《自动控制理论》(夏德钤 翁贻方版)第四版课后习题详细解答答案 下载本文

2 K ; 10

(4)Kp?limG(s)??,Kv?limsG(s)? s?0 s?0 K

,Ka?lims2G(s)?0 s?0200

3-3 设单位反馈系统的开环传递函数为 G(s)? 10 s(0.1s?1) 1 R2t2 2

若输入信号如下,求系统的给定稳态误差级数。

(1)r(t)?R0,(2)r(t)?R0?R1t,(3)r(t)?R0?R1t?解: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)

?? 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 s?0s?0 ds(0.1s2?s?10)3

(1)r(t)?R0,此时有rs(t)?R0,

?s(t)???rrs(t)?0,于是稳态误差级数为 esr?t??C0rs(t)?0,t?0

(2)r(t)?R0?R1t,此时有rs(t)?R0?R1t,级数为 ?s(t)?R1,??rr(t)?0,于是稳态误差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,r 22

?r?(t)?R2,于是稳态误差级数为 s ?s(t)?esr?t??C0rs(t)?C1r

3-4 设单位反馈系统的开环传递函数为 C2?r?s(t)?0.1(R1?R2t),t?0 2! G(s)?10 s(0.1s?1)

若输入为r(t)?sin5t,求此系统的给定稳态误差级数。 解: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?500 s(0.1s?1)?0s?0s?00.1s2?s?500

d500(0.2s?1)1C1?lim?e?s??lim?s?0dss?0(0.1s2?s?500)2500C0?lim?e?s??lim

d2100(0.1s2?s?500)?

1000(0.2s?1)298C2?lim2?e?s??lim?s?0dss?0(0.1s2?s?500)35002 ? 以及 rs(t)?sin5t ?s(t)?5cos5tr ?r?s(t)??25sin5t ?

则稳态误差级数为

C??esr?t???C0?2?25???sin5t??C1?5???cos5t 2?? ??4.9?10?4???sin5t??1?102???cos5t

3-6 系统的框图如图3-T-1a所示,试计算在单位斜坡输入下的稳态误差的终值。如在输入端加入一比例微分环节(参见图3-T-1b),试证明当适当选取a值后,系统跟踪斜坡输入的稳态误差可以消除。 a)

b) 图3-T-1

解:系统在单位斜坡输入下的稳态误差为:esr? 2? ?n

,加入比例—微分环节后 C?s???R?s??1?as??C?s??G?s?

2

??1?as??n1?as?G?s?C?s??R?s??2R?s?2 1?Gss?2??ns??n s2??2??a?n??ns E?s??R?s??C?s??R?s? s2?2??ns??n 2 R?s?? 1 s2 2??a?n esr?limsE?s?? s?0 ?n 可见取a? 2? ?n

,可使esr?0

3-7 单位反馈二阶系统,已知其开环传递函数为 2?n G(s)? s(s?2??n)

从实验方法求得其零初始状态下的阶跃响应如图3-T-2所示。经测量知,Mp?0.096,