②两业x大 §6.2串联超前(微分)校正(2) )RC超前网络4G()=+B10 H=20loy/T 1 olga 201ga 1 olga aT P(o)=arctan(aTo)-arctan (To)=arctan Tola-1) aT 1+ata ( dolo d ang/(o)]=0 90 d 1 d To(a-1)T(a-D)[l-ata 0 do 1+at a 1+aT2a2)2 a+ -1 1/√aT To(a-1) 2/a tan P(am) 1+aro 1+sin pm T 可解出 sin p Pm =am)=arctan arcsin 2√2 + L(Om=10lga§6.2 串联超前（微分）校正（2） a aT T H 20lg 1 1  20lg  (1) RC 超前网络 2 2 1 ( 1) ( ) arctan( ) arctan( ) arctan       aT T a aT T      m m a   1 sin 1 sin    L( m )  10lg a 1 1 ( )     Ts aTs a G s c (a  1) 0 ( )     d d  tan()  0 d d 0 (1 ) ( 1)[1 ] 1 ( 1) 2 2 2 2 2 2 2                  aT T a aT aT T a d d  m  1 aT a a aT T a aT m m 2 1 1 ( 1) tan ( ) 1 2 2            1 1 arcsin 2 2 1 ( ) arctan       a a a  m   m 可解出