二冲激偶的傅或叶变换 FTD()=160=÷Cmdo dt [S()]=2 (j@)elo 'do F7δ(t) dt dt 6()|=(01F7()=27()yn[6(o do二.冲激偶的傅立叶变换   −  =    t e d j t 2 1 ( )     −  =    t  j e d dt d j t ( ) ( ) 2 1  t j dt d FT =      ( ) n n n t j dt d FT  ( ) = ( )       ( ) 2 ( )  ()   n n n n d d FT t = j FT[ (t)] =1