南京邮电学院：《电路与信号分析》课程教学资源（PPT课件讲稿）第八章 傅里叶分析（8.1-8.3）

f(t) notat f(t)cosnootdt+kf(t)cos nootdt A cosmo tdt T 4A nCaz、2A.nO0 sin( sin( 2 1元 2 A、2A.nOnT f(t) sin(--)cosn@ot n=1 1丌 2

) 2 sin( 2 ) 2 sin( 4 0 0 0       n n n A n T A = = A n tdt T  = 2 0 0 cos 4     = = + 1 0 0 ) cos 2 sin( 2 ( ) n n t n n A T A f t      f t n tdt T a T n  = 0 ( ) cos 2  f t n tdt T f t n tdt T   = + − 2 0 0 0 2 0 ( ) cos 2 ( ) cos 2    

若A=1,T=2x,=丌, =1, 0 0n为偶数 1丌 S n=1.59… 1丌 1丌 n=3.7.11 b.=0 f(t)= +-(cost-cos 3t+=cos 5t--coS 7t +...)

若A =1,T = 2, =,          − = = = =   3,7,11, 2 1,5,9, 2 0 ) 2 sin( 2 n n n n n n n an     为偶数 , 2 1 1, 0 = a0 = = 0 n b cos7 ) 7 1 cos5 5 1 cos3 3 1 (cos 2 2 1  f (t) = + t − t + t − t + 

A=1 0n为偶数 0n≠3,71 4=(2mzxm为奇数2 丌n=3.711 12345 3兀 5丌 0 2345

− = =  ==  3,7,11, 0 3,7,11, 2 0 0 1 2 nn n n n AA n n    为奇数 为偶数 A n 0.5 2 32 52 0 1 2 3 4 5  − 1 2 3 4 5   n 0

例:一个周期信号可表示为 f(t)=2+3c0s2t+4sn2t+2si(3t+30)-c0s(7t+150 试画出其振幅谱和相位谱 解:先将含有相同频率的正弦项与余弦项合并为一个余弦项, 且所有项都表示为带正振幅的余弦项。 3c0s2t+4sin2t=5c0s(21-53.13) sin(3t+30)=cos(3t+30-90)=cos(3t-60) COs(7t+150)=cos(71+150-180)=cos(3t-30 f(t)=2+5c0s(2t-53.13)+2cos(3t-60)+cos(7t-30)

解：先将含有相同频率的正弦项与余弦项合并为一个余弦项， 试画出其振幅谱和相位谱 ( ) 2 3cos2 4sin 2 2sin( 3 30 ) cos(7 150 )   f t = + t + t + t + − t + 例：一个周期信号可表示为 且所有项都表示为带正振幅的余弦项。 cos(7 150 ) cos(7 150 180 ) cos(3 30 ) sin( 3 30 ) cos(3 30 90 ) cos(3 60 ) 3cos 2 4sin 2 5cos(2 53.13 )          − + = + − = − + = + − = − + = − t t t t t t t t t ( ) 2 5cos(2 53.13 ) 2cos(3 60 ) cos(7 30 )     f t = + t − + t − + t −

f(t)=2+5c0(21-53.13)+2cos(3t-60)+cos(7t-30) 2 234567 23 56 30 53.13

( ) 2 5cos(2 53.13 ) 2cos(3 60 ) cos(7 30 )    f t = + t − + t − + t − 0 1 2 3 4 5 6 7   − 30 − 53.13 − 60  n An 5 0 1 2 3 4 5 6 7  2 1