# FFT with normalized spatial frequency for image sensor MTF

44 views (last 30 days)
Ben Hendrickson on 19 Apr 2020
Commented: Ali Madani on 22 Oct 2020
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!
Ben Hendrickson on 21 Apr 2020
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 on 22 Oct 2020
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.

Jack on 22 Apr 2020
Edited: Jack on 22 Apr 2020
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])

### Categories

Find more on Graphics Performance in Help Center and File Exchange

### Community Treasure Hunt

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

Start Hunting!