Why N point FFT of a under-sampled signal in time domain is worse than only a N point FFT of the time domain signal (Not resampled)

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Tworit Dash
Tworit Dash 2021-4-14
回答： vidyesh 2024-2-20
I am trying to understand what happens when I try to reconstruct the spectrum of a signal with fewer samples. I tried 2 approaches. I resample the time domain signal based on the number of samples (N) I need and then took an FFT. In the second approach, I just did an N point FFT in which, as MATLAB says, the samples after N are ignored. Surprisingly, the second approach gives a good result. I don't understand why. For the ground truth, I simulate the time domain signal using a Gaussian-shaped frequency spectrum.
My MATLAB program is given below. You can change the value of N and see the results. Also you can change the interpolation method for the resampled signal case. I did all types of checks and I think the N point FFT performs better. However, one would think that the resampled signal should performed better when a FFT is applied.
clear;
close all;
% Simulation of ground truth
n = 512;
[sig_a, sig_f] = DS_simulator(10^(Inf/20), 1, 5, 0.2, n, 7.5); % sig_a is the time domain signal with 512 points and sig_f is the Gaussian spectrum from which sig is derived
figure; plot((abs(sig_f).^2)); % Original spectrum
%% Analysis
N = 256; % Number of fft points
idx = 1:n;
idxq = linspace(min(idx), max(idx), N);
sig_resampled = interp1(idx, sig_a, idxq, 'cubic'); % resampling the signal
figure; plot(1:1:n, abs(sig_a)); hold on; plot(1:n/N:n, abs(sig_resampled), '*'); % signal and resampled signal
sig_resamp_doppler = 1/N .* abs(fftshift(fft(sig_resampled, N))).^2; % FFT of resampled signal
sig_doppler = 1/N .* abs(fftshift(fft(sig_a, N))).^2; % FFT of original signal
figure; plot(abs(sig_doppler)); hold on; plot(abs(sig_resamp_doppler)); % Reconstructed spectrum
The code of ground truth signal generation whose spectrum is a Gaussian shape with mean as mu and standard deviation as sigma
function [data, data_f] = DS_simulator(SNR, m0, mu, sigma, n, v_amb)
vel_axis = linspace(-v_amb, v_amb, n);
X = rand(1, n);
Theta = 2 .* pi * rand(1, n);
if sigma < 0.02
[~, idx1] = min(abs(vel_axis - mu));
S = dirac(vel_axis - vel_axis(idx1));
idx = S == Inf;
S(idx) = 1;
else
S = m0/sqrt(2*pi*sigma^2) * exp(-(vel_axis - mu).^2/(2*sigma^2));
end
N = sum(S) ./ (n .* SNR); % Noise Power
P = -(S + N) .* log(X); % Power spectrum
data_f = sqrt(P);
data = ifft(fftshift(sqrt(n) .* sqrt(P) .* exp(1j .* Theta)));% complex time domain signal
end

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回答（1 个）

vidyesh
vidyesh 2024-2-20
Hi Tworit,
It is my understanding that you are trying to discern why an N-point FFT of an undersampled signal yields worse results than simply applying an N-point FFT directly to the original signal.
When reconstructing the spectrum of a signal with fewer samples, using an N-point FFT directly on the original signal without resampling can produce better results than resampling to 'N' points before applying the FFT. This is because the direct N-point FFT retains its original spectral content up to the Nyquist frequency.
In contrast, resampling involves interpolation, which can distort the signal and introduce aliasing, especially if the new sample rate isn't high enough to capture all the original frequency components. Therefore, the direct method avoids the potential issues introduced by resampling and more accurately reflects the original spectrum.
For more information on aliasing and its effects, refer to the following page
https://www.tutorialspoint.com/effects-of-undersampling-aliasing-and-anti-aliasing-filter
Hope this helps

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