请问各位大大们三题有关根轨迹设计的问题
第一题:
画出系统特徵方程式的跟轨迹,其中
L(s) = (s+1) / [s(s+1)(s+2)]
然後计算跟轨迹的增益使共轭复数极点有阻尼比0.5
第二题:
数值控制工具机的定位伺服机构有正规化和缩放後的转移函数
G(s) = 1/ [s(s+1)]
在单位负回授,若闭回路极点位於 s = -1±j√3 处,系统就可符合性能规格。
(a)证明只用比例控制 D(s) = Kp 时,不可能符合规定。
(b)设计能符合规格的超前补偿 D(s) = K[(s+z)/(s+p)]
第三题
假设单位负回授的闭回路系统有前馈转移函数
G(s) = 1 / [s(s+2)]
设计落後补偿使得闭回路系统的主极点位於 s = -1±j 处并且对单位斜坡
输入的稳态误差低於0.2
谢谢~
不好意思,我忘记要放答案
第一题答案
This must be a typo! The roots at -1 cancel and the second order system
will have damping of 0.5 at K = 4. A more interesting case occurs for
num = s+3. In this case, the roots are at ..42+j.7 and the gain is 0.47
第二题答案
(a) With proportional control, the poles have real part at s = ..5.
(b) To design a lead, we select the pole to be at p = .10 and compute
the zero and gain to be z = .3, k = 12.
第三题答案
The poles can be put in the desired location with proportional control
alone, with a gain of kp = 2 resulting in a Kv = 1. To get a Kv = 5, we
add a compensation with zero at 0.1 and a pole at 0.02.
D(s) = 2[(s + 0.1)/(s+ 0.02)]
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※ 编辑: luckznn 来自: 203.73.6.207 (01/09 14:28)
1F:推 pimday1125:有答案吗 (1)k=4 (3)补偿器为(s+0.05)/(s+0.01),K=2 01/09 14:45
※ 编辑: luckznn 来自: 203.73.66.74 (01/09 19:10)
2F:→ luckznn:大大,有过程吗?我只有解答,没过程\/ 01/09 19:11