自动控制原理复习资料——卢京潮版第三章

内容发布更新时间 : 2024/11/5 16:37:19星期一 下面是文章的全部内容请认真阅读。

解:依题可知

??h(?)?2? ?tp?0.75''???%?2.18?2?9%2?22?K1K2K1?n?K2??n?(s)?2??? (1)

s?as?K2s2?2??ns??n2?a?2??n?h(?)?lims.?(s).R(s)?lims.s?0s?0K1K21.?K1?2 (2)

s2?as?K2stp??1???n2?0.75 (4)

?%?e???1??2ln0.09?????0.7665?2??1???0.09 ?(5)

2?0.7665???0.60833(??52.55?)?21?0.7665?(5)?(4):

?n??0.751?0.6082?5.236弧秒 (6)

(5).(6)?(1)

?K2??n2?5.2362?27.4??a?2??n?2?0.608?5.236?6.37 ?K?2?1系统极点分布:

???arccos??52.5? ???5.236?n

3. 过阻尼二阶系统性能估算:

?n1?n◆ ?(s)?22?22Ts?2?Ts?1s?2??ns??nT?12(??1) (1)

s1,2??2??n?4?2?n?4?n?????n??n?2?1222?s??(??n?n2??1)s??n(????1).1;T2?122??2?.?1?n211(s?)(s?)T1T2 T1?T2 (2)

T1?12????1?n????1?n◆ 找出?,?n与 T1,T2 之间的关系:

?s2?2??ns??n?(s?211111 )(s?)?s2?(?)s?T1T2T1T2T1T2比较:?n?1 T1T21111T??1?1TTT21T1T21T2 ??1???fs(1) (3)

2?n2121T2T1T2T1T2? 求阶跃响应:

1(1T2T1?1)(?1)2r?1(t)?nTT11C(s)?????(s)???1?211ss(s?1)(s?1)ss?s?T1T2T1T2h(t)?1?1T(2?1)T1e?tT1

?1T(1?1)T2e?tT2求ts表达式:依ts定义:

0.95?1?1T(2?1)T1e?tsT1?1T(1?1)T2e?tsT2

tsT1(3)*解:?f()?f(?)

T1T2? 过阻尼二阶系统求ts思路:

22???n?T1T2K?nts?T1?1(????1)?nP86图3-17?(s)?2? ? ? ????????22T1?s?2??ns??n?????T2?1(????1)?n缺例

t ?ts?(s).T1T1????题: 例:

?(s)=1.61612??0.1s2?2??ns??n2s2?10s?16(s?2)(s?8)1? ??n?16?4T??0.5?T1?41?ts???T2p86 ????3.3?ts?33.7?1.65??2???10T1?1.25?1?T?1?0.125???1.25????2?2?4?8?? 注: 1)当

T1?5(??1.4)时,欠阻尼二阶系统ts?3T1—-近似用一阶系统代替 T22)过阻尼二阶系统零极点分布与动态性能之间的关系

i. ii.

s??1T极点对ts影响较大――主导极点(1?6时) T1T2tTT1值有关(s~f(1)) T2T1T2ts与T1值、

3)系统相当于两个惯性环节串联时的特性 欠阻尼二阶系统动态性能计算

习:

⑴ 极点的表示方法: ⑵ 动态性能计算公式:

??t?p?1??2?n??????1??2 ??%?e?3.5?ts????n??⑶ ?,?n变化时动态性能的变化规律

???%?2?实部:??n不变;虚部1???n????ts??极点直角表示法:????%??2虚部:1???不变;实部?????nn???ts??

???%??阻尼比:?不变;固有频率?n????ts?? 极点极坐标法:????%???固有频率?n不变;阻尼比??????ts??举例:系统如右图示,

求Ka分别取值为1500,200,13.5时的动态性能: 解:开环传递函数 G(s)? 开环增益

?(s)?5Ka

s(s?34.5)K?5Ka34.5

5Ka

s2?34.5s?5KaKa ?n?5Ka34.5??2?ntp??1???n2 ?%?e??1??2?ts?3.5ess?r?t??n1K 1500 86.2 0.2;??78.5? 0.037 52% 0.2 0.0046 200 31.6 0.545;??57?0.12 13% 0.2 0.0345 13.5 8.22 ?s1??2.082.1;??s2??32.4? 0 1.45 0.5111

h(t)响应曲线见右下图。

3 单位斜坡响应与ess讨论: 误差传递函数:

?e(s)?1 1?G(s)?s(s?34.5)

s2?34.5s?5KaE(s)??e(s).R(s)

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