武汉大学：《信号与系统》课程教学资源（PPT课件讲稿）第三章 习题倮

例三:P164.3-13 ①() i(t)=-i1(-t) o=0,an=0 -E(sin naotdt-sin nott) T 2 0 [1 coS(2n兀 T

( ) ( ) 0, 0 i 1 t = −i 1 −t a0 = an = i t n tdt T b T n 0 0 1 ( )sin 2 =   ( sin sin ) 2 0 0 0 n tdt n tdt T T   = −     [1 cos(2 )] 2 T n n    = − 0  T ( ) 1 i t ( ) 2 ( ) v t 1 i t 例三：P164.3-13

1.当z=5ST=10S 0 n even 2元 0 n(odd) fo 二二 103=100kHz T,10×10 6 v, (t=sin Oot x100kQ2=1273sin oot 元 2jf.z=10,T=20s f =50US.then.v,(t)=0 T20×10

1.当  = 5s T =10S bn = ( ) 4 n odd n T   2 0 = kHz T f 10 100 10 10 1 1 5 0 6 = =  = = − v t t k t 2 0 0 sin 100 127.3sin 4 ( )    =    2.if . = 10s,T = 20s 50 ... ... ( ) 0 20 10 1 1 6 = 2 =  = = − s then v t T f  0 n(even)

3..z=15s,7=30∠s f =-×100kHz T30×10 v3()=sn3t×100g≈42.4sin30t 3丌 三傅立叶变换的定义 F(jo)=f(t)e jo dt, F(0)=f(t)dt f (t)= F(ja)eloda, f(o) FOda 2兀 2丌

3.if . = 15s,T = 30s kHz T f 100 3 1 30 10 1 1 6 =   = = − v t t k t 3 0 0 sin 3 100 42.4sin 3 3 4 ( )    =    三.傅立叶变换的定义    −  − −  F(j) = f(t)e dt,F(0) = f(t)dt j t    −  − = =        f t F j e d f F j d j t ( ) 2 1 ( ) , (0) 2 1 ( )

典型信号的傳立叶变换 ()<>1 1<>2丌() (1)<>+(O) sgn(t)<> G(t)=E[(t+)-l(t--<>Eoa()

典型信号的 傅立叶变换 ( ) 1 ( )      + j u t j t 2 sgn( )  ) 2 )] ( 2 ) ( 2 ( ) [ (     G t = E u t + − u t −  E Sa 1 2 ( ) ( ) 1      t 

E Eeu> (为正实) C+7 Ee>2mEo(o-)(c,为正实数) Ecos0t<>Ex[(+0)+o(-) Esin Oot<>JETLS(o+0o-SO-OoI

   j E Ee t + −  2 ( ) 0 0      Ee  E − j t （ 0 为正实数） cos [ ( ) ( )] 0  E   +0 +   −0 E t sin [ ( ) ( )] 0     +0 −  −0 E t j E (  为正实数）