2015/11/14 环节的频率响应-积分环节 1 Gj0)==-j G(j@)1 1 L(@)=-201g@ ∠Gjo)=arctan(-o-） p(@)=-90° 0 0 20dB/dec Imt 40 0=0 0 e 悬 % -90 -180 0.1 10 100 0=0 w (radlsec) School of Mechanical Engineering ME369-lecture 7.2 Shanghal Jiao Tong University Fall 2015 环节的频率响应一微分环节 G(jo)=j@ G(j@)=@ p(o)=90 G(jo)=arctan( L(@)=201g@ 20dB/dec Imt 0 0=0 -20 180 1 90 0=0 45 0 Re 0 0.1 10 100 (radisec) School of Mechanical Engineering ME369-lecture 7.2 Shanghal Jiao Tong University Fall 2015 32015/11/14 3 BE315-Lecture 7.2 Fall 2011 ME369-lecture 7.2 Fall 2015 School of Mechanical Engineering Shanghai Jiao Tong University 1 1 G j j ( ) j       1 | ( ) | G j   1 ( ) arctan( ) 0 G j      L( ) 20lg      ( ) 90   环节的频率响应--积分环节 BE315-Lecture 7.2 Fall 2011 ME369-lecture 7.2 Fall 2015 School of Mechanical Engineering Shanghai Jiao Tong University G j j ( )    | ( ) | G j  ( ) arctan( ) 0 G j     L( ) 20lg     ( ) 90  环节的频率响应—微分环节