例2求8(t)的频谱F(jo) δ(t) 解:F(o)=F[(1)=()eot =d(t)1at=1 F(jO)↑ 6(t)<>1t  (t) 0 1 F(j) 0 ω 例2.求 (t)的频谱 F(j) t t e dt j t    − = = - F(j ) F[ ( )] ( )  解：    ( ) 1 1 - =  =     t dt  (t) 1