第五章 线性系统的频域分析与校正
习题与解答
5-1 试求题5-75图(a)、(b)网络的频率特性。
CR1R1urR2CucurR2uc
(a) (b)
图5-75 R-C网络
解 (a)依图:
1sCR2?1R1?sCU(j?)R2?j?R1R2CK(1?j?1?)??1 Ga(j?)?c
Ur(j?)R1?R2?j?R1R2C1?jT1?R1Uc(s)?Ur(s)R2?K1(?1s?1)T1s?1R2?K??1R?R12?? ??1?R1C?RRC?T1?12?R1?R2?U(s)? (b)依图:cUr(s)?1T2s?1R1?R2?sCU(j?)1?j?R2C1?j?2??? Gb(j?)?c
Ur(j?)1?j?(R1?R2)C1?jT2?
R2?1sC?2s?1??2?R2C ?T?(R?R)C12?2 5-2 某系统结构图如题5-76图所示,试根据频率特性的物理意义,求下列输入信号作用时,系统的稳态输出cs(t)和稳态误差es(t) (1) r(t)?sin2t
(2) r(t)?sin(t?30?)?2cos(2t?45?) 解 系统闭环传递函数为: ?(s)?1 图5-76 系统结构图 s?2参考类别# 77
12??频率特性: ?(j?)? ??jj??24??24??2幅频特性: ?(j?)?
124????相频特性: ?(?)?arctan()
21s?1?, 系统误差传递函数: ?e(s)?1?G(s)s?2则 ?e(j?)?
1??24??21,??e(j?)?arctan??arctan()
2(1)当r(t)?sin2t时, ??2,rm=1
?2?0.35, ?(j2)?arctan()??45?
285?e(j?)??2??0.79,8 2?e(j2)?arctan?18.4?6 css?rm?(j2)sin(2t???)?0.35sin(2t?45?)
则 ?(j?)??2? ess?rm?e(j2)sin(2t??e)?0.79sin(2t?18.4) (2) 当 r(t)?sin(t?30?)?2cos(2t?45?)时: ? ?(j1)????1?1,??2?2,rm1?1rm2?2
5?1?0.45?(j1)?arctan()??26.5? 52101?0.63?e(j1)?arctan()?18.4? ?e(j1)?53?? cs(t)?rm?(j1)?sin[t?30??(j1)]?rm?(j2)?cos[2t?45??(j2)]
?0.4sin(t?3.4)?0.7cos(2t?90)
es(t)?rm?e(j1)?sin[t?30??e(j1)]?rm?e(j2)?cos[2t?45??e(j2)] ?0.63sin(t?48.4)?1.58cos(2t?26.6)
5-3 若系统单位阶跃响应 h(t)?1?1.8e试求系统频率特性。
?4t???????0.8e?9t(t?0)
参考类别# 78
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