5k z=RC=100×10×5×103=0.5s 阶跃响应为 0.5U/s 100uF lc(=(1-e)E(t) du e 8(t)mA 由齐次性和叠加性得实际响应为 ic=5l-e8(t 1e2(-0.5) 5 e(t)-e200s)(t-0.5)mA100 10 5 10 0.5s 6 3 = = = − − RC ( ) mA 5 1 d d C 2 e t t u i C t C − = = 5US 0. + 5k - ic 100F ( ) (1 ) ( ) 2t u t e t C − = − 阶跃响应为: 由齐次性和叠加性得实际响应为: ( 0.5)] 5 1 ( ) 5 1 5[ 2 2( 0.5) = − − − − − i e t e t t t C ( ) ( 0.5) mA 2 2( 0.5) = − − − − − e t e t t t