华侨大学：《信号与系统》课程教学资源（PPT课件讲稿，机电版）傅里叶复指数级数证法2

傅里叶复指教级撤证法2 由欧拉公式 cos(n@ot+onsI (e1(y+1)+e-mn+) x()=4+∑4,cos(nat+g) x()=4+∑4(e 1 nj(not+on)+e j(not+g, n- =A+∑41(ee ern +e Jnaoto-j9

X 傅里叶复指数级数证法 2 0 0 ( ) 1 ( ) cos n n n x t A A n t    = = + +  •由欧拉公式： 0 0 j( ) j( ) 0 1 cos( ) (e e ) 2 n t n t n n n t n       + − + + = + 0 0 j( ) j( ) 0 1 1 ( ) (e e ) 2 n n n t n t n n x t A A      + − + = = + +  0 0 j j j j 0 1 1 (e e e e ) 2 n n n t n t n n A A      − − = = + + 

傅里叶复指教级撤证法2 A+∑A jna PoOjA, ee +∑4 Jnooto-j9 e 2 dger+2a, 2"lien+ Eans e i mw"e l. Abe j0bt。j0 e+ even+ n=1 n=0 令x 得 C0=4 e n≠0 CnFA1,n≠0

X 0 0 j j j j 0 1 1 1 1 e e e e 2 2 n n n t n t n n n n A A A       − − = = = + +   0 1 j e 2 n t A n  0 0 0 j j 0 0 1 1 1 e e 2 n j t n t j n n n A e e A     − = =− = + +   0 0 0 0 j t j A e e  = + 0 j( ) j 1 1 e e 2 n n t n n A  − − − = − − − −  0 j j 1 1 e e 2 n n t n n A    =  + 傅里叶复指数级数证法 2 j e n 0 j ( ) e n t n n x t c   =− =  0 0 , 0 , 0 2 n j n n j A e n c A e n   =  =     令： 得 0 0 | | 1 | | , 0 2 n n C A C A n = = 