自动控制理论(邹伯敏)第四章答案

确定根轨迹和虚轴的交点

令s?j?,特征方程D(s)?(j??2)3?K0?02????6??K0?8?0???0????23 ??3??（舍去）或?K??8???0?K0?64???12??0j?54321?-5-4-3-2-1-1-2-3-4-5

(1) 令s?j???????23 ??K0?64过s平面原点，和实轴负方向夹角为?60作射线，和根轨迹的交

?=0.5?arccos??0.5????60点即为主导极点。由图知，主导极点为?1?j3。又

pc1?pc2?pc3?p1?p2?p3??6?pc3??4

所以

(?4?2)3?K0?0?K0?8

Kp?limG(s)H(s)?lims?08?1

s?0(s?2)3(2) 系统的闭环传递函数可以近似为

C(s)88??2 R(s)(s?pc1)(s?pc2)s?2s?4??2??n?2??n?2??2?? ??n?4???0.5?

??1??2??Mp?e?100%?16.3%

?tp?????1.8138s

2?d?n1??3???5% ts???2% ts?4?n??3s

?n??4s