自動(dòng)控制原理課后答案

經(jīng)典word整理文檔，僅參考，雙擊此處可刪除頁(yè)眉頁(yè)腳。本資料屬于網(wǎng)絡(luò)整理，如有侵權(quán)，請(qǐng)聯(lián)系刪除，謝謝！.RK21R2解2111Rr1sC1R111212sC1Rs1RC2sC222Ts11RRr2212sC12c(t)e(t)ssrt)sint1解s21222j241)21s1,1G(s)s2e12,則2ee42m.>.12(jarctan()45(j)則2281rrt)sin(t30)2cos(2t45)11r2m2211.80.81解C(s),R(s)ss4s9s(s4)(s9)sC(s)R(s)(s)則(s4)(s9)(j)(jj9)KK解G(j)ej()2j3KKj()(3)G(j)e2(j)330.52))。和解G(j)H(j)j0.5)jj2217.8885(2)0.3835AA153.435(2)327.535G(s)(2s8ss)s2G(s)5G(j)2))22.>.G(j)00G(j)G(j)G(j)G(j)G(j)180012G(j)2,G(j)1800G(j),G(j)900G(j)0K(TsG(s)K,T,T0；2sTs1211G(j)180，當(dāng)G(j)G(j)。KTs解G(s)2sTs1KTsG(s)G(j)2sTs01G(s)G(s)0KlimsG(s)Ke1K1K1。，vs0jTjT)1TTjTT)TT)G(0.5121221121T1T1T222122T2T1T0.5K1，，。212.>.1jG(j)jj0.5)解G(j)00s2G(s)(2s8s；200G(s)；s(ss240(sG(s)s(s0.2)(ss220(3sG(s)s(6ss4s25)(10s228(sG(s)s(sss4s222G(s)(2s8s200G(s)s(ss240(s100(2sG(s)s(s0.2)(sss2s(ss20.220(3sG(s)s(6ss4s25)(10s220.81s18(s250.1G(s)s(sss4s22214s(ssss2525.>.G(s)為G(s)01K證。svK1Kvsv1KK11,故20lg0K,Kvv(j)v11v1圖5－785－11題圖KG(s)ss(12lgKL()40，K100則：G(s)11(ss12sK(G(s)K21s01Cs(22KsG(s)ss(23G(s)、G(s)和G(s)123L()20lgK11.>.G(s)K11KG(s)2s2s(0.8K2120lgK/20lg0K1,22L()20lgK20lgK0333GG1G(s)21GG42318G,G,G1G(s)將s(0.125s234解題號(hào)Z備注NPPK1123K23400012K(TT)0002122K567s312000051234K8901KTs11011.>.K12K；10KTK,TKKT)jT)解G(j)jjjT)21X)T1或TK132T；21；9K,T解G(j)0令G(j)ReG(j)0s2解(s40G(j)0.>.G(j)G(j)T0.21()1T20G(j)的繪出幅相特性曲線G(j)100G(s)s(0.2s50G(s)(0.2ss2)(s10G(s)s(s25ss100(2G(s)sss(s1020100100G(s)解s(0.2sss(55Cg5050G(s)(0.2ss2)(sss(2s52.>.6cccc110ggg1010ss400Cgh1s2sscg180()24.8ch0.3439.3(dB)解102即：a242cc00c1000ccca0.84，。c。Kh.>.解()1tgK18090180tg0010cchccKj)G(j)H(j)u)jv)hj(j2K)10(Ku();v()hh(2(2令v()0,則10(K21K02hh1KhK)又令u()1h(21K)(Khh91,K1KK10hhh10故當(dāng)K110hKesG(s)s1KG()ej1KG(j)1210.8tg()()ej11KG(j)1即12()G(j)0.81.>.由(2)式tg18K11[(0.8)]22.45,K2.65:c2G(j)1解令)221804tg1