自动控制理论(夏德钤)第3版 课后答案

《自动控制理论(夏德钤)第3版 课后答案》由会员上传分享，免费在线阅读，更多相关内容在教育资源-天天文库。

1、《自动控制理论第3版》夏德钤版习题参考答案第二章UsRRCSRRRCS12122212-1(a)U1sR1R2CSR1R2R1R2R1R2CS1RR12Us11(b)2UsRRCCs(RCRCRC)s121212111222U2sRCs1U2sR11U2sR1R12-2(a)(b)(c)Cs1U1sRCsU1sRRCs1U1sR442-3设激磁磁通Ki恒定ffsCmUas260sL

2、JsLfRJsRfCCaaaaem2CsKCAm2-4Rs3260iLJsiLfRJsiRfCCsKCaaaaemAm232-5i2.19100.084u0.2ddCsG1G2G3CsG1G2G3G42-8(a)(b)Rs1GHGRs1GGHGGGHGH1131212342132-9框图化简中间结果如图A-2-1所示。R(s)++1C(s)0.72s0.3s1_+0

3、.161.22sKs图A-2-1题2-9框图化简中间结果Cs0.7s0.4232Rss0.90.7ks1.180.42ks0.52CsG1G2G32-10G4Rs1GHGGHGGH211212322-11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CsGGG1123Rs1GGGGGGGHH124124512CsGGGGGH2124562Rs1GGGGGGGHH124124512Cs1CsR

4、abcdefagdefabcdiadgi22-12(a)(b)2Rs1cdhRsRCRCsRCRCRCs111221121222-13由选加原理,可得1CsGGRsGDsGDsGGHDs12212212131HGHG1122第三章3-1分三种情况讨论(a)当1时22s1,s11n2n221nt1nt21eectt2222

5、2n21n11(b)当01时22sj1,sj11n2n2ctt22entcos12t12entsin12tnn2nn1n21212tentsin12tarctgn22121nn(c)当1时s1,2n22cttent1nt设系统为单位反馈系统,有nn2ss2EsR

6、scsRsnr22s2nn1ss22n系统对单位斜坡输入的稳态误差为eimssr222s0ss2snnn3-2(1)K50,K0,K0(2)K,KK,K0pvapvaKK(3)K,K,K(4)K,K,K0pvapva102003-3首先求系统的给定误差传递函数E(s)1s(0.1s1)se2R(s)1G(s)0.1ss10误差系数可求得如下s(0.1s1)Climslim00e2

7、s0s00.1ss10d10(0.2s1)Climslim0.11s0dses0(0.1s2s10)2222d2(0.1ss10)20(0.2s1)Climslim02s02es023ds(0.1ss10)(1)r(t)R，此时有r(t)R,r(t)r(t)0，于是稳态误差级数为0s0ssetCr(t)0，t0sr0s(2)r(t)RRt，此时有r(t)RRt,r(t)R,r(t)0，于是稳态误差级数为01

8、s01s1setCr(t)Cr(t)0.1R，t0sr0s1s11212(3)r(t)RRtRt，此时有r(t)RRtRt,r(t)RRt，r(t)R，于是稳态误差级数为012s012s12s222CetCr(t)Cr(t)2r(t)0.1(RRt)，t0sr0s1ss122!3-4首先求系统的给定误差传递函数E(s)1s(0.1s1)se2R(s)1