計(jì)算機(jī)控制系統(tǒng)的狀態(tài)空間設(shè)(存儲版)
【正文】 因此由上式可知則如L ya p u n o vVrqpV .0,0,0,0 ?????? (3) 狀態(tài)觀測器方法 )1(23)25(3212))( 12 1(1???????????????xxxxmxxxxxfxxx?????將式 (1)寫成下列形式 )2()(000001101321321xfxxxxxx?????????? ???????????????????????????????????? ???????根據(jù)狀態(tài)觀測器方法設(shè)計(jì)同步系統(tǒng) )?(00)?( 100? 3? 2? 1001110? 3? 2? 1YYxfxxxxxx??????????? ???????????? ???????????????????????????????????? ????????)?(??),( 11 xfxKYxfKxY ????其中? ??????????????????????????????? ??????????????????????????32113121132100000110eeekkkeee ???????則誤差方程為? ?? ?為誤差。 L=(acker(A1,C1,P1))39。)。 B=[2,1,1]39。第五章 計(jì)算機(jī)控制系統(tǒng)的 狀態(tài)空間設(shè)計(jì)方法 53 狀態(tài)觀測器的設(shè)計(jì) )()()()()1(kCxkykBukAxkx????程為考慮被控對象的狀態(tài)方)(?)(?)()(?)1(?kxCkykBukxAkx????測系統(tǒng)人為構(gòu)造一個(gè)相同的量)(?)(?)()()(?)()()()(?)())(?)(()()(?)1(?kxCkykKCxkBukxKCAkKykBukxKCAkykyKkBukxAkx??????????????環(huán)系統(tǒng)誤差減小。1,1,0]。 system is observable 39。 P1=[0,0,0]。r a n kr a n k o ????????? CACW? ? ????????2121 llLkkK ，設(shè)狀態(tài)反饋陣)()(d et)d et (211222121??????????????????????llzllzlzlllzBLAzI對于閉環(huán)極點(diǎn)))(( 2 ????? zzzz相應(yīng)的閉環(huán)特征多項(xiàng)式 21 ?? ll ，解得)()2(1d e t)d e t (211221kkzkzzkkzKCAzI???????????????????對于狀態(tài)觀測器))(( 2 ??????? zzjzjz相應(yīng)的閉環(huán)特征多項(xiàng)式 21 ?? kk ，解得? ? ???????? KL即? ? )()()()()()()()()()1(kxkukykukxkKykBukxKCAkx?????????????????????????????控制器為54 降維狀態(tài)觀測器的設(shè)計(jì) )()()()()1(kCxkykBukAxkx????程為考慮被控對象的狀態(tài)方)(?)(?)()(?)1(?kxCkykBukxAkx????測系統(tǒng)人為構(gòu)造一個(gè)相同的量－由觀測器估計(jì)可直接測量)()()()()(kxkxkxkxkxbaba ????????)
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