L??c\ ??lg40?lg?1\ 40?20 40lg?c\ ??c\?0.4
40?102?100,
?c\ ?(0.4)?90?arctan00.40.4?arctan?840 5160查惯性环节表,在0.7?c'''?0.28处:???34
84?34?50
以?20dB/dec交0dB线于E:(?E?0.0028),得出滞后校正装置传递函数:
000s?10.28 Gc(s)? s?10.00280.40.4?0??arctan?arctan??34.59c??0.280.0028?\?0.4在c处: ?
1.744?L?20lgG?20lg??38.27dBcc?142.86??s?40??1??0.28? Gc(s)G(s)?
sss??????s??1???1???1??5??16??0.0028?验算:?g\?8.6
40?30.73?33.7dB h??20lgGcG(?g\??208.6?1.99?1.1353?30715.0.40.40.40.4??1800??GcG(0.4)?1800?900?arctan?arctan?arctan?arctan0.285160.0028
000000 ?90?55?4.57?1.432?89.6?50 (满足要求)
s?13.57s?10.28G(s)??因此确定: c
s357s?1?10.0028
5-36 设单位反馈系统的开环传递函数
G(s)?
Ks(s?1)(0.25s?1)
109
要求校正后系统的静态速度误差系数Kv≥5(rad/s),截止频率ωc≥2(rad/s),相角裕度γ≥45°,试设计串联校正装置。
解 在??2以后,系统相角下降很快,难以用超前校正补偿;迟后校正也不能奏效,故采用迟后-超前校正方式。根据题目要求,取
??2, K?Kv?5 ?c?)?180??arctan2?arctan原系统相角裕度 ??180???G(j?c最大超前角
?m???????5??45??0??5??50?
2?90??0? 4查教材图5-65(b) 得: a?8, 10lga?9dB
??2作BC,使BA?AC;过C作20dB/dec线并且左右延伸各3倍频程,定出D、过?cG,进而确定E、F点。各点对应的频率为:
5?2.5
22??0.1?2?0.2 ?E?0.1?c?0.67?0.2??0.0536 ?F??ED2.5?*??3?6 ?G??c?s??s??1?1)?????0.2??0.67?
有 Gc(s)??s??s??1???1???0.0536??6??s??s?5??1???1)??0.2??0.67? Gc(s)G(s)? sss??????s(s?1)??1???1???1??4??0.0536??6?
?*??c22?验算:
?)G(j?c?) ??180??Gc(j?c 110
?arctan2222?arctan?arctan?arctan?48.8