f=zeros（3030）;
f（5:2413:17）=1;
imshow（f?‘notruesize‘）;
F=fft2（f256256）;?%?快速傅立葉變換算法只能處矩陣維數為2的冪次，f矩陣不
??????????????????????%?是，通過對f矩陣進行零填充來調整???
F2=fftshift（F）;??????%?一般在計算圖3-1的圖形函數的傅立葉變換時，坐標原點在
??????????????????????%?函數圖形的中心位置處，而計算機在對圖像執行傅立葉變換
??????????????????????%?時是以圖像的左上角為坐標原點。所以使用函數fftshift進?
??????????????????????%行修正，使變換后的直流分量位于圖形的中心；
figureimshow（log（abs（F2））[-1?5]‘notruesize‘）;

?屬性????????????大小?????日期????時間???名稱
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?????文件???????6427??2002-07-05?02:44??chap3\3-22.jpg

?????文件????????544??2004-04-04?10:30??chap3\P0301.m

?????文件????????557??2004-04-04?11:05??chap3\P0302.m

?????文件????????126??2004-04-04?11:16??chap3\P0303.m

?????文件????????346??2004-04-04?11:24??chap3\P0304.m

?????文件????????367??2004-04-04?12:05??chap3\P0305.m

?????文件????????481??2004-04-04?12:04??chap3\P0306.m

?????文件????????433??2004-04-04?12:13??chap3\P0307.m

?????文件????????222??2004-04-04?12:12??chap3\P0308.m

?????文件????????600??2004-04-04?12:27??chap3\P0309.m

?????文件????????263??2004-04-04?12:33??chap3\P0310.m

?????文件????????662??2004-04-04?12:37??chap3\P0311.m

?????文件????????614??2004-04-04?12:43??chap3\P0312.m

?????文件????????600??2004-07-21?13:35??chap3\readme.txt

????...D..R?????????0??2009-07-03?11:50??chap3

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????????????????12242????????????????????15