2 sin ot f()= Jat d 2n ja 2 sgn( t) FOo=-2f(t)sin atdt -2 sin tdt O sin ctdt O
j t d t e d j f t j t 2 sgn( ) 2 1 sin 2 1 ( ) = = − − 1 sin 2 sin ( ) 2 ( )sin 0 0 0 = = − = − tdt tdt F j f t tdt
四,常数的付立叶变换 f(t) E EG(t)<>Et sa( z z FLET-limEzSae2=27E lim 2 sa( 2) P171-35 S(t)=lim[-sa(kt) FLE=2TES(O) F[=2丌o() 2
四.常数的付立叶变换 2 − 2 E ) 2 ( ) ( EG t E sa ) 2 ( 2 ) 2 2 [ ] ( lim lim F E E Sa E sa → → = = P17.1-35 ( ) [ ( )] lim sa k t k t k → = F[E] = 2E () f (t) E F[1] = 2 () t
五u()的付应叶变换 u(t) 方法 F[(O)]=x+[1+sgn(t) sgn(t) F(ja)=m()+ yO P168.3-30
五.u(t)的付立叶变换 u(t) [1 sgn( )] 2 1 F[u(t)] = + + t j F j 1 ( ) = ( ) + P168.3-30 方法一 sgn( ) 2 1 t 2 1 t t t
右法二:剝用单边指数画数取极限 u(t)=lim e(t>o) a→>0 . eu( t<> C+/ F(j0)= +10a-+ =A(0)+jB(0)
a j e u t u t e t a t a t a + = − − → 1 ( ) ( ) lim ( 0) 0 ( ) ( ) 1 ( ) 2 2 2 2 A j B a j a a a j F j e = + + − + = + = 方法二:利用单边指数函数取极限