7-5 解:H(s)= ,|H(A2) A=△O2|H(1Q2)=△ y+()2v041=0999 P A2=△H(2)=△2 1+(3)2√26△1≈0981△2 0.197 V(1)= Q2Cos(Q21-4e/3 (V) K 7-7 解:1)6(∞) △=0.5 △=0.005×106xrad/s Kp=10T 1/ 10·丌 Φ=-1g-1(2)=-1g-1(2)≈-0.197 设输出电压为H(0)=m:co20050m+Asn(2x×101+d)-(∞)7-5 解： S Kp Kp H s + ( ) = ， 2 1 ( ) 1 ( ) Kp H j  +  = 。 1 1 2 1 1 1 1 0.995 101 10 ) 10 1 1 ( 1  ( )  =    +  A =  H j =  ) 0.1 10 1 ( ) ( 1 1 1 1 = −  −   = − − − tg Kp tg 1 2 2 2 2 2 2 0.981 26 5 ) 5 1 1 ( 1  ( )  =    +  A =  H j =  ) 0.197 5 1 ( ) ( 1 2 1 2 = −  −   = − − − tg Kp tg                        +           −   + − = − − 2 0 0 1 0 1 1 ( ) cos cos( ) Kp Kp t t g Kp V t V t i o o o m i     （V） 7-7 解：1） ( ) = 0.5   = Kp e   rad   6  = 0.00510 rad/s  4 Kp =10 1/s 26 5 ) 5 1 1 ( 1 ) 10 2 10 1 ( 1 1 ( ) 1 ( ) 2 2 4 3 2 = +  + =  +  =   Kp H j 设输出电压为 ( ) cos2.005 10 sin (2 10 ) ( ) 6 3 Vo t =Vom  t + A   t +  −e  ) 0.197 5 1 ( ) ( 1 1 = −  −   = − − − tg Kp tg