Bode plot

2

1

2

1

27

Circuit of the Phase-lag controller

τsτs

sV

sVsG

in

oc

11

)(

)()(

2

212 R

RRC βRτ

28

1 1 )( :functiontransfer

controller lead-lag Phase )

sβτ

sτ

sατ

sτsG

b

b

a

ac 1

1

1

1

2

bb τ z

βτp

1111

aa τ z

τp

1122

j

1z 1p2z2p

Zero and pole


29

1 1

1

1

1

1)(

response Frequency

b

b

a

ac j

j

j

jjωG

Effects are similar to PID.

Compensation ideal:

First make the phase-lag compensation－ to satisfy ess

and compensate a part of γc .

second make the phase-lead compensation－ to satisfy the transitional requirements.

dBL( /)

dB/dec20

0

dB/dec 20

b1

a1

b1

a

1

)(

090

090

Bode plot

30

1

Circuit of the Phase lag-lead controller

31

6.2.3 Comparing active compensation controllers and passive compensation controllers

)1)(1(

)1)(1( : leadlag phase

)1(

)1

1(

:PID

1

1 : lagphase )1(

)1

1(

: PI

1

1 : leadphase )(1 : PD

2

s/ατsατ

sτsτ

s

ssK

sτsτ

K

s

s

s

sKsτ

K

s

ssτK

ba

ba

I

IDIp

DI

p

I

Ip

Ip

Dp


32

6.36.3 Cascade compensation by Root loci methodCascade compensation by Root loci method

6.3.1 Phase-lead compensation (P569)

21

s

KsGH )( :function transfer loop-open

system. the for oncompensati the determine

%2% overshoot 4s;t criterion) time(2% setting

:are system the of ionsspecificat ePerformanc

s 5

Example 6.3.1:

The root loci of the system shown in Fig.6.3.1

Fig.6.3.1

solution

Analysis: unstable. phase-lead compensation

33

),( :as roots

dominant desired the choose can we, % and to According

, snpnnd

s

tjS

t

2

21 1

0.44)( 21

: rootsdominant desired

144

%)4.25%(4.0%100% :of terms In

2,1

nn

1 2

jS

Choose

t

e

d

s


Fig.6.3.2

2121 jSd ,

34


0100

10

10

1

5321802180

180180

)()()(

)()()(

:have weloci-root of criterion phase the to According

tgpszs

sGHsGsGHG

cdcdc

dcdcdssc

1

,1

1

1

1)( :Applying

ccc

cc pz

ps

zs

s

s sG

Fig.6.3.3

czcp

c

1ds

There are two approaches to determinezc and pc .

(1) Maximum α method

)( c 2

1

(2) Method based on the open-loop gain

35


cd

c ctgsGH

ctg )(

csc

1

1

For this example we choose the Maximum α method:

0000

01

3263531802

1

632

)(

tg

Fig.6.3.3

czcp

c

1ds

In terms of the sine’s law:

24180

191180

10

10

. )sin(

)sin(

. )sin(

sin

cd

c

c

c

cd

c

ps

p

zs

z

36Root locus of the compensted system


12380

1840

s

ssGc .

.)( :have weSo

The root locus of the compensated system is shown in Fig.6.3.4

Fig.6.3.4

19.12.4

1ds

Steps of the cascade phase-lead Compensation:

(1) Determine the dominant roots based on the performance specifications of the

system: ),( , snpnnd tjS 221 1

(2) plot the root locus of the system and analyze what compensation device should be applied.

37


(3) Determine the angle φc to be compensated:

)()( ddcc sGHsG 00180

(4) calculate θ andγ:

cd

c ctgsGH

ctg )(

csc

1

1or )( c 2

1n

ntg

2

1 1

(6) plot the root locus of the compensated system and make validity check.

(5)

calculate zc and pc In terms of the sine’s law :

)sin(

)sin( ;

)sin(

sin

10

10 180180 d

c

c

c

d

cs

p

s

z

38

210 )s(s

KGH(s)

:system a of functiontransfer loop-open The

20:constanterror Speed

poles)dominant the0.707(for ratio Damping

:are system thefor ionsSpecificat

vK

Example 6.3.2:

Solution:

The root locus of the system is shown in Fig.6.3.5.

-10

045

Fig.6.3.5

6.3.2 Phase-lag compensation using the root locus (P577)

39


Fig 6.3.6

.at axis- the on

lie poles loop-closed two ,but

, :s Analysi

10

2000

200010

202

jjω

K

KK

Kv

. :poles theat gain the and

:poles the ,. When

236

9292

7070

21

K

.j.-s ,d

used. be should

oncompensati lag phase a andsly,synchronou

satisfied benot can . and 707020 vK

The detail of the root-loci is shown in Fig 6.3.6.

40

1. )(

:rcompensato lag-phase The

s

s

ps

zss G

c

cc 1

11

58236

2000.

: and of ratio the Make

uncompv

vdesire

c

c

cc

K

K

p

z

pz

05 )()( and cdcd pszs

.

.s(s)G

.8.5

0.1 and 0.1 :choose We

c

c

01180

10

01180

s

pz c

Fig 6.3.6


41

).(s)s(s

).K(s(s)GH(s)Gc

0118010

102

:dcompensate functiontransfer loop-open The

Validate……


Steps of the cascade phase-lag Compensation:

(1) Determine the dominant roots based on the performance specifications of the system:

),( , snpnnd tjS 221 1

(2) plot the root locus of the system and analyze what compensation device should be applied.

If the phase-lag Compensation be applied:

42


1. )( :rcompensato lag-phase The

s

s

ps

zssG

c

cc 1

11

uncompv

vdesire

c

ccc

K

K

p

zpz : and of ratio the Make (3)

satisfied. be to )()( and / :make to 05 cdcdcc pszszp : choose Rationally (4) cz

(5) plot the root locus of the compensated system and make validity check.

6.3.3 Phase lag-lead compensation by the root locus method

Basic ideal:

43

First: make the phase-lead compensation－ to satisfy the transitional requirements.

Second: make the phase-lag compensation－ to satisfy ess

requirements.

Make compensation using PD and PI for example 6.3.1 and example 6.3.2

Exercise:

6.3.3 Phase lag-lead compensation by the root locus method6.3.3 Phase lag-lead compensation by the root locus method

44

6.4.1 Phase-Lead Compensation using Bode diagram

d.compensate

be can angle Lead-phase maxmum , be to make :Ideal cm ).( is

system teduncompensa functiontransfer loop open :Assume

sGHK 00

6.46.4 Cascade compensation by frequency response methodCascade compensation by frequency response method

) (

)()( desire the fromget be can :

desire the fromget be can :

:oncompensati Active1.

(s)(s)GGHK

sGsGHτ

KKeK

s)τ(K(s)G

c

c

jsC

jωsCCD

pssp

DpC

1

180

1

1

00

00

0

45

.15~1060dB/dec,- ;10~5

40dB/dec,- ;5,at /20: ) of slope the If

.the to made be to cascade the

to due of increasing the of because angle dcompensate:

. desired :

15~5: -Get 3)

diagram. Bode the from and the Measure 2)

. desired the and ) functiontransfer loop-open the

to according system teduncompensa the of diagram Bode thePlot 1)

1)( τs1

τs1 :oncompensati Passive 2.

oooo

oc0

00

c

cd

oococd

coc

00

decdB(ωL

(s)GH K(s)G

φ

e(jωGHK

(s)G

C

c

m

ss

C

6.4 Cascade compensation by frequency response method

46

etc. the examining

the of diagram Bode theplot 6)

lglg

:fromget be can here

1 :formula the from get 5)

sin1

sin1get

1

1sin from 4)

m

m

m1m

CC

c

ωω

m

,γ ω

(s) G(s)GHK

α(jw)GHK

φ

φφ

m

00

00 1020


47

Example:

thefor ionsspecificat , functiontransfer loop-Open 21

S

KGH(s)

))((. ; ; :are system cc20

2

1101045 ttress


solution: －－ 40dB/dec

Fig.6.4.110

cGlg20

).( 10101 sseK

1631000 . ;

Fig.6.4.1. in shown is diagram Bode The

c c

τs

τs(s) Gc

1

1 : oncompensati lead-phase the usemust We

10 at angle add and c 45

48


－－ 40dB/dec

Fig.6.4.110

cGlg20)180)()((

450045 o

oooo0m

jGH

φ ccd

17.0 sin1

sin1

om 45

m

m

24.0 101

cm

dBdBjGH 4.15log2020)(log20but 17.010

)7.14.1520lg20(

7.1 and 041.01

24.01 make we So

cc

ccc

KK

Ks

sKG

1. Make validity check for this example.2. Make compensation using PD for this example.

Exercise:

49

6.4.2 Phase-Lag Compensation using Bode diagram

)s.s

K

ss

K GH(jω v

1502

()()

:functiontransfer teduncompensa The

r.compensato the Dtermine

. and ,45 margin Phase

:are system thefor ionsSpecificat

c 20 vK

1 ,1

1(s): oncompensati lag-phase the use can We

6.32at 20

:Fig.6.4.2 in shown is system teduncompensa the of diagram Bode The

cc

s

sGc

Example:

2

6.32－－ 20dB/dec

－－ 40dB/dec

020c

－－ 900

－－ 1800

－－ 20lgβ

solution:Fig.6.4.2

50

oo130)( where 1.5 frequency the locate We 50 cc φ

1020log20dB

20dB is at nattenuatio required The

.20)(lg 1.5, At

c

.c

dBjGH51

20

s

ssG

c

7.661

67.61)(67.6

r)compensato lag-phase thefor at error 5 Allowing(Consider

15.010

1 :Make

c


2

6.32－－ 20dB/dec

－－ 40dB/dec

020c

－－ 900

－－ 1800

－－ 20lgβ

Validate……

Fig.6.4.2

51


1. Make validity check for this example. 2. Make phase-lag compensation for γc=50o and Kv=20.

Steps of the phase-lag compensation:

etc. and ionsspecificat the

validate to system dcompensate the of diagram Bode thePlot )5

.get to )105:(1

Make )4

)(log200log2 :from Get 3)

t.requiremen the satisfy hich Find 2)

. desired the and ) functiontransfer loop-open the

to according system teduncompensa the of diagram Bode thePlot 1)

cc

cc

00

～kk

jGH

w

e(jωGHK

c

ss

c

Exercise:

52

6.4.4 Compensation according to the desired frequency response

Example

device. oncompensati series the determine

)1(

1

:is system teduncompensa the of functiontransfer loop-open the If

Fig.6.4.3. in shown is response frequency desired The

ssGH(s)

solution

10

100－－ 2020dB/dec

－－ 4040dB/dec

First: make the phase-lag compensation－ to satisfy ess

and compensate a part of γc .

Second: make the phase-lead compensation－ to satisfy the requirementsγc and ωc etc.

6.4.3 Phase-Lag-lead Compensation using Bode diagram

Fig.6.4.3

53

6.4.4 Compensation according to the desired frequency respons6.4.4 Compensation according to the desired frequency responsee

10

100－－ 2020dB/dec

－－ 4040dB/dec

In terms of the desired frequency response we have:

)101.0(

10)()()(

sssGHsGsGH cdesire

)()(

1

1

sssGH

).(

)(

)().()(

)()(

1010

110

11

101010

s

s

sssssGH

sGHsG desire

c

Fig.6.4.3

54

system.

teduncompensa the and system dcompensate

the between eperformanc the compare and

device oncompensati cascade the Determine

)(

and system) phase

(minimum Fig.6.4.4 in shown is system a of

response frequency loop-open desired The

1

1

ssGH(s)

10

100－－ 2020dB/dec

－－ 4040dB/dec

1

－－ 4040dB/dec

Exercise:

Fig. 6.4.4

55

6.5 Feedback compensation6.5 Feedback compensation

R(s) C(s))(sG0

CG－－

G’0(s)

R(s) C(s)10G 20G

CG－－

G’20(s)

6.5.1 The configuration of the Feedback compensation

6.5.2 The basic Feedback compensators

on;compensati feedback ation)al(accelerdifferenti2

])1(1[)(

on;compensati feedback al(speed)differenti])1(1[)(

on;compensati feedback n)al(positioproportion)(

2

th

sssG

sssG

sG

c

c

c

Fig. 6.5.1

56

2. Impair(weaken) the influences of the disturbance to the encircled elements. 3. make the performance of the encircled elements to be desired .

1. Decrease the time constant of the encircled elements → Quicken the response of the encircled elements－may be;

For example

)1/( alsobut ,1

11)( )( ,

1)(

2020'20

20

'

'

'20

20

20'20

2020

KKKK

TT

sT

K

KTs

KsGsG

Ts

KsG c

6.5.3 Function of the feedback:

202020'

'20

20

20'20

,)(

11)( )( ifBut

KKKTT

sT

K

sKTs

KsGssGc

57

6.5.4 The design procedure of the feedback compensator 1. Design the desired characteristics, such as the desired Bode diagram, of the encircled elements in terms of the system’s analysis. 2. Choose the appropriate feedback compensators to get the desired characteristics. Example R(s) C(s)

10G 20G

CG－－

G’20(s)

Fig. 6.5.2

For the system shownin Fig. 6.5.2, G20=10/s2, the desired G’20(jω) shown in Fig. 6.5.3. determine the Gc.

110－－ 2020dB/dec

－－ 4040dB/dec

0.1

－－ 4040dB/dec

Fig. 6.5.3

solution

11.0

9.9)1(

1)11.0(

)110(1.0)(

2

20'20

'2020

20'20

20

20

202

'20

s

s

GG

GGG

G

GG

GG

G

ss

ssG

c

c

1 chap 6 the compensation of the linear control systems p553

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