f(t) 2.有始信号分解为阶跃信号二 如图,同理 07 f()J0+1+f2+…k+… f 其中 f(0) 0Ax2AcA△C f0=f(0)U/() f1=[f(△z)-f(0)(t-△) f(△r)-f(0 △ (t-△r)△r U(t-△z)△ △ △ t=△r : f=[f(k△r)-f(0k-1)△z)U(t-kAr)= (t-k△z)△r △T t=kAT... ... (0) ( ) 0 f = f U t           −     −   =   − = =  − −  = ( ) ( ) ( ) (0) [ ( ) (0)] ( ) 1 U t f U t f f f f f U t t 2.有始信号分解为阶跃信号 其中 f(0) 如图，同理 f (t)  f 0 + f 1 + f 2 + f k +         −     =  − −  −  = =  [ ( ) (( 1) )] ( ) U(t k ) f f f k f k U t k t k k   2  k fk t f0 f1 f(t) 0