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二元傅立葉全息圖在matlab上的實現。體現了其五個實現步驟。結果為生成啁啾光纖光柵的程序

代碼片段和文件信息
clear;
N=1800;
y=linspace(-10.810.8N);
fy=0.5*(1+cos(11.8372*y));
A=fy
B=ones(381);
C=B*A;
figure;
imshow(C);
a=fft2(C38180);
b=fftshift(a);
figure;
imshow(b);
A1=abs(b);
B1=angle(b);
Amax=max(max(A1));
A2=A1/Amax;
B2=B1/(2*pi);
m=10/38;
n=21.6/180;
W=m/2;
figure;
axis([0?10?0?21.6]);
hold?on;
for?J=1:38
????x0=m/2+(J-1)*m;
????for?I=1:180
????????y0=n/2+(I-1)*n;
????????H=A2(JI)*n;
????????F=B2(JI)*m;
????????if?abs(F)<=m/4
????????????x2=x0-W/2;x3=x0+W/2;
????????????y2=y0-H/2;y3=y0+H/2;
????????????fill([x2x2x3x3][y2y3y3y2]‘k‘);
????????elseif?F>m/4
????????????x22=x0+F-W/2;x33=x0+m/2;
????????????y22=y0-H/2;y33=y0+H/2;
????????????fill([x22x22x33x33][y22y33y33y22]‘k‘);
????????????x222=x0-m/2;x333=x0+F+W/2-m;
????????????fill([x222x222x333x333][y22y33y33y22]‘k‘);
????????else?x22=x0-m/2;x33=x0+F+W/2;
????????????y2=y0-H/2;y3=y0+H/2;
????????????fill([x22x22x33x33][y2y3y3y2]‘k‘);
????????????x222=x0+F-W/2+m;x333=x0+m/2;
????????????fill([x222x222x333x333][y2y3y3y2]‘k‘);
????????end
????end
end
axis(‘equal‘);
axis?off;
?屬性????????????大小?????日期????時間???名稱
-----------?---------??----------?-----??----
?????文件??????17146??2008-05-15?20:08??F=0\hs_err_pid2876.log
?????文件???????1118??2008-05-15?21:43??F=0\M.asv
?????文件???????1132??2008-05-15?21:43??F=0\M.m
?????文件????????316??2008-05-15?19:54??F=0\m1.asv
?????文件????????311??2008-05-15?21:17??F=0\m1.m
?????文件????????308??2008-05-14?20:50??F=0\m2.m
?????文件????????994??2008-05-15?21:48??F=0\m3.asv
?????文件????????984??2008-05-14?21:24??F=0\m3.m
?????文件?????????48??2008-05-15?21:48??F=0\NN.asv
?????文件????????151??2008-05-15?21:11??F=0\NN.m
?????目錄??????????0??2008-05-15?21:48??F=0
-----------?---------??----------?-----??----
????????????????22726????????????????????12
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