四步相移法讨论

[qq_32754891]

%% 调制后的图像相位计算
for i=1:512
for j=1:512
if (I1(i,j)==I3(i,j))
if(I2(i,j)>I4(i,j))
I(i,j)=pi/2;
elseif(I2(i,j)<I4(i,j))
I(i,j)=3*pi/2;
else
I(i,j)=0;
end
elseif (I1(i,j)>I3(i,j))
if I2(i,j)>=I4(i,j)
I(i,j)=atan((I2(i,j)-I4(i,j))/(I1(i,j)-I3(i,j)));
else
I(i,j)=2*pi-atan((I4(i,j)-I2(i,j))/(I1(i,j)-I3(i,j)));
end
else
if I2(i,j)>=I4(i,j)
I(i,j)=pi-atan((I2(i,j)-I4(i,j))/(I3(i,j)-I1(i,j)));
else
I(i,j)=pi+atan((I4(i,j)-I2(i,j))/(I3(i,j)-I1(i,j)));
end
end
end
end

%% 参考图像相位计算
for i=1:512
for j=1:512
if (I5(i,j)==I7(i,j))
if(I6(i,j)>I8(i,j))
phase(i,j)=pi/2;
elseif(I6(i,j)<I8(i,j))
phase(i,j)=3*pi/2;
else
phase(i,j)=0;
end
elseif (I5(i,j)>I7(i,j))
if I6(i,j)>=I8(i,j)
phase(i,j)=atan((I6(i,j)-I8(i,j))/(I5(i,j)-I7(i,j)));
else
phase(i,j)=2*pi-atan((I8(i,j)-I6(i,j))/(I5(i,j)-I7(i,j)));
end
else
if I6(i,j)>=I8(i,j)
phase(i,j)=pi-atan((I6(i,j)-I8(i,j))/(I7(i,j)-I5(i,j)));
else
phase(i,j)=pi+atan((I8(i,j)-I6(i,j))/(I5(i,j)-I7(i,j)));
end
end
end
end
上面是所贴代码，楼主想法是这个代码的意思是不是根据所移动相位差，而提取从而构造新的一个相位图，也就是I（i,j）和phase(i,j),希望有人看到，指导一下

[种大王]

谢谢，很有用