2015/11/14 环节的频率响应一一阶微分环节（续） 惯性环节 1 G(jo)= 阶微分环节 G(j@)=jt@+1 jT@+1 30 (p)/o)7 20dB/dee 20 际幅须特性 10 实际城特性 20ddee∠ 30 渐近线 0 转折率 10 (radisec) 互为倒数的环节 -对数幅频特性曲线关于零分贝线对称 -一相频特性曲线关于零度线对称 School of Mechanical Engineering ME369-lecture 7.2 Shanghai Jiao Tong University Fall 2015 环节的频率响应一振荡环节 1 G(j0)= T2(jo2+2T5(jo)+1 )0 (0=0) Gjo)=1∠0 1 G(jo)--J2G =0.3 -0.7 0.2 (0=o) Gj@)=0∠-180 0=0 0.1 2 -1 Re School of Mechanical Engineering ME369-lecture 7.2 Shanghai Jiao Tong University Fall 2015 72015/11/14 7 BE315-Lecture 7.2 Fall 2011 ME369-lecture 7.2 Fall 2015 School of Mechanical Engineering Shanghai Jiao Tong University 惯性环节 一阶微分环节 G j j ( ) 1     1 ( ) 1 G j jT     互为倒数的环节 ----- 对数幅频特性曲线关于零分贝线对称 ----- 相频特性曲线关于零度线对称 环节的频率响应—一阶微分环节（续） BE315-Lecture 7.2 Fall 2011 ME369-lecture 7.2 Fall 2015 School of Mechanical Engineering Shanghai Jiao Tong University 2 2 1 ( ) ( ) 2 ( ) 1 G j T j T j        ( )   ( 0)   G j ( ) 1 0    1 ( ) n T     1 ( ) 2 G j j     G j ( ) 0 180     环节的频率响应—振荡环节