Remove harmonics background noise

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Sammy
Sammy on 8 Jul 2024
Commented: Mathieu NOE on 11 Jul 2024
Hi,
I have a signal sampled at 128 [kHz], it contains noisy spikes at harmonic multiples of 59.6 [Hz]. Tried different bandstop filtering techniques, but can't seem to get a proper result.
Would appreciate it if something with some experience could elaborate how to remove the humming noise in the background.
Cheers
Sammy
  2 Comments
Mathieu NOE
Mathieu NOE on 8 Jul 2024
hello
this is quite a strange signal - what is it suppose to represent ? looks like a pulse train rather than just a simple corruption by hum noise. How did you come with a 59.6 [Hz] frequency for the spikes ?
looking at different time scales , we have no simple repetitive spikes , these spikes seem to occur more randomly that I expected in first place. Maybe simply a low pass filter would suffice but that depends what you want to keep in your signal
Sammy
Sammy on 8 Jul 2024
Edited: Sammy on 8 Jul 2024
  1. This signal was acquired using a rolling shutter camera which has a very short row time of 8 [us] and a framerate of 59.6 FPS, which is the source of the noise. The signal itself contains the humming noise and in the background a sine sweep signal which should range between 200-2000 [Hz].
  2. The spikes I was referring to are in the frequency domain,(allthough there are also some spikes in the time domain), which you can see if you look at its PSD:
figure;
S=fft(short_sweep);
power_spectrum = abs(S).^2;
frequencies = linspace(0, Fs/2, length(power_spectrum)/2 + 1);
plot(frequencies,power_spectrum(1:length(frequencies)));
title('Power Spectrum of s(t) (resampled) - unfiltered');
xlim([0 10000]);

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Accepted Answer

Mathieu NOE
Mathieu NOE on 8 Jul 2024
I have some doubts we can recover a clean sinewave from that ... nevertheless I tried a few things ... first I decimated the signal by a factor of 20 as Fs = 128kHz is pretty fast for a signal that does not exceed 2 kHz
then bandpass filtered it - the spectrogram tells us that there is indeed some kind of sweeip burried in the noise , but it starts above 200 Hz obviously and also its amplitude is not constant with time
here we are for the moment :
then we will try to remove the horizontal lines with a notch filter - looped as many times as there are H lines (harmonics) .
But we need to make sure the frame rate is measures as accurately as possible. So I picked the peaks of the 1 second of signal and theit time indexes are checked to be regurarly spaced . But the computed frequency here is 62.0456 hz and not 59.6 Hz (this is due to the lack of frequency resolution of your fft computation). So with that info , we can run a notch filter a this frequency and its N harmonics
the new spectrogram appears "cleaned" from these H lines , but there is still an important background noise, so the time signal is still very noisy.
NB : my code uses the function peakseek PeakSeek - File Exchange - MATLAB Central (mathworks.com) that I provide in attachment. A simpler yet very effective alternative to findpeaks !
clc
clearvars
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% load signal
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
load('short_sweep.mat')
signal = s(:); % 1 channel of data
clear s
dt = 1/128e3; % sampling time interval
Fs = 1/dt; % sampling rate is 128kHz here
[samples,channels] = size(signal);
t = (0:samples-1)*dt;
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% compute "clicks" period on first second
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
t1 = t(1:Fs);
signal1 = signal(1:Fs);
% try with peakseek
minpeakdist = Fs/100; % in samples
minpeakh = 2;
[locs1, pks1]=peakseek(signal1,minpeakdist,minpeakh);
% plot(t1,signal1,t1(locs1),signal1(locs1),'dk');
period = mean(diff(t1(locs1)));
ff = 1/period;
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% decimate (if needed)
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% NB : decim = 1 will do nothing (output = input)
decim = 20;
if decim>1
Fs = Fs/decim;
for ck = 1:channels
newsignal(:,ck) = decimate(signal(:,ck),decim);
end
signal = newsignal;
end
samples = length(signal);
time = (0:samples-1)*1/Fs;
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% band pass filter section %%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
f_low = 100;
f_high = 2500;
N_bpf = 2; % filter order
[b,a] = butter(N_bpf,2/Fs*[f_low f_high]);
signal = filtfilt(b,a,signal);
%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% notch filter
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
fc_notch = ff; % 62.0456 Hz in fact
rho = 0.997; % something slightly less than 1 (will change the width and the depth of the notch filter).
N = 45; % how many harmonics of fc_notch do we want to remove ?
for k = 1:N+1
w0 = 2*pi*k*fc_notch/Fs;
% digital notch (IIR)
num1z=[1 -2*cos(w0) 1];
den1z=[1 -2*rho*cos(w0) rho^2];
% now let's filter the signal
signal = filtfilt(num1z,den1z,signal);
end
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%% FFT analysis
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
nfft = 512;
overlap = 0.75;
dB_range = 40; % display this dB range on y axis
[S,F,T] = myspecgram(signal, Fs, nfft, overlap); % overlap is 75% of nfft here
sg_dB = 20*log10(abs(S)); % expressed now in dB
% saturation of the dB range : the lowest displayed level is spectrogram_dB_scale dB below the max level
min_disp_dB = round(max(max(sg_dB))) - dB_range;
%% plots
figure(1);
plot(time,signal)
figure(2);
imagesc(T,F,sg_dB)
caxis([min_disp_dB min_disp_dB+dB_range]);
% ylim([100 Fs/2.56]);
hcb = colorbar('vert');
set(get(hcb,'Title'),'String','dB')
set(gca,'YDir','Normal')
xlabel('Time (secs)')
ylabel('Freq (Hz)')
title('Specgram')
colormap('jet');
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
function [ZxP,ZxN] = find_zc(x,y,threshold)
% put data in rows
x = x(:);
y = y(:);
% positive slope "zero" crossing detection, using linear interpolation
y = y - threshold;
zci = @(data) find(diff(sign(data))>0); %define function: returns indices of +ZCs
ix=zci(y); %find indices of + zero crossings of x
ZeroX = @(x0,y0,x1,y1) x0 - (y0.*(x0 - x1))./(y0 - y1); % Interpolated x value for Zero-Crossing
ZxP = ZeroX(x(ix),y(ix),x(ix+1),y(ix+1));
% negative slope "zero" crossing detection, using linear interpolation
zci = @(data) find(diff(sign(data))<0); %define function: returns indices of +ZCs
ix=zci(y); %find indices of + zero crossings of x
ZeroX = @(x0,y0,x1,y1) x0 - (y0.*(x0 - x1))./(y0 - y1); % Interpolated x value for Zero-Crossing
ZxN = ZeroX(x(ix),y(ix),x(ix+1),y(ix+1));
end
function [fft_specgram,freq_vector,time] = myspecgram(signal, Fs, nfft, Overlap)
% FFT peak spectrogram of signal (example sinus amplitude 1 = 0 dB after fft).
% signal - input signal,
% Fs - Sampling frequency (Hz).
% nfft - FFT window size
% Overlap - buffer overlap % (between 0 and 0.95)
signal = signal(:);
samples = length(signal);
% fill signal with zeros if its length is lower than nfft
if samples<nfft
s_tmp = zeros(nfft,1);
s_tmp((1:samples),:) = signal;
signal = s_tmp;
samples = nfft;
end
% window : hanning
window = hanning(nfft);
window = window(:);
% compute fft with overlap
offset = fix((1-Overlap)*nfft);
spectnum = 1+ fix((samples-nfft)/offset); % Number of windows
% % for info is equivalent to :
% noverlap = Overlap*nfft;
% spectnum = fix((samples-noverlap)/(nfft-noverlap)); % Number of windows
% main loop
fft_specgram = [];
for ci=1:spectnum
start = (ci-1)*offset;
sw = signal((1+start):(start+nfft)).*window;
fft_specgram = [fft_specgram abs(fft(sw))*4/nfft]; % X=fft(x.*hanning(N))*4/N; % hanning only
end
% one sidded fft spectrum % Select first half
if rem(nfft,2) % nfft odd
select = (1:(nfft+1)/2)';
else
select = (1:nfft/2+1)';
end
fft_specgram = fft_specgram(select,:);
freq_vector = (select - 1)*Fs/nfft;
% time vector
% time stamps are defined in the middle of the buffer
time = ((0:spectnum-1)*offset + round(nfft/2))/Fs;
end
  4 Comments
Mathieu NOE
Mathieu NOE on 9 Jul 2024
my pleasure !
if my answer has helped you , maybe you could consider accepting it ! :)
Mathieu NOE
Mathieu NOE on 11 Jul 2024
fyi , some digital filters matlab code in attachment

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