令F(s) s+6 K K (S+2)(S+5)s+2s+5 K1=F(s)(s+2 2+64 2+53 5+6 K2=F(s)·(s+5) 5+2 3 则F(s)=-3 3 e-3)E(t) s+2s+5 3 3( 2)( 5) 2 5 6 2. ( ) 1 2 + + + = + + + = s K s K s s s 令F s 3 1 5 2 5 6 ( ) ( 5) 3 4 2 5 2 6 ( ) ( 2) 2 5 1 2 = − − + − + =  + = = − + − + =  + = =− =− s s K F s s K F s s 则 e e t） s s F s t t ) ( 3 1 3 4 ( 5 3 1 2 3 4 ( ) 2 5  − −  − + − + =