自動(dòng)控制原理第三章 時(shí)域分析法

1、 0t0)0()0()0( ccc tAtttr0,0,0)(和和 1)(t t0)()(0tt 時(shí)時(shí)，當(dāng)當(dāng))(t 。即即積積分分面面積積為為，且且定定義義：11)(000)( dttttt t0)(t 1)( tL 0,0,0)(tAttrsAtrL )(stL1)( 1 A 0,0,0)(tAtttr2sAAtL 21stL 1t0A 0,210,0)(2tAtttr3221sAAtL t)(tx032121stL 21)( 1)(23322AtdtdAtdtdtAdtdtA tASintr )(為為角角頻頻率率。為為振振幅幅， A22sin sAtAL0tc(t)( c0tc(t)( c

2、dt0tc(t)( c2)( cdtrtrt0tc(t)( cststptpt)(ptc)(02.0 c)(05.0 c或或pM%100)()()( cctcMppstpt0tc(t)( c)(ptc)(02.0 c)(05.0 c或或A AB B%100 BAMpsrpdtttt,pMstpMAB超調(diào)量超調(diào)量Mp =AB100%)(02.0 c)(05.0 c或或)(sC-Ts1)(sE)(sR11111)()()( TsTsTssRsCsssRttr1)(),(1)( ,11111111)(TssTsTssTssC TteTssLtc 1111)(10tc(t)1632. 01)(1 et

3、cTeTtdtdctTtt11)(00 因因?yàn)闉?()(ttr TteTtctg 1)()(TtTeTttc )()(ttr )(T)(tc0t)(tc)(tr221)(ttr )1(21)(22TteTTtttc 閉環(huán)極點(diǎn)閉環(huán)極點(diǎn)（特征根）（特征根）：- -1/TtTeTtc11)( tTTeTttc1)( tTetc11)( )1(21)(22TteTTtttc )()(ttr )( 1)(ttr ttr )(221)(ttr )(sC-)2(2nnss )(sE)(sR222222)2(1)2()()()(nnnnnnnsssssssRsCs 2222)()()(nnnsssRsCs 無(wú)

4、無(wú)阻阻尼尼自自然然振振蕩蕩角角頻頻率率阻阻尼尼比比n 0222 nnss 122, 1 nns 122, 1 nns222221)()(nnnssssssC 0,cos1)( tttcn 0 njs 2, 1n n 122, 1 nns10 22, 11 nnjs122, 1 nns222222222121)(nnnnnnnnnssssssssssC 0, )11sin(11)(2122 ttgtetcntn 10 22, 11 nnjs0, )11sin(11)(2122 ttgtetcntn n 21 nd1 ns 2, 1122, 1 nns2222)(1121)(nnnnnnsssss

5、ssC )1 (1)(tetcntn 1 122, 1 nns122, 1 nnsssssCnnn1)1()1()(222 )1()1(1211)(2)1(2)1(222 ttnneetc1 122, 1 nns122, 1 nns 1 02468101200.20.40.60.811.21.41.61.820.30.30.50.5ntc( t) )10( n nncos 22, 11 nnjsn 21 ndn 21 nj21 njn j 0tc(t)( cstpt)(ptc)(02.0 c)(05.0 c或或rt%100)()()( cctcMpp)10( 0, )sin(11)(2 tte

6、tcdtn rtt 1)( rtc0)sin( rdt rdt 211 tgtdr，其其中中上上升升時(shí)時(shí)間間rt)10( 增大增大 或減小或減小 ，均能減小，均能減小 ，從而加快系統(tǒng)的初始響應(yīng)速度。從而加快系統(tǒng)的初始響應(yīng)速度。n rt0, )sin(11)(2 ttetcdtn pt0)cos(1)sin(1)(22 pddtpdtntttetedttdcpnpnp tgttgndpd 21)(,.)2 , 1 , 0( , nntpd 21 ndptpt0)cos()sin( pddpdntt增大增大 或減小或減小 ，均能減小，均能減小 ，從而加快系統(tǒng)的初始響應(yīng)速度。從而加快系統(tǒng)的初始響應(yīng)速

7、度。閉環(huán)極點(diǎn)離實(shí)軸越遠(yuǎn)，峰值時(shí)間越小。閉環(huán)極點(diǎn)離實(shí)軸越遠(yuǎn)，峰值時(shí)間越小。n pt 最大超調(diào)量最大超調(diào)量pM%100%100)()()(21 ecctcMpp)(tc21 ndpt221211)sin(11)( eetcp0, )sin(11)(2 ttetcdtn 21sin)sin( n 21 nj21 njn j 211)( etcp最大超調(diào)量只和阻尼比最大超調(diào)量只和阻尼比 有關(guān)，有關(guān)， 越大，越大， 越小。越小。 pM00.10.20.30.40.50.60.70.80.910102030405060708090100 pM%10021 eMp 調(diào)節(jié)時(shí)間調(diào)節(jié)時(shí)間st%)sin(12 te

8、dtn211 tne0, )sin(11)(2 ttetcdtn 10)(tctnst %)1ln(2 %12 snte10)(tctstst 211 tne 112 4912. 3)02. 0ln( 3996. 2)05. 0ln( nst %)1ln(2 時(shí)當(dāng)時(shí)當(dāng)52,3,4nnst8 . 00 調(diào)節(jié)時(shí)間近似與調(diào)節(jié)時(shí)間近似與 成反比。成反比。n 閉環(huán)極點(diǎn)離虛軸越遠(yuǎn)，閉環(huán)極點(diǎn)離虛軸越遠(yuǎn)， 越小。越小。st 1 st0 10 )10( %10021 eMppM 。，一定時(shí)，一定時(shí)，當(dāng)當(dāng)， snnnstt )3(4或8 . 04 . 0 707. 021 )(sR )1( TssK)(sCsTK

9、25. 016 ， pM%16 pMn st2222221)(nnnssTKsTsTKKsTsKs TTKnn122 25. 025. 0162121)/(825. 016KTsradTKn 0.5， 解得%,16%10021eMp45 . 025. 0414122 TK%44%10021 eMp )%5( ,5.1825.033)%2( ,2825.044誤誤差差帶帶取取誤誤差差帶帶取取sstnns 2222)()(nnnsssRsC tetcdtnn sin1)(2 ttcnn sin)( tnntetc 2)()(12)()1()1(222ttnnneetc 1)( sR21)(ssR

10、)2sin(12)(2 tettcdntnn)21(22)(tettcntnnn tntnnnneettc )1(222)1(222221212121212122)( )sin(1)(tttcnn )2()(2)1()()(222222nnnnnnsszzsssssRsC 10 1 zssR1)( )111sin(1)2(1)(221222 aatgteaaatcntnnza )111sin(1)2(1)(221222 aatgteaaatcntn0, )sin(11)(2 ttetcdtn )111sin(1)2(1)(221222 aatgteaaatcntn0, )sin(11)(2 t

11、tetcdtn )111sin(1)2(1)(221222 aatgteaaatcntn)2(2nnss )(sR)(sC 1 s 222222222)2( 2) 1(2) 1()2() 1(1)2() 1()()()(nnnnnnnnnnnnddssssssssssssssRsCs )(sCd2nd 2222)(nnnsss 222)2( 2) 1()(nnnndssss TKsTKsTKKsKTsKsKTssKsKTssKs 1)1()1(1)1()(22)1( TssKs )(sR)(sC5 .01 TKsTKsTKs 1)(2 TKTKn 12112n1則： KTTKn21 加加微微分

12、分負(fù)負(fù)反反饋饋前前 KTKTKn2111 n K 1%16%100%21111 eP25. 0 5 . 01 0625. 01611, 21 KK 求得：)%2()( 185 . 04411誤誤差差帶帶對(duì)對(duì)應(yīng)應(yīng)stns nnnnmmmmasasasabsbsbsbs 11101110.)(nnnsspszsksnjnknknkkjmiig 21112212,)2()()()(12 2112221021)(1)()(nknknkkknkknkkknjjjssCsBpsAsAsssC teCteBeAAtcknktnkkknktnkknjtpjnkknkkj2121101sin1cos)(221

13、nkkjp ,kkjCBA,kkjCBA,)(2()()(222pssszssnnn z p n dj dj 55 nnpz 以以及及)2()(222nnnsspzs pzpssszsssssCtcnnnsst )(2()(1lim)(lim)(lim22200 簡(jiǎn)簡(jiǎn)化化前前pzpsszssssCtcnnnsst )2(1lim)(lim)(lim22200 簡(jiǎn)化后簡(jiǎn)化后0)(lim tgt)()(.)()(11101110sDsBasasasabsbsbsbsRsCnnnnmmmm )()()()(110jjjjKikjijsjspsasB )sincos()(11tBtAeectcjjrj

14、jjtkitpiji kirjjjjjjjiijsjsspscsRsDsBsC11)()()()()()( j 0, 00110aasasa 10,aa0202112, 1212024, 0aaaaasasasa 210,aaa01110 nnnnasasasa01110 nnnnasasasa1321321321531420gdddcccbbbaaaaaa04321ssssssnnnnnaaaaaaaaabaaaaaaaaabaaaaaaaaab 1321321321531420gdddcccbbbaaaaaa04

15、321ssssssnnnnn 11231121311bababbbbaac 11351131512bababbbbaac 11471141713bababbbbaac 1321321321531420gdddcccbbbaaaaaa04321ssssssnnnnn 0322130 asasasa0123ssss003130213120aaaaaaaaaa 3210,aaaa03021 aaaa054322345 sssss012345ssssss0050093205905 . 15 . 0532411 - -1 3 0（ 2） 1 0 0（ ） 329 0122234 ssss 22 0 00

16、1002202)( 002211101234sssss 0161620128223456 ssssss3456ssss000001612201612216208108624 ss01243 ss033 ss0123ssss831830310161620128223456 ssssss0)4)(2(22 ss08624 ss2,24,32, 1jsjs 設(shè)系統(tǒng)特征方程為：設(shè)系統(tǒng)特征方程為：s4+5s3+7s2+5s+6=0勞勞 斯斯 表表s0s1s2s3s451756116601 勞斯表何時(shí)會(huì)出現(xiàn)零行勞斯表何時(shí)會(huì)出現(xiàn)零行?2 出現(xiàn)零行怎么辦出現(xiàn)零行怎么辦?3 如何求對(duì)稱(chēng)的根如何求對(duì)稱(chēng)的根? 由零

17、行的上一行構(gòu)成由零行的上一行構(gòu)成輔助方程輔助方程: 有大小相等符號(hào)相反的有大小相等符號(hào)相反的特征根時(shí)會(huì)出現(xiàn)零行特征根時(shí)會(huì)出現(xiàn)零行s2+1=0對(duì)其求導(dǎo)得零行系數(shù)對(duì)其求導(dǎo)得零行系數(shù): 2s1211繼續(xù)計(jì)算勞斯表繼續(xù)計(jì)算勞斯表1第一列全大于零第一列全大于零,所以系統(tǒng)穩(wěn)定所以系統(tǒng)穩(wěn)定錯(cuò)啦錯(cuò)啦! 求解輔助方程得求解輔助方程得: s1,2=j由綜合除法可得另兩由綜合除法可得另兩個(gè)根為個(gè)根為s3,4= -2,-30.1110 nnnnasasasa00 a1ananaaaaaaaaaaaaaaa000000000042053164207531 nn 043223140 asasasasa4203142031

18、000000aaaaaaaaaa 0, 030212031211 aaaaaaaaa0, 000431420313 aaaaaaa05432234 ssss56514253101234 sssss)40( , 0 iaiKsssKsssKsssKs 158)5)(3(1)5)(3()(23015823 Ksss)5)(3( sssK )(sR)(sCKKKssss812081510123 0 K1200012008120 KKKK有120 LK015823 Ksssaas aazs aa068523 sss022, 06) 1( 8) 1( 5) 1(2323 zzzzzz即022110123

19、zzzz1z1z0222 z12, 1jz 輔輔助助方方程程的的解解為為1 zssTsTsTTsTCsGmmme01. 0,048. 0) 1(1)(2 0,58 . 401. 0048. 0 TTTm) 1(1) 1(1)(2 sTCsTTsTCsGmemme) 1(12 sTTsTCmme1K)(sR)(sC sK2)1()(221 sTTsTsCKKsGmmeemmeCKKssTTsTCKKs/)(212321 emmeCKKssTTsTCKKs/)(212321 1001/21 TCKKeemeemmeCKKssTCKKCKKssTTsTCKKs/)(21221212321 120,0

20、1. 0,048. 0,121 KKTTCme50,01. 0,048. 0,121KKTTCme niinssssssssese1212121)(1)1()1)(1(11)(11111 )()()(0tctct )(0tc)(tc)(limttss ss - -)(s )(0sC)(1sG)(2sG)(sH)(sR)(sN-+)(sC)(sE)()()(tbtrte )(tr)(tb)(limteetss sse- -)(s )(0sC)(1sG)(2sG)(sH)(sR)(sN-+)(sC)(sE)(sB)(sE)()(0tctr )(trsssse - -)(s )(0sC)(1sG)(

21、2sG)(sH)(sR)(sN-+)(sC)(sE)(sB)(sE- -)(s )(0sC)(1sG)(2sG)(sH)(sR)(sN-+)(sC)(sE)(sB)(sE)()(0tctr )(trsssse )()()()()()()()()(0ssHsCsHsCsHsBsRsE )()()(0sCsHsR 0)( sE- -)(s )(0sC)(1sG)(2sG)(sH)(sR)(sN-+)(sC)(sE)(sB)(sE)(sE)(s )(sR)(sN)(sC)(2sG)(1sG)(sE)(sH)(sB)(11)()()(11)()()(21sGsHsGsGsRsEskE )(sH)(sR

22、)(sB)(sE)(1sG)(sC)(2sG)(sR)(sN)(sC)(2sG)(1sG)(sE)(sH)(sB)(1sG)(2sG)(sH)(sC)(sB)(sN)(sE1)()()(1)()()()()(212sHsGsGsHsGsNsEsNE )()()()()(sNssRssENEE )()()(1)()()()()()(1)(21221sHsGsGsNsHsGsHsGsGsR )(sR)(sN)(sC)(2sG)(1sG)(sE)(sH)(sB)()()(1)()()()(212sHsGsGsGsNsCsNN )(11)(21sRHGGsE )(1sG)(2sG)(sH)(sR)(s

23、N-+)(sC)(sE)(lim)(lim0ssEteestss ) 12)(1() 15 . 0(ssssK)(sR)(sC0)5 . 01(3223 KsKss60 K)15 . 0()12)(1()12)(1()(11)()()( sKsssssssGsRsEskE21)(ssR 21)15 . 0()12)(1()12)(1()(ssKsssssssE KssKsssssssssEessss11)15 . 0()12)(1()12)(1(lim)(lim200 61 sse1 . 0 sse)(lim)(lim0ssEteestss 22)( ssR)(11)(11)(21sRGsRH

24、GGsEk )(1)(lim)(lim)(lim00sGssRssEteeksstssr ssre)()12()1()12()1()(01211212121sGsKsTsTsTssssKsGnllllnjjmkkkkmiik K )(0sGnnnmmmG 212102,2,1)0( KssRssGsKssRsGssRevvsvsksssr )(lim)(1)(lim)(1)(lim10000)(sGkKssRssGsKssRsGssRevvsvsksssr )(lim)(1)(lim)(1)(lim10000012)(lim0sGKksp ,)(lim00KsKGKsp ，時(shí)當(dāng)1 ssR1)(

25、pksksssrKsGsGssRe 11)(lim11)(1)(lim00)(lim)(lim000sGsKsGKsksp ，時(shí)當(dāng)0 Kessr 11,)(lim00 sGsKKsp 0 ssrepKpKssepKKeKsKGKssrsp 11,)(lim000，時(shí)當(dāng) 0,)(lim100 ssrspesGsKK ，時(shí)當(dāng)1 0 KeKsKGKssrsp 11,)(lim000，時(shí)當(dāng) 1 0 21)(ssRvksksssrKsGssGssRe1)(lim1)(1)(lim00 ，時(shí)當(dāng)0 ，時(shí)當(dāng)1 ，時(shí)當(dāng)2 )(lim0ssGKksv )(lim)(lim)(lim010000sGsKsGssKs

26、sGKssksv , 0)(lim00 ssKGKsv ssre,)(lim00KsKGKsv Kessr1 ,)(lim010 sGsKKvsv0 ssre ssrsvessKGK,0)(lim000，時(shí)當(dāng) KeKsKGKssrsv1,)(lim100 ，時(shí)當(dāng) 0,)(lim2010 ssrvsvesGsKK，時(shí)當(dāng) vKssevKvKKeKsKGKssrsv1,)(lim00 型系統(tǒng)，型系統(tǒng)，21)(ssR31)(ssRaksksssrKsGssGssRe1)(lim1)(1)(lim200 ssrsaesKGsK,0)(lim0020，時(shí)當(dāng) KeKsKGKssrsa1,)(lim200 ，

27、時(shí)當(dāng) 0,)(lim3020 ssrvsaesGsKK，時(shí)當(dāng) )(lim20sGsKksa )(lim)(lim)(lim02002020sGsKsGsKssGsKssksa ssrsaesKGsK,0)(lim1010，時(shí)當(dāng) KeKsKGKssrsa1,)(lim200 ，時(shí)當(dāng) 0,)(lim300 ssrsaesGsKK，時(shí)當(dāng) aKsseaKaK ssrsaesKGsK,0)(lim0020，時(shí)當(dāng) ssrsaesKGsK,0)(lim1010，時(shí)當(dāng) KeKsKGKssrsa1,)(lim00 型系統(tǒng)，型系統(tǒng)，31)(ssR時(shí)，時(shí)，當(dāng)當(dāng)2)(2CtBtAtr avpssrKCKBKAe 1有有pKR 1vKRaKR)()(11)()()(sHsGsRs