角波发生器电路2(续): 0 01 D O OMI H ta 1+ 2 R +L OM (3ttrt2: o(t=-UoM, uo(t=U+L ∫undt=-1Uo+ UOM RC R 2 RC(t-tu) R tt2:u(t=t2)0+ R UoM 可解出: uan从-UoM→+UoM , =3 LRC R － + +  － + + A1 A2 uo uo1 R02 R01 R C R2 R1 u1+ (3)t= t1~ t2 : uo1 (t1 ) =-UOM , uo (t1 ) = U+L t=t2 : uo (t=t2 ) = OM 2 1 H U R R U+ = uo1从-UOM → +UOM 2 1 RC R R t2=3 可解出: UOM uo1 0 t - UOM t1 U+L U+H uo = uo (t1 u ) −  u dt= RC 1 o o1 + UOM RC (t-t1 OM ) 2 1 U R R − t1 t t2 三角波发生器电路2 (续) :