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    發(fā)布日期: 2021-04-26
  • 語(yǔ)言: Matlab
  • 標(biāo)簽: 相位提取??二維??

資源簡(jiǎn)介

二維傅里葉變換相位提取的matlab模擬程序。

資源截圖

代碼片段和文件信息

%%%?二維傅里葉變換相位提取

clear?all

N=512;??????%?抽樣點(diǎn)數(shù)
L=1;????????%?x,y方向長(zhǎng)度
a=1;????????%?背景強(qiáng)度
b=0.8;??????%?調(diào)制度
r0=1/16;
f_x=r0*N;????%?x方向載頻
f_y=r0*N;????%?y方向載頻
delta=L/N;???%?空域采樣間隔
delta_f=1/(N*delta);?%?頻域采樣間隔

[x?y]=meshgrid((-N/2:N/2-1)*delta);
[fx?fy]=meshgrid((-N/2:N/2-1)*delta_f);????

[XY]=meshgrid(1:N);

W=0.65e-6.*(x.^2+y.^2).^2;
PV=max(max(W))-min(min(W));
RMS=sqrt(sum(sum(W.^2)))/N;
lamda=532e-9;
k=2*pi/lamda;
phi_real=k.*W;
i=a+b.*cos(2*pi*f_x.*x+2*pi*f_y.*y+phi_real);
bg=sum(sum(i))/(N*N);
??
?i1=uint8(255*i);
?figureimagesc(i1);
?colormap?gray
?axis?square
?
flag_1=((sqrt((X-N/2).^2+(Y-N/2).^2))>N/2);
flag_2=((sqrt((X-N/2).^2+(Y-N/2).^2))flag_3=(X>(N/2-15))&(X<(N/2+15));
flag_4=(Y>(N/2-15))&(Y<(N/2+15));

flag=flag_1|flag_2;%|flag_3|flag_4;

flag1=not(flag);

?i_P=i.*flag1;
?
?i1=uint8(255/2*i_P);
?figureimagesc(i1);
?colormap?gray
?axis?square

?%%%%%%%%%%%%%%%%%%%%%%%%%?三?求出載頻值??%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
?
?I_P=ft2(i_Pdelta);
?
?[j?k]=find(abs(I_P)==max(max(abs(I_P))));?
?s0=j(1);??t0=k(1);???????????????????????%?找出零級(jí)頻譜坐標(biāo)
?
?[j?k]=find(abs(I_P(1:(N/2-N/128)1:(N/2-N/128)))...
?==max(max(abs(I_P(1:(N/2-N/128)1:(N/2-N/128))))));??%?找出負(fù)一級(jí)頻譜坐標(biāo)
?s1=j(1);t1=k(1);

fx0=s0-s1;???????????????????????????????????%?求出載頻值??????
fy0=t0-t1;

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%?四?條紋延拓?%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

????i_E=(i-bg).*flag1;
%???i1=uint8(255*abs(i_E));
%???figureimagesc(i1);
%???colormap?gray
%???axis?square

for?m=1:10
?
????I_E=ft2(i_Edelta);
%??
%??I1=uint8(255*abs(I_E));
%??figureimagesc(I1);
%??colormap?gray
%??axis?square
%??
?%%%%%%%%%%%%%%%%%%%?找出一級(jí)和負(fù)一頻譜坐標(biāo)?%%%%%%%%%%%%%%%%%%%%%%
?
[j?k]=find(abs(I_E(1:(N/2-N/128)1:(N/2-N/128)))...
==max(max(abs(I_E(1:(N/2-N/128)1:(N/2-N/128))))));??%?在第三象限內(nèi)找出一級(jí)頻譜坐標(biāo)
s11=j(1);??t11=k(1);?

[j?k]=find(abs(I_E((N/2+N/128):N(N/2+N/128):N))...
==max(max(abs(I_E((N/2+N/128):N(N/2+N/128):N)))));??%?在第二象限內(nèi)找出負(fù)一級(jí)頻譜坐標(biāo)
s22=j(1)+N/2+N/128-1;??t22=k(1)+N/2+N/128-1;?

%%%%%%%%%%%%%%%%%%%%%??濾波器設(shè)計(jì)?%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

r1=1/16;
H1=((sqrt((X-s11).^2+(Y-t11).^2))

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%??濾波??%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

I_E=I_E.*H1;

?%%%%%%%%%%%%%%%%%%%%%%%%%?反傅里葉變換?%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
?
i_E=ift2(I_Edelta_f);


?i_E=i_E.*flag+(i-bg).*flag1;
?
end
?
?i1=uint8(255*i_E);
?figureimagesc(i1);
?colormap?gray
?axis?square



?屬性????????????大小?????日期????時(shí)間???名稱(chēng)
-----------?---------??----------?-----??----

?????文件???????2668??2012-10-17?10:48??explam-fft.m

-----------?---------??----------?-----??----

?????????????????2668????????????????????1


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