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Se pon conrol n he sae space seng Nels Kjølsad Poulsen Deparmen of Informacs and Mahemacal Modellng he echncal Unversy of Denmark 29-3-. Absrac hs repor s nened as a supplemen or an eenson o he maeral used n connecon o or afer he courses Sochasc Adapve Conrol (242) and Sac and Dynamc Opmaon (27) gven a he Deparmen of Informacs and Mahemacal Modellng, he echncal Unversy of Denmark. he focus s n hs repor relaed o he problem of handlng a se pon or a consan reference n a sae space seng. In prncple jus abou any (sae space conrol) desgn mehodology may be appled. Here he presenaon s based on LQ desgn, bu oher ypes such as poleplacemen can be appled as well. hs s he Mone Perolo paper whch s a complaon of resuls gahered from he leraure. A major par of he resuls are colleced from he basc conrol leraure durng a sabacal year a Oford unversy and s furher compled and repored n Mone Perolo (Umbra, Ialy). Inroducon he focus n hs repor s se pon conrol or conrol of a dynamc sysem wh a pece wse consan reference. hs problem s sparsely handled n he leraure despe s praccal applcaon. hs s obvously due o s lack of heorecal conens and neres. We wll n hs repor ry o gve he smples presenaon and llusrae he eenon ncludng more general frameworks. For ha maer we wll n he frs par of he repor assume ha sysem s scalar (SISO or sngle npu sngle oupu).
2 2 he sandard regulaon problem he reference s n hs repor assumed o be consan or pece wse consan, whch off course nclude he sandard sep change n he reference. hs problem has wo major neres. he frs problem s he regulaon problem n whch he se pon s consan and he problem s o mach he sepon. he ne focus area s he closed loop properes n connecon o sepon changes. hs s he servo problem n connecon o a sep changes. Relaed problem whch are no addressed n hs repor are problems relaed o consan (or pece wse consan) dsurbances. Moreover, consrans n he conrol acons (or relaed sgnals) or saes are also omed n hs repor. Obvously, oher ypes of varaons (eg. harmonc) n he reference sgnal are also omed here. In hs repor we wll consder he problem of conrollng a sysem gven as: such ha oupu + A + Bu y C + Du mach a se pon (w). Ofen s no only he oupu ha eners no he objecve funcon. In general we can focus on C + D u he choce C I, D s que frequen. In he followng n wll denoe he sysem order and he number of sae, n u s he number of npus and n y s he number of oupus. 2 he sandard regulaon problem he objec n regulaon s o reduce he nfluence from dsurbances or smple o keep he sysem close o he orgn. he sandard regulaon problem has several formulaons. he mos common verson s an LQ formulaon n whch he cos funcon s quadrac n he saes and he conrol acons. In he H 2 formulaon he cos funcon s also a quadrac cos, bu n an augmened oupu vecor whch conans elemens of he errors (objecve) and he conrol acons (coss). 2. he LQ formulaon he sandard LQ problem s o fnd an npu sequence u ha ake he sysem + A + Bu from an nal sae along a rajecory such ha he cos nde J N Q + u Ru s mnmed. Evenually he horon, N, s nfne.e. N. he soluon o hs problem (see 2) can be formulaed as a sae feedback soluon u L ()
2.2 Oupu conrol 3 or as a me funcon u LΦ Φ A BL dependng only on he nal sae and age (.e. ). In a deermnsc seng he wo soluons are dencal, bu n an unceran envronmen he frs (he sae feed back verson) s more robus wh respec o uncerany and nose. he resuls can be found n Append D.. he LQ resuls are no resrced o he sandard presenaon gven above. I can, as ndcaed n Append D.2, also nclude cross erms n he cos funcon. J N Q + u Ru + 2 Su N u Q S S R u he resuls are que smlar o he ones n () ecep off course for he dependence on S. 2.2 Oupu conrol Ofen he problem s o conrol he oupu of a sysem y C wh due respec o he conrol effor. hs s ofen formulaed as a (LQ) problem n whch he cos funcons N J y 2 + ρ2 u 2 (2) has o be mnmed. Here ρ 2 s a desgn parameer. In he LQ framework hs s equvalen o Q C C R ρ 2 hs problem also emerge f we wan o mnme J N y 2 subjec o N u 2 C u C u s he desgn parameer. In ha case ρ 2 s a Lagrange mulpler. In pracce he wo approaches are much more relaed han epeced by a frs nspecon. he problem n (2) can also be formulaed as a mnmaon of J y u N 2 q q ρ 2 In general formulaon of he H 2 problem s a mnmaon of he cos funcon N J 2 q C + D u
4 2 he sandard regulaon problem here s a gh connecon beween C, D, q and Q, R, S. For eample, he cos funcon n (2) appears when C C D q ρ 2 In general he H 2 problem can be ransformed no he LQ problem hrough: Q S C S q C R D On he oher hand he H 2 problem can emerge from he LQ problem f Q S Q S R D s facored no C Q D q C D hs facoraon s by no means unque. A facor can for eample eher be n q or n C and D. In connuous me he problem s o conrol he sysem ẋ A + Bu such ha he cos funcon J 2 P + Z 2 Q + u Ru d s mnmed. he soluon o hs problem s n saonary ( ) gven by: u L : SA + A S + Q SBR B S and L R B S If he cos funcon has a cross erm Z J P + Q + Q + 2 Su d hen he soluon s : u R (B S + S ) S A + A S + Q (S B + S)R (B S + S ) he realon beween he LQ and H 2 problem s he same n connuous me as n dscree me.»» Q S C S R D W ˆ C D As n he dscree me case hs wll ypcally be he case f he problem arse from a mnmaon of Z J y 2 W d whch s a weghed (W) negral square of he oupu.e. he H 2 problem. y C + D u
5 3 Feed forward Le us now urn he he focus area of hs repor, namely conrol of a sysem such ha s oupu vecor mach he reference vecor. We assume ha a se pon w ess, s pecewse consan and he dmenson equals n. Probably he mos obvous way of nroducng he reference (see eg. ) s o nclude a feed forward erm (M, n u n y ) n he conrol. ha s o use conrol law In hs case he closed loop becomes u Mw L + Φ + BMw Φ A BL he feed forward erm, M, can be chosen such ha he oupu mach he reference n saonary,.e. y C(I Φ) B + D Mw Le us denoe he DC gan hrough he sysem as κ C(I Φ) B + D and assume ha s non ero (.e. nverble). In he normal case n u n y we can use: M κ If n u > n y we have an era flebly and can use M κ (κκ ) On he oher hand f n u < n u we can fulfll our objecve. If we wll mnme he dsance beween our objecve (w) and our possbly (KMw) hen we can use: M (κ κ) κ I s well known ha hs soluon s sensve o modellng errors. If he DC gan n he sysem model s wrong, hen he closed loop wll have a smlar error. In connuous me he dscrpon s ẋ A + Bu y C + Du and we have he same resul jus s he DC gan s gven by κ CA B + D
6 4 arge values 4 arge values Anoher very classcal way of handlng he se pon s o consder he sandard problem as a devaon away from a saonary pon. hs mehod has been eensvely been used n 3. We wll he denoe hs pon as a arge pon. ha s o consder he problem as a redefnon of orgn accordng o he reference. ha means he conrol s gven by: u u L( ) he arge pon, u has o be a saonary pon.e. o fulfll A + Bu he arge pon also have o mach he se pon,.e. o fulfll w C + Du In oal he saonary pon should sasfy he followng equaon A I B w C D u I hs ndcaes ha boh and u s proporonal o he se pon and Mw u + L M was dscussed n he prevous secon. hs soluon has, as he feed forward mehod, a hgh sensvy o modellng errors. he classcal way of handlng hs problem s use negral acon as we wll reurn o n Secon 9. Here we wll pursu anoher ype of approach. If we look a he problem of modellng errors, we can handle n saonary. One way o ackle he problem s o nclude a (consan) dsurbance n he npu or he oupu. + A + B(u + d) y C + D(u + d) he npu dsurbance has be esmaed wh an observer or a Kalman fler. In ha way, he dsurbance wll epress he modellng error a DC. Anoher, bu smlar, approach s o nclude en oupu dsurbance and operae wh a desgn model + A + Bu y C + Du + d Also here he dsurbance has o be esmaed. he wo approaces can be merged no he general form + A + Bu + Gd (3) y C + Du + Hd G and H are marces of approprae dmensons. In ha case we can fnd he arge values from: A I B G w C D H u + d I
7 d has o be esmaed. If we assume ha he dsurbance closely s consan we can model by he followng model d + d + ξ ξ N d (, R d ) and augmen he sochasc verson of (3). A G B v + u d I d + ξ + y C H + Du d + e hen d (and sae ) can be esmaed by means of an observer or a Kalman fler. (4) In connuous me he sysem s descrbed by d A + Bu + Gd d y C + Du + Hd and he arge values can be deermned by:»»» A B G + C D u H» d I w he sae varable whch s no neccesarely known can be esmaed from he descrpon» d d d» A G I d d d ξ» B +» v u + ξ 5 Opmal rackng In hs secon we wll eend he sandard dscree me LQ conrol problem o nclude a reference sgnal. hs approach s based on he presenaon and resuls n 2. In hs presenaon we wl resrc he approach and assume ha he reference sgnal s consan (e. s a se pon). Consder he problem of conrollng a dynamc sysem n dscree me such ha he cos funcon s mnmed. + A + Bu (5) N J y N r N 2 P + C r 2 Q + u 2 R In seady sae (N, r r) he soluon s gven by u L + Kv L R + B SB B SA K R + B SB B
8 6 Inernal model prncple (IMP) and S s he soluon o he algebrac Rca equaon; and v found from S C QC + A SA A SB R + B SB B SA v (I Φ ) C Qr Φ A BL Unless he sysem has (or has been mposed) a pure negraon, hs sraegy wll resul n a seady sae error (for R > ). In connuous me he problem s o conrol he sysem such ha he cos funcon ẋ A + Bu Z J y r 2 P + C r 2 Q + u 2 R d s mnmed. In seady sae (, r r) he soluon s gven by L R B S u L + Kv and S s he soluon o he algebrac Rca equaon; and v found from K R B SA + A S + C QC SBR B S v Φ C Qr Φ A BL Consder he problem of conrollng a dynamc sysem n connuous me gven by (2) such ha he cos funcon Z J C r 2 P + r 2 Q d C + Du s mnmed. In seady sae ( ) he soluon s gven by u L + Kv + Mr L ˆD QD (D QC + B S ) K ˆD QD B and S s he soluon o he algebrac Rca equaon; and v found from M ˆD QD D Q S A + A S + C QC (C QD + SB)(D QD) (D QC + B S ) v Φ (C DL) Qr 6 Inernal model prncple (IMP) Ye anoher way of ransformng he se pon problem (and oher ype reference problems) no a sandard regulaon problem s o use he socalled Inernal Model Prncple (IMP). I ha case a model of he se pon varance s creaed and buld no an augmened sysem descrpon.
9 he model r + r + ξ ξ s a whe nose sequence, s a suable model for se pons varaon (wh unknown changes and unknown nsans of changes). In ha case he oal descrpon becomes r + A r B + u + y C r One obvous problem n hs approach s ha he sysem descrpon s no conrolable. ha ofen resuls ofen resuls n problems when usng sandard sofware for solvng he Rca equaon. he soluon ess due o he fac ha he unconrollable par of he sae space s no vsble n he cos funcon. he soluon u L L r can be compared wh he soluon n secon 5. ξ In connuous me he sysem d d A + Bu has o be conrolled such he oupu s close o he reference. he model d r ξ d ξ s whe nose s a suable model for a sepon sgnal. he oal model can be gven as:»»»»» d A B + u d r r + ξ and he ask s o mnme he error y ˆ C» r (n some sense and wh respec o he conrol power). he soluon s u ˆ» L L r whch s a feedback from he sae and a feedforward from he se pon sgnal. 7 Conrol moves In hs secon we wll dscuss conrol formulaed n erms conrol moves. hs can be regarded as f he decson varable s he velocy of he conrol. I can be eended o nclude he dervave of he conrol acon o any order. If an opmal conrol sraegy from secon 5 or 6 s appled on a sysem (whou an negral acon) hen a non ero se pon wll resul n a seady sae error. hs s due a conflc beween seady sae error and seady sae conrol acon. One remedy s o dscharge he DC componen of he conrol acon n he cos funcon. he mos drec and smples way of don hs s o consder he conrol moves (conrol velocy) raher han he conrol acon self.
7 Conrol moves 7. Velocy conrol he problem of conrollng a dynamc sysem + A + Bu y C such ha he oupu s close o a ceran (consan) se pon, r, has some challenges. In order o avod he problem of a seady sae error we can formulae he objecve n erms of he conrol moves v. If he conrol moves are consan beween samples hen: u + s v + + s v and he descrpon of he sysem can be wren as: A B Bs + v (6) s + y C hs s more or less he same as nroducng (a dscree me) negraon n fron of he sysem. he problem s hen o conrol he sysem n (6) accordng o he mehod descrbed n secon 5 and 6. In ha case he cos s changed and he conrol relaed cos n he objecve funcon s shfed n frequency and he wegh s pu on he conrol velocy. he conrol acon can also be 7.2 Acceleraon conrol he mehod can be eended and he problem formulaed such ha he decson varable s he acceleraon, a, hen (f he acceleraon s consan beween samples): u + s v + 2 2 s a + + s v + 2 2 s a v + v + s a and he sysem can be wren as: A B B s s + v v + y C v 2 B 2 s 2 2 s s hs s he same as nroducng wo negraon n fron of he sysem. a (7)
7.3 General velocy conrol 7.3 General velocy conrol hese deas can be eended o jus abou any order. Le us llusrae hs n a forh order case. Le u be he conrol acon and le u 4 be he forh order dervave. hen (f u 4 s consan beween samples): u + s u + 2 2 s u 2 + 6 3 s u 3 + 24 4 s u 4 u u 2 u 3 or saed oherwse: u u 2 u 3 + + + s u + 2 2 s u 2 + 6 3 s u 3 + 24 4 s u 4 u + s u 2 + 2 2 s u 3 + 6 3 s u 4 u 2 + s u 3 + 2 2 s u 4 u 3 + s u 4 6 3 s s 2 s 2 s 2 s 2 s he oal sysem descrpon becomes: A B B s s u u 2 u 3 + 2 B 2 s 2 2 s 6 B s 3 6 s 3 s 2 2 s s y C hs mehod can be eened o any order. u u 2 u 3 u u 2 u 3 u u 2 u 3 + 24 4 s 6 3 s 2 2 s s + u 4 24 B 4 s 24 4 s 6 3 s 2 2 s s u 4 7.4 Genereled velocy conrol In he prevuous secons we solved he problem of machng he se pon by means of usng a hgher order dervave as he decson varable. In hs secon we wll use a combnaons of hgher order dervaves as decson varables. ha means he conrol decson s a blend of dfferen frequences. 7.4. par Le us frs focus on he problem when we use he conrol move, v and he conrol level, ũ, as decson varable. In ha case: u ũ + + s v + + s v or smply: + A B y C B Bs ũ + s v
2 7 Conrol moves 7.4.2 Par 2 If he decson varables are he acceleraon, he level of he conrol move (velocy) and he conrol level, hen we can nroduce v + v + ṽ + s a + + s (v + ṽ ) + 2 2 s a ũ and ṽ are perurbaons on he conrol and conrol move respecvely. he decson consss of he vecor ũ ṽ a In ha case: u ũ + + s (v + ṽ ) + 2 2 s a Consequenly, he acceleraon model n (7) can be ransformed no A B B s s + B B s 2 B s 2 s 2 2 ũ s ṽ v v s a + y C v (8) 7.4.3 Par 3 he mehod can, as he smple mehod from he prevous secons, be eended o any fne order. For n 4 we have he resulng sysem descrpon: A B B s 2 B 2 s 6 B s 3 s 2 u 2 s 6 3 s s 2 u 2 s 2 u s u 2 u 3 u 3 + B B s 2 B 2 s 6 B 3 s 24 B s 4 s 2 + s 2 6 s 3 24 s 4 s 2 s 2 6 s 3 s 2 s 2 s y C u u 2 u 3 ũ ũ ũ 2 ũ 3 u 4 In connuous me he problem s o conrol he sysem d d A + Bu y C
7.4 Genereled velocy conrol 3 such ha he oupu mach he reference. Velocy conrol: If we wan o use he conrol velocy d u v d raher han he conrol sgnal self as he decson varable hen he sysem can be descrbed as:»»»» d A B + v d u u y ˆ C» u Acceleraon conrol: If we wan o use he conrol acceleraon d d u v d v a d as he decson varable hen he sysem can be descrbed as: 2 d 4 3 2 u 5 4 A B 3 2 5 4 3 u 5 d v v 2 3 y ˆ C General velocy conrol: As n descree me hese dears can be generaled o any order. Le he decson varable be v u (n) hen for n 4 we have he descrpon: 2 3 2 d u d 6 u () 7 4 u (2) 5 6 4 u (3) Generaled velocy conrol: u (n) dn d n u A B 4 u v y ˆ C 6 4 3 2 7 5 6 4 2 5 u u () u (2) u (3) + u u () u (2) u (3) 3 2 7 5 4 3 5 a 3 2 7 + 6 5 4 If we wll use a hgh order dervave as well as perubaons of s negrals, we can use he followng agmenaon scheme. he for n 4 we have he descrpon: 2 3 2 3 2 3 A B 2 3 2 3 B ũ d u d 6 u () 7 4 u (2) 5 u 6 7 4 5 6 u () 7 4 u (2) 5 + ũ () 6 7 4 5 6 ũ (2) 7 4 ũ (3) 5 u (3) u (3) ũ (4) 2 y ˆ C 6 4 u u () u (2) u (3) 3 7 5 3 7 5 v
4 8 Frequency weghng 8 Frequency weghng he reason for nroducng velocy conrol and s relaed s o le he conrol sgnal eners no he performance n such a way ha he resulng conroller can compensae for he sepon change n an approprae manner. Now assume he he sgnals n he performance nde are flered verson of he orgnal sgnals, such as he oupu, y, conrol u. J E { N ( y f y f H y (q)y ) ( u f O O 2 y f O2 O 2 u f H u (q)u hese frequency weghs or flers has a sae space represenaon such as: u f ) } y + Ay y + B y y u + Au u + Bu u If he sysem s gven by: y f C y y + D y y + A + Bu y C + Du u f C u u + D u u hen he sysem descrpon can be augmened o nclude he flered versons of he sgnals: y u + For he flhered oupu we have In a smlar way s If he augmened sae vecor A B y C A y A u y u y f D y C + C y y + D y Du ( D y C C y D y D ) u f ( C u D ) u y u y u u + y u u B B y D B u s nroduced he we can wre: y f ( ) D u f y C C y D y D C u D u u u
5 Insead of mnmng he orgnal cos we can mnme { N J E ( u ) ( Q Q )} 2 Q 2 Q 2 u C (D y ) (C y ) (C u ) D (D y ) (D y ) Q Q 2 Q 2 Q 2 O O 2 D y C C y D y D O2 O 2 C u D u Now he connuous me verson of he problem jus menoned. Assume he he sgnals n he performance nde are flered verson of he orgnal sgnals, such as he oupu, y, conrol u. ( Z»! ) J E y f u f O O 2 y f O2 O 2 u f d y f Hy(p)y uf Hu(p)u hese frequency weghs or flers has a sae space represenaon such as: d d y A y y + By y d d u A u u + B u u y f Cy y + Dy y u f Cu u + Du u If he sysem s gven by: d A + Bu d y C + Du hen he sysem descrpon can be augmened o nclude he flered versons of he sgnals: 2 d 4 y d u 3 5 For he flhered oupu we have In a smlar way s If he augmened sae vecor s nroduced he we can wre: " # y f u f 2 4 B y C A y A A u 3 2 3 5 4 y 5 + u y f D y C + C y y + Dy Du ` D y C C y D y D 6 4 2 u f ` C u D u 6 4 2 4 y u 3 5 2 y u u 2 4 B B y D B u y u u 3 7 5 3 7 5 D y C C y D y D C u D u u 3 5 u Insead of mnmng he orgnal cos we can mnme j Z» J E u Q Q 2 Q 2 Q 2 «ff d u
6 9 Inegral acon 2 6 4 C (D y ) (C y ) (C u ) D (D y ) (D y )» Q Q 2 Q 2 Q 2 3» 7 O O 2 5 O2 O 2» D y C C y D y D C u D u 9 Inegral acon Whn he conrol area here are a few mehods o avod a seady sae errors. One of he sandard rcks n conrol s o nroduce an negral acon. In he sae space seng nvolves a sae whch negrae he error, e. + + (r y ) ha resuls n an augmened sysem gven by A B + C + y C u + r If he conroller s gven by u u L hen he closed loop s gven by A BL BL C + B + B u + L + r hs can also be formulaed as: whch resuls n + u Mr L A BL BL C e C + r BM + he problem s o deermne M such ha he response s opmal. r Whn he conrol area here are a few mehods o avod a seady sae errors. One of he sandard rcks n conrol s o nroduce an negral acon. In he sae space seng nvolves a sae whch negrae he error, e. d r y d
7 ha resuls n an augmened sysem gven by»»»» d A B + d C y ˆ C»» u + r If he conroller s gven by» u u L hen he closed loop s gven by»»»» d A BL BL B + d C» B u +»» L + r hs can also be formulaed as: whch resuls n» d d» u Mr L» A BL BL C» e ˆ C» + r he problem s o deermne M such ha he response s opmal.» BM + r Concluson In hs paper we have revwed dfferen conrol mehods for handlng se pons n a sae space seng. he mehods covers area from feed forward and arge values o frequency weghs and negral acon. he paper s held n dscree me, bu he resuls s also gven n connuous me. Acknowledgemen hs research was graefully suppored by a gran from he Dansh Research Councl for echncal Scence, SVF (Saens eknsk-vdenskabelge Forsknngsråd) under lcense no. 98576. References H. Kwakernaak and R. Svan. Lnear Opmal Conrol Sysems. Wleynerscence, 972. 2 F.L. Lews. Opmal Conrol. Wley, 986. 3 Gabrele Pannoccha, Nabl Laach, and James B. Rawlngs. A canddae o replace pd conrol: Sso-consraned lq conrol. AIChE Journal, 25.
8 B Quadrac forms II A Quadrac opmaon I Consder he problem of mnmng a quadrac cos funcon J u h h 2 h 2 h 22 u h + 2 h 2 u + u h 22 u I s que elemenary o fnd he dervave of he cos funcon and he saonary pon mus fulfll he saonary pon d d uj 2 h 2 + 2u h 22 (9) h 2 + h 22 u u h 22 h 2 s a mnmum o he cos funcon f h 22 s possve defne. Furhermore, he mnmum of he cos funcon s quadrac n. If we use (9) and complee he square hen: J h (u ) h 22 u ( h h 2 h 22 h 2) S S h h 2 h 22 h 2 B Quadrac forms II Le us now consder he more comple problem of mnmng + g gu or smply: J u h h 2 h 2 h 22 u u J h + 2 h 2 u + u h 22 u + g + g u u + σ A sandard resul gves he mnmum as or Applyng () gves he opmal cos: + σ 2 h 2 + 2u h 22 + g u () u h 22 h 2 + 2 g u J h + g + σ (u ) h 22 u (h h 2 h 22 h 2 ) + (g g u h 22 h 2 ) + σ 4 g u h 22 g u h + g + σ
9 h h h 2 h 22 h 2 g g h 2 h 22 g u σ σ 4 g u h 22 g u C Feed forward hs append s a srah forward eenon of secon 3 and s vald boh n a descree and connuous me framework. In closed loop and n saonary we have ȳ Kū K R ny nu s he DC gan hrough he sysem. Our objecve s o fnd ū such ha In he feedforward sraegy we use M R nu ny. ȳ w ū Mw C. he balanced problem If n y n u and f K s nonsngular hen s rval ha ū K w or M K C.2 he overfleble problem Frsly, le n u > n y. ha means we have a surplus of flebly o acheve our objecve. We can hen choce o fnd ha saonary conrol whch has he lowes se (and sll acheve our objecve). hs can be formulaed as mnmng subjec o J 2ū ū w Kū he Lagrange funcon for hs problem s J L 2ū ū + λ (Kū w)
2 D he Dscree me LQ conrol problem whch s saonary wr. ū for ū + λ K or for ū K λ In order o acheve our objecve we mus have λ (KK ) w or hs resuls n ū K (KK ) w M K (KK ) C.3 he resrced problem Le us hen focus on he suaon n u < n y e. when we have less conrol flebly. In ha case we can acheve our objecve (ȳ w) bu have o fnd a ū (or a M) such ha he dsance beween he objecve and he possble s mnmed. In oher word we wll fnd an ū such ha s mnmed. hs s obaned for J 2 ε ε ε w Kū I ε K (w Kū) K or f hs resuls n: ū (K K) K w M (K K) K D he Dscree me LQ conrol problem In hs secon we wll revew he resuls relaed o conrol of a lnear me nvaran dynamc sysem + A + Bu () such ha a quadrac cos funcon s mnmed. D. he Sandard DLQ Conrol Problem Le us frs focus on he sandard problem. In hs cone we wll conrol he sysem n () such ha he (sandard LQ) cos funcon N J NP N + Q + u Ru (2) s mnmed. he Bellman equaon wll n hs case be V ( ) mn u Q + u Ru + V + ( + ) (3)
D.2 DLQ and cross erms 2 wh he end pon consrans If we es he canddae funcon V N ( N ) N P N V ( ) S hen he nner par of he mnmaon n (3) wll be I u Q + A S + A A S + B B S + A R + B S + B u he mnmum for he hs funcon s accordng o Append A gven by u L L R + B S + B B S + A and he canddae funcon s n fac a soluon o he Bellman equaon n (3) f S Q + A S + A A S + B R + B S + B B S + A S N P If he gan, L, s used n he recurson for S S Q + A S + (A BL ) S N P As a smple mplcaon from he proof we have ha V ( ) J S whch s usefull n connecon o an nerpreaon of S. D.2 DLQ and cross erms In order o connec he (very) relaed LQ formulaon and H 2 formulaon we wll augmen he sandard problem wh cross erms n he cos funcon. Assume ha a dscree me (LI) sysem s gven as n () and he cos funcon (nsead of (2)) s: N J N P N + Q + u Ru + 2 Su or J N P N + N u Q S S R u he suaon becomes a b more complcaed. he cross erms especally occurs f he conrol problem s formulaed as a problem n whch (he square of) an oupu sgnal y C + Du s mnmed,.e. In ha case Q S S R J N y 2 W C D W C D
22 D he Dscree me LQ conrol problem he Bellman equaon becomes n he specal case V ( ) mn u u Q S S R u + V + ( + ) V N ( N ) NP N and agan we wll ry he follwong canddae funcon V ( ) S hs can be solved head on or by ransformng he problem no he sandard one. D.2. Usng ransformaon echnque If R s nverble hen we can noduce a new decson varable, v, gven by: u v R S he nsanuous loss erm (frs erm n he Bellman equaon) can be epressed as: u Q S S R u Q + v Rv In smlar way we fnd for he dynamcs Q Q SR S + A + Bu (A BR S) + Bv Ā + Bv Ā A BR S For he fuure cos o go (he second erm n he Bellman equaon) we have: V + ( + ) + S + + (Ā + Bv ) S + (Ā + Bv ) We have now ransformed he problem o he sandard form and he nner mnmaon n he Bellman equaon V ( ) mn u Q + u Ru + V + ( + ) s hen smply: I wh he soluon v Q + Ā S + Ā Ā S + B B S + Ā R + B S + B v v L L R + B S + B B S + Ā he canddae funcon s a soluon o he Bellman equaon f S Q + Ā S + Ā Ā S + B R + B S + B B S + Ā (4) Ā S + (Ā B L ) + Q hs means ha u L + R S L R + B S + B B S + Ā
D.3 Opmal rackng 23 D.2.2 Drec mehod If R s no nverble hen we are forced o use a more drec approach whch resuls n he followng nner mnmaon (mnmaon of he nner par n he Bellman equaon): I wh he soluon and a Rcca equaon u Q + Ā S + Ā S + Ā S + B S + B S + Ā R + B S + B u u L L R + B S + B S + B S + A S Q + A S + A S + A S + B R + B S + B S + B S + A (5) Noce, ha (4) s he sandard Rcca equaon, as (5) conans (drecly) he cross erm S. he ransformaon mehod do requre ha R s nverble. D.3 Opmal rackng In hs secon we wll eend he sandard dscree me LQ conrol problem o nclude a reference sgnal. Laer on n hs presenaon, we wl resrc he approach and assume ha he reference sgnal s consan (e. s a se pon). Consder he problem of conrollng a dynamc sysem n dscree me gven by () such ha he cos funcon N J y N r N 2 P + C r 2 Q + u 2 R s mnmed. hs equvalen o he LQ cos funcon N J (C N r N ) P(C N r N ) + (C r ) Q(C r ) + u Ru (6) s o be mnmed. he Bellman equaon wll n case be V ( ) mn u (C r ) Q(C r ) + u Ru + V + ( + ) (7) wh he end pon consrans If we es he canddae funcon V N ( N ) (C N r N ) P(C N r N ) V ( ) S 2v + σ S N C PC v N C Pr N σ N r N Pr N hen he nner par of he mnmaon n (7) wll be I (C r ) Q(C r ) + u Ru +(A + Bu ) S + (A + Bu ) 2v +(A + Bu ) + σ + (C QC + A S + A) + u (R + B S + B)u + r Qr +2 A S + Bu 2(r QC + v + A) 2v + Bu + σ +
24 D he Dscree me LQ conrol problem he mnmum for he hs funcon s accordng o Append B gven by u R + B S + B B S + A B v + and he canddae funcon s n fac a soluon o he Bellman equaon n (7) f S C QC + A S + A A S + B R + B S + B B S + A v A v + + C Qr A S + B R + B S + B B v + σ σ + + r Qr v+b R + B S + B B v + If we nroduce he gans L R + B S + B B S + A K R + B S + B B hen he conrol law can be wren as u L + K v + and he Rca equaons becomes S C QC + A S + (A BL ) S N P v (A BL ) v + + C Qr v N C Pr N σ σ + + r Qr v + BK v + σ N r N Pr N In seady sae (N, r r) he soluon s gven by u L + Kv L R + B SB B SA K R + B SB B and S s he soluon o he algebrac Rca equaon; and v found from S C QC + A SA A SB R + B SB B SA v (I Φ ) C Qr Φ A BL Unless he sysem has (or has been mposed) a pure negraon, hs sraegy wll resul n a seady sae error (for R > ). D.4 Reference conrol wh cross erm In hs secon we wll eend he resul from he prevous secon a b furher. Consder he problem of conrollng a dynamc sysem n dscree me gven as n () such ha he cos funcon J C N N r N 2 P + r 2 Q
D.4 Reference conrol wh cross erm 25 C + Du s mnmed. hs equvalen o he (sandard LQ) cos funcon J ( C N r N ) P( C N r N ) (8) N + (C + Du r ) Q(C + Du r ) s o be mnmed. he Bellman equaon wll n hs case be V ( ) mn u (C + Du r ) Q(C + Du r ) + V + ( + ) (9) wh he end pon consrans If we es he canddae funcon V N ( N ) ( C N r N ) P( C N r N ) V ( ) S 2v + σ S N C P C v N C Pr N σ N r NPr N hen he nner par of he mnmaon n (9) wll be I (C + Du r ) Q(C + Du r ) +(A + Bu ) S + (A + Bu ) 2v +(A + Bu ) + σ (C QC + A SA) + r Qr + u (D QD + B SB)u + 2 (C QD + A SB)u 2(r QD + v +B)u 2v +A + σ + he mnmum for he hs funcon s accordng o Append B gven by u D QD + B SB ( (D QC + B SA) B v D Qr ) If we apply he resuls n Append B hen we can see ha he canddae funcon s n fac a soluon o he Bellman equaon n (25) f S A SA + C QC ( C SD + A SB ) D QD + B SB ( D QC + B SA ) v A v + ( C SD + A SB ) D QD + B SB ( B v + + D Qr ) σ σ + + r Qr ( r QD + v + B) D QD + B SB ( D Qr + B v + ) he nal (or raher ermnal) condons are: If we nroduce he gans S N C PC v N C Pr σ N r N Pr N L D QD + B SB ( D QC + B SA ) K D QD + B SB B M D QD + B SB D Q hen he conrol law can be wren as u L + K v + M r
26 E he Connuous me LQ Conrol and he Rca equaons becomes S v A S(A BL) + C QC C SDL (A BL ) v + L DQr σ σ + + r Qr ( r QD + v + B) D QD + B SB ( D Qr + B ) v + In seady sae (, r r) he soluon s gven by u L + Kv + Mr L D QD + B SB ( D QC + B SA ) K D QD + B SB B M D QD + B SB D Q and S s he soluon o he algebrac Rca equaon; S A SA + C QC ( C SD + A SB ) D QD + B SB ( D QC + B SA ) and v found from v I Φ L DQr Φ A BL E he Connuous me LQ Conrol In hs secon we wll revew he resuls gven n he prevous secons bu n connuous me. We wll sar wh he sandard LQ problem and hen n order o connec wh he H 2 formulaon revew he LQ problem wh a cross erm n he cos funcon. he problem s o conrol he LI sysem gven n connuous me such ha some objecves are mee. d d A + Bu (2) E. he Sandard CLQ Conrol problem Consder he problem of conrollng a connuous me LI sysem n (2) such ha he performance nde J P + Q + Q d s mnmed. he Bellman equaon s for hs suaon wh d d V ( ) mn u Q + u Ru + d d V ( ) (A + Bu ) V P as boundary condon. For he canddae funcon V ( ) S
E.2 CLQ and cross erms 27 hs (Bellman) equaon becomes hs s fulflled for Ṡ mn u Q + u Ru + 2 S A + 2 S Bu u R B S he canddae funcon s ndead a Bellman funcon f S s he soluon o he Rcca equaon Ṡ S A + A S + Q S BR B S S P In erms of he gan L R B S he Rcca equaon can also be epressed as Ṡ S A + A S + Q L RL Ṡ S (A BL ) + A S + Q (A BL ) S + S A + Q Ṡ S (A BL ) + (A BL ) S + Q + L RL I can be shown ha J S E.2 CLQ and cross erms Le us now focus on he problem performance nde has a cross erm,.e. J P + Q + Q + 2 Su d As n he dscree me case hs wll ypcally be he case f he problem arse from a mnmaon of J y 2 W d whch s a weghed (W) negral square of he oupu y C + Du.e. he H 2 problem. In ha case Q S C S R D W C D he Bellman equaon s now for hs suaon wh d d V ( ) mn u u Q S S R V P + d u d V ( ) (A + Bu ) as boundary condon. Agan we can go drecly for a soluon, bu f R s nverble, we can ranform he problem o he sandard form. (2)
28 E he Connuous me LQ Conrol E.2. Usng ransformaon echnque If we use he same mehod as n he dscree me and nroduce a new decson varable, v hrough u v R S hen nsananuous loss erm s rewren o u Q S S R u Q + v Rv Q Q SR S Furhermore he dynamcs s ransformed o A + Bu (A BR S ) + Bv Ā + Bv Ā (A BR S ) he Bellman equaon s now n he newly nroduced varable d d V ( ) mn Q + v Rv + dd V ( ) ( ) Ā + Bv v wh V P as boundary condon. For he canddae funcon hs Bellman equaons becomes V ( ) S Ṡ mn v Q + v Rv + 2 S Ā + 2 S Bv he soluon o hs problem s v L L R B S Ṡ S Ā + Ā S + Q S BR B S S P (22) he las equaon ensures ha he canddae funcon ndeed s a soluon. he oal soluon s consequenly gven as u ( L + R S ) L R B S or smpy as u R (B S + S ) Noce, ha (22) s he same Rcca equaon ha arse from he sandard problem ecep for he ransformaon of A and Q. Furhermore L s he same as arse from he sandard problem.
E.3 Reference conrol 29 E.2.2 Drec mehod If R s no nverble hen (2) mus be solved drecly. For he canddae funcon he Bellman equaon, (2), becomes V ( ) S Ṡ mn u Q + u Ru + 2 Su + 2 S A + 2 S Bu whch s mnmed for u R (B S + S ) Ṡ S A + A S + Q (S B + S)R (B S + S ) S P (23) I s que easy o check ha (for R beng nverble) he soluons o (23) and (22) are dencal. E.3 Reference conrol In hs secon we wll eend he sandard dscree me LQ conrol problem o nclude a reference sgnal. Consder he problem of conrollng a dynamc sysem n connuous me gven by (2) such ha he cos funcon J y r 2 P + C r 2 Q + u 2 R s mnmed. hs equvalen o he (sandard LQ) cos funcon J (C r ) P(C r ) + (C r ) Q(C r ) + u Ru d (24) s o be mnmed. he Bellman equaon wll n hs case be d d V ( ) mn u wh he end pon consrans If we es he canddae funcon (C r ) Q(C r ) + u Ru + d d V ( )(A + Bu ) V ( ) (C r ) P(C r ) V ( ) S 2v + σ S C PC v C Pr σ r Pr hen he nner par of he mnmaon n (25) wll be I (C r ) Q(C r ) + u Ru + 2 S v (A + Bu) (C QC + AS + SA ) + r Qr + u Ru 2 r QC + v A + 2( S v )Bu d (25)
3 E he Connuous me LQ Conrol he mnmum for he hs funcon s accordng o Append A gven by u R B S v he canddae funcon s n fac a soluon o he Bellman equaon n (25) f Ṡ S A + A S + C QC S BR B S S C PC v (A BR B S) v + C Qr v C Pr σ r Qr v BR B v σ r Pr If we nroduce he gans L R B S K R B hen he conrol law can be wren as and he Rca equaons becomes u L + K v Ṡ S A + A S + C QC L RL S C PC v (A BL ) v + C Qr v C Pr σ r Qr v K RK v σ r Pr In seady sae (, r r) he soluon s gven by u L + Kv L R B S K R B and S s he soluon o he algebrac Rca equaon; SA + A S + C QC SBR B S and v found from v Φ C Qr Φ A BL E.4 Reference conrol wh cross erm In hs secon we wll eend he resuls relaed o reference conrol and nclude a possble cross erm. Consder he problem of conrollng a dynamc sysem n connuous me gven by (2) such ha he cos funcon J C r 2 P + r 2 Q C + Du s mnmed. hs equvalen o he (sandard LQ) cos funcon J ( C r ) P( C r ) + d (C + Du r ) Q(C + Du r ) d (26)
E.4 Reference conrol wh cross erm 3 s o be mnmed. he Bellman equaon wll n case be d d V ( ) mn u wh he end pon consrans If we es he canddae funcon (C + Du r ) Q(C + Du r ) + d d V ( )(A + Bu ) V ( ) ( C r ) P( C r ) V ( ) S 2v + σ S C P C v C Pr σ r Pr hen he nner par of he mnmaon n (27) wll be I (C + Du r ) Q(C + Du r ) + 2 S v (A + Bu) (C QC + AS + SA ) + r Qr + u D QDu + 2 C QDu 2r QDu 2 r QC + v A + 2( S v )Bu (C QC + AS + SA ) + r Qr 2 r QC + v A +u D QDu + 2 C QDu 2r QDu + 2( S v )Bu he mnmum for he hs funcon s accordng o Append A gven by u D QD ( (D QC + B S ) B v D Qr ) he canddae funcon s n fac a soluon o he Bellman equaon n (25) f Ṡ S A + A S + C QC (C QD + SB)(D QD) (D QC + B S ) v (A B(D QD) (D QC + B S )) v + C Qr (C QD + SB)(D QD) D Qr σ r Qr (v B + r QD)(D QD) (B v + D Qr ) he ermnal condons are: If we nroduce he gans S C PC v C Pr σ r Pr L D QD (D QC + B S ) (27) K D QD B M D QD D Q hen he conrol law can be wren as and he Rca equaons becomes u L + K v + M r
32 E he Connuous me LQ Conrol Ṡ S A + A S + C QC L (D QD)L S C PC v (A BL ) v + (C DL) Qr v C Pr σ r Qr (v K + r M )(D QD)(K v + M r ) σ r Pr In seady sae (, r r) he soluon s gven by u L + Kv + Mr L D QD (D QC + B S ) K D QD B M D QD D Q and S s he soluon o he algebrac Rca equaon; S A + A S + C QC (C QD + SB)(D QD) (D QC + B S ) and v found from v Φ (C DL) Qr Φ A BL