FFT with normalized spatial frequency for image sensor MTF

Question

Ben Hendrickson 2020-4-19

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https://ww2.mathworks.cn/matlabcentral/answers/519124-fft-with-normalized-spatial-frequency-for-image-sensor-mtf

评论： Ali Madani 2020-10-22

I'm attempting to use the slanted edge method to calculate the MTF for a camera system according to Harvest Imaging (https://harvestimaging.com/blog/?p=1328), and struggling to plot the result correctly. The method involves taking the Fourier transform of a line spread function and plotting it from DC to the sampling frequency. I have chosen the sampling frequency to equal 2, since that is the minimum number of pixels required to record contrast. Here is my code:

LSF = [];
LSF(1:250) = 0;
LSF(51:70) = 140;
Fs = 2;             % Sampling frequency   (pixels/lp)                 
T = 1/Fs;           % Sampling period     (lp/pixel) 
L = length(LSF);    % Length of signal    (length of line in pixels)
P2 = abs(fft(LSF/L));
P1 = P2(1:L/2+1);
P1(2:end-1) = 2*P1(2:end-1);
f = Fs*(0:(L/2))/L;
nP1 = P1-min(P1);
nP1 = nP1./max(nP1);
plot(f,nP1,'linewidth',3) 
title('MTF')
xlabel('Normalized Spatial Frequency')

According to the example, I should reach my first minimum of the sinc function at x=1

But, I appear to be off by a factor of 10.

Can anyone point out where I'm going wrong? I'm sure it's a simple fix.

Thanks much!

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Ben Hendrickson 2020-4-21

Appreciate the links! I agree that it should be obtained in cycles/pixel, then coverted to lp/mm. The problem may have to do with the say the initial function (SFR) is plotted. The idea is to super-resolve the edge, which means it should be plotted against fractions of a pixel. Mine is not. I'll keep banging my head against it and let you know when I've figured it out.

Thanks for talking it out with me!

Ali Madani 2020-10-22

Hi Ben, did you figure out how to normalized the x-axis? I've been having the same question and cannot figure out how to do this.

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Answer 1

Jack 2020-4-22

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https://ww2.mathworks.cn/matlabcentral/answers/519124-fft-with-normalized-spatial-frequency-for-image-sensor-mtf#answer_427540

编辑：Jack 2020-4-22

在 MATLAB Online 中打开

Hi Ben,

You are off by a factor of 10 because there are 10 samples in one of your 'pixels'. The result of the FT is in cycles/set, which can also be expressed in other units as shown below (see here for a detailed explanation). I used similar figures to yours that however result in the sinc having samples at MTF nulls.

Jack

n = 256;                                %samples per set
px = 16;                                %samples per pixel
LSF = zeros(1,n);                 
LSF(1:px)=140;                          %the pixel has intensity 140
MTF = 1/sum(LSF) * abs(fft(LSF));   
f = 0:n-1;                              %cycles per set
figure; plot(f,MTF);                    %cycles per set
xlabel('cycles/set'); ylabel('MTF')
figure; plot(f/n,MTF);                  %cycles per sample
xlabel('cycles/sample'); ylabel('MTF')
figure; plot (f/n*px,MTF)               %cycles per pixel
xlabel('cycles/pixel'); ylabel('MTF'); xlim([0 1])