解H(s)=C1(s) T(s 1(s) 4S+4 U(s)2U2(s) 4 +1+ s+5s+6 2+2s H,(s)=()_2sU(s) S Is(s)2+2ss2+5s+6 U()=H1()(=~4+4 S(S2+5s+6) 4s U2(s)=H2(S)() S(2++6) S(t)=2+2e2-eS(t)=4e2-4 3 理步文通大浮解 4/s 2s 2 1 I(s) U1(s) + + - -U2(s) I1(s) 5 6 4 4 2 2 1 1 4 1 ( ) ( ) ( ) 1 1 + + + = + + + = = s s s s s I s U s H s 2 S 5 6 4 2 2 2 ( ) ( ) ( ) ( ) 2 2 2 + + = + = = s ss s sU s I s U s H s 2 S ( 5 6 ) 4 4 ( ) ( ) ( ) 1 1 + + + = = s s s s U s H s I s 2 5 6 ) 4 ( ) ( ) ( ) 2 2 + + = = s(s ss U s H s I s 2 2t 3t 1 e 38 2e 32 S t − − ( ) = + − 2t 3t S2 t 4e 4e − − ( ) = −