自动控制理论第三版课后习题答案 夏德钤翁贻方版.docxVIP

自动控制理论第三版课后习题答案 夏德钤翁贻方版 自动控制理论第三版课后习题答案lpar;夏德钤翁贻方版rpar; 《自动控制理论 第3版》习题参考答案 2-1 (a) U2?s?R1R2CS?R2R2R1CS?1 RRU1sR1R2CS?R1?R2R1?R212 U2sR1R2C1C2s2?(R1C1?R1C2?R2C2)s?1 2-2 (a) U?s?RU2?s?RCs?1U?s?R?R1? ??1? (b) 2 (c) 2???1?1Cs?1? U1sRRU1sRCsU1sR?4?Cs?1 2-3 设激磁磁通??Kfif恒定 Cm???s?? 60Uas?? s?LaJs2??Laf?RaJ?s?Raf?Ce?Cm?? RsKACm? iLaJs3?i?Laf?RaJ?s2?i?Raf?Ce?Cm??s?KACm? 2-5 id?2.19?10?3?0.084?ud?0.2? 2-8 (a) ?G1?G2?G3G1?G2G3?G4?C?s?C?s? (b) ?? Rs1?G1?H1G3Rs1?G1G2H1?G2G3?G4H2?G1H32-9 框图化简中间结果如图A-2-1所示。 图A-2-1 题2-9框图化简中间结果 C?s?0.7s?0.42 Rss?0.9?0.7ks2?1.18?0.42ks?0.52 G1G2G3C?s???G4 Rs1?G2H1?G1G2H1?G2G3H2 2-11 系统信号流程图如图A-2-2所示。 图A-2-2 题2-11系统信号流程图 C1?s?G1G2G3Rs? 1?G1G2?G4?G1G2G4G5H1H2C 2?s?G1G2G4G5G6H2Rs? 1?G1G2?G4?G1G2G4G5H1H2 ?abcdef?agdef?abcdi?adgi?C?s?Rs? R2R 1C1R2C2s2?R1C1?R2C1?R2C2s?1 2-13 由选加原理,可得 1?HG?G1G2R?s??G2D1?s??G2D2?s??G1G2H1D3?s?? 11?H2G2 3-1 分三种情况讨论 (a) 当??1时 s1?????2?1???2n,s2?????1? ???????1?????????????1???ntc?t?t?2?1?nt???ee?? ?? 21?2?2n2?2n????2?1???1?? (b) 当0???1时 s1?????j??2??n,s2??? ??j??2? ?nc?t??t?2? 1?2?2??nt?? ?cos???nt? ??2?e?sin??2?nt ???sin?nt ??nt?arctg?2 (c) 当??1时 s1,2???n c?t??t? e??nt??1??n? ?t?n?2?(b) 设系统为单位反馈系统,有 Er?s??R?s??c?s??R?s? s?s?2??n? 22 s?2??n??n s?s?2??n?12? s2s2?2??ns??n2?n 系统对单位斜坡输入的稳态误差为 esr??ims? 3-2 (1) Kp?50,Kv?0,Ka?0 (2) Kp??,Kv?K,Ka?0 (3) Kp??,Kv??,Ka? ,Ka?0 (4) Kp??,Kv? 3-3 首先求系统的给定误差传递函数 误差系数可求得如下 E(s)1s(0.1s?1) ?? R(s)1?G(s)0.1s2?s?10 s(0.1s?1) ?02s?0s?0 0.1s?s?10 d10(0.2s?1) C1?lim?e?s??lim?0.122s?0s?0 ds(0.1s?s?10)C0?lim?e?s??lim d22(0.1s2?s?10)?20(0.2s?1)2 C2?lim2?e?s??lim?0 ds(0.1s2?s?10)3 (1) r(t)?R0，此时有rs(t)?R0, ?s(t)??rr?(t)?0，于是稳态误差级数为 s esr?t??C0rs(t)?0，t?0 (2) r(t)?R0?R1t，此时有rs(t)?R0?R1t,?s(t)?R