故 I(t)+S(t)+x(t)=n (1) H+)=1(t)=BI()S(t-x(+At)=x(] 即 d0=BI(0s0- dx(t) (2) dt dt x(t+△t)-x(t)=I(t)△t 即 dx (3) dt 从上面(1)2)3)中消去(t),可得 ds =-BIS dt d =-BIS-AS dt 14 14 故 I t S t x t n ( ) ( ) ( ) + + = (1) I t t I t I t S t t x t t x t ( ) ( ) ( ) ( ) [ ( ) ( )] +  − =  − +  −  即 ( ) ( ) ( ) ( ) dI t dx t I t S t dt dt = −  (2) x t t x t I t t ( ) ( ) ( ) +  − =   即 dx I dt =  (3) 从上面(1)(2)(3)中消去x t( ) ，可得 dS IS dt dI IS S dt     = −    = − − 