ˆ x ESTIMATOR. state vector estimate

Transcription

1 hapte 9 ontolle Degn wo Independent Step: Feedback Degn ontol Law =- ame all tate ae acceble a lot of eno ae necea Degn of Etmato alo called an Obeve whch etmate the ente tate vecto gven the otpt and npt PLN ONROL LW - mat of contant ˆ ESIMOR tate vecto etmate OMPENSION - ontol Law Degn med tem fo contol law degn =- =- fo an nth ode tem thee ae n feedback gan,, n. choce of the oot can geneall be placed anwhee

2 chaactetc eqaton I Placng Root Eample: Undamped ocllato wth feq. tate pace decpton REVIEW let' place the oot both at - j j j n -j - n j n we want to doble the natal feqenc and nceae dampng fom = to = co deed chaactetc eqaton d det 4 4 I det o 4 eqatng ame coeffcent: 4 4 4

3 Ue of anoncal Fom mple wa of calclatng the gan when ode geate than thee to e pecal canoncal fom of the tate eqaton. he pecal tcte of the tem mat efeed to a companon fom. Eample: hd ode cae: he chaactetc eqaton a a a Recall the phae vaable fom a lowe companon fom G b n... b n b n n a n... a n.... : a a a n n :... b b b n n D he cloed loop tem mat hd ode cae: a a a a a a haactetc eqaton I a a a

4 4 f the deed pole locaton elt n the chaactetc eqaton d eqatng lke coeffcent of and a a a Eample: Dll poblem D9. page 65 of tet 7 6 k k k Fnd k to place the cloed-loop tem pole at = -, -4, -5 NS deed chaactetc polnomal d , 47, 6 we have k a 7 5 k a k a 6 57

5 5 Degn Pocede Gven, and deed d, tanfom to c, c and olve fo gan We then need to tanfom gan back to ognal tate pace note: the pole can onl be placed abtal f the tem fll contollable h pocede encaplated n ckemann fomla ckemann Fomla... M k d whee M n... contollablt mat whee n the ode of the tem o the nmbe of tate and d defned a I n n n n d... whee the ' ae the coeffcent of the deed chaactetc polnomal d n n... n Eample ppl the fomla fo the ndamped ocllato 4 4 d 4, 4 ecall d contol canoncal companon fom.e. phae vaable fom

6 alo M c M c whch the ame a the elt pevol obtaned. ackng Poblem Fo tep npt: Wll fnd N to ene zeo tead-tate eo to tep npt now N N N N N now tead-tate eo e - table nvee et N N to get zeo tead tate eo N 6

7 Integal ontol ed to get zeo tead tate eo R E X U X - Y - ntegato n the fowad path Integato nceae the tem ode b one,.e. agment the plant model b an added tate vaable edt dt dt gmented tem become zeo matce compatble dmenon ontol law he degn now poceed a befoe. Eample Doble Integato G gment the plant 7

8 8 Select pole of the cloed loop tem to be at j,5 N.. pole becae of eta tate 7 Stead tate otpt to a nt tep npt can be deved a follow dt d n tead tate

9 9 Fo the eample 7 7 det adj 7 onl tem of nteet

10 Obeve Degn t, t t, ˆ t ˆ t - L We wll etmate tate athe than meae them PLN U Y - L ˆ X Obeve mlate the ognal tem Ognal Stem Obeve ˆ ˆ ˆ ˆ L ˆ

11 Eo between tate and the etmate ˆ ~ ˆ ˆ L ˆ L ~ ~ ~ Obeve eo wll go to zeo amptotcall -L table.e. egenvale ae n the LHP note: the egenvale of -L ae the obeve pole, whch can be placed abtal f the tem obevable th can be done b choce of the obeve gan L a colmn vecto fo ngle otpt tem Dalt Defnton: a tem detectable f the ntable mode ae obevable Eample a tem tablzable f the ntable mode ae contollable ontol Etmaton ontol Etmaton M M o L M... n dalt M o : n n o n M : n

12 ckemann fomla to fnd obeve gan : e M o L DERIVION USING DULIY ckemann Fomla fo contol poblem... M Dalt o e e o M M L :... : e M o L Eample Degn an obeve fo G Note the tem completel obevable Degn the obeve wth pole at j ctal chaactetc eqaton: l l l l L I

13 Deed chaactetc eqaton: j j 4 8 Eqatng coeffcent l 4 l 8 4 L 8 he obeve eqaton ae ˆ ˆ ˆ 4 ˆ 8 ˆ PLN ˆ ˆ SRUURE OF OSERVER

14 ontol Ung Obeve Plant: Obeve: ˆ ˆ L ˆ ˆ ˆ Etmated tate feedback: ˆ clong the loop ˆ ˆ L ˆ ˆ L L ˆ L ˆ L L ˆ 4

15 Sepaaton Pncple Intodce tanfomaton z I N N P whee P ˆ w IN I N note P P z I N N w I ˆ ˆ N I N heefoe ng th tanfomaton the old agmented tate vecto compng the plant tate and ˆ the etmato ate now become and the etmato eo. he new tem mat P P note ~ block-tangla L ~ Egenvale of a block-tangla mat ae eqal to the egenvale of the dagonal block. So the egenvale of the fll tem compe the egenvale of the plant.e. egenvale of - and the obeve.e. egenvale of -L. ltenatve ppoach ˆ ˆ L L ˆ ˆ ˆ L L ˆ L L ˆ ~ L ~ ˆ now ˆ ˆ ~ ~ 5

16 ompenato anfe Fncton H U Y ompenato otpt, whch the plant npt ompenato npt, whch the plant otpt PLN ompenato Otpt - ˆ OSERVER ompenato Inpt OMPENSOR we have ˆ ˆ ˆ L ˆ ˆ ˆ L Lˆ L ˆ L Xˆ I L LY U I L LY H Degn Ie Poblem wth pole placement that thee no contol ove compento pole and zeo Optmm choce fo obeve ntal condton ˆ '' hoce of obeve pole:. hooe them to be fate than contolle pole. ltenatvel, chooe them to be at plant zeo f the tem ha RHP zeo, e the LHP mage. 6

17 Redced-Ode Obeve Degn If tem ha n tate and m meaement, then an obeve of ode n-m wll be ffcent. If we wll ame ha the tcte: mm I I R, R m I m m nm meaed tate nmeaed tate m m m known npt dnamc of the nmeaed tate alo m known meaement otpt elatonhp Smmazng: known meaement m known npt a b Recall fo fll-ode obeve: Plant: Obeve: ˆ L ˆ ˆ ˆ compang and how the coepondence ˆ 7

18 m 4 btttng 4 nto we get the edced ode eqaton ˆ ˆ ˆ m L 5 let now defne the etmato eo we get ˆ theefoe btactng 5 fom a and ng b.e. ~ L ~ 6 Degn poceed b, gven an and we chooe an L to place etmato pole. Rewtng 5 we have ˆ L ˆ L L L 7 he peence of the devatve of the meaement.e. not good nce th amplfe the noe. o get aond th we ntodce a new tate z whee z ˆ L 8 ˆ z L 9 the eo dnamc ae gven b th eqaton btttng 9 nto 7 lead to the fnal fom of the edced-ode obeve z Dz F G ˆ z L whee D L F DL L G L z the tate of the etmato 8

19 he block dagam of the edced ode obeve hown below ^ Eample Doble ntegato G ^ z Dz F G ˆ whee z L D L F DL L G L Dnamc of edced ode obeve ~ L ~ 9

20 Reqe obeve pole at I L I L whee L D F 4 G ckeman fomla fo edced ode obeve gan L e M : M : n fo eample: e e M L

21 Redced-Ode anfe Fncton m m ˆ z L ˆ now alo z Dz F G z z L L z Dz F G z D G z F G L G L anfe fncton: U Y 'I ' 'D' whee ' D G ' F G G L ' D' L When not of the fom I? D Qz Qz Qz z Q Qz Q Qz D o fnd a tanfomaton Q o that Q of fom I let Q Q Q o Q Q Q I

22 let P be nonngla abta mat f PQ I Q Q I PQ Q Q Q Q I Q P