2015/10/21 阶系统的自由响应 i+aj+aoy=0 y0)=A 特征方程 s2+a3+a=0 不等实根S,≠S2 ymee(t)=Ae+e 相等实根S1=S2 yree(t)=Ae+Ate 共轭复根： ,2=a±jB yiee(t)=Ae cos(Bt+) School of Mechanical Engineering ME369-lecture 6.5 Shanghai Jiao Tong University Fall 2015 阶系统的自由响应（不等实根） yie()）=Ae+A,e (馬<0)&(52<0) (s<0)&(32=0) (5<0)&(52>0) ↑mg ↑mg. ↑1g Real Real Real y(1) (0 0 Time(f) Time(f) Time(f School of Mechanical Engineering ME369-lecture 6.5 Shanghai Jiao Tong University Fall 2015 22015/10/21 2 ME369-lecture 6.5 Fall 2015 School of Mechanical Engineering Shanghai Jiao Tong University 1 0 y a y a y    0 2 1 0 s a s a    0 1 2 1,2 1 0 1 4 2 2 a      s a ay A (0)  二阶系统的自由响应 特征方程 1 2 1 2 ( )   s t s t free 不等实根 y t Ae A e 1 1 1 2 ( )   s t s t free y t Ae A te ( ) cos( )     at free y t Ae t 共轭复根 s j 1,2     s s 1 2  相等实根 s s 1 2  ME369-lecture 6.5 Fall 2015 School of Mechanical Engineering Shanghai Jiao Tong University 二阶系统的自由响应（不等实根） 1 2 1 2 ( )   s t s t free y t Ae A e 1 2 ( 0) & ( 0) s s   1 2 ( 0) & ( 0) s s   1 2 ( 0) & ( 0) s s  