第三章自控系统的时域分析 - 图文(8)

?c0?t??r?t?

???t??e?t?ess?lime?t??lim?r?t??c?t??t??t??

<2>. 非单位反馈系统(图P85)

??t??e?t??E?s??R?s??B?s??R?s??H?s?C?s?

究，即 ?用偏差代替误差进行研ess?lime?t??lim?r?t??b?t??

t??t??或ess?lime?t??limsE?s?t??s?0

3.误差传递函数. 一般系统

1R?s? E?s??1?G?s?H?s? ??e?s?R?s?

E?s?1?e?s????误差传递函数R?s?1?G?s?H?s?sR?s?sE?s??lim ess?lim s?0s?01?G?s?H?s?ess与输入R?s?及开环传递函数G?s?H?s?有关.4.系统按稳态误差划分的类型 设系统开环传递函数为

K??1s?1???2s?1????ms?1?G?s?H?s???s?T1s?1??T2S?1???Tns?1?

sess?limsE?s??lim?R?s?s?0s?01?G?s?H?s?

s??1?T1s?1???Tns?1??lim??R?s?s?0s?Ts?1???Ts?1??K??1s?1????ms?1?1n?lims?0s???1

s?KR?s? 按稳定误差划分的型： 当?

二.给定输入信号下的稳态误差

1.阶跃输入信号下的稳态误差与静态位置误差系数Kp r?t??R0?1?t?,R?s???0,1,2,?R时,分别称为0型,1型,2型?系统..

R0 ssR0sE?s??lim? ess?lim s?0s?01?G?s?H?s?s ? 令KpR0

1?limG?s?H?s?s?0?limG?s?H?s??静态位置误差系数 s?0

R0 ?ess?1?KpR01?K

??1,Kp??,ess?0??0,Kp?K,ess?

结论: 0型系统在阶跃输入作用下有误差,常称有差系统.

要使ess?,可?K.对阶跃输入,要使ess?0,必须??1.

2.斜坡输入信号下的稳态误差与静态速度误差系数K? r?t??v0?t,R?s??v0 s2 ess?lim ?lims?0svv0?0?lim s?01?G?s?H?s?s2s?0s?sG?s?H?s?v0

????sGsHs令K??limsG?s?H?s??静态速度误差系数.s?0?ess? ?v0K?

?0,K??0,ess??v0??1,K??K,ess?K ??2,K???,ess?0

结论:0型系统不能跟踪斜坡输入；1型可跟踪，但有与K有关的误差；2型及以上在斜坡输入下的ess

3.抛物线输入信号下的稳态误差与静态加速度误差系数Ka r?t???0。

a02a0t,R?s??3 2ssa0a0?3?lim2 s?01?G?s?H?s?ss?0sG?s?H?s? ess?lim2令K?limsG?s?H?s??静态加速度误差系数 as?0 ess?a0 Kaess??a ??2,K?K,e?0assK??3,Ka??,ess?0??0,??1,Ka?0,Ka?0,ess??

结论: 0型和1型不能跟踪r?t??有准确跟踪.

表3-1，2 P88 总结

4.复合输入下的稳态误差

2r?t??R0?v0t?1at时20a02t,2型可跟踪但有误差,3型及以上才2 ess?R0va?0?0

1?KpK?Ka 至少?

?2,ess才能满足要求,但?大会降低系统稳定性.

例1. 系统框图如右，设

10?1?.G?s??

s?s?4? ?2?.G?s?? 当输入rR(SE(S- G(sC(S10?s?1?

s2?s?4??t??4?6t?3t2.时，分别求ess。

102.5?解: <1> G?s??（此处应写成一般形式：时间常数形

????ss?4s0.25s?1式） ???1,K?2.5 则

Kp??,K??2.5,Ka?0 ?ess466????? 1?KpK?Ka10?s?1?2.5?s?1?? s2?s?4?s2?0.25s?1? <2>. G?s?????2,K?2.5,则Kp??,K???,Ka?2.5466?ess????2.41?KpK?Ka

三.扰动输入引起的稳态误差.

??sG2?s?????e?limsEs?limDs sfn? s?0s?0?1?G?s?G?s?H?s?12??

四.减小稳态误差的措施

在保证稳定的前提下提高稳态精度的措施：

<1>在增大G1的开环放大倍数K1的同时，附加校正装置. <2>增加前向通道积分环节个数的同时,也要对系统进行校正.

<3>采用复合控制.

1）按扰动补偿的复合控制

分析扰动时，令R（S）=0 R(S) E(S) — GD D(S) + C(S) G1 + G2 ?1?G1GD?G2D?s? E?s???C?s???1?G1G2若GD??

2)按给定补偿的复合控制.

1，则E?s??0 -----扰动作用实现完全不变性的条件 G1Gr?s? E(sR(s-

+ G1?s?

+ G2?s? C(s

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