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 = 0.00510 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 ( ) cos2.005 10 sin (2 10 ) ( ) 6 3 Vo t =Vom t + A t + −e ) 0.197 5 1 ( ) ( 1 1 = − − = − − − tg Kp tg
05|H(19)=0.5 5 o()= om cos[2005×106m+0.49sin(2m×103t-0.197)-05] 2)环路3dB带宽为K 79求解 S72+1 已知:KP=KK4=5×10rad/sH1(s)s(r+r2)+1 06rad/s,9=200rad/s.oo=1.005×10°rad/ △O0=Oo-00=-5×103rad/s 环路锁定后的稳态相差:0()=sin(2)=-0,47rad 闭环传递函数:H(s) 02(s)K,H Hc(/9)=1.18 输出信号: (1)= Yom cosl10°1-0.(∞)+05×|Hc(2)×sn(200+9)() 7-13求解 根据题意,得知K=2V,K。=104H=/V,f=1M, 则 K=K K e2r=4 10 (rad/s) 而|△O|=|oo-ono=4x×10(rad/s)=Kp 故平衡点在=sm-(△o/KP)=90 所以,当输入信号的初始相角为30°时,可以锁定。 稳态相差为(∞)=sin(△o/Kp)=90 控制电压为vp=△O/K。=2
( ) cos[2.005 10 0.49sin( 2 10 0.197) 0.5] 0.49 26 5 0.5 ( ) 0.5 6 3 = + − − = = Vo t Vom t t A H j 2) 环路 3dB 带宽为 Kp. 7-9 求解: 已知:K K K rad s P d 5 10 / 4 = = 闭环传递函数: ( ) 1 1 ( ) 1 2 2 + + + = s s H s F 10 / , 200 / , 6 0 rad s rad s i = = 1.005 10 / , 6 0 rad s o = ( ) ( ) ( ) ( ) ( ) 1 2 s K H s K H s s s H s p F p F c + = = rad s i o 5 10 / 3 0 = 0 − 0 = − rad KP e ( ) sin ( ) 0.4 7 1 = − = − 环路锁定后的稳态相差: 输出信号: ( ) cos[10 ( ) 0.5 ( ) sin( 200 )] ( ) 6 vo t = Vom t − e + HC j t + V H ( j) =1.118 C 7-13 求解 根据题意,得知 K V K Hz V f MHz d 2 , 10 / , 0 1 4 = = = , 则 2 4 10 ( / ) 4 K K K rad s P = d • • = 而 | |= | | 4 10 ( / ) 4 0 0 rad s i −o = = KP 故平衡点在 sin ( / ) 90 1 = = − e KP 所以,当输入信号的初始相角为 30 时,可以锁定。 稳态相差为 ( ) sin ( / ) 90 1 = = − e KP 控制电压为 VP = / K = 2V