自动控制理论_第三版_课后答案

《自动控制理论 第3版》习题参考答案

第二章

2-1 (a)

U2?s?R1R2CS?R2R2R1CS?1 ???U1?s?R1R2CS?R1?R2R1?R2R1R2CS?1R1?R2(b)

U1?s?1 ?2U2?s?R1R2C1C2s?(R1C1?R1C2?R2C2)s?12-2 (a)

U?s?RU2?s?RCs?1U?s?R?R1???1? (b) 2 (c) 2???1?1Cs?1? U1?s?RRU1?s?RCsU1?s?R?4?Cs?142-3 设激磁磁通??Kfif恒定

Cm???s??

60Ua?s???s?LaJs2??Laf?RaJ?s?Raf?Ce?Cm??2???2-4

C?s??R?s?KACm?

60??iLaJs3?i?Laf?RaJ?s2?i?Raf?Ce?Cm??s?KACm?2???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) ??R?s?1??G1?H1?G3R?s?1?G1G2H1??G2G3?G4??H2?G1H3?R(s) + 2-9 框图化简中间结果如图A-2-1所示。

0.7_ + + 1s2?0.3s?10.161.2?2sC(s) Ks

图A-2-1 题2-9框图化简中间结果

C?s?0.7s?0.42?3 2R?s?s??0.9?0.7k?s??1.18?0.42k?s?0.522-10

G1G2G3C?s???G4

R?s?1?G2H1?G1G2H1?G2G3H22-11 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

G1G2G3C1?s??R?s?1?G1G2?G4?G1G2G4G5H1H2G1G2G4G5G6H2C2?s??R?s?1?G1G2?G4?G1G2G4G5H1H22-12 (a)

R2C?s?1?abcdef?agdef?abcdi?adgi? (b) C?s??? 2??Rs1?cdhR?s?R1C1R2C2s??R1C1?R2C1?R2C2?s?12-13 由选加原理,可得

C?s??

1?G1G2R?s??G2D1?s??G2D2?s??G1G2H1D3?s??

1??H1G1?H2?G2第三章

3-1 分三种情况讨论 (a) 当??1时

s1?????2?1?n,s2?????2?1?n?2???????1??nt????????2?1????nt? ??2?1ee??c?t??t???222??n2?2?1?n????2?1????1???????????(b) 当0???1时

s1????j1??2?n,s2????j1??2?nc?t??t??t?2?2?????n??2??n12e???ntcos1???nt?2???nt1?2?21??2?ne???ntsin1??2?nt

?????n1???ne2?2?1??2sin?1???nt?arctg2?1?2??(c) 当??1时

s1,2???nc?t??t?2?n????e??nt?1?nt??n2??2

设系统为单位反馈系统,有

Er?s??R?s??c?s??R?s?s?s?2??n? 22s?2??n??n系统对单位斜坡输入的稳态误差为 esr??ims?s?0s?s?2??n?12? ??222ss?2??ns??n?n3-2 (1) Kp?50,Kv?0,Ka?0 (2) Kp??,Kv?K,Ka?0

(3) Kp??,Kv??,Ka?KK,Ka?0 (4) Kp??,Kv?102003-3 首先求系统的给定误差传递函数

?e?s??误差系数可求得如下

E(s)1s(0.1s?1) ??2R(s)1?G(s)0.1s?s?10s(0.1s?1)?02s?0s?00.1s?s?10d10(0.2s?1)C1?lim?e?s??lim?0.1s?0s?0ds(0.1s2?s?10)2C0?lim?e?s??limd22(0.1s2?s?10)?20(0.2s?1)2C2?lim2?e?s??lim?0s?0s?0ds(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 sesr?t??C0rs(t)?0，t?0

(2) r(t)?R0?R1t，此时有rs(t)?R0?R1t,?s(t)?R1,??rr(t)?0，于是稳态误差级数为 s?s(t)?0.1R1，t?0 esr?t??C0rs(t)?C1r(3) r(t)?R0?R1t?11?s(t)?R1?R2t，?R2t2，此时有rs(t)?R0?R1t?R2t2,rr?(t)?R2，于是稳态误差级数为 s22?s(t)?esr?t??C0rs(t)?C1r3-4 首先求系统的给定误差传递函数

C2?t?0 r?s(t)?0.1(R1?R2t)，2!?e?s??误差系数可求得如下

E(s)1s(0.1s?1)?? 2R(s)1?G(s)0.1s?s?500s(0.1s?1)?020.1s?s?500d500(0.2s?1)1C1?lim?e?s??lim?s?0s?0ds(0.1s2?s?500)2500C0?lim?e?s??lims?0s?0d2100(0.1s2?s?500)?1000(0.2s?1)298C2?lim2?e?s??lim?s?0s?0ds(0.1s2?s?500)35002??

rs(t)?sin5t?s(t)?5cos5tr?r?(t)??25sin5ts稳态误差级数为

C??esr?t???C0?2?25???sin5t??C1?5???cos5t 2????4.9?10?4???sin5t??1?102???cos5t3-5 按技术条件（1）~（4）确定的二阶系统极点在s平面上的区域如图A-3-1 (a) ~ (d)的阴影区域。

图A-3-1 二阶系统极点在s平面上的分布区域

3-6 系统在单位斜坡输入下的稳态误差为 esr?加入比例—微分环节后

2??n

C?s???R?s??1?as??C?s??G?s??1?as?G?s?R?s???1?as??C?s??R?s?1?G?s?s?2??s??s??2??a???sE?s??R?s??C?s??R?s?

2n22nn2ns?2??ns??n2n2R?s??1s22??a?nesr??imsE?s??s?0?n可见取a?2??n，可使esr?0

3-7

??0.598,?n?19.588

3-8 G?s??4 2ss?4s?6??3-9 按照条件（2）可写出系统的特征方程

(s?1?j)(s?1?j)(s?a)?(s2?2s?2)(s?a)?s?(2?a)s?(2?2a)s?2a?0将上式与1?G(s)?0比较，可得系统的开环传递函数

32

G(s)?根据条件（1），可得

2a

s?s2?(2?a)s?(2?2a)?Kv?解得a?1，于是由系统的开环传递函数为

12a ?0.5?esr2?2aG(s)?2

s?s2?3s?4?

?1?M3-10

?2?Mpp?46.6%,ts?7.99s?2%?,(?n?2.12rad/s,??0.24)?16.3%,ts?8s?2%?,(?n?1rad/s,??0.5)?3?ts?15s,(?n?0.4rad/s,??1.25)，过阻尼系统，无超调。

3-11 （1）当a = 0时，??0.354,?n?22。

（2）?n不变，要求??0.7，求得a = 0.25 3-12 1. 单位脉冲响应 (a) 无零点时 c?t???n1??2e???ntsin1??2?nt,?t?0?

（b）有零点z??1时

c?t??1?2??n??n??n1??22e???nt2?1???n2sin?1???nt?arctg?1???n???,?t?0? ??1??2?n比较上述两种情况，可见有z??1零点时，单位脉冲响应的振幅较无零点时小，而且产生相移，相移角为arctg。

1???n2．单位阶跃响应 (a) 无零点时

c?t??1?11??2e???nt2?1??2sin?1???nt?arctg?????,?t?0? ??（b）有零点z??1时

c?t??1?1?2??n??n1??222?1??e???ntsin?1??2?nt?arctg??n?????,?t?0? ??

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