Lead &Lag Compensators

7/27/2019 Lead &Lag Compensators

1/17

Chapter 10

Compensator

10.1 Introduction

Fig 1.1-1: Compensat or k. bs as ++ ;

phase l ead or phase l ag compensat or .

Compensat i ng networks are used i n cl osed- l oop syst em t o i mproveper f ormance. The compensat ors shown ar e made- up of el ect r i c r esi st orsand capaci t ors, whi ch are passi ve el ement s.


2/17

220 Compensator.

The t r ansf er f unct i ons devel oped ar e based on no l oadi ng ef f ectupon t he out put . Al l t he tr ansf er f unct i ons are expr essed i n nondi mensi onal f orm.

From t he f i l t er i ng standpoi nt , t he hi gh pass f i l t er i s of t en

r ef er r ed t o as a phase- l ead cont r ol l er si nce posi t i ve phase i si nt r oduced t o t he syst em over some appr opr i ate f r equency range. Thel ow- pass f i l t er i s al so known as a phase- l ag cont r ol l er , si nce t hecorr espondi ng phase i nt r oduced i s negat i ve.

10.2 Phase-Lag Compensator

Fig 10.2-1:Typi cal l ag compensat or ci r cui t .

The ci r cui t shown i s a t ypi cal l ag or i nt egral compensat or . Theout put si gnal i s pr opor t i onal t o t he sum of t he i nput si gnal and i t si nt egral .

The desi gnat i on l ag appl i ed t o t hi s networ k i s based on t hest eady- st at e si nusoi dal r esponse. The si nusoi dal r esponse E2 wi t h a

si nusoi dal i nput E1.

I ni t i al condi t i ons ar e consi der ed t o be zer o.

sT1

sT!

sC)RR(1

sCR1

)RR) (s(I

)R) (s(I=

)s(E

)s(E

2

1

112

11

s1C1

12

s1C1

1

1

2

++

=++

+=

++

+

; wher e T1 = R1C1 and T2 = ( R1+R2) C1

=T/1s+1

T/1s+1=

T/1s

1

T/1s1 1

2

1

12

1+

+

+( 10. 1)

wher e T1 < T2

sl ope - 1 +1

T sl ope - 1 0

Fr eq. 1/ T2 1/ T1


3/17

221 Compensator.

Sever al general ef f ect s of l ag compensat i on ar e:

1. The bandwi dt h of t he syst em i s usual l y decr eased.

2. The predomi nant t i me const ant of t he system i s usual l yi ncreased, produci ng a more sl uggi sh syst em.

3. For a gi ven r el at i ve st abi l i t y, t he val ue of t he er r or const ant i si ncr eased.

4. For a gi ven val ue of er r or const ant , r el at i ve st abi l i t y i si mproved.

Example 10. 2 - 1:

The open- l oop t r ansf er f unct i on of t he or i gi nal syst em and t heper f or mance speci f i cat i ons are gi ven as f ol l ows:

100k;

)s2.1) (s1.1(s

k=)s(G 1vp=

++

sec ( 10. 2)

r el at i ve dampi ng r at i o = . 707

)10s) (5s(sk50=)s(Gp ++

( 10. 3)

The r oot l ocus of compensat ed syst em, f or kv = 100, k = 100 ,whi ch corr esponds t o an unst abl e syst em:


4/17

222 Compensator.

( 10. 4))( k100=k=kl i m=)s(Gsl i m=k v0s

0s

v

Char act er i st i c equat i on i s:

s( s+5)( s+10) + 50k = 0. ( 10. 5)

The graph i s shown bel ow i n ( Fi g 10. 2- 2) :

Fig 10.2-2

When k = 1. 635 ; t he uncompensat ed char act er i st i c equat i on r ootsar e: - 11. 118 , - 1. 91+j 1. 91 and - 1. 91- j 1. 91, whi ch cor r esponds t o ar el at i ve dampi ng r at i o of . 707.

The compensat or equat i on:

Ts1aTs1=

sT1

sT1=G

2

1c +

+++

( 10. 6)

21

221 RR

R=aandaTT;1


5/17

223 Compensator.

01.s6135.s. 01635=

)01.s(100

)6135.s(635.1=

s1001s635.11=)s(G

++

++

++

and the open l oop t r ansf er f unct i on of t he compensated syst em i s:

100=kwher e;)01.s) (10s) (5s(s

)6135.s(k815.=)s(G)s(G=)s(G pc +++

+

When k = 100 , t he r oots of t he compensat ed char act er i st i c

equat i on ar e at - 11. 13, - 1. 198, - 1. 341 j 1. 397. The r el at i vedampi ng r at i o of t he compl ex r oot s i s sl i ght l y l ess t han 0. 7. Thi scan be i mproved by sel ect i ng a smal l er val ue f or a or a l arge T.

Fig 10.2-1: Uni t st ep r esponse of t he syst emwi t h phase- l ag cont r ol .

The uni t st ep r esponse of t he syst em wi t h t he phase- l ag cont r ol l er asi s shown. The peak overshoot of t he r esponse i s appr oxi matel y 36per cent .

10.3 Phase Lead Compensator

The ci r cui t shown i s a l ead compensat or . t he out put si gnal i spr opor t i onal t o t he sum of t he i nput si gnal and i t s der i vat i ve. Thel ead desi gnat i on of t hi s net wor k i s based on the st eady- st at esi nusoi dal r esponse. The si nusoi dal out put E2 l eads t he si nusoi dali nput E1.

Fig 10.3-1: Passi ve phase- l ead.


6/17

224 Compensator.

1TT=Q&RR CRRT;CRTwher e

T/1s1

T/1s+1Q=

sT1

sT1

T

T=

)s(E

)s(E

12

212212221

1

1

1

22

1

1

2

1

2

a;T/1saT/1s

++

( 10. 8)

wher e

21

21

2

21

RR

RR=Tand

R

RR=a

++

C

The l ead networ k provi des compensat i on by vi r t ue of i t s phase- l ead

pr oper t y i n t he l ow t o medi um f r equency range and i t s negl i gi bl eat t enuat i on at hi gh f r equenci es.

The l ow t o medi um f r equency r ange i s def i ned as t he vi ci ni t y of

t he r esonant f r equency p

Al t hough t he net wor k may be si mpl i f i ed f ur t her , and st i l l be

r epr esent i ng a l ow- pass f i l t er by el i mi nat i ng R2 .


7/17

225 Compensator.

Example 10.3-1:

The bl ock di agram shown bel ow descr i bes t he components of a sum-seeker cont r ol syst em.

Fig 10.3-1:Bl ock di agr am of a sum- seeker cont r ol syst em.

The syst em may be mount ed on a space vehi cl e so t hat i t wi l l t r ack

t he sun wi t h hi gh accur acy. The var i abl e r r epr esent s t he r ef erenceangl e of t he sol ar r ay and o denot es t he vehi cl e axi s. The obj ect i vef or t he sun- seeker cont r ol system i s t o mai nt ai n t he er r or bet ween r , o and near zero.

The paramet er s of t he syst em ar e gi ven as:

Rf = 10000

kb = . 0125 V/rad/sec

ki = . 0125 N.m/A

Ra = 6. 25

J = 10- 6 kg-m2

ks = . 1 A/rad

k : t o be determi nedB = 0

n = 800

The open- l oop t r ansf er f unct i on of t he uncompensat ed syst em i s:

skkJ sR

n/kkRk

)s(

)s(

bi2

a

iFso

+=

( 10. 9)


8/17

226 Compensator.

Subst i t ut i ng t he numeri cal val ues of t he syst em paramet er s:

)25s(sk2500

)s(

)s(o+

=

( 10. 10)

The speci f i cat i ons of t he syst em are gi ven as f ol l ows:

1. The st eady- st at e val ue of ( t ) due t o a uni t r amp f unct i on i nputf or r ( t ) shoul d be l ess than or equal t o . 01 r ad per r ad/ sec oft he f i nal st eady- st at e out put vel oci t y. I n ot her wor ds, t hest eady- st ate err or due t o a r amp i nput shoul d be l ess t han orequal t o 1 percent .

2. The peak over shoot shoul d be l ess t han 10 percent .The l oop gai n of t he syst em i s deter mi ned f r om t he st eady- st at e

er r or r equi r ement .

Appl yi ng t he f i nal - val ue t heor em t o ( t ) , we have:

+

==

)s(

)s(1

)s(sl i m)s(sl i m)t(l i m

o

r

0s0st( 10. 11)

For t he uni t r amp i nput :

k01.)t(l i m

s

1)s(0s2

r ==

( 10. 12)

Thus f or t he st eady- st at e er r or t o be l ess t han or equal t o . 01 , kmust be gr eat er t han or equal t o 1. For k = 1, t he worst case, t hechar act er i st i c equat i on of t he uncompensated syst emi s:

s2

+ 25s + 2500 = 0 ( 10. 13)

Thus t he dampi ng r at i o of t he uncompensat ed syst em i s mer el y 25percent , whi ch cor r esponds t o a peak overshoot of over 44. 4 percent .

The open- l oop t r ansf er f unct i on of t he syst em wi t h t he phase- l eadcompensator i s:

)Ts1) (25s(s)aTs1(2500

=)s(G++

+( 10. 14)

wher e ramp er r or const ant kv = 100.

The char act er i st i c equat i on becomes:

s( s+25) ( 1+Ts) + 2500 = 0 ( 10. 15)

0=2500s25s

)25s(Ts+1

2

2

++

+ ( 10. 16)

t hi s equat i on i s of t he f or m 1+ G1( s) = 0


9/17

227 Compensator.

The char act er i st i c equat i on of t he compensat ed syst em now becomes:

s( s+25) ( 1+Ts) + 2500 ( 1+aTs) = 0 ( 10. 17)

0=2500)Ts1) (25s(s

aTs2500+1+++

( 10. 18)

t hi s equat i on i s of t he f or m 1+ G2( s) = 0

2500)Ts1) (25s(saTs2500=( s)G2 +++

( 10. 19)

For ver y smal l T, t he t er m ( 1+Ts) i n t he denomi nat or can benegl ected :

s2

+ 25( 1+100aT) s + 2500 = 0 ( 10. 20)

Let us sel ect t he dampi ng rat i o f or t he appr oxi mat i ng second- ordersyst em t o be . 707. Then gi ves:

= . 707 = 25aT + . 25 aT = . 0183

I f we sel ect a val ues f or T a overshoot % char acter i st i cequat i on r oot s

. 00588 5. 83 7. 3 - 30. 93 , - 82. 07 j 83. 73

. 0015 12 6. 6 - 433. 6 , - 45. 68 j 28. 21

. 0022 8 13. 8 - 460. 3 , - 32. 35 j 40. 85

I f we sel ect i n t hi s case wi t h a = 5. 83 , t hen T = . 00588

The t r ansf er f unct i on of phase- l ead compensat or i s:

170s17.29s=

)T/1(sT)a/1(s

=)s(Gc ++

++

( 10. 21)

10.4 Phase Lead-Lag Compensator

The ci r cui t shown i s a l ead- l ag compensat or , whi ch combi nes t he

char act er i st i cs of t he l ag and t he l ead compensat or s. Thi s i s cal l eda l ead- l ag net work because t he phase of t he si nusoi dal r esponse E2,

compar ed wi t h t he si nusoi dal i nput E1, var i es f rom a l ag t o a l eadangl e as t he f r equency i s i ncreased f r om zer o to i nf i ni t y. The phaseangl e can be det er mi ned f r om t he st eady- st at e sol ut i on of t hedi f f er ent i al equat i on.


10/17

228 Compensator.

221431

21

1

2

sTTs)TTT(1

)sT1) (sT1(=

)s(E

)s(E

++++

++

+

+

+

+

41

32

s1s1

s1s1

= ( 10. 22)

wher e T1 = R1C1 ; T2 = R2C2 ; T3 = R1C2 , and T4 = R2C1

1

4

+1

3

+1

2

- 1

11

2 s1s1s1s1=)s(E

)s(E

+

+

+

+

= Gc ( s) ( 10. 23)


11/17

229 Compensator.

+

+ 222

111

2

2

1

1c CR=bT

CR=aT

)sT1(

)sbT+( 1

)sT1(

)saT+( 1=)s(Gor ( 10. 24)

lead lag

T1T2 = R1R2C1C2 ; abT1T2 = R1R2C1C2

ab = 1.

whi ch means t hat a and b can t be speci f i ed i ndependent l y.

Design Sample:

1- Gi ven t he bl ock di agr am:

The speci f i cat i ons are:

i ) Asympt ot i cal l y stabl e.

i i ) kv 1000

i i i ) c 200 rad/sec

Tkd v

= = 1 11000

10 3 ( 10. 25)

Tpc

= =3 3200

0015

. ( c = p ) ( 10. 26)


12/17

230 Compensator.

Desi gn pr ocedures:

1- Bode pl ot

G( s) = 111000

s110s1

s1000

+

+ ( 10. 27)

cr ossover 1000

10 103

sl ope - 1 - 1 - 1

-1 -2 -3

When, we have sl ope = - 2 t he syst em i s unst abl e, so we shoul dmake i t st abl e.

Gc( s)G( s) =211

1000s1

50s1

10s1

s1000

+

+

+

+ ( 10. 28)

1000s+1

50s+1

=1000

s150s1)s(G

11

c

+

+

+= ( 10. 29)


13/17


14/17

232 Compensator.

. 0292-=c1)-c+c( q. 0898=c2q

2. 3887=1+c-q 22

2

Hence 2. 3887 2

+ . 0898 - . 0292 = 0.

Sol vi ng = . 09336 i s posi t i ve r oot .

b =

2/12c

1c

c

= 3. 295

and t he t r ansf er f unct i on i s obt ai ned as :

. 3076+s. 3076+s09336.=

b+s

b)+( s=)s(Gc

10.5 Digital Implementation of Compensators

Si nce di gi t al cont r ol syst ems have many advant ages overcont i nuous- dat a syst ems, compensat ors t hat are desi gned i n t he anal ogdomai n are i mpl ement ed di gi t al l y.

The bl ock di agram of t he anal og PI D cont r ol l er , where PI D st andsf or Pr opor t i onal , I nt egr al , and Der i vat i ve, i s shown bel ow.

Fig 10.5-1: PI D Cont r ol l er .

kp i s i mpl ement ed di gi t al l y by a gai n el ement . Si nce a di gi t alcomput er or pr ocessor al ways has a f i ni t e di gi t al word l engt h. Mostdi gi t al comput ers are based on t he bi nar y- number syst em.

The t i me der i vat i ve of a f unct i on f ( t ) at t = kT can beappr oxi mat ed numer i cal l y by use of t he val ues of f ( t ) measur ed at t =kT and t = ( k+1) T , t hat i s

1)T)]-f ( ( k-[ f (kT)T1

dt)t(df

kTt=

=( 10. 31)


15/17

233 Compensator.

To f i nd t he z- t r ansf er f unct i on of t he der i vat i ve oper at i ondescr i bed numer i cal l y above we take t he z- t r ansf orm on both si des, wehave:

Z )z(FTz1z=)s(F)z-( 1T1=dt)t(df

1-kTt

= ( 10. 32)

Thus, t he z- t r ansf er f unct i on of t he di gi t al di f f erent i at or i swr i t t en:

GD( z) = KD Tz1z ( 10. 33)

Wher e KD i s t he pr opor t i onal const ant of t he der i vat i ve

compensator . Repl aci ng z by eTs

. So when the sampl i ng per i od T

appr oaches zero, GD( z) appr oaches KD s , whi ch i s t he t r ansf erf unct i on of t he anal og der i vat i ve compensator.

For t he i ntegr ator , we normal l y have a number of choi ces ofdi gi t al appr oxi mat i on. Thi s appr oxi mat i on i s equi val ent t o t hesampl e- and- hol d ( zero or der) oper at i on. The bl ock di agr amr epr esent at i on of r ect angul ar i nt egr at i on.

(a) Rect angul ar i nt egr ati on.

(b) Equi val ent r ect angul ar i nt egr ati on wi t h sampl e-and- hol d.

Fig 10.5-2: ( a) & ( b)

The z - t r ansf er f unct i on of t he di gi t al i nt egrat or can be wr i t t en

GI ( z) = Kz . Z 1z

Tk=

s1

se1 ITs

( 10. 34)

Agai n, as T appr oaches zero, GI ( z) appr oaches KI / s , t he t ransf erf uncti on of t he anal og i nt egr al cont r ol l er .


16/17

234 Compensator.

I n pr act i ce, t her e ar e other numer i cal i nt egr at i on r ul es. Such ast he t r apezoi dal i nt egr at i on, Si mpson s r ul e, and so on.

The bl ock of a di gi t al PI D cont r ol l er i s shown bel ow:

Fig 10.5-3: Bl ock di agr amof a di gi t al PI D compensator .

Once t he t r ansf er f unct i on of a di gi t al compensat or i sdet ermi ned, t he compensat or can be i mpl ement ed by a di gi t al processoror comput er. The oper at or z- 1 i s i nt er pr eted as a t i me del ay of Tseconds, where T i s t he sampl i ng peri od. The t i me del ay i si mpl ement ed by st or i ng a var i abl e i n some conveni ent st oragel ocat i on i n t he comput er and t hen t aki ng i t out af t er T has cl asped.For t he di gi t al di f f erent i at or, t he t ransf er f uncti on i s wr i t t en:

GD( z ) = )z1(T

K 1D ( 10. 35)

f or t he di gi tal i ntegrator , i t i s :

GI ( z ) = 1

1I

z1

TzK

( 10. 36)

Any cont i nuous dat a compensat or can be made i nto a di gi t alcompensator si mpl y by addi ng sampl e and hol d uni t s at t he i nput andt he out put t er mi nal s. The sampl i ng per i od T shoul d be suf f i ci ent l ysmal l so t hat t he dynami c char act er i st i c of t he cont i nuous- dat acompensat or ar e not l ost t hr ough the di gi t i zat i on. The syst em shownbel ow actual l y suggest s t hat gi ven t he cont i nuous- dat a compensat or

G ( s), t he equi val ent Gc( z) can be obt ai ned as shown.c

Fig 10.5-4: Real i zat i on of di gi t al cont r ol l er by an anal og cont r ol l er wi t hsampl e- and- hol d.


17/17

235 Compensator.

Fig 10.5-5: Di gi t al programr eal i zat i on.

Consi der t hat t he cont i nuous- data compensat or i s r epr esent ed byt he t r ansf er f unct i on:

Gc( s ) = 61.1s1s

++ ( 10. 37)

The t r ansf er f unct i on Gc( z) i s wr i t t en:

Gc( z) = ( 1- z- 1) Z

2.z5.z=

)61.1s(s1s

++ ( 10. 38)

Documents

Lead &Lag Compensators