2H 例： uc(0)=1Vi(0)=0.5A uc÷0.5F 求电流响应() ￡[i()=I(S)￡[5]=5/s 2i+2票+id=5 (2+2s+子s)-1+号= 15 S+4 i() 2S2+2S+2uC 0.5F 2 + - i 2H 5V + - £ [i(t)]=I(S) £ [5]=5/s 2i+2 + ³ idx di dt 0.5 1 -’ t =5 I(S)= S + 4 2S2+2S+2 uC(0– )=1V i (0– )=0.5A 1 S 2 S ( 2+2S+ )I(S) – 1 + = 5 S ≲⭥⍱૽ᓄi(t) ?? i(t) ֻ˖