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反校正馈 - PowerPoint PPT Presentation

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PowerPoint Slideshow about '反校正馈' - nairi

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Presentation Transcript

G2(s)的特性!!

• 由于|G2(jω)Hc(jω)| >> 1，故在受校正频段

Hc(ω) + L2(ω) ＝ Ls(ω) － Lds(ω) > 0

Ls(ω) － Lds(ω)越大，校正装置精度越高。

• 局部反馈回路必须稳定。

• （分析法）预先选择参数待定的反馈校正装置，根据性能要求通过分析法确定参数。

（1）系统在最大跟踪速度 及最大跟踪加速度 时,系统的最大误差 ；

（2） 在单位阶跃信号作用下，系统的瞬态响应时间

（1）绘制系统固有对数频率特性Ls()

Ls

Lds

L2

Ls-Lds

（2）绘制希望特性Lds()

c=7.4s-1 ;考虑到一

Ls

Lds

L2

Ls-Lds

2=2.4s-1

3=23.8s-1

(==Ls()的一个转角频率)

Ls

Lds

L2

Ls-Lds

Ls

Ls

Lds

L2

Ls-Lds

(3) 绘制L2()+20lg|Hc(j )|

Ls

Lds

L2

Ls-Lds