《信号与系统》课程教学课件（PPT讲稿）帕塞瓦尔定理证明

帕塞瓦尔定理证明 设{g,(为完备的正交函数集，即f()=lim∑c,&,() 误差函数 r-u-0 即 (a-2a+(a-0 因为c,= 一sna 代入f0di-22c,cg,0d+立egd1=0 即 么r0t=2cg0)a-2te,8,a业

X 第 1 帕塞瓦尔定理证明 页 设 gr (t) 为完备的正交函数集，即  → = = n r r r n f t c g t 1 ( ) ( ) lim 误差函数   =       − − =  = 2 1 ( ) ( ) d 0 1 ( ) 2 2 1 1 2 e t t r r r f t c g t t t t f t 即       =  = − + = 2 1 2 1 2 1 1 2 2 1 2 ( )d 2 ( ) ( )d ( )d 0 t t r t t r r r r r t t f t t c g t f t t c g t t 因为   = 2 1 2 1 ( )d ( ) ( )d 2 t t r t t r r g t t f t g t t c   = 2 1 2 1 ( ) ( )d ( )d 2 t t r r t t r f t g t t c g t t 代入 ( )d 2 ( )d ( )d 0 1 2 2 1 2 2 2 1 2 1 2 1 − +  =      =  = r t t r r r t t r r r t t f t t c c g t t c g t t ( ) ( )         =  = = = 1 2 1 2 2 2 2 1 2 1 2 1 d d ( ) d r t t r r r t t r r t t 即 f t t c g t t c g t t