2015/11/14 环节的频率响应一二阶微分环节 GU@)=t2Uo)}2+2r5(U@)+1 lG(j@)F(1-r(joYY+4r'o ∠G(jo)=arctan 2to I-ro) 0=0 ↑m (o=0)Gjo)=1∠0 1 (@=-)Gjo)=25∠90 25,0=1/π (o=o)G(jo)=o∠180 G@) 0=0 Re School of Mechanical Engineering ME369-lecture 7.2 Shanghai Jiao Tong University Fall 2015 环节的频率响应一延滞(transport lag)环节 G(jo)=el IG(j@)=1 L(@)=201g|G(j@)0 p(@)=-or(rad)=-57.3 @r 10 (gp)/()7 0 lm↑ 2 0=0 Re 500 60 0.1 o (rad's) School of Mechanical Engineering ME369-lecture 7.2 Shanghai Jiao Tong University Fall 2015 102015/11/14 10 BE315-Lecture 7.2 Fall 2011 ME369-lecture 7.2 Fall 2015 School of Mechanical Engineering Shanghai Jiao Tong University 2 2 G j j j ( ) ( ) 2 ( ) 1         2 2 2 2 2 2 | ( ) | (1 ( ) ) 4 G j j          2 2 2 ( ) arctan( ) 1 G j        ( )   ( 0)   G j ( ) 1 0    1 ( )    G j ( ) 2 90     G j ( ) 180    环节的频率响应—二阶微分环节 BE315-Lecture 7.2 Fall 2011 ME369-lecture 7.2 Fall 2015 School of Mechanical Engineering Shanghai Jiao Tong University ( ) j G j e         ( ) ( ) 57.3     rad | ( ) | 1 G j  L G j ( ) 20lg | ( ) | 0     环节的频率响应—延滞（transport lag）环节