Integral Control via Bias Estimation

Transcription

1 1 Integal Contol via Bias stimation Consie the sstem ẋ = A + B +, R n, R p, R m = C +, R q whee is an nknown constant vecto. It is possible to view as a step istbance: (t) = 0 1(t). (If in fact (t) vaies with time, bt moe slowl than the othe sstem namics, then moelling the istbance as a step ma be a goo appoimation.) Let s consie as the state of a namical sstem with nknown initial conitions: ḋ = 0, 0 nknown (It is also possible to teat othe namics in this wa; I have aske o to o so in Poblems 2 an 3 of Poblem Set 3.) Sppose that we wish to foce ( t) to asmptoticall tack a constant efeence signal espite the nknown istbance. We have seen two appoaches to this poblem. In each case, we se state feeback to stabilize the sstem; this entails no loss of genealit becase we can alwas se an obseve to econstct the state. (In Poblem 1 of Poblem Set 3, o ae aske to veif that this pocee woks.) O two appoaches ae, fist, to se a constant gain pecompensato: N B ( s I A ) 1 C

2 2 If the close loop sstem is stable, then we can calclate that the esponse to a step comman (t) = 0 1(t) an step istbance (t) = 0 1(t) satisfies (t) ss, whee ss := C( A + B) 1 BN 0 + [C( A + B) 1 + ] 0 If C( A + B) 1 B is ight invetible, then setting N = [C( A + B) 1 B] # iels ss := 0 + [C( A + B) 1 + ] 0 The isavantages of this scheme ae that it cannot fo the effects of istbances an small paamete vaiations. An altenate appoach that oes accont fo istbances an paamete vaiations is to se integal contol: B ( si A ) 1 C I w I / s e We have seen in class that if the close loop sstem is stable, then the stea state esponse to a step comman an a step istbance satisfies (t) ss, whee ss := 0. Hence, pefect comman tacking an istbance ejection ae possible; moeove, the sstem is insensitive to small moelling eos. Of cose, in pactice we won't sall be able to mease the states of the plant. Hence, we mst se an obseve to estimate these states,

3 3 as epicte in the following iagam. In Poblem 1 of Poblem Set 3, o ae aske to show that this sstem is stable if the state feeback sstem in the pevios iagam is stable, an if the obseve is stable. B ( si A ) 1 C ^ Obseve I w I / s e Note that the contol inpt in this configation oes not epen eplicitl pon the comman signal. It is often possible to mease, an hence we can consie sing this measement in a feefowa contol scheme as we i in Poblem 3 of Poblem Set 2. The avantage of the eslting Two Degee of eeom (2DO) contol topolog is that it affos geate feeom in achieving esign taeoffs.

4 4 Altenatel, let s sppose fo the sake of agment that we can mease the state of the istbance. We can then consie sing this measement in a feefowa/feeback contol law = ˆ I w + G as shown below: G B ( si A ) 1 C ^ Obseve I w I / s e Pesmabl, b sing feefowa fom the istbance, we can obtain a faste esponse to the istbance than othewise. Thee is one poblem: we will geneall not be able to mease the istbance! Recall, howeve, that a step istbance ma be viewe as the otpt of a namical sstem consisting solel of integatos: ḋ = 0, 0 nknown

5 5 Since we nee to se an obseve anwa, sppose that we esign the obseve, if possible, to estimate the state of the istbance as well as that of the plant, an se a contol law = ˆ I w + G ˆ B ( si A ) 1 C G ^ ^ Obseve I w I / s e Becase a step istbance ma be viewe as an nknown bias, the obseve in the above iagam is sometimes teme an "nknown bias estimato". A simila pocee ma be se to estimate the state of othe sots of istbances. We shall now consie the poblem of esigning an obseve to estimate the istbance. Becase the concepts we nee fo this o not epen pon the se of integal contol in the above iagam, we will instea eplain the pocee base pon the simple contol scheme with constant gain pecompensato. Once we know how to esign an obseve fo the istbance, the eslt can be applie to the poblem pose above. The following iscssion is aapte fom W. J. Rgh, Linea Sstem Theo, PenticeHall, 1993

6 6 Consie the sstem epicte below: N B ( s I A ) 1 C G ^ ^ Obseve The state eqations fo this configation ae: Plant: ẋ = A + B + = C + Obseve: ˆẋ ˆ = A ˆ 0 0 ˆ + B 0 + L ŷ ŷ = [ C ] ˆˆ L 2 ( ), L = L 1 Contol: = ˆ + G ˆ + N We mst answe seveal qestions: (1) When can a stable obseve fo an w be esigne? (2) When is the close loop sstem stable? (3) Does (t) 0 as esie? We also nee to fin vales fo, L, N, an G.

7 7 Theoem: Given the sstem escibe above, efine P (s) = C( si A) 1 B an P (s) = C(sI A) 1 + Assme that an (i) ( A, B) is contollable, (ii) ( A,C) is obsevable, (iii) nomal ank P (s) = q (iv) ( A, B,C) has no zeos at s = 0 (v) (vi) (vii) ( A, ) is contollable, nomal ank P (s) = m ( A,,C, ) has no zeos at s = 0 Then an L ma be chosen to stabilize the eslting close loop sstem, an N an G ma be chosen so that the stea state esponse to a step comman (t) = 0 1(t) an a step istbance (t) = 0 1(t) satisfies (t) ss = 0. Notes: (1) If A has no eigenvales at s = 0, then conitions (iv) an (vi) ma be eplace b ank P (0) = q an ank P w (0) = m, espectivel. (2) Conition (iii) implies that thee ae at least as man contol inpts as contolle otpts, an that these otpts ae inepenent. (3) Conitions (v)(vii) essentiall inse that all the istbance states affect the stea state otpts of the sstem. These conitions will gaantee that we can constct an obseve fo the istbance states.

8 8 Poof: ist sbstitte the eqation fo ŷ into the obseve; this iels: ˆẋ ˆ = ( à L C ) ˆˆ + B 0 + L = A L 1 C L 1 ˆ L 2 C L 2 ˆ + B 0 + L 1 L 2 whee à = A 0 0, C = [ C ]. Define the estimation eo states := ˆ an := ˆ. Then the estimation eo namics ae given b: ẋ ḋ = A L1C L 1 L 2 C L 2. It follows that we can esign a stable obseve if the matices constitte an obsevable pai. Using the feeback contol ( Ã, C ) = ˆ + G ˆ + N iels the following state space esciption fo the close loop sstem: ẋ = A + B( ˆ + Gŵ + N) + w ˆẋ ˆ = A L1C L 2 C ( ) + L 1 L 1 ˆ L 2 ˆ + B 0 ˆ + G ˆ + N L C + 2 ( ) Combining these eqations iels: ẋ A B BG BN ˆẋ = L 1 C A B L 1 C + BG L 1 L 2 ˆ + BN + L 1 ˆ L 2 C L 2 C L 2 ˆ 0 L 2

9 9 = [ C 0 0 ] ˆ + ˆ As sal, it is ifficlt to etemine an close loop popeties in these cooinates. Let s instea change cooinates to. Using the fact that ḋ = ˆḋ (becase is a constant) iels (afte some algeba) that: ẋ A B B BG BN + BG ẋ = 0 A L 1 C L ḋ 0 L 2 C L = [ C 0 0 ] + It is clea fom the above eqations that (a) Thee is a sepaation pinciple: the close loop eigenvales ae the eigenvales of the state feeback, A B, pls those of the obseve fo ˆ an ˆ. It follows that if ( A, B) is contollable, an Ã, C ( ) is obsevable, then we ma alwas obtain a stable close loop sstem. (b) If the close loop sstem is stable, then the stea state esponse to a step comman (t) = 0 1(t) an a step istbance (t) = 0 1(t) appoaches a constant vale. Hence, the eivatives of the state vaiables asmptoticall appoach zeo. Setting the left han sie of the above eqation eqal to zeo ths iels that an ss = 0 ss = 0 ( ) 1 BN 0 + C( A + B) C( A + B) 1 BG ss := C A + B [ ] 0

10 Sppose that C( A + B) 1 B is ight invetible. (Right invetibilit is gaantee b assmptions (iii) an (iv).) Then setting 10 an N = [ C( A + B) 1 B] # G = N[ C( A + B) 1 + ] iels ss = 1. The onl thing that emains to be poven is that o assmptions gaantee that Ã, C ( ) is obsevable. We o this b showing that ank λi à λi A = ank 0 λi = n + m, C C whee λ is eithe eqal to an eigenvale of A, o λ = 0. Case 1: Sppose that λ 0 is an eigenvale of A. Then λi A ank 0 λi = n + m C if an onl if this mati has n + m lineal inepenent colmns. Becase λ 0, it is clea that the last m colmns ae lineal inepenent among themselves an ae also lineal inepenent fom the fist n colmns. Hence λi A ank 0 λi = n + m ank λi A C = n C if an onl if λ is an obsevable eigenvale of A.

11 11 Case 2: Consie λ = 0. The ank test eces to eqiing that ank A C = n + m This conition is implie b assmptions (ii) an (v)(vii). ### To smmaize: Choose so that A B is stable Choose L 1 an L 2 so that A 0 0 L 1 L [ 2 C ] is stable Set N = [ C( A + B) 1 B] # an G = N[ C( A + B) 1 + ] Then ( t) 0, 0, 0. It is inteesting to note that this contol scheme is also a fom of integal contol; it is jst that the integatos ae bie in the obseve. Inee, we have ˆ = L 2 ŷ ( ) o, in block iagam fom ^ L 2. ^ I / s ^