he mo va mang noron, matlab
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TS. NGUYN NH HIN & TS. LI KHC LI
H M & NRONTRONG K THUT IU KHINSch Chuyn kho dng cho o to Sau i hc ngnh iu khin & T ng ho
NH XUT BN KHOA HC T NHIN V CNG NGH H NI 2007
MC LCLI NI U.................................................................................................. 6 Chng 1: LGIC M..................................................................................... 1 1.1. TNG QUAN V LGIC M............................................................. 1 1.1.1. Qu trnh pht trin ca 1gic m .................................................. 1 1.1.2. C s ton hc ca 1gic m.......................................................... 1 1.1.3. Lgic m l 1gic ca con ngi ................................................... 2 1.2. KHI NIM V TP M.................................................................... 3 1.2.1. Tp kinh in .................................................................................. 3 1.2.3. Cc thng s c trng cho tp m ................................................ 4 1.2.4. Cc dng hm lin thuc ca tp m .............................................. 5 1.3. CC PHP TON TRN TP M..................................................... 5 1.3.1. Php hp hai tp m ....................................................................... 5 1.3.2. Php giao ca hai tp m ................................................................ 6 1.3.3. Php b ca mt tp m ................................................................. 8 1.4. BIN NGN NG V GI TR CA BIN NGN NG ............... 8 1.5. LUT HP THNH M ..................................................................... 9 1.5.1. Mnh hp thnh ......................................................................... 9 1.5.2. M t mnh hp thnh ............................................................... 9 1.5.3. Lut hp thnh m........................................................................ 10 1.5.4. Cc cu trc c bn ca lut hp thnh ........................................ 11 1.5.5. Lut hp thnh n c cu trc SISO........................................... 12 1.5.7. Lut ca nhiu mnh hp thnh............................................... 19 1.5.7. Lut hp thnh SUM-MIN v SUM-PROD................................. 22 1.6. GII M ............................................................................................. 23 2.6.1. Phng php cc i .................................................................... 24 Chng 2: IU KHIN M........................................................................ 29 2.1. CU TRC CA B IU KHIN M.......................................... 29 2.1.1. S khi b iu khin m........................................................ 29 2.1.2. Phn loi b iu khin m .......................................................... 30 2.1.3. Cc bc tng hp b iu khin m........................................... 31 2.2. B IU KHIN M TNH.............................................................. 32
2.2.1. Khi nim...................................................................................... 32 2.2.2. Thut ton tng hp mt b iu khin m tnh .......................... 32 2.2.3. Tng hp b iu khin m tuyn tnh tng on........................ 33 2.3. B IU KHIN M NG............................................................ 35 2.4. THIT K H IU KHIN M BNG PIIN MM MATLAB . 37 2.4.1. Gii thiu hp cng c lgic m .................................................. 37 2.3.2. V d thit k h m ..................................................................... 41 2.5. H IU KHIN M LAI (F-PID) ................................................... 45 2.6. H IU KHIN THCH NGHI M ................................................ 46 2.6.1. Khi nim...................................................................................... 46 2.6.2. Tng hp b iu khin thch nghi m n nh............................ 48 2.7. TNG HP B IU KHIN M THCH NGHI TRN C S L THUYT THCH NGHI KINH IN....................................................... 58 2.7.1. t vn ..................................................................................... 58 2.7.2. M hnh ton hc ca b iu khin m ...................................... 60 2.7.3. Xy dng c cu thch nghi cho b iu khin m ...................... 66 2.7.4. Mt s ng dng iu khin cc i tng cng nghip.............. 70 Chng 3: TNG QUAN V MNG NRON............................................ 75 3.1. NRON SINH HC ........................................................................... 75 3.1.1. Chc nng, t chc v hot ng ca b no con ngi .............. 75 3.1.2. Mng nron sinh hc .................................................................... 76 3.2. MNG NRON NHN TO ............................................................ 77 3.2.1. Khi nim...................................................................................... 77 3.2.2. M hnh nron .............................................................................. 80 3.3. CU TRC MNG ............................................................................ 83 3.3.1. Mng mt lp................................................................................ 83 3.3.2. Mng nhiu lp............................................................................. 84 3.4. CU TRC D LIU VO MNG ................................................. 87 3.4.1. M t vct vo i vi mng tnh............................................... 88 3.4.2. M t vct vo lin tip trong mng ng .................................. 89 3.5. HUN LUYN MNG ...................................................................... 92 3.5.1. Hun luyn gia tng ...................................................................... 92 3.5.2 Hun luyn mng theo gi............................................................. 94 Chng 4: MNG PERCEPTRONS ............................................................. 98 4.1. M U ............................................................................................. 98
4.1.1. M hnh nron perceptron ............................................................ 98 4.1.2. Kin trc mng perceptron ......................................................... 100 4.2. THIT LP V M PHNG PERCEPTRON TRONG MATLAB100 4.2.1 Thit lp ....................................................................................... 100 4.2.2. M phng (sim) .......................................................................... 102 4.2.3. Khi to ...................................................................................... 103 4.3. CC LUT HC .............................................................................. 104 4.3.1. Khi nim.................................................................................... 104 4.3.2. Lut hc Perceptron (learnp) ...................................................... 105 4.3.3. Hun luyn mng (train)............................................................. 107 4.4. CC HN CH CA PERCEPTRON ............................................ 111 4.5. S DNG GIAO DIN HA KHO ST MNG NRON .................................................................................................................. 112 4.5.1. Gii thiu v GUI ....................................................................... 112 4.5.2. Thit lp mng Perceptron (nntool)............................................ 113 4.5.3. Hun luyn mng........................................................................ 115 4.5.4. Xut kt qu Perceptron ra vng lm vic.................................. 116 4.5.5. Xo ca s d liu mng (Network/Data Window) ................... 117 4.5.6 Nhp t dng lnh ....................................................................... 117 4.5.7. Ct bin vo file v np li n .................................................... 118 Chng 5: MNG TUYN TNH ............................................................... 119 5.1. M U ........................................................................................... 119 5.1.1. Khi nim.................................................................................... 119 5.1.2. M hnh nron ............................................................................ 119 5.2. CU TRC MNG .......................................................................... 120 5.2.1. Cu trc....................................................................................... 120 5.2.2. Khi to nron tuyn tnh (Newlin) ........................................... 121 5.3. THUT TON CC TIU TRUNG BNH BNH PHNG SAI LCH........................................................................................................ 122 5.4. THIT K H TUYN TNH .......................................................... 123 5.5. MNG TUYN TNH C TR ....................................................... 123 5.5.1 Mt tr.......................................................................................... 123 5.5.2. Thut ton LMS (learnwh) ......................................................... 123 5.5.3. S phn loi tuyn tnh (train) .................................................... 125 5.6. MT S HN CH CA MNG TUYN TNH.......................... 126
Chng 6: H M - NRON (FUZZY-NEURAL).................................... 128 6.1 S KT HP GIA LOGIC M V MNG NRON .................. 128 6.1.1 Khi nim..................................................................................... 128 6.1.2. Kt hp iu khin m v mng nron ...................................... 129 6.2. NRON M...................................................................................... 133 6.3. HUN LUYN MNG NRON-M ............................................. 135 6.4. S DNG CNG C ANFIS TRONG MATLAB THIT K H M - NRON (ANFIS and the ANFIS Editor GUI)............................... 139 6.4.1. Khi nim.................................................................................... 139 6.4.2. M hnh hc v suy din m thng qua ANFIS (Model Learning and Inferencc Through ANFIS)............................................................ 140 6.4.3. Xc nhn d liu hun luyn (Familiarity Brecds Validation)... 141 6.5. S DNG B SON THO ANFIS GUI ...................................... 143 6.5.1. Cc chc nng ca ANFIS GUI ................................................. 143 6.5.2. Khun dng d liu v b son tho ANFIS GUI: kim tra v hun luyn (Data Formalities and the ANFIS Editor GUI: Checking and Training) ............................................................................................... 144 6.5.3. Mt s v d ................................................................................ 145 6.6. SON THO ANFIS T DNG LNH ......................................... 153 6.7. THNG TIN THM V ANFIS V B SON THO ANFIS EDITOR GUI............................................................................................ 157 6.7.1. D liu hun luyn (Training Data)............................................ 158 6.7.2. Cu trc u vo FIS (Input FIS Structure)................................ 158 6.7.3. Cc ty chn hun luyn (Training Options) ............................. 159 6.7.4 Tu chn hin th Display Options.............................................. 159 6.7.5. Phng php hun luyn (Method) ............................................ 160 6.7.6. Cu trc u ra FIS cho d liu hun 1uyn............................... 160 6.7.7. Sai s hun luyn ........................................................................ 160 6.7.8. Bc tnh (Step-size) .................................................................. 160 6.7.9. D liu kim tra (Checking Data)............................................... 161 6.7.10. Cu trc u ra FIS cho d liu kim tra (Output FIS Structure for Checking Data) ............................................................................... 162 6.7.11. Sai s kim tra (Checking Error) .............................................. 162 TI LIU THAM KHO ............................................................................ 163
LI NI UNgy nay, cc h thng m v mng n ron ngy cng c ng dng rng ri trong nhiu lnh vc ca i sng x hi. c bit, trong lnh vc iu khin v t ng ho, h m v mng n ron ngy cng chim u th v mang li nhiu li ch to ln. Vi u im c bn l c th x l vi chnh xc cao nhng thng tin "khng chnh xc" h m v mng nron l c s ca h "iu khin thng minh" v "tr tu nhn to". p ng nhu cu tm hiu v ng dng lgic m v mng n ron ca ng o bn c, c s c v v ng vin ca BGH trng i hc K thut Cng nghip, chng ti mnh dn vit cun sch "H m v nron trong k thut iu khin". Cun sch c vit da trn cc bi ging v h thng iu khin thng minh cho hc vin cao hc ngnh T ng ho trng i hc K thut Cng nghip. Cun sch khng phn tch qu su nhng vn l thuyt phc tp m ch cung cp cho bn c nhng ni dung rt c bn v H m, mng n ron nhn to v h M-nron. Mc tiu cao hn l gip bn c bit cch khai thc nhng cng c sn c ca phn mm MATLAB phn tch, thit k cc b iu khin m, nron nhm iu khin cc i tng trong cng nghip. Mi phn u c cc v d c th hng dn thit k. Cun sch l ti liu tham kho cho hc vin cao hc, sinh vin ngnh iu khin, cc k s ngnh in, Cng ngh thng tin v cc nghin cu sinh quan tm n lnh vc iu khin m v mng nron. Trong qu trnh bin son, khng trnh khi cn nhiu sai st. Chng ti mong nhn c s ng gp kin cc ca ng nghip v bn c gn, xa. Xin chn thnh cm n! Thi Nguyn, ngy 01 thng 12 nm 2006 Cc tc gi
Chng 1
LGIC M1.1. TNG QUAN V LGIC M 1.1.1. Qu trnh pht trin ca 1gic m T nm 1965 ra i mt l thuyt mi l l thuyt tp m (Fuzzy set theory) o gio s Lofti A. Zadeh trng i hc Califonia - M a ra. T khi l thuyt ra i n c pht trin mnh m qua cc cng trnh khoa hc ca cc nh khoa hc nh: Nm 1972 GS Terano v Asai thit lp ra c s nghin cu h thng iu khin m Nht, nm 1980 hng Smith Co. bt u nghin cu iu khin m cho l hi... Nhng nm u thp k 90 cho n nay h thng iu khin m v mng nron (Fuzzy system and neural network) c cc nh khoa hc, cc k s v sinh vin trong mi lnh vc khoa hc k thut c bit quan tm v ng dng trong sn xut v i sng. Tp m v lgic m da trn cc thng tin "khng y , v i tng iu khin y v i tng mt cch chnh xc. Cc cng ty ca Nht bt u dng lgic m vo k thut iu khin t nm 1980. Nhng do cc phn cng chun tnh ton theo gii thut 1gic m rt km nn hu ht cc ng dng u dng cc phn cng chuyn v lgic m. Mt trong nhng ng dng dng lgic m u tin ti y l nh my x l nc ca Fuji Electric vo nm 1983, h thng xe in ngm ca Hitachi vo nm 1987. 1.1.2. C s ton hc ca 1gic m Lgic m v xc xut thng k u ni v s khng chn chn. Tuy nhin mi lnh vc nh ngha mt khi nim khc nhau v i tng. Trong xc sut thng k s khng chc chn lin quan n s xut hin ca mt s kin chc chn" no . V d: Xc sut vin n trng ch l 0, Bn thn ca s kin "trng ch" c nh ngha r rng, s khng 1
chc chn y l c trng ch hay khng v c nh lng bi mc xc sut (trong trng hp ny l 0,8). Loi pht biu ny c th c x l v kt hp vi cc pht biu khc bng phng php thng k, nh l xc sut c iu kin chng hn. S khng chc chn trong ng ngha, lin quan n ngn ng ca con ngi, l s khng chnh xc trong cc t ng m con ngi dng c lng vn v rt ra kt lun. V d nh cc t m t nhit "nng", "lnh", "m"s khng c mt gi tr chnh xc no gn cho cc t ny, cc khi nim ny cng khc nhau i vi nhng ngi khc nhau (l lnh i vi ngi ny nhng khng lnh i vi ngi khc). Mc d cc khi nim khng c nh ngha chnh xc nhng con ngi vn c th s dng chng cho cc c lng v quyt nh phc tp. Bng s tru tng v c suy ngh, con ngi c th gii quyt cu ni mang ng cnh phc tp m rt kh c th m hnh bi ton hc chnh xc. S khng chc chn theo ng vng: Nh ni trn, mc d dng nhng pht biu khng mang tnh nh lng nhng con ngi vn c th thnh cng trong cc c lng phc tp. Trong nhiu trng hp, con ngi dng s khng chc chn ny tng thm linh hot. Nh trong hu ht x hi, h thng lut php bao gm mt s lut, mi lut m t mt tnh hung. V d mt lut quy nh ti trm xe phi b t 2 nm, mt lut khc li gim nh trch nhim. V trong mt phin ta, chnh n phi quyt nh s ngy pht t ca tn trm da trn mc ru trong ngi, trc y c tin n hay tin s khng,... t kt hp li a ra mt quyt nh cng bng. 1.1.3. Lgic m l 1gic ca con ngi Trong thc t, ta khng nh ngha mt lut cho mt trng hp m nh ngha mt s lut cho cc trng hp nht nh. Khi nhng lut ny l nhng im ri rc ca mt tp cc trng hp lin tc v con ngi xp x chng. Gp mt tnh hung c th, con ngi s kt hp nhng lut m t cc tnh hung tng t. S xp x ny da trn s linh hot ca cc t ng cu to nn lut, cng nh s tru tng v s suy ngh da trn s linh hot trong lgic ca con ngi. thc thi lgic ca con ngi trong k thut cn phi c mt m hnh ton hc ca n. T lgic m ra i nh mt m hnh ton hc cho php 2
m t cc qu trnh quyt nh v c lng ca con ngi theo dng gii thut. D nhin cng c gii hn, l lgic m khng th bt chc tr tng tng v kh nng sng to ca con ngi. Tuy nhin, lgic m cho php ta rt ra kt lun khi gp nhng tnh hung khng c m t trong lut nhng c s tng ng. V vy, nu ta m t nhng mong mun ca mnh i vi h thng trong nhng trng hp c th vo lut th lgic m s to ra gii php da trn tt c nhng mong mun . 1.2. KHI NIM V TP M 1.2.1. Tp kinh in Khi nim tp hp c hnh thnh trn nn tng lgic v c nh ngha nh l s sp xp chung cc i tng c cng tnh cht, c gi l phn t ca tp hp . Cho mt tp hp A, mt phn t x thuc A c k hiu: x A. Thng thng ta dng hai cch biu din tp hp kinh in, l: Lit k cc phn t ca tp hp, v d tp A1 = {xe p, xe my, xe ca, xe ti}; - Biu din tp hp thng qua tnh cht tng qut ca cc phn t, v d: tp cc s thc (R), Tp cc s t nhin (N). biu din mt tp hp A trn tp nn X, ta dng hm thuc A(x), vi: 1 khi x A A(x) ch nhn mt trong 2 gi tr "1" 0 khi x A hoc "0" k hiu = {x X| x tho mn mt s tnh cht no }. Ta ni: Tp A c nh ngha trn tp nn X. A(x) = Hnh 1.1 m t hm ph thuc A(x) ca tp cc s thc t -5 n 5. A = {xR|5 x 5} 1.2.2. nh ngha tp m Trong khi nim tp hp kinh in hm ph thuc A(x) ca tp A, ch c mt trong hai gi tr l "1" nu xA hoc "0" nu xA. 3
Cch biu din hm ph thuc nh trn s khng ph hp vi nhng tp c m t "m" nh tp B gm cc s thc gn bng 5: B = {x R| x 5}. Khi ta khng th khng nh chc chn s 4 c thuc B hay khng? m ch c th ni n thuc B gao nhiu phn trm. tr li c cu hi ny, ta phi coi hm ph thuc B(x) c gi tr trong khong t 0 n 1 tc l: 0 B(x) 1. T phn tch trn ta c nh ngha: Tp m B xc nh trn tp kinh in M l mt tp m mt phn t ca n c biu din bi mt cp gi tr (x,B(x)). Trong x M v B(x) l nh x. nh x B(x) c gi l hm lin thuc ca tp m B. Tp kinh in M c gi l c s ca tp m B. 1.2.3. Cc thng s c trng cho tp m Cc thng s c trng cho tp m l cao, min xc nh v min tin cy (hnh 1.3) + cao ca mt tp m B (nh ngha trn c s M) l gi tr ln nht trong cc gi tr ca hm lin thuc: H = SUP B (x)xM
Mt tp m c t nht mt phn t c ph thuc bng 1 c gi l tp m chnh tc (H = 1). Ngc li, mt tp m B vi H < 1 gi l tp m khng chnh tc.+ Min xc nh ca tp m B (nh ngha trn c s M) c k hiu bi S l tp con ca M c gi tr hm lin thuc khc khng: 4
S = {x M| B(x) > 0}.+ Min tin cy ca tp m B (nh ngha trn c s M) c k hiu bi T, l tp con ca M c gi tr hm lin thuc bng 1:
T= {x M| B(X) = 1}.1.2.4. Cc dng hm lin thuc ca tp m
C rt nhiu cch khc nhau biu din hm lin thuc ca tp m. Di y l mt s dng hm lin thuc thng dng: + Hm lin thuc hnh tam gic (hnh 1.4a); + Hm lin thuc hnh thang (hnh 1.4b); + Hm lin thuc dng Gauss (hnh l.4c); + Hm lin thuc dng Sign (hnh 1.4d); + Hm Sigmoidal (hnh 1.4e); + Hm hnh chung (hnh 1.4f).
Hnh 1.4. Cc dng hm lin thuc ca tp m1.3. CC PHP TON TRN TP M
Trn tp m c 3 php ton c bn l php hp, php giao, v php b.1.3.1. Php hp hai tp m
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a/ Hp ca hai lp m c cng c s
Hnh 1.5. Hp ca hai tp m c cng c s theo quy tc Max (a), theo Lukasiewwiez (b) Hp ca hai tp m A v B c cng c s M l mt tp m cng xc nh trn c s M vi hm lin thuc c xc nh theo mt trong cc cng thc sau: 1. A B(x) = Max{A(x), B(x)};2. A B(x) = min {1, A(x) + B(x)} php hp Lukasiewiez);
3. A B(x) = 4. A B(x) =
max{A(x), B(x)} khi min{A(x), B(x)}=0 1 khi min{A(x), B(x)} 0 A(x) + B(x) (Tng Einstein) 1 + A(x) + B(x)
5. A B(x) = A(x) = B(x) - A(x)A(x) (tng trc tip). Ch : C nhiu cng thc khc nhau c dng tnh hm lin thuc A B(x) ca hai tp m. Song trong k thut iu khin m ta ch yu dng 2 cng thc hp, l ly Max v php hp Lukasiewiez. b/ Hp hai tp m khc c s thc hin php hp 2 tp m khc c s, v nguyn tc ta phi a chng v cng mt c s. Xt tp m A vi hm lin thuc A(x) c nh ngha trn c s M v B vi hm lin thuc B(x) c nh ngha trn c s N, hp ca 2 tp m A v B l mt tp m xc nh trn c s MxN vi hm lin thuc: A B(x, y) = Max {A(x, y), B(x, y)} Vi A(x, y) = A(x) vi mi y N v B(x, y) = B(y) vi mi x M.1.3.2. Php giao ca hai tp m
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a/ Giao hai tp m cng c s
Hnh 1.6. Giao ca hai tp m c cng c s theo quy tc Min (a) v theo tch i s (b) Giao ca hai tp m A v B c cng c s M l mt tp m cng xc nh trn c s M vi hm lin thuc A B(x) c tnh: 1, A B(x) = Min {A(x), B(x)}; 2. A B(x) = A(x).B(x) (tch i s);
cng ging nh trong php hp, trong k thut iu khin ch yu ta s dng cng thc 1 v cng thc 2 thc hin php giao 2 tp m. b/ Giao hai tp m khc c s thc hin php giao 2 tp m khc c s, ta cn phi a v cng c s. Khi , giao ca tp m A c hm lin thuc A(x) nh ngha trn c s M vi tp m B c hm lin thuc B(x) nh ngha trn c s N l mt tp m xc nh trn c s M x N c hm lin thuc c tnh: A B(x, y) = MIN{A(x, y), B(x, y)} 7
Trong : A(x, y) = A(x) vi mi y N v B(x, y) = B(x) vi mi x M.1.3.3. Php b ca mt tp m
B ca tp m A c c s M v hm lin thuc A(x) l mt tp m AC xc nh trn cng c s M vi hm lin thuc: A(x) = 1- A(x)1.4. BIN NGN NG V GI TR CA BIN NGN NG
Thc t hng ngy chng ta lun dng cc t ng, li ni m t cc bin. V d khi ta ni: "in p cao qu", "xe chy nhanh qu",... nh vy bin "in p", bin "Tc xe",... nhn cc gi tr t "nhanh" n "chm", t "cao" n "thp". dng tng minh, cc bin ny nhn cc gi tr c th (r) nh in p bng 200 V, 250 V...; tc xe bng 60 km/h, 90 km/h... Khi cc bin nhn cc gi tr khng r rng nh "cao", "rt cao" "nhanh", "hi nhanh"... ta khng th dng cc gi tr r m t c m phi s dng mt s khi nim mi m t gi l bin ngn ng. M bin c th gn bi cc t trong ngn ng t nhin lm gi tr ca n gi l bin ngn ng. Mt bin ngn ng thng bao gm 4 thng s: X, T, U, M. Vi: + X: Tn ca bin ngn ng; + T: Tp ca cc gi tr ngn ng; + U: Khng gian nn m trn bin ngn ng X nhn cc gi tr r; + M: Ch ra s phn b ca T trn U. V d: Bin ngn ng "Tc xe" c tp cc gi tr ngn ng l rt chm, chm, trung bnh, nhanh, rt nhanh, khng gian nn ca bin l tp cc s thc dng. Vy bin tc xe c 2 min gi tr khc nhau: - Min cc gi tr ngn ng N = [rt chm, chm, trung bnh, nhanh, rt nhanh]. - Min cc gi tr vt l V = {x R (x0)}. Mi gi tr ngn ng (mi phn t ca Ni c tp nn l min gi tr vt l V. T mt gi tr vt l ca bin ngn ng ta c c mt vct gm cc ph thuc ca x: 8
X T = [rt chm chm trung bnh nhanh rt nhanh] nh x trn c gi l qu trnh fuzzy ho gi tr r x. V d: ng vi tc 50 km/h ta c
1.5. LUT HP THNH M 1.5.1. Mnh hp thnh
Xt hai bin ngn ng v ; Bin nhn gi tr (m) A c hm lin thuc A(x) v nhn gi tr (m) B c hm lin thuc B(x) th hai biu thc: = A; = B c gi l hai mnh . Lut iu khin: nu = A th = B c gi l mnh hp thnh.Trong = A gi l mnh iu kin v = B gi l mnh kt lun. Mt mnh hp thnh c th c nhiu mnh iu kin v nhiu mnh kt lun, cc mnh lin kt vi nhau bng ton t "v". Da vo s mnh iu kin v s mnh kt lun trong mt mnh hp thnh m ta phn chng thnh cc cu trc khc nhau: - Cu trc SISO (mt vo, mt ra): Ch c mt mnh iu kin v mt mnh kt lun. V d: nu = A th = B. - Cu trc MISO (Nhiu vo, mt ra): C t 2 mnh iu kin tr ln v mt mnh kt lun. V d: nu 1 = A1 v 2 = A2 th = B. - Cu trc MIMO (Nhiu vo, nhiu ra): C t nht 2 mnh iu kin v 2 mnh kt lun. V d: nu 1 = A1 v 2 = A2 th 1 = B1 v 2 = B21.5.2. M t mnh hp thnh
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Xt mnh hp thnh: nu = A th - B; T mt gi tr x0 c ph thuc A(x0) i vi tp m A ca mnh iu kin, ta xc nh c tho mn mnh kt lun. Biu din tho mn ca mnh kt lun nh mt tp m B cng c s vi B th mnh hp thnh chnh l nh x: A(x0) B(y). nh x ny ch ra rng mnh hp thnh l mt tp m mi phn t l mt gi tr (A(x0), B(y)) tc l mi phn t l mt tp m. M t mnh hp thnh tc l m t nh x trn. nh x (A(x0), B(y)) c gi l hm lin thuc ca lut hp thnh. xy dng B(y) c rt nhiu kin khc nhau. Trong k thut iu khin ta thng s dng nguyn tc ca Mamdani " ph thuc ca kt lun khng c ln hn ph thuc ca iu kin"? T nguyn tc ta c hai cng thc xc nh hm lin thuc cho mnh hp thnh A => B: 1. cng thc MIN: A=>B(x, y) = MIN{A(x), B(y)} 2. cng thc PROD: A=>B(x, y) = A(x)B(xy1.5.3. Lut hp thnh m
Lut hp thnh l tn chung gi m hnh R biu din (mt hay nhiu) hm lin thuc A=>B(x, y) cho (mt hay nhiu) mnh hp thnh A B. Mt lut hp thnh ch c 1 mnh hp thnh gi l lut hp thnh n, c t 2 mnh hp thnh tr ln gi l lut hp thnh phc. Xt lut hp thnh R gm 3 mnh hp thnh: R1: Nu x = A1 Th y = B1 hoc R2: Nu x = A2 Th y = B2 hoc R3: Nu x = A3 Th y = B3 hoc
Hnh 1.9. M t hm lin thuc ca lut hp thnh 10
Vi mi gi tr r x0 ca bin ngn ng u vo, ta c 3 tp m ng vi 3 mnh hp thnh R1, R2, R3 ca lut hp thnh R. Gi hm lin thuc ca cc tp m u ra l: B' ( y ) ; B' ( y ) ; B' ( y ) th gi tr ca lut hp thnh R1 2 3
ng vi x0 l tp m B thu c qua php hp 3 tp m: B = B1 B2 B3. Tu theo cch thu nhn cc hm lin thuc B' ( y ) ; B' ( y ) ; B' ( y ) v1 2 3
phng php thc hin php hp nhn tp m B m ta c tn gi cc lut hp thnh khc nhau: - Lut hp thnh MAX-MIN nu B' ( y ) ; B' ( y ) ; B' ( y ) thu c qua1 2 3
php ly Min cn php hp thc hin theo lut Max; - Lut hp thnh MAX-PROD nu B' ( y ) ; B' ( y ) ; B' ( y ) thu c qua1 2 3
php PROD cn php hp thc hin theo lut Max; - Lut hp thnh SUM-MIN nu B' ( y ) ; B' ( y ) ; B' ( y ) thu c qua1 2 3
php ly Min cn php hp thc hin theo lut SUM; - Lut hp thnh SUM - PROD nu B' ( y ) ; B' ( y ) ; B' ( y ) thu c1 2 3
qua php ly PROD cn php hp thc hin theo Lukasiewicz. Vy, xc nh hm lin thuc B(y) ca gi tr u ra B ca lut hp thnh c n mnh hp thnh R1, R2, ta thc hin theo cc bc sau: + Xc nh tho mn hj. + Tnh B' ( y ) ; B' ( y ) ; B' ( y ) theo quy tc min hoc Prod1 2 3
B ( y ) = Min{uA(x0), B ( y ) }= Min {hj, B ( y ) }' ji j
hoc B ( y ) = A(x0). B ( y ) = hj. B ( y ) .j j j
+ xc nh B(y) bng cch thc hin php hp cc B' ( y )j
1.5.4. Cc cu trc c bn ca lut hp thnh
Ta s kho st hai cu trc c bn ca lut hp thnh, l cu trc SISO v cu trc MISO. 11
+ Cu trc SISO l cu trc trong lut hp thnh c cc mnh iu kin v mnh kt lun l cc mnh n.
V d:
R1: nu = Al th = B1 hoc
R2: nu = A2 th = B2.+ Cu trc MISO l cu trc trong lut hp thnh c cc mnh iu kin l mnh phc v mnh kt lun l mnh n.
V d:
R1: nu 1 = A1 v 2 = B1 th = C1 hoc
R2: nu 1 = A2 v 2 = B2 th = C2.1.5.5. Lut hp thnh n c cu trc SISO
a) Lut hp thnh MIN Lut hp thnh MIN l tn gi m hnh (ma trn) R ca mnh hp thnh A B khi hm lin thuc A=>B(x, y) ca n c xy dng theo quy tc MIN. Xt lut hp thnh ch c 1 mnh : Nu = A th = B xy dng R, trc tin hai hm lin thuc A(x) v B(y) c ri rc ho vi tn s ri rc nh khng b mt thng tin. V d: A(x), B(y) c ri rc ho ti cc im: x {10, 20, 30, 40, 50} y {0.5, 0.6, 0.7, 0.8, 0.9}. Vi cc im ri rc ny th theo A=>B(20; 0.7) = R(20; 0.7)=MIN{A(20),b(0.7)}=MIN{0.5; 1}= 0.5 A=>B(30; 0.7) = R(30; 0.7)=MIN{A(30),b(0.7)}= MIN{1; 1}= 1 .
Hnh 1.10. Ri rc ho cc hm lin thuc 12
Nhm tt c cc gi tr A=>B(x, y) = R(x,y) gm 5 x 5= 25 gi tr, thnh ma trn R (c gi l ma trn hp thnh MIN) gm 5 hng 5 ct.
Khi tn hiu u vo l mt gi tr r x0 = 20, tn hiu u ra B c hm lin thuc: B(y) = R(20, y) = {0; 0.5; 0.5; 0.5; 0}. thun tin cho vic xc nh hm lin thuc ca tn hiu ra di dng nhn ma trn, ta nh ngha mt ma trn T = {a1 a2} ma trn ny ch c mt phn t bng 1 cn cc phn t khc u bng 0. V d vi tp 5 phn t cho tn hiu u vo x {10; 20; 30; 40; 50} th ng vi x0 = 20 (phn t th hai) ta c: a = (0 1 0 0 0) V khi B(y) = R(x0, y) = aT. R = {0 0.5 0.5 0.5 0}. Tng qut cho mt gi tr r x0 bt k x0 X = {10 20 30 40 50} ti u vo vct chuyn v c dng: aT = (a1, a2, a3, a4, a5) trong ch c mt phn t a; duy nht c ch s i l ch s ca x0 trong X c gi tr bng 1, cc phn t cn li u bng 0. Hm lin thuc mB'(y) di dng ri rc c xc nh:
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Ch : Trong biu thc (1.1) tnh B'(y) ta cn ci t thut ton nhn ma trn ca i s tuyn tnh, do tc x l chm. khc phc nhc im ny, php nhn ma trn (1.1) c thay bi lut MAX-MIN ca Zadeh vi MAX (php ly cc i) thay vo v tr php cng v MIN (php ly cc tiu) thay vo v tr php nhn. Khi :
lK = max min {ai rki}1i 5
Kt qu hai php tnh (1.1) v (1.2) vi u vo l mt gi tr r hon ton ging nhau. Cng t l do trn m lut hp thnh MIN cn c tn gi l lut hp thnh MAX-MIN. b/ Lut hp thnh PROD Tng t nh lm vi lut hp thnh MIN, ma trn R ca lut hp thnh PROD c xy dng gm cc hng l m gi tr ri rc ca u ra B'(y1), B'(y2), B'(ym) cho n gi tr r u vo xn, xn,., xn Nh Vy ma trn R s c n hng v m ct. Xt v d trn cho 5 gi tr u vo: {x1, x2, x3, x4, x5} = {10 20 30 40 50} th vi tng gi tr xi, 5 gi tr ca hm lin thuc u ra tng ng B'(0.5), B'(0.6), B'(0.7), B'(0.8), B'(0.9) c lit k trong ma trn R c gi l ma trn hp thnh PROD. T ma trn R trn, hm lin thuc B'(y) ca gi tr u ra khi u vo l gi tr r x4 cng c xc nh bng cng thc: aT = (0, 0, 0, 1, 0) B'(y) = R(x4, y) = aT .R = {0, 0.25, 0.5, 0.25, 0}. rt ngn thi gian tnh v cng m rng cng thc trn cho trng hp u vo l gi tr m, php nhn ma trn T.R cng c thay bng lut MAX- PROD ca Zadeh nh lm cho lut hp thnh MIN. Trong php nhn c thc hin bnh thng cn php ly cc i thay vo v tr ca php cng.14
R
0.50 0 0 0 0
0.60 0.25 0.5 0.25 0
0.70 0.5 1 0.5 0
0.80 0.25 0.5 0.25 0
0.90 0 0 0 0
i=1 i=2 i=3 i=4 i=5
10 20 30 40 50
c) Thut ton xy dng R
T cc phn tch trn, ta rt ra thut ton xy dng R cho lut hp thnh n c cu trc SISO (Nu = A Th = B) nh sau: 1- Ri rc ho A(x) ti n im x1, x2,,xn ti m im y1, y2,,yn (n c th khc m) 2- Xy dng ma trn R gm n hng v m ct:
3- Xc nh hm lin thuc B'(y) ca u ra ng vi gi tr r du vo xk theo biu thc:
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trong : lK = max min {ai rki}, k = 1,2,.., m nu s dng cng thc1i n
MAX-MIN v lK = max prod {ai rki}, k = 1,2,.., m nu s dng cng thc1i n
MAX-PROD. 4- Xc nh B'(y) theo cng thc: B'(y) = ( l1, l2,,lm).Ch :
Trong trng hp u vo l gi tr m A' vi hm lin thuc A'(y) th hm lin thuc B'(y) ca gi tr u ra B': B'(y) = ( l1, l2,,lm) cng c tnh theo cng thc (2.4) v lk = max min {ai rki}, k = 1, 2,, m1i n
trong a l vct gm cc gi tr ri rc ca hm lin thuc A'(x) ca A' ti cc im: x X = {x1, x2,,xn} tc l aT = (A'(x1), A'(x2),, A'(xn)). Gi thit c n im ri rc x1, x2,,xn ca c s A v m im ri rc y1, y2,,ym ca c s B ta c hai vct: AT={A(x1), A(x2),, A(xn)} v AT={B(y1), B(y2),, B(xm)} theo Zadeh ta c th xc inh ngay c R thng qua tch dyadic, tc l tch ca mt vct vi mt vct chuyn v: R = A.BT Trong nu quy tc p dng l MAX - MIN th php nhn phi c thay bng php tnh ly cc tiu (min), vi quy tc MAX - PROD th thc hin php nhn nh bnh thng.V d: Lut iu khin: Nu = A Th = B. Hy xy dng ma trn R ca lut AB(x, y). 16
Vi 5 im ri rc ca X (c s ca A) ta c: {x1, x2, x3, x4, x5} = {10, 20, 30, 40, 50} tng ng AT = {0; 0.5; 1; 0.5; 0} V Vi 5 im ri rc ca Y (c s ca B) {y1, y2, y3,yx4, y5} = {0.5, 0.6, 0.7, 0.8, 0.9} Tng ng BT = {0; 0.5; l; 0.5; 0}. Nu s dng quy tc MAX-MIN (php nhn c thay bng min) ma trn hp thnh R s nh sau:
Nu s dng quy tc MAX-PROD (php nhn thc hin bnh thng) ta c ma trn hp thnh R l:
1.5.6. Lut hp thnh n c cu trc MISO Xt mt mnh hp thnh vi d mnh iu kin:Nu 1 = A1 v 2 = A2 v v d = Ad th = B
Bao gm d bin ngn ng u vo 1, 2,, d v mt bin u ra . Vic m hnh ho mnh trn cng c thc hin tng t nh vic m hnh ho mnh hp thnh c mt iu kin, trong lin kt v gia cc mnh (hay gi tr m) c thc hin bng php giao cc tp m A1, 17
A2,,An Vi nhau theo cng thc: A1 A2(x) = min {A1(x), A2(x)}. Kt qu ca php giao s l tho mn H ca lut (hnh 1-12). Cc bc xy dng lut hp thnh R nh sau: 1- Ri rc ho min xc nh hm lin thuc A1(x1), A2(x2),, Ad(xd), B(y) ca cc mnh iu kin v mnh kt lun. 2- Xc nh tho mn H cho tng vct cc gi tr r u vo l vct t hp d im mu thuc min xc nh ca cc hm lin thuc A(x), (i = 1, 2,.., d). Chng hn vi mt vct cc gi tr r u vo:
x H=
=
trong ci (i= 1,2,...,d) l mt trong cc im mu trong min xc nh ca Ai(x) th: MIN{A1(c1), A2(c2),, Ad(cd)}
Hnh 1.13. Xy dng R cho lut hp thnh hai mnh iu kin
3- Lp R gm cc hm lin thuc gi tr m u ra cho tng vct cc gi tr u vo theo nguyn tc: B(y)= MIN {H, B(y)} Nu s dng quy tc MAX-MIN B(y)= H, B(y) Nu s dng quy tc MAX-PROD.
Ch : i vi lut hp thnh R c d mnh iu kin khng th biu din di dng ma trn c na m thnh mt li trong khng gian d + 1 chiu.
Tht vy, xt mt mnh hp thnh vi hai mnh iu kin: 18
Nu = A v = B th = C
Lut hp thnh R ca n c dng nh hnh 2.12: R: A^B CCc bc xy dng R nh sau:
1. Ri rc ho cc hm lin thuc: - Hm lin thuc A(x) c ri rc ho ti 5 im: x {1; 2; 3; 4; 5}. - Hm lin thuc B(y) c ri rc ho tt 5 im: y {3; 4; 5; 6; 7}. - Hm lin thuc C(z) c ri rc ho ti 5 im: z {5; 6; 7; 8; 9}. 2. Lp R gm cc hm lin thuc cho tng vect gi tr u vo v ng vi tng cp im u vo l mt hm lin thuc C'(z) ca bin m u ra C (hnh 1.14).
1.5.7. Lut ca nhiu mnh hp thnh
Trong thc t hu nh khng b iu khin m no ch lm vic vi mt mnh hp thnh m thng thng vi nhiu mnh hp thnh? hay cn gi l mt tp cc lut iu khin Rk. sau y ta s trinh by cch lin kt cc lut iu khin ring r Rk li vi nhau trong mt b iu khin chung v qua m nu bt c ngha ca k hiu "MAX" s dng trong tn gi lut hp thnh nh MAX- MIN hay MAX-PROD.a) Lut hp thnh ca hai mnh hp thnh
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Xt lut iu khin gm hai mnh hp thnh:R1: Nu = A1 th = B1 hoc R2: Nu = A2 th = B2
Hm lin thuc ca cc tp m c m t trong hnh 2.15. K hiu R l lut hp thnh chung ca b iu khin, ta c: R = R1 R2 K hiu hm lin thuc ca R1 l R1(x, y) v ca R2 l R2(x, y), th theo cng thc A B(x) = max {A(x), B(x)}. Hm lin thuc ca R s c xc nh: R(x, y) = max {R1(x, y), R2(x, y)}. Vi mt gi tr r x0 ti u vo, ta c tho mn ca cc mnh iu kin nh sau: i vi lut iu khin R1: - tho mn: H1 = A1(x0) - Gi tr m u ra B1: B1(y) = min{H1, B1(y)}(hnh 2.l5a). i vi lut iu khin R2: - tho mn: H2 = A2(x0) - Gi tr m u ra B2: B2(y) = min{H2, B2(y)}(hnh 2.l5b). T y ta c: R(x0, y) = MAX{B1(y), B2(y)}
Hnh 2.15. hm lin thuc ca lut iu khin theo quy tc MAX-MIN
a) Xc nh hm lin thuc u ra ca lut iu khin th nht. 20
b) Xc nh hm lin thuc u ra ca lut iu khin th hai. c) Hm lin thuc u ra ca lut hp thnh. chnh l hm lin thuc ca gi tr m u ra B ca b iu khin gm hai lut iu khin R = R1 R2 khi u vo l mt gi tr r x0 (hnh 2.15c). xc nh lut hp thnh chung R, trc ht hai c s X v Y ca cc gi tr A1, A2 v B1, B2 c ri rc ho, gi s ti cc im: X = {x1, x2, x3,,xn} (n im mu) Y = {y1, y2, y3,,ym} (m im mu). Gi tr ca cc hm lin thuc A1(x), A2(x), B1(y), B2(y) sau khi ri rc ho l
T y suy ra:
v do lut hp thnh chung s l:
b) Lut hp thnh ca nhiu mnh hp thnh Xt lut iu khin R gm p mnh hp thnh: 21
trong cc gi tr m A1, A2,, Ap c cng c s X v B1, B2,, Bp c cng c s Y. Gi hm lin thuc ca Ak v Bk l Ak(x) v Bk(y) vi k = 1, 2,..., p. Thut ton trin khai: R = R1 R2 Rp c thc hin theo cc bc sau:Bc 1: Ri rc ho X ti n im (x1, x2, x3,, xn) V Y ti m im (y1, y2, y3,, yn) Bc 2: Xc nh cc vct Ak v Bk (k = 1, 2,..,p) ti cc im ri rc theo biu thc:
TAk = {Ak(x1), Ak(x2),, Ak(xn)} TBk = {Bk(y1), Bk(y2),, Bk(yn)}Bc 3: Xc nh m hnh (ma trn) Rk cho mnh th k
Rk = Ak.TBk = (rkij), i = 1, 2,, n v j = 1, 2,,m trong php (.) c thay bng php tnh ly cc tiu min khi s dng nguyn tc MAX-MIN v s dng php nhn bnh thng khi s dng nguyn tc MAX- PROD. Bc 4: Xc nh lut hp thnh R = Max (rkij) vi k = 1, 2,..., p}.1.5.7. Lut hp thnh SUM-MIN v SUM-PROD
phn trn, chng ta tm hiu phng php xy dng lut hp thnh chung R cho mt tp gm nhiu mnh hp thnh Rk c lin kt vi nhau bng php hp theo biu thc: A B(x) = max{A(x), B(x)}. Kiu lin kt ny khng c tnh thng k. V d khi a s cc mnh hp thnh Rk c cng mt gi tr u ra nhng khng phi l gi tr ln nht s khng c ti v b mt trong kt qu chung. khc phc nhc im ny php hp Lukasiewicz theo biu: 22
A B(x) = min{1, A(x) + B(x)} thay cho A B(x) = max{ A(x), B(x)} lin kt cc lut iu khin Rk li vi nhau thnh lut hp thnh chung R
trong php ly cc tiu min c thc hin gia s 1 v tng phn t ca ma trn tng. cng thc ny, R c xc nh bng cch cng cc Rk Ca cc mnh hp thnh nn lut hp thnh chung R theo lin kt Lukasiewicz s c tn gi l SUM-MIN hoc SUM-PROD.
Hnh 2.16. Hm lin thuc ca hp hai lut iu khin theo quy tc SUM-MIN
Thut ton trin khai R theo quy tc SUM-MIN hay SUM-PROD cng bao gm cc bc nh khi trin khai vi quy tc MAX-MIN hoc MAXPROD trnh by mc trn ch khc bc 4 ta s dng cng thc: R = n min 1, Rk k =1
Hnh 1.16 l mt v d v m hnh ho R gm hai mnh hp thnh theo quy tc SUM-MIN.1.6. GII M
T mt gi tr r x0 u vo, sau khi qua khi lut hp thnh ta c tp 23
m u ra B'. Vn t ra l cn phi xc nh gi tr r y0 t tp m u ra . Mun vy ta cn thc hin vic gii m. Gii m l qu trnh xc nh mt gi tr r y0 no c th chp nhn c t hm lin thuc B(y) ca gi tr m B (tp m B). C hai phng php gii m chnh l phng php cc i v phng php im trng tm.2.6.1. Phng php cc i
gii m theo phng php cc i, ta cn thc hin 2 bc:- Xc nh min cha gi tr r y0 (min G): l min m ti hm lin thuc B(y) t gi tr cc i ( cao H ca tp m B), tc l min:
G = {y Y| B(y) = H}- Xc nh y0 c th chp nhn c t G.
Hnh 1.17 l tp m u ra ca mt lut hp thnh gm 2 mnh hp thnh: R1: Nu = A1 Th = B1 R2: Nu = A2 Th = B2 Min cha gi tr r G l khong [y1, y2] ca min gi tr ca tp m u ra B2 ca lut iu khin: R2: Nu = A2 Th = B2 vi y1 l im cn tri ca G y1 = inf ( y ) v y2 l im cn phi ca G yG y1 = sup( y ) . Khi , lut R2 c gi l lut iu khin quyt nh. yG Vy lut iu khin quyt nh l lut Rk, k{1, 2,, p} m gi tr m u ra ca n c cao ln nht (Bng cao H ca B).
D xc nh y0 trong khong [y1, y2] ta c th p dng theo mt trong ba nguyn l: Nguyn l trung bnh; nguyn l cn tri v nguyn l cn phi.
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Hnh 1.17a.b.c. Cc nguyn l gii m theo phng php cc di
a) Nguyn l trung binh Gi tr r y1 s l trung bnh cng ca y1 v y2
b) Nguyn l cn tri Gi tr r y0 c ly bng cn tri y1 ca G
c) Nguyn l cn phi Gi tr r y0 c ly bng cn phi y2 ca G
Nhn xt:
+ Gi tr r y0 ly theo nguyn l trung bnh s khng ph thuc vo tho mn ca lut iu khin quyt nh nu tp m B' l tp u (hnh 1.17a), cn theo nguyn l cn tri v cn phi, gi tr r y0 Ph thuc tuyn tnh vo tho mn ca lut iu khin quyt nh (hnh 1.17b,c).
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Hnh 1.18. a) y0 vi cc nguyn tc chn khc nhau
b) Hm lin thuc B c min G khng lin thng+ Sai lch ca ba gi tr r, xc nh theo nguyn l trung bnh, cn tri hay cn phi s cng ln nu tho mn H ca lut iu khin cng nh (hnh 1.18a). + Khi min G l min khng lin thng s dng phng php cc i s khng chnh xc (hnh 2.18b). + i vi lut hp thnh MAXPROD, min G ch c mt im duy nht, do kt qu gii m theo c 3 nguyn l ging nhau (hnh 1.19).
1.6.2. Phng php im trng tm Gii m theo phng php im trng tm s cho ra kt qu y' l honh ca im trng tm min c bao bi trc honh v ng B(y) (hnh 1.20). Cng thc xc nh y0 theo phng php im trng tm nh sau:
Vi s l min xc nh ca tp m B'.a) Phng php im trng tm cho lut hp thnh SUM-MIN Gi s c q lut iu khin c trin khai. Khi mi gi tr m B ti u ra ca b iu khin s l tng ca 26
q gi tr m u ra ca tng lut hp thnh. K hiu gi tr m u ra ca lut iu khin th k l BK(y) vi k = 1,2,...,q. Vi quy tc SUM- MIN, hm lin thuc B(x) s l:
sau khi bin i, ta c:
b) Phng php cao S dng cng thc:
Cho c hai lut hp thnh MAX-MIN v SUM-MIN vi thm mt gi thit l mi tp m BK(y) c xp x bng mt cp gi tr (yk, Hk) duy nht (singleton), trong Hk l cao ca BK(y) v yk l mt im mu trong min gi tr ca BK(y)
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Hnh 1.21. So snh cc phng php gii mCh : Tu hnh dng hm lin thuc B m sai khc gia cc phng php gii m c khc nhau. Hnh 1.21 cho bit kt qu cc phng php gii m ng vi mt hm lin thuc B c th.
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Chng 2
IU KHIN M2.1. CU TRC CA B IU KHIN M 2.1.1. S khi b iu khin m
Hot ng ca mt b iu khin m ph thuc vo kinh nghim v phng php rt ra kt lun theo t duy ca con ngi sau c ci t vo my tnh trn c s logic m. Mt b iu khin m bao gm 3 khi c bn: Khi m ho, thit b hp thnh v khi gii m. Ngoi ra cn c khi giao din vo v giao din ra (hnh 2.1).
Hnh 2.1. Cc khi chc nng ca b iu khin m- Khi m ho c chc nng chuyn mi gi tri r ca bin ngn ng u vo thnh vct c s phn t bng s tp m u vo. -Thit b hp thnh m bn cht ca n s trin khai lut hp thnh R c xy dng trn c s lut iu khin. - Khi gii m c nhim v chuyn tp m u ra thnh gi tr r y0 (ng vi mi gi tri r x0 iu khin i tng. - Giao din u vo thc hin vic tng hp v chuyn i tin hiu vo (t tng t sang s), ngoi ra cn c th c thm cc khu ph tr thc hin bi ton ng nh tch phn, vi phn.... - Giao din u ra thc hin chuyn i tn hiu ra (t s sang tng t) iu khin i tng.
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Nguyn tc tng hp mt b iu khin m hon ton da vo nhng phng php ton hc trn c s nh ngha cc bin ngn ng vo/ra v s la chn nhng lut iu khin. Do cc b iu khin m c kh nng x l cc gi tr vo/ra biu din di dng du phy ng vi chnh xc cao nn chng hon ton p ng c cc yu cu ca mt bi ton iu khin "r rng" v "chnh xc"2.1.2. Phn loi b iu khin m
Cng ging nh iu khin kinh in, b iu khin m c phn loi da trn cc quan im khc nhau: Theo s lng u vo v u ra ta phn ra b iu khin m "Mt vo mt ra" (SISO); "Nhiu vo - mt ra" (MISO); "Nhiu vo - nhiu ra" (MIMO) (hnh 2.2a,b,c).
Hnh 2.2a,b,c. Cc b iu khin m
B iu khin m MIMO rt kh ci t thit b hp thnh. Mt khc, mt b iu khin m c m u ra d dng ci t thnh m b iu khin m ch c mt u ra v vy b iu khin m MIMO ch c ngha v l thuyt, trong thc t khng dng. - Theo bn cht ca tn hiu a vo b iu khin ta phn ra b iu khin m tnh v b iu khin m ng. B iu khin m tnh ch c kh nng x l cc tn hiu hin thi, b iu khin m ng c s tham gia ca cc gi tr o hm hay tch phn ca tn hiu, chng c ng dng cho cc bi ton iu khin ng. B iu khin m tnh ch c kh nng x l cc gi tr tn hiu hin thi. m rng min ng dng ca chng vo cc bi ton iu khin ng, cc khu ng hc cn thit s c ni thm vo b iu khin m tnh nhm cung cp cho b iu khin cc gi tr o hm hay tch phn ca tn hiu. Cng vi nhng khu ng hc b sung ny, b iu khin tnh s tr thnh b iu khin m ng. 30
2.1.3. Cc bc tng hp b iu khin m
Cu trc tng qut ca mt h iu khin m c ch ra trn hnh 2.3.
Hnh 2.3. Cu trc tng qut mt h m
Vi mt min compact X Rn (n l s u vo) cc gi tr vt l ca bin ngn ng u vo v mt ng phi tuyn g(x) tu nhng lin tc cng cc o hm ca n trn X th bao gi cng tn ti mt b iu khin m c bn c quan h:
Sup |y(x) g(x)|< xX
vi l mt s thc dng bt k cho trc.
iu cho thy k thut iu khin m c th gii quyt c mt bi ton tng hp iu khin (tnh) phi tuyn bt k. tng hp c cc b iu khin m v cho n hot ng mt cch hon thin ta cn thc hin qua cc bc sau: 1- Kho st i tng, t nh ngha tt c cc bin ngn ng vo, ra v min xc nh ca chng. Trong bc ny chng ta cn ch mt s c im c bn ca i tng iu khin nh: i tng bin i nhanh hay chm? c tr hay khng? tnh phi tuyn nhiu hay t?,... y l nhng thng tin rt quan trng quyt nh min xc nh ca cc bin ngn ng u vo, nht l cc bin ng hc (vn tc, gia tc,...). i vi tn hiu bin thin nhanh cn chn min xc nh ca vn tc v gia tc ln v ngc li. 2- M ho cc bin ngn ng vo/ra: Trong bc ny chng ta cn xc nh s lng tp m v hnh dng cc hm lin thuc cho mi bin ngn ng. S lng cc tp m cho mi bin ngn ng c chn tu . Tuy nhin nu chn t qu th vic iu chnh s khng mn, chn nhiu qu s kh khn 31
khi ci t lut hp thnh, qu trnh tnh ton lu, h thng d mt n nh. Hnh dng cc hm lin thuc c th chn hnh tam gic, hnh thang, hm Gaus,... 3- Xy dng cc lut iu khin (mnh hp thnh): y l bc quan trng nht v kh khn nht trong qu trnh thit k b iu khin m. Vic xy dng lut iu khin ph thuc rt nhiu vo tri thc v kinh nghim vn hnh h thng ca cc chuyn gia. Hin nay ta thng s dng mt vi nguyn tc xy dng lut hp thnh h thng lm vic, sau m phng v chnh nh dn cc lut hoc p dng mt s thut ton ti u (c trnh by phn sau). 4- Chn thit b hp thnh (MAX-MIN hoc MAX-PROD hoc SUMMIN hoc SUM-PRROD) v chn nguyn tc gii m (Trung bnh, cn tri, cn phi, im trng tm, cao). 5- Ti u h thng: Sau khi thit k xong b iu khin m, ta cn m hnh ho v m phng h thng kim tra kt qu, ng thi chnh nh li mt s tham s c ch lm vic ti u. Cc tham s c th iu chnh trong bc ny l. Thm, bt lut iu khin; Thay i trng s cc lut; Thay i hnh dng v min xc nh ca cc hm lin thuc.2.2. B IU KHIN M TNH 2.2.1. Khi nim
B iu khin tnh l b iu khin m c quan h vo/ra y(x), vi x l u vo v y l u ra, theo dng mt phng trnh i s (tuyn tnh hoc phi tuyn). B iu khin m tnh khng xt ti cc yu t "ng" ca i tng (vn tc, gia tc,). Cc b iu khin tnh in hnh l b khuch i P, b iu khin re lay hai v tr, ba v tr,...2.2.2. Thut ton tng hp mt b iu khin m tnh
Cc bc tng hp b iu khin m tnh v c bn ging cc bc chung tng hp b iu khin m nh trnh by trn. hiu k hn ta xt v d c th sau:V d: Hy thit k b iu khin m tnh SISO c hm truyn t y = f(x) trong khong x = [a1,a2] tng ng vi y trong khong y [1, 2].
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Bc 1: nh ngha cc tp m vo, ra
- nh ngha N tp m u vo: A1, A2,, An trn khong [a1,a2] ca x c hm lin thuc Ai (x) (i = 1, 2,..., Ni dng hnh tam gic cn. - nh ngha N tp m u ra: B1, B2,, BN trn khong [1, 2] ca y c hm lin thuc Bj(x) (j = 1, 2,..., N) dng hnh tam gic cn.Bc 2: Xy dng lut iu khin
Vi N hm lin thuc u vo ta s xy dng c N lut iu khin theo cu trc:Ri: nu = Ai; th = Bi. Bc 3: Chn thit b hp thnh
Gi thit chn nguyn tc trin khai SUM-PROD cho mnh hp thnh, v cng thc Lukasiewicz cho php hp th tp m u ra B khi u vo l mt gi tr r x0 s l:
v Bi(y) l mt hm Kronecker Bi(y)Ai(x0) = Ai(x0) khi :
Bc 4: Chn phng php gii m
Chn phng php cao gii m, ta c:
Quan h truyn t ca b iu khin m c dng:
2.2.3. Tng hp b iu khin m tuyn tnh tng on
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Trong k thut nhiu khi ta cn phi thit k b iu khin m vi c tnh vo - ra cho trc tuyn tnh tng on. Chng hn, cn thit k b iu khin m c c tnh vo - ra nh hnh 2.4. Thut ton tng hp b iu khin ny ging nh thut ton tng hp b iu khin m vi hm truyn t y(x) bt k. Tuy nhin, cc on c tnh thng v ni vi nhau mt cch lin tc ti cc nt th cn tun th mt s nguyn tc sau:+ Mi gi tri r u vo phi lm tch cc 2 lut iu khin. + Cc hm lin thuc u vo c dng hnh tam gic c nh l mt im nt k, c min xc inh l khong [xk-1, xk+1] (hnh 2.5a).
Hnh 2.4. c tnh vo - ra cho trc + Cc hm lin thuc u ra c dng singleton ti cc im nt yk (hnh 2.5b). + Ci t lut hp thnh Max-Min vi lut iu khin tng qut:Rk: nu = Ak; th = Bk.
+ Gii m bng phng php cao.
Hnh 2.5 a.b. hm lin thuc ca cc bin ngn ng vo, ra
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2.3. B IU KHIN M NG
B iu khin m ng l b iu khin m m u vo c xt ti cc trng thi ng ca i tng nh vn tc, gia tc, do hm ca gia tc,.... V d i vi h iu khin theo sai lch th u vo ca b iu khin m ngoi tn hiu sai lch e theo thi gian cn c cc o hm ca sai lch gip cho b iu khin phn ng kp thi vi cc bin ng t xut ca i tng. Cc b iu khin m ng hay c dng hin nay l b iu khin m theo lut t l tch phn(PI), t l vi phn (PD) v t l vi tch phn (PID). Mt b iu khin m theo lut I c th thit k t mt b m theo lut P (b iu khin m tuyn tnh) bng cch mc ni tip mt khu tch phn vo trc hoc sau khi m . Do tnh phi tuyn ca h m, nn vic mc khu tch phn trc hay sau h m hon ton khc nhau (hnh 3.2 a,b).
Hnh 2.6a,b. h iu khin m theo lut PI
Khi mc thm mt khu vi phn u vo ca mt b iu khin m theo lut t l s c c mt b iu khin m theo lut t l vi phn PD (hnh 2.4).
Hnh 2.7. h iu khin m theo lut PD
Cc thnh phn ca b iu khin ny cng ging nh b iu khin theo lut PD thng thng bao gm sai lch gia tn hiu ch o v tn hiu ra ca h thng e v o hm ca sai lch e'. Thnh phn vi phn gip cho h 35
thng phn ng chnh xc hn vi nhng bin i ln ca sai lch theo thi gian. Trong k thut iu khin kinh in, b iu khin PID c bit n nh l mt gii php a nng v c min ng dng rng ln. inh ngha v b iu khin theo lut PID kinh in trc y vn c th s dng cho mt b iu khin m theo lut PID. B iu khin m theo lut PID c thit k theo hai thut ton: - Thut ton chnh nh PID; - Thut ton PID tc . B iu khin m c thit k theo thut ton chnh nh PID c 3 u vo gm sai lch e gia tn hiu ch o v tn hiu ra, o hm v tch phn ca sai lch. u ra ca b iu khin m chnh l tn hiu iu khin rt).
Vi thut ton PID tc , b iu khin PID c 3 u vo: sai lch e gia tn hiu u vo v tn hiu ch o, o hm bc nht e' v o hm du ca tn hiu iu bc hai e" ca sai lch. u ra ca h m l o hm dt khin u(t).
Do trong thc t thng c mt hoc hai thnh phn trong (3.6), (3.7) c b qua, nn thay v thit k mt b iu khin PID hon chnh ngi ta li thng tng hp cc b iu khin PI hoc PD.
Hnh 2.8. H iu khin m theo lut PID
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B iu khin PID m c thit k trn c s ca b iu khin PD m bng cch mc ni tip u ra ca b iu khin PD m mt khu tch phn (hnh 2.6). Hin nay c rt nhiu dng cu trc khc nhau ca PID m c nghin cu. Cc dng cu trc ny thng c thit lp trn c s tch b iu chnh PID thnh hai b iu chnh PD v PI (hoc I). Vic phn chia ny ch nhm mc ch thit lp cc h lut cho PD v PI (hoc I) gm hai (hoc 1) bin vo, mt bin ra, thay v phi thit lp 3 bin vo. H lut cho b iu chnh PID m kiu ny thng da trn ma trn do Mac Vicar-whelan xut. Cu trc ny khng lm gim s lut m ch n gin cho vic tnh ton.2.4. THIT K H IU KHIN M BNG PIIN MM MATLAB 2.4.1. Gii thiu hp cng c lgic m
Hp cng c Lgic m (The Fuzzy Logic Toolbox) l t hp cc hm c xy dng trn nn Matlab gip cho vic thit k, m phng, kim tra v hiu chnh b iu khin m mt cch d dng. thit k b iu khin m trong hp cng c ny, ta c th thc hin thng qua dng lnh hoc thng qua giao din ho. Trong khun kh cun sch ny ch gii thiu nhng thao tc c bn thit k b iu khin m thng qua giao din ho. Phn thit k thng qua dng lnh, ta c th c trong phn "Fuzzy Logic Toolbox" ca Malab.
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Hnh 2. 9
Sau khi c cu trc ca b iu khin m, ta tin hnh son tho cc hm lin thuc vo, hm lin thuc ra, cc lut iu khin.
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Hnh 2.11
Kch p chut vo biu tng Input (Hnh 2.11)Chn Edit, v chn Add MFs hoc Add Custom MF thm hm lin thuc, chn Remov Select MF g b mt hm lin thuc no , nu chn Remov All MFs s g b tt c cc hm lin thuc ca bin chn. Theo mc nh, s hm lin thuc l 3 c dng tam gic, ta c th thay i s lng cng nh hnh dng hm lin thuc. thay i hnh dng mt hm lin thuc no , ta kch chut vo hm , n s chuyn sang mu , sau kch chut vo hp thoi nh ch ra hnh 2.12 chn hm lin thuc mong mun. Trn Range v Display Range ta c nhp cc gi tr v min xc nh v min hin th ca bin ngn ng, mc nh ca cc min l t 0 n 1. Trn Name v Params (hnh 2.12) ta c th t tn v min xc nh cho tng tp m.
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son tho lut hp thnh, ta n Edit, Rules trn mn hnh hin ra ca s hnh 2.13. Sau mi ln son xong mt lut ta n Add rule xc nhn. thay i mt lut hp thnh ta n Change rule. xo mt lut iu khin ta n Delete rules. Mun quan st hot ng ca cc lut ta n View Rules. n View Surface quan st quan h vo ra ca b iu khin (hnh 2.14a, b). Sau khi thit k xong b iu khin, ta cn t tn v lu chng bng cch n File, Export To Disk ct vo a hoc to Workspase lu vo vng lm vic ca Matlab.
Mun m mt b iu khin m lu trn a, n File, Export To Disk sau n Import from disk, chn file cn m. Sau khi thit k xong b iu khin m bng ca s Edit GUI, ta chuyn v ca s m phng SIMULINK, m mt file mi vi ui '.mat', xy dng m hnh m phng cho h, tin hnh chy m phng v hiu chnh h thng.
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Hnh 2. 14a.b. a) Quan st hot ng ca cc lut
b) Quan h vo-ra ca b iu khin2.3.2. V d thit k h m
minh ho cho nhng vn trnh by trn, sau y chng ta tin hnh phn tch, thit k b iu khin m iu khin i tng nhit l in tr c hm s truyn l:
Bit in p cp cho l c gi tr nh mc l 230 V. S khi ca h c ch ra trn hnh 2.15.
Hnh 2.15. S khi h iu khin nhit l in trBc 1: Tm hiu h thng
L in tr dng gia nhit chi tit bng kim loi cho cc cng on nh ti, ram... L in tr c nung nng bng dy in tr, ngun in 41
cung cp cho l l ngun p c th iu chnh c. Vic iu khin nhit l c thc hin thng qua iu khin in p cung cp cho l. Trong k thut iu khin, ngi ta m t l bng mt khu qun tnh bc nht c tr c hm s truyn:
Trong , hng s thi gian T v thi gian tr T c gi tri tu vo loi l v cng sut l. B iu khin in p c in p iu chnh c v bin thin trong khong t 100V: 230V, c m t gn ng bng mt khu c hm s truyn: w(s) = ke~2s vi k = 23, ~ = O,05(s). Cm bin nhit c coi l 1 khu t l vi h s:
in p t c gi tr ln nht l 10 V. Khu so snh lm nhim v so snh in p t v in p phn hi ly t u ra ca khi cm bin, u ra ca khu so snh l sai lch e = U ucb. L din tr ni ring, cng nh i tng nhit ni chung thng khng cho php c qu iu chnh, do e bin thin trong khong t 10 n 0.Bc 2: Chn cc bin ngn ng vo, ra
Gi thit ta iu khin l in tr theo quy lut PI, khi bin ngn ng u vo b iu khin m l sai lch (k hiu l E) v tch phn sai lch (k hiu l TE). u ra b iu khin m l in p (k hiu l U). Min gi tr ca cc bin ngn ng c chn nh sau: E = [010]; TE = [01500]; U = [020]; ngn ng c chn nh hnh 2.16a,b,c hm lin thuc ca cc bin
ET = [E1(x) E2(x) E3(x) E4(x) E5(x)] (hnh 2.16a); TET = [TE1(x) TE2(x) TE3(x) TE4(x) TE5(x)] (hnh 2.16b); UT = [U1(x) U2(x) U3(x) U4(x) U5(x)] (hnh 2.16a); 42
Hnh 2.16a,b,c. Hnh dng cc hm lin thuc u vo v u raBc 3: Xy dng lut hp thnh: Vi 5 tp m ca mi u vo, ta xy dng c 5 x 5 = 25 lut iu khin. Cc lut iu khin ny c xy dng theo 2 nguyn tc sau:
- Sai lch cng ln th tc ng iu khin cng ln. - Tch phn sai lch cng ln th tc ng iu khin cng ln.R1: R2: R3: R4: R5: R6: R7: R8: R9: R10: R11: R12: R13: R14: R15: R16: R17: R18: R19: R20: R21: R22: R23: R24: R25: Nu Nu Nu Nu Nu Nu Nu Nu Nu Nu Nu Nu Nu Nu Nu Nu Nu Nu Nu Nu Nu Nu Nu Nu Nu E E E E E E E E E E E E E E E E E E E E E E E E E = = = = = = = = = = = = = = = = = = = = = = = = = E1 E2 E3 E4 E5 E1 E2 E3 E4 E5 E1 E2 E3 E4 E5 E1 E2 E3 E4 E5 E1 E2 E3 E4 E5 v v v v v v v v v v v v v v v v v v v v v v v v v TE TE TE TE TE TE TE TE TE TE TE TE TE TE TE TE TE TE TE TE TE TE TE TE TE = = = = = = = = = = = = = = = = = = = = = = = = = TE1 TE1 TE1 TE1 TE1 TE2 TE2 TE2 TE2 TE2 TE3 TE3 TE3 TE3 TE3 TE4 TE4 TE4 TE4 TE4 TE5 TE5 TE5 TE5 TE5 th th th th th th th th th th th th th th th th th th th th th th th th th U U U U U U U U U U U U U U U U U U U U U U U U U = = = = = = = = = = = = = = = = = = = = = = = = = U1 U2 U3 U4 U5 U2 U3 U4 U5 U5 U3 U4 U5 U5 U5 U4 U5 U5 U5 U5 U5 U5 U5 U5 U5 hoc hoc hoc hoc hoc hoc hoc hoc hoc hoc hoc hoc hoc hoc hoc hoc hoc hoc hoc hoc hoc hoc hoc hoc
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Bc 4: Chn lut hp thnh Max-Min, gii m bng phng php trng tm, ta quan st c s tc ng ca cc lut v quan h vo - ra ca b iu khin nh hnh 2.17a,b. Bc 5: M phng h thng: S m phng h thng c ch ra trn hnh 2.18. Kt qu m phng c ch ra trn hnh 2.19.
H'nh 2.17a, b. Quan h vo - Ra ca b iu khin
Hnh 2.18. S m phng h thng
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2.5. H IU KHIN M LAI (F-PID)
H m lai vit tt l F-PID l h iu khin trong thit b iu khin gm 2 thnh phn: Thnh phn iu khin kinh in v thnh phn iu khin m. B iu khin F-PID c th thit lp da trn hai tn hiu l sai lch e(t) v o hm ca n e(t). B iu khin m c c tnh rt tt vng sai lch ln, vi c tnh phi tuyn ca n c th to ra phn ng ng rt nhanh. Khi qu trnh ca h tin gn n im t (sai lch e(t) v o hm ca n e(t) xp xi bng 0) vai tr ca b iu khin m (FLC) b hn ch nn b iu khin s lm vic nh mt b iu chnh PID bnh thng. Trn hnh 2.20 th hin tng thit lp b iu khin m lai F-PID v phn vng tc ng ca chng.
Hnh 2.21. Vng tc ng ca cc b iu khin
S chuyn i gia cc vng tc ng ca FLC v PID c th thc hin nh kho m hoc dng chnh FLC. Nu s chuyn i dng FLC th ngoi nhim v l b iu chnh FLC cn lm nhim v gim st hnh vi ca h thng thc hin s chuyn i. Vic chuyn i tc ng gia FLC v PID c th thc hin nh lut n gin sau: 45
if |e(t) dng ln v | e (t)| dng ln th u l FLC if |e(t) dng nh v | e (t)| dng nh th u l PID.
.
(2.8) (2.9)
thc hin chuyn i m gia cc mc FLC v b chuyn i PID, ta c th thit lp nhiu b iu chnh PIDi (i = 1,2... n) m mi b c chn ti u cht lng theo mt ngha no to ra c tnh tt trong 1 vng gii hn ca bin vo (hnh 2.21). Cc b iu chnh ny c chung thng tin u vo v s tc ng ca chng ph thuc vo gi tr u vo. Trong trng hp ny, lut chuyn i c th vit theo h m nh sau:Nu (trng thi ca h) l Ei th (tn hiu iu khin) = ui
Trong i = 1, 2,..., n; Ei l bin ngn ng ca tn hiu vo, ui l cc hm vi cc tham s ca tc ng iu khin. Nu ti mi vng iu chnh, tc ng iu khin l do b iu chnh PIDi vi:
Nh vy, cc h s ca b iu chnh PIDi mi ph thuc cc tn hiu u vo tng qut hn l ph thuc vo trng thi ca h. Nu coi cc h s Kpi, KDi V Kli chnh l kt qu gii m theo phng php trung bnh trng tm t ba h m hm: H m hm tnh h s Kp vi h lut: Ru(i): if E is Ei and DE is DEi then Kp = Kpi. H m hm tnh h s KD vi h lut: Ru(i): if E is Ei and DE is DEi then KD = KDi. H m hm tnh h s K1 vi h lut: Ru(i): if E is Ei and DE is DEi then KI = KIi.2.6. H IU KHIN THCH NGHI M 2.6.1. Khi nim a/ nh ngha: H iu khin thch nghi m l h iu khin thch nghi c xy dng trn c s ca h m
(2. 11)
(2. 12)
(2. 13)
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So vi h iu khin thch nghi kinh in, h iu khin thch nghi m c min tham s chnh nh rt ln. Bn cnh cc tham s Kp, TI, TD ging nh b iu khin PID thng thng, b iu khin m ta cn c th chnh nh cc tham s khc nh hm lin thuc, cc lut hp thnh, cc php ton OR, AND, NOT, nguyn l gii m v.v... Trong thc t, h iu khin thch nghi c s dng ngy cng nhiu v n c cc u im ni bt so vi h thng thng. Vi kh nng t chnh nh li cc tham s ca b iu chnh cho ph hp vi i tng cha bit r a h thch nghi m tr thnh mt h iu khin thng minh.b/ Phn loi
Mt cch tng qut, h iu khin thch nghi m c th phn thnh 2 loi: - B iu khin m t chnh l b iu khin m c kh nng chnh nh cc tham s ca cc tp m (cc hm lin thuc); - B iu khin m t thay i cu trc l b iu khin m c kh nng chnh nh li cc lut iu khin. i vi loi ny h thng c th bt u lm vic vi mt vi lut iu khin c chnh nh trc hoc cha cc lut.c/ Cc phng php iu khin thch nghi m
Cc b iu khin thch nghi r v m u c mch vng thch nghi c xy dng trn c s ca 2 phng php:
Hnh 2.22. Cu trc phng php iu khin thch nghi trc tip Phng php trc tip (hnh 2.22) thc hin thng qua vic nhn dng thng xuyn cc tham s ca i tng trong h kn. Qu trnh nhn dng 47
thng s ca i tng c th thc hin bng cch thng xuyn o trng thi ca cc tn hiu vo/ra ca i tng v chn 1 thut ton nhn dng hp l, trn c s m hnh i tng bit trc hoc m hnh m; Phng php gin tip (hnh 2.23) thc hin thng qua phim hm mc tiu ca h kn xy dng trn cc ch tiu cht lng. Phim hm mc tiu c th c xy dng trn c s cc ch tiu cht lng ng ca h thng nh qu iu chnh, thi gian qu hay cc ch tiu tch phn sai lch... B iu khin thch nghi m c th chia thnh 2 loi:
Hnh 2.23. Cu trc phng php iu khin thch nghi gin tip2.6.2. Tng hp b iu khin thch nghi m n nh a. C s l thuyt
Xt 1 h phi tuyn SISO c m t bi phng trnh: y(n) = f(y, y,, y(n-1)) + bu; y = x l bin trng thi. y(n) = f(y) + bu (2.14)
Trong u l u vo, y l u ra, hm phi tuyn f(.) v hng s b c gi thit cha bit, y = [y, y,... y(n-1)]T. Mc tiu l thit k b iu khin m to ra tn hiu iu khin u sao cho tn hiu ra y(t) ca h thng bm theo qu o yd cho trc no . Nu bit trc f(y) v b, ta c th tng hp c b iu khin theo cc phng php kinh in [9], [55], b iu khin c tn hiu u ra l:
48
Cc h s k1, k2, kn c chn sao cho tt c cc nghim ca phng trnh: pn + knpn-1 +... + k1 = 0 nm na tri mt phng phc. Tc l cc nghim pk c phn thc m:
Do c iu kin (2.17) nn nghim ca e(t) chc chn tho mn iu kin:
Ta thy rng bi ton tng hp trn ch c ngha khi bit chnh xc m hnh ton hc ca h thng, hay ni cch khc l trong (2.1) ta bit f(y) v b. iu ny khng ph hp vi nhiu bi ton thc t. V vy mc tiu iu khin ra l phi xc nh b iu khin m u = u(x, ) v lut iu khin vct tham s sao cho tho mn cc iu kin sau: - H kn phi n nh ton cc trong phm vi ca cc bin y(t), u(x, ). (t) ) v
Tc l: |x(t)| Mx < ; | (t)| M0 < ; |u(x, )| Mu < vi mi t 0. Trong Mx, M0, Mu l cc tham s do ngi thit k t ra. - sai lch e = yd - y cng nh cng tt. Khi f(.) v b bit th ta d dng tng hp c b iu khin: 49
Trong , u* c coi l ti u. Nhng v f(.) v b cha bit nn u* khng th thc hin c, ta s thit k b iu khin m xp x ho iu khin ti u ny. Gi thit b iu khin u l t hp 2 b iu khin: B iu khin m uf(x, ) v b iu khin gim st us(x): u = uf+ us (2.20)
Trong uf(x, ) l b iu khin m c cp trong tng kt 2.1.Tng kt 2.l: Xt mt h logic m MISO c n u vo x v 1 u ra y (x
= (x1, x2,, xn)T Rn v y R). nh ngha Nj tp m Ai jj vi cc hm lin thuc A j bao ph min xc nh ca cc bin ngn ng u vo (j = 1,, nij
i l s u vo). Lut iu khin Ru1 ...in c dng:
if e1 = Ai11 and e2 = Ai2 andand = Ain then u = Bi1 ...in (2.21) 2 n trong i1 = 1, 2..., N1;... in = 1, 2,..., Nn l s hm lin thuc cho mi bin u vo, Bi1 ...in l tp m u ra. S dng lut hp thnh PROD, m ho theo ng singleton v gii m bng phng php trung bnh trng tm, ta thu c b iu khin m:
trong (x) l vct hm m c s.
50
Thay (2.20) vo (2.14) ta c:
d n yd T (3.29) ta rt ra: f(x) = -bu * + + KTe thay vo (3.35) n dt
y(n) = -bu* + yd(n) + KTe + b[uf(x, ) + uS(x)]. Sau khi bin i ta c: e(n) = -KTe + b [u* - uf(x, ) - us(x)]. Hoc vit di dng phng trnh trng thi: = Ae + B[u*-uf(x, ) us(x)] Trong : (2.27) (2.26)
1 T e pe. Trong P l ma trn dng i xng 2 c xc nh t phng trnh Lyapunov:
Chn hm Lyapunov V =
ATP + PA = - Q (Q > 0). o hm V ta c:
(2.29)
Thay (2.27), (2.29) vo (2.30) ta c:
ta cn phi tm hm us sao cho V 0. Gi thit ta xc nh c hm fu (x) v hng s bL tho mn iu kin: |f(x)| fu (x) v 0 < bL < b th hm iu khin gim st us(x) c xy dng 51
nh sau:
Trong :
( l 1 hng s c chn bi ngi thit k). V b > 0, sugn(eTPB) c th xc nh, hn na tt c cc thnh phn trong (2.32) c th xc nh c, v vy b iu khin gim st us l hon ton xc nh. Thay (2.32) v (2.19) vo (2.31) v xt cho trng hp I1* = 1 ta c:
vy s dng us theo (2.32) ta lun nhn c V V . T (2.32) ta thy rng us ch xut hin khi khng tho mn iu kin: V V.
Do vy trong khong sai s nh (ngha l V V ) th ch c b iu khin m uf lm vic cn b iu khin gim st khng lm vic (us = 0). Khi h thng c khuynh hng mt n nh (V > V ) th b iu khin gim st bt u lm vic hng cho V V .* Nu chn I1 1 th t (2.33) ta cn phi m bo khng ch gii hn ca
vct trng thi m cn phi m bo cho e hi t v 0. Ta khng chn phng n ny v us thng rt ln. Tht vy, t (2.33) ta thy us t l vi gii hn trn ca fu m gii hn ny thng rt ln. Tn hiu iu khin ln c th gy phin phc do c lm tng 52
thm chi ph ph. Bi vy ta chn us lm vic theo kiu gim st. tm lut iu khin thch nghi vct tham s ta thay uf (x, ) = (x). t * l vct tham s ti u:
Chn hm Lyapunov dng:
Vi l mt hng s dng, ta c:
Gi Pn l ct cui cng ca ma trn P, t (2.28) ta c:
eTpB = To b. (2.37) Thay (2.37) vo(2.36) ta c:
Chn lut thch nghi:
53
th (2.38) tr thnh:
trong : e PBu s 0 y l iu tt nht ta c th t c.b) Thut ton tng hp b iu khin m thch nghi
T
tng hp b iu khin m thch nghi, ta c th tin hnh theo 2 bc: Bc 1 l chn cu trc ca b iu khin m, bc 2 l xc nh thch nghi cc vct tham s.+ Chn cu trc ca b iu khin m
Cu trc ca b iu khin m thch nghi nh hnh 2.24. trong i tng iu khin l 1 h phi tuyn bt k c m t tng qut bng biu thc (2.1). B iu khin m thch nghi c th c nhiu u vo gm sai lch v cc o hm ca chng. Mc ch ca vic thit k b iu khin m l to ra tn hiu iu khin u, sao cho qu o u ra ca i tng (y) bm theo qu o cho trc (yd), cho d c s thay i thng s v cu trc ca i tng.
Hnh 2.24: S cu trc b iu khin m thch nghi+ Cc bc thc hin thut ton
54
Trong trng hp tng qut, b iu khin m c n u vo, thut ton tng hp c tm tt theo cc bc sau: - Bc 1. Xc nh hm lin thuc ca cc bin ngn ng u vo.
nh ngha min xc nh ca cc thnh phn ej l:
j j , Ch rng, gi tri thc ca ej c th bn ngoi khong min max j j chn, y , l khong m ej ri vo nhiu nht. min max
Hnh 2.25. Hm lin thuc vi 7 tp mnh ngha Nj tp m A1j... AnJ trn min j j , , min max
hm lin thuc
ca cc tp m c th chn l hnh tam gic, hnh thang, hm Gaus, hm sigmoid v.v... Chn hm lin thuc kiu hnh tam gic v hnh thang c u im l n gin, song c nhc im l iu chnh khng trn. Hnh 2.25 l v d v hm lin thuc kiu Gaus gia v kiu sigmoid 2 bn i vi 1 bin ngn ng u vo.
55
- Bc 2. Xy dng b iu khin m u t tch N1... Nn lut sau y:
Lut Ru i1 ...in if e1 = Ai1 and e2 = Ai2 andand en = Ain then u = Bi1 ...in (2.44) 1 2 n Trong i1 = 1, 2..., N1;... in = 1, 2,..., Nn l s hm lin thuc cho mi bin u vo
Bi1 ...in l tp m u ra s c xc inh.Vic thit k b iu khin m by gi chuyn sang vic xc nh cc thng s Bi1 ...in S dng lut hp thnh PROD, m ho theo ng singleton v gii m bng phng php trung bnh trng tm ta thu c b iu khin m:
trong : (e) l tp hp hm m c s bit.
lu thut ton tng hp hm m c s xe) nh hnh 2.26.
yi1 ...in l im trng tm ca Bi1 ...in chng s c chnh nh theo lut thchnghi cho ph hp vi i tng. 56
l mt vct gm tp hp cc yi1 ...in vi i1 = 1... N1;... in = 1 Nn
Cc thng s e c chnh nh nh s dng lut thch nghi sau:
Trong l 1 hng s dng xc inh tc ca thut ton cn pn l ct cui cng ca ma trn P, vi P l nghim ca phng trnh Lyapunov. ATP + pa = -Q trong Q l ma trn dng xc nh tu , A l ma trn (n x n) (2.50)
Hnh 2.26. Lu thut ton tng hp hm m c s (e)vi cc hng s k1, k2 c chn sao cho tt c cc nghim ca phng trnh: Pn + knPn-1 +... + k1 = 0 nm bn na tri mt phng phc. Vi cch tng hp nh vy h thng chc chn tho mn iu kin Lim e(t) = 0 .t
57
T cc tp m u vo (2.41)... (2.43) v cc thng s . Pn c xc nh trn ta tin hnh xy dng b iu khin m theo trnh t sau: - nh ngha cc hm lin thuc (2.41)... (2.43). - Xy dng hm m c s s (2.47). - Xc inh lut thch nghi - Xy dng b iu khin (2.46).Ch :
- H s y trong (2.49) ni ln tc hi t ca thut ton thch nghi. N c chn v sau c kim nghim thng qua m phng, nu y chn qu nh thut ton thch nghi hi t chm, y chn ln, qu trnh hi t nhanh nhng nu y chn qu ln h thng s mt n nh. - Cc gi tr P1, P2 c Xc nh t phng trnh Lyapunov (2.40), Tuy nhin ln ca n cng nh hng ng k n cht lng ca h thng. V vy sau khi thit k xong cn chnh nh li cc gi tr ca chng sao cho m bo cht lng tt trong ton di thay i ca cc thng s ca i tng.2.7. TNG HP B IU KHIN M THCH NGHI TRN C S L THUYT THCH NGHI KINH IN 2.7.1. t vn
Mt cu trc thng dng nht ca h iu khin logic m (FLC - Fuzzy Logic Control) l cu trc kiu phn hi sai lch. S nh hnh 2.27. Trong k1, l cc h s khuch i u vo, K l h s khuch i u ra. Thc tin cho thy vic chnh nh FLC kh khn hn nhiu so vi chnh nh b iu khin kinh in, mt trong nhng l do chnh l tnh mm do ca vng nhn bit c bn ca b iu khin m v s mc ni cc thng s ca chng. Tuy nhin khng c mt cch h thng ho no a ra tt c nhng thng s ny.
58
Hnh 2.27. Cu trc c bn ca b iu khin m 2 u voHin nay trong cng nghip cc b iu khin logic m (FLC) thng c thit k theo kinh nghim v s hiu bit nh tnh i tng ca cc chuyn gia. Vic chnh nh FLC c thc hin thng qua chnh nh cc hm lin thuc u vo v u ra v mang nhiu tnh cht "m mm". Do khng ph hp cho vic chun ho cht lng v kh tr thnh mt phng php lun c h thng. Trong mc ny chng ta s tip cn kiu thit k hn hp theo hng kt hp c hai cch tip cn nh tnh v tip cn nh lng. u tin ta xy dng m hnh c bn ca b iu khin m bao gm cc hm lin thuc v cc lut hp thnh, chng c th to ra mt p ng hp l mt mc no . Lut hp thnh c bn c chn l mt lut hp thnh tuyn tnh, cn hm lin thuc c th c xc nh theo hnh tam gic, hnh thang hoc hm Gaus. Sau khi xc nh c hm lin thuc v lut hp thnh c bn, ta s dng chng tm ra h s khuch i t l. C th s dng nhiu phng php nh lng khc nhau, vic xc nh cc h s khuch i t l ng rt quan trng i vi s hot ng ca FLC. Trong iu khin kinh in, ta bit mt Algorithm iu khin thch nghi theo m hnh mu s dng phng php gradient hay phng php Lyapunov rt thch hp cho vic iu khin mt qu trnh khng nhn bit c, c bit i vi h phi tuyn. Mt b iu khin m vi mt lut hp thnh tuyn tnh v cc hm lin hp thuc tam gic c th xp x tuyn tnh xung quanh trng thi cn bng. Do ta s dng tng ca b iu khin thch nghi kinh in p dng cho b iu khin m thch nghi vi mt vi s xp x no . Mc tiu chnh ca mc ny l: Tm ra cch tip cn nh lng xc nh m hnh ton hc ca b iu khin m vi mt vi s xp x no . Xy dng b iu khin m thch nghi cho nhng h thng phi tuyn v h thng bin i theo thi gian trn c s l thuyt thch nghi kinh in. B iu khin ny c th s dng iu khin i tng nh l b thch 59
nghi trc tuyn, hoc dng lm c s cho vic tng hp b iu khin m thng thng. n gin ta tin hnh xy dng c ch thch nghi cho b iu khin m hai u vo t kt qu c th d dng m rng cho nhng b iu khin m c nhiu u vo khc. Cu trc ca cc b iu khin m thch nghi da trn c s l thuyt Lyapunov v phng php Gradien kinh in.2.7.2. M hnh ton hc ca b iu khin m
Xt b iu khin m hai u vo nh hnh 2.27. xy dng m hnh ton hc ca n ta thc hin theo cc bc sau:a/ Chn cc hm lin thuc
Cc tp m u vo c chn m ho l E v R. Ta chn s lng cc tp m vo v ra bng nhau v bng N, cc hm lin thuc s b chn hnh tam gic vi mi hm lin thuc bao ph khng gian trng thi 2A cho mi u vo v 2B cho u ra. Gi s chn j hm lin thuc m cho E, R, U, chn j hm lin thuc dng cho E, R, U v 1 hm lin thuc bng zero cho E, R, U (hnh 2.28). Nh vy s lng cc hm lin thuc ca mi bin vo/ra l: N = 2j + 1. n gin cho vic xy dng lut hp thnh, thay v s dng cc ngn ng nh "m nhiu", "dng nhiu" v.v... ta s dng cc ch s l s, v d -1 (x), -2(x), -0(x), 1 (x) Ta thy rng, mc du s dng cc hm lin thuc ging nhau m t 2 tp m u vo nhng thng qua cc h s k1 v (hnh 2.27) chng thc s l cc hm lin thuc khc nhau.
Hnh 2.28. Minh ho vic nh ngha hm lin thuccho cc bin u vo v u ra 60
b/ Chn lut iu khin
Vi b iu khin m 2 u vo, mi u vo c N tp m ta s c N2 lut iu khin miu t tt c cc kh nng kt hp ca Ei v Rj Dng tng qut ca lut hp thnh l:Nu E = Ei v R = Rj th U = uk Vi k = f(i, j) nh ngha 1: Cc lut iu khin ca mt b iu khin m c gi l tuyn tnh nu f(i,j) l 1 hm tuyn tnh i vi i v j.
V d: f = i + j; f = I + j + 1 vvTrong f(i, j) l quy lut sinh ra cc lut iu khin. Vi cc f(i, j) khc nhau s cho cc lut iu khin khc nhau. Vic chn lut iu khin c th coi l mt ngh thut v ph thuc rt nhiu vo kin thc v kinh nghim ca cc chuyn gia. Trong mc ny tc gi cp n vic chun ha v n gin ha vic chn lut iu khin nhm to iu kin thun li cho ngi thit k h iu khin m.
Hnh 2.29 minh ho lut iu khin tuyn tnh vi f(i,j) = i +j cho b iu khin m 2 u vo 1 u ra vi 7 hm lin thuc cho mi bin vo v ra. Bng 2.1 v Hnh 2.30 l quan h vo-ra ca lut hp thnh tuyn tnh.
Bng 2.1I+j Uk-l -3 3 -2 2 1 -1 0 0 1 1 2 2 3 3
inh ngha 2: B iu khin m c s (Basis Fuzzy Controll - BFC) l b iu khin m c 2 u vo v 1 u ra, s tp m ca cc u vo v u ra bng nhau, lut hp thnh c s dng l lut hp thnh tuyn tnh c/ Phn tch lut c s thnh suy lun
61
Cc lut c s chia vng lm vic ca b iu khin m c bn thnh nhiu vung, vi u ra ca lut trn 4 gc nh hnh 2.29. V tt c cc thao tc m u c th c tnh ton trn cc ny nn chng c gi l suy lun [33], [55]. Mt cch tng qut ta c th chn suy lun IC(i, j) phn tch. ny c to bi cc hm lin thuc i(E), i+1(E), j(R) v j+1(R) cc ng cho ca chia chng ra thnh 4 vng (ICI... IC4) (hnh 2.3 l). Vi tr tuyt i ca 1 suy lun IC(i, j) trong lut c bn l t [iA, jA] n [(i+1)A,, (j + 1)A], v tr tng i ca mi vng trong IC(i,j) l t [0, 0] n [A, A]. Cc d liu vo (E, R) trong lut c bn lun lun c nh x n d liu vo tng i (e*, r* trong IC(i, j) theo cng thc [22]:
Tt c nhng thao tc m bao gm "M ho", "suy din m" v "gii m" u c th c thc hin trong suy lun IC.
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Hnh 2.29. S hnh thnh suy lun t lut hp thnhd/ Cc thao tc m trong suy lun
Trong suy lun ta c th thc hin cc thao tc m nh: M ho, suy din m v gii m. S dng phng php suy lun Max-Min ca Mamdani, cc thao tc c trnh by nh sau:
+ M ho: T cc biu thc (2.52) v (2.53) ta thy trong mt IC(i,j) cc u vo (E, R) c xc nh bi (e*, r*) vi cc gi tr hm lin thuc ca e* l i v i+1, cc gi tr hm lin thuc ca r* l i v i+1V lun tn ti quan h: i + i+1 = 1; j v j+1 = 1 do gi tr cc hm lin thuc u vo trong suy lun:
63
+ Suy din m
T lut hp thnh c s: Nu E = Ei v R = Rj th U = uk vi k = f(i,j) = i + j. (2.55)
Hm lin thuc ca cc tp m u ra c biu din trong Hnh 2.29 vi gi tr u ra l: uk = k.B. (2.56)
Ti mi vng ca suy lun ta thu c cc gi tr 1, 2, 3 (bng 2.2) thng qua php ly Max-min [21] vi:
+ Gii m
Dng phng php trung bnh trng tm [20] ta c tn hiu ra:
64
trong I = 1, 2, 3, 4 l cc vng tng ng ca suy lun. e/ Xy dng biu thc ton hc ca b iu khin m Qua cc phn tch trn ta thy rng cc tn hiu vo khc nhau (e*, r*) c th ri trn cc vng khc nhau ca suy lun t IC1 - IC4, l do kt qu ca php ly Max- min.
+ Xt vng IC1:T (2.54) v bng 2.2 ta c:
T bng (2.2), (2.54) v (2.58) ta c:
T ta rt ra:
Tng t vi cc suy lun khc, cui cng ta thu c [10]: 65
1 (I = 1, 2, 3, 4) l tham s phi tuyn trong vng IC1. Ta thy iu khin m vi lut hp thnh tuyn tnh thc s l iu khin phi tuyn nh biu thc (2.61). N s tr thnh iu khin tuyn tnh trng thi cn bng. Trong biu thc (2.61) ta cn phi xc nh cc h s khuch i t l u vo k1, v u ra K. Gi tr danh nh ca cc h s khuch i u vo k1 v c th c xc nh theo phng php ca H.X. Li [10]. Thng thng vic xc nh h s khuch i u ra K ng l rt kh khn.2.7.3. Xy dng c cu thch nghi cho b iu khin m a/ H iu khin thch nghi theo m hnh mu (MRAS) dng l thuyt thch nghi kinh in
Xt mt i tng iu khin c m t bi phng trnh:
M hnh mu c phng trnh:
Tn hiu iu khin:
vi sai s: = y ym
Biu thc cha tham s iu chnh. Ta cn tm ra c cu thch nghi iu chnh cc tham s l v 2 ti gi tr mong mun sao cho sai s tin ti 0. tm ra c cu thch nghi ny ta c th dng l thuyt n nh Lyapunov 66
hoc phng php Gradient theo cc b sau:B 2.1: (lut thch nghi theo Lyapunov)
Gi thit b > 0 v v chn hm Lyapunov c dng:
th quy lut iu chnh cc tham s l, 2 cho 0 l:
Nu ch c 1 tham s bin thin, lut iu chnh thch nghi tham s tr thnh:
B 2.2: (Lut thch nghi theo Gradient)
Gi thit l mt vct tham s cn c xc nh, v ph thuc sai lch gia u ra ca i tng (y) v u ra ca m hnh (ym). Tiu chun sai lch p ng ca h c chn:
th quy lut iu chnh theo hng ca gradient ca J l:
Trong iu khin thch nghi kinh in, ni chung khng cn mt m hnh mu hon ho, tuy nhin s sai khc gia m hnh v i tng cng nh tnh phi tuyn ca n ch nm trong gii hn no , nu qu gii hn ny b iu chnh s khng lm vic hiu qu na. khc phc nhc im , trong cun sch ny cc tc gi xut s dng h iu khin m thch nghi theo m hnh.b/ iu chnh thch nghi h s khuch i u ra ca b iu khin m
Tn hiu u ra ca b iu khin m (2.60) c vit: 67
B iu khin m 2 u vo trong biu thc (2.60) vi h s khuch i u ra K, c th c biu din nh l F. e cng thm 1 gii hn tr T nh biu thc (2.69) (hnh 2.31) gii hn tr T s tin ti zero khi h thng tin n im cn bng [11], [12].
Hnh 2.31. S khi b iu khin m vi h s khuch i u ra K
Ta s p dng phng php Lyapunov v phng php Gradient chnh nh thch nghi h s khuch i u ra K ca b iu khin m. Qu trnh iu chnh c thc hin theo 2 cu trc chnh c gi chung l iu khin thch nghi m theo m hnh mu (MRAFC) (Model Reference Adaptive Fuzzy Controller). Ta tin hnh kho st 2 s l s phn hi u ra v s iu khin thch nghi m theo m hnh hiu chnh trc (FMRAFC) (Feedfonvard Model Reference Adaptive Fuzzy Controller).c/ S o iu khin thch nghi m theo m hnh mu (MRAFC)
Xt mt cu trc iu khin m thch nghi theo m hnh c biu din trn hnh 2.32 [19], [20]. Trong : i tng iu khin c hm s truyn G, m hnh mu c hm truyn Gm, b iu khin m bao gm b iu khin m c bn kt hp vi b khuch i K. Cn phi tm ra quy lut chnh nh h s K sao cho sai 68
lch gia m hnh v i tng tin n 0 ( 0).
Xp x 1 trong (2.61) thnh mt hng s, h thng vng kn xung quanh trng thi cn bng tr thnh tuyn tnh vi phng trnh ca vng kn l:
V
KFG Gm . Khi quy 1 + KFG lut iu chnh thch nghi cho h s khuch i u ra ca FLC c th xc nh t (2.68) l:Gi thit y tin n ym th ta c th xp x .
xt n nh ca s trn, ta chn hm Lyapunov:
d/ S iu khin thch nghi m kiu truyn thng (FMRAFC)
Mt cu trc khc ca b iu khin thch nghi m c biu din trn 69
hnh 2.33 s ny gi l s thch nghi m truyn thng (Feedforward Model Reference Adaptive Fuzzy Controller - FMRAFC) [ 11]. Trong s ny sai lch gia tn hiu t v tin hiu u ra ca i tng c thay th bng gi tr sai lch gia i tng v m hnh
trong : +KFG) 1.
ym = v gi thit rng khi y tin n ym th KFG/(1 1 + KFG
T (2.68) ta rt ra quy lut thch nghi cho h s khuch i u ra l:
Ta thy do hm truyn ca m hnh khng c mt lut thch nghi (2.73) v (2.74) nn cu trc thch nghi ny chu ng tt i vi gii hn ln sai lch gia m hnh v i tng. Trong thc t n ch cn mt m hnh xp x bm gn ng v d m hnh mu bc nht: Gm = cng c th p dng am + S cho phn ln cc i tng iu khin.2.7.4. Mt s ng dng iu khin cc i tng cng nghip
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Mc ch ca phn ny l thng qua m phng trnh by tnh hiu qu ca b iu khin thch nghi m c tng hp trn c s l thuyt iu khin thch nghi kinh in. ng thi thng qua (MRAFC) ta xc nh c h s khuch i u ra cho b iu khin m, lm c s cho vic xy dng thut ton tng hp b iu khin m. Cc ng dng c xy dng cho 3 lp i tng in hnh trong cng nghip: i tng tuyn tnh bc hai trong c khu tch phn c m t bi:
i tng tuyn tnh bc 3 vi nhng tham s khng bit, c cho bi cu trc gn ng sau?
Mt i tng phi tuyn vi cc thng s bin thin theo thi gian c m t gn ng bng phng trnh:
Hnh 2.34. S cu trc h MRAFC vi lut iu khin theo Lyapunov M hnh mu l khu qun tnh bc nht c hm truyn:
vi am = bm = 1. Tn hiu t UC l sng hnh vung. 71
S cu trc h MRAFC vi lut iu khin theo Lyapunov c biu din trn hnh 2.34 v theo Gradient c biu din trn hnh 2.35.
a/ Kt qu m phng
Cc kt qu m phng c ch ra trn cc hnh t hnh 2.36 n hnh 2.44. tin so snh ta a ra p ng tng ng vi 3 cu trc MRAC, FMRAFC theo Lyapunov v FMRAFC theo Phng php Gradient.b/ Nhn xt
T kt qu m phng trn ta rt ra mt s nhn xt sau: p ng ca h FMRAFC theo phng php Lyapunov v phng php Gradient gn ging nhau v c biu din trn cc hnh t hnh 2.36 n hnh 2.41. Ta thy: i vi i tng tuyn tnh bc hai c khu tch phn p ng ca FMRAFC trong hnh 2.36 v hnh 2.37 t cht lng ng tt, qu trnh lm vic s bm theo m hnh mt cch nhanh chng. i vi i tng tuyn tnh bc 3 p ng ca FMRAFC trong hnh 2.38 v hnh 2.39 gn ging vi i tng bc nht. i vi i tng khng tuyn tnh bin i theo thi gian, p ng ca FMRAFC hnh 2.40 v hnh 2.41 khng thay i nhiu so vi i tng bc 2. Vy h iu khin thch nghi m (MRAFC) c th t c p ng tt hn rt nhiu so vi h iu khin thch nghi kinh in (MRAC), c bit cho 72
nhng i tng bin i theo thi gian v khng m hnh ho c. Bn cnh ch ra kh nng to ln ca b iu khin m thch nghi lm vic vi cc qu trnh khng nhn bit c. T nhng kt qu trn, ta c th tip tc pht trin theo hng ny xy dng cc b iu khin m t chnh trc tuyn m c th t c p ng ti u mt cch t ng cho mt gii hn rng hn cc qu trnh.
Hnh 2.36: p ng ca FMRAFC vi lp i tng bc hai trong c khu tch phn theo Liapunov ng vi 2 gi tr ca K= 2; 5 v T = 0,1; 0,3
Hnh 2.37: p ng ca FMRAFC vi lp i tng bc hai trong c khu tch phn theo Gradient ng vi K=2; 5 v T=0,2; 0,3
Hnh 2.38: p ng ca FMRAFC vi lp i tng bc 3 theo Liapunov ng vi K= 2; 5; T1=0,003; 0,005 v T2 = 0,1; 0,5
Hnh 2.39: p ng ca FMRAFC vi lp i tng bc 3 theo Gradient ng vi K=2; 5; T1=0,003; 0,005 v T2 = 0,1; 0,5
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Hnh 2.40: p ng h FMRAFC vi i tng phi tuyn theo Liapunov
Hnh 2.41: p ng ca FMRAFC vi i tng phi tuyn theo Gradient
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Chng 3
TNG QUAN V MNG NRON3.1. NRON SINH HC 3.1.1. Chc nng, t chc v hot ng ca b no con ngi
B no ngi c chc nng ht sc quan trng trong i sng ca con ngi. N gn nh kim sot hu ht mi hnh vi ca con ngi t cc hot ng c bp n gin n nhng hot ng phc tp nh hc tp, nh, suy lun, t duy, sng to,... B no ngi c hnh thnh t s lin kt ca khong 1011 phn t (t bo), trong c khong 1010 phn t l nron, s cn li khong 9*1010 phn t l cc t bo thn kinh m v chng c nhim v phc v cng nh h tr cho cc nron. Thng thng mt b no trung bnh cn nng khong 1,5 kg v c th tch l 235 cm3, cho n nay ngi ta vn cha thc s bit r cu to chi tit ca b no. Tuy vy v i th th cu to no b c phn chia ra thnh nhiu vng khc nhau. Mi vng c th kim sot mt hay nhiu hot ng ca con ngi. B no c cu trc nhiu lp. Lp bn ngoi thng thy nh l cc np nhn, l lp c cu to phc tp nht. y l ni kim sot v pht sinh cc hnh ng phc tp nh nghe, nhn, t duy,... Hot ng ca b no ni ring v ca h thn kinh ni chung c con ngi quan tm nghin cu t lu nhng cho n nay ngi ta vn cha hiu r thc s v hot ng ca b no v h thn kinh. c bit l trong cc hot ng lin quan n tr c nh suy ngh, nh, sng to,... Tuy th cho n nay, ngi ta cng c nhng hiu bit cn bn v hot ng cp thp ca no. Mi nron lin kt vi khong 104 nron khc, cho nn khi hot ng th b no hot ng mt cch tng lc v t hiu qu cao. Ni mt cch khc l cc phn t ca no hot ng mt cch song song v tng tc ht sc tinh vi phc tp, hiu qu hot ng thng rt cao, nht l trong cc vn phc tp, v tc x l ca b no ngi rt nhanh mc d tc x l ca 75
mi nron (c th xem nh phn t x l hay phn t tnh) l rt chm so vi x l ca cc cng logic silicon trong cc chp vi x l (10-3 giy so vi 1 0-10 giy). Hot ng ca c h thng thn kinh bao gm no b v cc gic quan nh sau: Trc ht con ngi b kch thch bi gic quan t bn ngoi hoc trong c th. S kch thch c bin thnh cc xung in bi chnh cc gic quan tip nhn kch thch. Nhng tn hiu ny c chuyn v trung ng thn kinh l no b x l. Trong thc t no b lin tc nhn thng tin x l, nh gi v so snh vi thng tin lu tr a ra cc quyt nh thch ng. Nhng mnh lnh cn thit c pht sinh v gi n nhng b phn thi hnh thch hp nh cc c tay, chn,... Nhng b phn thi hnh bin nhng xung in thnh d liu xut ca h thng.Tm li: b no ngi c chc nng ht sc quan trng i vi i sng ca con ngi. Cu to ca n rt phc tp, tinh vi bi c to thnh t mng nron c hng chc t t bo vi mc lin kt gia cc nron l rt cao. Hn na, n cn c chia thnh cc vng v cc lp khc nhau. B no hot ng da trn c ch hot ng song song ca cc nghn to nn n. 3.1.2. Mng nron sinh hc a/ Cu to
Nron l phn t c bn to nn b no con ngi. S cu to ca mt nron sinh hc c ch ra nh trong hnh 3.1. Mt nron in hnh c 3 phn chnh:
Hnh 3.1. M hnh 2 nron sinh hc 76
- Thn nron (so ma): Nhn ca nron c t y. - Cc nhnh (dendrite): y chnh l cc mng dng cy ca cc dy thn kinh ni cc soma vi nhau.
- Si trc (Axon): y l mt ni kt, hnh tr di v mang cc tn hiu t ra ngoi. Phn cui ca axon c chia thnh nhiu nhnh nh (c ca dendrite v axon) kt thc trong mt c quan nh hnh c hnh c gi l synapte m ti y cc nron a cc tn hiu ca n vo cc nron khc. Nhng im tip nhn vi cc synapte trn cc nron khc c th cc dendrite hay chnh soma.b/ Hot ng
Cc tn hiu a ra bi mt synapte v c nhn bi cc dendrite l cc kch thch in t. Vic truyn tn hiu nh trn lin quan n mt qu trnh ha hc phc tp m trong cc cht truyn c trng c gii phng t pha gi ca ni tip ni. iu ny lm tng hay gim in th bn trong thn ca nron nhn. Nron nhn tn hiu s kch hot (fire) nu in th vt khi mt ngng no v mt xung (hoc in th hot ng) vi mnh (cng ) v thi gian tn ti c nh c gi ra ngoi thng qua axon ti phn nhnh ca n ri ti cc ch ni synapte vi cc nron khc. Sau khi kch hot, nron s ch trong mt khong thi gian c gi l chu k, trc khi n c th c kch hot li. Synapses l Hng phn (excitatory) nu chng cho php cc kch thch truyn qua gy ra tnh.trng kch hot (fire) i vi nron nhn. Ngc li, chng l c ch (inhibitory) nu cc kch thch truyn qua lm ngn tr trng thi kch hot (fire) ca nron nhn.3.2. MNG NRON NHN TO 3.2.1. Khi nim
Nron nhn to l s sao chp nron sinh hc ca no ngi, n c nhng c tnh sau: - Mi nron c mt s u vo, nhng kt ni (Synaptic) v mt u ra (axon) - Mt nron c th hot ng (+35 mV) hoc khng hot ng (-0,75 mV) 77
- Ch c mt u ra duy nht ca mt nron c ni vi cc u vo khc nhau ca nron khc. iu kin nron c kch hot hay khng kch hot ch ph thuc nhng u vo hin thi ca chnh n. Mt nron tr nn tch cc nu u vo ca n vt qua ngng mt mc nht nh.. C nhiu kiu nron nhn to khc nhau. Hnh 3.2 biu din mt kiu rt n gin. Cc u vo c hm trng Wj v b tng. u ra ca b tng c s dng quyt nh mt gi tr ca u ra thng qua hm chuyn. C nhiu kiu hm chuyn khc nhau (s c cp phn sau). Tng t nron sinh hc ca con ngi, nron s c kch hot nu tng gi tr vo vt qu ngng v khng c kch hot nu tng gi tr vo thp hn ngng. S lm vic nh vy ca nron gi l s kch hot nhy bc.
Hnh 3.2. M hnh nron n gin
Hnh 3.3. Mng nron 3 lp Kt ni mt vi nron ta c mng nron. Hnh 3.3 l mt mng nron gm 3 lp: lp vo, lp n v lp ra. Cc nron lp vo trc tip nhn tn hiu u vo, mi nron ch c mt tn hiu vo. Mi nron lp n c ni vi tt c cc nron lp vo v lp ra. Cc nron lp ra c u vo c ni vi tt c cc nron 78
lp n, chng l u ra ca mng. Cn ch rng mt mng nron cng c th c nhiu lp n. Cc mng nron trong mi nron ch c lin h vi tt c cc nron lp k tip v tt c cc mi lin kt ch c xy dng t tri sang phi c gi l mng nhiu lp truyn thng (perceptrons). Thng thng mng nron c iu chnh hoc c hun luyn hng cc u vo ring bit n ch u ra. Cu trc hun luyn mng c ch ra trn hnh 3.4. y, hm trng ca mng c iu chnh trn c s so snh u ra vi ch mong mun (taget) cho ti khi u ra mng ph hp vi ch. Nhng cp vo/ch (input/taget) c dng gim st cho s hun luyn mng. c c mt s cp vo/ra, mi gi tr vo c gi n mng v gi tr ra tng ng c thc hin bng mng l s xem xt v so snh vi gi tr mong mun. Bnh thng tn ti mt sai s bi l gi tr mong mun khng hon ton ph hp vi gi tr thc. Sau mt ln chy, ta c tng bnh phng ca tt c cc sai s. Sai s ny c s dng xc nh cc hm trng mi.
Hnh 3.4. Cu trc hun luyn mng nron Sau mi ln chy, hm trng ca mng c sa i vi c tnh tt hn tng ng vi c tnh mong mun. Tng cp gi tr vo/ra phi c kim tra v trng lng c iu chnh mt vi ln. S thay i cc hm trng ca mng c dng li nu tng cc bnh phng sai s nh hn mt gi tri t trc hoc chy mt s ln chy xc nh (trong trng hp ny mng c th khng tho mn yu cu t ra do sai lch cn cao). C 2 phng php c bn hun luyn mng nron: Hun luyn gia tng (tin dn) v hun luyn theo gi. S hun luyn theo gi ca mng nhn c bng vic thay i hm trng v dc trong mt tp (batch) ca vct u vo. Hun luyn tin dn l thay i hm trng v dc ca 79
mng sau mi ln xut hin ca mt phn t vct u vo. Hun luyn tin dn i khi c xem nh hun luyn trc tuyn hay hun luyn thch nghi. Mng nron c hun luyn thc hin nhng hm phc tp trong nhiu lnh vc ng dng khc nhau nh trong nhn dng, phn loi sn phm, x l ting ni, ch vit v iu khin h thng. Thng thng hun luyn mng nron, ngi ta s dng phng php hun luyn c gim st, nhng cng c mng thu c t s hun luyn khng c gim st. Mng hun luyn khng gim st c th c s dng trong trng hp ring xc inh nhm d liu. Mng nron bt u xut hin t 50 nm nhng mi chi tm thy cc ng dng t khong 10 nm tr li y v vn ang pht trin nhanh chng. Nh vy, r rng c s khc bit vi nhng h thng iu khin hoc ti u ho, ni m cc thut ng, c s ton hc v th tc thit k c thit lp chc chn v c ng dng t nhiu nm.3.2.2. M hnh nron a/ Nron n gin: mt nron vi mt u vo v hng v khng c dc c ch ra trn hnh 1.5a,b.
Hnh 3.5a,b. M hnh nron n gin Tn hiu vo v hng p thng qua trng lin kt v hng w tr thnh wp cng l i lng v hng. y wp l i s duy nht ca hm truyn f, tn hiu u ra l i lng v hng a. Hnh l.5b l nron c dc b. Ta c th hiu b nh l php cng n gin vo tch wp hoc nh l mt s thng ging ca hm f hnh a i mt lng b. dc c xem nh mt trng lng, ch c iu u vo l mt hng s bng 1. Tn hiu vo hm truyn mng l n l tng ca trng u vo wp v c b, p ng ra a 80
c coi l i s ca hm chuyn f. Hm chuyn f c th l hm bc nhy, hm sigmoid... Hnh 3.6 di y gii thiu mt s dng hm chuyn ca nron.
Hnh 3.6. Mt s dng hm chuyn ca mng nron Ch rng w v b u l cc tham s iu chnh v hng ca nron. tng c bn ca mng nron iu chnh cc tham s ny nh th no mng t c mt ch mong mun hay mt hnh vi no . Nh vy ta c th hun luyn mng lm mt cng vic no bng cch iu chnh cc trng lin kt v dc, hoc mng c th t iu chnh cc tham s ny t c cc kt qu mong mun.Ch :
- Tt c cc nron u cho sn mt dc (b), tuy nhin chng ta c th b i khi cn thit. - dc b l mt tham s iu chnh v hng ca nron, n khng phi l mt u vo, song hng s 1 phi dc xem nh u vo v n cn c coi nh vy khi xem xt ph thuc tuyn tnh ca cc vc l u vo. b/ Nron vi nhiu u vo (vc t vo) Nron vi vct vo gm R phn t c chi ra trn hnh 3.7. Trong cc u vo l p1, p2,, pR c nhn vi cc trng lin kt w1,1, w1,2, w1,R cc trng lin kt c biu din bng ma trn hng, vct p l ma trn ct, khi ta c:
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Trong W l ma trn trng lin kt c kch thc 1 x R, P l vct vo gm R phn t. Cch biu din trn s rt kh khn khi m t mng gm nhiu nron v c nhiu lp. n gin ta s dng k hiu nh hnh 3.8. Trong vct u vo c biu din bi thanh m bn tri. Kch thc ca p c ch ra bn di k hiu p l R x 1.(ta s dng ch vit hoa R ch kch thc ca mt vct). Nh vy p l mt vct gm R phn t vo, cc u vo ny nhn vi ma trn W (1xR). Ging nh phn trn, y hng s 1 a vo nron nh mt u vo v c nhn vi dc b. Hm chuyn ca mng l f. u vo hm chuyn l n bng tng ca dc b v tch Wp. Tng ny c i qua hm chuyn f c u ra ca nron l a. Trong trng hp ny a l mt i lng v hng. Ch rng nu c t 2 nron tr ln th u ra s l mt vct.
Hnh 3.9. mt s hm chuyn thng dng Mt lp mng c nh ngha nh hnh 3.8, l s kt hp gia cc trng lin kt, php nhn, php cng, dc b v hm chuyn f. Trong kch thc ca ma trn c ch r bn di tn bin ma trn ca chng. Khi mt hm chuyn c th c s dng th trn hnh v biu tng ca hm chuyn s thay th f trn. Hnh 3.9 l mt vi v d v cc hm 82
chuyn thng dng.3.3. CU TRC MNG
Nhiu nron kt hp vi nhau to thnh mng nghn, mng nron c th c mt lp hoc nhiu lp.3.3.1. Mng mt lp
Mt cu trc mng 1 lp vi R u vo v S nron c ch ra trn hnh 3.10. Trong : - Vc t vo p c R phn t pT = [p1 p2... PR]. - Vct vo n c s phn t nT = [n1 n2... ns]. - Vct vo a c s phn t aT = [a1 a2... as]. Trong mng ny mi phn t ca vct vo p lin h vi u vo mi nron thng qua ma trn trng lin kt W. B cng ca nron th i thu thp cc trng lin kt u vo v dc to thnh mt u ra v hng n;. Cc ni tp hp vi nhau to thnh s phn t ca vct vo n. Cui cng lp ra nron ta thu c vct a gm s phn t.Ch : Nhn chung s u vo ca mt lp khc vi s nron, tc l R S. Trong mt lp, khng bt buc phi c s u vo bng s nron ca n.
Hnh 3.10. Cu trc mng nron 1 83
Ta c th thit lp lp n ca cc nron c cc hm chuyn khc nhau mt cch d dng bi l hai mng c t song song. Tt c cc mng c th c chung u vo v mi mng c th thit lp mt vi u ra. Cc phn t ca vct u vo c a vo mng thng qua ma trn trng W, vi:
Trong : Ch s hng trong cc phn t ca ma trn W cho bit nron ni n cn ch s ct cho bit ni xut pht ca trng lin kt. V d: w12 ni ln s c mt ca tn hiu vo t phn t th hai n nron th nht vi trng lin kt l w12. Tng t nh trnh by vi 1 nron, n gin ta k hiu mng mt lp gm S nron, R u vo nh hnh v 3.11.Trong : vct vo P c kch thc R, ma trn trng lin kt W c kch thc S x R cn a v b l cc vct c kch thc S. Nh chng ta bit, mt lp mng bao gm ma trn trng lin kt, ton t nhn, vct dc b, b tng v hp hm truyn.3.3.2. Mng nhiu lp a/ K hiu quy c cho mt lp mng
kho st mng nhiu lp trc ht chng ta cn a ra cc k hiu quy c cho mt lp mng. c bit ta cn phi phn bit s khc nhau gia ma trn trng lin kt u vo v cc ma trn trng lin kt gia cc lp v nm vng k hiu ngun v ch ca ma trn trng lin kt. Ta gi ma trn trng lin kt ni vi u vo l cc trng vo (input weights) v cc ma trn n t lp ra l trng lin kt lp (layer weights). Ta s dng cc ch s vit bn trn phn bit ngun (ch s th hai) v ch (ch s th nht) cho cc trng lin kt v cc phn t khc ca mng.
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Hnh 3.11. K hiu mng R u vo v S nron
Hnh 3.12. K hiu mt lp mng minh ho, ta xt mt lp mng c nhiu u vo nh hnh 3.12. Trong R l s phn t lp vo v Sl l s nron ca lp 1. Ta thy ma trn trng lin kt vi vct vo P l ma trn trng vo (IW1,1) c ngun l 1 (ch s th 2) v ch l 1 (ch s th nht). ng thi cc phn t ca 1 lp nh dc, tn hiu vo hm chuyn, u ra c ch s vit trn l 1 ni rng chng c lin kt vi lp th nht (b1, n1, a1). phn sau ta s s dng ma trn trng lin kt lp (LW) ging nh ma trn trng vo (IW). Vi mt mng c th c ma trn trng IW1,1 c k hiu:IW1,1 net.IW{1, 1}
Nh vy, ta c th vit k hiu thu c mng nhp vo cho hm chuyn nh sau:n{1} = net.IW{1, 1}*p + net.b{1};
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Mt mng nron c th c mt vi lp. Mi lp c ma trn trng lin kt W, vct dc b v u ra a. phn bit cc ma trn trng lin kt vct vo cho mi lp mng trong s , ta thm con s ch lp vit pha trn cho bin s quan tm. Hnh 3.13 l k hiu s mng 3 lp. Trong c R1 u vo, S1 nron lp 1, S2 nron lp 2... Thng thng, cc lp khc nhau c s nron khc nhau. Ch rng u ra ca mi lp trung gian l u vo ca lp tip theo. Nh vy lp 2 c th c xem nh mng 1 lp vi S1 u vo, S2 nron v S2 x S1 trng lin kt ca ma trn W2. u vo ca lp 2 l vct a1, u ra l vct a2. Khi c k hiu ca tt c cc vct v ma trn ca lp 2 ta c th coi n nh l mng 1 lp. Cch tip cn