自动控制原理第五章课后习题答案（免费）
5－１设单位反馈系统的开环传递函数为
G0(s)?Ks(s?1)对系统进行串联校正，满足开环增益 K ? 12 s? 1 及 ??40?解:
① 首先确定开环增益K,Kv?limSGs?00(S)?k?12
② 未校正系统开环传函为:G(s)?012s(s?1)
Bode DiagramGm = 70.5 dB (at 200 rad/sec) , Pm = 16.5 deg (at 3.39 rad/sec)100Magnitude (dB)Phase (deg)500-50-100-90-135-18010-210-1100101102Frequency (rad/sec)
c③ 绘制未校正系统的开环对数频率特性,得到幅穿频率?对应相位角?G(j?c)??164,???160???'??3.4,
,采用超前校正装置,最大相角
???m???(180??G0(j?c))????40?16?6?30
④
?m?sin?1??1??1,???3
⑤ 在已绘图上找出?10lg?令?⑥
c??10lg3??4.77的频率?m?4.4弧度/秒
??m
1?m??T?T?1??m?0.128/s,??T?0.385/s1?0.385s1?0.128s
校正装置的传函为:G(s)?1??Ts1?Ts?
12(1?0.39s)校正后的开环传函为:G(s)?G校正后??180?137?43?40????(s)?Gc(s)?0s(s?1)(1?0.13s)
,满足指标要求.
Bode DiagramGm = 99.2 dB (at 1.82e+003 rad/sec) , Pm = 42.4 deg (at 4.53 rad/sec)100Magnitude (dB)Phase (deg)500-50-100-90-135-18010-210-1100101102103Frequency (rad/sec)5－２设单位反馈系统的开环传递函数为
G0(s)?Ks(s?1)(0.25s?1)要求 ??41?,K?5s?1,?b?1.02rad/s设计串联迟后校正装置。
解: 由题意,取K?Kv?5,待校正系统的传函为:G0(s)?5s(s?1)(0.25s?1)
① 绘制未校正系统的开环对数频率特性,得出:
Bode DiagramGm = 0.0388 dB (at 2 rad/sec) , Pm = 0.102 deg (at 2 rad/sec)100Magnitude (dB)Phase (deg)500-50-100-90-135-180-225-27010-210-1100101102Frequency (rad/sec)
?'c?2rad/s,??180??G0(?c)?0''?'''0'?,采用滞后校正装置.
?'''''''?② 选取?(?选取?''c)??9,而要求??41,??(?c)????(?c)?50c'''
?0.6rad/s,于是可测得L(?c)?15.8dB'''c
③ 由20lgb??L(?),?b?0.1622,令
1bT?0.1?c?T?102.754,?bT?16.67''④ 串联滞后校正装置的传函为:G(s)?1?bTs1?Ts?1?16.67s1?102.754s
⑤ 校正后系统的开环传函 为:G(s)?G校正后,?(s)?Gc(s)?05(16.67s?1)s(s?1)(0.25s?1)(102.754s?1)??
?180?136?44?41??,满足性能指标要求.
Bode DiagramGm = 15.2 dB (at 1.94 rad/sec) , Pm = 42.4 deg (at 0.666 rad/sec)10050Magnitude (dB)Phase (deg)0-50-100-150-90-135-180-225-27010-410-310-210-1100101102Frequency (rad/sec)
5-3、请分别写出超前校正网络和滞后校正网络的传递函数，画出的
它们的Bode图，写出最大超前相位角和滞后相位角的关系式。
解:
超前校正网络:G(s)?1??Ts,?1?Ts?1
L20dB/dec1/αT1/Tφω9045?m?arcsin??1??1ω
G(s)?1?bTs1?Ts,0?b?1
滞后校正网络:
L20dB/dec1/αT1/Tωω?m?arcsinφ45901?b1?b
自动控制原理第五章课后习题答案(免费)
