Fourrier transform on image
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How can we compute numerically the features orientation and spacing from a FFT2 applied on image with oriented grid of objects?
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Meg Noah
2020년 1월 12일
편집: Meg Noah
2020년 1월 12일
Here's an example - the grid of features is space every 20 meters, the frequency found at 0.05/m corresponds to the frequency of dots.
% Online references for FFT's
% https://www.gaussianwaves.com/2015/11/interpreting-fft-results-complex-dft-frequency-bins-and-fftshift/
% https://blogs.uoregon.edu/seis/wiki/unpacking-the-matlab-fft/
clc
close all
clear all
% generate spatial frame data and coordinates
dx_m = 1; % [m]
dy_m = 1; % [m]
nx = 101;
ny = nx;
% a grid of values
imgData = zeros(1,nx);
imgData(10:20:nx-9) = 1;
imgData = imgData.*imgData';
if (mod(nx,2) == 0)
X1D = dx_m.*[-nx/2:1:nx/2-1];
else
X1D = dx_m.*[-(nx-1)/2:1:(nx-1)/2];
end
if (mod(ny,2) == 0)
Y1D = dy_m.*[-ny/2:1:ny/2-1];
else
Y1D = dy_m.*[-(ny-1)/2:1:(ny-1)/2];
end
% visualize spatial data
figure('Color','white');
subplot(2,1,1)
imagesc(X1D,Y1D,imgData);
title({'Image Data'},'fontsize',12);
axis equal
axis tight
colorbar
set(gca,'fontweight','bold');
xlabel('X [m]'); ylabel('Y [m]');
% now to show the power spectrum
% with frequency space grid
dfy = 1/(ny*dy_m);
dfx = 1/(nx*dx_m);
if (mod(nx,2) == 0)
FX = dfy.*[-nx/2:1:nx/2-1];
else
FX = dfy.*[-(nx-1)/2:1:(nx-1)/2];
end
if (mod(ny,2) == 0)
FY = dfy.*[-ny/2:1:ny/2-1];
else
FY = dfy.*[-(ny-1)/2:1:(ny-1)/2];
end
subplot(2,1,2)
% imagesc(FX,FY,20*log10(abs(fftshift(fft2(imgData)))));
imagesc(FX,FY,(abs(fftshift(fft2(imgData)))));
axis equal; axis tight; colorbar
set(gca,'fontweight','bold');
xlabel('Frequency [1/m]'); ylabel('Frequency [1/m]');
title('FFT2D Output','fontsize',12);
Adding the ability to rotate the image
clc
close all
clear all
% generate spatial frame data and coordinates
dx_m = 1; % [m]
dy_m = 1; % [m]
nx = 101;
ny = nx;
rotation = 30;
% a grid of values
imgData = zeros(1,nx);
imgData(10:20:nx-9) = 1;
imgData = imgData.*imgData';
imgData = imrotate(imgData,rotation,'crop');
if (mod(nx,2) == 0)
X1D = dx_m.*[-nx/2:1:nx/2-1];
else
X1D = dx_m.*[-(nx-1)/2:1:(nx-1)/2];
end
if (mod(ny,2) == 0)
Y1D = dy_m.*[-ny/2:1:ny/2-1];
else
Y1D = dy_m.*[-(ny-1)/2:1:(ny-1)/2];
end
% visualize spatial data
figure('Color','white');
subplot(2,1,1)
imagesc(X1D,Y1D,imgData);
title({'Image Data'},'fontsize',12);
axis equal
axis tight
colorbar
set(gca,'fontweight','bold');
xlabel('X [m]'); ylabel('Y [m]');
% now to show the power spectrum
% with frequency space grid
dfy = 1/(ny*dy_m);
dfx = 1/(nx*dx_m);
if (mod(nx,2) == 0)
FX = dfy.*[-nx/2:1:nx/2-1];
else
FX = dfy.*[-(nx-1)/2:1:(nx-1)/2];
end
if (mod(ny,2) == 0)
FY = dfy.*[-ny/2:1:ny/2-1];
else
FY = dfy.*[-(ny-1)/2:1:(ny-1)/2];
end
subplot(2,1,2)
% imagesc(FX,FY,20*log10(abs(fftshift(fft2(imgData)))));
imagesc(FX,FY,(abs(fftshift(fft2(imgData)))));
axis equal; axis tight; colorbar
set(gca,'fontweight','bold');
xlabel('Frequency [1/m]'); ylabel('Frequency [1/m]');
title('FFT2D Output','fontsize',12);
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Meg Noah
2020년 1월 17일
What is the spatial dimension of the x-axis and y-axis - in other words, how many meters is this image from pixel center to pixel center along a row and along a column?
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