FFT with normalized spatial frequency for image sensor MTF

39 次查看（过去 30 天）
Ben Hendrickson
Ben Hendrickson 2020-4-19
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!
  7 个评论
显示 5更早的评论
Ben Hendrickson
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
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.

请先登录，再进行评论。

请先登录，再回答此问题。

回答（1 个）

Jack
Jack 2020-4-22
编辑：Jack 2020-4-22
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])

类别

在 Help Center 和 File Exchange 中查找有关 Fourier Analysis and Filtering 的更多信息

标签

产品

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by