发布时间 : 星期三 文章机械控制工程基础课后答案-董玉红、徐莉萍主编更新完毕开始阅读dc3462812a160b4e767f5acfa1c7aa00b42a9d69
(2)f(t)?0.03?0.03cos2t1ss2?4-s20.12F(s)?L(f(t))?0.03?0.032?0.03?ss?22s(s2?4）s(s2?4）(3)f(t)?sin(5t??3)?sin（5t??15)?s5?515F(s)?L(f(t))?2t??e15s?25s2?25
(4)cos12t及复位移0.4s?0.4F(s)?L(f(t))?2（s?0.4）?1442.5'??(t)??K1?x3(t)?x2(t)??Bx3?(t)?x2m1x3(t)\'?(t)?x2m2x2(t)?K1?x3(t)?x2(t)??Bx3(t)?K2?x1(t)?x2(t)?????两边进行拉氏变换：m1s2X3(s)??k1X3(s)?k1X2(s)?BsX3(s)?BsX2(s)m2s2X2(s)?k1X3(s)?k1X2(s)?BsX3(s)?BsX2(s)?K2X1(s)-K2X2(s)m1s2X3(s)?k1X3(s)?BsX3(s)?k1X2(s)?BsX2(s)（1）k1X3(s)?BsX3(s)?K2X1(s)?m2s2X2(s)?BsX2(s)?K2X2(s)?k1X2(s)（2）由（1）(m1s2?Bs?k1)X3(s)X2(s)?Bs?k1代入（2）(m2s2?Bs?K2?k1)(m1s2?Bs?k1)X3(s)k1X3(s)?BsX3(s)?K2X1(s)?Bs?k1(m2s2?Bs?K2?k1)(m1s2?Bs?k1)X3(s)（-k1?Bs）X3(s)?K2X1(s)Bs?k1?X3(s)K2?X1(s)(m2s2?Bs?K2?k1)(m1s2?Bs?k1)（-k1?Bs）Bs?k1K2(Bs?k1)(m2s2?Bs?K2?k1)(m1s2?Bs?k1)（-k1?Bs）(Bs?k1)
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3．1 时间响应：在输入作用下，系统的输出（响应）在时域的表现形式。
3.2，脉冲响应的拉氏变换等于传递函数乘以脉冲信号的拉氏变换，由于脉冲信号的拉氏变换为1，所以脉冲响应的拉氏变换等于传递函数。
t?3.3 一阶系统的阶跃响应为
xo(t)?1?eT,t?0 2?n 3.4典型二阶系统G(s)?22s?2??s??nn
阻尼比及系统的无阻尼固有频率是系统的两个重要参数。它们决定着系统的时间响应特性。二阶系统的阻尼比决定了其振荡特性。
3-5 欠阻尼二阶系统单位阶跃响应曲
???nte
c(t)?1?sin(?dt??),t?0 21?? 1??2??arctg?arccos? ???1??2?dn
2 1.8 1.6 1.4
?=0.2 ?=0.4 ?=0.6 ?=0.8 xo(t) 1.2
1 0.8 0.6 0.4 0.2 0
tp 5 t
10 15 欠阻尼二阶系统单位阶跃响应曲线
?＝1
c(t)?1?(1??nt)e??nt,t?0
?＞1
4-1
c(t)?1?12(1???2?1??2)?122e?(??e?2?1)?nt?(???2?1)?nt2(1????1??)，t?0系统传递函数：1/(s?1)1?(s)??1?1/(s?1)s?2频率特性：?(j?)??A(?）?1j??21
?(?）?-arctg?2?4?2
(1)、当r(t)?sin2t时
1
?(j?)??2??0.35, 8
?1?2)??45? ?(j2)?tg(2
??2c(t)?0.35sin(2t?45)
（见PPT） 4-2