求K,v,T??
解:
?s?a?开环增益?a G(s)?v ?r?1(t)s(Ts?1)??v:系统型别ess?0:v?1K(s?a)K(s?a)sv(Ts?1) ?(s)??vs?as(Ts?1)?s?a1?vs(Ts?1)D(s)?Tsv?1?sv?s?a 系统响应收敛、稳定,则v?3
由图:h(?)?10?lims..s?01K(s?a)?K?K?10 vss(Ts?1)?s?aK(s?a)sv(Ts?1)?s?ah?(0)?k(0)?10?lims.?(s)?lims.s?0s?? ?limKs?Kass??Tsv?1?sv?s?aKs?10,v?1Ts22
当T?0时得:lims???v?1? ?KK?10?10????T?1??T?v?2Ks2?10 当T?0时得:lim ?s??svT?0??得: ??(1)K?10, v?1, T?1
?(2)K?10, v?2, T?0例 系统如右 求:
?r(t)?1(t)(1) ?时,ess??
K?1?1(2) n(t)?1(t)时,能否选择K1使essn??0.099? 解:
10K1(1) 开环传函 G(s)?(0.1s?1)(0.2s?1)(0.5s?1)?K?10K1 ?v?0?利用静态误差系数法;
K1?1A11ess???1?K1?10K111?10E(s)?10(0.1s?1)(0.2s?1)(0.5s?1) (2) ?en(s)???10KN(s)1?(0.1s?1)(0.2s?1)(0.5s?1)?10K11(0.1s?1)(0.2s?1)(0.5s?1) ??1000
(s?10)(s?5)(s?2)?1000K1D(s)?(s?10)(s?5)(s?2)?1000K1?s3?17s2?80s?100(10K1?1)
劳斯:
s3 1 80s2 17 100(10K1?117?80?100(10K1?1) ?K1?1.2617?1s0 100(10K1?1) ?K1???0.110s1 ?K1稳定范围?0.1?K1?1.26
1?1000?10令ess?lims.???0.099 s?0s(s?10)(s?5)(s?2)?1000K11?10K110?10.099解出:K1=?10——不在10稳定范围之内,故不能选K1使
ess=?0.099
第三章 小结复习题
1. 系统结构图如图 求:
C(s)C(s) ,R1(s)R2(s)解:
??1??Li?L1L313 ?1???G1?G1G2?G2G3?(?G1)(?G2G3)? ?1?G1?G1G2?G2G3?G1G2G3对于R1?C:
P1?G1 ?1?1?G2G3 P2??G1G2 ?2?1 P3??G1G2G3 ?3?1
?CG1(1?G2G3)?G1G2?G1G2G3G1?G1G2 ??R1??对于R2?C:
P1?G2 ?1?1 P2?G2G3 ?2?1?G1
?CG2?G2G3(1?G1)G2?G2G3?G1G2G3?? R2??2. 已知系统结构图,及单位阶
?tp?1''跃响应的指标:?
?%?16.3%?求(1)G(s)?? (2)?(s)??
(3)由tp,?%确定K和?
(4)r(t)?1.5t时ess?? 解:依图:
1010K10K①G(s)?K.s(s?1)? (1) ?开环增益K0?10?1?10?s(s?1)?10?s1?s(s?1)② ?(s)?③ 得
G(s)10K ?21?G(s)s?(1?10?)s?10K??n?10K K??n210? ?2??n?1 (2) (1?10?)?? ???10210K?依题:tp??1??2?n?1??n=?1??2=3.14=3.63(弧/秒) 0.866 ?%?16.3%??=60???=0.5
3.632?1.32 代入(2)式:K?10 ??2?0.5?3.63?1?0.263''
10④ 由⑴知:v?1—一型系统
K0?10K10?1.32??3.64 1?10?1?10?0.263又?r(t)?1.5t A?1.5
Kv?K0
?ess?A1.5??0.413 Kv3.643. 系统结构图如右
1) 希望系统所有特征根位于s??2的左侧区域,且??0.5,画出特征根在
s平面的分布范围(用阴影线表
示)
2) 求出满足上述条件的K,T的取值范围 3) r(t)?t,求ess??
4) 为使上述稳态误差为0,加一比例微分环节,试求使ess?0的Kc?? 解:①
?(s)?KKT? (1)1Ks(Ts?1)?Ks2?s?
TT ? 代换:s*?s?2 s?s*?2②
D(s)?s2?1K1Ks??(s*?2)2?(s*?2)? TTTT1K?2?s*2?(?4)s*?(?4) (2)
TT满足s落在s=-2左侧时,应有:
11?4?0?T? T4K?2?4?0?K?4T?2?0 T K??4T?2
???n=依⑴: ??????2?KT (3)
1?0.5KT由⑶:KT?1?K?
?v?1?AA1?? ③ 依结构图,对系统:??ess?KvKK?K?K开环增益?1T对输入:A?1
④ 依题:E的位置应在右图所
示位置,此时不能直接用误