第二节非正弦周期函数的分解 傅里叶级数:若周期为T,角频率o=2π/T的周期函数,满足狄里赫利条 件,则的可展开为 f(t=ao +a, cost+b, sin at +a, cos2ot+b, sin2ot+.+ar coskat+b, sin kat+ ao+2(a coskot+ b, sin kot 7(h f(t)cos kott f(t)cos kotd(at) f(sin kott f(tsin katd(ot) a coskat+b sin kat=A, sin(kat +pn) ..f(t)=A+A sin( ot+V1)+A, sin( 2ot+v2)..+Ak sin( kat+Vk)+ =A+∑4 k-l sin( kot +VR)第二节 非正弦周期函数的分解 傅里叶级数：若周期为T，角频率ω=2π/T的周期函数，满足狄里赫利条 件，则的可展开为 ∵  = T f t dt T a 0 0 ( ) 1   = =      2 0 0 ( ) cos ( ) 1 ( ) cos 2 f t k tdt f t k t d t T a T k   = =      2 0 0 ( )sin ( ) 1 ( )sin 2 f t k tdt f t k t d t T b T k   = = + + = + + + + + + + + 1 0 0 1 1 2 2 ( cos sin ) ( ) cos sin cos2 sin 2 cos sin k k k k k a a k t b k t f t a a t b t a t b t a k t b k t           cos sin A sin(k t )  +  = k  + k a k t b k t k k f (t) = A0 + A1 sin(t +1 ) + A2 sin( 2t + 2 ) ++ Ak sin( kt + k ) +   = = + + 1 0 sin( ) k k k A A kt  ∴