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【文章內容簡介】 Example ?20 ?DTFT][)(][ nunx n?????? jeX 1)(?Example (cont.) ][nx21?NkXkExample (cont.) kX][nx21?NkmNnn ?? 1 10 1 ??? Nn1m o d, nNn ?1n1nNn m o d,取模運算 如果 ， m為整數；則有： 此運算符表示 n被 N除，商為 m，余數為 。 是 的解 ,或稱作取余數 ,或說作 n對N取模值 , 或簡稱為取模值 ,n模 N。 取模運算示例 79m od,252792259,251?????????nNnNn59m od,555949,4?????????????NnNnCircular Time Shift (圓周移位） 1,1,0 ?? Nn ?0 40 4周期延拓 5?N],[nxnnCircular Time Shift ]m o d,2[ Nnx ?]m o d,2[ Nnx ?0 40 4nnTime reversal (時間反轉） nn1,1,0 ?? Nn ?5?N],[nx0 4]m o d,[ Nnx ?0 4Linear Convolution (線性卷積） If both and are zero, for and ][nx ][nv 0?n Nn?? ?????10][][][*][][Niinvixnvnxny is zero for and ][ny 0?n 12 ?? NnCircular Convolution (圓周卷積） Because Npoint DFT of will not incorporate the values for , it is necessary to define a convolution operation so that the convolved signal is zero outside the range ][*][][ nvnxny ?Nn?? ?????10]m o d,[][][][][NiNinvixnvnxny1,1,0 ?? Nn ?][*][][ nvnxny ?NExample of linear convolution and circular convolution ????????????????????????3],0[]3[]1[]2[]2[]1[]3[]0[2],3[]3[]0[]2[]1[]1[]2[]0[1],2[]3[]3[]2[]0[]1[]1[]0[0],1[]3[]2[]2[]3[]1[]0[]0[nvxvxvxvxnvxvxvxvxnvxvxvxvxnvxvxvxvx][][ nvnx 4????????????????????????????6],3[]3[5],2[]3[]3[]2[4],1[]3[]2[]2[]3[]1[3],0[]3[]1[]2[]2[]1[]3[]0[2],0[]2[]1[]1[]2[]0[1],0[]1[]1[]0[0],0[]0[nvxnvxvxnvxvxvxnvxvxvxvxnvxvxvxnvxvxnvx][*][ nvnxThe relationship between linear circular convolution and circular convolution Setting , an Lstep interval linear convolution and circular convolution give the same result, that is ][*][][ nvnxny ?12 ?? NL],[][ nvnx? 1,1,0 ?? Ln ?LProperties of the DFT Linearity Circular time shift Time reversal Multiplication by a plex exponential Circular convolution Multiplication in the time domain Parseval’s theorem P331 System analysis via the DTFT and DFT ],[nx],[nhThe (N+Q)point DFT of is The (N+Q)point DFT of is given by Nn ??0Qn ??0][nx?????? 10)/(2][NnQNknjk enxX?][nh?????QnQNknjk enhH0
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