归首士学 NORTHWESTERN POLYTFCHNICAL UNTVERSITY §5.1 频率特性的基本概念(3) 例2系统结构图如图所示,r(t)=3sin(2t+30°),求ct,e(t)。re ua 解.Φ(S) +j@ 1+o S+1 ∠d(o)=- arctan o=-63.4°=∠c,(t)-∠r()=∠c,(t)-30° c()=3/ 3 (D)=sin(2t-33.4°) ∠c,()=-634°+30°=-33.4° 2 J 1+jol√1+ 22(j0)=90°- arctan a=90°-63.4°=∠e,(t)-30° ()=—=sin(21+56.6°) ∠e,(t)=26.6°+30°=566°§5.1 频率特性的基本概念 (3) 例2 系统结构图如图所示, r(t)=3sin(2t+30º), 求 cs(t), es(t)。 1 1 ( ) s 解. s 3 ( ) 5 1 1 1 1 1 ( ) 2 2 c t j j s w w w w ( ) arctan 63.4 ( ) ( ) ( ) 30 2 j c t r t c t s s w w w cs (t) 3 5 c (t) 63.4 30 33.4 s sin(2 33.4 ) 5 3 c (t) t s 1 ( ) s s s e 3 ( ) 5 2 1 1 ( ) 2 2 e t j j j s e w w w w w w ( ) 90 arctan 90 63.4 ( ) 30 2 j e t e s w w w es (t) 6 5 e (t) 26.6 30 56.6 s sin(2 56.6 ) 5 6 e (t) t s