自動控制原理期末考試題3

1、第三章 時域分析法習題及解答111min從 T 1min, T=0.25min1h(t )=1-e t 0.1, t =T ln0.9T 1111t， ln0.1h(t )=0.9=1-etTT 222t t t TT r21r(t)t)tC(t) (t 0.9)0.9e 10t1 C(s) =ss s s 221 1 s1 s ss22C(s)(s)R(s) st001234567t( )hK(s)1 設K11C(s)(s)R(s) K( )1s1)ssT1ht) K Ke th() K 6T161ht)66e TT66 T (s)3.2s1a KKTsT Ka s11KC(s)(s)R(s

2、) s T Ka)s1111s1)11K(1-e)ta0當T2ns22 s2n1 0.707，n4 2nn0.7071, 2n00211s(ssb (s)s 102s21s(sD(s)= s 102s102 KG(s) s(0.01s2 sK 和 D(s) s(0.01s s1)k 02D(s) s s 100s100k 032Routh: s31100100ks220 0 100k10s0k 0 0 0kk k k 由Routh 0得 3rad snaa 和 、 KK1KK1K1s2 s(sa) (s2)(sa) s2 s(sa)D(s) s(s2)(sa)s(sa)K(sa)KsK s (

3、3a)s as3K 022D( ) j a j a K 03 332nnnn ( 3 0 D jaK2nnImD( 03nnna318K t)、t和2tG(s) ss7(sG(s) s(s4)(s2 2s2)8(0.5sG(s) s2(0.1sK 1010G(s)ss1)0v D(s)(0.1s1)(0.5s1)100rt)t)A1e1K rt)t rt)t2e e73 217(s3)KG(s)42 8s(s4)(s22s2)1vD(s) s(s4)(s 2s2)7(s3)028(0.5s1)K 8G(s)s s1)22vrt)t) rt)tee0rt)t22 1 K 8 4Ae G(s) s

4、sr(t)12tK 100100G(s)ss1)1vD(s) s(0.1s1)100 s s10002e 0ss1A21e 2K 1e e e 12 E(s) R(s)C(s)b s b sm1 b sbm(s) mm110s a s a sann1n mn110a b abE(s) R(s)C(s)和。1001E(s) R(s) C(s)1(s) (s)R(s) R(s) R(s)eb s b s L bs bmm1 ( ) 1 ( ) 1 s s mm110s a s L as aenn1n110s a s L b s b smm1a b s a bL ( ) nn1n1mm11100s

5、a s L as ann1n1101rt)tR(s)s2e sE(s)s (s)R(s)es0s0s a s1L b s b smm1a b s a bL ( ) nn1lim n1mm11100ss a s L as ann1s0n110e 0ssa b 0,a b 0a b ,a b11100100 n(t)的控制系統(tǒng)如圖3-56n(t) N t)K2E(s)N(s)T s1K Ts(s)221K1( 1 )T s2TsK111Tss1Nnt) N t) N(s)00se lims (s)N(s)s0K Ts1)NK Nlims21020Ts1K T s1) sK 1s01121 E(s)

6、R(s)1s(0.5s ( )s 0.5s se211s(0.5sE(s)N(s)1s(0.5s ( )s 0.5s s2s(0.5ss s N se e e lims (s)R(s) ( ) ( )12es0s0s(0.5s 1 s0.5s s ss(0.5s s00.5s s s22s0s0s1s(s2(s)11s s12es(s1)1s(s(s)1s s121s(ss(s 1s(s 1 1s1 s2e e e s limss1 ss212s22s0s0C(t) e 0Kv T 、 與 saG(s)s vK(sa)s K(sa)v(s)sa v1 s s a1vs vv2K(sa)1C()s(s)R(s)s K s sa sv1vs0s01s Ts1)v(s)sa Tsv1 s s aev1s Ts1)vs 1ve lims 0 s sa sv1vs0v1 n(t)5(t)rt)t(rad s)0.2(rad)。G (s) G (s)cc(ss(s)G (s)(ssG (s)1cc(ss105e lims (s)N(s)lims(ss2)10G (s) sssnens0s0cksG (s) e 0 D，ssn(s) s 3s 2s10k32取kc1(ss(s)G (s)(ssG (s)e1cc(sse lims (s)R(s)es0(ss2)2lims