按第二种频率抽样,得 N 1,0≤k≤hnt 2丌 N N-1 0.13<k 25 则F滤波器的频率响应: sin 51 兀\k+ sin i5 k H k=051sin k+ 51sin a+=k+则FIR滤波器的频率响应： 按第二种频率抽样，得 ( ) 1, 0 12 2 1 0, 13 25 2 c N k Int N H k N k            − =          =   −   =  ( ) 12 25 0 1 1 sin 51 sin 51 2 51 2 2 51 2 1 1 51sin 51sin 2 51 2 2 51 2 j j k k k H e e k k           − =                         − + + +                 = +                   − + + +               