【正文】
?? ?1 ()1NkNNkNaWRkaW??? ( ) ( )n Nx n a R n??2021/6/14 數(shù)字信號處理 ( ) [ ( )] { I m [ ( )]}Y k D F T y n D F T f n??1 ()opFkj?*1 ( ) ( ( ) ) ( )2 NNF k F N k R kj ??? ? ??? *1 1 1 1 1 ()2 1 1 1 1N N N NNk k N k N kN N N Na b a bj j R kj aW bW aW bW????? ? ? ?? ? ? ??? ??? ? ? ???? ? ? ?**1 1 1 1 1 ()2 1 1 11N N N NNkk kkNN NNa b a bj j R kj aW bW a W b W??? ? ? ?? ? ? ??? ????1 ()1NNkNb RkbW???10()Nn k nNNnb W R k??? ?? ?1 ()1NkNNkNbWRkbW??? ( ) ( )n Ny n b R n??2021/6/14 數(shù)字信號處理 ? ? *1 1 1 ( )2 Nj N j N R k??? ? ? ???? ?1 1 1 ( )2 Nj N j N R k? ? ? ?()NRk?( ) ( )x n n???( ) [ ( )] { R e[ ( )]}X k D F T x n D F T f n??()epFk? *1 ( ) (( )) ( )2 NNF k F N k R k??? ? ???? ? ? ?21F k jN??2021/6/14 數(shù)字信號處理 ? ? *1 1 1 ( )2 NjN jN R kj ??? ? ? ???? ?1 1 1 ( )2 NjN jN R kj? ? ? ?()NN R k?( ) ( )y n N n???( ) [ ( )] { I m [ ( )]}Y k D F T y n D F T f n??1 ()opFkj? *1 ( ) (( )) ( )2 NNF k F N k R k? ? ?2021/6/14 數(shù)字信號處理 20. 已知序列 現(xiàn)對于 x(n) 的 變換在單位圓上 等分抽樣,抽樣值為 試求有限長序列 , 點(diǎn)?,F(xiàn)將長度變成 rN點(diǎn)的有限長序列 ()xn ? ?( ) ( )X k D F T x n?()yn( ) , 0 1()0 , 1x n n NynN n rN? ? ???? ? ? ??試求 rN點(diǎn) 與 的關(guān)系。 {1,1,3, 2}()xn5(( ))xn? 66( ( ) ) ( )x n R n? 33( ( ) ) ( )x n R n6(( ))xn 55( ( 3 ) ) ( )x n R n? 77( ( ) ) ( )x n R n2021/6/14 數(shù)字信號處理 5(( ))xn?6(( ))xn66(( )) ( )x n R n?2021/6/14 數(shù)字信號處理 55(( 3 )) ( )x n R n?33(( )) ( )x n R n77(( )) ( )x n R n2021/6/14 數(shù)字信號處理 5. 試求以下有限長序列的 點(diǎn) (閉合形式表達(dá)式): N DFT0( ) co s ( ) ( )Nx n a n R n??(1) 10( ) ( ) ( )NnkNNnX k x n W R k??? ?解:002101 ( ) ( )2N j nkj n j n NNna e e e R k???? ????????????2100c o s( ) ( )N j nkNNna n e R k??? ??? ?002211( ) ( )001 ()2NN j k n j k nNNNnna e e R k??????? ? ? ???????2021/6/14 數(shù)字信號處理 000022( ) ( )1 1 1()211j N j NNj k j kNNeea R kee????????? ? ? ????????????? 0 0 00 0 02 2 21 2 1 2 1 2( ) ( ) ( )2 2 21 ( )2()N N Nj j jj k j k j kN N Ne e eae e e? ? ?? ? ?? ? ???? ? ? ? ????????? 0 0 00 0 02 2 21 2 1 2 1 2( ) ( ) ( )2 2 2()()()N N Nj j jNj k j k j kN N Ne e eRke e e? ? ?? ? ?? ? ???? ? ? ? ??? ???? ??0000112200sin( ) sin