择:(1)设零输入响应为r(),零状态响应为r(),则有 ()=r()+r2()=2e”+sim(2)y(2) ()+2rn(1)=|e+2si()(t) 解得:()=3e( r2()=-e+sin(2)(t) r3()=r(t)+r2(t-t) 3eu(t)+ 3(t-t0) +sin(2-20)(t-t) r()=2-:(1)+0572( =233)+0.5--+in(2 =55+05s02)()(1) 设零输入响应为r (t) zi ，零状态响应为r (t) 解： zs ，则有 解得: ( ) ( ) ( ) [2 sin(2 )] ( ) 3 1r t r t r t e t u t t = zi + zs = + − ( ) ( ) 2 ( ) [ 2sin(2 )] ( ) 3 2r t r t r t e t u t t = zi + zs = + − ( ) 3 ( ) 3 r t e u t t zi − = ( ) [ sin(2 )] ( ) 3 r t e t u t t zs = − + − ( ) ( ) ( ) 3 0 r t r t r t t ∴ = zi + zs − 3 ( ) [ sin(2 2 )] ( ) 0 0 3 3( ) 0 e u t e t t u t t t t t = + − + − − − − − ( ) e t u( )t e u t e t u t r t r t r t t t t zi zs [5.5 0.5sin(2 )] 2[3 ( )] 0.5[ sin(2 )] ( ) 2 ( ) 0.5 ( ) 3 3 3 4 = + = + − + ∴ = + − − −