2S 解』(t)=L (s+1)2+100 s+100 100 L 2 (s+1)2+1002(s+1)2+100 2e [cos 100t-sin 100t) 2v()的轰达式可展开为 v1(t)=coS(100)+cos(101t)+co(991 H(j0)= J 2ia (jm+1)2+1002(jo)2+2j+1002
2 [cos100 sin 100 ) ] ( 1) 100 100 ( 1) 100 100 2[ ] ( 1) 100 2 1. ( ) [ 2 2 2 2 1 2 2 1 e t t s s s L s s h t L t = − + + − + + + = + + = − − 解 ： − 2 2 2 2 1 1 ( ) 2 100 2 ( 1) 100 2 ( ) cos(99 ) 2 1 cos(101 ) 2 1 ( ) cos(100 ) 2. ( ) + + + + = = + + j j j j j H j v t t t t v t 的表达式可展开为：
2 2+j (+100(0-100 考虑到所研的频率洄仅在=100附近 取近条件+100≈2;于是有 H(o)≈ 1+(OD-10OH(100)=1 2 H(101) 45 e 2 2 4 H(99)=e
( 100)( 100) 2 2 + − + = j 考虑到所研究的频率范围仅在＝100附近 1 ( 100) 1 ( ) 100 2 + − j H j 取近似条件 ＋ ；于是有 0 0 4 5 4 5 2 2 (99) 2 2 (101) (100) 1 j j H e H e H = = = −
V2(t)=cos(100)+ cos(101t-45)+cos(99+45 =cos(100)+cos(100)cos(t-450) X n [1 cos(t-45 )cos 100t v(t)=(1+100)cos100t
cos(99 45 ) 2 2 cos(101 45 ) 2 2 [ 2 1 ( ) cos(100 ) 0 0 v2 t = t + t − + t + t t t t t cos( 45 ) cos100 2 2 [1 cos(100 ) cos( 45 ) 2 2 cos(100 ) 0 0 = + − = + − v (t) (1 100t) cos100t 1 = +