魯棒迭代學(xué)習(xí)控制及其在注塑成型工藝中的應(yīng)用畢業(yè)課程設(shè)計(jì)外文文獻(xiàn)翻譯、中中英文翻譯、外文翻譯
魯棒迭代學(xué)習(xí)控制及其在注塑成型工藝中的應(yīng)用畢業(yè)課程設(shè)計(jì)外文文獻(xiàn)翻譯、中中英文翻譯、外文翻譯,魯棒迭代,學(xué)習(xí),控制,節(jié)制,及其,注塑,成型,工藝,中的,應(yīng)用,利用,運(yùn)用,畢業(yè),課程設(shè)計(jì),外文,文獻(xiàn),翻譯,中英文
to 00100091)0033923 4,874 644 ao 0 of in to to 6 (2001) 7025–?,16024, s is an in of to ? LC)is or k to a to is in an 1978)1984). on +852+F. of & 1996; 999)998)LC as an it 1998). a 1996) by LC on an of an s is to LC of to In in be to 1/$ 2001 1)00339. et 6 (2001) 7025–7034is to 996)to A to is on of of s t+1)=t)+t)+t);06k=0;1;2;:::;yk(t)=t)+!k(t);n;m;p;(1)k to yk(t) is of at 06t)k(t)q. (1) be ; B; to on q. (1) yk(t)=t?1?l)+t); (2)t)=)+t?1?l)+!k(t): (3)in 1996) to a In 2)t);uk(t)t)(4)?????))..)??????;?????)).. ?1)??????;?????))..)??????;G=?????? ::: 0B ::: 0...........B ::: ????:(t), a a be In it is ,1996). If (1), , B=0, a as in 996)969)996)a t); 16)is to (996)is of to at k+1)is k +1)th in ;t]of ; B et 6 (2001) 7025–7034 70271):?xk(t+1)=t)+t);06k=0;1;2;:::;?yk(t)=t); ?n;m; ?p; (5):’by q. (1) in or r(t), k+1)(t):=[r(t)? ?(t)]TQ(t)[r(t)? ?(t)]+N?1[(t)]TR(t)[(t)]; (6)(t)=(t) ? uk(t), (t) (t) t. q. (6) =[r? ?]TQ[r? ?]+(7)=(1);Q(2);:::;Q(N)};R=(0);R(1);:::;R(N ?1)},???????)?)..)???????;r=???????r(1)r(2)..)???????: (8)7),=??1r? ?]: (9)9),?(t)(996)9)(t)=t+1){I ?B[t+1)B+R(t+1)]?1×t+1)}A+t+1)C;t=0;1;:::;N?1; S(N)=0; (10)(t)=[I +S(t)(t)1[(t+1)+t+1)?ek(t+1)];t=0;1;:::;N?1; (N)=0 (11)ek(t+1)=r(t+1)? ?yk(t+1)(t)=?uk(t)?[t)B+R(t)]?1(t)A[?(t)? ?xk(t)]+R?1(t)(t):(12)a q.(5)4),by (10)–(12) (1). a =?1; (13)(t)=[I +S(t)(t)1×[(t+1)+t+1)ek(t+1)];t=0;1;:::;N?1; (N)=0; (14)(t)=uk(t)?[t)B+R(t)]?1(t)A[(t)?xk(t)]+R?1(t)(t);(15)t)10)10),3)–(15)15),(t)t)15)) a 3rd q. (15)],1996), a on of a to be An 028 F. et 6 (2001) 7025–7034a of of q.(10), 13)–(15) to 1) is + +R?1R?1; (16): (17)q. (13) in 4)r?=R?1?; (18)=?=(I +1I +1(19)=r?13)4),=(I +R?11I +R?11R?1r?): (20) (10); 13)–(15) q. (1); in to 16)7)); t)k(t)k; =(1r?; (21)=0; (22)C1?is If it q. (3)C1?20)9),=(I +R?1(I +R?1r??l);(23)=(I +(I +?l; (24)=(I+R?11[I?(I+R?11]?1×R?1r?=(R?11R?1r?=(1r?;21)=024),22)=of as (16) 17). A (16) 17) AK YN TG at 13)–(15)=(I + +r??l); (25)=(I +(I +?l:(26)F. et 6 (2001) 7025–7034 7029a be in of a be by (14) 15) t)=t+1){I ?B[t+1)B+?1×t+1)}A+t=0;1;:::;N?1; S(N)=0; (27)(t)=S(t)1×[(t+1)+t+1)];t=0;1;:::;N?1; (N)=0; (28)(t)=uk(t)?[t)B+1(t)A[(t)?xk(t)]+(t): (29)(t)(t),to of A (3), (19) 20) a a (0) k →∞. (10) 28)t)=t+1){I ?B[t+1)B+?1×t+1)}A+t=0;1;:::;N?1; )=0; (30)(t)=Sk(t)1×[(t+1)+t+1)];t=0;1;:::;N?1; (N)=0: (31)It is k(t) → 0 Sk(t) → 0 ∞, a t)t)15) is in It As a is to in in by of of is it is by 1999; 000)LC to of an in of is as by to of 13)–(15) is to to an a is . et 6 (2001) 7025–70341. of 000 2. of overal
收藏